Some Methods in the Calculus of Variations

Size: px
Start display at page:

Download "Some Methods in the Calculus of Variations"

Transcription

1 CHAPTER 6 Some Methods in the Clculus of Vritions 6-. If we use the vried function ( α, ) α sin( ) + () Then d α cos ( ) () d Thus, the totl length of the pth is d S + d d α cos ( ) + α cos ( ) d Setting ( ) u, the epression for S becomes S α cosu + α cos u du (4) The integrl cnnot be performed directl since it is, in fct, n elliptic integrl. Becuse α is smll quntit, we cn epnd the integrnd nd obtin + 8 du S α cos u α cos u α cos u α cos u If we keep the terms up to co s u nd perform the integrtion, we find (5) which gives S 6 + α (6) 65

2 66 CHAPTER 6 Therefore S α 8 α (7) S α (8) nd S is minimum when α. 6-. The element of length on plne is ds d + d () from which the totl length is If S is to be minimum, f is identified s ( ) d S d + d + (, ) d, d () f d + d Then, the Euler eqution becomes d d d d + (4) where d. (4) becomes d or, d d + + (5) constnt C (6) from which we hve Then, C C constnt (7) + b (8) This is the eqution of stright line.

3 SOME METHODS IN THE CALCULUS OF VARIATIONS The element of distnce in three-dimensionl spce is ds d + d + dz () Suppose,, z depends on the prmeter t nd tht the end points re epressed b ( ( t), ( t), z( t ) ), ( ), ( ), ( ) ( ) t t z t. Then the totl distnce is The function f is identified s t t d d dz S + + dt dt dt dt () f + + z f f f Since, the Euler equtions become z from which we hve d dt d dt d dt f f f z constnt C + + z constnt C + + z z constnt C z (4) (5) From the combintion of these equtions, we hve C C C z C 3 If we integrte (6) from t to the rbitrr t, we hve (6)

4 68 CHAPTER 6 C C z z C C 3 On the other hnd, the integrtion of (6) from to t gives from which we find the constnts t C C z z C C 3 C, C, nd C 3. Substituting these constnts into (7), we find z z (9) z z This is the eqution epressing stright line in three-dimensionl spce pssing through the,,,, z. two points ( z ), ( ) (7) (8) 6-4. z φ ρ ds The element of distnce long the surfce is ds d + d + dz () In clindricl coordintes (,,z) re relted to (ρ,φ,z) b ρ cos φ ρ sin φ z z () from which d ρsin φ dφ d ρcos φ dφ dz dz

5 SOME METHODS IN THE CALCULUS OF VARIATIONS 69 Substituting into () nd integrting long the entire pth, we find φ S ρ dφ + dz ρ + z dφ (4) φ where dz z. If S is to be minimum, dφ f ρ + z must stisf the Euler eqution: f f z φ z (5) f Since, the Euler eqution becomes z z φ ρ + z (6) from which or, z ρ + z constnt C (7) Since ρ is constnt, (8) mens C z C ρ (8) dz constnt dφ nd for n point long the pth, z nd φ chnge t the sme rte. The curve described b this condition is heli The re of strip of surfce of revolution is z (, ) ds (, ) da ds d + d () Thus, the totl re is

6 7 CHAPTER 6 d () A + where d. In order to mke A minimum, f d f + f + + must stisf eqution (6.39). Now Substituting into eqution (6.39) gives d d d + d ( )( ) + dd + + Multipling b Integrtion gives + nd rerrnging gives d d ( + ) ln + ln ln + where ln is constnt of integrtion. Rerrnging gives Integrting gives or ( ) b cosh + which is the eqution of ctenr. cosh b

7 SOME METHODS IN THE CALCULUS OF VARIATIONS θ θ (, ) (, ) If we use coordintes with the sme orienttion s in Emple 6. nd if we plce the minimum point of the ccloid t (,) the prmetric equtions re ( cosθ ) + ( θ + sin θ) Since the prticle strts from rest t the point (, ) 6.9] Then, the time required to rech the point (, ), the velocit t n elevtion is [cf. Eq. ( ) v g () is [cf. Eq. 6.] + t g( ) d Using () nd the derivtives obtined therefrom, cn be written s θ + cosθ t dθ g cosθ cosθ θ (4) () Now, using the trigonometric identit, θ θ + cos cos, we hve t g θ θ cos d θ θ θ cos cos g θ θ cos d θ (5) θ θ sin sin Mking the chnge of vrible, z sin θ, the epression for t becomes t g θ sin dz (6) θ sin z The integrl is now in stndrd form:

8 7 CHAPTER 6 Evluting, we find d sin (7) Thus, the time of trnsit from (, ) the strting point. t (8) g to the minimum point does not depend on the position of 6-7. θ θ n n (n > n ) The time to trvel the pth shown is (cf. Emple 6.) b d ds t v θ v v + v c n c n d () Although we hve v v(), we onl hve dv d when. The Euler eqution tells us d d v + Now use v c n nd tn θ to obtin () n sin θ const. This proves the ssertion. Alterntivel, Fermt s principle cn be proven b the method introduced in the solution of Problem To find the etremum of the following integrl (cf. Eqution 6.) (, ) J f d we know tht we must hve from Euler s eqution f This implies tht we lso hve

9 SOME METHODS IN THE CALCULUS OF VARIATIONS 73 J f d giving us modified form of Euler s eqution. This m be etended to severl vribles nd to include the imposition of uilir conditions similr to the derivtion in Sections 6.5 nd 6.6. The result is J g + λ j when there re constrint equtions of the form i j ( i ) j ( ) g, ) The volume of prllelepiped with sides of lengths, b, c is given b j i V b c () We wish to mimize such volume under the condition tht the prllelepiped is circumscribed b sphere of rdius R; tht is, + b + c R 4 We consider, b, c s vribles nd V is the function tht we wnt to mimize; () is the constrint condition: Then, the equtions for the solution re { } g, b, c () V g + λ V g + λ b b V g + λ c c (4) from which we obtin bc + λ c + λb (5) Together with (), these equtions ield b + λc Thus, the inscribed prllelepiped is cube with side b c R (6) 3 3 R.

10 74 CHAPTER 6 b) In the sme w, if the prllelepiped is now circumscribed b n ellipsoid with semies, b, c, the constrint condition is given b b c (7) 4 4b 4z where, b, c re the lengths of the sides of the prllelepiped. Combining (7) with () nd (4) gives Then, b c (8) b c, b b, c c (9) 6-9. The verge vlue of the squre of the grdient of (,, 3) is epressed s I φ V ( ) d d d3 φ within certin volume V φ φ φ v V + + d d d3 () 3 In order to mke I minimum, must stisf the Euler eqution: If we substitute f into (), we hve which is just Lplce seqution: f φ φ φ f f φ i i φ () i 3 φ i i i φ (4) Therefore, φ must stisf Lplce s eqution in order tht I hve minimum vlue.

11 } SOME METHODS IN THE CALCULUS OF VARIATIONS This problem lends itself to the method of solution suggested in the solution of Problem 6-8. The volume of right clinder is given b V R H The totl surfce re A of the clinder is given b bses side ( ) A A + A R + RH R R+H () We wish A to be minimum. () is the constrint condition, nd the other equtions re where g V R H. The solution of these equtions is A g + λ R R A g + λ H H R H (4) () 6-. R θ ds The constrint condition cn be found from the reltion ds Rdθ (see the digrm), where ds is the differentil rc length of the pth: ( ) ds d + d Rdθ () which, using, ields + 4 d Rdθ () If we wnt the eqution of constrint in other thn differentil form, () cn be integrted to ield ( ) A+ R θ ln where A is constnt obtined from the initil conditions. The rdius of curvture of prbol,, is given t n point (,) b r. The condition for the disk to roll with one nd onl one point of contct with the prbol is R< r ; tht is,

12 76 CHAPTER 6 R < (4) 6-. The pth length is given b nd our eqution of constrint is () + + s ds z d ( ) g,, z + + z ρ () The Euler equtions with undetermined multipliers (6.69) tell us tht d dg λ λ d + + z d with similr eqution for z. Eliminting the fctor λ, we obtin This simplifies to d d z d z zd z ( ) ( ) ( ) ( ) z + + z + z z z + + z z + z z nd using the derivtive of (), ( zz ) z z ( + zz ) z z + + (6) ( ) ( ) z z z (7) This looks to be in the simplest form we cn mke it, but is it plne? Tke the eqution of plne pssing through the origin: A+ B z (8) nd mke it differentil eqution b tking derivtives (giving A + B z nd B z ) nd eliminting the constnts. The substitution ields (7) ectl. This confirms tht the pth must be the intersection of the sphere with plne pssing through the origin, s required. (4) (5) 6-3. For the reson of convenience, without lost of generlit, suppose tht the closed curve psses through fied points A(-,) nd B(,) (which hve been chosen to be on is O). We denote the prt of the closed curve bove nd below the O is s ( ) nd ( ) respectivel. (note tht > nd < ) The enclosed re is ( ) J(, ) ( ) d ( ) d ( ) ( ) d f(, ) d

13 SOME METHODS IN THE CALCULUS OF VARIATIONS 77 The totl length of closed curve is { } ( ) ( ) + ( ) + + ( ) + ( ) + + ( ) K, d d d g, d Then the generlized versions of Eq. (6.78) (see tetbook) for this cse re f d f g d g d + λ λ () d d d + ( ) f d f g d g d + λ λ d d d + ( ) () Anlogousl to Eq. (6.85); from () we obtin ( ) ( ) from () we obtin ( ) ( ) A + A λ B + B λ (4) where constnts A s, B s cn be determined from 4 initil conditions ( ±, ) nd ( ±, ) We note tht < nd >, so ctull nd (4) ltogether describe circulr pth of rdius λ. And this is the sought configurtion tht renders mimum enclosed re for given pth length It is more convenient to work with clindricl coordintes (r,φ,z) in this problem. The constrint here is z r, then dz dr ( ) ds dr + r dφ + dz dr + r dβ φ where we hve introduced new ngulr coordinte β In this form of ds, we clerl see tht the spce is -dimensionl Eucliden flt, so the shortest line connecting two given points is stright line given b: r r r cos ( β β ) φ φ cos this line psses through the endpoints (r, φ ± ), then we cn determine unmbiguousl the shortest pth eqution cos r( φ) φ cos nd z r

14 78 CHAPTER 6 Accordingl, the shortest connecting length is dr dφ r sin dφ l d d I [ ] d ) Treting I[] s mechnicl ction, we find the corresponding Euler-Lgrnge eqution d ( ) d Combining with the boundr conditions (, ) nd (, ), we cn determine unmbiguousl the functionl form of ( ) (sin ) (sin). b) The corresponding minimum vlue of the integrl is d I [ ] d dcos cot ().64 d sin c) If then I[] ( 3 ) ) S is rc length d dz d 9 S d + d + dz d + + d + + Ld d d d 4 Treting S nd L like mechnicl ction nd Lgrngin respectivel, we find the cnonicl momentum ssocited with coordinte d δl p d d δ 9 d d d Becuse L does not depend on eplicitl, then E-L eqution implies tht p is constnt (i.e. dp d ), then the bove eqution becomes 9 3 d + 4 p 9 9 p d + A + d p p + B 4 4 where A nd B re constnts. Using boundr conditions (, ) nd (, ) one cn determine the rc eqution unmbiguousl

15 SOME METHODS IN THE CALCULUS OF VARIATIONS ( ) nd z 3 b) z ) Eqution of ellipse which implies + b b becuse + b b so the miml re of the rectngle, whose corners lie on tht ellipse, is This hppens when b) The re of the ellipse is A M[A] M[4] ib. nd b b ; so the frction of rectngle re to ellipse re is then M[ A] A 6-8. One cn see tht the surfce z is locll smmetric with respect to the line z where >, <, z <. This line is prbol. This implies tht if the prticle strts from point (,-,-) (which belongs to the smmetr line) under grvit idell will move downwrd long this line. Its velocit t ltitude z (z < ) cn be found from the conservtion of energ. vz ( ) gz ( + )

16 8 CHAPTER 6

Eigen Values and Eigen Vectors of a given matrix

Eigen Values and Eigen Vectors of a given matrix Engineering Mthemtics 0 SUBJECT NAME SUBJECT CODE MATERIAL NAME MATERIAL CODE : Engineering Mthemtics I : 80/MA : Prolem Mteril : JM08AM00 (Scn the ove QR code for the direct downlod of this mteril) Nme

More information

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: olumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge

More information

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: Volumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge

More information

50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS

50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS 68 CHAPTE MULTIPLE INTEGALS 46. e da, 49. Evlute tn 3 4 da, where,. [Hint: Eploit the fct tht is the disk with center the origin nd rdius is smmetric with respect to both es.] 5. Use smmetr to evlute 3

More information

We divide the interval [a, b] into subintervals of equal length x = b a n

We 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 information

ES.182A Topic 32 Notes Jeremy Orloff

ES.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 information

Reference. Vector Analysis Chapter 2

Reference. Vector Analysis Chapter 2 Reference Vector nlsis Chpter Sttic Electric Fields (3 Weeks) Chpter 3.3 Coulomb s Lw Chpter 3.4 Guss s Lw nd pplictions Chpter 3.5 Electric Potentil Chpter 3.6 Mteril Medi in Sttic Electric Field Chpter

More information

13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes

13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce

More information

4. Calculus of Variations

4. Calculus of Variations 4. Clculus of Vritions Introduction - Typicl Problems The clculus of vritions generlises the theory of mxim nd minim. Exmple (): Shortest distnce between two points. On given surfce (e.g. plne), nd the

More information

(b) Let S 1 : f(x, y, z) = (x a) 2 + (y b) 2 + (z c) 2 = 1, this is a level set in 3D, hence

(b) Let S 1 : f(x, y, z) = (x a) 2 + (y b) 2 + (z c) 2 = 1, this is a level set in 3D, hence Problem ( points) Find the vector eqution of the line tht joins points on the two lines L : r ( + t) i t j ( + t) k L : r t i + (t ) j ( + t) k nd is perpendiculr to both those lines. Find the set of ll

More information

Physics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016

Physics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016 Physics 333, Fll 16 Problem Set 7 due Oct 14, 16 Reding: Griffiths 4.1 through 4.4.1 1. Electric dipole An electric dipole with p = p ẑ is locted t the origin nd is sitting in n otherwise uniform electric

More information

Week 10: Line Integrals

Week 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 information

Total Score Maximum

Total Score Maximum Lst Nme: Mth 8: Honours Clculus II Dr. J. Bowmn 9: : April 5, 7 Finl Em First Nme: Student ID: Question 4 5 6 7 Totl Score Mimum 6 4 8 9 4 No clcultors or formul sheets. Check tht you hve 6 pges.. Find

More information

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.

Space Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space. Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)

More information

Mathematics. Area under Curve.

Mathematics. Area under Curve. Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding

More information

Partial Derivatives. Limits. For a single variable function f (x), the limit lim

Partial 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 information

13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes

13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce

More information

Lecture XVII. Vector functions, vector and scalar fields Definition 1 A vector-valued function is a map associating vectors to real numbers, that is

Lecture XVII. Vector functions, vector and scalar fields Definition 1 A vector-valued function is a map associating vectors to real numbers, that is Lecture XVII Abstrct We introduce the concepts of vector functions, sclr nd vector fields nd stress their relevnce in pplied sciences. We study curves in three-dimensionl Eucliden spce nd introduce the

More information

Thomas Whitham Sixth Form

Thomas Whitham Sixth Form Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos

More information

A. Limits - L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. -1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1.

A. Limits - L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. -1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1. A. Limits - L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( ) lim where lim f or lim f limg. c g = c limg( ) = c = c = c How to find it: Try nd find limits by

More information

PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by

PROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by PROPERTES OF RES Centroid The concept of the centroid is prol lred fmilir to ou For plne shpe with n ovious geometric centre, (rectngle, circle) the centroid is t the centre f n re hs n is of smmetr, the

More information

Chapter 9. Arc Length and Surface Area

Chapter 9. Arc Length and Surface Area Chpter 9. Arc Length nd Surfce Are In which We ppl integrtion to stud the lengths of curves nd the re of surfces. 9. Arc Length (Tet 547 553) P n P 2 P P 2 n b P i ( i, f( i )) P i ( i, f( i )) distnce

More information

HOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016

HOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016 HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M

More information

Not for reproduction

Not for reproduction AREA OF A SURFACE OF REVOLUTION cut h FIGURE FIGURE πr r r l h FIGURE A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundry of solid of revolution of the type

More information

, MATHS H.O.D.: SUHAG R.KARIYA, BHOPAL, CONIC SECTION PART 8 OF

, MATHS H.O.D.: SUHAG R.KARIYA, BHOPAL, CONIC SECTION PART 8 OF DOWNLOAD FREE FROM www.tekoclsses.com, PH.: 0 903 903 7779, 98930 5888 Some questions (Assertion Reson tpe) re given elow. Ech question contins Sttement (Assertion) nd Sttement (Reson). Ech question hs

More information

CONIC SECTIONS. Chapter 11

CONIC SECTIONS. Chapter 11 CONIC SECTIONS Chpter. Overview.. Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig..). Fig.. Suppose we rotte the line m round

More information

Solutions to Problems Integration in IR 2 and IR 3

Solutions to Problems Integration in IR 2 and IR 3 Solutions to Problems Integrtion in I nd I. For ec of te following, evlute te given double integrl witout using itertion. Insted, interpret te integrl s, for emple, n re or n verge vlue. ) dd were is te

More information

along the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate

along the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate L8 VECTOR EQUATIONS OF LINES HL Mth - Sntowski Vector eqution of line 1 A plne strts journey t the point (4,1) moves ech hour long the vector. ) Find the plne s coordinte fter 1 hour. b) Find the plne

More information

A LEVEL TOPIC REVIEW. factor and remainder theorems

A LEVEL TOPIC REVIEW. factor and remainder theorems A LEVEL TOPIC REVIEW unit C fctor nd reminder theorems. Use the Fctor Theorem to show tht: ) ( ) is fctor of +. ( mrks) ( + ) is fctor of ( ) is fctor of + 7+. ( mrks) +. ( mrks). Use lgebric division

More information

Calculus 2: Integration. Differentiation. Integration

Calculus 2: Integration. Differentiation. Integration Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is

More information

Classical Mechanics. From Molecular to Con/nuum Physics I WS 11/12 Emiliano Ippoli/ October, 2011

Classical Mechanics. From Molecular to Con/nuum Physics I WS 11/12 Emiliano Ippoli/ October, 2011 Clssicl Mechnics From Moleculr to Con/nuum Physics I WS 11/12 Emilino Ippoli/ October, 2011 Wednesdy, October 12, 2011 Review Mthemtics... Physics Bsic thermodynmics Temperture, idel gs, kinetic gs theory,

More information

Chapter 2. Constraints, Lagrange s equations

Chapter 2. Constraints, Lagrange s equations Chpter Constrints, Lgrnge s equtions Section Constrints The position of the prticle or system follows certin rules due to constrints: Holonomic constrint: f (r. r,... r n, t) = 0 Constrints tht re not

More information

ES.182A Topic 30 Notes Jeremy Orloff

ES.182A Topic 30 Notes Jeremy Orloff ES82A opic 3 Notes Jerem Orloff 3 Non-independent vribles: chin rule Recll the chin rule: If w = f, ; nd = r, t, = r, t then = + r t r t r t = + t t t r nfortuntel, sometimes there re more complicted reltions

More information

Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Jackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The two-dimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero

More information

a < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1

a < 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 information

Indefinite Integral. Chapter Integration - reverse of differentiation

Indefinite 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 information

1 Part II: Numerical Integration

1 Part II: Numerical Integration Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble

More information

Math 32B Discussion Session Session 7 Notes August 28, 2018

Math 32B Discussion Session Session 7 Notes August 28, 2018 Mth 32B iscussion ession ession 7 Notes August 28, 28 In tody s discussion we ll tlk bout surfce integrls both of sclr functions nd of vector fields nd we ll try to relte these to the mny other integrls

More information

A. Limits - L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C.

A. Limits - L Hopital s Rule ( ) How to find it: Try and find limits by traditional methods (plugging in). If you get 0 0 or!!, apply C.! 1 6 C. A. Limits - L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( x) lim where lim f x x! c g x ( ) = or lim f ( x) = limg( x) = ". ( ) x! c limg( x) = 0 x! c x! c

More information

x 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx

x 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx . Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute

More information

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus

ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry and basic calculus ES 111 Mthemticl Methods in the Erth Sciences Lecture Outline 1 - Thurs 28th Sept 17 Review of trigonometry nd bsic clculus Trigonometry When is it useful? Everywhere! Anything involving coordinte systems

More information

MA1104 Multivariable Calculus Lecture Notes 1. Wong Yan Loi

MA1104 Multivariable Calculus Lecture Notes 1. Wong Yan Loi MA114 Multivrible lculus Lecture Notes 1 Wong Yn Loi eprtment of Mthemtics, Ntionl Universit of ingpore, ingpore 11976 1 This notes is written eclusivel for students tking the modules MA114, Multivrible

More information

( β ) touches the x-axis if = 1

( β ) touches the x-axis if = 1 Generl Certificte of Eduction (dv. Level) Emintion, ugust Comined Mthemtics I - Prt B Model nswers. () Let f k k, where k is rel constnt. i. Epress f in the form( ) Find the turning point of f without

More information

Further integration. x n nx n 1 sinh x cosh x log x 1/x cosh x sinh x e x e x tan x sec 2 x sin x cos x tan 1 x 1/(1 + x 2 ) cos x sin x

Further integration. x n nx n 1 sinh x cosh x log x 1/x cosh x sinh x e x e x tan x sec 2 x sin x cos x tan 1 x 1/(1 + x 2 ) cos x sin x Further integrtion Stndrd derivtives nd integrls The following cn be thought of s list of derivtives or eqully (red bckwrds) s list of integrls. Mke sure you know them! There ren t very mny. f(x) f (x)

More information

1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.

1.2. Linear Variable Coefficient Equations. y + b ! = a y + b  Remark: The case b = 0 and a non-constant can be solved with the same idea as above. 1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt

More information

The Basic Functional 2 1

The Basic Functional 2 1 2 The Bsic Functionl 2 1 Chpter 2: THE BASIC FUNCTIONAL TABLE OF CONTENTS Pge 2.1 Introduction..................... 2 3 2.2 The First Vrition.................. 2 3 2.3 The Euler Eqution..................

More information

Chapter 6 Techniques of Integration

Chapter 6 Techniques of Integration MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln

More information

P 1 (x 1, y 1 ) is given by,.

P 1 (x 1, y 1 ) is given by,. MA00 Clculus nd Bsic Liner Alger I Chpter Coordinte Geometr nd Conic Sections Review In the rectngulr/crtesin coordintes sstem, we descrie the loction of points using coordintes. P (, ) P(, ) O The distnce

More information

5.7 Improper Integrals

5.7 Improper Integrals 458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the

More information

CET MATHEMATICS 2013

CET MATHEMATICS 2013 CET MATHEMATICS VERSION CODE: C. If sin is the cute ngle between the curves + nd + 8 t (, ), then () () () Ans: () Slope of first curve m ; slope of second curve m - therefore ngle is o A sin o (). The

More information

Math 113 Exam 1-Review

Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

More information

Math 115 ( ) Yum-Tong Siu 1. Lagrange Multipliers and Variational Problems with Constraints. F (x,y,y )dx

Math 115 ( ) Yum-Tong Siu 1. Lagrange Multipliers and Variational Problems with Constraints. F (x,y,y )dx Mth 5 2006-2007) Yum-Tong Siu Lgrnge Multipliers nd Vritionl Problems with Constrints Integrl Constrints. Consider the vritionl problem of finding the extremls for the functionl J[y] = F x,y,y )dx with

More information

Geometric and Mechanical Applications of Integrals

Geometric and Mechanical Applications of Integrals 5 Geometric nd Mechnicl Applictions of Integrls 5.1 Computing Are 5.1.1 Using Crtesin Coordintes Suppose curve is given by n eqution y = f(x), x b, where f : [, b] R is continuous function such tht f(x)

More information

Year 12 Mathematics Extension 2 HSC Trial Examination 2014

Year 12 Mathematics Extension 2 HSC Trial Examination 2014 Yer Mthemtics Etension HSC Tril Emintion 04 Generl Instructions. Reding time 5 minutes Working time hours Write using blck or blue pen. Blck pen is preferred. Bord-pproved clcultors my be used A tble of

More information

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

x = 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 information

1.9 C 2 inner variations

1.9 C 2 inner variations 46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for

More information

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved. Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite

More information

Topic 1 Notes Jeremy Orloff

Topic 1 Notes Jeremy Orloff Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble

More information

10 Vector Integral Calculus

10 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 information

( ) Same as above but m = f x = f x - symmetric to y-axis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists.

( ) Same as above but m = f x = f x - symmetric to y-axis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists. AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find

More information

US01CMTH02 UNIT Curvature

US01CMTH02 UNIT Curvature Stu mteril of BSc(Semester - I) US1CMTH (Rdius of Curvture nd Rectifiction) Prepred by Nilesh Y Ptel Hed,Mthemtics Deprtment,VPnd RPTPScience College US1CMTH UNIT- 1 Curvture Let f : I R be sufficiently

More information

Section 14.3 Arc Length and Curvature

Section 14.3 Arc Length and Curvature Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in

More information

Math 100 Review Sheet

Math 100 Review Sheet Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s

More information

First midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009

First midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009 Mth 3B Review Steve 18 Mrch 2009 About the finl Fridy Mrch 20, 3pm-6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No

More information

2. VECTORS AND MATRICES IN 3 DIMENSIONS

2. VECTORS AND MATRICES IN 3 DIMENSIONS 2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2-dimensionl Vectors x A point in 3-dimensionl spce cn e represented y column vector of the form y z z-xis y-xis z x y x-xis Most of the

More information

Section - 2 MORE PROPERTIES

Section - 2 MORE PROPERTIES LOCUS Section - MORE PROPERTES n section -, we delt with some sic properties tht definite integrls stisf. This section continues with the development of some more properties tht re not so trivil, nd, when

More information

Time : 3 hours 03 - Mathematics - March 2007 Marks : 100 Pg - 1 S E CT I O N - A

Time : 3 hours 03 - Mathematics - March 2007 Marks : 100 Pg - 1 S E CT I O N - A Time : hours 0 - Mthemtics - Mrch 007 Mrks : 100 Pg - 1 Instructions : 1. Answer ll questions.. Write your nswers ccording to the instructions given below with the questions.. Begin ech section on new

More information

Show that the curve with parametric equations x t cos t, ; Use a computer to graph the curve with the given vector

Show that the curve with parametric equations x t cos t, ; Use a computer to graph the curve with the given vector 7 CHAPTER VECTOR FUNCTIONS. Eercises 2 Find the domin of the vector function.. 2. 3 4 Find the limit. 3. lim cos t, sin t, t ln t t l 4. lim t, e t l rctn 2t, ln t t 5 Mtch the prmetric equtions with the

More information

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions

AQA Further Pure 2. Hyperbolic Functions. Section 2: The inverse hyperbolic functions Hperbolic Functions Section : The inverse hperbolic functions Notes nd Emples These notes contin subsections on The inverse hperbolic functions Integrtion using the inverse hperbolic functions Logrithmic

More information

k ) and directrix x = h p is A focal chord is a line segment which passes through the focus of a parabola and has endpoints on the parabola.

k ) and directrix x = h p is A focal chord is a line segment which passes through the focus of a parabola and has endpoints on the parabola. Stndrd Eqution of Prol with vertex ( h, k ) nd directrix y = k p is ( x h) p ( y k ) = 4. Verticl xis of symmetry Stndrd Eqution of Prol with vertex ( h, k ) nd directrix x = h p is ( y k ) p( x h) = 4.

More information

Partial Differential Equations

Partial Differential Equations Prtil Differentil Equtions Notes by Robert Piché, Tmpere University of Technology reen s Functions. reen s Function for One-Dimensionl Eqution The reen s function provides complete solution to boundry

More information

4 The dynamical FRW universe

4 The dynamical FRW universe 4 The dynmicl FRW universe 4.1 The Einstein equtions Einstein s equtions G µν = T µν (7) relte the expnsion rte (t) to energy distribution in the universe. On the left hnd side is the Einstein tensor which

More information

Lecture 13 - Linking E, ϕ, and ρ

Lecture 13 - Linking E, ϕ, and ρ Lecture 13 - Linking E, ϕ, nd ρ A Puzzle... Inner-Surfce Chrge Density A positive point chrge q is locted off-center inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

R(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of

R(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of Higher Mthemtics Ojective Test Prctice ook The digrm shows sketch of prt of the grph of f ( ). The digrm shows sketch of the cuic f ( ). R(, 8) f ( ) f ( ) P(, ) Q(, ) S(, ) Wht re the domin nd rnge of

More information

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

More information

Appendix 3, Rises and runs, slopes and sums: tools from calculus

Appendix 3, Rises and runs, slopes and sums: tools from calculus Appendi 3, Rises nd runs, slopes nd sums: tools from clculus Sometimes we will wnt to eplore how quntity chnges s condition is vried. Clculus ws invented to do just this. We certinly do not need the full

More information

Note 16. Stokes theorem Differential Geometry, 2005

Note 16. Stokes theorem Differential Geometry, 2005 Note 16. Stokes theorem ifferentil Geometry, 2005 Stokes theorem is the centrl result in the theory of integrtion on mnifolds. It gives the reltion between exterior differentition (see Note 14) nd integrtion

More information

Chapter 6 Notes, Larson/Hostetler 3e

Chapter 6 Notes, Larson/Hostetler 3e Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn

More information

Mathematics Extension 2

Mathematics Extension 2 00 HIGHER SCHOOL CERTIFICATE EXAMINATION Mthemtics Extension Generl Instructions Reding time 5 minutes Working time hours Write using blck or blue pen Bord-pproved clcultors m be used A tble of stndrd

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties 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 information

4.4 Areas, Integrals and Antiderivatives

4.4 Areas, Integrals and Antiderivatives . res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order

More information

Student Handbook for MATH 3300

Student Handbook for MATH 3300 Student Hndbook for MATH 3300 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 0.5 0 0.5 0.5 0 0.5 If people do not believe tht mthemtics is simple, it is only becuse they do not relize how complicted life is. John Louis

More information

10.5. ; 43. The points of intersection of the cardioid r 1 sin and. ; Graph the curve and find its length. CONIC SECTIONS

10.5. ; 43. The points of intersection of the cardioid r 1 sin and. ; Graph the curve and find its length. CONIC SECTIONS 654 CHAPTER 1 PARAETRIC EQUATIONS AND POLAR COORDINATES ; 43. The points of intersection of the crdioid r 1 sin nd the spirl loop r,, cn t be found ectl. Use grphing device to find the pproimte vlues of

More information

Math 211/213 Calculus III-IV. Directions. Kenneth Massey. September 17, 2018

Math 211/213 Calculus III-IV. Directions. Kenneth Massey. September 17, 2018 Mth 211/213 Clculus -V Kenneth Mssey Crson-Newmn University September 17, 2018 C-N Mth 211 - Mssey, 1 / 1 Directions You re t the origin nd giving directions to the point (4, 3). 1. n Mnhttn: go est 4

More information

APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING

APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING 6 Courtes NASA APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING Clculus is essentil for the computtions required to lnd n stronut on the Moon. In the lst chpter we introduced

More information

Engg. Math. I (Unit-II)

Engg. Math. I (Unit-II) Dr. Stish Shukl of 7 Engg. Mth. I Unit-II) Integrl Clculus iemnn Integrl) The ide. Suppose, f be continuous function defined on [, b nd we wnt to clculte the re bounded by this function with the -is from

More information

df dt f () b f () a dt

df dt f () b f () a dt Vector lculus 16.7 tokes Theorem Nme: toke's Theorem is higher dimensionl nlogue to Green's Theorem nd the Fundmentl Theorem of clculus. Why, you sk? Well, let us revisit these theorems. Fundmentl Theorem

More information

AP Calculus Multiple Choice: BC Edition Solutions

AP 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 information

Problem set 1: Solutions Math 207B, Winter 2016

Problem set 1: Solutions Math 207B, Winter 2016 Problem set 1: Solutions Mth 27B, Winter 216 1. Define f : R 2 R by f(,) = nd f(x,y) = xy3 x 2 +y 6 if (x,y) (,). ()Show tht thedirectionl derivtives of f t (,)exist inevery direction. Wht is its Gâteux

More information

and that at t = 0 the object is at position 5. Find the position of the object at t = 2.

and that at t = 0 the object is at position 5. Find the position of the object at t = 2. 7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we

More information

Simple Harmonic Motion I Sem

Simple Harmonic Motion I Sem Simple Hrmonic Motion I Sem Sllus: Differentil eqution of liner SHM. Energ of prticle, potentil energ nd kinetic energ (derivtion), Composition of two rectngulr SHM s hving sme periods, Lissjous figures.

More information

Math 0230 Calculus 2 Lectures

Math 0230 Calculus 2 Lectures Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two

More information

df dx There is an infinite number of different paths from

df dx There is an infinite number of different paths from Integrl clculus line integrls Feb 7, 18 From clculus, in the cse of single vrible x1 F F x F x f x dx, where f x 1 x df dx Now, consider the cse tht two vribles re t ply. Suppose,, df M x y dx N x y dy

More information

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim

Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)

More information

Year 12 Trial Examination Mathematics Extension 1. Question One 12 marks (Start on a new page) Marks

Year 12 Trial Examination Mathematics Extension 1. Question One 12 marks (Start on a new page) Marks THGS Mthemtics etension Tril 00 Yer Tril Emintion Mthemtics Etension Question One mrks (Strt on new pge) Mrks ) If P is the point (-, 5) nd Q is the point (, -), find the co-ordintes of the point R which

More information

Exercise Qu. 12. a2 y 2 da = Qu. 18 The domain of integration: from y = x to y = x 1 3 from x = 0 to x = 1. = 1 y4 da.

Exercise Qu. 12. a2 y 2 da = Qu. 18 The domain of integration: from y = x to y = x 1 3 from x = 0 to x = 1. = 1 y4 da. MAH MAH Eercise. Eercise. Qu. 7 B smmetr ( + 5)dA + + + 5 (re of disk with rdius ) 5. he first two terms of the integrl equl to becuse is odd function in nd is odd function in. (see lso pge 5) Qu. b da

More information

CHM Physical Chemistry I Chapter 1 - Supplementary Material

CHM Physical Chemistry I Chapter 1 - Supplementary Material CHM 3410 - Physicl Chemistry I Chpter 1 - Supplementry Mteril For review of some bsic concepts in mth, see Atkins "Mthemticl Bckground 1 (pp 59-6), nd "Mthemticl Bckground " (pp 109-111). 1. Derivtion

More information

Math 190 Chapter 5 Lecture Notes. Professor Miguel Ornelas

Math 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 information

KINEMATICS OF RIGID BODIES

KINEMATICS OF RIGID BODIES KINEMTICS OF RIGID ODIES Introduction In rigid body kinemtics, e use the reltionships governing the displcement, velocity nd ccelertion, but must lso ccount for the rottionl motion of the body. Description

More information