Problem Set 2 Solutions
|
|
- Adam Kelly
- 7 years ago
- Views:
Transcription
1 University of Cliforni, Berkeley Spring 2012 EE 42/100 Prof. A. Niknej Prolem Set 2 Solutions Plese note tht these re merely suggeste solutions. Mny of these prolems n e pprohe in ifferent wys. 1. In prolems like this, you my fin it helpful to rerw the iruit to ignore ll the irrelevnt terminls n suh tht everything is retngulr. Note tht sine we re fining eq from terminls n in eh of these, we must simplify the iruits towr these terminls. () Terminls n re onnete together, while e n f re irrelevnt. If you rerw the iruit s elow, you n see tht the two resistors in the first olumn re in prllel s, they re onnete together t oth pirs of terminls (ue to the short). This is lso true for the seon olumn of resistors. So eh of the pirs n e rerwn s one resistor with 1 2. e f /2 /2 It is esy to see tht ll the resistors re in series, so eq = 3. () Terminls e n f re shorte. Intuitively, eq shoul only inlue the two inner resistors, sine the other four re shorte out. Tht is, ny urrent flowing from to woul only go through 2 of resistne n proee etween e n f vi the short. e f e f 2 2 () The short etween n e gets ri of the one resistor etween the two noes. After tht, everything else simplifies s efore.
2 e f e f 2 8/3 2. The key to this prolem is tht the ler is infinite. Using this ft, we n onlue tht the equivlent resistne in shoul not hnge if we remove one rung of the ler. So we n reple the ler, sve for the first rung, with one resistor in. in in Now we n write n eqution in terms of n in n solve. in in = ( in ) = 2 in 2 in 2 in 2 2 = 0 Sine this is qurti eqution, there re two solutions. We know tht resistne nnot e negtive, so we isr the negtive solution. The positive solution is in = (1 3) 3. We hve two unknown noes, so we ll hve two nol equtions, one t noe 1 n the other t noe 2. V 1 V 2 V = 0 V 2 V 1 V = 0 Solving this system gives us V 1 = V n V 2 = 74.4 V. We n lso o this using superposition. Here we split the prolem into two suiruits, eh one with ifferent urrent soure zeroe out. ell tht the zeroe out urrent soure eomes n open iruit. Ω 1 2 8A 30Ω 20Ω 2
3 V 1 V 2 V 2 V 1 V = 0 V 2 20 = 0 This first iruit gives us V 1 = V n V 2 = 19.2 V. Ω Ω 20Ω -3A V 1 V 2 V 2 V 1 V 1 30 = 0 V = 0 The solutions re V 1 = 7.2 V n V 2 = 55.2 V. Clerly, the sum of these with the solutions for the first suiruit yiels the sme nswers s the originl iruit. 4. There re five noes in this iruit, lele elow. Pling the referene noe t v E reues the numer of unknowns to three, sine now v E = 0 V n v D = 5 V. Ω A B C Ω D 5V 18Ω 75Ω 1.2A E 30Ω The three nol equtions re the following: V A V C V A V B 18 V A 5 = 0 43V A 25V B 9V C = 45 V B V A 18 V C V A V B V C 75 V C V B = 0 25V A 31V B 6V C = 540 V C 30 = 0 3V A 2V B 10V C = 0 The solutions re V A = 32.8 V, V B = 47.6 V, n V C = 19.3 V. The rnh urrents n e foun using Ohm s lw n KCL: I AC = 0.27 A, I AB = 0.82 A, I BC = 0.38 A, I AD = 0.56 A, I CE = 0.64 A, I DE = 0.56 A. 3
4 5. The iruit hs four unknown noes, lele elow. 100Ω 0Ω 1V V A V B V C 5V 10 V x 80Ω V x 5Ω V x 90Ω In ition, we must lso write supernoe eqution t the 1 oure. The epenent urrent soure oes not present prolem, sine we tret V x s one of the unknowns. The nol equtions t V A n V x re V A V A 80 V A V B 0 10V x = 0 V x V C 5 The two equtions t the supernoe re V x 90 = 0 V B V A 0 V C V x 5 V B V C = 1 The solution to these equtions is V A = V, V B = V, V C = V, V x = V. 6. Note tht the left prt of the iruit is inepenent of the right prt. Sine there is no omplete loop onneting the two prts, no urrent n flow etween them; the wire onneting them only serves s ommon groun. The equivlent resistne on the left is = 9 Ω. So the totl urrent from is just Vs. Also, we hve tht I 9 x is just hlf of tht, sine the two prllel rnhes hve equl resistne. So I x = Vs. 18 On the right hlf, the two resistors re in prllel, so the equivlent resistne is just 2 Ω. Sine the urrent going through them is given y AI x = AVs, the voltge ross 18 them is V out = I eq = AVs Vout. Solving for A n plugging in 9 V in = 9, we hve tht A = The steps to fining the equivlent iruits re to fin V o, I s, n/or eq. Sine there is epenent soure in this iruit, it is esier to fin the first two. V o is simply the voltge ifferene etween n. By inspetion, this is just = 0 V o = (100 kω)βi x = (100 kω)β 10 kω = 10β I s is foun y shorting the terminls n solving for I. Note tht shorting out the 100 kω resistor will leve us with I = βi x = β Vs (note the sign!). 10 kω 4
5 Thus, our esire vlues re V T h = 10β n I N = βvs. Also, the Thévenin n 10 kω Norton resistne is just given y T h = N = V T h I N = 100 kω. 8. () The mximum power trnsfer theorem tells us tht the lo resistne shoul e equl to the Thévenin resistne of the iruit. In this se, it is trivil to see tht tht L = T h = s. If you wnt to review the erivtion of this result, plese onsult the leture notes or textook. () We wnt the power ross the lo, P L, to e equl to 80% of the power elivere y the voltge soure, P s. The former n e foun using P L = I 2 L, wheres the ltter is P s = I. In this iruit, I = So putting everything together, we hve Solving for L gives us L = 4 s. s L Vs 2 L ( s L ) = 0.8 Vs 2 2 s L () Power is epenent on oth voltge n urrent. While the nswer to prt () gives us higher voltge ross L n greter frtion of the totl power, the totl power is tully muh smller in this se. This is euse the urrent is 5 times smller ue to the greter resistne, while the voltge ross L is only 1.6 times lrger. 9. Intuitively we know tht losing the swith puts 2 n 3 in prllel, lowering the effetive resistne. So it must e in the open se tht v o tkes the greter voltge vlue, 26 V. In this senrio, the iruit looks like the following: Ω 39V 2 v o = 26V Solving for this iruit, using voltge ivier, nol nlysis, or otherwise, gives us 2 = 100 Ω. Now if we lose the swith, v o = 24 V. The vlue of 2 remins unhnge, so ll tht s remining is to solve for 3. Ω 39V 100 Ω v o 3 5
6 We n write nol eqution t v o = 24 V: The solution is 3 = 400 Ω. v o 39 v o 100 v o 3 = () When the rige is lne, no urrent flows ross the glvnometer. By KCL, the urrent ross 1 n 2 re the sme, s well s the urrent ross 3 n 4. In ition, the voltges v n v must e equl ue to the 0 urrent. 1 G 3 I 1 I 2 2 x By KVL, the voltge rops ross 1 n 3 must then e the sme. The sme resoning goes for 2 n x. I 1 1 = I 2 3 I 1 2 = I 2 x Sustituting for the urrents n solving for x, we hve x = () Given our vlues for 1 n 3, we n mesure x = 2 2. Sine the highest vlue tht 2 n tke is 1000 Ω, the mximum resistne tht n e mesure for x is just 0 Ω. As 2 omes in inrements of 10 Ω, the mesurement n e me with n ury to within 20 Ω. () The Thévenin equivlent is foun y fining the voltge n equivlent resistne ross the open terminls n. Note tht here we nnot ssume the rige is lne, s we re looking t the generl se. v o is simply the ifferene in potentil etween noes n, oth of whih n e foun vi voltge ivier: 2 v = 1 2 v = 3 x ( 2 V T h = v v = ) x x The equivlent resistne n e foun y zeroing out the voltge soure, mking it short, n fining eq. 6 x
7 1 3 2 x We hve 1 n 2 in prllel with eh other, n 3 n x in prllel with eh other. These two pirs re in series, so the equivlent resistne is T h = ( 1 2 ) ( 3 x ). 7
Maximum area of polygon
Mimum re of polygon Suppose I give you n stiks. They might e of ifferent lengths, or the sme length, or some the sme s others, et. Now there re lots of polygons you n form with those stiks. Your jo is
More informationCHAPTER 31 CAPACITOR
. Given tht Numer of eletron HPTER PITOR Net hrge Q.6 9.6 7 The net potentil ifferene L..6 pitne v 7.6 8 F.. r 5 m. m 8.854 5.4 6.95 5 F... Let the rius of the is R re R D mm m 8.85 r r 8.85 4. 5 m.5 m
More informationRatio and Proportion
Rtio nd Proportion Rtio: The onept of rtio ours frequently nd in wide vriety of wys For exmple: A newspper reports tht the rtio of Repulins to Demorts on ertin Congressionl ommittee is 3 to The student/fulty
More informationLesson 2.1 Inductive Reasoning
Lesson.1 Inutive Resoning Nme Perio Dte For Eerises 1 7, use inutive resoning to fin the net two terms in eh sequene. 1. 4, 8, 1, 16,,. 400, 00, 100, 0,,,. 1 8, 7, 1, 4,, 4.,,, 1, 1, 0,,. 60, 180, 10,
More informationAnswer, Key Homework 10 David McIntyre 1
Answer, Key Homework 10 Dvid McIntyre 1 This print-out should hve 22 questions, check tht it is complete. Multiple-choice questions my continue on the next column or pge: find ll choices efore mking your
More informationModule 5. Three-phase AC Circuits. Version 2 EE IIT, Kharagpur
Module 5 Three-hse A iruits Version EE IIT, Khrgur esson 8 Three-hse Blned Suly Version EE IIT, Khrgur In the module, ontining six lessons (-7), the study of iruits, onsisting of the liner elements resistne,
More informationAngles 2.1. Exercise 2.1... Find the size of the lettered angles. Give reasons for your answers. a) b) c) Example
2.1 Angles Reognise lternte n orresponing ngles Key wors prllel lternte orresponing vertilly opposite Rememer, prllel lines re stright lines whih never meet or ross. The rrows show tht the lines re prllel
More informationChapter. Contents: A Constructing decimal numbers
Chpter 9 Deimls Contents: A Construting deiml numers B Representing deiml numers C Deiml urreny D Using numer line E Ordering deimls F Rounding deiml numers G Converting deimls to frtions H Converting
More information50 MATHCOUNTS LECTURES (10) RATIOS, RATES, AND PROPORTIONS
0 MATHCOUNTS LECTURES (0) RATIOS, RATES, AND PROPORTIONS BASIC KNOWLEDGE () RATIOS: Rtios re use to ompre two or more numers For n two numers n ( 0), the rtio is written s : = / Emple : If 4 stuents in
More informationThe remaining two sides of the right triangle are called the legs of the right triangle.
10 MODULE 6. RADICAL EXPRESSIONS 6 Pythgoren Theorem The Pythgoren Theorem An ngle tht mesures 90 degrees is lled right ngle. If one of the ngles of tringle is right ngle, then the tringle is lled right
More informationMATH PLACEMENT REVIEW GUIDE
MATH PLACEMENT REVIEW GUIDE This guie is intene s fous for your review efore tking the plement test. The questions presente here my not e on the plement test. Although si skills lultor is provie for your
More informationc b 5.00 10 5 N/m 2 (0.120 m 3 0.200 m 3 ), = 4.00 10 4 J. W total = W a b + W b c 2.00
Chter 19, exmle rolems: (19.06) A gs undergoes two roesses. First: onstnt volume @ 0.200 m 3, isohori. Pressure inreses from 2.00 10 5 P to 5.00 10 5 P. Seond: Constnt ressure @ 5.00 10 5 P, isori. olume
More informationCS99S Laboratory 2 Preparation Copyright W. J. Dally 2001 October 1, 2001
CS99S Lortory 2 Preprtion Copyright W. J. Dlly 2 Octoer, 2 Ojectives:. Understnd the principle of sttic CMOS gte circuits 2. Build simple logic gtes from MOS trnsistors 3. Evlute these gtes to oserve logic
More information1 Fractions from an advanced point of view
1 Frtions from n vne point of view We re going to stuy frtions from the viewpoint of moern lger, or strt lger. Our gol is to evelop eeper unerstning of wht n men. One onsequene of our eeper unerstning
More informationVectors Summary. Projection vector AC = ( Shortest distance from B to line A C D [OR = where m1. and m
. Slr prout (ot prout): = osθ Vetors Summry Lws of ot prout: (i) = (ii) ( ) = = (iii) = (ngle etween two ientil vetors is egrees) (iv) = n re perpeniulr Applitions: (i) Projetion vetor: B Length of projetion
More informationRotating DC Motors Part II
Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors
More information1. Definition, Basic concepts, Types 2. Addition and Subtraction of Matrices 3. Scalar Multiplication 4. Assignment and answer key 5.
. Definition, Bsi onepts, Types. Addition nd Sutrtion of Mtries. Slr Multiplition. Assignment nd nswer key. Mtrix Multiplition. Assignment nd nswer key. Determinnt x x (digonl, minors, properties) summry
More information, and the number of electrons is -19. e e 1.60 10 C. The negatively charged electrons move in the direction opposite to the conventional current flow.
Prolem 1. f current of 80.0 ma exists in metl wire, how mny electrons flow pst given cross section of the wire in 10.0 min? Sketch the directions of the current nd the electrons motion. Solution: The chrge
More information2 DIODE CLIPPING and CLAMPING CIRCUITS
2 DIODE CLIPPING nd CLAMPING CIRCUITS 2.1 Ojectives Understnding the operting principle of diode clipping circuit Understnding the operting principle of clmping circuit Understnding the wveform chnge of
More informationFAULT TREES AND RELIABILITY BLOCK DIAGRAMS. Harry G. Kwatny. Department of Mechanical Engineering & Mechanics Drexel University
SYSTEM FAULT AND Hrry G. Kwtny Deprtment of Mechnicl Engineering & Mechnics Drexel University OUTLINE SYSTEM RBD Definition RBDs nd Fult Trees System Structure Structure Functions Pths nd Cutsets Reliility
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More informationAppendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:
Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you
More informationGENERAL OPERATING PRINCIPLES
KEYSECUREPC USER MANUAL N.B.: PRIOR TO READING THIS MANUAL, YOU ARE ADVISED TO READ THE FOLLOWING MANUAL: GENERAL OPERATING PRINCIPLES Der Customer, KeySeurePC is n innovtive prout tht uses ptente tehnology:
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationSECTION 7-2 Law of Cosines
516 7 Additionl Topis in Trigonometry h d sin s () tn h h d 50. Surveying. The lyout in the figure t right is used to determine n inessile height h when seline d in plne perpendiulr to h n e estlished
More informationVolumes by Cylindrical Shells: the Shell Method
olumes Clinril Shells: the Shell Metho Another metho of fin the volumes of solis of revolution is the shell metho. It n usull fin volumes tht re otherwise iffiult to evlute using the Dis / Wsher metho.
More informationHomework 3 Solutions
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.
More informationExample 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More informationUnit 5 Section 1. Mortgage Payment Methods & Products (20%)
Unit 5 Setion 1 Mortgge Pyment Methos & Prouts (20%) There re tully only 2 mortgge repyment methos ville CAPITAL REPAYMENT n INTEREST ONLY. Cpitl Repyment Mortgge Also lle Cpitl & Interest mortgge or repyment
More informationBinary Representation of Numbers Autar Kaw
Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy
More informationLectures 8 and 9 1 Rectangular waveguides
1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationPractice Test 2. a. 12 kn b. 17 kn c. 13 kn d. 5.0 kn e. 49 kn
Prtie Test 2 1. A highwy urve hs rdius of 0.14 km nd is unnked. A r weighing 12 kn goes round the urve t speed of 24 m/s without slipping. Wht is the mgnitude of the horizontl fore of the rod on the r?
More informationSolution to Problem Set 1
CSE 5: Introduction to the Theory o Computtion, Winter A. Hevi nd J. Mo Solution to Prolem Set Jnury, Solution to Prolem Set.4 ). L = {w w egin with nd end with }. q q q q, d). L = {w w h length t let
More informationSection 5-4 Trigonometric Functions
5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More informationOn Equivalence Between Network Topologies
On Equivlene Between Network Topologies Tre Ho Deprtment of Eletril Engineering Cliforni Institute of Tehnolog tho@lteh.eu; Mihelle Effros Deprtments of Eletril Engineering Cliforni Institute of Tehnolog
More information1.2 The Integers and Rational Numbers
.2. THE INTEGERS AND RATIONAL NUMBERS.2 The Integers n Rtionl Numers The elements of the set of integers: consist of three types of numers: Z {..., 5, 4, 3, 2,, 0,, 2, 3, 4, 5,...} I. The (positive) nturl
More informationWords Symbols Diagram. abcde. a + b + c + d + e
Logi Gtes nd Properties We will e using logil opertions to uild mhines tht n do rithmeti lultions. It s useful to think of these opertions s si omponents tht n e hooked together into omplex networks. To
More informationIf two triangles are perspective from a point, then they are also perspective from a line.
Mth 487 hter 4 Prtie Prolem Solutions 1. Give the definition of eh of the following terms: () omlete qudrngle omlete qudrngle is set of four oints, no three of whih re olliner, nd the six lines inident
More informationWHAT HAPPENS WHEN YOU MIX COMPLEX NUMBERS WITH PRIME NUMBERS?
WHAT HAPPES WHE YOU MIX COMPLEX UMBERS WITH PRIME UMBERS? There s n ol syng, you n t pples n ornges. Mthemtns hte n t; they love to throw pples n ornges nto foo proessor n see wht hppens. Sometmes they
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More informationRegular Sets and Expressions
Regulr Sets nd Expressions Finite utomt re importnt in science, mthemtics, nd engineering. Engineers like them ecuse they re super models for circuits (And, since the dvent of VLSI systems sometimes finite
More informationMath 135 Circles and Completing the Square Examples
Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More informationMath 314, Homework Assignment 1. 1. Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.
Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint
More informationCS 316: Gates and Logic
CS 36: Gtes nd Logi Kvit Bl Fll 27 Computer Siene Cornell University Announements Clss newsgroup reted Posted on we-pge Use it for prtner finding First ssignment is to find prtners P nd N Trnsistors PNP
More informationBoğaziçi University Department of Economics Spring 2016 EC 102 PRINCIPLES of MACROECONOMICS Problem Set 5 Answer Key
Boğziçi University Deprtment of Eonomis Spring 2016 EC 102 PRINCIPLES of MACROECONOMICS Prolem Set 5 Answer Key 1. One yer ountry hs negtive net exports. The next yer it still hs negtive net exports n
More informationForensic Engineering Techniques for VLSI CAD Tools
Forensi Engineering Tehniques for VLSI CAD Tools Jennifer L. Wong, Drko Kirovski, Dvi Liu, Miorg Potkonjk UCLA Computer Siene Deprtment University of Cliforni, Los Angeles June 8, 2000 Computtionl Forensi
More informationWeek 11 - Inductance
Week - Inductnce November 6, 202 Exercise.: Discussion Questions ) A trnsformer consists bsiclly of two coils in close proximity but not in electricl contct. A current in one coil mgneticlly induces n
More informationLINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES
LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of
More informationThe art of Paperarchitecture (PA). MANUAL
The rt of Pperrhiteture (PA). MANUAL Introution Pperrhiteture (PA) is the rt of reting three-imensionl (3D) ojets out of plin piee of pper or ror. At first, esign is rwn (mnully or printe (using grphil
More informationSOLVING QUADRATIC EQUATIONS BY FACTORING
6.6 Solving Qudrti Equtions y Ftoring (6 31) 307 In this setion The Zero Ftor Property Applitions 6.6 SOLVING QUADRATIC EQUATIONS BY FACTORING The tehniques of ftoring n e used to solve equtions involving
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationHow To Find The Re Of Tringle
Heron s Formul for Tringulr Are y Christy Willims, Crystl Holom, nd Kyl Gifford Heron of Alexndri Physiist, mthemtiin, nd engineer Tught t the museum in Alexndri Interests were more prtil (mehnis, engineering,
More informationExperiment 6: Friction
Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht
More informationPhysics 43 Homework Set 9 Chapter 40 Key
Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nm-wide region t x
More information1 GSW IPv4 Addressing
1 For s long s I ve een working with the Internet protools, people hve een sying tht IPv6 will e repling IPv4 in ouple of yers time. While this remins true, it s worth knowing out IPv4 ddresses. Even when
More informationSOLVING EQUATIONS BY FACTORING
316 (5-60) Chpter 5 Exponents nd Polynomils 5.9 SOLVING EQUATIONS BY FACTORING In this setion The Zero Ftor Property Applitions helpful hint Note tht the zero ftor property is our seond exmple of getting
More informationMA 15800 Lesson 16 Notes Summer 2016 Properties of Logarithms. Remember: A logarithm is an exponent! It behaves like an exponent!
MA 5800 Lesson 6 otes Summer 06 Rememer: A logrithm is n eponent! It ehves like n eponent! In the lst lesson, we discussed four properties of logrithms. ) log 0 ) log ) log log 4) This lesson covers more
More information4.11 Inner Product Spaces
314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces
More informationOrthopoles and the Pappus Theorem
Forum Geometriorum Volume 4 (2004) 53 59. FORUM GEOM ISSN 1534-1178 Orthopoles n the Pppus Theorem tul Dixit n Drij Grinerg strt. If the verties of tringle re projete onto given line, the perpeniulrs from
More information. At first sight a! b seems an unwieldy formula but use of the following mnemonic will possibly help. a 1 a 2 a 3 a 1 a 2
7 CHAPTER THREE. Cross Product Given two vectors = (,, nd = (,, in R, the cross product of nd written! is defined to e: " = (!,!,! Note! clled cross is VECTOR (unlike which is sclr. Exmple (,, " (4,5,6
More informationH SERIES. Area and Perimeter. Curriculum Ready. www.mathletics.com
Are n Perimeter Curriulum Rey www.mthletis.om Copyright 00 3P Lerning. All rights reserve. First eition printe 00 in Austrli. A tlogue reor for this ook is ville from 3P Lerning Lt. ISBN 78--86-30-7 Ownership
More informationand thus, they are similar. If k = 3 then the Jordan form of both matrices is
Homework ssignment 11 Section 7. pp. 249-25 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If
More informationUnderstanding Basic Analog Ideal Op Amps
Appliction Report SLAA068A - April 2000 Understnding Bsic Anlog Idel Op Amps Ron Mncini Mixed Signl Products ABSTRACT This ppliction report develops the equtions for the idel opertionl mplifier (op mp).
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More informationEnd of term: TEST A. Year 4. Name Class Date. Complete the missing numbers in the sequences below.
End of term: TEST A You will need penil nd ruler. Yer Nme Clss Dte Complete the missing numers in the sequenes elow. 8 30 3 28 2 9 25 00 75 25 2 Put irle round ll of the following shpes whih hve 3 shded.
More informationClause Trees: a Tool for Understanding and Implementing Resolution in Automated Reasoning
Cluse Trees: Tool for Understnding nd Implementing Resolution in Automted Resoning J. D. Horton nd Brue Spener University of New Brunswik, Frederiton, New Brunswik, Cnd E3B 5A3 emil : jdh@un. nd spener@un.
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationAREA OF A SURFACE OF REVOLUTION
AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.
More informationInterior and exterior angles add up to 180. Level 5 exterior angle
22 ngles n proof Ientify interior n exterior ngles in tringles n qurilterls lulte interior n exterior ngles of tringles n qurilterls Unerstn the ie of proof Reognise the ifferene etween onventions, efinitions
More informationCalculus of variations. I = F(y, y,x) dx (1)
MT58 - Clculus of vritions Introuction. Suppose y(x) is efine on the intervl, Now suppose n so efines curve on the ( x,y) plne. I = F(y, y,x) x (1) with the erivtive of y(x). The vlue of this will epen
More informationFurther applications of area and volume
2 Further pplitions of re n volume 2A Are of prts of the irle 2B Are of omposite shpes 2C Simpson s rule 2D Surfe re of yliners n spheres 2E Volume of omposite solis 2F Error in mesurement Syllus referene
More informationMathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100
hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by
More informationCalculating Principal Strains using a Rectangular Strain Gage Rosette
Clulting Prinipl Strins using Retngulr Strin Gge Rosette Strin gge rosettes re used often in engineering prtie to determine strin sttes t speifi points on struture. Figure illustrtes three ommonly used
More informationUNCORRECTED SAMPLE PAGES
6 Chpter Length, re, surfe re n volume Wht you will lern 6A Length n perimeter 6B Cirumferene of irles n perimeter of setors 6C Are of qurilterls n tringles 6D Are of irles 6E Perimeter n re of omposite
More informationN V V L. R a L I. Transformer Equation Notes
Tnsfome Eqution otes This file conts moe etile eivtion of the tnsfome equtions thn the notes o the expeiment 3 wite-up. t will help you to unestn wht ssumptions wee neee while eivg the iel tnsfome equtions
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More informationAlgebra Review. How well do you remember your algebra?
Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then
More informationJCM TRAINING OVERVIEW Multi-Download Module 2
Multi-Downlo Moule 2 JCM Trining Overview Mrh, 2012 Mrh, 2012 CLOSING THE MDM2 APPLICATION To lose the MDM2 Applition proee s follows: 1. Mouse-lik on the 'File' pullown Menu (See Figure 35 ) on the MDM2
More information1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator
AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.
More informationBasic Research in Computer Science BRICS RS-02-13 Brodal et al.: Solving the String Statistics Problem in Time O(n log n)
BRICS Bsic Reserch in Computer Science BRICS RS-02-13 Brodl et l.: Solving the String Sttistics Prolem in Time O(n log n) Solving the String Sttistics Prolem in Time O(n log n) Gerth Stølting Brodl Rune
More informationp-q Theory Power Components Calculations
ISIE 23 - IEEE Interntionl Symposium on Industril Eletronis Rio de Jneiro, Brsil, 9-11 Junho de 23, ISBN: -783-7912-8 p-q Theory Power Components Clultions João L. Afonso, Memer, IEEE, M. J. Sepúlved Freits,
More informationTHE LONGITUDINAL FIELD IN THE GTEM 1750 AND THE NATURE OF THE TERMINATION.
THE LONGITUDINAL FIELD IN THE GTEM 175 AND THE NATURE OF THE TERMINATION. Benjmin Guy Loder Ntionl Physil Lbortory, Queens Rod, Teddington, Middlesex, Englnd. TW11 LW Mrtin Alexnder Ntionl Physil Lbortory,
More informationCURVES ANDRÉ NEVES. that is, the curve α has finite length. v = p q p q. a i.e., the curve of smallest length connecting p to q is a straight line.
CURVES ANDRÉ NEVES 1. Problems (1) (Ex 1 of 1.3 of Do Crmo) Show tht the tngent line to the curve α(t) (3t, 3t 2, 2t 3 ) mkes constnt ngle with the line z x, y. (2) (Ex 6 of 1.3 of Do Crmo) Let α(t) (e
More informationUnit 6: Exponents and Radicals
Eponents nd Rdicls -: The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N): - counting numers. {,,,,, } Whole Numers (W): - counting numers with 0. {0,,,,,, } Integers (I): -
More informationWarm-up for Differential Calculus
Summer Assignment Wrm-up for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:
More informationIntegration by Substitution
Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is
More informationLesson 4.1 Triangle Sum Conjecture
Lesson 4.1 ringle um onjecture Nme eriod te n ercises 1 9, determine the ngle mesures. 1. p, q 2., y 3., b 31 82 p 98 q 28 53 y 17 79 23 50 b 4. r, s, 5., y 6. y t t s r 100 85 100 y 30 4 7 y 31 7. s 8.
More informationCOMPONENTS: COMBINED LOADING
LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of
More informationAngles and Triangles
nges nd Tringes n nge is formed when two rys hve ommon strting point or vertex. The mesure of n nge is given in degrees, with ompete revoution representing 360 degrees. Some fmiir nges inude nother fmiir
More informationBasic Laws Circuit Theorems Methods of Network Analysis Non-Linear Devices and Simulation Models
EE Modul 1: Electric Circuits Theory Basic Laws Circuit Theorems Methods of Network Analysis Non-Linear Devices and Simulation Models EE Modul 1: Electric Circuits Theory Current, Voltage, Impedance Ohm
More informationtr(a + B) = tr(a) + tr(b) tr(ca) = c tr(a)
Chapter 3 Determinant 31 The Determinant Funtion We follow an intuitive approah to introue the efinition of eterminant We alreay have a funtion efine on ertain matries: the trae The trae assigns a numer
More informationEcon 4721 Money and Banking Problem Set 2 Answer Key
Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in
More informationChapter. Fractions. Contents: A Representing fractions
Chpter Frtions Contents: A Representing rtions B Frtions o regulr shpes C Equl rtions D Simpliying rtions E Frtions o quntities F Compring rtion sizes G Improper rtions nd mixed numers 08 FRACTIONS (Chpter
More information