5. Fundamental fracture mechanics

Size: px
Start display at page:

Download "5. Fundamental fracture mechanics"

Transcription

1 5. Fudmetl frcture mechic

2

3 Stre cocetrtio Frcture New free urfce formtio Force Mteril Eviromet Frcture mechic Ect crck growth drivig force? Reitce to frcture of mteril? Iititio Growth Stre cocetrtio Notch Cro ectio uddel chge P P Stre lie Stre lie bpth Thick tre lie Stre cocetrtio P P

4 Stre cocetrtio fctor Stre cocetrtio fctor S tre cocetrtio fctor K t m ( 5.3) Stre cocetrtio of hole Ifiite plte = Teio of plte with hole At poit(r,θ), tre of θdirectio 4 3 θ co θ 4 r r i mimum t hole edge m co 0 3 (θ=0 =r) Stre cocetrtio K t m 3 K t 3 ( 5.)

5 Stre cocetrtio fctorⅡ Stre cocetrtio fctor of ellipe hole = Ifiit plte with ellipe hole( d b) pplied re Stre cocetrtio fctor (ellipe hole) m K t + b ρ ( 5.) Ifiite plte For circle hole = ρ K t + 3 ρ

6 Stre cocetrtio fctorⅢ Applictio ρ= Coider ellipe hole Stre cocetrtio fctor ( = 0 ρ=) 0 K t 0 ρ 7.3 = 0 Ifiite plte Stre cocetrtio fctor K t A ρ A

7 Stre cocetrtio fctorⅣ Eercie Obti the tre cocetrtio fctor for followed otch Equtio K t A ρ A 4 Ifiite plte 4 A K t ρ= A 3.83 ρ

8 Stre cocetrtiofctorⅤ Stre cocetrtio fctor P=(b-) = 3 (Rdiu ) Ifiite plte (Rdiu ) K t =3 Fiite plte () Kt of circle hole For rdiu d, Sme K t = 3 (b) Stre cocetrtio fctor for fiite plte Stre i omil tre, t Miimum cro ectio

9 Deformtio mode of crck Compoet with crck form uder pplied lod. z z z Crck Crck Crck Mode ModeⅡ ModeⅢ Deformtio mode of crck

10 Stre cocetrtio fctor of ellipe hole 3 3 ( 5.4) m 3 ρ ( 5.) ρ K t m 0 Notch tip of ellipe =0

11 Stre ditributio t crck tip Stre i viciit of crck, Stre / 0 0 ρ/=0 K t = (Crck) ρ/=0.0 K t = ρ/=0. K t =7.3 = b Kt ρ ρ/=0.5 K t =3.83 K t m K t ρ For ce of crck (ρ= 0), ( 5.) Ditce from crck tip / K t (Idepedet of crck legth d crck tpe) Stre ditributio t ellipe otch tip

12 Stre iteit fctor Stre t otch tip OK Differet otch d tre Kt M. tre t otch root Stre cocetrtio fctor Epre tre t otch root NO For crck Mimum tre (Idepedet of crck tpe d legth) Stre iteit fctor K Iteit of tre field i viciit of crck Stre iteit fctor of ifiite plte with crck K π Net

13 Stre iteit fctorⅡ Stre ditributio t crck tip ( 5.4) 3 3 For crck Notch rdiu ρ 0 0 ρ 3 ( 5.5) Proportio to qure root Ivere proportio ~ Stre ditributio i viciit of crck ~ I viciit of crck i er 0

14 Stre iteit fctorⅢ Stre iteit fctor of ifiite plte with crck ( 0) π π K :tre iteit fctor K π Mode I K π K π Oe crck d ifiite plte [ Uit MP m ] For fiite plte d three dimeio Stre ditributio i viciit of crck tip d tre iteit fctor K π F Corrected fctor deped o crck geometr

15 Stre iteit fctor Ⅳ Crck legth d tre ditributio Log crck legth tre K π Log crck Short crck < Short crck legth tre K ' ' π A K equl to K Sme iteit of tre field Stre ditributio of differet crck legth Iteit of tre i viciit of crck i decided b ol tre iteit fctor

16 Stre iteit fctorⅤ Summr Stre t crck tip i Stre i viciit of crck i ivere proportio to qure root ( ditce from crck tip) Iteit of tre t crck tip i decided b ol tre iteit fctor (No reltio betwee eterl force, pecime dimeio d crck legth)

17 Stre iteit fctor of differet tpe crck

18 Smll cle ieldig Eltic bod ~ Uder pplied lod, tre i Itt frcture Prctice No lier ivlid of K Smll cle ieldig Pltic zoe i ver mller th crck legth I eltic zoe roud pltic zoe, it c coider the me o pltic deformtio. Stre c be evluted b tre iteit fctor, K. It clled mll cle ieldig tte. Pltic deformed zoe Vlid K

19 Smll cle ieldigⅡ Pltic zoe t crck tip Yield tre E A r p B Pltic zoe corrected eltic tre ditributio r p F Eltic tre ditributio φ O O D R=r p Crck Imgir eltic crck C Stre ditributio After ieldig Smll cle ieldig i ple tre tte K ( 5.6) π K π r p K π Eltic perfect pltic bod Applied lod i ot chged b pltic deformtio. Pltic zoe eted util two re re the me. After correctio, pltic zoe ize R R r p K K π K π ( 5.3)

20 Smll cle ieldig Ⅲ Crck opeig diplcemet E Stre ditributio corrected b pltic zoe C Stre ditributio fter B Yield tre pltic deformtio A r p r p F Eltic tre ditributio φ O O D R=r p Crck Imgir eltic crck Crck opeig diplcemet b pltic deformtio 4K φ πe K Coditio of mll cle ieldig Pltic zoe ize R d Crck tip opeig diplcemet φ Proportio to K over two

21 Pltic zoe ize = R/ Smll cle ieldig Ⅳ To dipper igulrit, deciio of R/ (Dgdle model ) R π ec S For mll cle ieldig ( 5.5) Rge of mll Scl ieldig 応力比 = / Smll cle ieldig π K 8 Pltic zoe ize t crck tip Applied tre < ield tre S Whe <<, R ec π 8 S S Equl to ( 5.3) Smll cle ield ( 5.6) Reltive pplied lod / S = 0.4 Reltive pltic zoe R/ = 0.

22 Smll cle ieldig Ⅵ Stre tte Ple tre (thi plte) dimeio Ple tri (thick plte) 3dimeio Surfce, ple tre δ crck Pltic zoe Thicke B r p K π Crck opeig diplcemet Ple tre Ple tri Pltic zoe i thick plte Iide mteril, ple tri r K 6 π p

23 Frcture toughe Frcture toughe? For crcked bod to pltic deform, Whe tre iteit fctor i over the criticl vlue, Crck uddel propgte d frcture occur Frcture toughe me the reitce to crck propgtio of mteril uder ttic lod

24 Frcture tougheⅡ Frcture toughe K C Ple tre regio Regio() Stble growth Tritio regio Slt Regio(Ⅱ) Ple tri frcture toughe Thicke B K C Ple tri regio Regio(Ⅲ) Verticl Utble growth Sher lip Ftigue crck otck

25 Frcture toughe Ⅲ Frcture toughe K C Ple tri tte + Smll cle ieldig B, Ple tri Frcture toughe Thicke B K.5 S K C C Thi plte Thick plte Ple tre i pltic zoe Stble growth, Slt tpe frcture urfce High frcture toughe Ple tri i pltic zoe Utble crck growth Verticl frcture urfce A cott frcture toughe

26 Frcture toughe Ⅳ K C t room temperture 材料 アルミニウム合金 04-T T65 降伏応力 (MP) 平面ひずみ破壊靭性 K C チタン合金 Ti-6Al-4V 鋼 AISI 4340 A B

27 Stre d diplcemet Stre i the viciit of crck tip(r,θ) Plte r E,ι θ τ K θ θ 3θ co i i π r K θ θ 3θ co i i π r K π r θ θ 3θ co i i τ v u τ, directio diplcemet u K G Ple tre Ple tri r θ co κ i θ π 3ν ν 3 4ν v K G r θ i κ co θ π

28 Smll ieldigⅤ (Pltic zoe i viciit of crck tip) Stre / m R Smll ellipe =ρ/4 Circle =ρ Due to m Regio of me Pltic zoe Lrge ellipe =4ρ Pltic zoe R/ρ m (Eltic mimum tre) ρ (Notch rdii) Sme R/ 0.4 Applied tre i the me

29 Cocept of lier frcture mechic K = K ρ=0 Pltic zoe ()Sme eltic tre field (b)eltic-pltic tre filed Cocept of frcture mechic For differet crck legth, if tre iteit fctor i the me, Eltic tre i the me, d the eltic d pltic tre i lo the me At crck tip, the me frcture pheomeo occur

30 Cocept of lier otch mechic m = m ρ =ρ ρ = ρ ρ 一定 t t t t Pltic zoe ()Sme eltic tre field Cocept of lier otch mechic (b)sme eltic pltic tre field For two otche otch rdii ρd eltic m. tre re the me Additio to eltic tre, eltic-pltic tre i lo the me. At ech otch tip, the me frcture pheomeo occur.

31 Stri eerg relee rte I(Griffith theor) Eerg relee rte clculted tri eerg Crck grow Stri eerg whe crck grow uit legth π E Free urfce Stri eerg relee rte ρ G π E π E K E Griffith equtio Eγ π K G E γ

32 Stri eerg relee rte II (Irwi tud) Eerg relee rte clculted from tre t crck tip 0 K π 0 v K 4 E Δ Δ π () (b) Chge of tri eerg with icreig crck growth Crck grow Eltic tri eerg relee ΔU Releed eltic Stri eerg ΔU = After crckig, lod pplie to Δ Ad revered diplcemet, v, occur Before crckig, workig,δw

33 Stre d Stri eerg relee rte III Eerg relee rte clculted from tre t crck tip After growth Coider crck upper ide Before growth 0 (c) Dplcemet d Workig t =Stri eerg (d) Stri eerg chge with icreig crck growth U 0 Δ K π 4K E K Δ Δ d πe 0 K U Δ GΔ E Δ π G ; Stri eerg relee rte (Drivig force of crck) d

34 Stre cocetrtio 0 Ulimited plte = (Notch) Teile tre bed o I the me compreive tre bed o Pi = + P P () (b) (c) (b)for o hole plte ppled + Stre cocetrtio (Teile tre) (c)circumferece of mll hole m cocetrted force, Pi ditribute (Compreive tre) Stre ditributio b cocetrted force

35 Stre ditributio i the viciit of crck tip Smll ellipe =ρ/4 Circle =ρ Agreemet of tre Lrge ellipe =4ρ 0.3 ρ Ditce from crck tip /ρ If ρd m re the me For differet otch, Stre ditributio i equl Whe Notch rdii ρd m re the me, Stre ditributio Stre ditributio log i of ellipe pplied 3 3

S(x)along the bar (shear force diagram) by cutting the bar at x and imposing force_y equilibrium.

S(x)along the bar (shear force diagram) by cutting the bar at x and imposing force_y equilibrium. mmetric leder Bems i Bedig Lodig Coditios o ech ectio () pplied -Forces & z-omets The resultts t sectio re: the bedig momet () d z re sectio smmetr es the sher force () [ for sleder bems stresses d deformtio

More information

9.6 Blend-Out Repairs

9.6 Blend-Out Repairs 9.6 Bled-Out Repirs Oe of the ccepted procedures for removig smll mout of crck dmge i the field is through the use of bled-out repirs. These repirs re efficietly ccomplished d for the most prt, retur the

More information

F x = 2x λy 2 z 3 = 0 (1) F y = 2y λ2xyz 3 = 0 (2) F z = 2z λ3xy 2 z 2 = 0 (3) F λ = (xy 2 z 3 2) = 0. (4) 2z 3xy 2 z 2. 2x y 2 z 3 = 2y 2xyz 3 = ) 2

F x = 2x λy 2 z 3 = 0 (1) F y = 2y λ2xyz 3 = 0 (2) F z = 2z λ3xy 2 z 2 = 0 (3) F λ = (xy 2 z 3 2) = 0. (4) 2z 3xy 2 z 2. 2x y 2 z 3 = 2y 2xyz 3 = ) 2 0 微甲 07- 班期中考解答和評分標準 5%) Fid the poits o the surfce xy z = tht re closest to the origi d lso the shortest distce betwee the surfce d the origi Solutio Cosider the Lgrge fuctio F x, y, z, λ) = x + y + z

More information

Area, Volume, Rotations, Newton s Method

Area, Volume, Rotations, Newton s Method Are, Volume, Rottio, Newto Method Are etwee curve d the i A ( ) d Are etwee curve d the y i A ( y) yd yc Are etwee curve A ( ) g( ) d where ( ) i the "top" d g( ) i the "ottom" yd Are etwee curve A ( y)

More information

Limit of a function:

Limit of a function: - Limit of fuctio: We sy tht f ( ) eists d is equl with (rel) umer L if f( ) gets s close s we wt to L if is close eough to (This defiitio c e geerlized for L y syig tht f( ) ecomes s lrge (or s lrge egtive

More information

HIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 3 UNIT (ADDITIONAL) AND 3/4 UNIT (COMMON) Time allowed Two hours (Plus 5 minutes reading time)

HIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 3 UNIT (ADDITIONAL) AND 3/4 UNIT (COMMON) Time allowed Two hours (Plus 5 minutes reading time) HIGHER SCHOOL CERTIFICATE EXAMINATION 999 MATHEMATICS 3 UNIT (ADDITIONAL) AND 3/ UNIT (COMMON) Time llowed Two hours (Plus 5 miutes redig time) DIRECTIONS TO CANDIDATES Attempt ALL questios. ALL questios

More information

Content: Essential Calculus, Early Transcendentals, James Stewart, 2007 Chapter 1: Functions and Limits., in a set B.

Content: Essential Calculus, Early Transcendentals, James Stewart, 2007 Chapter 1: Functions and Limits., in a set B. Review Sheet: Chpter Cotet: Essetil Clculus, Erly Trscedetls, Jmes Stewrt, 007 Chpter : Fuctios d Limits Cocepts, Defiitios, Lws, Theorems: A fuctio, f, is rule tht ssigs to ech elemet i set A ectly oe

More information

Graphing Review Part 3: Polynomials

Graphing Review Part 3: Polynomials Grphig Review Prt : Polomils Prbols Recll, tht the grph of f ( ) is prbol. It is eve fuctio, hece it is smmetric bout the bout the -is. This mes tht f ( ) f ( ). Its grph is show below. The poit ( 0,0)

More information

Add Maths Formulae List: Form 4 (Update 18/9/08)

Add Maths Formulae List: Form 4 (Update 18/9/08) Add Mths Formule List: Form 4 (Updte 8/9/08) 0 Fuctios Asolute Vlue Fuctio f ( ) f( ), if f( ) 0 f( ), if f( ) < 0 Iverse Fuctio If y f( ), the Rememer: Oject the vlue of Imge the vlue of y or f() f()

More information

STRESS INTENSITY FACTORS AND FATIGUE CRACK GROWTH OF IRREGULAR PLANAR CRACKS SUBJECTED TO ARBITRARY MODE I STRESS FIELDS

STRESS INTENSITY FACTORS AND FATIGUE CRACK GROWTH OF IRREGULAR PLANAR CRACKS SUBJECTED TO ARBITRARY MODE I STRESS FIELDS STRESS INTENSITY FCTORS ND FTIGUE CRCK GROWTH OF IRREGULR PLNR CRCKS SUBJECTED TO RBITRRY MODE I STRESS FIELDS Grzegorz GLINK University of Wterloo Deprtment of Mechnicl nd Mechtronics Engineering Wterloo.

More information

BC Calculus Review Sheet

BC Calculus Review Sheet BC Clculus Review Sheet Whe you see the words. 1. Fid the re of the ubouded regio represeted by the itegrl (sometimes 1 f ( ) clled horizotl improper itegrl). This is wht you thik of doig.... Fid the re

More information

Approximate Integration

Approximate Integration Study Sheet (7.7) Approimte Itegrtio I this sectio, we will ler: How to fid pproimte vlues of defiite itegrls. There re two situtios i which it is impossile to fid the ect vlue of defiite itegrl. Situtio:

More information

[ 20 ] 1. Inequality exists only between two real numbers (not complex numbers). 2. If a be any real number then one and only one of there hold.

[ 20 ] 1. Inequality exists only between two real numbers (not complex numbers). 2. If a be any real number then one and only one of there hold. [ 0 ]. Iequlity eists oly betwee two rel umbers (ot comple umbers).. If be y rel umber the oe d oly oe of there hold.. If, b 0 the b 0, b 0.. (i) b if b 0 (ii) (iii) (iv) b if b b if either b or b b if

More information

Crushed Notes on MATH132: Calculus

Crushed Notes on MATH132: Calculus Mth 13, Fll 011 Siyg Yg s Outlie Crushed Notes o MATH13: Clculus The otes elow re crushed d my ot e ect This is oly my ow cocise overview of the clss mterils The otes I put elow should ot e used to justify

More information

is continuous at x 2 and g(x) 2. Oil spilled from a ruptured tanker spreads in a circle whose area increases at a

is continuous at x 2 and g(x) 2. Oil spilled from a ruptured tanker spreads in a circle whose area increases at a . Cosider two fuctios f () d g () defied o itervl I cotiig. f () is cotiuous t d g() is discotiuous t. Which of the followig is true bout fuctios f g d f g, the sum d the product of f d g, respectively?

More information

First assignment of MP-206

First assignment of MP-206 irt igmet of MP- er to quetio - 7- Norml tre log { : MP Priipl tree: I MP II MP III MP Priipl iretio: { I { II { III Iitill uppoe tht i tre tte eribe i the referee tem ' i the me tre tte but eribe i other

More information

ALGEBRA II CHAPTER 7 NOTES. Name

ALGEBRA II CHAPTER 7 NOTES. Name ALGEBRA II CHAPTER 7 NOTES Ne Algebr II 7. th Roots d Rtiol Expoets Tody I evlutig th roots of rel ubers usig both rdicl d rtiol expoet ottio. I successful tody whe I c evlute th roots. It is iportt for

More information

Discrete Mathematics I Tutorial 12

Discrete Mathematics I Tutorial 12 Discrete Mthemtics I Tutoril Refer to Chpter 4., 4., 4.4. For ech of these sequeces fid recurrece reltio stisfied by this sequece. (The swers re ot uique becuse there re ifiitely my differet recurrece

More information

Qn Suggested Solution Marking Scheme 1 y. G1 Shape with at least 2 [2]

Qn Suggested Solution Marking Scheme 1 y. G1 Shape with at least 2 [2] Mrkig Scheme for HCI 8 Prelim Pper Q Suggested Solutio Mrkig Scheme y G Shpe with t lest [] fetures correct y = f'( ) G ll fetures correct SR: The mimum poit could be i the first or secod qudrt. -itercept

More information

, so the state may be taken to be l S ÅÅÅÅ

, so the state may be taken to be l S ÅÅÅÅ hw4-7.b: 4/6/04:::9:34 Hoework: 4-7 ü Skuri :, G,, 7, 8 ü. bel the eigefuctio of the qure well by S, with =,,. The correpodig wve fuctio re x = px i Iitilly, the prticle i kow to be t x =, o the tte y

More information

=> PARALLEL INTERCONNECTION. Basic Properties LTI Systems. The Commutative Property. Convolution. The Commutative Property. The Distributive Property

=> PARALLEL INTERCONNECTION. Basic Properties LTI Systems. The Commutative Property. Convolution. The Commutative Property. The Distributive Property Lier Time-Ivrit Bsic Properties LTI The Commuttive Property The Distributive Property The Associtive Property Ti -6.4 / Chpter Covolutio y ] x ] ] x ]* ] x ] ] y] y ( t ) + x( τ ) h( t τ ) dτ x( t) * h(

More information

S. Socrate 2013 K. Qian

S. Socrate 2013 K. Qian S. Socrte 213 K. Qi odig Coditios o ech Sectio () pplied lodig oly log the is () of the br. The oly iterl resultt t y sectios is the il force N() Fid N()log the br (il force digrm) by cuttig the br t ech

More information

1. The 0.1 kg particle has a speed v = 10 m/s as it passes the 30 position shown. The coefficient of kinetic friction between the particle and the

1. The 0.1 kg particle has a speed v = 10 m/s as it passes the 30 position shown. The coefficient of kinetic friction between the particle and the 1. The 0.1 kg pticle h peed v = 10 m/ it pe the 30 poitio how. The coefficiet of kietic fictio betwee the pticle d the veticl ple tck i m k = 0.0. Detemie the mgitude of the totl foce exeted by the tck

More information

Tranformations. Some slides adapted from Octavia Camps

Tranformations. Some slides adapted from Octavia Camps Trformtio Some lide dpted from Octvi Cmp A m 3 3 m m 3m m Mtrice 5K C c m ij A ij m b ij B A d B mut hve the me dimeio m 3 Mtrice p m m p B A C H @K?J H @K?J m k kj ik ij b c A d B mut hve A d B mut hve

More information

334 MATHS SERIES DSE MATHS PREVIEW VERSION B SAMPLE TEST & FULL SOLUTION

334 MATHS SERIES DSE MATHS PREVIEW VERSION B SAMPLE TEST & FULL SOLUTION MATHS SERIES DSE MATHS PREVIEW VERSION B SAMPLE TEST & FULL SOLUTION TEST SAMPLE TEST III - P APER Questio Distributio INSTRUCTIONS:. Attempt ALL questios.. Uless otherwise specified, ll worig must be

More information

MASSACHUSETTS INSTITUTE of TECHNOLOGY Department of Mechanical Engineering 2.71/ OPTICS - - Spring Term, 2014

MASSACHUSETTS INSTITUTE of TECHNOLOGY Department of Mechanical Engineering 2.71/ OPTICS - - Spring Term, 2014 .7/.70 Optic, Spri 04, Solutio for Quiz MASSACHUSETTS INSTITUTE of TECHNOLOGY Deprtmet of Mechicl Eieeri.7/.70 OPTICS - - Spri Term, 04 Solutio for Quiz Iued Wed. 03//04 Problem. The ive opticl ytem i

More information

8.1 Arc Length. What is the length of a curve? How can we approximate it? We could do it following the pattern we ve used before

8.1 Arc Length. What is the length of a curve? How can we approximate it? We could do it following the pattern we ve used before 8.1 Arc Legth Wht is the legth of curve? How c we pproximte it? We could do it followig the ptter we ve used efore Use sequece of icresigly short segmets to pproximte the curve: As the segmets get smller

More information

Name: A2RCC Midterm Review Unit 1: Functions and Relations Know your parent functions!

Name: A2RCC Midterm Review Unit 1: Functions and Relations Know your parent functions! Nme: ARCC Midterm Review Uit 1: Fuctios d Reltios Kow your pret fuctios! 1. The ccompyig grph shows the mout of rdio-ctivity over time. Defiitio of fuctio. Defiitio of 1-1. Which digrm represets oe-to-oe

More information

( ) dx ; f ( x ) is height and Δx is

( ) dx ; f ( x ) is height and Δx is Mth : 6.3 Defiite Itegrls from Riem Sums We just sw tht the exct re ouded y cotiuous fuctio f d the x xis o the itervl x, ws give s A = lim A exct RAM, where is the umer of rectgles i the Rectgulr Approximtio

More information

Spherical refracting surface. Here, the outgoing rays are on the opposite side of the surface from the Incoming rays.

Spherical refracting surface. Here, the outgoing rays are on the opposite side of the surface from the Incoming rays. Sphericl refrctig urfce Here, the outgoig ry re o the oppoite ide of the urfce from the Icomig ry. The oject i t P. Icomig ry PB d PV form imge t P. All prxil ry from P which trike the phericl urfce will

More information

Chem 253A. Crystal Structure. Chem 253B. Electronic Structure

Chem 253A. Crystal Structure. Chem 253B. Electronic Structure Chem 53, UC, Bereley Chem 53A Crystl Structure Chem 53B Electroic Structure Chem 53, UC, Bereley 1 Chem 53, UC, Bereley Electroic Structures of Solid Refereces Ashcroft/Mermi: Chpter 1-3, 8-10 Kittel:

More information

ELEC 372 LECTURE NOTES, WEEK 6 Dr. Amir G. Aghdam Concordia University

ELEC 372 LECTURE NOTES, WEEK 6 Dr. Amir G. Aghdam Concordia University ELEC 37 LECTURE NOTES, WEE 6 Dr mir G ghdm Cocordi Uiverity Prt of thee ote re dpted from the mteril i the followig referece: Moder Cotrol Sytem by Richrd C Dorf d Robert H Bihop, Pretice Hll Feedbck Cotrol

More information

A GENERAL METHOD FOR SOLVING ORDINARY DIFFERENTIAL EQUATIONS: THE FROBENIUS (OR SERIES) METHOD

A GENERAL METHOD FOR SOLVING ORDINARY DIFFERENTIAL EQUATIONS: THE FROBENIUS (OR SERIES) METHOD Diol Bgoo () A GENERAL METHOD FOR SOLVING ORDINARY DIFFERENTIAL EQUATIONS: THE FROBENIUS (OR SERIES) METHOD I. Itroductio The first seprtio of vribles (see pplictios to Newto s equtios) is ver useful method

More information

Chapter Real Numbers

Chapter Real Numbers Chpter. - Rel Numbers Itegers: coutig umbers, zero, d the egtive of the coutig umbers. ex: {,-3, -, -,,,, 3, } Rtiol Numbers: quotiets of two itegers with ozero deomitor; termitig or repetig decimls. ex:

More information

Algebra 2 Important Things to Know Chapters bx c can be factored into... y x 5x. 2 8x. x = a then the solutions to the equation are given by

Algebra 2 Important Things to Know Chapters bx c can be factored into... y x 5x. 2 8x. x = a then the solutions to the equation are given by Alger Iportt Thigs to Kow Chpters 8. Chpter - Qudrtic fuctios: The stdrd for of qudrtic fuctio is f ( ) c, where 0. c This c lso e writte s (if did equl zero, we would e left with The grph of qudrtic fuctio

More information

Name of the Student:

Name of the Student: Egieerig Mthemtics 5 NAME OF THE SUBJECT : Mthemtics I SUBJECT CODE : MA65 MATERIAL NAME : Additiol Prolems MATERIAL CODE : HGAUM REGULATION : R UPDATED ON : M-Jue 5 (Sc the ove QR code for the direct

More information

Chem 253B. Crystal Structure. Chem 253C. Electronic Structure

Chem 253B. Crystal Structure. Chem 253C. Electronic Structure Chem 5, UC, Berele Chem 5B Crstl Structure Chem 5C Electroic Structure Chem 5, UC, Berele 1 Chem 5, UC, Berele Electroic Structures of Solid Refereces Ashcroft/Mermi: Chpter 1-, 8-10 Kittel: chpter 6-9

More information

Sharjah Institute of Technology

Sharjah Institute of Technology For commets, correctios, etc Plese cotct Ahf Abbs: hf@mthrds.com Shrh Istitute of Techolog echicl Egieerig Yer Thermofluids sheet ALGERA Lws of Idices:. m m + m m. ( ).. 4. m m 5. Defiitio of logrithm:

More information

ALGEBRA. Set of Equations. have no solution 1 b1. Dependent system has infinitely many solutions

ALGEBRA. Set of Equations. have no solution 1 b1. Dependent system has infinitely many solutions Qudrtic Equtios ALGEBRA Remider theorem: If f() is divided b( ), the remider is f(). Fctor theorem: If ( ) is fctor of f(), the f() = 0. Ivolutio d Evlutio ( + b) = + b + b ( b) = + b b ( + b) 3 = 3 +

More information

Vectors. Vectors in Plane ( 2

Vectors. Vectors in Plane ( 2 Vectors Vectors i Ple ( ) The ide bout vector is to represet directiol force Tht mes tht every vector should hve two compoets directio (directiol slope) d mgitude (the legth) I the ple we preset vector

More information

Taylor Polynomials. The Tangent Line. (a, f (a)) and has the same slope as the curve y = f (x) at that point. It is the best

Taylor Polynomials. The Tangent Line. (a, f (a)) and has the same slope as the curve y = f (x) at that point. It is the best Tylor Polyomils Let f () = e d let p() = 1 + + 1 + 1 6 3 Without usig clcultor, evlute f (1) d p(1) Ok, I m still witig With little effort it is possible to evlute p(1) = 1 + 1 + 1 (144) + 6 1 (178) =

More information

Chapter #3 EEE Subsea Control and Communication Systems

Chapter #3 EEE Subsea Control and Communication Systems EEE 87 Chter #3 EEE 87 Sube Cotrol d Commuictio Sytem Cloed loo ytem Stedy tte error PID cotrol Other cotroller Chter 3 /3 EEE 87 Itroductio The geerl form for CL ytem: C R ', where ' c ' H or Oe Loo (OL)

More information

Intrinsic Carrier Concentration

Intrinsic Carrier Concentration Itrisic Carrier Cocetratio I. Defiitio Itrisic semicoductor: A semicoductor material with o dopats. It electrical characteristics such as cocetratio of charge carriers, deped oly o pure crystal. II. To

More information

Limits and an Introduction to Calculus

Limits and an Introduction to Calculus Nme Chpter Limits d Itroductio to Clculus Sectio. Itroductio to Limits Objective: I this lesso ou lered how to estimte limits d use properties d opertios of limits. I. The Limit Cocept d Defiitio of Limit

More information

INFINITE SERIES. ,... having infinite number of terms is called infinite sequence and its indicated sum, i.e., a 1

INFINITE SERIES. ,... having infinite number of terms is called infinite sequence and its indicated sum, i.e., a 1 Appedix A.. Itroductio As discussed i the Chpter 9 o Sequeces d Series, sequece,,...,,... hvig ifiite umber of terms is clled ifiite sequece d its idicted sum, i.e., + + +... + +... is clled ifite series

More information

FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures

FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING Lectures MODULE 0 FURTHER CALCULUS II. Sequeces d series. Rolle s theorem d me vlue theorems 3. Tlor s d Mcluri s theorems 4. L Hopitl

More information

Chapter 23. Geometric Optics

Chapter 23. Geometric Optics Cpter 23 Geometric Optic Ligt Wt i ligt? Wve or prticle? ot Geometric optic: ligt trvel i trigt-lie pt clled ry Ti i true if typicl ditce re muc lrger t te wvelegt Geometric Optic 2 Wt it i out eome ddreed

More information

National Quali cations AHEXEMPLAR PAPER ONLY

National Quali cations AHEXEMPLAR PAPER ONLY Ntiol Quli ctios AHEXEMPLAR PAPER ONLY EP/AH/0 Mthemtics Dte Not pplicble Durtio hours Totl mrks 00 Attempt ALL questios. You my use clcultor. Full credit will be give oly to solutios which coti pproprite

More information

Mathematical Notation Math Calculus & Analytic Geometry I

Mathematical Notation Math Calculus & Analytic Geometry I Mthemticl Nottio Mth - Clculus & Alytic Geometry I Nme : Use Wor or WorPerect to recrete the ollowig ocumets. Ech rticle is worth poits c e prite give to the istructor or emile to the istructor t jmes@richl.eu.

More information

Options: Calculus. O C.1 PG #2, 3b, 4, 5ace O C.2 PG.24 #1 O D PG.28 #2, 3, 4, 5, 7 O E PG #1, 3, 4, 5 O F PG.

Options: Calculus. O C.1 PG #2, 3b, 4, 5ace O C.2 PG.24 #1 O D PG.28 #2, 3, 4, 5, 7 O E PG #1, 3, 4, 5 O F PG. O C. PG.-3 #, 3b, 4, 5ce O C. PG.4 # Optios: Clculus O D PG.8 #, 3, 4, 5, 7 O E PG.3-33 #, 3, 4, 5 O F PG.36-37 #, 3 O G. PG.4 #c, 3c O G. PG.43 #, O H PG.49 #, 4, 5, 6, 7, 8, 9, 0 O I. PG.53-54 #5, 8

More information

Failure Theories Des Mach Elem Mech. Eng. Department Chulalongkorn University

Failure Theories Des Mach Elem Mech. Eng. Department Chulalongkorn University Failure Theories Review stress trasformatio Failure theories for ductile materials Maimum-Shear-Stress Theor Distortio-Eerg Theor Coulomb-Mohr Theor Failure theories for brittle materials Maimum-Normal-Stress

More information

EXERCISE a a a 5. + a 15 NEETIIT.COM

EXERCISE a a a 5. + a 15 NEETIIT.COM - Dowlod our droid App. Sigle choice Type Questios EXERCISE -. The first term of A.P. of cosecutive iteger is p +. The sum of (p + ) terms of this series c be expressed s () (p + ) () (p + ) (p + ) ()

More information

Pre-Calculus - Chapter 3 Sections Notes

Pre-Calculus - Chapter 3 Sections Notes Pre-Clculus - Chpter 3 Sectios 3.1-3.4- Notes Properties o Epoets (Review) 1. ( )( ) = + 2. ( ) =, (c) = 3. 0 = 1 4. - = 1/( ) 5. 6. c Epoetil Fuctios (Sectio 3.1) Deiitio o Epoetil Fuctios The uctio deied

More information

CITY UNIVERSITY LONDON

CITY UNIVERSITY LONDON CITY UNIVERSITY LONDON Eg (Hos) Degree i Civil Egieerig Eg (Hos) Degree i Civil Egieerig with Surveyig Eg (Hos) Degree i Civil Egieerig with Architecture PART EXAMINATION SOLUTIONS ENGINEERING MATHEMATICS

More information

0 otherwise. sin( nx)sin( kx) 0 otherwise. cos( nx) sin( kx) dx 0 for all integers n, k.

0 otherwise. sin( nx)sin( kx) 0 otherwise. cos( nx) sin( kx) dx 0 for all integers n, k. . Computtio of Fourier Series I this sectio, we compute the Fourier coefficiets, f ( x) cos( x) b si( x) d b, i the Fourier series To do this, we eed the followig result o the orthogolity of the trigoometric

More information

GRAPHING LINEAR EQUATIONS. Linear Equations. x l ( 3,1 ) _x-axis. Origin ( 0, 0 ) Slope = change in y change in x. Equation for l 1.

GRAPHING LINEAR EQUATIONS. Linear Equations. x l ( 3,1 ) _x-axis. Origin ( 0, 0 ) Slope = change in y change in x. Equation for l 1. GRAPHING LINEAR EQUATIONS Qudrt II Qudrt I ORDERED PAIR: The first umer i the ordered pir is the -coordite d the secod umer i the ordered pir is the y-coordite. (, ) Origi ( 0, 0 ) _-is Lier Equtios Qudrt

More information

: : 8.2. Test About a Population Mean. STT 351 Hypotheses Testing Case I: A Normal Population with Known. - null hypothesis states 0

: : 8.2. Test About a Population Mean. STT 351 Hypotheses Testing Case I: A Normal Population with Known. - null hypothesis states 0 8.2. Test About Popultio Me. Cse I: A Norml Popultio with Kow. H - ull hypothesis sttes. X1, X 2,..., X - rdom smple of size from the orml popultio. The the smple me X N, / X X Whe H is true. X 8.2.1.

More information

1. (25 points) Use the limit definition of the definite integral and the sum formulas to compute. [1 x + x2

1. (25 points) Use the limit definition of the definite integral and the sum formulas to compute. [1 x + x2 Mth 3, Clculus II Fil Exm Solutios. (5 poits) Use the limit defiitio of the defiite itegrl d the sum formuls to compute 3 x + x. Check your swer by usig the Fudmetl Theorem of Clculus. Solutio: The limit

More information

Probabilistic Fatigue Life Prediction Method for Notched Specimens Based on the Weakest-link theory

Probabilistic Fatigue Life Prediction Method for Notched Specimens Based on the Weakest-link theory 213 2 32 2 Mechnicl Science nd Technology for erospce Engineering Februry Vol. 32 213 o. 2 2116 Weibull Weibull Weibull - Weibull TC4 5% 1% 9% Weibull O346. 3 13-8728 213 2-164-6 Probbilistic Ftigue Life

More information

Mathematical Notation Math Calculus & Analytic Geometry I

Mathematical Notation Math Calculus & Analytic Geometry I Mthemticl Nottio Mth - Clculus & Alytic Geometry I Use Wor or WorPerect to recrete the ollowig ocumets. Ech rticle is worth poits shoul e emile to the istructor t jmes@richl.eu. Type your me t the top

More information

Solution Manual. for. Fracture Mechanics. C.T. Sun and Z.-H. Jin

Solution Manual. for. Fracture Mechanics. C.T. Sun and Z.-H. Jin Solution Mnul for Frcture Mechnics by C.T. Sun nd Z.-H. Jin Chpter rob.: ) 4 No lod is crried by rt nd rt 4. There is no strin energy stored in them. Constnt Force Boundry Condition The totl strin energy

More information

Force and Motion. Force

Force and Motion. Force Force d Motio Cocept of Force Newto s hree Lws ypes of Forces Free body lysis Equilibrium Noequilibrium Frictio Problem Solvig Force A Force is push or pull tht is exerted o object by some other object.

More information

Summer Math Requirement Algebra II Review For students entering Pre- Calculus Theory or Pre- Calculus Honors

Summer Math Requirement Algebra II Review For students entering Pre- Calculus Theory or Pre- Calculus Honors Suer Mth Requireet Algebr II Review For studets eterig Pre- Clculus Theory or Pre- Clculus Hoors The purpose of this pcket is to esure tht studets re prepred for the quick pce of Pre- Clculus. The Topics

More information

Theorem 5.3 (Continued) The Fundamental Theorem of Calculus, Part 2: ab,, then. f x dx F x F b F a. a a. f x dx= f x x

Theorem 5.3 (Continued) The Fundamental Theorem of Calculus, Part 2: ab,, then. f x dx F x F b F a. a a. f x dx= f x x Chpter 6 Applictios Itegrtio Sectio 6. Regio Betwee Curves Recll: Theorem 5.3 (Cotiued) The Fudmetl Theorem of Clculus, Prt :,,, the If f is cotiuous t ever poit of [ ] d F is tiderivtive of f o [ ] (

More information

The Reimann Integral is a formal limit definition of a definite integral

The Reimann Integral is a formal limit definition of a definite integral MATH 136 The Reim Itegrl The Reim Itegrl is forml limit defiitio of defiite itegrl cotiuous fuctio f. The costructio is s follows: f ( x) dx for Reim Itegrl: Prtitio [, ] ito suitervls ech hvig the equl

More information

z line a) Draw the single phase equivalent circuit. b) Calculate I BC.

z line a) Draw the single phase equivalent circuit. b) Calculate I BC. ECE 2260 F 08 HW 7 prob 4 solutio EX: V gyb' b' b B V gyc' c' c C = 101 0 V = 1 + j0.2 Ω V gyb' = 101 120 V = 6 + j0. Ω V gyc' = 101 +120 V z LΔ = 9 j1.5 Ω ) Drw the sigle phse equivlet circuit. b) Clculte

More information

Chapter 10: The Z-Transform Adapted from: Lecture notes from MIT, Binghamton University Dr. Hamid R. Rabiee Fall 2013

Chapter 10: The Z-Transform Adapted from: Lecture notes from MIT, Binghamton University Dr. Hamid R. Rabiee Fall 2013 Sigls & Systems Chpter 0: The Z-Trsform Adpted from: Lecture otes from MIT, Bighmto Uiversity Dr. Hmid R. Rbiee Fll 03 Lecture 5 Chpter 0 Lecture 6 Chpter 0 Outlie Itroductio to the -Trsform Properties

More information

Chapter 30: Reflection and Refraction

Chapter 30: Reflection and Refraction Chpter 30: Reflectio d Refrctio The ture of light Speed of light (i vcuum) c.9979458 x 0 8 m/s mesured ut it is ow the defiitio Michelso s 878 Rottig Mirror Experimet Germ Americ physicist A.A. Michelso

More information

Composite Structures

Composite Structures Compoite Structure 5 (8) 49 57 Cotet lit vilble t ScieceDirect Compoite Structure jourl homepge: www.elevier.com/locte/comptruct teril tilorig for reducig tre cocetrtio fctor t circulr hole i fuctiolly

More information

Unit 1. Extending the Number System. 2 Jordan School District

Unit 1. Extending the Number System. 2 Jordan School District Uit Etedig the Number System Jord School District Uit Cluster (N.RN. & N.RN.): Etedig Properties of Epoets Cluster : Etedig properties of epoets.. Defie rtiol epoets d eted the properties of iteger epoets

More information

18.01 Calculus Jason Starr Fall 2005

18.01 Calculus Jason Starr Fall 2005 18.01 Clculus Jso Strr Lecture 14. October 14, 005 Homework. Problem Set 4 Prt II: Problem. Prctice Problems. Course Reder: 3B 1, 3B 3, 3B 4, 3B 5. 1. The problem of res. The ciet Greeks computed the res

More information

In an algebraic expression of the form (1), like terms are terms with the same power of the variables (in this case

In an algebraic expression of the form (1), like terms are terms with the same power of the variables (in this case Chpter : Algebr: A. Bckgroud lgebr: A. Like ters: I lgebric expressio of the for: () x b y c z x y o z d x... p x.. we cosider x, y, z to be vribles d, b, c, d,,, o,.. to be costts. I lgebric expressio

More information

BITSAT MATHEMATICS PAPER. If log 0.0( ) log 0.( ) the elogs to the itervl (, ] () (, ] [,+ ). The poit of itersectio of the lie joiig the poits i j k d i+ j+ k with the ple through the poits i+ j k, i

More information

Chapter 7. , and is unknown and n 30 then X ~ t n

Chapter 7. , and is unknown and n 30 then X ~ t n Chpter 7 Sectio 7. t-ditributio ( 3) Summry: C.L.T. : If the rdom mple of ize 3 come from ukow popultio with me d S.D. where i kow or ukow, the X ~ N,. Note: The hypothei tetig d cofidece itervl re built

More information

Frequency-domain Characteristics of Discrete-time LTI Systems

Frequency-domain Characteristics of Discrete-time LTI Systems requecy-domi Chrcteristics of Discrete-time LTI Systems Prof. Siripog Potisuk LTI System descriptio Previous bsis fuctio: uit smple or DT impulse The iput sequece is represeted s lier combitio of shifted

More information

Instructor(s): Acosta/Woodard PHYSICS DEPARTMENT PHY 2049, Fall 2015 Midterm 1 September 29, 2015

Instructor(s): Acosta/Woodard PHYSICS DEPARTMENT PHY 2049, Fall 2015 Midterm 1 September 29, 2015 Instructor(s): Acost/Woodrd PHYSICS DEPATMENT PHY 049, Fll 015 Midterm 1 September 9, 015 Nme (print): Signture: On m honor, I hve neither given nor received unuthorized id on this emintion. YOU TEST NUMBE

More information

National Quali cations SPECIMEN ONLY

National Quali cations SPECIMEN ONLY AH Ntiol Quli ctios SPECIMEN ONLY SQ/AH/0 Mthemtics Dte Not pplicble Durtio hours Totl mrks 00 Attempt ALL questios. You my use clcultor. Full credit will be give oly to solutios which coti pproprite workig.

More information

82A Engineering Mathematics

82A Engineering Mathematics Clss Notes 9: Power Series /) 8A Egieerig Mthetics Secod Order Differetil Equtios Series Solutio Solutio Ato Differetil Equtio =, Hoogeeous =gt), No-hoogeeous Solutio: = c + p Hoogeeous No-hoogeeous Fudetl

More information

Algebra II, Chapter 7. Homework 12/5/2016. Harding Charter Prep Dr. Michael T. Lewchuk. Section 7.1 nth roots and Rational Exponents

Algebra II, Chapter 7. Homework 12/5/2016. Harding Charter Prep Dr. Michael T. Lewchuk. Section 7.1 nth roots and Rational Exponents Algebr II, Chpter 7 Hrdig Chrter Prep 06-07 Dr. Michel T. Lewchuk Test scores re vilble olie. I will ot discuss the test. st retke opportuit Sturd Dec. If ou hve ot tke the test, it is our resposibilit

More information

n 2 + 3n + 1 4n = n2 + 3n + 1 n n 2 = n + 1

n 2 + 3n + 1 4n = n2 + 3n + 1 n n 2 = n + 1 Ifiite Series Some Tests for Divergece d Covergece Divergece Test: If lim u or if the limit does ot exist, the series diverget. + 3 + 4 + 3 EXAMPLE: Show tht the series diverges. = u = + 3 + 4 + 3 + 3

More information

Force and Motion. Force. Classifying Forces. Physics 11- Summer /21/01. Chapter 4 material 1. Forces are vector quantities!

Force and Motion. Force. Classifying Forces. Physics 11- Summer /21/01. Chapter 4 material 1. Forces are vector quantities! Force d Motio Cocept of Force Newto s hree Lws ypes of Forces Free body lysis Equilibrium Noequilibrium Frictio Problem Solvig Force A Force is push or pull tht is exerted o object by some other object.

More information

Note 7 Root-Locus Techniques

Note 7 Root-Locus Techniques Lecture Note of Cotrol Syte I - ME 43/Alyi d Sythei of Lier Cotrol Syte - ME862 Note 7 Root-Locu Techique Deprtet of Mechicl Egieerig, Uiverity Of Sktchew, 57 Cpu Drive, Sktoo, S S7N 5A9, Cd Lecture Note

More information

Chapter #2 EEE Subsea Control and Communication Systems

Chapter #2 EEE Subsea Control and Communication Systems EEE 87 Chpter # EEE 87 Sube Cotrol d Commuictio Sytem Trfer fuctio Pole loctio d -ple Time domi chrcteritic Extr pole d zero Chpter /8 EEE 87 Trfer fuctio Lplce Trform Ued oly o LTI ytem Differetil expreio

More information

EVALUATING DEFINITE INTEGRALS

EVALUATING DEFINITE INTEGRALS Chpter 4 EVALUATING DEFINITE INTEGRALS If the defiite itegrl represets re betwee curve d the x-xis, d if you c fid the re by recogizig the shpe of the regio, the you c evlute the defiite itegrl. Those

More information

Solving Systems of Equations

Solving Systems of Equations PGE : Formultio d Solutio i Geosystems Egieerig Dr. Blhoff Solvig Systems of Equtios Numericl Methods with MTLB, Recktewld, Chpter 8 d Numericl Methods for Egieers, Chpr d Cle, 5 th Ed., Prt Three, Chpters

More information

Student Success Center Elementary Algebra Study Guide for the ACCUPLACER (CPT)

Student Success Center Elementary Algebra Study Guide for the ACCUPLACER (CPT) Studet Success Ceter Elemetry Algebr Study Guide for the ACCUPLACER (CPT) The followig smple questios re similr to the formt d cotet of questios o the Accuplcer Elemetry Algebr test. Reviewig these smples

More information

POWER SERIES R. E. SHOWALTER

POWER SERIES R. E. SHOWALTER POWER SERIES R. E. SHOWALTER. sequeces We deote by lim = tht the limit of the sequece { } is the umber. By this we me tht for y ε > 0 there is iteger N such tht < ε for ll itegers N. This mkes precise

More information

Addendum. Addendum. Vector Review. Department of Computer Science and Engineering 1-1

Addendum. Addendum. Vector Review. Department of Computer Science and Engineering 1-1 Addedum Addedum Vetor Review Deprtmet of Computer Siee d Egieerig - Coordite Systems Right hded oordite system Addedum y z Deprtmet of Computer Siee d Egieerig - -3 Deprtmet of Computer Siee d Egieerig

More information

Solutions to RSPL/1. log 3. When x = 1, t = 0 and when x = 3, t = log 3 = sin(log 3) 4. Given planes are 2x + y + 2z 8 = 0, i.e.

Solutions to RSPL/1. log 3. When x = 1, t = 0 and when x = 3, t = log 3 = sin(log 3) 4. Given planes are 2x + y + 2z 8 = 0, i.e. olutios to RPL/. < F < F< Applig C C + C, we get F < 5 F < F< F, $. f() *, < f( h) f( ) h Lf () lim lim lim h h " h h " h h " f( + h) f( ) h Rf () lim lim lim h h " h h " h h " Lf () Rf (). Hee, differetile

More information

Week 13 Notes: 1) Riemann Sum. Aim: Compute Area Under a Graph. Suppose we want to find out the area of a graph, like the one on the right:

Week 13 Notes: 1) Riemann Sum. Aim: Compute Area Under a Graph. Suppose we want to find out the area of a graph, like the one on the right: Week 1 Notes: 1) Riem Sum Aim: Compute Are Uder Grph Suppose we wt to fid out the re of grph, like the oe o the right: We wt to kow the re of the red re. Here re some wys to pproximte the re: We cut the

More information

MATRIX ALGEBRA, Systems Linear Equations

MATRIX ALGEBRA, Systems Linear Equations MATRIX ALGEBRA, Systes Lier Equtios Now we chge to the LINEAR ALGEBRA perspective o vectors d trices to reforulte systes of lier equtios. If you fid the discussio i ters of geerl d gets lost i geerlity,

More information

UNIT #5 SEQUENCES AND SERIES COMMON CORE ALGEBRA II

UNIT #5 SEQUENCES AND SERIES COMMON CORE ALGEBRA II Awer Key Nme: Dte: UNIT # SEQUENCES AND SERIES COMMON CORE ALGEBRA II Prt I Quetio. For equece defied by f? () () 08 6 6 f d f f, which of the followig i the vlue of f f f f f f 0 6 6 08 (). I the viul

More information

A New Estimator Using Auxiliary Information in Stratified Adaptive Cluster Sampling

A New Estimator Using Auxiliary Information in Stratified Adaptive Cluster Sampling Ope Jourl of ttitic, 03, 3, 78-8 ttp://d.doi.org/0.436/oj.03.3403 Publied Olie eptember 03 (ttp://www.cirp.org/jourl/oj) New Etimtor Uig uilir Iformtio i trtified dptive Cluter mplig Nippor Cutim *, Moc

More information

BME 207 Introduction to Biomechanics Spring 2018

BME 207 Introduction to Biomechanics Spring 2018 April 6, 28 UNIVERSITY O RHODE ISAND Deprtment of Electricl, Computer nd Biomedicl Engineering BME 27 Introduction to Biomechnics Spring 28 Homework 8 Prolem 14.6 in the textook. In ddition to prts -e,

More information

Section IV.6: The Master Method and Applications

Section IV.6: The Master Method and Applications Sectio IV.6: The Mster Method d Applictios Defiitio IV.6.1: A fuctio f is symptoticlly positive if d oly if there exists rel umer such tht f(x) > for ll x >. A cosequece of this defiitio is tht fuctio

More information

Introduction to Modern Control Theory

Introduction to Modern Control Theory Itroductio to Moder Cotrol Theory MM : Itroductio to Stte-Spce Method MM : Cotrol Deig for Full Stte Feedck MM 3: Etitor Deig MM 4: Itroductio of the Referece Iput MM 5: Itegrl Cotrol d Rout Trckig //4

More information

(200 terms) equals Let f (x) = 1 + x + x 2 + +x 100 = x101 1

(200 terms) equals Let f (x) = 1 + x + x 2 + +x 100 = x101 1 SECTION 5. PGE 78.. DMS: CLCULUS.... 5. 6. CHPTE 5. Sectio 5. pge 78 i + + + INTEGTION Sums d Sigm Nottio j j + + + + + i + + + + i j i i + + + j j + 5 + + j + + 9 + + 7. 5 + 6 + 7 + 8 + 9 9 i i5 8. +

More information

Definite Integral. The Left and Right Sums

Definite Integral. The Left and Right Sums Clculus Li Vs Defiite Itegrl. The Left d Right Sums The defiite itegrl rises from the questio of fidig the re betwee give curve d x-xis o itervl. The re uder curve c be esily clculted if the curve is give

More information

THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING

THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING OLLSCOIL NA héireann, CORCAIGH THE NATIONAL UNIVERSITY OF IRELAND, CORK COLÁISTE NA hollscoile, CORCAIGH UNIVERSITY COLLEGE, CORK SUMMER EXAMINATION 2005 FIRST ENGINEERING MATHEMATICS MA008 Clculus d Lier

More information

EE105 - Fall 2006 Microelectronic Devices and Circuits

EE105 - Fall 2006 Microelectronic Devices and Circuits EE15 - Fll 6 Microelectroic evice Circuit Prof. J M. Rbey (@eec Lecture 4: Ccitor P Juctio Overview Lt lecture iffuio curret Overview of IC fbrictio roce Review of electrottic Thi lecture Ccitce Juctio

More information