( ) ( ) ( ) ( ) ( ) ( y )


 Eileen Benson
 4 years ago
 Views:
Transcription
1 8. Lengh of Plne Curve The mos fmous heorem in ll of mhemics is he Pyhgoren Theorem. I s formulion s he disnce formul is used o find he lenghs of line segmens in he coordine plne. In his secion you ll see how his sme heorem, when comined wih he clculus concep of limis, llows us o find he lenghs of coninuous plne curves defined s grphs of funcions or s grphs of prmeric equions. We ll do his y wy of he Riemnn inegrl you were inroduced o in prior clculus course. Consider piece of curve defined y funcion y f ( x) over n inervl x. To pproxime he lengh of his curve, one cn sudivide he inervl [, ] ino suinervls (o he righ we ve used four suinervls). This gives us he following poins on he curve: P ( x, y ), P ( x, y ). P x, y, P x, y, P x, y, Here s usul, y f x i i. for ech of hese poins. The curve lengh is hen pproximely he sum of he lenghs of he line segmens connecing hese poins: ( dis P, P dis P, P dis P, P dis P, P 1 1 ( ) ( y ) x x + y y + x x + y y + x x + y y + x x + y Using summion noion his is shorer: ( x x ) + ( y y ) Lengh k+ 1 k k+ 1 k k Of course if we use more suinervls he pproximion will e even closer o he rue lengh of he curve. Going from suinervls o n suinervls is very esy using summion noion: ) k ( x x ) ( y y ) Lengh + k+ 1 k k+ 1 k
2 To mke hings more precise nd specific, we divide he inervl [, ] ino n equl suinervls, ech of lengh xn Δ x. Then xk + k Δ x for k, 1,,, n. Defined his wy, x nd n. Since ech suinervl is of lengh x k k+ 1 k Δ, we hve h xk 1 x Δ y y y, so our lengh pproximion formul ecomes: + k x Δ for ll k. Also, les define xk xk yk yk x yk k k ( + 1 ) ( + 1 ) Δ Lengh + Δ + If we now fcor ou he Δ x, we ge: Δy k Δyk Lengh Δ x +Δyk 1+ x 1 Δ + Δ k k Δx k Δx x Tking he limi s he numer of sudivisions n goes o infiniy, our summion ecomes Riemnn inegrl nd he pproximion ecomes he exc curve lengh. In fc, h s how we will define curve lengh: Δyk dy Lengh lim 1+ x 1 n Δ + k Δx dy Lengh ( f '( x) ) Exmple 1. Find he lengh of he curve defined y x x y for x. Since 1 1 x x dy y x x x, we hve 1 dy Lengh x 1 + x Mking he susiuion u 1 + x du, we ge
3 1 u Lengh udu u du 1 ( ) 1 ( 1) Exmple. Find he lengh of he prolic rc defined y x y for x 1. Since 1 x dy x y x, we hve 1 1 dy Lengh x Mking he susiuion x n sec d θ θ θ, we ge Lengh 1+ n θ sec θ dθ secθ sec θ dθ sec θ dθ Invoking ulr inegrion: + sec x sec x sec x n x n x We ge: sec x sec xn x sec xn x sec xn x sec x sec x 1 sec x sec x + sec xn x sec x x x x+ x x x+ ln sec x + n x sec sec n sec sec n Thus: sec Lengh sec xn x+ ln sec x+ n x x + C sec θ dθ secθ nθ + ln secθ + nθ θ θ
4 1 Lengh sec n ln sec n sec n ln sec n ln 1+ ( 1+ ln ln 1+ ) x Exmple. Find he lengh of he piece of cenry curve defined y y cosh for x. Since x dy x 1 x y cosh sinh sinh, we hve x sinh dy x x Lengh 1 1 sinh cosh sinh sinh ( sinh1 sinh ) ( sinh1 ) sinh1 ( e ) or 1 1 e e e e e e 1 1 e e e Exmple. Find he rc lengh of he curve defined y x + 8 y for x. x x x x + 8 x + 8 dy y x x dy x 16 8x Hence: ( x 16) ( x ( x + 8 ) ) x 8 ( x 16) x 8 8 ( x + ) x x 8 8 6x x x x + x 16 dy 1 6x x 6x dy x 16 x x x + x x x 6x 6x 6x x x
5 And so: dy x + 16 x 16 1 x x 8x 8x 8x 8 Lengh x x ( ) ( ) ( 1) 1 1 Even finding he lengh of he grph of pr of simple nd degree polynomil wsn so esy. Unforunely mos rc lengh inegrls re no solvle, excep numericlly. Bu if you cn les se up he correc rc lengh inegrl, you cn find numericl pproximions o he exc nswer h re s ccure s you like, using he pproprie sofwre (or clculor). If you ke look ny clculus ex s coverge of he opic of rc lengh, you ll noice h here re no mny exmples; nor re here mny homework prolems. No only his, u you ll see he sme prolems! There jus ren mny rc lengh inegrls h cn e solved in he usul sense. We ll ge few more using prmeric equions! Le curve e defined y prmeric equions () x f y g, for. To pproxime he curve s lengh, we ll divide he inervl ino n equl suinervls, ech of lengh Δ. Then we define k + k Δ, for n k, 1,,, n. We hve he poins P ( x, y ), P ( x, y ). P x, y, P x, y, P x, y, 1 1 1
6 Here x f ( ) nd y g( ) for ech of hese poins. The curve lengh is hen pproximely he k k k k sum of he lenghs of he line segmens connecing hese poins: ( dis P, P dis P, P dis P, P dis P, P 1 1 ( ) ( y ) x x + y y + x x + y y + x x + y y + x x + y Using summion noion his is shorer: ( x x ) + ( y y ) k Lengh k k k k As efore, if we use more suinervls he pproximion will e even closer o he rue lengh of he curve. And s efore, going from suinervls o n suinervls is very esy using summion noion: k ( x x ) ( y y ) Lengh + k+ 1 k k+ 1 k ) Leing Δ ( nd y g( ) g( ) x f f k k k + 1 ) Δ for k, 1,,,, he ove formul k k+ 1 k ecomes: Lengh Δ x +Δy k k k Nex we muliply y Δ Δ : Δ Δ xk +Δyk Δxk Δyk k k k Δ k Δ k Δ Δ Lengh Δ x +Δy Δ + Δ Tking he limi s he numer of sudivisions n goes o infiniy, our summion ecomes Riemnn inegrl nd he pproximion ecomes he exc curve lengh. In fc, h s how we will define curve lengh for prmericlly defined curves: Δxk Δy k dy lim n k Δ Δ d d Lengh + Δ + d dy Lengh + d ( f '() ) + ( g '() ) d d d
7 Exmple 5. Find he rc lengh of he curve prmericlly defined y x y for 1. dy dy nd + ( ) + ( ) d d d d And so: 1 1 dy Lengh d d d d Mking he susiuion u 9 + du 18 d du d, our inegrl ecomes: u d 18 u du 18 u du 18 Lengh Exmple 6. Find he lengh of he curve deermined y x rsin y rcos for. This prmerizion rces ou circle of rdius r once (clockwise nd sring 1:), so he lengh of he rc is simply he circumference of h circle. Since he circumference of circle wih rdius r is r, h s he lengh of he curve. Bu h ws oo esy! Le s do i gin, using he defining prmeric equions: dy dy rcos nd rsin + rcos + rsin d d d d r cos + sin r 1 r And so: dy Lengh + d r d r d r r ( ) r d d
8 Exmple 7. Find he lengh of he curve deermined y ( ) ( ) x e cos y e sin for 5. dy e cos e sin nd e sin + e cos d d ( ) ( ) ( ) ( ) dy d d ( e cos( ) e sin ( ) ) e sin ( ) e cos( ) ( cos( ) sin ( )) sin ( ) + cos( ) e + ( ) ( ) e C S S C + + Noice in h ls line how he noion is simplified. This kind of susiuion mkes he lger much esier, nd eses he wrier s crmp! To coninue e C CS + S + S + SC + C e C + S + ( S + C ) e 1+ Thus, dy Lengh + d e 1+ d 1+ e d 1+ e d d 5 ( e e ) ( e ) Exmple 8. Find he lengh of he curve deermined y x + 1 y n ln 1 for d ln + 1 d ln ( + 1) d ln ( + 1) d ln ( + 1) d d d d d n + + dy d 1 dy d d 1+ d d Hence, 1 1 dy d Lengh + d d d + 1
9 Mking he susiuion n d sec d θ θ θ, our inegrl ecomes: 1 d sec θ dθ sec θ dθ Lengh + 1 n θ + 1 secθ θ θ sec + sec n secθ dθ sec + n θ θ θ secθ dθ dθ ln sec + n secθ + nθ secθ + nθ θ θ ( + ) ln sec + n ln sec + n ln + 1 ln 1+ ln 1 Exmple 9. Find n inegrl h equls he lengh of he sine curve over one period. One period of he sine curve cn e prmerized y x y sin for. Thus: Hence, dy dy 1 nd cos ( cos) 1+ cos d d d d dy Lengh + d 1+ cos d d d This inegrl cn e solved y ny elemenry inegrion echniques one will ever encouner in Clculus I. ( Bu in cse you re ineresed, h lengh is pproximely ) In he ls secion of his chper you ll lern some echniques for evluing inegrls numericlly. Numericl echniques re essenil for evluing inegrls like he one we jus encounered in our ls exmple. And such inegrls cully end o e he rule rher hn he excepion. Bu sophisiced numericl inegrion lgorihms now hrdwired ino mny inexpensive clculors, nd powerful sofwre like Mple re lso ville o esily del wih reclcirn inegrls!
MTH 146 Class 11 Notes
8. Are of Surfce of Revoluion MTH 6 Clss Noes Suppose we wish o revolve curve C round n is nd find he surfce re of he resuling solid. Suppose f( ) is nonnegive funcion wih coninuous firs derivive on he
More information4.8 Improper Integrals
4.8 Improper Inegrls Well you ve mde i hrough ll he inegrion echniques. Congrs! Unforunely for us, we sill need o cover one more inegrl. They re clled Improper Inegrls. A his poin, we ve only del wih inegrls
More informatione t dt e t dt = lim e t dt T (1 e T ) = 1
Improper Inegrls There re wo ypes of improper inegrls  hose wih infinie limis of inegrion, nd hose wih inegrnds h pproch some poin wihin he limis of inegrion. Firs we will consider inegrls wih infinie
More informationFM Applications of Integration 1.Centroid of Area
FM Applicions of Inegrion.Cenroid of Are The cenroid of ody is is geomeric cenre. For n ojec mde of uniform meril, he cenroid coincides wih he poin which he ody cn e suppored in perfecly lnced se ie, is
More information0 for t < 0 1 for t > 0
8.0 Sep nd del funcions Auhor: Jeremy Orloff The uni Sep Funcion We define he uni sep funcion by u() = 0 for < 0 for > 0 I is clled he uni sep funcion becuse i kes uni sep = 0. I is someimes clled he Heviside
More informationENGR 1990 Engineering Mathematics The Integral of a Function as a Function
ENGR 1990 Engineering Mhemics The Inegrl of Funcion s Funcion Previously, we lerned how o esime he inegrl of funcion f( ) over some inervl y dding he res of finie se of rpezoids h represen he re under
More information5.1The InitialValue Problems For Ordinary Differential Equations
5.The IniilVlue Problems For Ordinry Differenil Equions Consider solving iniilvlue problems for ordinry differenil equions: (*) y f, y, b, y. If we know he generl soluion y of he ordinry differenil
More informationCHAPTER 11 PARAMETRIC EQUATIONS AND POLAR COORDINATES
CHAPTER PARAMETRIC EQUATIONS AND POLAR COORDINATES. PARAMETRIZATIONS OF PLANE CURVES., 9, _ _ Ê.,, Ê or, Ÿ. 5, 7, _ _.,, Ÿ Ÿ Ê Ê 5 Ê ( 5) Ê ˆ Ê 6 Ê ( 5) 7 Ê Ê, Ÿ Ÿ $ 5. cos, sin, Ÿ Ÿ 6. cos ( ), sin (
More informationMathematics 805 Final Examination Answers
. 5 poins Se he Weiersrss Mes. Mhemics 85 Finl Eminion Answers Answer: Suppose h A R, nd f n : A R. Suppose furher h f n M n for ll A, nd h Mn converges. Then f n converges uniformly on A.. 5 poins Se
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum SquredError Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > for ll smples y i solve sysem of liner inequliies MSE procedure y i = i for ll smples
More informationMinimum Squared Error
Minimum Squred Error LDF: Minimum SquredError Procedures Ide: conver o esier nd eer undersood prolem Percepron y i > 0 for ll smples y i solve sysem of liner inequliies MSE procedure y i i for ll smples
More informationf t f a f x dx By Lin McMullin f x dx= f b f a. 2
Accumulion: Thoughs On () By Lin McMullin f f f d = + The gols of he AP* Clculus progrm include he semen, Sudens should undersnd he definie inegrl s he ne ccumulion of chnge. 1 The Topicl Ouline includes
More informationINTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).
INTEGRALS JOHN QUIGG Eercise. Le f : [, b] R be bounded, nd le P nd Q be priions of [, b]. Prove h if P Q hen U(P ) U(Q) nd L(P ) L(Q). Soluion: Le P = {,..., n }. Since Q is obined from P by dding finiely
More informationAn integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples.
Improper Inegrls To his poin we hve only considered inegrls f(x) wih he is of inegrion nd b finie nd he inegrnd f(x) bounded (nd in fc coninuous excep possibly for finiely mny jump disconinuiies) An inegrl
More informationChapter 2. Motion along a straight line. 9/9/2015 Physics 218
Chper Moion long srigh line 9/9/05 Physics 8 Gols for Chper How o describe srigh line moion in erms of displcemen nd erge elociy. The mening of insnneous elociy nd speed. Aerge elociy/insnneous elociy
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 informationMAT 266 Calculus for Engineers II Notes on Chapter 6 Professor: John Quigg Semester: spring 2017
MAT 66 Clculus for Engineers II Noes on Chper 6 Professor: John Quigg Semeser: spring 7 Secion 6.: Inegrion by prs The Produc Rule is d d f()g() = f()g () + f ()g() Tking indefinie inegrls gives [f()g
More informationMotion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.
Moion Accelerion Pr : Consn Accelerion Accelerion Accelerion Accelerion is he re of chnge of velociy. = v  vo = Δv Δ ccelerion = = v  vo chnge of velociy elpsed ime Accelerion is vecor, lhough in onedimensionl
More information3. Renewal Limit Theorems
Virul Lborories > 14. Renewl Processes > 1 2 3 3. Renewl Limi Theorems In he inroducion o renewl processes, we noed h he rrivl ime process nd he couning process re inverses, in sens The rrivl ime process
More informationREAL ANALYSIS I HOMEWORK 3. Chapter 1
REAL ANALYSIS I HOMEWORK 3 CİHAN BAHRAN The quesions re from Sein nd Shkrchi s e. Chper 1 18. Prove he following sserion: Every mesurble funcion is he limi.e. of sequence of coninuous funcions. We firs
More information1.0 Electrical Systems
. Elecricl Sysems The ypes of dynmicl sysems we will e sudying cn e modeled in erms of lgeric equions, differenil equions, or inegrl equions. We will egin y looking fmilir mhemicl models of idel resisors,
More informationSeptember 20 Homework Solutions
College of Engineering nd Compuer Science Mechnicl Engineering Deprmen Mechnicl Engineering A Seminr in Engineering Anlysis Fll 7 Number 66 Insrucor: Lrry Creo Sepember Homework Soluions Find he specrum
More informationHow to Prove the Riemann Hypothesis Author: Fayez Fok Al Adeh.
How o Prove he Riemnn Hohesis Auhor: Fez Fok Al Adeh. Presiden of he Srin Cosmologicl Socie P.O.Bo,387,Dmscus,Sri Tels:96377679,735 Emil:hf@scsne.org Commens: 3 ges SubjClss: Funcionl nlsis, comle
More informationHow to prove the Riemann Hypothesis
Scholrs Journl of Phsics, Mhemics nd Sisics Sch. J. Phs. Mh. S. 5; (B:56 Scholrs Acdemic nd Scienific Publishers (SAS Publishers (An Inernionl Publisher for Acdemic nd Scienific Resources *Corresonding
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationgraph of unit step function t
.5 Piecewie coninuou forcing funcion...e.g. urning he forcing on nd off. The following Lplce rnform meril i ueful in yem where we urn forcing funcion on nd off, nd when we hve righ hnd ide "forcing funcion"
More informationMATH 124 AND 125 FINAL EXAM REVIEW PACKET (Revised spring 2008)
MATH 14 AND 15 FINAL EXAM REVIEW PACKET (Revised spring 8) The following quesions cn be used s review for Mh 14/ 15 These quesions re no cul smples of quesions h will pper on he finl em, bu hey will provide
More informationHonours Introductory Maths Course 2011 Integration, Differential and Difference Equations
Honours Inroducory Mhs Course 0 Inegrion, Differenil nd Difference Equions Reding: Ching Chper 4 Noe: These noes do no fully cover he meril in Ching, u re men o supplemen your reding in Ching. Thus fr
More informationSome Inequalities variations on a common theme Lecture I, UL 2007
Some Inequliies vriions on common heme Lecure I, UL 2007 Finbrr Hollnd, Deprmen of Mhemics, Universiy College Cork, fhollnd@uccie; July 2, 2007 Three Problems Problem Assume i, b i, c i, i =, 2, 3 re rel
More informationProperties of Logarithms. Solving Exponential and Logarithmic Equations. Properties of Logarithms. Properties of Logarithms. ( x)
Properies of Logrihms Solving Eponenil nd Logrihmic Equions Properies of Logrihms Produc Rule ( ) log mn = log m + log n ( ) log = log + log Properies of Logrihms Quoien Rule log m = logm logn n log7 =
More informationChapter Direct Method of Interpolation
Chper 5. Direc Mehod of Inerpolion Afer reding his chper, you should be ble o:. pply he direc mehod of inerpolion,. sole problems using he direc mehod of inerpolion, nd. use he direc mehod inerpolns o
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationIntegral Transform. Definitions. Function Space. Linear Mapping. Integral Transform
Inegrl Trnsform Definiions Funcion Spce funcion spce A funcion spce is liner spce of funcions defined on he sme domins & rnges. Liner Mpping liner mpping Le VF, WF e liner spces over he field F. A mpping
More informationA Kalman filtering simulation
A Klmn filering simulion The performnce of Klmn filering hs been esed on he bsis of wo differen dynmicl models, ssuming eiher moion wih consn elociy or wih consn ccelerion. The former is epeced o beer
More informationQuestion Details Int Vocab 1 [ ] Question Details Int Vocab 2 [ ]
/3/5 Assignmen Previewer 3 Bsic: Definie Inegrls (67795) Due: Wed Apr 5 5 9: AM MDT Quesion 3 5 6 7 8 9 3 5 6 7 8 9 3 5 6 Insrucions Red ody's Noes nd Lerning Gols. Quesion Deils In Vocb [37897] The chnge
More informationUnit 5. Integration techniques
18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd
More informationContraction Mapping Principle Approach to Differential Equations
epl Journl of Science echnology 0 (009) 4953 Conrcion pping Principle pproch o Differenil Equions Bishnu P. Dhungn Deprmen of hemics, hendr Rn Cmpus ribhuvn Universiy, Khmu epl bsrc Using n eension of
More informationPHYSICS 1210 Exam 1 University of Wyoming 14 February points
PHYSICS 1210 Em 1 Uniersiy of Wyoming 14 Februry 2013 150 poins This es is opennoe nd closedbook. Clculors re permied bu compuers re no. No collborion, consulion, or communicion wih oher people (oher
More informationAverage & instantaneous velocity and acceleration Motion with constant acceleration
Physics 7: Lecure Reminders Discussion nd Lb secions sr meeing ne week Fill ou Pink dd/drop form if you need o swich o differen secion h is FULL. Do i TODAY. Homework Ch. : 5, 7,, 3,, nd 6 Ch.: 6,, 3 Submission
More informationCollision Detection and Bouncing
Collision Deecion nd Bouncing Collisions re Hndled in Two Prs. Deecing he collision Mike Biley mj@cs.oregonse.edu. Hndling he physics of he collision collisionouncing.ppx If You re Lucky, You Cn Deec
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationPhysics 2A HW #3 Solutions
Chper 3 Focus on Conceps: 3, 4, 6, 9 Problems: 9, 9, 3, 41, 66, 7, 75, 77 Phsics A HW #3 Soluions Focus On Conceps 33 (c) The ccelerion due o grvi is he sme for boh blls, despie he fc h he hve differen
More informationPositive and negative solutions of a boundary value problem for a
Invenion Journl of Reerch Technology in Engineering & Mngemen (IJRTEM) ISSN: 24553689 www.ijrem.com Volume 2 Iue 9 ǁ Sepemer 28 ǁ PP 7383 Poiive nd negive oluion of oundry vlue prolem for frcionl, difference
More informationSolutions to Problems from Chapter 2
Soluions o Problems rom Chper Problem. The signls u() :5sgn(), u () :5sgn(), nd u h () :5sgn() re ploed respecively in Figures.,b,c. Noe h u h () :5sgn() :5; 8 including, bu u () :5sgn() is undeined..5
More informationf(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2
Impope Inegls To his poin we hve only consideed inegls f() wih he is of inegion nd b finie nd he inegnd f() bounded (nd in fc coninuous ecep possibly fo finiely mny jump disconinuiies) An inegl hving eihe
More informationTMatch: Matching Techniques For Driving YagiUda Antennas: TMatch. 2a s. Z in. (Sections 9.5 & 9.7 of Balanis)
3/0/018 _mch.doc Pge 1 of 6 TMch: Mching Techniques For Driving YgiUd Anenns: TMch (Secions 9.5 & 9.7 of Blnis) l s l / l / in The TMch is shunmching echnique h cn be used o feed he driven elemen
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationdt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.
Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies
More informationThe Evaluation Theorem
These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for inclss presenttion nd should not
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 informationChapter 2. First Order Scalar Equations
Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions.
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos(  1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin(  1 ) = π 2 6 2 6 Cn you do similr problems?
More informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationIX.1.1 The Laplace Transform Definition 700. IX.1.2 Properties 701. IX.1.3 Examples 702. IX.1.4 Solution of IVP for ODEs 704
Chper IX The Inegrl Trnform Mehod IX. The plce Trnform November 4, 7 699 IX. THE APACE TRANSFORM IX.. The plce Trnform Definiion 7 IX.. Properie 7 IX..3 Emple 7 IX..4 Soluion of IVP for ODE 74 IX..5 Soluion
More informationu(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.
Lecture 4 Complex Integrtion MATHGA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex
More informationThe problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests.
ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion
More information14.4. Lengths of curves and surfaces of revolution. Introduction. Prerequisites. Learning Outcomes
Lengths of curves nd surfces of revolution 4.4 Introduction Integrtion cn be used to find the length of curve nd the re of the surfce generted when curve is rotted round n xis. In this section we stte
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 informationMagnetostatics Bar Magnet. Magnetostatics Oersted s Experiment
Mgneosics Br Mgne As fr bck s 4500 yers go, he Chinese discovered h cerin ypes of iron ore could rc ech oher nd cerin mels. Iron filings "mp" of br mgne s field Crefully suspended slivers of his mel were
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.
Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos(  1 2 ) = rcsin( 1 2 ) = rcsin(  1 2 ) = Cn you do similr problems? Review of Bsic Concepts
More informationSection 7.1 Area of a Region Between Two Curves
Section 7.1 Are of Region Between Two Curves White Bord Chllenge The circle elow is inscried into squre: Clcultor 0 cm Wht is the shded re? 400 100 85.841cm White Bord Chllenge Find the re of the region
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationHermiteHadamardFejér type inequalities for convex functions via fractional integrals
Sud. Univ. BeşBolyi Mh. 6(5, No. 3, 355 366 HermieHdmrdFejér ype inequliies for convex funcions vi frcionl inegrls İmd İşcn Asrc. In his pper, firsly we hve eslished Hermie HdmrdFejér inequliy for
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its crosssection in plne pssing through
More informationThe solution is often represented as a vector: 2xI + 4X2 + 2X3 + 4X4 + 2X5 = 4 2xI + 4X2 + 3X3 + 3X4 + 3X5 = 4. 3xI + 6X2 + 6X3 + 3X4 + 6X5 = 6.
[~ o o : o o ill] i 1. Mrices, Vecors, nd GussJordn Eliminion 1 x y = =  z= The soluion is ofen represened s vecor: n his exmple, he process of eliminion works very smoohly. We cn elimine ll enries
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 informationA 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m
PHYS : Soluions o Chper 3 Home Work. SSM REASONING The displcemen is ecor drwn from he iniil posiion o he finl posiion. The mgniude of he displcemen is he shores disnce beween he posiions. Noe h i is onl
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationr 0 ( ) cos( ) r( )sin( ). 1. Last time, we calculated that for the cardioid r( ) =1+sin( ),
Wrm up Recll from lst time, given polr curve r = r( ),, dx dy dx = dy d = (r( )sin( )) d (r( ) cos( )) = r0 ( )sin( )+r( ) cos( ) r 0 ( ) cos( ) r( )sin( ).. Lst time, we clculted tht for crdioid r( )
More informationSpring 2017 Exam 1 MARK BOX HAND IN PART PIN: 17
Spring 07 Exm problem MARK BOX points HAND IN PART 0 555=x5 0 NAME: Solutions 3 0 0 PIN: 7 % 00 INSTRUCTIONS This exm comes in two prts. () HAND IN PART. Hnd in only this prt. () STATEMENT OF MULTIPLE
More informationIX.1.1 The Laplace Transform Definition 700. IX.1.2 Properties 701. IX.1.3 Examples 702. IX.1.4 Solution of IVP for ODEs 704
Chper IX The Inegrl Trnform Mehod IX. The plce Trnform November 6, 8 699 IX. THE APACE TRANSFORM IX.. The plce Trnform Definiion 7 IX.. Properie 7 IX..3 Emple 7 IX..4 Soluion of IVP for ODE 74 IX..5 Soluion
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More information7. Indefinite Integrals
7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
More informationWeek #13  Integration by Parts & Numerical Integration Section 7.2
Week #3  Inegraion by Pars & Numerical Inegraion Secion 7. From Calculus, Single Variable by HughesHalle, Gleason, McCallum e. al. Copyrigh 5 by John Wiley & Sons, Inc. This maerial is used by permission
More informationSome basic notation and terminology. Deterministic Finite Automata. COMP218: Decision, Computation and Language Note 1
COMP28: Decision, Compuion nd Lnguge Noe These noes re inended minly s supplemen o he lecures nd exooks; hey will e useful for reminders ou noion nd erminology. Some sic noion nd erminology An lphe is
More information14. The fundamental theorem of the calculus
4. The funmenl heorem of he clculus V 20 00 80 60 40 20 0 0 0.2 0.4 0.6 0.8 v 400 200 0 0 0.2 0.5 0.8 200 400 Figure : () Venriculr volume for subjecs wih cpciies C = 24 ml, C = 20 ml, C = 2 ml n (b) he
More informationMATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.
MATH 3B: MIDTERM REVIEW JOE HUGHES. Inegraion by Pars. Evaluae 3 e. Soluion: Firs make he subsiuion u =. Then =, hence 3 e = e = ue u Now inegrae by pars o ge ue u = ue u e u + C and subsiue he definiion
More informationChapter 2: Evaluative Feedback
Chper 2: Evluive Feedbck Evluing cions vs. insrucing by giving correc cions Pure evluive feedbck depends olly on he cion ken. Pure insrucive feedbck depends no ll on he cion ken. Supervised lerning is
More informationEXERCISE  01 CHECK YOUR GRASP
UNIT # 09 PARABOLA, ELLIPSE & HYPERBOLA PARABOLA EXERCISE  0 CHECK YOUR GRASP. Hin : Disnce beween direcri nd focus is 5. Given (, be one end of focl chord hen oher end be, lengh of focl chord 6. Focus
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationFall 2017 Exam 1 MARK BOX HAND IN PART PIN: 17
Fll 7 Exm problem MARK BOX points HAND IN PART 35=x5 NAME: Solutions PIN: 7 % INSTRUCTIONS This exm comes in two prts. () HAND IN PART. Hnd in only this prt. () STATEMENT OF MULTIPLE CHOICE PROBLEMS.
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 informationRESPONSE UNDER A GENERAL PERIODIC FORCE. When the external force F(t) is periodic with periodτ = 2π
RESPONSE UNDER A GENERAL PERIODIC FORCE When he exernl force F() is periodic wih periodτ / ω,i cn be expnded in Fourier series F( ) o α ω α b ω () where τ F( ) ω d, τ,,,... () nd b τ F( ) ω d, τ,,... (3)
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 informationES.182A Topic 32 Notes Jeremy Orloff
ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In
More informationExploring parametric representation with the TI84 Plus CE graphing calculator
Exploring prmetric representtion with the TI84 Plus CE grphing clcultor Richrd Prr Executive Director Rice University School Mthemtics Project rprr@rice.edu Alice Fisher Director of Director of Technology
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
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 informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More informationECE Microwave Engineering
EE 537635 Microwve Engineering Adped from noes y Prof. Jeffery T. Willims Fll 8 Prof. Dvid R. Jcson Dep. of EE Noes Wveguiding Srucures Pr 7: Trnsverse Equivlen Newor (N) Wveguide Trnsmission Line Model
More informationChapters Five Notes SN AA U1C5
Chpters Five Notes SN AA U1C5 Nme Period Section 5: Fctoring Qudrtic Epressions When you took lger, you lerned tht the first thing involved in fctoring is to mke sure to fctor out ny numers or vriles
More informationECE Microwave Engineering. Fall Prof. David R. Jackson Dept. of ECE. Notes 10. Waveguides Part 7: Transverse Equivalent Network (TEN)
EE 537635 Microwve Engineering Fll 7 Prof. Dvid R. Jcson Dep. of EE Noes Wveguides Pr 7: Trnsverse Equivlen Newor (N) Wveguide Trnsmission Line Model Our gol is o come up wih rnsmission line model for
More informationGENERALIZATION OF SOME INEQUALITIES VIA RIEMANNLIOUVILLE FRACTIONAL CALCULUS
 TAMKANG JOURNAL OF MATHEMATICS Volume 5, Number, 75, June doi:5556/jkjm555 Avilble online hp://journlsmhkueduw/    GENERALIZATION OF SOME INEQUALITIES VIA RIEMANNLIOUVILLE FRACTIONAL CALCULUS MARCELA
More informationMath 190 Chapter 5 Lecture Notes. Professor Miguel Ornelas
Mth 19 Chpter 5 Lecture Notes Professor Miguel Ornels 1 M. Ornels Mth 19 Lecture Notes Section 5.1 Section 5.1 Ares nd Distnce Definition The re A of the region S tht lies under the grph of the continuous
More informationMath 113 Exam 2 Practice
Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = 2.
Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot
More information