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Deep Dive into Mathematica's Numerics: Applications and Tips
Andrew Moylan
In this course from the Wolfram Mathematica Virtual Conference 2011, you'll learn how to best use Mathematica's numerics functions in advanced settings. Topics include techniques and best practices for using multiple numerics functions together, advanced numeric features, and understanding precision and accuracy
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1111990166
9781111990169
College Algebra:Learn to think mathematically and develop genuine problem-solving skills with Stewart, Redlin, and Watson's COLLEGE ALGEBRA, Sixth Edition. This straightforward and easy-to-use algebra book will help you learn the fundamentals of algebra in a variety of practical ways. The book features new tools to help you succeed, such as learning objectives before each section to prepare you for what you're about to learn, and a list of formulas and key concepts after each section that help reinforce what you've learned. In addition, the book includes many real-world examples that show you how mathematics is used to model in fields like engineering, business, physics, chemistry, and biology.
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Rent College Algebra 6th edition today, or search our site for James textbooks. Every textbook comes with a 21-day "Any Reason" guarantee. Published by CENGAGE Learning.
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Apply and extend previous understandings of numbers to the system of rational numbers. Expressions and Equations
Apply and extend previous understandings of arithmetic to algebraic expressions. Reason about and solve one-variable equations and inequalities.
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Passage to Abstract Mathematics facilitates the transition from introductory mathematics courses to the more abstract work that occurs in advanced courses. This text covers logic, proofs, numbers, sets, induction, functions, and more–material which instructors of upper-level courses often…CREATE YOUR OWN SYNCHRONIZED ROBOT ARMY! PLAN, DESIGN, ASSEMBLE, AND PROGRAM ROBOT SQUADS THAT COMMUNICATE and cooperate with each other to accomplish together what they can't do individually. Build Your Own Teams of Robots with LEGO MINDSTORMS NXT and Bluetooth shows you how to construct a…
Statistics in Action teaches a modern approach to statistics that uses graphing calculators or statistical software and allows readers to place more emphasis on statistical concepts and data analysis than on following recipes for calculations. The text is intended to be used as much as a dialog as…
This book summarizes, and highlights main messages from, a February 2007 Global Forum convened by the World Bank to discuss strategies, programs, and policies for building science, technology and innovation (STI) capacity to promote sustainable growth and poverty reduction in developing countries.
Across the Board is the definitive work on chessboard problems. It is not simply about chess but the chessboard itself--that simple grid of squares so common to games around the world. And, more importantly, the fascinating mathematics behind it. From the Knight's Tour Problem and Queens…
Birdwatchers often come across bird names that include a person's name, either in the vernacular (English) name or latinised in the scientific nomenclature. Such names are properly called eponyms, and few people will not have been curious as to who some of these people were (or are).Names such as…
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Welcome to Free-Academy.com
Please enjoy our ever growing database of free video lectures. It is our goal at Free-Academy.com to eventually build a website where you can learn anything that can be taught. From Introductory Psychology to advanced Differential Equations, and from cooking dinner to changing a tire Free-Academy.com will become the one place on the web for reputable rich media educational materials!
Please enjoy our first online course in Introductory Calculus. All our lectures are supported with a frequently asked question list submitted by our students. Additionally any question is answered in 24 hours in video form!
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We have Prepared over 26.5 HOURS of FREE Calculus lectures for our website! We are in the process of encoding the videos and uploading them to our server. Given the scale of this project it will take us until April 10th to complete. In the mean time we have added the names of all the lectures to be added to our Free Calculus CourseCheck back regularly as we bring these videos online and add the links. And don't forget, if you see anything you would like added email us at [email protected] and we will be sure to include it!
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Lecture: Definition of the Derivative
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Lecture: Rigorous Derivative of Sin x
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Lecture: Definition of the Derivative
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MERLOT Search - category=2543&community=3022
A search of MERLOT materialsCopyright 1997-2014 MERLOT. All rights reserved.Fri, 19 Sep 2014 13:17:28 PDTFri, 19 Sep 2014 13:17:28 PDTMERLOT Search - category=2543&community=3022
4434Cut-the-Knot!
This site is the parent site of an extensive collection of interactive mathematics authored by Alexander Bogomolny and includes an interactive monthly column . The content is accessible to the casual reader but offers much depth along with links to other high-quality resources. Altogether, this site is a mathematician's delight.Graph Theory Lessons
The applets contain topics typically found in undergraduate graph theory and discrete structures classes like null graphs, the handshaking lemma, isomorphism, complete graphs, subgraphs, regular graphs, platonic graphs, adjacency matrices, graph coloring, bipartite graphs, simple circuits, Euler and Hamilton circuits, trees, unions and sums of graphs, complements of graphs, line graphs, spanning trees, plane graphs, shortest paths, minimal spanning trees. The applet utilizes Petersen software written by the author. Peterson can draw, edit and manipulate simple graphs, examine properties of the graphs, and demonstrate them using computer animation.Eric Weisstein's World of MathematicsThe Math Forum - Internet Mathematics Library
A huge list of resources in various topics of mathematics and math education.Discrete Math Resources
This site consists of examples, exercises, games, and other learning activities associated with the textbook, Discrete Mathematics: Mathematical Reasoning and Proof with Puzzles, Patterns and Games by Doug Ensley and Winston Crawley. Requires Adobe Flash player.Graph Theory: Trees
This applet is part of a larger collection of lessons on graph theory. The focus of this particular applet is Trees. The user will explore full binary, ternary and n-ary trees, the relationship of parent and child, and depth-first and breadth-first traversal.Tower of Hanoi
No description givenLP Explorer
LP Explorer is a Java applet which enables the simplex method to be applied to a linear programming (LP) problem and the sensitivity of the solution to changes in the problem data to be examined.An Introduction to Percolation and Many-body Physics by Computer
A nice textual introduction to percolation theory, along with an applet that allows students to investigate the concepts involved experimentally.Euler Circuits, Hamilton Circuits, Directed Graphs
This applet is a component of a larger site on Graph Theory. It introduces Euler Circuits and Hamilton Circuits and provides opportunities to investigate their properties using Peterson software.
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-disc CD-ROM multimedia presentation of the most important principles of the algebra- and trigonometry-based physics course. Content screens provide in-depth coverage of abstract and often difficult principles, building connections between physical concepts and mathematics. The presentation contains more than 350 movies"oboth animated and live video"oincluding laboratory demonstrations, "real world" examples, graphic models, and step-by-step explanations of essential mathematics. An accompanying workbook contains practical physics problems directly related to the presentation, along with worked solutions. This CD can be used with any algebra/trig based introductory physics text.
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Also understand the meaning of the derivative in terms of a rate of change and local linear approximation and they should be able to use derivatives to solve a variety of problems. This course is designed to emphasize the study of the properties and applications of common geometric figures in two and three dimensions. It includes the study of transformations and right triangle trigonometry
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PrefaceBasic Engineering Mathematics 5th Edition intro-duces and then consolidates basic mathematical princi-ples and promotes awareness of mathematical conceptsfor students needing a broad base for further vocationalstudies.In thisfifth edition,newmaterial hasbeen added to manyof the chapters, particularly some of the earlier chap-ters, together with extra practical problems interspersedthroughout the text. The extent of this fifth editionis such that four chapters from the previous editionhave been removed and placed on the easily availablewebsite chapters removed to the website are 'Numbersequences', 'Binary, octal and hexadecimal', 'Inequali-ties' and 'Graphs with logarithmic scales'.The text is relevant to:• 'Mathematics for Engineering Technicians' forBTEC First NQF Level 2 – Chapters 1–12, 16–18,20, 21, 23 and 25–27 are needed for this module.• The mandatory 'Mathematics for Technicians' forBTEC National Certificate and National Diploma inEngineering, NQF Level 3 – Chapters 7–10, 14–17,19, 20–23, 25–27, 31, 32, 34 and 35 are neededand, in addition, Chapters 1–6, 11 and 12 are helpfulrevision for this module.• Basic mathematics for a wide range of introduc-tory/access/foundation mathematics courses.• GCSE revision and for similar mathematics coursesin English-speaking countries worldwide.Basic Engineering Mathematics 5th Edition provides alead into Engineering Mathematics 6th Edition.Each topic considered in the text is presented in away that assumes in the reader little previous know-ledge of that topic. Each chapter begins with a briefoutline of essential theory, definitions, formulae, lawsand procedures; however, these are kept to a minimumas problem solving is extensively used to establish andexemplify the theory. It is intended that readers will gainreal understanding through seeing problems solved andthen solving similar problems themselves.This textbook contains some 750 worked problems,followed by over 1550 further problems (all withanswers at the end of the book) contained within some161 Practice Exercises; each Practice Exercise fol-lows on directly from the relevant section of work. Inaddition, 376 line diagrams enhance understanding ofthe theory. Where at all possible, the problems mirrorpotential practical situations found in engineering andscience.Placed at regular intervals throughout the text are14 Revision Tests (plus another for the websitechapters) to check understanding. For example, Revi-sion Test 1 covers material contained in Chapters 1and 2, Revision Test 2 covers the material containedin Chapters 3–5, and so on. These Revision Tests donot have answers given since it is envisaged that lec-turers/instructors could set the tests for students toattempt as part of their course structure. Lecturers/in-structors may obtain a complimentary set of solu-tions of the Revision Tests in an Instructor's Manual,available from the publishers via the internet – see the end of the book a list of relevant formulae con-tained within the text is included for convenience ofreference.The principle of learning by example is at the heart ofBasic Engineering Mathematics 5th Edition.JOHN BIRDRoyal Naval School of Marine EngineeringHMS Sultan, formerly University of Portsmouthand Highbury College, Portsmouth
AcknowledgementsThe publisher wishes to thank CASIO Electronic Co.Ltd, London for permission to reproduce the image ofthe Casio fx-83ES calculator on page 23.The publishers also wish to thank the AutomobileAsso-ciation for permission to reproduce a map of Portsmouthon page 131.
Instructor's ManualFull worked solutions and mark scheme for all theAssignments are contained in this Manual which isavailable to lecturers only.To download the Instructor's Manual visit
4 Basic Engineering Mathematics(i) 8 × 4 = 32. Place the 2 in the units column andcarry 3 into the tens column.(ii) 8 × 6 = 48;48 + 3 (carried) = 51. Place the 1 inthe tens column and carry the 5 into the hundredscolumn.(iii) 8 × 7 = 56;56 + 5 (carried) = 61. Place 1 in thehundredscolumn and 6 in thethousandscolumn.(iv) Place 0 in the units column under the 2.(v) 3 × 4 = 12. Place the 2 in the tens column andcarry 1 into the hundreds column.(vi) 3 × 6 = 18;18 + 1 (carried) = 19. Place the 9 inthe hundreds column and carry the 1 into thethousands column.(vii) 3 × 7 = 21;21 + 1 (carried) = 22. Place 2 in thethousands column and 2 in the ten thousandscolumn.(viii) 6112+ 22920 = 29032Hence, 764 × 38 = 29032Again, knowing multiplication tables is ratherimportantwhen multiplying such numbers.It is appreciated, of course, that such a multiplicationcan,and probably will,beperformed using a calculator.However, there are times when a calculator may not beavailable and it is then useful to be able to calculate the'long way'.Problem 9. Multiply 178 by −46When the numbers have different signs, the result willbe negative. (With this in mind, the problem can nowbe solved by multiplying 178 by 46). Following theprocedure of Problem 8 gives1 7 8× 4 61 0 6 87 1 2 08 1 8 8Thus, 178 × 46 = 8188 and 178 × (−46) = −8188Problem 10. Determine 1834 ÷ 72627 1834(i) 7 into 18 goes 2, remainder 4. Place the 2 abovethe 8 of 1834 and carry the 4 remainder to thenext digit on the right, making it 43.(ii) 7 into 43 goes 6, remainder 1. Place the 6 abovethe 3 of 1834 and carry the 1 remainder to thenext digit on the right, making it 14.(iii) 7 into 14 goes 2, remainder 0. Place 2 above the4 of 1834.Hence, 1834÷ 7 = 1834/7 =18347= 262.The method shown is called short division.Problem 11. Determine 5796 ÷ 1248312 5796489996363600(i) 12 into 5 won't go. 12 into 57 goes 4; place 4above the 7 of 5796.(ii) 4 × 12 = 48; place the 48 below the 57 of 5796.(iii) 57 − 48 = 9.(iv) Bring down the 9 of 5796 to give 99.(v) 12 into 99 goes 8; place 8 above the 9 of 5796.(vi) 8 × 12 = 96; place 96 below the 99.(vii) 99 − 96 = 3.(viii) Bring down the 6 of 5796 to give 36.(ix) 12 into 36 goes 3 exactly.(x) Place the 3 above the final 6.(xi) Place the 36 below the 36.(xii) 36 − 36 = 0.Hence, 5796 ÷ 12 = 5796/12 =579612= 483.The method shown is called long division.
Basic arithmetic 5Now try the following Practice ExercisePracticeExercise 2 Further problems onmultiplication and division (answers onpage 340)Determine the values of the expressions given inproblems 1 to 9, without using a calculator.1. (a) 78 × 6 (b) 124 × 72. (a) £261 × 7 (b) £462 × 93. (a) 783kg × 11 (b) 73kg × 84. (a) 27mm × 13 (b) 77mm × 125. (a) 448 × 23 (b) 143 × (−31)6. (a) 288m ÷ 6 (b) 979m ÷ 117. (a)18137(b)896168. (a)2142413(b) 15900 ÷ 159. (a)8873711(b) 46858 ÷ 1410. A screw has a mass of 15grams. Calculate,in kilograms, the mass of 1200 such screws(1kg = 1000g).1.4 Highest common factors andlowest common multiplesWhen two or more numbers are multiplied together, theindividualnumbers are called factors. Thus, a factoris anumber which divides into another number exactly. Thehighest common factor (HCF) is the largest numberwhich divides into two or more numbers exactly.For example, consider the numbers 12 and 15.The factors of 12 are 1, 2, 3, 4, 6 and 12 (i.e. all thenumbers that divide into 12).The factors of 15 are 1, 3, 5 and 15 (i.e. all the numbersthat divide into 15).1 and 3 are the only common factors; i.e., numberswhich are factors of both 12 and 15.Hence, the HCF of 12 and 15 is 3 since 3 is the highestnumber which divides into both 12 and 15.A multiple is a number which contains another numberan exact number of times. The smallest number whichis exactly divisible by each of two or more numbers iscalled the lowest common multiple (LCM).For example, the multiples of 12 are 12, 24, 36, 48,60, 72,... and the multiples of 15 are 15, 30, 45,60, 75,...60 is a common multiple (i.e. a multiple of both 12 and15) and there are no lower common multiples.Hence, the LCM of 12 and 15 is 60 since 60 is thelowest number that both 12 and 15 divide into.Here are some further problems involving the determi-nation of HCFs and LCMs.Problem 12. Determine the HCF of the numbers12, 30 and 42Probably the simplest way of determining an HCF is toexpress each number in terms of its lowest factors. Thisis achieved by repeatedly dividing by the prime numbers2, 3, 5, 7, 11, 13, … (where possible) in turn. Thus,12 = 2 × 2 × 330 = 2 × 3 × 542 = 2 × 3 × 7The factors which are common to each of the numbersare 2 in column 1 and 3 in column 3, shown by thebroken lines. Hence, the HCF is 2 × 3; i.e., 6. That is,6 is the largest number which will divide into 12, 30and 42.Problem 13. Determine the HCF of the numbers30, 105, 210 and 1155Using the method shown in Problem 12:30 = 2 × 3 × 5105 = 3 × 5 × 7210 = 2 × 3 × 5 × 71155 = 3 × 5 × 7 × 11The factors which are common to each of the numbersare 3 in column 2 and 5 in column 3. Hence, the HCFis 3 × 5 = 15.Problem 14. Determine the LCM of the numbers12, 42 and 90
Chapter 2Fractions2.1 IntroductionA mark of 9 out of 14 in an examination may be writ-ten as914or 9/14.914is an example of a fraction. Thenumber above the line, i.e. 9, is called the numera-tor. The number below the line, i.e. 14, is called thedenominator.When the value of the numerator is less than thevalue of the denominator, the fraction is called aproper fraction.914is an example of a properfraction.When thevalueofthenumeratorisgreaterthan thevalueof the denominator, the fraction is called an improperfraction.52is an example of an improper fraction.A mixed number is a combination of a whole numberand a fraction. 212is an example of a mixed number. Infact,52= 212.There are a number of everyday examples in whichfractions are readily referred to. For example, threepeople equally sharing a bar of chocolate would have13each. A supermarket advertises15off a six-pack ofbeer; if the beer normally costs £2 then it will nowcost £1.60.34of the employees of a company arewomen; if the company has 48 employees, then 36 arewomen.Calculators are able to handle calculations with frac-tions. However, to understand a little more about frac-tions we will in this chapter show how to add, subtract,multiply and divide with fractions without the use of acalculator.Problem 1. Change the following improperfractions into mixed numbers:(a)92(b)134(c)285(a)92means 9 halves and92= 9 ÷ 2, and 9 ÷ 2 = 4and 1 half, i.e.92= 412(b)134means 13 quarters and134= 13 ÷ 4, and13 ÷ 4 = 3 and 1 quarter, i.e.134= 314(c)285means 28 fifths and285= 28 ÷ 5, and 28 ÷ 5 =5 and 3 fifths, i.e.285= 535Problem 2. Change the following mixed numbersinto improper fractions:(a) 534(b) 179(c) 237(a) 534means 5 +34. 5 contains 5 × 4 = 20 quarters.Thus, 534contains 20 + 3 = 23 quarters, i.e.534=234DOI: 10.1016/B978-1-85617-697-2.00002-8
10 Basic Engineering MathematicsThe quick way to change 534into an improperfraction is4 × 5 + 34=234.(b) 179=9 × 1 + 79=169.(c) 237=7 × 2 + 37=177.Problem 3. In a school there are 180 students ofwhich 72 are girls. Express this as a fraction in itssimplest formThe fraction of girls is72180.Dividing both the numerator and denominator by thelowest prime number, i.e. 2, gives72180=3690Dividing both the numerator and denominator again by2 gives72180=3690=18452 will not divide into both 18 and 45, so dividing both thenumerator and denominator by the next prime number,i.e. 3, gives72180=3690=1845=615Dividing both the numerator and denominator again by3 gives72180=3690=1845=615=25So72180=25in its simplest form.Thus,25of the students are girls.2.2 Adding and subtracting fractionsWhen the denominators of two (or more) fractions tobe added are the same, the fractions can be added 'onsight'.For example,29+59=79and38+18=48.In the latter example, dividing both the 4 and the 8 by4 gives48=12, which is the simplified answer. This iscalled cancelling.Additionand subtraction of fractions is demonstratedin the following worked examples.Problem 4. Simplify13+12(i) Make the denominators the same for each frac-tion. The lowest number that both denominatorsdivideinto is called the lowest commonmultipleor LCM (see Chapter 1, page 5). In this example,the LCM of 3 and 2 is 6.(ii) 3 divides into 6 twice. Multiplying both numera-tor and denominator of13by 2 gives13=26=(iii) 2 dividesinto 6, 3 times. Multiplyingboth numer-ator and denominator of12by 3 gives12=36=(iv) Hence,13+12=26+36=56+ =Problem 5. Simplify34−716(i) Make the denominators the same for each frac-tion. The lowest common multiple (LCM) of 4and 16 is 16.(ii) 4 divides into 16, 4 times. Multiplying bothnumerator and denominator of34by 4 gives34=1216=(iii)716already has a denominator of 16.
Chapter 3Decimals3.1 IntroductionThe decimal system of numbers is based on the digits0 to 9.Thereareanumberofeveryday occurrencesin which weuse decimal numbers. For example, a radio is, say, tunedto 107.5MHz FM; 107.5 is an example of a decimalnumber.In a shop, a pair of trainers cost, say, £57.95; 57.95 isanotherexampleofadecimal number.57.95 isadecimalfraction, where a decimal point separates the integer, i.e.57, from the fractional part, i.e. 0.9557.95 actually means (5 × 10)+ (7 × 1)+ 9 ×110+ 5 ×11003.2 Converting decimals to fractionsand vice-versaConverting decimals to fractions and vice-versa isdemonstrated below with worked examples.Problem 1. Convert 0.375 to a proper fraction inits simplest form(i) 0.375 may be written as0.375 × 10001000i.e.0.375 =3751000(ii) Dividing both numerator and denominator by 5gives3751000=75200(iii) Dividing both numerator and denominator by 5again gives75200=1540(iv) Dividing both numerator and denominator by 5again gives1540=38Since both 3 and 8 are only divisible by 1, we cannot'cancel' any further, so38is the 'simplest form' of thefraction.Hence, the decimal fraction 0.375 =38as a properfraction.Problem 2. Convert 3.4375 to a mixed number(i) 0.4375 may be written as0.4375 × 1000010000i.e.0.4375 =437510000(ii) Dividing both numerator and denominator by 25gives437510000=175400(iii) Dividing both numerator and denominator by 5gives175400=3580(iv) Dividing both numerator and denominator by 5again gives3580=716Since both 5 and 16 are only divisible by 1, wecannot 'cancel' any further, so716is the 'lowestform' of the fraction.(v) Hence, 0.4375 =716Thus, the decimal fraction 3.4375=3716as a mixednumber.DOI: 10.1016/B978-1-85617-697-2.00003-X
Decimals 17Problem 3. Express78as a decimal fractionTo convert a proper fraction to a decimal fraction, thenumerator is divided by the denominator.0.8 7 58 7.0 0 0(i) 8 into 7 will not go. Place the 0 above the 7.(ii) Place the decimal point above the decimal pointof 7.000(iii) 8 into 70 goes 8, remainder 6. Place the 8 abovethe first zero after the decimal point and carry the6 remainder to the next digit on the right, makingit 60.(iv) 8 into 60 goes 7, remainder 4. Place the 7 abovethe next zero and carry the 4 remainder to the nextdigit on the right, making it 40.(v) 8 into 40 goes 5, remainder 0. Place 5 above thenext zero.Hence, the proper fraction78= 0.875 as a decimalfraction.Problem 4. Express 51316as a decimal fractionFor mixed numbers it is only necessary to convert theproper fraction part of the mixed number to a decimalfraction.0.8 1 2 516 13.0 0 0 0(i) 16 into 13 will not go. Place the 0 above the 3.(ii) Place the decimal point above the decimal pointof 13.0000(iii) 16 into 130 goes 8, remainder 2. Place the 8 abovethe first zero after the decimal point and carry the2 remainder to the next digit on the right, makingit 20.(iv) 16 into 20 goes 1, remainder 4. Place the 1 abovethe next zero and carry the 4 remainder to the nextdigit on the right, making it 40.(v) 16 into 40 goes 2, remainder 8. Place the 2 abovethe next zero and carry the 8 remainder to the nextdigit on the right, making it 80.(vi) 16 into 80 goes 5, remainder 0. Place the 5 abovethe next zero.(vii) Hence,1316= 0.8125Thus, the mixed number 51316= 5.8125 as a decimalfraction.Now try the following Practice ExercisePracticeExercise 8 Converting decimals tofractions and vice-versa (answers onpage 341)1. Convert 0.65 to a proper fraction.2. Convert 0.036 to a proper fraction.3. Convert 0.175 to a proper fraction.4. Convert 0.048 to a proper fraction.5. Convert the following to proper fractions.(a) 0.65 (b) 0.84 (c) 0.0125(d) 0.282 (e) 0.0246. Convert 4.525 to a mixed number.7. Convert 23.44 to a mixed number.8. Convert 10.015 to a mixed number.9. Convert 6.4375 to a mixed number.10. Convert the following to mixed numbers.(a) 1.82 (b) 4.275 (c) 14.125(d) 15.35 (e) 16.212511. Express58as a decimal fraction.12. Express 61116as a decimal fraction.13. Express732as a decimal fraction.14. Express 11316as a decimal fraction.15. Express932as a decimal fraction.3.3 Significant figures and decimalplacesA number which can be expressed exactly as adecimal fraction is called a terminating decimal.
18 Basic Engineering MathematicsFor example,3316= 3.1825 is a terminating decimalA number which cannot be expressed exactly as a deci-mal fraction is called a non-terminating decimal. Forexample,157= 1.7142857... is a non-terminating decimalThe answer to a non-terminating decimal may beexpressed in two ways, depending on the accuracyrequired:(a) correct to a number of significant figures, or(b) correct to a number of decimal places i.e. thenumber of figures after the decimal point.The last digit in the answer is unaltered if the next digiton the right is in the group of numbers 0, 1, 2, 3 or 4.For example,1.714285... = 1.714 correct to 4 significant figures= 1.714 correct to 3 decimal placessince the next digit on the right in this example is 2.The last digit in the answer is increased by 1 if the nextdigit on the right is in the group of numbers 5, 6, 7, 8 or9. For example,1.7142857... = 1.7143 correct to 5 significant figures= 1.7143 correct to 4 decimal placessince the next digit on the right in this example is 8.Problem 5. Express 15.36815 correct to(a) 2 decimal places, (b) 3 significant figures,(c) 3 decimal places, (d) 6 significant figures(a) 15.36815 = 15.37 correct to 2 decimal places.(b) 15.36815 = 15.4 correct to 3 significant figures.(c) 15.36815 = 15.368 correct to 3 decimal places.(d) 15.36815 = 15.3682 correct to 6 significantfigures.Problem 6. Express 0.004369 correct to(a) 4 decimal places, (b) 3 significant figures(a) 0.004369 = 0.0044 correct to 4 decimal places.(b) 0.004369 = 0.00437 correct to 3 significantfigures.Note that the zeros to the right of the decimal point donot count as significant figures.Now try the following Practice ExercisePracticeExercise 9 Significant figures anddecimal places (answers on page 341)1. Express 14.1794 correct to 2 decimal places.2. Express 2.7846 correct to 4 significant figures.3. Express 65.3792 correct to 2 decimal places.4. Express 43.2746 correct to 4 significantfigures.5. Express 1.2973 correct to 3 decimal places.6. Express 0.0005279 correct to 3 significantfigures.3.4 Adding and subtracting decimalnumbersWhen adding or subtracting decimal numbers, careneeds to be taken to ensure that the decimal pointsare beneath each other. This is demonstrated in thefollowing worked examples.Problem 7. Evaluate 46.8 + 3.06 + 2.4 + 0.09and give the answer correct to 3 significant figuresThe decimal points are placed under each other asshown. Each column is added, starting from theright.46.83.062.4+0.0952.3511 1(i) 6 + 9 = 15. Place 5 in the hundredths column.Carry 1 in the tenths column.(ii) 8 + 0 + 4 + 0 + 1 (carried) = 13. Place the 3 inthe tenths column. Carry the 1 into the unitscolumn.(iii) 6 + 3 + 2 + 0 + 1 (carried) = 12. Place the 2 intheunitscolumn.Carry the1 into thetenscolumn.
Chapter 4Using a calculator4.1 IntroductionIn engineering, calculations often need to be performed.For simple numbers it is useful to be able to use men-tal arithmetic. However, when numbers are larger anelectronic calculator needs to be used.There are several calculators on the market, many ofwhich will be satisfactory for our needs. It is essentialto have a scientific notationcalculator which will haveall the necessary functions needed and more.This chapter assumes you have a CASIO fx-83EScalculator, or similar, as shown in Figure 4.1.Besides straightforward addition, subtraction, multipli-cation and division, which you will already be ableto do, we will check that you can use squares, cubes,powers, reciprocals, roots, fractions and trigonomet-ric functions (the latter in preparation for Chapter 21).There are several other functions on the calculatorwhich we do not need to concern ourselves with at thislevel.4.2 Adding, subtracting, multiplyingand dividingInitially, after switching on, press Mode.Of the three possibilities, use Comp, which is achievedby pressing 1.Next, press Shift followed by Setup and, of the eightpossibilities, use Mth IO, which is achieved by press-ing 1.By all means experiment with the other menu options –refer to your 'User's guide'.All calculators have +, −, × and ÷ functions andthese functions will, no doubt, already have been usedin calculations.Problem 1. Evaluate 364.7 ÷ 57.5 correct to 3decimal places(i) Type in 364.7(ii) Press ÷.(iii) Type in 57.5(iv) Press = and the fraction3647575appears.(v) Press the S ⇔ D functionand the decimal answer6.34260869... appears.Alternatively, after step (iii) press Shift and = and thedecimal will appear.Hence, 364.7 ÷ 57.5 = 6.343 correct to 3 decimalplaces.Problem 2. Evaluate12.47 × 31.5970.45 × 0.052correct to 4significant figures(i) Type in 12.47(ii) Press ×.(iii) Type in 31.59(iv) Press ÷.(v) The denominator must have brackets; i.e. press (.(vi) Type in 70.45 × 0.052 and complete the bracket;i.e. ).(vii) Press = and the answer 107.530518... appears.Hence,12.47 × 31.5970.45 × 0.052= 107.5 correct to 4 significantfigures.DOI: 10.1016/B978-1-85617-697-2.00004-1
26 Basic Engineering Mathematics3. Evaluate√34528correct to 2 decimal places.4. Evaluate√0.69 correct to 4 significantfigures.5. Evaluate√0.025 correct to 4 decimal places.6. Evaluate 3√17 correct to 3 decimal places.7. Evaluate 4√773 correct to 4 significantfigures.8. Evaluate 5√3.12 correct to 4 decimal places.9. Evaluate 3√0.028 correct to 5 significantfigures.10. Evaluate6√2451−4√46 correct to 3 decimalplaces.Express the answers to questions 11 to 15 inengineering form.11. Evaluate 5 × 10−3 × 7 × 10812. Evaluate3 × 10−48 × 10−913. Evaluate6 × 103 × 14× 10−42 × 10614. Evaluate56.43 × 10−3 × 3 × 1048.349 × 103correct to3 decimal places.15. Evaluate99 × 105 × 6.7 × 10−336.2 × 10−4correct to 4significant figures.4.3.4 FractionsLocate the and functions on your calculator(the latter function is a Shift function found abovethe function) and then check the following workedexamples.Problem 14. Evaluate14+23(i) Press the function.(ii) Type in 1(iii) Press ↓ on the cursor key and type in 4(iv)14appears on the screen.(v) Press → on the cursor key and type in +(vi) Press the function.(vii) Type in 2(viii) Press ↓ on the cursor key and type in 3(ix) Press → on the cursor key.(x) Press = and the answer1112appears.(xi) Press the S ⇔ D function and the fractionchanges to a decimal 0.9166666...Thus,14+23=1112= 0.9167 as a decimal, correct to4 decimal places.It is also possible to deal with mixed numbers on thecalculator. Press Shift then the function andappears.Problem 15. Evaluate 515− 334(i) Press Shift then the function and appearson the screen.(ii) Type in 5 then → on the cursor key.(iii) Type in 1 and ↓ on the cursor key.(iv) Type in 5 and 515appears on the screen.(v) Press → on the cursor key.(vi) Typein – and then pressShift then the functionand 515− appears on the screen.(vii) Type in 3 then → on the cursor key.(viii) Type in 3 and ↓ on the cursor key.(ix) Type in 4 and 515− 334appears on the screen.(x) Press = and the answer2920appears.(xi) Press S ⇔ D function and the fraction changes toa decimal 1.45Thus, 515− 334=2920= 1920= 1.45 as a decimal.
Using a calculator 27Now try the following Practice ExercisePracticeExercise 16 Fractions (answers onpage 341)1. Evaluate45−13as a decimal, correct to4 decimal places.2. Evaluate23−16+37as a fraction.3. Evaluate 256+ 158as a decimal, correct to4 significant figures.4. Evaluate 567− 318as a decimal, correct to4 significant figures.5. Evaluate13−34×821as a fraction.6. Evaluate38+56−12as a decimal, correct to4 decimal places.7. Evaluate34×45−23÷49as a fraction.8. Evaluate 889÷ 223as a mixed number.9. Evaluate 315× 113− 1710as a decimal, cor-rect to 3 decimal places.10. Evaluate415− 123314× 235−29as a decimal, cor-rect to 3 significant figures.4.3.5 Trigonometric functionsTrigonometric ratios will be covered in Chapter 21.However, very briefly, there are three functions on yourcalculator that are involved with trigonometry. They are:sin which is an abbreviation of sinecos which is an abbreviation of cosine, andtan which is an abbreviation of tangentExactly what these mean will be explained inChapter 21.There are two main ways that angles are measured, i.e.in degrees or in radians. Pressing Shift, Setup and 3shows degrees, and Shift, Setup and 4 shows radians.Press 3 and your calculator will be in degrees mode,indicated by a small D appearing at the top of the screen.Press 4 and your calculator will be in radian mode,indicated by a small R appearing at the top of thescreen.Locate the sin, cos and tan functions on your calculatorand then check the following worked examples.Problem 16. Evaluate sin38◦(i) Make sure your calculator is in degrees mode.(ii) Press sin function and sin( appears on thescreen.(iii) Type in 38 and close the bracket with) and sin (38)appears on the screen.(iv) Press = and the answer 0.615661475... appears.Thus, sin38◦ = 0.6157, correct to 4 decimal places.Problem 17. Evaluate 5.3 tan (2.23 rad)(i) Make sure your calculator is in radian mode bypressing Shift then Setup then 4 (a small R appearsat the top of the screen).(ii) Type in 5.3 then press tan function and 5.3 tan(appears on the screen.(iii) Type in 2.23 and close the bracket with) and5.3 tan (2.23) appears on the screen.(iv) Press = and the answer −6.84021262... appears.Thus, 5.3tan(2.23rad) = −6.8402, correct to 4 dec-imal places.Now try the following Practice ExercisePracticeExercise 17 Trigonometricfunctions (answers on page 341)Evaluate the following, each correct to 4 decimalplaces.1. Evaluate sin67◦2. Evaluate cos43◦3. Evaluate tan71◦4. Evaluate sin15.78◦5. Evaluate cos63.74◦6. Evaluate tan39.55◦ − sin52.53◦7. Evaluate sin(0.437 rad)
28 Basic Engineering Mathematics8. Evaluate cos(1.42 rad)9. Evaluate tan(5.673 rad)10. Evaluate(sin42.6◦)(tan 83.2◦)cos13.8◦4.3.6 π and e x functionsPress Shift and then press the ×10x function key and πappears on the screen. Either press Shift and = (or =and S ⇔ D) and the value of π appears in decimal formas 3.14159265...Press Shift and then press the ln function key and eappears on the screen. Enter 1 and then press = ande1 = e = 2.71828182...Now check the following worked examples involvingπand ex functions.Problem 18. Evaluate 3.57π(i) Enter 3.57(ii) Press Shift and the ×10x key and 3.57π appearson the screen.(iii) Either press Shift and = (or = and S ⇔ D)and the value of 3.57π appears in decimal as11.2154857...Hence, 3.57 π = 11.22 correct to 4 significantfigures.Problem 19. Evaluate e2.37(i) Press Shift and then press the ln function key ande appears on the screen.(ii) Enter 2.37 and e2.37 appears on the screen.(iii) Press Shiftand = (or= and S ⇔ D) and the valueof e2.37 appears in decimal as 10.6973922...Hence, e 2.37 = 10.70 correct to 4 significant figures.Now try the following Practice ExercisePracticeExercise 18 π and ex functions(answers on page 341)Evaluatethefollowing,each correct to 4 significantfigures.1. 1.59π 2. 2.7(π− 1)3. π2√13 − 1 4. 3eπ5. 8.5e−2.5 6. 3e2.9 − 1.67. 3e(2π−1) 8. 2πeπ39.5.52π2e−2 ×√26.7310.⎡⎣ e2−√3π ×√8.57⎤⎦4.4 Evaluation of formulaeThe statement y = mx + c is called a formula for y interms of m, x and c.y, m, x and c are called symbols.When given values of m, x and c we can evaluate y.There are a large number of formulae used in engineer-ing and in this section we will insert numbers in placeof symbols to evaluate engineering quantities.Just four examples of important formulae are:1. A straight line graph is of the form y = mx + c (seeChapter 17).2. Ohm's law states that V = I × R.3. Velocity is expressed as v = u + at.4. Force is expressed as F = m × a.Here are some practical examples. Check with yourcalculator that you agree with the working and answers.Problem 20. In an electrical circuit the voltage Vis given by Ohm's law, i.e. V = IR. Find, correct to4 significant figures, the voltage when I = 5.36Aand R = 14.76V = IR = (5.36)(14.76)Hence, voltage V = 79.11V, correct to 4 significantfigures.Problem 21. The surface area A of a hollow coneis given by A = πrl. Determine, correct to 1decimal place, the surface area when r = 3.0cmand l = 8.5cmA = πrl = π(3.0)(8.5)cm2Hence, surface area A = 80.1cm2, correct to 1 deci-mal place.
Using a calculator 29Problem 22. Velocity v is given by v = u + at. Ifu = 9.54m/s, a = 3.67m/s2 and t = 7.82s, find v,correct to 3 significant figuresv = u + at = 9.54 + 3.67 × 7.82= 9.54 + 28.6994 = 38.2394Hence, velocity v = 38.2m/s, correct to 3 significantfigures.Problem 23. The area, A, of a circle is given byA = πr2. Determine the area correct to 2 decimalplaces, given radius r = 5.23mA = πr2= π(5.23)2= π(27.3529)Hence, area, A = 85.93m2, correct to 2 decimalplaces.Problem 24. Density =massvolume. Find the densitywhen the mass is 6.45kg and the volume is300 × 10−6 cm3Density =massvolume=6.45kg300 × 10−6 m3= 21500kg/m3Problem 25. The power, P watts, dissipated in anelectrical circuit is given by the formula P =V 2R.Evaluate the power, correct to 4 significant figures,given that V = 230 V and R = 35.63P =V2R=(230)235.63=5290035.63= 1484.70390...Press ENG and 1.48470390...× 103 appears on thescreen.Hence, power, P = 1485W or 1.485kW correct to 4significant figures.Now try the following Practice ExercisePracticeExercise 19 Evaluation offormulae (answers on page 341)1. The area A of a rectangle is given by theformula A = lb. Evaluate the area whenl = 12.4cm and b = 5.37cm.2. The circumference C of a circle is given bytheformula C = 2πr. Determine the circum-ference given r = 8.40mm.3. A formula used in connection with gasesis R =PVT. Evaluate R when P = 1500,V = 5 and T = 200.4. Thevelocity ofabody isgiven byv = u + at.The initial velocity u is measured when timet is 15 seconds and found to be 12m/s. If theacceleration a is 9.81m/s2calculate the finalvelocity v.5. Calculate the current I in an electrical circuit,where I = V/R amperes when the voltage Vis measured and found to be 7.2 V and theresistance R is 17.7 .6. Find the distance s, given that s =12gt2when time t = 0.032seconds and accelera-tion due to gravity g = 9.81m/s2. Give theanswer in millimetres.7. The energy stored in a capacitor is givenby E =12CV 2 joules. Determine the energywhen capacitance C = 5 × 10−6 farads andvoltage V = 240V.8. Find the area A of a triangle, given A =12bh,when the base length l is 23.42m and theheight h is 53.7m.9. Resistance R2 is given by R2 = R1(1 + αt).Find R2, correct to 4 significant figures, whenR1 = 220, α = 0.00027 and t = 75.610. Density =massvolume. Find the density whenthe mass is 2.462kg and the volume is173cm3. Give the answer in units of kg/m3.11. Velocity = frequency × wavelength. Findthe velocity when the frequency is 1825Hzand the wavelength is 0.154m.12. Evaluate resistance RT , given1RT=1R1+1R2+1R3when R1 = 5.5 ,R2 = 7.42 and R3 = 12.6 .Here are some further practical examples. Again, checkwith your calculator that you agree with the workingand answers.
Using a calculator 317. Energy, E joules, is given by the formulaE =12L I2. Evaluate the energy whenL = 5.5 and I = 1.28. The current I amperes in an a.c. circuitis given by I =V(R2 + X2). Evaluate thecurrent when V =250, R=11.0 and X=16.29. Distance s metres is given by the formulas = ut +12at2. If u = 9.50, t = 4.60 anda = −2.50, evaluate the distance.10. The area, A, of any triangle is givenby A =√[s(s − a)(s − b)(s − c)] wheres =a + b + c2. Evaluate the area, givena = 3.60cm, b = 4.00cm and c = 5.20cm.11. Given that a = 0.290, b = 14.86, c = 0.042,d = 31.8 and e = 0.650, evaluate v given thatv =abc−de12. Deduce the following information from thetrain timetable shown in Table 4.1.(a) At what time should a man catch a trainat Fratton to enable him to be in LondonWaterloo by 14.23h?(b) A girlleaves Cosham at 12.39h and trav-els to Woking. How long does the jour-ney take? And, if the distance betweenCosham and Woking is 55miles, calcu-late the average speed of the train.(c) A man living at Havant has a meetingin London at 15.30h. It takes around25minutes on the underground to reachhis destination from London Waterloo.What train should he catch from Havantto comfortably make the meeting?(d) Nine trains leave Portsmouth harbourbetween 12.18h and 13.15h. Whichtrain should be taken for the shortestjourney time?
Chapter 5Percentages5.1 IntroductionPercentages are used to give a common standard. Theuse of percentages is very common in many aspectsof commercial life, as well as in engineering. Interestrates, sale reductions, pay rises, exams and VAT are allexamples of situations in which percentages are used.For this chapter you will need to know about decimalsand fractions and be able to use a calculator.We are familiar with the symbol for percentage, i.e. %.Here are some examples.• Interest rates indicate the cost at which we can bor-row money. If you borrow £8000 at a 6.5% interestrate for a year, it will cost you 6.5% of the amountborrowed to do so,which will need to berepaid alongwith the original money you borrowed. If you repaythe loan in 1 year, how much interest will you havepaid?• A pair of trainers in a shop cost £60. They are adver-tised in a sale as 20% off. How much will youpay?• If you earn £20000 p.a. and you receive a 2.5%pay rise, how much extra will you have to spend thefollowing year?• A book costing £18 can be purchased on the internetfor 30% less. What will be its cost?When we have completed his chapter on percentagesyou will be able to understand how to perform the abovecalculations.Percentages are fractions having 100 as their denom-inator. For example, the fraction40100is written as 40%and is read as 'forty per cent'.The easiest way to understand percentages is to gothrough some worked examples.5.2 Percentage calculations5.2.1 To convert a decimal to a percentageA decimal number is converted to a percentage bymultiplying by 100.Problem 1. Express 0.015 as a percentageTo express a decimal number as a percentage, merelymultiply by 100, i.e.0.015 = 0.015 × 100%= 1.5%Multiplying a decimal number by 100 means movingthe decimal point 2 places to the right.Problem 2. Express 0.275 as a percentage0.275 = 0.275 × 100%= 27.5%5.2.2 To convert a percentage to a decimalA percentage is converted to a decimal number bydividing by 100.Problem 3. Express 6.5% as a decimal number6.5% =6.5100= 0.065Dividing by 100 means moving the decimal point 2places to the left.DOI: 10.1016/B978-1-85617-697-2.00005-3
34 Basic Engineering MathematicsProblem 4. Express 17.5% as a decimal number17.5% =17.5100= 0.1755.2.3 To convert a fraction to a percentageA fraction is converted to a percentage by multiplyingby 100.Problem 5. Express58as a percentage58=58× 100% =5008%= 62.5%Problem 6. Express519as a percentage, correctto 2 decimal places519=519× 100%=50019%= 26.3157889...by calculator= 26.32% correct to 2 decimal placesProblem 7. In two successive tests a studentgains marks of 57/79 and 49/67. Is the secondmark better or worse than the first?57/79 =5779=5779× 100% =570079%= 72.15% correct to 2 decimal places49/67 =4967=4967× 100% =490067%= 73.13% correct to 2 decimal placesHence, the second test is marginally better than thefirst test. This question demonstrates how much easierit is to compare two fractions when they are expressedas percentages.5.2.4 To convert a percentage to a fractionA percentage is converted to a fraction by dividing by100 and then, by cancelling, reducing it to its simplestform.Problem 8. Express 75% as a fraction75% =75100=34The fraction75100is reduced to its simplest form by can-celling, i.e. dividing both numerator and denominatorby 25.Problem 9. Express 37.5% as a fraction37.5% =37.5100=3751000by multiplying both numerator anddenominator by 10=1540by dividing both numerator anddenominator by 25=38by dividing both numerator anddenominator by 5Now try the following Practice ExercisePracticeExercise 21 Percentages (answerson page 342)In problems 1 to 5, express the given numbers aspercentages.1. 0.0032 2. 1.7343. 0.057 4. 0.3745. 1.2856. Express 20% as a decimal number.7. Express 1.25% as a decimal number.8. Express1116as a percentage.9. Express513as a percentage, correct to 3decimal places.
Percentages 3510. Express as percentages, correct to 3 significantfigures,(a)733(b)1924(c) 1111611. Place the following in order of size, the small-est first, expressing each as a percentagecorrect to 1 decimal place.(a)1221(b)917(c)59(d)61112. Express 65% as a fraction in its simplest form.13. Express 31.25% as a fraction in its simplestform.14. Express 56.25% as a fraction in its simplestform.15. Evaluate A to J in the following table.Decimal number Fraction Percentage0.5 A BC14DE F 30G35HI J 855.3 Further percentage calculations5.3.1 Finding a percentage of a quantityTo find apercentageofaquantity,convert thepercentageto a fraction (by dividing by 100) and remember that 'of'means multiply.Problem 10. Find 27% of £6527% of £65 =27100× 65= £17.55 by calculatorProblem 11. In a machine shop, it takes 32minutes to machine a certain part. Using a new tool,the time can be reduced by 12.5%. Calculate thenew time taken12.5% of 32 minutes =12.5100× 32= 4 minutesHence, new time taken = 32 − 4 = 28 minutes.Alternatively, if the time is reduced by 12.5%, it nowtakes 100% − 12.5% = 87.5% of the original time, i.e.87.5% of 32 minutes =87.5100× 32= 28 minutesProblem 12. A 160 GB iPod is advertised ascosting £190 excluding VAT. If VAT is added at17.5%, what will be the total cost of the iPod?VAT = 17.5% of £190 =17.5100× 190 = £33.25Total cost of iPod = £190 + £33.25 = £223.25A quicker method to determine the total cost is: 1.175 ×£190 = £223.255.3.2 Expressing one quantity as apercentage of another quantityTo express one quantity as a percentage of another quan-tity, divide the first quantity by the second then multiplyby 100.Problem 13. Express 23cm as a percentage of72cm, correct to the nearest 1%23cm as a percentage of 72cm =2372× 100%= 31.94444...%= 32% correct tothe nearest 1%Problem 14. Express 47 minutes as a percentageof 2 hours, correct to 1 decimal placeNote that it is essential that the two quantities are in thesame units.Working in minute units, 2 hours = 2 × 60= 120 minutes47 minutes as a percentageof 120 min =47120× 100%= 39.2% correct to1 decimal place
36 Basic Engineering Mathematics5.3.3 Percentage changePercentage change is given bynew value−original valueoriginal value× 100%.Problem 15. A box of resistors increases in pricefrom £45 to £52. Calculate the percentage changein cost, correct to 3 significant figures% change =new value − original valueoriginal value× 100%=52 − 4545× 100% =745× 100= 15.6% = percentage change in costProblem 16. A drilling speed should be set to400rev/min. The nearest speed available on themachine is 412rev/min. Calculate the percentageoverspeed% overspeed =available speed−correct speedcorrect speed×100%=412 − 400400×100% =12400×100%= 3%Now try the following Practice ExercisePracticeExercise 22 Further percentages(answers on page 342)1. Calculate 43.6% of 50kg.2. Determine 36% of 27m.3. Calculate, correct to 4 significant figures,(a) 18% of 2758 tonnes(b) 47% of 18.42 grams(c) 147% of 14.1 seconds.4. When 1600 bolts are manufactured, 36 areunsatisfactory. Determinethepercentagethatis unsatisfactory.5. Express(a) 140kg as a percentage of 1t.(b) 47s as a percentage of 5min.(c) 13.4cm as a percentage of 2.5m.6. A block of Monel alloy consists of 70%nickel and 30% copper. If it contains 88.2gof nickel,determine the mass of copper in theblock.7. An athlete runs 5000m in 15 minutes 20seconds. With intense training, he is able toreduce this time by 2.5%. Calculate his newtime.8. A copper alloy comprises 89% copper, 1.5%iron and the remainder aluminium. Find theamount of aluminium, in grams, in a 0.8kgmass of the alloy.9. A computer is advertised on the internet at£520, exclusive of VAT. If VAT is payable at17.5%, what is the total cost ofthe computer?10. Express 325mm as a percentage of 867mm,correct to 2 decimal places.11. A child sleeps on average 9 hours 25 minutesper day. Express this as a percentage of thewhole day, correct to 1 decimal place.12. Express 408g as a percentage of 2.40kg.13. When signing a new contract, a Premiershipfootballer's pay increases from £15500 to£21500 per week. Calculate the percentagepay increase, correct to 3 significant figures.14. A metal rod 1.80m long is heated and itslength expands by 48.6mm. Calculate thepercentage increase in length.15. 12.5% of a length of wood is 70cm. What isthe full length?16. A metal rod, 1.20m long, is heated andits length expands by 42mm. Calculate thepercentage increase in length.5.4 More percentage calculations5.4.1 Percentage errorPercentage error =errorcorrect value× 100%Problem 17. The length of a component ismeasured incorrectly as 64.5mm. The actual length
Percentages 37is 63mm. What is the percentage error in themeasurement?% error =errorcorrect value× 100%=64.5 − 6363× 100%=1.563× 100% =15063%= 2.38%The percentage measurement error is 2.38% too high,which is sometimes written as + 2.38% error.Problem 18. The voltage across a component inan electrical circuit is calculated as 50V usingOhm's law. When measured, the actual voltage is50.4V. Calculate, correct to 2 decimal places, thepercentage error in the calculation% error =errorcorrect value× 100%=50.4 − 5050.4× 100%=0.450.4× 100% =4050.4%= 0.79%The percentage error in the calculation is 0.79% toolow, which is sometimes written as −0.79% error.5.4.2 Original valueOriginal value =new value100 ± % change× 100%Problem 19. A man pays £149.50 in a sale for aDVD player which is labelled '35% off'. What wasthe original price of the DVD player?In this case, it is a 35% reduction in price, so weusenew value100 − % change× 100, i.e. a minus sign in thedenominator.Original price =new value100 − % change× 100=149.5100 − 35× 100=149.565× 100 =1495065= £230Problem 20. A couple buys a flat and make an18% profit by selling it 3 years later for £153400.Calculate the original cost of the houseIn this case, it is an 18% increase in price, so weusenew value100 + % change× 100, i.e. a plus sign in thedenominator.Original cost =new value100 + % change× 100=153400100 + 18× 100=153400118× 100 =15340000118= £130 000Problem 21. An electrical store makes 40% profiton each widescreen television it sells. If the sellingprice of a 32 inch HD television is £630, what wasthe cost to the dealer?In this case, it is a 40% mark-up in price, so weusenew value100 + % change× 100, i.e. a plus sign in thedenominator.Dealer cost =new value100 + % change× 100=630100 + 40× 100=630140× 100 =63000140= £450The dealer buys from the manufacturer for £450 andsells to his customers for £630.5.4.3 Percentage increase/decrease andinterestNew value =100 + % increase100× original valueProblem 22. £3600 is placed in an ISA accountwhich pays 6.25% interest per annum. How much isthe investment worth after 1 year?
38 Basic Engineering MathematicsValue after 1 year =100 + 6.25100× £3600=106.25100× £3600= 1.0625 × £3600= £3825Problem 23. The price of a fully installedcombination condensing boileris increased by 6.5%.It originally cost £2400. What is the new price?New price =100 + 6.5100× £2,400=106.5100× £2,400 = 1.065 × £2,400= £2,556Now try the following Practice ExercisePracticeExercise 23 Further percentages(answers on page 342)1. A machine part has a length of 36mm. Thelength is incorrectly measured as 36.9mm.Determine the percentage error in the mea-surement.2. When a resistor is removed from an electri-cal circuit the current flowing increases from450μA to 531μA. Determine the percentageincrease in the current.3. In a shoe shop sale, everything is advertisedas '40% off'. If a lady pays £186 for a pair ofJimmy Choo shoes, what was their originalprice?4. Over a four year period a family homeincreases in value by 22.5% to £214375.What was the value of the house 4 years ago?5. An electrical retailer makes a 35% profit onall its products. What price does the retailerpay for a dishwasher which is sold for £351?6. The cost of a sports car is £23500 inclusiveof VAT at 17.5%. What is the cost of the carwithout the VAT added?7. £8000 is invested in bonds at a building soci-ety which is offering a rate of 6.75% perannum. Calculate the value of the investmentafter 2 years.8. An electrical contractor earning £36000 perannum receives a pay rise of 2.5%. Hepays 22% of his income as tax and 11%on National Insurance contributions. Calcu-late the increase he will actually receive permonth.9. Five mates enjoy a meal out. With drinks, thetotal bill comes to £176. They add a 12.5%tip and divide the amount equally betweenthem. How much does each pay?10. In December a shop raises the cost of a 40inch LCD TV costing £920 by 5%. It doesnot sell and in its January sale it reducesthe TV by 5%. What is the sale price ofthe TV?11. A man buys a business and makes a 20%profit when he sells it three years later for£222000. What did he pay originally for thebusiness?12. A drilling machine should be set to250rev/min. The nearest speed available onthe machine is 268rev/min. Calculate thepercentage overspeed.13. Two kilograms of a compound contain 30%of element A, 45% of element B and 25% ofelement C. Determine the masses of the threeelements present.14. A concrete mixture contains seven parts byvolume of ballast, four parts by volume ofsand and two parts by volume of cement.Determine the percentage of each of thesethree constituents correct to the nearest 1%and the mass of cement in a two tonne drymix, correct to 1 significant figure.15. In a sample of iron ore, 18% is iron. Howmuch ore is needed to produce 3600kg ofiron?16. A screw's dimension is 12.5 ± 8%mm. Cal-culate the maximum and minimum possiblelength of the screw.17. The output power of an engine is 450kW. Iftheefficiency oftheengineis75%,determinethe power input.
Chapter 6Ratio and proportion6.1 IntroductionRatio is a way of comparing amounts of something; itshows how much bigger one thing is than the other.Some practical examples include mixing paint, sandand cement, or screen wash. Gears, map scales, foodrecipes, scale drawings and metal alloy constituents alluse ratios.Two quantities are in direct proportion when theyincrease or decrease in the same ratio. There are sev-eral practical engineering laws which rely on directproportion. Also, calculating currency exchange ratesand converting imperial to metric units rely on directproportion.Sometimes, as one quantity increases at a particularrate, another quantity decreases at the same rate; this iscalled inverse proportion. For example, the time takento do a job is inversely proportional to the number ofpeople in a team: double the people, half the time.When we have completed this chapter on ratio andproportion you will be able to understand, and confi-dently perform, calculations on the above topics.For this chapter you will need to know about decimalsand fractions and to be able to use a calculator.6.2 RatiosRatios are generally shown as numbers separated by acolon (:) so the ratio of 2 and 7 is written as 2:7 and weread it as a ratio of 'two to seven.'Some practical examples which are familiar include:• Mixing 1 measure of screen wash to 6 measures ofwater; i.e., the ratio of screen wash to water is 1:6• Mixing 1 shovel of cement to 4 shovels of sand; i.e.,the ratio of cement to sand is 1:4• Mixing 3 parts of red paint to 1 part white, i.e., theratio of red to white paint is 3:1Ratio is the number of parts to a mix. The paint mix is4 parts total, with 3 parts red and 1 part white. 3 partsred paint to 1 part white paint means there is34red paint to14white paintHere are some worked examples to help us understandmore about ratios.Problem 1. In a class, the ratio of female to malestudents is 6:27. Reduce the ratio to its simplestform(i) Both 6 and 27 can be divided by 3.(ii) Thus, 6:27 is the same as 2:9.6:27 and 2:9 are called equivalent ratios.It is normal to express ratios in their lowest, or simplest,form. In this example, the simplest form is 2:9 whichmeans for every 2 females in the class there are 9 malestudents.Problem 2. A gear wheel having 128 teeth is inmesh with a 48-tooth gear. What is the gear ratio?Gear ratio = 128:48A ratio can be simplified by finding common factors.(i) 128 and 48 can both be divided by 2, i.e. 128:48is the same as 64:24(ii) 64 and 24 can both be divided by 8, i.e. 64:24 isthe same as 8:3(iii) There is no number that divides completely intoboth 8 and 3 so 8:3 is the simplest ratio, i.e. thegear ratio is 8:3DOI: 10.1016/B978-1-85617-697-2.00006-5
Ratio and proportion 41Thus, 128:48 is equivalent to 64:24 which is equivalentto 8:3 and 8:3 is the simplest form.Problem 3. A wooden pole is 2.08m long. Divideit in the ratio of 7 to 19(i) Since the ratio is 7:19, the total number of partsis 7 + 19 = 26 parts.(ii) 26 parts corresponds to 2.08m = 208cm, hence,1 part corresponds to20826= 8.(iii) Thus, 7 parts corresponds to 7 × 8 = 56 cm and19 parts corresponds to 19 × 8 = 152 cm.Hence, 2.08m divides in the ratio of 7:19 as 56 cmto 152 cm.(Check: 56 + 152must add up to 208, otherwise an errorwould have been made.)Problem 4. In a competition, prize money of£828 is to be shared among the first three in theratio 5:3:1(i) Since the ratio is 5:3:1 the total number of partsis 5 + 3 + 1 = 9 parts.(ii) 9 parts corresponds to £828.(iii) 1 part corresponds to8289= £92, 3 parts cor-responds to 3 × £92 = £276 and 5 parts corre-sponds to 5 × £92 = £460.Hence, £828 divides in the ratio of 5:3:1 as £460 to£276 to £92. (Check: 460 + 276 + 92 must add up to828, otherwise an error would have been made.)Problem 5. A map scale is 1:30000. On the mapthe distance between two schools is 6cm.Determine the actual distance between the schools,giving the answer in kilometresActual distance between schools= 6 × 30000 cm = 180000 cm=180,000100m = 1800 m=18001000m = 1.80 km(1mile ≈ 1.6km, hence the schools are just over 1mileapart.)Now try the following Practice ExercisePracticeExercise 24 Ratios (answers onpage 342)1. In a box of 333 paper clips, 9 are defective.Express the number of non-defective paperclipsasaratio ofthenumberofdefectivepaperclips, in its simplest form.2. A gear wheel having 84 teeth is in mesh witha 24-tooth gear. Determine the gear ratio in itssimplest form.3. In a box of 2000 nails, 120 are defective.Express the number of non-defective nails asa ratio of the number of defective ones, in itssimplest form.4. A metal pipe 3.36m long is to be cut into twoin the ratio 6 to 15. Calculate the length of eachpiece.5. The instructions for cooking a turkey say thatit needs to be cooked 45 minutes for everykilogram. How long will it take to cook a 7kgturkey?6. In a will, £6440 is to be divided among threebeneficiaries in the ratio 4:2:1. Calculate theamount each receives.7. A local map has a scale of 1:22500. The dis-tance between two motorways is 2.7km. Howfar are they apart on the map?8. Prize money in a lottery totals £3801 and isshared among three winners in the ratio 4:2:1.How much does the first prize winner receive?Here are some further worked examples on ratios.Problem 6. Express 45p as a ratio of £7.65 in itssimplest form(i) Changing both quantities to the same units, i.e. topence, gives a ratio of 45:765(ii) Dividing both quantities by 5 gives45:765 ≡ 9:153(iii) Dividing both quantities by 3 gives9:153 ≡ 3:51(iv) Dividing both quantities by 3 again gives3:51 ≡ 1:17
42 Basic Engineering MathematicsThus, 45 p as a ratio of £7.65 is 1:1745:765,9:153,3:51 and 1:17 are equivalent ratiosand 1:17 is the simplest ratio.Problem 7. A glass contains 30 ml of whiskywhich is 40% alcohol. If 45ml of water is added andthe mixture stirred, what is now the alcohol content?(i) The 30 ml of whisky contains 40%alcohol =40100× 30 = 12 ml.(ii) After 45ml of water is added we have 30 + 45= 75ml of fluid, of which alcohol is 12 ml.(iii) Fraction of alcohol present =1275(iv) Percentage of alcohol present =1275× 100%= 16%.Problem 8. 20 tonnes of a mixture of sand andgravel is 30% sand. How many tonnes of sand mustbe added to produce a mixture which is 40% gravel?(i) Amount of sand in 20 tonnes = 30% of 20 t=30100× 20 = 6t.(ii) If the mixture has 6t of sand then amount ofgravel = 20 − 6 = 14t.(iii) We want this 14t of gravel to be 40% of thenew mixture. 1% would be1440t and 100% of themixture would be1440× 100 t = 35t.(iv) If there is 14t of gravel then amount of sand= 35 − 14 = 21t.(v) We already have 6t of sand, so amount of sandto be added to produce a mixture with 40%gravel = 21 − 6 = 15t.(Note 1tonne = 1000 kg.)Now try the following Practice ExercisePracticeExercise 25 Further ratios(answers on page 342)1. Express 130 g as a ratio of 1.95kg.2. In a laboratory, acid and water are mixed in theratio 2:5. How much acid is needed to make266ml of the mixture?3. A glass contains 30 ml of gin which is 40%alcohol. If 18ml of water is added and themixture stirred, determine the new percentagealcoholic content.4. A wooden beam 4m long weighs 84kg. Deter-mine the mass of a similar beam that is 60 cmlong.5. An alloy is made up of metals P and Q in theratio 3.25:1 by mass. How much of P has tobe added to 4.4kg of Q to make the alloy?6. 15000 kg of a mixture of sand and gravel is20% sand. Determine the amount of sand thatmust be added to produce a mixture with 30%gravel.6.3 Direct proportionTwo quantities are in direct proportion when theyincrease or decrease in the same ratio. For example,if 12 cans of lager have a mass of 4kg, then 24 cans oflager will have a mass of 8kg; i.e., if the quantity of cansdoubles then so does the mass. This is direct proportion.In the previous section we had an example of mixing1 shovel of cement to 4 shovels of sand; i.e., the ratioof cement to sand was 1:4. So, if we have a mix of 10shovels ofcement and 40 shovels of sand and we wantedto double the amount of the mix then we would needto double both the cement and sand, i.e. 20 shovels ofcement and 80 shovels of sand. This is another exampleof direct proportion.Here are three laws in engineering which involve directproportion:(a) Hooke's law states that, within the elastic limit ofa material, the strain ε produced is directly propor-tional to the stress σ producing it, i.e. ε ∝ σ (notethan '∝' means 'is proportional to').(b) Charles's law states that, for a given mass of gasat constant pressure, the volume V is directly pro-portional to its thermodynamic temperature T , i.e.V ∝ T.(c) Ohm's law states that the current I flowingthrough a fixed resistance is directly proportionalto the applied voltage V , i.e. I ∝ V .
Ratio and proportion 43Here are some worked examples to help us understandmore about direct proportion.Problem 9. 3 energy saving light bulbs cost£7.80. Determine the cost of 7 such light bulbs(i) 3 light bulbs cost £7.80(ii) Therefore, 1 light bulb costs7.803= £2.60Hence, 7 light bulbs cost 7 × £2.60 = £18.20Problem 10. If 56litres of petrol costs £59.92,calculate the cost of 32 litres(i) 56litres of petrol costs £59.92(ii) Therefore, 1 litre of petrol costs59.9256= £1.07Hence, 32 litres cost 32 × 1.07 = £34.24Problem 11. Hooke's law states that stress, σ, isdirectly proportional to strain, ε, within the elasticlimit of a material. When, for mild steel, the stressis 63MPa, the strain is 0.0003. Determine (a) thevalue of strain when the stress is 42 MPa, (b) thevalue of stress when the strain is 0.00072(a) Stress is directly proportional to strain.(i) When the stress is 63MPa, the strain is0.0003(ii) Hence, a stress of 1MPa corresponds to astrain of0.000363(iii) Thus,the value of strain when the stress is42 MPa =0.000363× 42 = 0.0002(b) Strain is proportional to stress.(i) When the strain is 0.0003, the stress is63MPa.(ii) Hence, a strain of 0.0001 corresponds to633MPa.(iii) Thus,the value of stress when the strain is0.00072 =633× 7.2 = 151.2 MPa.Problem 12. Charles's law states that for a givenmass of gas at constant pressure, the volume isdirectly proportional to its thermodynamictemperature. A gas occupies a volume of 2.4litresat 600 K. Determine (a) the temperature when thevolume is 3.2 litres, (b) the volume at 540 K(a) Volume is directly proportionalto temperature.(i) When the volume is 2.4litres, the tempera-ture is 600 K.(ii) Hence, a volume of 1 litre corresponds to atemperature of6002.4K.(iii) Thus,the temperature when the volume is3.2 litres =6002.4× 3.2 = 800K.(b) Temperature is proportional to volume.(i) When the temperature is 600 K, the volumeis 2.4litres.(ii) Hence, a temperature of 1K corresponds toa volume of2.4600litres.(iii) Thus, the volume at a temperature of540 K =2.4600× 540 = 2.16 litres.Now try the following Practice ExercisePracticeExercise 26 Direct proportion(answers on page 342)1. 3 engine parts cost £208.50. Calculate the costof 8 such parts.2. If 9litres of gloss white paint costs £24.75,calculate the cost of 24litres of the same paint.3. The total mass of 120 household bricks is57.6kg. Determine the mass of 550 suchbricks.4. A simple machine has an effort:load ratio of3:37. Determine the effort, in grams, to lift aload of 5.55kN.5. If 16 cans of lager weighs 8.32 kg, what will28 cans weigh?6. Hooke's law states that stress is directly pro-portional to strain within the elastic limit ofa material. When, for copper, the stress is60 MPa, the strain is 0.000625. Determine(a) the strain when the stress is 24MPa and(b) the stress when the strain is 0.0005
44 Basic Engineering Mathematics7. Charles's law states that volume is directlyproportional to thermodynamic temperaturefor a given mass of gas at constant pressure.A gas occupies a volume of 4.8litres at 330 K.Determine (a) the temperature when the vol-ume is 6.4litres and (b) the volume when thetemperature is 396K.Here are some further worked examples on directproportion.Problem 13. Some guttering on a house has todecline by 3mm for every 70 cm to allow rainwaterto drain. The gutter spans 8.4m. How much lowershould the low end be?(i) The guttering has to decline in the ratio 3:700 or3700(ii) If d is the vertical drop in 8.4m or 8400 mm, thenthe decline must be in the ratio d :8400 ord8400(iii) Nowd8400=3700(iv) Cross-multiplyinggives 700×d =8400×3 fromwhich, d =8400 × 3700i.e. d = 36mm, which is how much the lower endshould be to allow rainwater to drain.Problem 14. Ohm's law state that the currentflowing in a fixed resistance is directly proportionalto the applied voltage. When 90 mV is appliedacross a resistor the current flowing is 3A.Determine (a) the current when the voltage is60 mV and (b) the voltage when the current is 4.2 A(a) Current is directly proportional to the voltage.(i) When voltage is 90 mV, the current is 3A.(ii) Hence, a voltage of 1mV corresponds to acurrent of390A.(iii) Thus, when the voltage is 60 mV, thecurrent = 60 ×390= 2A.(b) Voltage is directly proportional to the current.(i) When current is 3A, the voltage is 90 mV.(ii) Hence, a current of 1A corresponds to avoltage of903mV = 30 mV.(iii) Thus, when the current is 4.2 A, thevoltage = 30 × 4.2 = 126mV.Problem 15. Some approximate imperial tometric conversions are shown in Table 6.1. Use thetable to determine(a) the number of millimetres in 12.5inches(b) a speed of 50 miles per hour in kilometresper hour(c) the number of miles in 300 km(d) the number of kilograms in 20 pounds weight(e) the number of pounds and ounces in56kilograms (correct to the nearest ounce)(f) the number of litres in 24 gallons(g) the number of gallons in 60 litresTable 6.1length 1inch = 2.54cm1mile = 1.6kmweight 2.2 lb = 1kg(1lb = 16oz)capacity 1.76 pints = 1litre(8 pints = 1 gallon)(a) 12.5inches = 12.5 × 2.54cm = 31.75cm31.73cm = 31.75 × 10 mm = 317.5 mm(b) 50 m.p.h. = 50 × 1.6km/h = 80 km/h(c) 300 km =3001.6miles = 186.5 miles(d) 20 lb =202.2kg = 9.09 kg(e) 56kg = 56 × 2.2 lb = 123.2 lb0.2 lb = 0.2 × 16oz = 3.2 oz = 3oz, correct tothe nearest ounce.Thus, 56kg = 123 lb 3 oz, correct to the nearestounce.(f) 24 gallons = 24 × 8 pints = 192 pints192 pints =1921.76litres = 109.1 litres
Ratio and proportion 45(g) 60 litres = 60× 1.76 pints = 105.6 pints105.6 pints =105.68gallons = 13.2 gallonsProblem 16. Currency exchange rates for fivecountries are shown in Table 6.2. Calculate(a) how many euros £55 will buy(b) the number of Japanese yen which can bebought for £23(c) the number of pounds sterling which can beexchanged for 6405kronor(d) the number of American dollars which can bepurchased for £92.50(e) the number of pounds sterling which can beexchanged for 2925 Swiss francsTable 6.2France £1 = 1.25 eurosJapan £1 = 185 yenNorway £1 = 10.50kronorSwitzerland £1 = 1.95francsUSA £1 = 1.80 dollars(a) £1 = 1.25 euros, hence £55 = 55 × 1.25 euros= 68.75 euros.(b) £1 = 185 yen, hence £23 = 23 × 185 yen= 4255 yen.(c) £1 = 10.50kronor, hence 6405 lira = £640510.50= £610.(d) £1 = 1.80 dollars, hence£92.50 = 92.50 × 1.80 dollars = $166.50(e) £1 = 1.95 Swiss francs, hence2925 pesetas = £29251.95= £1500Now try the following Practice ExercisePracticeExercise 27 Further directproportion (answers on page 342)1. Ohm's law states that current is proportionalto p.d. in an electrical circuit. When a p.d. of60 mV is applied across a circuit a current of24μAflows. Determine(a)thecurrent flowingwhen the p.d. is 5V and (b) the p.d. when thecurrent is 10 mA.2. The tourist rate for the Swiss franc is quoted ina newspaper as £1 = 1.92 fr. How many francscan be purchased for £326.40?3. If 1inch = 2.54cm, find the number of mil-limetres in 27inches.4. If 2.2 lb = 1kg and 1lb = 16oz, determine thenumber of poundsand ounces in 38kg (correctto the nearest ounce).5. If 1litre = 1.76 pints and 8 pints = 1 gallon,determine (a) the number of litres in 35 gallonsand (b) the number of gallons in 75litres.6. Hooke's law states that stress is directly pro-portional to strain within the elastic limit of amaterial. When for brass the stress is 21MPa,the strain is 0.00025. Determine the stresswhen the strain is 0.00035.7. If 12 inches = 30.48cm, find the number ofmillimetres in 23inches.8. The tourist rate for the Canadian dollar isquoted in a newspaper as £1 = 1.84fr. Howmany Canadian dollars can be purchased for£550?6.4 Inverse proportionTwo variables, x and y, are in inverse proportion to oneanother if y is proportional to1x, i.e. y α1xor y =kxork = xy where k is a constant, called the coefficient ofproportionality.Inverse proportion means that, as the value ofone vari-able increases, the value of another decreases, and thattheir product is always the same.For example, the time for a journey is inversely propor-tional to the speed of travel. So, if at 30m.p.h. a journeyis completed in20 minutes, then at 60m.p.h. the journeywould be completed in 10 minutes. Double the speed,half the journey time. (Note that 30 × 20 = 60× 10.)In another example, the time needed to dig a hole isinversely proportional to the number of people digging.So, if 4 men take 3 hours to dig a hole, then 2 men
46 Basic Engineering Mathematics(working at the same rate) would take 6 hours. Half themen, twice the time. (Note that 4 × 3 = 2 × 6.)Here are some worked examples on inverse proportion.Problem 17. It is estimated that a team of fourdesigners would take a year to develop anengineering process. How long would threedesigners take?If 4 designers take 1 year, then 1 designer would take4 yearsto develop theprocess.Hence, 3 designerswouldtake43years, i.e. 1 year 4 months.Problem 18. A team of five people can deliverleaflets to every house in a particular area in fourhours. How long will it take a team of three people?If 5 people take 4 hours to deliver the leaflets, then1 person would take 5 × 4 = 20 hours. Hence, 3 peo-ple would take203hours, i.e. 623hours, i.e. 6 hours40 minutes.Problem 19. The electrical resistance R of apiece of wire is inversely proportional to thecross-sectional area A. When A = 5mm2,R = 7.02 ohms. Determine (a) the coefficient ofproportionality and (b) the cross-sectional areawhen the resistance is 4ohms(a) Rα1A, i.e. R =kAor k = RA. Hence, whenR = 7.2 and A = 5, thecoefficient of proportionality,k = (7.2)(5) = 36(b) Since k = RA then A =kR. Hence, when R = 4,the cross sectional area, A =364= 9 mm2Problem 20. Boyle's law states that, at constanttemperature, the volume V of a fixed mass of gas isinversely proportional to its absolute pressure p. Ifa gas occupies a volume of 0.08m3at a pressure of1.5 × 106pascals, determine (a) the coefficient ofproportionality and (b) the volume if the pressure ischanged to 4 × 106 pascals(a) V ∝1pi.e. V =kpor k = pV . Hence, thecoefficient of proportionality, k= (1.5 × 106)(0.08) = 0.12 × 106(b) Volume,V =kp=0.12 × 1064 × 106= 0.03m3Now try the following Practice ExercisePracticeExercise 28 Further inverseproportion (answers on page 342)1. A 10 kg bag of potatoes lasts for a week with afamily of 7 people. Assuming all eat the sameamount, how longwill the potatoeslast ifthereare only two in the family?2. If 8 men take 5 days to build a wall, how longwould it take 2 men?3. If y is inversely proportional to x andy = 15.3 when x = 0.6, determine (a) thecoefficient of proportionality, (b) the value ofy when x is 1.5 and (c) the value of x when yis 27.24. A car travelling at 50 km/h makes a journey in70 minutes. How long will the journey take at70 km/h?5. Boyle's law states that, for a gas at constanttemperature, the volume of a fixed mass ofgas is inversely proportional to its absolutepressure. If a gas occupies a volume of 1.5m3at a pressure of 200 × 103 pascals, determine(a) the constant of proportionality, (b) thevolume when the pressure is 800 × 103 pas-cals and (c) the pressure when the volume is1.25m3.
Chapter 8Units, prefixes andengineering notation8.1 IntroductionOf considerable importance in engineering is a knowl-edge ofunitsof engineering quantities,the prefixes usedwith units, and engineering notation.We need to know, for example, that80kV = 80 × 103 V, which means 80000 voltsand 25mA = 25 × 10−3A,which means 0.025 amperesand 50nF = 50 × 10−9F,which means 0.000000050 faradsThis is explained in this chapter.8.2 SI unitsThe system of units used in engineering and scienceis the Système Internationale d'Unités (InternationalSystem of Units), usually abbreviated to SI units, and isbased on the metric system. This was introduced in 1960and has now been adopted by the majority of countriesas the official system of measurement.The basic seven units used in the SI system are listed inTable 8.1 with their symbols.There are, of course, many units other than these seven.These other units are called derived units and aredefined in terms of the standard units listed in the table.For example, speed is measured in metres per second,therefore using two of the standard units, i.e. length andtime.Table 8.1 Basic SI unitsQuantity Unit SymbolLength metre m (1m = 100 cm= 1000 mm)Mass kilogramkg (1kg = 1000 g)Time second sElectric current ampere AThermodynamictemperaturekelvin K (K = ◦C + 273)Luminousintensitycandela cdAmount ofsubstancemole molSome derived units are given special names. For exam-ple, force = mass × acceleration has units of kilogrammetre per second squared, which uses three of the baseunits, i.e. kilograms, metres and seconds. The unit of kgm/s2 is given the special name of a Newton.Table 8.2 contains a list of some quantities and theirunits that are common in engineering.8.3 Common prefixesSI units may be made larger or smaller by using prefixeswhich denote multiplication or division by a particularamount.DOI: 10.1016/B978-1-85617-697-2.00008-9
54 Basic Engineering MathematicsTable 8.2 Some quantities and their units that are common in engineeringQuantity Unit SymbolLength metre mArea square metre m2Volume cubic metre m3Mass kilogram kgTime second sElectric current ampere ASpeed, velocity metre per second m/sAcceleration metre per second squared m/s2Density kilogram per cubic metre kg/m3Temperature kelvin or Celsius K or ◦CAngle radian or degree rad or ◦Angular velocity radian per second rad/sFrequency hertz HzForce newton NPressure pascal PaEnergy, work joule JPower watt WCharge, quantity of electricity coulomb CElectric potential volt VCapacitance farad FElectrical resistance ohmInductance henry HMoment of force newton metre NmThe most common multiples are listed in Table 8.3.Aknowledgeofindicesisneeded sinceall oftheprefixesare powers of 10 with indices that are a multiple of 3.Here are some examples of prefixes used with engineer-ing units.A frequency of 15 GHz means 15 × 109 Hz, which is15000000000 hertz,i.e. 15 gigahertz is written as 15GHz and is equal to 15thousand million hertz.(Instead of writing 15000000000 hertz, it is muchneater, takes up less space and prevents errors causedby having so many zeros, to write the frequency as15GHz.)A voltage of 40 MV means 40 × 106 V, which is40000000 volts,i.e. 40 megavolts is written as 40MV and is equal to 40million volts.An inductance of 12 mH means 12 × 10−3 H or12103H or121000H, which is 0.012H,i.e. 12 millihenrys is written as 12mH and is equal to12 thousandths of a henry.
Units, prefixes and engineering notation 55Table 8.3 Common SI multiplesPrefix Name MeaningG giga multiply by 109i.e. × 1000000000M mega multiply by 106 i.e. × 1000000k kilo multiply by 103 i.e. × 1000m milli multiply by 10−3 i.e. ×1103=11000= 0.001μ micro multiply by 10−6 i.e. ×1106=11000000= 0.000001n nano multiply by 10−9 i.e. ×1109=11000000000= 0.000 000001p pico multiply by 10−12 i.e. ×11012=11000000000000=0.000000000001A time of 150 ns means 150 × 10−9 s or150109s, whichis 0.000000150s,i.e. 150 nanoseconds is written as 150ns and is equal to150 thousand millionths of a second.A force of 20 kN means 20 × 103 N, which is 20000newtons,i.e. 20 kilonewtons is written as 20kN and is equal to20 thousand newtons.A charge of 30 μC means 30 × 10−6 C or30106C, whichis 0.000030C,i.e. 30 microcoulombs is written as 30μC and is equalto 30 millionths of a coulomb.A capacitance of 45pF means 45 × 10−12 F or451012F,which is 0.000000000045F,i.e. 45 picofarads is written as 45pF and is equal to 45million millionths of a farad.In engineering it is important to understand what suchquantities as 15GHz, 40MV, 12mH, 150ns, 20kN,30μC and 45pF mean.Now try the following Practice ExercisePracticeExercise 32 SI units and commonprefixes (answers on page 343)1. State the SI unit of volume.2. State the SI unit of capacitance.3. State the SI unit of area.4. State the SI unit of velocity.5. State the SI unit of density.6. State the SI unit of energy.7. State the SI unit of charge.8. State the SI unit of power.9. State the SI unit of angle.10. State the SI unit of electric potential.11. State which quantity has the unit kg.12. State which quantity has the unit symbol .13. State which quantity has the unit Hz.14. State which quantity has the unit m/s2.15. State which quantity has the unit symbol A.16. State which quantity has the unit symbol H.17. State which quantity has the unit symbol m.18. State which quantity has the unit symbol K.19. State which quantity has the unit Pa.20. State which quantity has the unit rad/s.21. What does the prefix G mean?22. What isthe symbol and meaning of the prefixmilli?
Chapter 11Solving simple equations11.1 Introduction3x − 4 is an example of an algebraic expression.3x − 4 = 2 is an example of an algebraic equation (i.e.it contains an '=' sign).An equation is simply a statement that two expressionsare equal.Hence, A = πr2 (where A is the area of a circleof radius r)F =95C + 32 (which relates Fahrenheit andCelsius temperatures)and y = 3x + 2 (which is the equation of astraight line graph)are all examples of equations.11.2 Solving equationsTo 'solve an equation' means 'to find the value of theunknown'. For example, solving 3x − 4 = 2 means thatthe value of x is required.In this example, x = 2. How did we arrive at x = 2?This is the purpose of this chapter – to show how tosolve such equations.Many equations occur in engineering and it is essentialthat we can solve them when needed.Here are some examples to demonstrate how simpleequations are solved.Problem 1. Solve the equation 4x = 20Dividing each side of the equation by 4 gives4x4=204i.e. x = 5 by cancelling, which is the solution to theequation 4x = 20.The same operation must be applied to both sides of anequation so that the equality is maintained.We can do anything we like to an equation, as longas we do the same to both sides. This is, in fact, theonly rule to remember when solving simple equations(and also when transposing formulae, which we do inChapter 12).Problem 2. Solve the equation2x5= 6Multiplying both sides by 5 gives 52x5= 5(6)Cancelling and removing brackets gives 2x = 30Dividing both sides of the equation by 2 gives2x2=302Cancelling gives x = 15which is the solution of the equation2x5= 6.Problem 3. Solve the equation a − 5 = 8Adding 5 to both sides of the equation givesa − 5 + 5 = 8 + 5i.e. a = 8 + 5i.e. a = 13which is the solution of the equation a − 5 = 8.Note that adding 5 to both sides of the above equationresults in the −5 moving from the LHS to the RHS, butthe sign is changed to +.Problem 4. Solve the equation x + 3 = 7Subtracting 3 from both sides gives x + 3 − 3 = 7 − 3i.e. x = 7 − 3i.e. x = 4which is the solution of the equation x + 3 = 7.DOI: 10.1016/B978-1-85617-697-2.00011-9
Solving simple equations 817. An alloy contains 60% by weight of copper,the remainder being zinc. How much coppermust be mixed with 50kg of this alloy to givean alloy containing 75% copper?8. A rectangular laboratory has a length equal toone and a half times its width and a perimeterof 40m. Find its length and width.9. Applying the principle of moments to a beamresults in the following equation:F × 3 = (5 − F) × 7where F is the force in newtons. Determinethe value of F.
Chapter 12Transposing formulae12.1 IntroductionIn the formula I =VR, I is called the subject of theformula.Similarly, in the formula y = mx + c, y is the subject ofthe formula.When a symbol other than the subject is required tobe the subject, the formula needs to be rearranged tomake a new subject. This rearranging process is calledtransposing the formula or transposition.For example, in the above formulae,if I =VRthen V = IRand if y = mx + c then x =y − cmHow did we arrive at these transpositions? This is thepurpose of this chapter — to show how to transpose for-mulae. A great many equations occur in engineering andit is essential that we can transpose them when needed.12.2 Transposing formulaeThere are no new rules for transposing formulae.The same rules as were used for simple equations inChapter 11 are used; i.e., the balance of an equationmust be maintained: whatever is done to one side ofan equation must be done to the other.It is best that you cover simple equations before tryingthis chapter.Here are some worked examples to help understandingof transposing formulae.Problem 1. Transpose p = q +r + s to make rthe subjectThe object is to obtain r on its own on the LHS of theequation. Changing the equation around so that r is onthe LHS givesq +r + s = p (1)From Chapter 11 on simple equations, a term can bemoved from one side of an equation to the other side aslong as the sign is changed.Rearranging gives r = p − q − s.Mathematically, we have subtracted q + s from bothsides of equation (1).Problem 2. If a + b = w − x + y, express x asthe subjectAs stated in Problem 1, a term can be moved from oneside of an equation to the other side but with a changeof sign.Hence, rearranging gives x = w + y − a − bProblem 3. Transpose v = f λ to make λ thesubjectv = f λ relates velocity v, frequency f and wave-length λRearranging gives f λ = vDividing both sides by f givesf λf=vfCancelling gives λ =vfProblem 4. When a body falls freely through aheight h, the velocity v is given by v2= 2gh.Express this formula with h as the subjectDOI: 10.1016/B978-1-85617-697-2.00012-0
Solving quadratic equations 10514.3 Solution of quadratic equationsby 'completing the square'An expression such as x2 or (x + 2)2 or (x − 3)2 iscalled a perfect square.If x2 = 3 then x = ±√3If (x + 2)2 = 5 then x + 2 = ±√5 and x = −2 ±√5If (x − 3)2 = 8 then x − 3 = ±√8 and x = 3 ±√8Hence, if a quadratic equation can be rearranged so thatone side of the equation is a perfect square and the otherside of the equation is a number, then the solution ofthe equation is readily obtained by taking the squareroots of each side as in the above examples. The processof rearranging one side of a quadratic equation into aperfect square before solving is called 'completing thesquare'.(x + a)2= x2+ 2ax + a2Thus,inordertomakethequadraticexpressionx2 + 2axinto a perfect square, it is necessary to add (half thecoefficient of x)2, i.e.2a22or a2For example, x2 + 3x becomes a perfect square byadding322, i.e.x2+ 3x +322= x +322The method of completing the square is demonstratedin the following worked problems.Problem 15. Solve 2x2 + 5x = 3 by completingthe squareThe procedure is as follows.(i) Rearrangetheequation so that all termsareon thesame side of the equals sign (and the coefficientof the x2 term is positive). Hence,2x2+ 5x − 3 = 0(ii) Make the coefficient of the x2 term unity. In thiscase thisis achieved by dividingthroughoutby 2.Hence,2x22+5x2−32= 0i.e. x2+52x −32= 0(iii) Rearrange the equations so that the x2and xterms are on one side of the equals sign and theconstant is on the other side. Hence,x2+52x =32(iv) Add to both sides of the equation (half the coeffi-cient of x)2. In this case the coefficient of x is52Half the coefficient squared is therefore542Thus,x2+52x +542=32+542The LHS is now a perfect square, i.e.x +542=32+542(v) Evaluate the RHS. Thus,x +542=32+2516=24 + 2516=4916(vi) Take the square root of both sides of the equation(remembering that the square root of a numbergives a ± answer). Thus,x +542=4916i.e. x +54= ±74(vii) Solve the simple equation. Thus,x = −54±74i.e. x = −54+74=24=12or 0.5and x = −54−74= −124= −3Hence, x = 0.5 or x = −3; i.e., the roots of theequation 2x2 + 5x = 3 are 0.5 and −3.Problem 16. Solve 2x2 + 9x + 8 = 0, correct to 3significant figures, by completing the squareMaking the coefficient of x2 unity givesx2 +92x + 4 = 0Rearranging gives x2 +92x = −4
Solving quadratic equations 109Hence, the mass will reach a height of 16m after0.59s on the ascent and after 5.53s on the descent.Problem 25. A shed is 4.0m long and 2.0m wide.A concrete path of constant width is laid all the wayaround the shed. If the area of the path is 9.50m2,calculate its width to the nearest centimetreFigure 14.1 shows a plan view of the shed with itssurrounding path of width t metres.t2.0m4.0m (4.0 12t)SHEDtFigure 14.1Area of path = 2(2.0 × t) + 2t(4.0 + 2t)i.e. 9.50 = 4.0t + 8.0t + 4t2or 4t2+12.0t − 9.50 = 0Hence,t =−(12.0) ± (12.0)2 − 4(4)(−9.50)2(4)=−12.0 ±√296.08=−12.0 ± 17.204658i.e. t = 0.6506m or − 3.65058m.Neglecting the negative result, which is meaningless,the width of the path, t = 0.651m or 65cm correct tothe nearest centimetre.Problem 26. If the total surface area of a solidcone is 486.2cm2 and its slant height is 15.3cm,determine its base diameter.From Chapter 27, page 245, the total surface area A ofa solid cone is given by A = πrl + πr2, where l is theslant height and r the base radius.If A = 482.2 and l = 15.3, then482.2 = πr(15.3) + πr2i.e. πr2+ 15.3πr − 482.2 = 0or r2+ 15.3r −482.2π= 0Using the quadratic formula,r =−15.3 ± (15.3)2 − 4−482.2π2=−15.3 ±√848.04612=−15.3 ± 29.121232Hence, radius r = 6.9106cm (or −22.21cm, which ismeaningless and is thus ignored).Thus, the diameter of the base = 2r = 2(6.9106)= 13.82cm.Now try the following Practice ExercisePracticeExercise 57 Practical problemsinvolving quadratic equations (answers onpage 346)1. The angle a rotating shaft turns through in tseconds is given by θ = ωt +12αt2. Deter-mine the time taken to complete 4 radians ifω is 3.0 rad/s and α is 0.60 rad/s2.2. The power P developed in an electrical cir-cuit is given by P = 10I − 8I2, where I isthe current in amperes. Determine the currentnecessary to produce a power of 2.5 watts inthe circuit.3. The area of a triangle is 47.6cm2 and itsperpendicular height is 4.3cm more than itsbase length. Determine the length of the basecorrect to 3 significant figures.4. The sag, l, in metres in a cable stretchedbetween two supports, distance x m apart, isgiven by l =12x+ x. Determine the distancebetween the supports when the sag is 20 m.5. Theacid dissociation constant Ka ofethanoicacid is 1.8 × 10−5 moldm−3 for a particu-lar solution. Using the Ostwald dilution law,
Chapter 15Logarithms15.1 Introduction to logarithmsWith the use of calculators firmly established, logarith-mic tables are now rarely used forcalculation. However,the theory of logarithms is important, for there are sev-eral scientific and engineering laws that involve the rulesof logarithms.From Chapter 7, we know that 16 = 24.The number 4 is called the power or the exponent orthe index. In the expression 24, the number 2 is calledthe base.In another example, we know that 64 = 82.In this example, 2 is the power, or exponent, or index.The number 8 is the base.15.1.1 What is a logarithm?Consider the expression 16 = 24.An alternative, yet equivalent, way of writing thisexpression is log2 16 = 4.This is stated as 'log to the base 2 of 16 equals 4'.We see that the logarithm is the same as the poweror index in the original expression. It is the base inthe original expression that becomes the base of thelogarithm.The two statements 16 = 24andlog216 = 4 are equivalentIf we write either of them, we are automatically imply-ing the other.In general, if a number y can be written in the form ax,then the index x is called the 'logarithm of y to the baseof a', i.e.if y = axthen x = loga yIn another example, if we write down that 64 = 82 thentheequivalent statement using logarithmsislog8 64 = 2.In another example, if we write down that log3 27 = 3then the equivalent statement using powers is 33 = 27.So the two sets of statements, one involving powers andone involving logarithms, are equivalent.15.1.2 Common logarithmsFrom the above, if we write down that 1000 = 103, then3 = log10 1000. This may be checked using the 'log'button on your calculator.Logarithms having a base of 10 are called commonlogarithms and log10 is usually abbreviated to lg. Thefollowing values may be checked using a calculator.lg27.5 = 1.4393...lg378.1 = 2.5776...lg0.0204 = −1.6903...15.1.3 Napierian logarithmsLogarithms having a base of e (where e is a mathemat-ical constant approximately equal to 2.7183) are calledhyperbolic, Napierian or natural logarithms, andloge is usually abbreviated to ln. The following valuesmay be checked using a calculator.ln3.65 = 1.2947...ln417.3 = 6.0338...ln0.182 = −1.7037...Napierian logarithms are explained further in Chapter16, following.DOI: 10.1016/B978-1-85617-697-2.00015-6
Exponential functions 119In problems 3 and 4, evaluate correct to 5 decimalplaces.3. (a)17e3.4629 (b) 8.52e−1.2651(c)5e2.69213e1.11714. (a)5.6823e−2.1347(b)e2.1127 − e−2.11272(c)4 e−1.7295 − 1e3.68175. The length of a bar, l, at a temperature, θ,is given by l = l0eαθ , where l0 and α areconstants. Evaluate l, correct to 4 significantfigures, where l0 = 2.587, θ = 321.7 andα = 1.771 × 10−4.6. When a chain of length 2L is suspended fromtwo points, 2D metres apart on the same hor-izontal level, D = k lnL +√L2 + k2k.Evaluate D when k = 75m and L = 180m.16.2 The power series for exThe value of ex can be calculated to any required degreeof accuracy since it is defined in terms of the followingpower series:ex= 1 + x+x 22!+x 33!+x 44!+ ··· (1)(where 3!= 3 × 2 × 1 and is called 'factorial 3').The series is valid for all values of x.The series is said to converge; i.e., if all the terms areadded, an actual value for ex (where x is a real number)is obtained. The more terms that are taken, the closerwill be the value of ex to its actual value. The value ofthe exponent e, correct to say 4 decimal places, may bedetermined by substituting x = 1 in the power series ofequation (1). Thus,e1= 1 + 1 +(1)22!+(1)33!+(1)44!+(1)55!+(1)66!+(1)77!+(1)88!+ ···= 1 + 1 + 0.5 + 0.16667 + 0.04167 + 0.00833+ 0.00139 + 0.00020+ 0.00002 + ···= 2.71828i.e. e = 2.7183, correct to 4 decimal places.The value of e0.05, correct to say 8 significant figures, isfound by substituting x = 0.05 in the power series forex . Thus,e0.05= 1 + 0.05 +(0.05)22!+(0.05)33!+(0.05)44!+(0.05)55!+ ···= 1 + 0.05 + 0.00125 + 0.000020833+ 0.000000260+ 0.0000000026i.e. e 0.05= 1.0512711, correct to 8 significant figures.In this example, successive terms in the series growsmaller very rapidly and it is relatively easy to deter-mine the value of e0.05 to a high degree of accuracy.However, when x is nearer to unity or larger than unity,a very large number of terms are required for an accurateresult.If, in the series of equation (1), x is replaced by −x,thene−x= 1 + (−x) +(−x)22!+(−x)33!+ ···i.e. e−x= 1 − x +x 22!−x 33!+ ···In a similar manner the power series for ex maybe used to evaluate any exponential function of the formaekx , where a and k are constants.In the series of equation (1), let x be replaced by kx.Thenaekx= a 1 + (kx) +(kx)22!+(kx)33!+ ···Thus, 5e2x= 5 1 + (2x) +(2x)22!+(2x)33!+ ···= 5 1 + 2x +4x22+8x36+ ···i.e. 5e2x= 5 1 + 2x + 2x2+43x3+ ···Problem 4. Determine the value of 5e0.5, correctto 5 significant figures, by using the power seriesfor ex
Exponential functions 127(a) θ1, correct to the nearest degree, when θ2 is50◦C, t is 30s and τ is 60s and(b) the time t, correct to 1 decimal place, for θ2 tobe half the value of θ1(a) Transposing the formula to make θ1 the subjectgivesθ1 =θ21 − e−t/τ=501 − e−30/60=501 − e−0.5=500.393469...i.e. θ1 = 127◦C correct to the nearest degree.(b) Transposing to make t the subject of the formulagivesθ2θ1= 1 − e−tτfrom which, e−tτ = 1 −θ2θ1Hence, −tτ= ln 1 −θ2θ1i.e. t = −τ ln 1 −θ2θ1Since θ2 =12θ1t = −60ln 1 −12= −60ln0.5= 41.59sHence, the time for the temperature θ2 to be one halfof the value of θ1 is 41.6 s, correct to 1 decimal place.Now try the following Practice ExercisePracticeExercise 66 Laws of growth anddecay (answers on page 347)1. The temperature, T ◦C, of a cooling objectvaries with time, t minutes, accordingto the equation T = 150e−0.04t. Deter-mine the temperature when (a) t = 0,(b) t = 10 minutes.2. The pressure p pascals at height h metresabove ground level is given by p = p0e−h/C,where p0 is the pressure at ground leveland C is a constant. Find pressure p whenp0 = 1.012 × 105 Pa, height h = 1420m andC = 71500.3. The voltage drop, v volts, across an inductorL henrys at time t seconds is given byv = 200e−RtL , where R = 150 andL = 12.5 × 10−3 H. Determine (a) thevoltage when t = 160 × 10−6 s and (b) thetime for the voltage to reach 85V.4. The length l metres of a metal bar at tem-perature t ◦C is given by l = l0eαt , where l0and α are constants. Determine (a) the valueof l when l0 = 1.894, α = 2.038 × 10−4 andt = 250◦C and (b) the value of l0 whenl = 2.416,t =310◦C and α =1.682×10−4.5. The temperature θ2◦C of an electrical con-ductor at time t seconds is given byθ2 = θ1(1 − e−t/T ), where θ1 is the ini-tial temperature and T seconds is a con-stant. Determine (a) θ2 when θ1 = 159.9◦C,t = 30s and T = 80s and (b) the time t forθ2 to fall to half the value of θ1 if T remainsat 80s.6. Abelt isin contact with apulley forasectorofθ = 1.12 radians and the coefficient of fric-tion between these two surfaces is μ = 0.26.Determine the tension on the taut side of thebelt, T newtons, when tension on the slackside is given by T0 = 22.7newtons, giventhat these quantities are related by the lawT = T0eμθ .7. The instantaneous current i at time t isgiven by i = 10e−t/CRwhen a capacitoris being charged. The capacitance C is7×10−6 farads and the resistance R is0.3×106 ohms. Determine (a) the instanta-neous current when t is 2.5seconds and (b)thetimefortheinstantaneouscurrent to fall to5amperes. Sketch a curve of current againsttime from t = 0 to t = 6seconds.8. The amount of product x (in mol/cm3)found in a chemical reaction startingwith 2.5mol/cm3of reactant is given byx = 2.5(1 − e−4t ) where t is the time, inminutes, to form product x. Plot a graphat 30second intervals up to 2.5minutes anddetermine x after 1minute.
128 Basic Engineering Mathematics9. The current i flowing in a capacitor attime t is given by i = 12.5(1 − e−t/CR),where resistance R is 30k and thecapacitance C is 20μF. Determine (a)the current flowing after 0.5seconds and(b) the time for the current to reach10amperes.10. The amount A after n years of a sum investedP is given by the compound interest lawA = Pern/100, when the per unit interest rater is added continuously. Determine, correctto the nearest pound, the amount after 8 yearsfor a sum of £1500 invested if the interestrate is 6% per annum.
Chapter 17Straight line graphs17.1 Introduction to graphsA graph is a visual representation of information,showing how one quantity varies with another relatedquantity.We often see graphs in newspapers or in businessreports, in travel brochures and government publica-tions. For example, a graph of the share price (in pence)over a six month period for a drinks company, FizzyPops, is shown in Figure 17.1.Generally, we see that the share price increases to a highof400p in June,but dipsdown to around 280p in Augustbefore recovering slightly in September.A graph should convey information more quickly to thereader than if the same information was explained inwords.When this chapter is completed you should be able todraw up a table of values, plot co-ordinates, determinethe gradient and state the equation of a straight linegraph. Some typical practical examples are included inwhich straight lines are used.400Fizzy Pops350300250Apr 07 May 07 Jun 07 Jul 07 Aug 07 Sep 07Figure 17.117.2 Axes, scales and co-ordinatesWe are probably all familiar with reading a map to locatea town, or a local map to locate a particular street. Forexample, a street map of central Portsmouth is shown inFigure 17.2. Notice the squares drawn horizontally andvertically on the map; this is called a grid and enablesus to locate a place of interest or a particular road. Mostmaps contain such a grid.We locate places ofinterest on the map by stating a letterand a number – this is called the grid reference.For example, on the map, the Portsmouth & Southseastation is in square D2, King's Theatre is in square E5,HMS Warrior is in square A2, Gunwharf Quays is insquare B3 and High Street is in square B4.Portsmouth & Southsea station is located by movinghorizontally along the bottom of the map until thesquare labelled D is reached and then moving verticallyupwards until square 2 is met.The letter/number, D2, is referred to as co-ordinates;i.e., co-ordinates are used to locate the position ofDOI: 10.1016/B978-1-85617-697-2.00017-X
Straight line graphs 13112345A B C D E FFigure 17.2 Reprinted with permission from AA Media Ltd.a point on a map. If you are familiar with using amap in this way then you should have no difficultieswith graphs, because similar co-ordinates are used withgraphs.As stated earlier, a graph is a visual representationof information, showing how one quantity varies withanother related quantity. The most common method ofshowing a relationship between two sets of data is touse a pair of reference axes – these are two lines drawnat right angles to each other (often called Cartesian orrectangular axes), as shown in Figure 17.3.The horizontal axis is labelled the x-axis and the ver-tical axis is labelled the y-axis. The point where x is 0and y is 0 is called the origin.x values have scales that are positive to the right ofthe origin and negative to the left. y values have scalesthat are positive up from the origin and negative downfrom the origin.Co-ordinates are written with brackets and a commain between two numbers. For example, point A is shownwith co-ordinates (3, 2) and is located by starting at theA(3, 2)424 23 22 21 032121222324OriginC(23, 22)B (24, 3)yx1 3 42Figure 17.3origin and moving 3 units in the positive x direction(i.e. to the right) and then 2 units in the positive ydirection (i.e. up).When co-ordinates are stated the first number isalways the x value and the second number is always
132 Basic Engineering Mathematicsthe y value. In Figure 17.3, point B has co-ordinates(−4,3) and point C has co-ordinates (−3,−2).17.3 Straight line graphsThe distances travelled by a car in certain peri-ods of time are shown in the following table ofvalues.Time(s) 10 20 30 40 50 60Distance travelled (m) 50 100 150 200 250 300We will plot time on the horizontal (or x) axis with ascale of 1cm = 10s.We will plot distance on the vertical (or y) axis with ascale of 1cm = 50m.(When choosing scales it is better to choose ones such as1cm = 1unit,1cm = 2unitsor 1cm = 10unitsbecausedoing so makes reading values between these valueseasier.)With the above data, the (x, y) co-ordinates become(time, distance) co-ordinates; i.e., the co-ordinates are(10, 50), (20, 100), (30, 150), and so on.The co-ordinates are shown plotted in Figure 17.4using crosses. (Alternatively, a dot or a dot and circlemay be used, as shown in Figure 17.3.)A straight line is drawn through the plotted co-ordinates as shown in Figure 17.4.300DistanceTravelled(m)Time (s)2502001501005010 20 30 40 50 60Distance/time graphFigure 17.4Student taskThe following table gives the force F newtons which,when applied to a lifting machine, overcomes acorresponding load of L newtons.F (Newtons) 19 35 50 93 125 147L (Newtons) 40 120 230 410 540 6801. Plot L horizontally and F vertically.2. Scales are normally chosen such that the graphoccupies as much space as possible on thegraph paper. So in this case, the followingscales are chosen.Horizontal axis (i.e. L): 1cm = 50NVertical axis (i.e. F): 1cm = 10N3. Draw the axes and label them L (newtons) forthe horizontal axis and F (newtons) for thevertical axis.4. Label the origin as 0.5. Writeonthehorizontal scalingat 100,200,300,and so on, every 2cm.6. Write on the vertical scaling at 10, 20, 30, andso on, every 1cm.7. Plot on the graph the co-ordinates (40, 19),(120, 35), (230, 50), (410, 93), (540, 125) and(680, 147), marking each with a cross or a dot.8. Using aruler,drawthebest straight linethroughthe points. You will notice that not all of thepointslieexactly on a straight line.This is quitenormal with experimental values. In a practi-cal situation it would be surprising if all of thepoints lay exactly on a straight line.9. Extend the straight line at each end.10. From the graph, determine the force appliedwhen the load is 325N. It should be closeto 75N. This process of finding an equivalentvalue within the given data is called interpola-tion. Similarly, determine the load that a forceof 45N will overcome. It should be close to170N.11. From the graph, determine the force needed toovercome a 750N load. It should be close to161N. This process of finding an equivalent
Straight line graphs 133value outside the given data is called extrapo-lation.To extrapolateweneed to haveextendedthestraight linedrawn.Similarly,determinetheforce applied when theload iszero. It shouldbeclose to 11N. The point where the straight linecrosses the vertical axis is called the vertical-axis intercept. So, in this case, the vertical-axisintercept = 11N at co-ordinates (0, 11).The graph you have drawn should look somethinglike Figure 17.5 shown below.101000 200 300 400 500 600 700 800L (Newtons)F(Newtons)2030405060708090100110120130140150160Graph of F against LFigure 17.5In another example, let the relationship between twovariables x and y be y = 3x + 2.When x = 0, y = 0 + 2 = 2When x = 1, y = 3 + 2 = 5When x = 2, y = 6 + 2 = 8, and so on.The co-ordinates (0, 2), (1, 5) and (2, 8) have beenproduced and are plotted, with others, as shown inFigure 17.6.When the points are joined together a straight linegraph results, i.e. y = 3x + 2 is a straight line graph.17.3.1 Summary of general rules to beapplied when drawing graphs(a) Give the graph a title clearly explaining what isbeing illustrated.(b) Choose scales such that the graph occupies asmuch space as possible on the graph paperbeing used.1 2yϭ3xϩ 2yx02468Ϫ1Figure 17.6 Graph of y/x(c) Choose scales so that interpolation is made aseasy as possible. Usually scales such as 1cm =1unit,1cm = 2 units or 1cm = 10 units are used.Awkward scalessuch as1cm = 3 unitsor1cm = 7units should not be used.(d) The scales need not start at zero, particularly whenstarting at zero produces an accumulation ofpointswithin a small area of the graph paper.(e) The co-ordinates, or points, should be clearlymarked. This is achieved by a cross, or a dot andcircle, or just by a dot (see Figure 17.3).(f) A statement should be made next to each axisexplaining the numbers represented with theirappropriate units.(g) Sufficient numbers should be written next to eachaxis without cramping.Problem 1. Plot the graph y = 4x + 3 in therange x = −3 to x = +4. From the graph, find(a) the value of y when x = 2.2 and (b) the valueof x when y = −3Whenever an equation is given and a graph is required,a table giving corresponding values of the variable isnecessary. The table is achieved as follows:When x = −3, y = 4x + 3 = 4(−3) + 3= −12 + 3 = −9When x = −2, y = 4(−2) + 3= −8 + 3 = −5,and so on.Such a table is shown below.x −3 −2 −1 0 1 2 3 4y −9 −5 −1 3 7 11 15 19The co-ordinates (−3,−9), (−2,−5), (−1,−1), andso on, are plotted and joined together to produce the
134 Basic Engineering Mathematics20151011.85Ϫ5Ϫ3100Ϫ2 Ϫ1Ϫ1.5Ϫ3 12.22 43 xyFigure 17.7straight line shown in Figure 17.7. (Note that the scalesused on the x and y axes do not have to be the same.)From the graph:(a) when x = 2.2, y = 11.8, and(b) when y = −3, x = −1.5Now try the following Practice ExercisePracticeExercise 67 Straight line graphs(answers on page 347)1. Assuming graph paper measuring 20cm by20cm is available, suggest suitable scales forthe following ranges of values.(a) Horizontal axis: 3V to 55V; verticalaxis: 10 to 180 .(b) Horizontal axis: 7m to 86m; verticalaxis: 0.3V to 1.69V.(c) Horizontal axis: 5N to 150N; verticalaxis: 0.6mm to 3.4mm.2. Corresponding values obtained experimen-tally for two quantities arex −5 −3 −1 0 2 4y −13 −9 −5 −3 1 5Plot a graph of y (vertically) against x (hori-zontally)to scales of 2cm = 1 for the horizon-tal x-axis and 1cm = 1 for the vertical y-axis.(This graph will need the whole of the graphpaper with the origin somewhere in the centreof the paper).From the graph, find(a) the value of y when x = 1(b) the value of y when x = −2.5(c) the value of x when y = −6(d) the value of x when y = 73. Corresponding values obtained experimen-tally for two quantities arex −2.0 −0.5 0 1.0 2.5 3.0 5.0y −13.0 −5.5 −3.0 2.0 9.5 12.0 22.0Use a horizontal scale for x of 1cm = 12 unitand a vertical scale for y of 1cm = 2 units anddraw a graph of x against y. Label the graphand each of its axes. By interpolation, findfrom the graph the value of y when x is 3.5.4. Draw a graph of y − 3x + 5 = 0 over a rangeof x = −3 to x = 4. Hence determine(a) the value of y when x = 1.3(b) the value of x when y = −9.25. The speed n rev/min of a motor changes whenthe voltage V across the armature is varied.The results are shown in the following table.n (rev/min) 560 720 900 1010 1240 1410V (volts) 80 100 120 140 160 180It is suspected that one of the readings taken ofthe speed is inaccurate. Plot a graph of speed(horizontally) against voltage (vertically) andfind this value. Find also(a) the speed at a voltage of 132V.(b) the voltage at a speed of 1300rev/min.17.4 Gradients, intercepts andequations of graphs17.4.1 GradientsThe gradient or slope of a straight line is the ratio ofthe change in the value of y to the change in the value ofx between any two points on the line. If, as x increases,(→), y also increases, (↑), then the gradient is positive.
Straight line graphs 14111. A piece of elastic is tied to a support so that ithangs vertically and a pan, on which weightscan be placed, is attached to the free end. Thelength of the elastic is measured as variousweights are added to the pan and the resultsobtained are as follows:Load, W (N) 5 10 15 20 25Length, l (cm) 60 72 84 96 108Plot a graph of load (horizontally) againstlength (vertically) and determine(a) the value of length when the load is 17N.(b) the value of load when the length is74cm.(c) its gradient.(d) the equation of the graph.12. The following table gives the effort P to lifta load W with a small lifting machine.W (N) 10 20 30 40 50 60P (N) 5.1 6.4 8.1 9.6 10.9 12.4Plot W horizontally against P vertically andshow that the values lie approximately on astraight line. Determine the probable rela-tionship connecting P and W in the formP = aW + b.13. In an experiment the speeds N rpm of a fly-wheel slowly coming to rest were recordedagainst the time t in minutes. Plot the resultsand show that N and t are connected byan equation of the form N = at + b. Findprobable values of a and b.t (min) 2 4 6 8 10 12 14N (rev/min) 372 333 292 252 210 177 13217.5 Practical problems involvingstraight line graphsWhen a set of co-ordinate values are given or areobtained experimentally and it is believed that theyfollow a law of the form y = mx + c, if a straight linecan be drawn reasonably close to most oftheco-ordinatevalues when plotted, this verifies that a law of theform y = mx + c exists. From the graph, constantsm (i.e. gradient) and c (i.e. y-axis intercept) can bedetermined.Here are some worked problems in which practicalsituations are featured.Problem 12. The temperature in degrees Celsiusand the corresponding values in degrees Fahrenheitare shown in the table below. Construct rectangularaxes, choose suitable scales and plot a graph ofdegrees Celsius (on the horizontal axis) againstdegrees Fahrenheit (on the vertical scale).◦C 10 20 40 60 80 100◦F 50 68 104 140 176 212From the graph find (a) the temperature in degreesFahrenheit at 55◦C, (b) the temperature in degreesCelsius at 167◦F, (c) the Fahrenheit temperature at0◦C and (d) the Celsius temperature at 230◦FThe co-ordinates (10, 50), (20, 68), (40, 104), and soon are plotted as shown in Figure 17.17. When theco-ordinates are joined, a straight line is produced.Since a straight line results, there is a linear relationshipbetween degrees Celsius and degrees Fahrenheit.0403280120DegreesFahrenheit(8F)13116016720024023020 40 55Degrees Celsius (8C)7560 80 120110100yDEFGxABFigure 17.17(a) To find the Fahrenheit temperature at 55◦C, a verti-cal line AB is constructed from the horizontal axisto meet the straight line at B. The point where the
142 Basic Engineering Mathematicshorizontalline BD meets the vertical axis indicatesthe equivalent Fahrenheit temperature.Hence, 55◦C is equivalent to 131◦F.This process of finding an equivalent value inbetween the given information in the above tableis called interpolation.(b) To find the Celsius temperature at 167◦F, ahorizontal line EF is constructed as shown inFigure 17.17. The point where the vertical line FGcuts the horizontal axis indicates the equivalentCelsius temperature.Hence, 167◦F is equivalent to 75◦C.(c) If the graph is assumed to be linear even outside ofthe given data, the graph may be extended at bothends (shown by broken lines in Figure 17.17).From Figure 17.17, 0◦C corresponds to 32◦F.(d) 230◦F is seen to correspond to 110◦C.The process of finding equivalent values outsideof the given range is called extrapolation.Problem 13. In an experiment on Charles's law,the value of the volume of gas, V m3, was measuredfor various temperatures T◦C. The results areshown below.V m325.0 25.8 26.6 27.4 28.2 29.0T ◦C 60 65 70 75 80 85Plot a graph of volume (vertical) againsttemperature (horizontal) and from it find (a) thetemperature when the volume is 28.6m3 and (b) thevolume when the temperature is 67◦CIf a graph is plotted with both the scales starting at zerothen the result is as shown in Figure 17.18. All of thepoints lie in the top right-hand corner of the graph,making interpolation difficult. A more accurate graphis obtained if the temperature axis starts at 55◦C andthe volume axis starts at 24.5m3. The axes correspond-ing to these values are shown by the broken lines inFigure 17.18 and are called false axes, since the originis not now at zero. A magnified version of this relevantpart of the graph is shown in Figure 17.19. From thegraph,(a) When the volume is 28.6m3, the equivalent tem-perature is 82.5◦C.(b) When the temperature is 67◦C, the equivalentvolume is 26.1m3.30252015Volume(m3)1050 20 40 60 80 100Temperature (8C)yxFigure 17.182928.62827Volume(m3)262555 60 65 67 70 75 80 82.5 8526.1Temperature (ЊC)yxFigure 17.19Problem 14. In an experiment demonstratingHooke's law, the strain in an aluminium wire wasmeasured for various stresses. The results were:Stress (N/mm2) 4.9 8.7 15.0Strain 0.00007 0.00013 0.00021Stress (N/mm2) 18.4 24.2 27.3Strain 0.00027 0.00034 0.00039
Straight line graphs 143Plot a graph of stress (vertically) against strain(horizontally). Find (a) Young's modulus ofelasticity for aluminium, which is given by thegradient of the graph, (b) the value of the strain at astress of 20N/mm2 and (c) the value of the stresswhen the strain is 0.00020The co-ordinates (0.00007, 4.9), (0.00013, 8.7), andso on, are plotted as shown in Figure 17.20. Thegraph produced is the best straight line which can bedrawn corresponding to these points. (With experimen-tal results it is unlikely that all the points will lie exactlyon a straight line.) The graph, and each of its axes, arelabelled. Since the straight line passes through the ori-gin, stress is directly proportional to strain for the givenrange of values.Stress(N/mm2)0.000050481214162024280.00015 0.00025Strain0.0002850.00035yxABCFigure 17.20(a) The gradient of the straight line AC is given byABBC=28 − 70.00040 − 0.00010=210.00030=213 × 10−4=710−4= 7 × 104= 70000N/mm2Thus, Young's modulus of elasticity for alu-minium is70 000 N/mm2.Since1m2 = 106 mm2,70000N/mm2 is equivalent to 70000 ×106 N/m2,i.e. 70 ×109N/m2 (or pascals).From Figure 17.20,(b) The value of the strain at a stress of 20N/mm2 is0.000285(c) The value of the stress when the strain is 0.00020is 14 N/mm2.Problem 15. The following values of resistanceR ohms and corresponding voltage V volts areobtained from a test on a filament lamp.R ohms 30 48.5 73 107 128V volts 16 29 52 76 94Choose suitable scales and plot a graph with Rrepresenting the vertical axis and V the horizontalaxis. Determine (a) the gradient of the graph, (b) theR axis intercept value, (c) the equation of the graph,(d) the value of resistance when the voltage is 60Vand (e) the value of the voltage when the resistanceis 40 ohms. (f) If the graph were to continue in thesame manner, what value of resistance would beobtained at 110V?The co-ordinates (16, 30), (29, 48.5), and so on areshown plotted in Figure 17.21, where the best straightline is drawn through the points.147140120858060100ResistanceR(ohms)401020240 20 40 60 80 100 110 120Voltage V (volts)yCBAxFigure 17.21(a) The slope or gradient of the straight line AC isgiven byABBC=135 − 10100 − 0=125100= 1.25
144 Basic Engineering Mathematics(Note that the vertical line AB and the horizon-tal line BC may be constructed anywhere alongthe length of the straight line. However, calcula-tions are made easier if the horizontal line BC iscarefully chosen; in this case, 100.)(b) The R-axis intercept is at R = 10ohms (by extra-polation).(c) The equation of a straight line is y = mx + c, wheny is plotted on the vertical axis and x on the hori-zontal axis. m represents the gradient and c they-axis intercept. In this case, R corresponds to y,V corresponds to x, m = 1.25 and c = 10. Hence,the equation of the graph is R = (1.25V+ 10) .From Figure 17.21,(d) When the voltage is 60V, the resistance is 85 .(e) When theresistanceis40ohms,thevoltageis24V.(f) By extrapolation, when the voltage is 110V, theresistance is 147 .Problem 16. Experimental tests to determine thebreaking stress σ of rolled copper at varioustemperatures t gave the following results.Stress σ (N/cm2) 8.46 8.04 7.78Temperature t(◦C) 70 200 280Stress σ (N/cm2) 7.37 7.08 6.63Temperature t(◦C) 410 500 640Show that the values obey the law σ = at + b,where a and b are constants, and determineapproximate values for a and b. Use the law todetermine the stress at 250◦C and the temperaturewhen the stress is 7.54N/cm2.The co-ordinates (70, 8.46), (200, 8.04), and so on, areplotted as shown in Figure 17.22. Since the graph is astraight line then the values obey the law σ = at + b,and the gradient of the straight line isa =ABBC=8.36 − 6.76100 − 600=1.60−500= −0.0032Vertical axis intercept, b = 8.68Hence, the law of the graph is σ = 0.0032t + 8.68When the temperature is 250◦C, stress σ is given byσ = −0.0032(250) + 8.68 = 7.88N/cm2Temperature t (8C)Stress(N/cm2)8.688.368.508.007.507.006.500 100 200 300 400 500 600 7006.76yxCBAFigure 17.22Rearranging σ = −0.0032t + 8.68 gives0.0032t = 8.68 − σ, i.e. t =8.68 − σ0.0032Hence, when the stress, σ = 7.54N/cm2,temperature, t =8.68 − 7.540.0032= 356.3◦CNow try the following Practice ExercisePracticeExercise 69 Practical problemsinvolving straight line graphs (answers onpage 347)1. The resistance R ohms of a copper windingis measured at various temperatures t ◦C andthe results are as follows:R (ohms) 112 120 126 131 134t◦C 20 36 48 58 64Plot a graph of R (vertically) against t (hori-zontally) and find from it (a) the temperaturewhen the resistance is 122 and (b) theresistance when the temperature is 52◦C.2. The speed of a motor varies with armaturevoltage as shown by the following experi-mental results.
Straight line graphs 145n (rev/min) 285 517 615V (volts) 60 95 110n (rev/min) 750 917 1050V (volts) 130 155 175Plot a graph of speed (horizontally) againstvoltage(vertically) and draw the best straightline through the points. Find from the graph(a) the speed at a voltage of 145V and (b) thevoltage at a speed of 400rev/min.3. The following table gives the force Fnewtons which, when applied to a liftingmachine, overcomes a corresponding load ofL newtons.Force F (newtons) 25 47 64Load L (newtons) 50 140 210Force F (newtons) 120 149 187Load L (newtons) 430 550 700Choose suitable scales and plot a graph ofF (vertically) against L (horizontally).Drawthe best straight line through the points.Determine from the graph(a) the gradient,(b) the F-axis intercept,(c) the equation of the graph,(d) the force applied when the load is310N, and(e) the load that a force of 160N willovercome.(f) If the graph were to continue in thesame manner, what value of force willbe needed to overcome a 800N load?4. The following table gives the results of testscarried out to determine the breaking stressσ of rolled copper at varioustemperatures, t.Stress σ (N/cm2) 8.51 8.07 7.80Temperature t(◦C) 75 220 310Stress σ (N/cm2) 7.47 7.23 6.78Temperature t(◦C) 420 500 650Plot a graph of stress (vertically) againsttemperature (horizontally). Draw the beststraight linethrough the plotted co-ordinates.Determine the slope of the graph and thevertical axis intercept.5. The velocity v of a body after varying timeintervals t was measured as follows:t (seconds) 2 5 8v (m/s) 16.9 19.0 21.1t (seconds) 11 15 18v (m/s) 23.2 26.0 28.1Plot v vertically and t horizontally and drawa graph of velocity against time. Determinefrom the graph (a) the velocity after 10s,(b) the time at 20m/s and (c) the equationof the graph.6. The mass m of a steel joist varies with lengthL as follows:mass, m (kg) 80 100 120 140 160length, L (m) 3.00 3.74 4.48 5.23 5.97Plot a graph of mass (vertically) againstlength (horizontally). Determine the equa-tion of the graph.7. The crushing strength of mortar varies withthe percentage of water used in its prepara-tion, as shown below.Crushing strength, 1.67 1.40 1.13F (tonnes)% of water used, w% 6 9 12Crushing strength, 0.86 0.59 0.32F (tonnes)% of water used, w% 15 18 21
146 Basic Engineering MathematicsPlot a graph of F (vertically) against w(horizontally).(a) Interpolate and determine the crushingstrength when 10% water is used.(b) Assuming the graph continues in thesame manner, extrapolate and deter-mine the percentage of water used whenthe crushing strength is 0.15 tonnes.(c) What is the equation of the graph?8. In an experiment demonstrating Hooke'slaw,the strain in a copper wire was measured forvarious stresses. The results wereStress 10.6 × 106 18.2 × 106 24.0 × 106(pascals)Strain 0.00011 0.00019 0.00025Stress 30.7 × 106 39.4 × 106(pascals)Strain 0.00032 0.00041Plot a graph of stress (vertically) againststrain (horizontally). Determine(a) Young's modulus of elasticity for cop-per, which is given by the gradient ofthe graph,(b) the value of strain at a stress of21 × 106 Pa,(c) the value of stress when the strain is0.00030,9. An experiment with a set of pulley blocksgave the following results.Effort, E (newtons) 9.0 11.0 13.6Load, L (newtons) 15 25 38Effort, E (newtons) 17.4 20.8 23.6Load, L (newtons) 57 74 88Plot a graph of effort (vertically)against load(horizontally). Determine(a) the gradient,(b) the vertical axis intercept,(c) the law of the graph,(d) the effort when the load is 30N,(e) the load when the effort is 19N.10. The variation of pressure p in a vessel withtemperature T is believed to follow a law ofthe form p = aT + b, where a and b are con-stants. Verify this law for the results givenbelow and determine the approximate valuesof a and b. Hence, determine the pressuresat temperatures of 285K and 310K and thetemperature at a pressure of 250kPa.Pressure, p (kPa) 244 247 252Temperature, T (K) 273 277 282Pressure, p (kPa) 258 262 267Temperature, T (K) 289 294 300
Chapter 18Graphs reducing non-linearlaws to linear form18.1 IntroductionIn Chapter 17 we discovered that the equation of astraight line graph is of the form y = mx + c, wherem is the gradient and c is the y-axis intercept. Thischapterexplainshowthelawofagraph can still bedeter-mined even when it is not of thelinear form y = mx + c.The method used is called determination of law and isexplained in the following sections.18.2 Determination of lawFrequently, the relationship between two variables, sayx and y, is not a linear one; i.e., when x is plottedagainsty a curve results. In such cases the non-linear equationmay be modified to the linear form, y = mx + c, so thatthe constants, and thus the law relating the variables, canbe determined. This technique is called 'determinationof law'.Some examples of the reduction of equations to linearform include(i) y = ax2 + b compares with Y = mX + c, wherem = a, c = b and X = x2.Hence, y is plotted vertically against x2 horizon-tally to produce a straight line graph of gradienta and y-axis intercept b.(ii) y =ax+ b, i.e. y = a1x+ by is plotted vertically against1xhorizontally toproduce a straight line graph of gradient a andy-axis intercept b.(iii) y = ax2 + bxDividing both sides by x givesyx= ax + b.Comparing with Y = mX + c shows thatyxisplotted vertically against x horizontally to pro-duce a straight line graph of gradient a andyxaxis intercept b.Here are some worked problems to demonstrate deter-mination of law.Problem 1. Experimental values of x and y,shown below, are believed to be related by the lawy = ax2 + b. By plotting a suitable graph, verifythis law and determine approximate values ofa and bx 1 2 3 4 5y 9.8 15.2 24.2 36.5 53.0If y is plotted against x a curve results and it is notpossible to determine the values of constants a and bfrom the curve.Comparing y = ax2 + b with Y = mX + c shows thaty is to be plotted vertically against x2horizontally. Atable of values is drawn up as shown below.x 1 2 3 4 5x2 1 4 9 16 25y 9.8 15.2 24.2 36.5 53.0DOI: 10.1016/B978-1-85617-697-2.00018-1
154 Basic Engineering MathematicsNow try the following Practice ExercisePractice Exercise 71 Determination of lawinvolving logarithms (answers on page 348)In problems 1 to 3, x and y are two related vari-ables and all other letters denote constants. For thestated laws to be verified it is necessary to plotgraphs of the variables in a modified form. Statefor each, (a) what should be plotted on the verticalaxis, (b) what should be plotted on the horizon-tal axis, (c) the gradient and (d) the vertical axisintercept.1. y = bax 2. y = kxL 3.ym= enx4. The luminosity I of a lamp varies withthe applied voltage V and the relationshipbetween I and V is thought to be I = kV n.Experimental results obtained areI(candelas) 1.92 4.32 9.72V (volts) 40 60 90I(candelas) 15.87 23.52 30.72V (volts) 115 140 160Verify that the law is true and determinethe law of the graph. Also determine theluminosity when 75V is applied across thelamp.5. The head of pressure h and the flow veloc-ity v are measured and are believed to beconnected by the law v = ahb, where a andb are constants. The results are as shownbelow.h 10.6 13.4 17.2 24.6 29.3v 9.77 11.0 12.44 14.88 16.24Verify that the law is true and determine valuesof a and b.6. Experimental values of x and y are measuredas follows.x 0.4 0.9 1.2 2.3 3.8y 8.35 13.47 17.94 51.32 215.20The law relating x and y is believed to be ofthe form y = abx, where a and b are constants.Determine the approximate values of a and b.Hence, find the value of y when x is 2.0 andthe value of x when y is 100.7. The activity of a mixture of radioactive iso-topes is believed to vary according to the lawR = R0t−c, where R0 and c are constants.Experimental results are shown below.R 9.72 2.65 1.15 0.47 0.32 0.23t 2 5 9 17 22 28Verify that the law is true and determineapproximate values of R0 and c.8. Determine the law of the form y = aekxwhichrelates the following values.y 0.0306 0.285 0.841 5.21 173.2 1181x –4.0 5.3 9.8 17.4 32.0 40.09. The tension T in a belt passing round a pul-ley wheel and in contact with the pulley overan angle of θ radians is given by T = T0eμθ ,where T0 and μ are constants. Experimentalresults obtained areT (newtons) 47.9 52.8 60.3 70.1 80.9θ (radians) 1.12 1.48 1.97 2.53 3.06Determine approximate values of T0 and μ.Hence, find the tension when θ is 2.25 radiansand the value of θ when the tension is 50.0newtons.
Graphical solution of equations 157(a)yy5 x2x2121 0 1(b)y53x2yx2121 0 1yx2121 0 1(c)y 5 x221Figure 19.4Graphs of y = −x2, y = −3x2 and y = −12x2 areshown in Figure 19.5. All have maximum valuesat the origin (0, 0).(a)yy 52x2x21212201(b)y523x22122yx21012122yx2101(c)y52 x221Figure 19.5When y = ax2,(i) curves are symmetrical about the y-axis,(ii) the magnitude of a affects the gradient of thecurve, and(iii) the sign of a determines whether it has amaximum or minimum value.(b) y = ax2 + cGraphs of y = x2 + 3, y = x2 − 2, y = −x2 + 2and y = −2x2 − 1 are shown in Figure 19.6.When y = ax2 + c,(i) curves are symmetrical about the y-axis,(ii) the magnitude of a affects the gradient of thecurve, and(iii) the constant c is the y-axis intercept.(c) y = ax2 + bx + cWhenever b has a value other than zero the curveis displaced to the right or left of the y-axis.When b/a is positive, the curve is displaced b/2ato the left of the y-axis, as shown in Figure19.7(a).When b/a is negative, the curve is displacedb/2a to the right of the y-axis, as shown inFigure 19.7(b).Quadratic equations of the form ax2 + bx + c = 0may be solved graphically by(a) plotting the graph y = ax2 + bx + c, and(b) noting the points of intersection on the x-axis (i.e.where y = 0).yy 5x213y 52x21 2y 522x221y 5 x222x21 0(a)31yx2221 1(b)20yx21(c)021yx212124(d)01Figure 19.6(a)25 24 23 22 21 1yx12106420y5 x216x111(b)y 5x2 25x1421yx0246221 2 3 48Figure 19.7The x values of the points of intersection give therequired solutions since at these points both y = 0 andax2 + bx + c = 0.The number of solutions, or roots, of a quadratic equa-tion depends on how many times the curve cuts thex-axis. There can be no real roots, as in Figure 19.7(a),one root, as in Figures 19.4 and 19.5, or two roots, as inFigure 19.7(b).Here are some worked problems to demonstrate thegraphical solution of quadratic equations.Problem 3. Solve the quadratic equation4x2 + 4x − 15 = 0 graphically, given that thesolutions lie in the range x = −3 to x = 2.Determine also the co-ordinates and nature of theturning point of the curve
Revision Test 7 : GraphsThis assignment covers the material contained in Chapters 17–19. The marks available are shown in brackets at theend of each question.1. Determine the value of P in the following table ofvalues.x 0 1 4y = 3x − 5 −5 −2 P (2)2. Assuming graph paper measuring 20cm by 20cmis available, suggest suitable scales for the follow-ing ranges of values.Horizontal axis: 5N to 70N; vertical axis: 20mmto 190mm. (2)3. Corresponding valuesobtained experimentally fortwo quantities are:x −5 −3 −1 0 2 4y −17 −11 −5 −2 4 10Plot a graph of y (vertically) against x (hori-zontally) to scales of 1cm = 1 for the horizontalx-axis and 1cm = 2 for the vertical y-axis. Fromthe graph, find(a) the value of y when x = 3,(b) the value of y when x = −4,(c) the value of x when y = 1,(d) the value of x when y = −20. (8)4. If graphs of y against x were to be plotted for eachof the following, state (i) the gradient, and (ii) they-axis intercept.(a) y = −5x + 3 (b) y = 7x(c) 2y + 4 = 5x (d) 5x + 2y = 6(e) 2x −y3=76(10)5. Theresistance R ohmsofacopperwinding ismea-sured at various temperatures t◦C and the resultsare as follows.R ( ) 38 47 55 62 72t (◦C) 16 34 50 64 84Plot a graph of R (vertically) against t (horizon-tally) and find from it(a) the temperature when the resistance is 50 ,(b) the resistance when the temperature is 72◦C,(c) the gradient,(d) the equation of the graph. (10)6. x and y are two related variables and all otherletters denote constants. For the stated laws tobe verified it is necessary to plot graphs of thevariables in a modified form. State for each(a) what should be plotted on the vertical axis,(b) what should be plotted on the horizontal axis,(c) the gradient,(d) the vertical axis intercept.(i) y = p +rx2(ii) y =ax+ bx (4)7. The following results give corresponding valuesof two quantities x and y which are believed to berelated by a law of the form y = ax2 + bx wherea and b are constants.y 33.9 55.5 72.8 84.1 111.4 168.1x 3.4 5.2 6.5 7.3 9.1 12.4Verify the law and determine approximate valuesof a and b.Hence determine (i) the value of y when x is 8.0and (ii) the value of x when y is 146.5 (18)8. By taking logarithms of both sides of y = k xn ,show that lg y needs to be plotted vertically andlgx needs to be plotted horizontally to producea straight line graph. Also, state the gradient andvertical-axis intercept. (6)9. By taking logarithms of both sides of y = aek xshow that ln y needs to be plotted vertically andx needs to be plotted horizontally to produce astraight line graph. Also, state the gradient andvertical-axis intercept. (6)10. Show from the following results of voltage V andadmittance Y of an electrical circuit that the lawconnecting the quantities is of the form V =kYnand determine the values of k and n.VoltageV (volts) 2.88 2.05 1.60 1.22 0.96Admittance,Y(siemens) 0.52 0.73 0.94 1.23 1.57(12)
Chapter 20Angles and triangles20.1 IntroductionTrigonometry is a subject that involves the measurementof sides and angles of triangles and their relationship toeach other. This chapter involves the measurement ofangles and introduces types of triangle.20.2 Angular measurementAn angle is the amount of rotation between two straightlines. Angles may be measured either in degrees or inradians.If a circle is divided into 360 equal parts, then each partis called 1 degree and is written as 1◦i.e. 1 revolution = 360◦or 1 degree is1360th of a revolutionSome angles are given special names.• Any angle between 0◦and 90◦is called an acuteangle.• An angle equal to 90◦ is called a right angle.• Any angle between 90◦ and 180◦ is called an obtuseangle.• Any angle greater than 180◦ and less than 360◦ iscalled a reflex angle.• An angle of 180◦ lies on a straight line.• If two angles add up to 90◦ they are called comple-mentary angles.• If two angles add up to 180◦ they are called supple-mentary angles.• Parallel lines are straight lineswhich are in thesameplane and never meet. Such lines are denoted byarrows, as in Figure 20.1.• A straight line which crosses two parallel lines iscalled a transversal (see MN in Figure 20.1).PRMNgh efadbcQSFigure 20.1With reference to Figure 20.1,(a) a = c, b = d, e = g and f = h. Such pairs ofangles are called vertically opposite angles.(b) a = e, b = f , c = g and d = h. Such pairs ofangles are called corresponding angles.(c) c = e and b = h. Such pairs of angles are calledalternate angles.(d) b + e = 180◦ and c + h = 180◦. Such pairs ofangles are called interior angles.20.2.1 Minutes and secondsOne degree may be sub-divided into 60 parts, calledminutes.i.e. 1 degree = 60 minuteswhich is written as 1◦= 60 .DOI: 10.1016/B978-1-85617-697-2.00020-X
Angles and triangles 17120.3 TrianglesA triangle is a figure enclosed by three straight lines.The sum of the three angles of a triangle is equalto 180◦.20.3.1 Types of triangleAn acute-angled triangle is one in which all the anglesare acute; i.e., all the angles are less than 90◦. Anexample is shown in triangle ABC in Figure 20.15(a).A right-angled triangle is one which contains a rightangle; i.e., one in which one of the angles is 90◦. Anexample is shown in triangle DEF in Figure 20.15(b).678 598508408548AB C(a) (b)DE FFigure 20.15An obtuse-angled triangle is one which contains anobtuse angle; i.e., one angle which lies between 90◦and 180◦. An example is shown in triangle PQR inFigure 20.16(a).An equilateral triangleis one in which all the sides andall the angles are equal; i.e., each is 60◦. An example isshown in triangle ABC in Figure 20.16(b).PQ R1318278228(a)608608 608AB C(b)Figure 20.16An isosceles triangle is one in which two angles andtwo sides are equal. An example is shown in triangleEFG in Figure 20.17(a).A scalene triangle is one with unequal angles andtherefore unequal sides. An example of an acuteangled scalene triangle is shown in triangle ABC inFigure 20.17(b).(a) (b)ABC438828558FE G308758758Figure 20.17BACc baFigure 20.18With reference to Figure 20.18,(a) Angles A, B and C are called interior angles ofthe triangle.(b) Angle θ is called an exterior angle of the triangleand is equal to the sum of the two opposite interiorangles; i.e., θ = A + C.(c) a + b + c is called the perimeter of the triangle.AC BbacFigure 20.19
172 Basic Engineering MathematicsA right-angled triangle ABC is shown in Figure 20.19.The point of intersection of two lines is called a vertex(plural vertices); the three vertices of the triangle arelabelled as A, B and C, respectively. The right angleis angle C. The side opposite the right angle is giventhe special name of the hypotenuse. The hypotenuse,length AB in Figure 20.19, is always the longest side ofa right-angled triangle. With reference to angle B, ACis the opposite side and BC is called the adjacent side.With reference to angle A, BC is the opposite side andAC is the adjacent side.Often sides of a triangle are labelled with lower caseletters, a being the side opposite angle A,b being theside opposite angle B and c being the side oppositeangle C. So, in the triangle ABC, length AB = c, lengthBC = a and length AC = b. Thus, c is the hypotenusein the triangle ABC.∠ is the symbol used for 'angle'. For example, in thetriangle shown, ∠C = 90◦. Another way of indicatingan angle is to use all three letters. For example, ∠ABCactually means ∠B; i.e., we take the middle letter as theangle. Similarly, ∠BAC means ∠A and ∠ACB means∠C.Here are some worked examples to help us understandmore about triangles.Problem 18. Name the types of triangle shown inFigure 20.2022.62.82.52.52.12.139810785182(b)(d) (e)(c)(a)2Figure 20.20(a) Equilateral triangle (since all three sides areequal).(b) Acute-angled scalene triangle (since all theangles are less than 90◦).(c) Right-angled triangle (39◦ + 51◦ = 90◦; hence,the third angle must be 90◦, since there are 180◦in a triangle).(d) Obtuse-angled scalene triangle (since one of theangles lies between 90◦ and 180◦).(e) Isosceles triangle (since two sides are equal).Problem 19. In the triangle ABC shown inFigure 20.21, with reference to angle θ, which sideis the adjacent?BACFigure 20.21The triangle is right-angled; thus, side AC is thehypotenuse. With reference to angle θ, the opposite sideis BC. The remaining side, AB, is the adjacent side.Problem 20. In the triangle shown inFigure 20.22, determine angle θ568Figure 20.22The sum of the three angles of a triangle is equal to180◦.The triangle is right-angled. Hence,90◦ + 56◦ + ∠θ = 180◦from which, ∠θ = 180◦− 90◦− 56◦= 34◦.Problem 21. Determine the value of θ and α inFigure 20.23
174 Basic Engineering Mathematics2. Find the angles a to f in Figure 20.27.10581058def(c)328114°b c(b)578838 a(a)Figure 20.273. In the triangle DEF of Figure 20.28, whichside is the hypotenuse? With reference toangle D, which side is the adjacent?388DE FFigure 20.284. In triangle DEF of Figure 20.28, determineangle D.5. MNO is an isosceles triangle in whichthe unequal angle is 65◦ as shown inFigure 20.29. Calculate angle θ.658ONMPFigure 20.296. Determine ∠φ and ∠x in Figure 20.30.AB CEDx588198Figure 20.307. In Figure 20.31(a) and (b), find angles w,x, yand z. What is the name given to the types oftriangle shown in (a) and (b)?(a) (b)1108 1108x y708z2cm2cmFigure 20.318. Find the values of angles a to g inFigure 20.32(a) and (b).
Angles and triangles 175(a) (b)568299148419ab6881318cd egfFigure 20.329. Find the unknown angles a to k inFigure 20.33.9981258228 dbacke g hfijFigure 20.3310. Triangle ABC has a right angle at B and∠BAC is 34◦. BC is produced to D. If thebisectors of ∠ABC and ∠ACD meet at E,determine ∠BEC.11. Ifin Figure20.34 triangleBCD isequilateral,find the interior angles of triangle ABE.978ABCDEFigure 20.3420.4 Congruent trianglesTwo triangles are said to be congruent if they are equalin all respects; i.e., three angles and three sides in onetriangle are equal to three angles and three sides in theother triangle. Two triangles are congruent if(a) the three sides of one are equal to the three sidesof the other (SSS),(b) two sides of one are equal to two sides of the otherand the angles included by these sides are equal(SAS),(c) two angles of the one are equal to two angles ofthe other and any side of the first is equal to thecorresponding side of the other (ASA), or(d) their hypotenuses are equal and one other side ofone is equal to the corresponding side of the other(RHS).Problem 24. State which of the pairs of trianglesshown in Figure 20.35 are congruent and name theirsequence(c)MNPOQR(d)STUVXW(e)FD EBAC(a) (b)ABCE DF KLJHGIFigure 20.35(a) Congruent ABC, FDE (angle, side, angle; i.e.,ASA).(b) Congruent GIH, JLK (side, angle, side; i.e., SAS).(c) Congruent MNO, RQP (right angle, hypotenuse,side; i.e., RHS).(d) Not necessarily congruent. It is not indicated thatany side coincides.(e) Congruent ABC, FED (side, side, side; i.e.,SSS).
178 Basic Engineering MathematicsProblem 29. A rectangular shed 2m wide and3m high stands against a perpendicular building ofheight 5.5m. A ladder is used to gain access to theroof of the building. Determine the minimumdistance between the bottom of the ladder and theshedA side view is shown in Figure 20.43, where AFis the minimum length of the ladder. Since BDand CF are parallel, ∠ADB = ∠DFE (correspond-ing angles between parallel lines). Hence, trianglesBAD and EDF are similar since their angles are thesame.AB = AC − BC = AC − DE = 5.5 − 3 = 2.5mBy proportion:ABDE=BDEFi.e.2.53=2EFHence, EF = 232.5= 2.4m = minimum distancefrom bottom of ladder to the shed.3m2m 5.5mShedDE CBAFFigure 20.43Now try the following Practice ExercisePracticeExercise 80 Similar triangles(answers on page 349)1. In Figure 20.44, find the lengths x and y.111Њ32Њ32Њ 37Њ25.69mm4.74mm7.36mm14.58mmxyFigure 20.442. PQR is an equilateral triangle of side 4cm.When PQ and PR are produced to S and T,respectively, ST is found to be parallel withQR. If PS is 9cm, find the length of ST. Xis a point on ST between S and T such thatthe line PX is the bisector of ∠SPT. Find thelength of PX.3. In Figure 20.45, find(a) the length of BC when AB = 6cm,DE = 8cm and DC = 3cm,(b) the length of DE when EC = 2cm,AC = 5cm and AB = 10cm.D ECBAFigure 20.454. In Figure 20.46, AF = 8m,AB = 5m andBC = 3m. Find the length of BD.DECBA FFigure 20.46
Angles and triangles 17920.6 Construction of trianglesTo construct any triangle, the following drawing instru-ments are needed:(a) ruler and/or straight edge(b) compass(c) protractor(d) pencil.Here are some worked problems to demonstrate triangleconstruction.Problem 30. Construct a triangle whose sides are6cm,5cm and 3cmDFEC GBA 6cmFigure 20.47With reference to Figure 20.47:(i) Draw a straight line of any length and, with a pairof compasses, mark out 6cm length and label itAB.(ii) Set compass to 5cm and with centre at A describearc DE.(iii) Set compass to 3cm and with centre at B describearc FG.(iv) The intersection of the two curves at C is the ver-tex of the required triangle. Join AC and BC bystraight lines.It may be proved by measurement that the ratio of theangles of a triangle is not equal to the ratio of the sides(i.e., in this problem, the angle opposite the 3cm side isnot equal to half the angle opposite the 6cm side).Problem 31. Construct a triangle ABC such thata = 6cm,b = 3cm and ∠C = 60◦With reference to Figure 20.48:CAB a56cm608b53cmFigure 20.48(i) Draw a line BC,6cm long.(ii) Using a protractor centred at C, make an angle of60◦ to BC.(iii) From C measure a length of 3cm and label A.(iv) Join B to A by a straight line.Problem 32. Construct a triangle PQR given thatQR = 5cm,∠Q = 70◦ and ∠R = 44◦Q RPQ9R9708 4485cmFigure 20.49With reference to Figure 20.49:(i) Draw a straight line 5cm long and label it QR.(ii) Use a protractor centred at Q and make an angleof 70◦. Draw QQ .(iii) Use a protractor centred at R and make an angleof 44◦. Draw RR .(iv) The intersection of QQ and RR forms the vertexP of the triangle.Problem 33. Construct a triangle XYZ given thatXY = 5cm, the hypotenuse YZ = 6.5cm and∠X = 90◦
180 Basic Engineering MathematicsA9B A XQCP SZUVRYFigure 20.50With reference to Figure 20.50:(i) Draw a straight line 5cm long and label it XY.(ii) Produce XY any distance to B. With compass cen-tred at X make an arc at A and A . (The length XAand XA is arbitrary.) With compass centred at Adraw the arc PQ. With the same compass settingand centred at A , draw the arc RS. Join the inter-section of the arcs, C to X, and a right angle toXY is produced at X. (Alternatively, a protractorcan be used to construct a 90◦ angle.)(iii) Thehypotenuseisalwaysoppositetheright angle.Thus, YZ is opposite ∠X. Using a compasscentred at Y and set to 6.5cm, describe thearc UV.(iv) The intersection of the arc UV withXC produced,forms the vertex Z of the required triangle. JoinYZ with a straight line.Now try the following Practice ExercisePracticeExercise 81 Construction oftriangles (answers on page 349)In the following, construct the triangles ABC forthe given sides/angles.1. a = 8cm,b = 6cm and c = 5cm.2. a = 40mm,b = 60mm and C = 60◦.3. a = 6cm,C = 45◦ and B = 75◦.4. c = 4cm, A = 130◦ and C = 15◦.5. a = 90mm, B = 90◦,hypotenuse = 105mm.
Chapter 21Introduction to trigonometry21.1 IntroductionTrigonometry is a subject that involves the measurementof sides and angles of triangles and their relationship toeach other.The theorem of Pythagoras and trigonometric ratiosare used with right-angled triangles only. However,there are many practical examples in engineeringwhere knowledge of right-angled triangles is veryimportant.In this chapter, three trigonometric ratios – i.e. sine,cosine and tangent – are defined and then evaluatedusing a calculator. Finally, solving right-angled trian-gle problems using Pythagoras and trigonometric ratiosis demonstrated, together with some practical examplesinvolving angles of elevation and depression.21.2 The theorem of PythagorasThe theorem of Pythagoras states:In any right-angled triangle, the square of thehypotenuse is equal to the sum of the squares of theother two sides.In the right-angled triangle ABC shown in Figure 21.1,this meansb2= a2+ c2(1)BACabcFigure 21.1If the lengths of any two sides of a right-angled triangleare known, the length of the third side may be calculatedby Pythagoras' theorem.From equation (1): b = a2 + c2Transposing equation (1) for a gives a2 = b2 − c2, fromwhich a =√b2 − c2Transposing equation (1) for c gives c2 = b2 − a2, fromwhich c =√b2 − a2Here are some worked problems to demonstrate thetheorem of Pythagoras.Problem 1. In Figure 21.2, find the length of BCBACab54cmc53 cmFigure 21.2From Pythagoras, a2= b2+ c2i.e. a2= 42+ 32= 16 + 9 = 25Hence, a =√25 = 5cm.√25 = ±5 but in a practical example likethisan answerof a = −5cm has no meaning, so we take only thepositive answer.Thus a = BC = 5cm.DOI: 10.1016/B978-1-85617-697-2.00021-1
182 Basic Engineering MathematicsPQR is a 3, 4, 5 triangle. There are not many right-angled triangles which have integer values (i.e. wholenumbers) for all three sides.Problem 2. In Figure 21.3, find the length of EFE d FDf55cme513cmFigure 21.3By Pythagoras' theorem, e2= d2+ f 2Hence, 132= d2+ 52169 = d2+ 25d2= 169 − 25 = 144Thus, d =√144 = 12cmi.e. d = EF = 12cmDEF is a 5, 12, 13 triangle, another right-angledtriangle which has integer values for all three sides.Problem 3. Two aircraft leave an airfield at thesame time. One travels due north at an averagespeed of 300km/h and the other due west at anaverage speed of 220km/h. Calculate their distanceapart after 4hoursAfter 4 hours, the first aircraft has travelled4 × 300 = 1200km due northand the second aircraft has travelled4 × 220 = 880km due west,as shown in Figure 21.4. The distance apart after4hours = BC.ABNEWSC1200km880kmFigure 21.4From Pythagoras' theorem,BC2= 12002+ 8802= 1440000+ 774400 = 2214400and BC =√2214400 = 1488km.Hence, distance apart after 4 hours = 1488km.Now try the following Practice ExercisePracticeExercise 82 Theorem ofPythagoras (answers on page 350)1. Find the length of side x in Figure 21.5.41cm40cmxFigure 21.52. Find the length of side x in Figure 21.6(a).3. Find the length of side x in Figure 21.6(b),correct to 3 significant figures.25 m(a)7 m4.7 mm8.3 mm(b)xxFigure 21.64. In a triangle ABC, AB = 17cm, BC = 12 cmand ∠ABC = 90◦. Determine the length ofAC, correct to 2 decimal places.5. A tent peg is 4.0m away from a 6.0m hightent. What length of rope, correct to the
Introduction to trigonometry 183nearest centimetre, runs from the top of thetent to the peg?6. In a triangle ABC, ∠B is a right angle,AB = 6.92 cm and BC = 8.78cm. Find thelength of the hypotenuse.7. In a triangle CDE, D = 90◦,CD = 14.83mm and CE = 28.31mm.Determine the length of DE.8. Show that if a triangle has sides of 8, 15 and17cm it is right-angled.9. Triangle PQR is isosceles, Q being a rightangle. If the hypotenuse is 38.46cm find (a)the lengths of sides PQ and QR and (b) thevalue of ∠QPR.10. Aman cycles24kmduesouth and then 20kmdue east. Another man, starting at the sametime as the first man, cycles 32km due eastand then 7km due south. Find the distancebetween the two men.11. A ladder 3.5m long is placed against a per-pendicular wall with its foot 1.0m from thewall. How far up the wall (to the nearest cen-timetre) does the ladder reach? If the foot ofthe ladder is now moved 30cm further awayfrom the wall, how far does the top of theladder fall?12. Two ships leave a port at the same time. Onetravels due west at 18.4knots and the otherdue south at 27.6knots. If 1knot = 1 nauticalmile per hour, calculate how far apart the twoships are after 4 hours.13. Figure 21.7 shows a bolt rounded off at oneend. Determine the dimension h.R5 45mmhr516mmFigure 21.714. Figure 21.8 shows a cross-section of a com-ponent that is to be made from a round bar.If the diameter of the bar is 74mm, calculatethe dimension x.72mm74mmxFigure 21.821.3 Sines, cosines and tangentsWith reference to angle θ in the right-angled triangleABC shown in Figure 21.9,sineθ =opposite sidehypotenuse'Sine' is abbreviated to 'sin', thus sinθ =BCACHypotenuseAdjacentABCOppositeFigure 21.9Also, cosineθ =adjacent sidehypotenuse'Cosine' is abbreviated to 'cos', thus cosθ =ABACFinally, tangentθ =opposite sideadjacent side'Tangent' is abbreviated to 'tan', thus tanθ =BCABThese three trigonometric ratios only apply to right-angled triangles. Remembering these three equationsis very important and the mnemonic 'SOH CAH TOA'is one way of remembering them.
Introduction to trigonometry 1913. A ladder rests against the top of the perpendi-cular wall of a building and makes an angle of73◦ with the ground. If the foot of the ladder is2m from the wall, calculate the height of thebuilding.4. Determine the length x in Figure 21.26.x56810mmFigure 21.2621.6 Angles of elevation anddepressionIf, in Figure21.27, BC represents horizontal ground andAB a vertical flagpole, the angle of elevation of the topof the flagpole, A, from the point C is the angle that theimaginary straight line AC must be raised (or elevated)from the horizontal CB; i.e., angle θ.ABCFigure 21.27PQ RFigure 21.28If, in Figure 21.28, PQ represents a vertical cliff andR a ship at sea, the angle of depression of the shipfrom point P is the angle through which the imaginarystraight line PR must be lowered (or depressed) fromthe horizontal to the ship; i.e., angle φ. (Note, ∠PRQ isalso φ − alternate angles between parallel lines.)Problem 23. An electricity pylon stands onhorizontal ground. At a point 80m from the base ofthe pylon, the angle of elevation of the top of thepylon is 23◦. Calculate the height of the pylon to thenearest metreFigure 21.29 shows the pylon AB and the angle ofelevation of A from point C is 23◦.80 m238ABCFigure 21.29tan23◦=ABBC=AB80Hence, height of pylon AB = 80tan23◦= 80(0.4245) = 33.96m= 34m to the nearest metre.Problem 24. A surveyor measures the angle ofelevation of the top of a perpendicular building as19◦. He moves 120m nearer to the building andfinds the angle of elevation is now 47◦. Determinethe height of the buildingThe building PQ and the angles of elevation are shownin Figure 21.30.PQhxRS120478 198Figure 21.30In triangle PQS, tan19◦ =hx + 120Hence, h = tan19◦(x + 120)i.e. h = 0.3443(x + 120) (1)In triangle PQR, tan47◦=hxHence, h = tan47◦(x) i.e. h = 1.0724x (2)Equating equations (1) and (2) gives0.3443(x + 120) = 1.0724x0.3443x + (0.3443)(120) = 1.0724x(0.3443)(120) = (1.0724 − 0.3443)x41.316 = 0.7281xx =41.3160.7281= 56.74m
192 Basic Engineering MathematicsFrom equation (2), height of building,h = 1.0724x = 1.0724(56.74) = 60.85m.Problem 25. The angle of depression of a shipviewed at a particular instant from the top of a 75mvertical cliff is 30◦. Find the distance of the shipfrom the base of the cliff at this instant. The ship issailing away from the cliff at constant speed and 1minute later its angle of depression from the top ofthe cliff is 20◦. Determine the speed of the ship inkm/hFigure 21.31 shows the cliff AB, the initial position ofthe ship at C and the final position at D. Since the angleof depression is initially 30◦, ∠ACB = 30◦ (alternateangles between parallel lines).x75 m308208308208AB DCFigure 21.31tan30◦=ABBC=75BChence,BC =75tan30◦= 129.9m = initial positionof ship from base of cliffIn triangle ABD,tan20◦ =ABBD=75BC + CD=75129.9 + xHence, 129.9 + x =75tan20◦= 206.06mfrom which x = 206.06 − 129.9 = 76.16mThus, the ship sails 76.16m in 1 minute; i.e., 60s,Hence, speed of ship =distancetime=76.1660m/s=76.16 × 60× 6060 × 1000km/h = 4.57km/h.Now try the following Practice ExercisePracticeExercise 86 Angles of elevationand depression (answers on page 349)1. A vertical tower stands on level ground. Ata point 105m from the foot of the tower theangle of elevation of the top is 19◦. Find theheight of the tower.2. If the angle of elevation of the top of a vertical30m high aerial is 32◦, how far is it to theaerial?3. From the top of a vertical cliff 90.0m highthe angle of depression of a boat is 19◦50 .Determine the distance of the boat from thecliff.4. From the top of a vertical cliff 80.0m high theangles of depression of two buoys lying duewest of the cliff are 23◦and 15◦, respectively.How far apart are the buoys?5. From a point on horizontal ground a surveyormeasures the angle of elevation of the top ofa flagpole as 18◦40 . He moves 50m nearerto the flagpole and measures the angle of ele-vation as 26◦22 . Determine the height of theflagpole.6. A flagpole stands on the edge of the top of abuilding. At a point 200m from the buildingthe angles of elevation of the top and bot-tom of the pole are 32◦ and 30◦ respectively.Calculate the height of the flagpole.7. From a ship at sea, the angles of elevation ofthe top and bottom of a vertical lighthousestanding on the edge of a vertical cliff are31◦and 26◦, respectively. If the lighthouse is25.0m high, calculate the height of the cliff.8. From a window 4.2m above horizontal groundthe angle of depression of the foot of a buildingacrosstheroad is24◦ and theangleofelevationof the top of the building is 34◦. Determine,correct to the nearest centimetre, the width ofthe road and the height of the building.9. The elevation of a tower from two points, onedue west of the tower and the other due eastof it are 20◦ and 24◦, respectively, and the twopoints of observation are 300m apart. Find theheight of the tower to the nearest metre.
196 Basic Engineering Mathematics22.2 Angles of any magnitudeFigure 22.2 shows rectangular axes XX and YYintersecting at origin 0. As with graphical work, mea-surements made to the right and above 0 are positive,while those to the left and downwards are negative.Let 0A be free to rotate about 0. By convention,when 0A moves anticlockwise angular measurement isconsidered positive, and vice versa.360827081808908X9 XY9Y08AQuadrant 2Quadrant 3 Quadrant 4Quadrant 102 11212Figure 22.2Let 0A be rotated anticlockwise so that θ1 is anyangle in the first quadrant and let perpendicular ABbe constructed to form the right-angled triangle 0ABin Figure 22.3. Since all three sides of the trian-gle are positive, the trigonometric ratios sine, cosineand tangent will all be positive in the first quadrant.(Note: 0A is always positive since it is the radius of acircle.)Let 0A be further rotated so that θ2 is any angle in thesecond quadrant and let AC be constructed to form the1808908270836080823 41Quadrant 2Quadrant 4Quadrant 3Quadrant 10ABAACDEA22 2111111 1Figure 22.3right-angled triangle 0AC. Then,sinθ2 =++= + cosθ2 =−+= −tanθ2 =+−= −Let 0A be further rotated so that θ3 is any angle in thethird quadrant and let AD be constructed to form theright-angled triangle 0AD. Then,sinθ3 =−+= − cosθ3 =−+= −tanθ3 =−−= +Let 0A be further rotated so that θ4 is any angle in thefourth quadrant and let AE be constructed to form theright-angled triangle 0AE. Then,sinθ4 =−+= − cosθ4 =++= +tanθ4 =−+= −The above results are summarized in Figure 22.4, inwhich all three trigonometric ratios are positive in thefirst quadrant, only sine is positive in the second quad-rant, only tangent is positive in the third quadrant andonly cosine is positive in the fourth quadrant.The underlined letters in Figure 22.4 spell the wordCAST when starting in the fourth quadrant and movingin an anticlockwise direction.90818082708360808SineTangent CosineAll positiveFigure 22.4It is seen that, in the first quadrant of Figure 22.1,all of the curves have positive values; in the second onlysine is positive; in the third only tangent is positive;and in the fourth only cosine is positive – exactly assummarized in Figure 22.4.
Trigonometric waveforms 197A knowledge of angles of any magnitude is neededwhen finding, for example, all the angles between 0◦and 360◦ whose sine is, say, 0.3261. If 0.3261 is enteredinto a calculator and then the inverse sine key pressed (orsin−1 key) the answer 19.03◦ appears. However, thereis a second angle between 0◦ and 360◦ which the cal-culator does not give. Sine is also positive in the secondquadrant (either from CAST or from Figure 22.1(a)).The other angle is shown in Figure 22.5 as angle θ,where θ = 180◦ − 19.03◦ = 160.97◦. Thus, 19.03◦and 160.97◦ are the angles between 0◦ and 360◦ whosesine is 0.3261 (check that sin160.97◦ = 0.3261 on yourcalculator).19.038 19.03818082708360808908S AT CFigure 22.5Be careful! Your calculator only gives you one ofthese answers. The second answer needs to be deducedfrom a knowledge of angles of any magnitude, as shownin the following worked problems.Problem 1. Determine all of the angles between0◦ and 360◦ whose sine is −0.4638The angles whose sine is −0.4638 occur in the thirdand fourth quadrants since sine is negative in thesequadrants – see Figure 22.6.1.021.020.46380 908 1808 2708 3608332.378207.638xy 5sinxyFigure 22.6From Figure 22.7, θ = sin−1 0.4638 = 27.63◦. Mea-sured from 0◦, the two angles between 0◦ and 360◦whose sine is −0.4638 are 180◦+ 27.63◦i.e. 207.63◦and 360◦ – 27.63◦, i.e. 332.37◦. (Note that if a calcu-lator is used to determine sin−1(−0.4638) it only givesone answer: −27.632588◦.)TS AC90818082708360808Figure 22.7Problem 2. Determine all of the angles between0◦ and 360◦ whose tangent is 1.7629A tangent is positive in the first and third quadrants –see Figure 22.8.1.762960.448 240.4480 360827081808908y 5tan xyxFigure 22.8From Figure 22.9, θ = tan−1 1.7629 = 60.44◦. Mea-sured from 0◦, the two angles between 0◦ and 360◦whose tangent is 1.7629 are 60.44◦and 180◦+ 60.44◦,i.e. 240.44◦18082708360890808CTS AFigure 22.9Problem 3. Solve the equationcos−1(−0.2348) = α for angles of α between 0◦and 360◦
Trigonometric waveforms 19920.50.521.01.0ySRT S9O9y 5cosxAngle x8308 608 1208 1808 2408 3008 36084580815833083158285825580822582108180815081208908608OFigure 22.1222.3.1 Sine wavesThe vertical component TS may be projected across toT S , which is the corresponding value of 30◦ on thegraph of y against angle x◦. If all such vertical compo-nents as TS are projected on to the graph, a sine waveis produced as shown in Figure 22.11.22.3.2 Cosine wavesIf all horizontal components such as OS are projectedon to a graph of y against angle x◦, a cosine wave isproduced. It is easier to visualize these projections byredrawing the circle with the radius arm OR initially ina vertical position as shown in Figure 22.12.It is seen from Figures 22.11 and 22.12 that a cosinecurve is of the same form as the sine curve but is dis-placed by 90◦ (or π/2 radians). Both sine and cosinewaves repeat every 360◦.22.4 Terminology involved with sineand cosine wavesSine waves are extremely important in engineering, withexamples occurring with alternating currents and volt-ages – the mains supply is a sine wave – and with simpleharmonic motion.22.4.1 CycleWhen a sine wave has passed through a completeseries of values, both positive and negative, it is saidto have completed one cycle. One cycle of a sinewave is shown in Figure 22.1(a) on page 195 and inFigure 22.11.22.4.2 AmplitudeThe amplitude is the maximum value reached in a halfcycle by a sine wave. Another name for amplitude ispeak value or maximum value.A sine wave y = 5sinx has an amplitude of 5, asine wave v = 200sin314t has an amplitude of 200 andthe sine wave y = sin x shown in Figure 22.11 has anamplitude of 1.22.4.3 PeriodThe waveforms y = sin x and y = cosx repeat them-selves every 360◦. Thus, for each, the period is 360◦.A waveform of y = tan x has a period of 180◦ (fromFigure 22.1(c)).A graph of y = 3sin2A, as shown in Figure 22.13, hasan amplitude of 3 and period 180◦.A graph of y = sin3A, as shown in Figure 22.14, has anamplitude of 1 and period of 120◦.A graph of y = 4cos2x, as shown in Figure 22.15, hasan amplitude of 4 and a period of 180◦.In general, if y = Asinpx or y = Acospx,amplitude = A and period =360◦py3230 A8y 53 sin 2A2708 36081808908Figure 22.13
200 Basic Engineering Mathematicsy1.0Ϫ1.00 90Њ 270Њ AЊ180Њ 360Њyϭsin 3AFigure 22.14y90Њ 180Њ 270Њ 360Њ xЊ0Ϫ44 y ϭ4 cos 2xFigure 22.15Problem 4. Sketch y = 2sin35A over one cycleAmplitude = 2; period =360◦35=360◦ × 53= 600◦A sketch of y = 2sin35A is shown in Figure 22.16.180Њ 360Њ 540Њ 600ЊyAЊ0Ϫ22yϭ2 sin A35Figure 22.1622.4.4 Periodic timeIn practice, the horizontal axis of a sine wave will betime. The time taken for a sine wave to complete onecycle is called the periodic time, T.In the sine wave of voltage v (volts) against time t (mil-liseconds) shown in Figure 22.17, the amplitude is 10Vand the periodic time is 20ms; i.e., T = 20 ms.v10 20 t (ms)021010Figure 22.1722.4.5 FrequencyThe number of cycles completed in one second is calledthe frequency f and is measured in hertz, Hz.f =1Tor T =1fProblem 5. Determine the frequency of the sinewave shown in Figure 22.17In the sine wave shown in Figure 22.17, T = 20 ms,hencefrequency, f =1T=120 × 10−3= 50HzProblem 6. If a waveform has a frequency of200kHz, determine the periodic timeIf a waveform has a frequency of 200kHz, the periodictime T is given byperiodic time, T =1f=1200 × 103= 5 × 10−6s = 5μs22.4.6 Lagging and leading anglesA sine or cosine curve may not always start at 0◦.To show this, a periodic function is represented byy = Asin(x ± α) where α is a phase displacementcompared with y = Asin x. For example, y = sin A isshown by the broken line in Figure 22.18 and, onthe same axes, y = sin(A − 60◦) is shown. The graphy = sin(A − 60◦) is said to lag y = sinA by 60◦.In another example, y = cos A is shown by the bro-ken line in Figure 22.19 and, on the same axes, y =cos(A+45◦) is shown. The graph y = cos(A+45◦) issaid to lead y = cosA by 45◦.
204 Basic Engineering Mathematics3. v = 300sin(200πt − 0.412)V4. A sinusoidal voltage has a maximum value of120V and a frequency of 50Hz. At time t = 0,the voltage is (a) zero and (b) 50V. Expressthe instantaneous voltage v in the formv = Asin(ωt ± α).5. An alternating current has a periodic time of25ms and a maximum value of 20A. Whentime = 0, current i = −10 amperes. Expressthe current i in the form i = Asin(ωt ± α).6. An oscillating mechanismhasamaximumdis-placement of 3.2m and a frequency of 50Hz.At time t = 0 the displacement is 150cm.Express the displacement in the general formAsin(ωt ± α)7. The current in an a.c. circuit at any time tseconds is given byi = 5sin(100πt − 0.432) amperesDetermine the(a) amplitude, frequency, periodic time andphase angle (in degrees),(b) value of current at t = 0,(c) value of current at t = 8ms,(d) time when the current is first a maximum,(e) time when the current first reaches 3A.Also,(f ) sketch one cycle of the waveform show-ing relevant points.
Chapter 23Non-right-angled trianglesand some practicalapplications23.1 The sine and cosine rulesTo 'solve a triangle' means 'to find the values ofunknown sides and angles'. If a triangle is right-angled,trigonometric ratios and the theorem of Pythagorasmay be used for its solution, as shown in Chapter 21.However, for a non-right-angled triangle, trigono-metric ratios and Pythagoras' theorem cannot be used.Instead, two rules, called the sine rule and the cosinerule, are used.23.1.1 The sine ruleWith reference to triangle ABC of Figure 23.1, the sinerule statesasinA=bsinB=csinCabcB CAFigure 23.1The rule may be used only when(a) 1 side and any 2 angles are initially given, or(b) 2 sides and an angle (not the included angle) areinitially given.23.1.2 The cosine ruleWith reference to triangle ABC of Figure 23.1, thecosine rule statesa2= b2+ c2− 2bc cos Aor b2= a2+ c2− 2ac cos Bor c2= a2+ b2− 2ab cosCThe rule may be used only when(a) 2 sides and the included angle are initiallygiven, or(b) 3 sides are initially given.23.2 Area of any triangleThe area of any triangle such as ABC of Figure 23.1is given by(a)12× base × perpendicular heightDOI: 10.1016/B978-1-85617-697-2.00023-5
210 Basic Engineering Mathematics458V25100VV1540 VABOFigure 23.11Angle OBA = 180◦− 45◦= 135◦Applying the cosine rule,OA2= V 21 + V 22 − 2V1V2 cosOBA= 402+ 1002− {2(40)(100)cos135◦}= 1600 + 10000− {−5657}= 1600 + 10000+ 5657 = 17257Thus, resultant,OA =√17257 = 131.4 VApplying the sine rule131.4sin135◦=100sinAOBfrom which sinAOB =100sin135◦131.4= 0.5381Hence, angle AOB = sin−1 0.5381 = 32.55◦ (or147.45◦, which is not possible)Hence, the resultant voltage is 131.4 volts at 32.55◦to V1Problem 10. In Figure 23.12, PR represents theinclined jib of a crane and is 10.0 m long. PQ is4.0 m long. Determine the inclination of the jib tothe vertical and the length of tie QR.PQR12084.0 m 10.0mFigure 23.12Applying the sine rule,PRsin120◦=PQsin Rfrom which sin R =PQsin120◦PR=(4.0)sin 120◦10.0= 0.3464Hence, ∠R = sin−10.3464 = 20.27◦(or 159.73◦,which is not possible)∠P = 180◦ − 120◦ − 20.27◦ = 39.73◦, which is theinclination of the jib to the verticalApplying the sine rule,10.0sin120◦=QRsin39.73◦from which length of tie,QR =10.0sin39.73◦sin120◦= 7.38mNow try the following Practice ExercisePracticeExercise 92 Practical situationsinvolving trigonometry (answers onpage 350)1. A ship P sails at a steady speed of 45km/hin a direction of W 32◦ N (i.e. a bearing of302◦) from a port. At the same time anothership Q leaves the port at a steady speed of35km/h in a direction N 15◦ E (i.e. a bearingof 015◦). Determine their distance apart after4 hours.2. Two sides of a triangular plot of land are52.0 m and 34.0 m, respectively. If the areaof the plot is 620 m2, find (a) the lengthof fencing required to enclose the plot and(b) the angles of the triangular plot.3. A jib crane is shown in Figure 23.13. If thetie rod PR is 8.0 m long and PQ is 4.5m long,determine (a) the length of jib RQ and (b) theangle between the jib and the tie rod.R1308 PQFigure 23.134. A building site is in the form of a quadri-lateral, as shown in Figure 23.14, and itsarea is 1510 m2. Determine the length of theperimeter of the site.
Non-right-angled triangles and some practical applications 21128.5m34.6m72852.4m 758Figure 23.145. Determine the length of members BF and EBin the roof truss shown in Figure 23.15.4 m2.5 m5084m2.5m5085 m 5 mEA B CDFFigure 23.156. A laboratory 9.0 m wide has a span roof whichslopes at 36◦on one side and 44◦on the other.Determine the lengths of the roof slopes.7. PQ and QR are the phasors representing thealternating currents in two branches of a cir-cuit. Phasor PQ is 20.0 A and is horizontal.Phasor QR (which is joined to the end of PQto form triangle PQR) is 14.0 A and is at anangle of 35◦ to the horizontal. Determine theresultant phasor PR and the angle it makeswith phasor PQ.23.6 Further practical situationsinvolving trigonometryProblem 11. A vertical aerial stands onhorizontal ground. A surveyor positioned due eastof the aerial measures the elevation of the top as48◦. He moves due south 30.0 m and measures theelevation as 44◦. Determine the height of the aerialIn Figure 23.16, DC represents the aerial, A is the initialposition of the surveyor and B his final position.From triangle ACD, tan48◦ =DCACfrom whichAC =DCtan48◦Similarly, from triangle BCD, BC =DCtan44◦BCDA30.0 m448488Figure 23.16For triangle ABC, using Pythagoras' theorem,BC2= AB2+ AC2DCtan44◦2= (30.0)2+DCtan48◦2DC2 1tan2 44◦−1tan2 48◦= 30.02DC2(1.072323 − 0.810727) = 30.02DC2=30.020.261596= 3440.4Hence, height of aerial, DC =√3340.4 = 58.65 m.Problem 12. A crank mechanism of a petrolengine is shown in Figure 23.17. Arm OA is10.0 cm long and rotates clockwise about O. Theconnecting rod AB is 30.0 cm long and end B isconstrained to move horizontally.(a) For the position shown in Figure 23.17,determine the angle between the connectingrod AB and the horizontal, and the lengthof OB.(b) How far does B move when angle AOBchanges from 50◦ to 120◦?508O30.0cm10.0 cmABFigure 23.17
Non-right-angled triangles and some practical applications 213BCA408Figure 23.204. From Figure 23.20, determine how far Cmoves, correct to the nearest millimetre, whenangle CAB changes from 40◦ to 160◦, Bmoving in an anticlockwise direction.5. A surveyor standing W 25◦S of a tower mea-sures the angle of elevation of the top ofthe tower as 46◦30 . From a position E 23◦Sfrom the tower the elevation of the top is37◦15 .Determinetheheight ofthetowerifthedistance between the two observationsis75m.6. Calculate, correct to 3 significant figures, theco-ordinates x and y to locate the hole centreat P shown in Figure 23.21.100 mm1168 1408PxyFigure 23.217. An idler gear, 30 mm in diameter, has to befitted between a 70 mm diameter driving gearand a 90 mm diameter driven gear, as shownin Figure 23.22. Determine the value of angleθ between the centre lines.90mmdia30mmdia70mmdia99.78mmFigure 23.228. 16 holes are equally spaced on a pitch circleof 70 mm diameter. Determine the length ofthe chord joining the centres of two adjacentholes.
Chapter 24Cartesian and polarco-ordinates24.1 IntroductionThere are two ways in which the position of a point ina plane can be represented. These are(a) Cartesian co-ordinates, i.e. (x, y).(b) Polar co-ordinates, i.e. (r, θ), where r is a radiusfrom a fixed point and θ is an angle from a fixedpoint.24.2 Changing from Cartesian topolar co-ordinatesIn Figure 24.1, if lengths x and y are known thenthe length of r can be obtained from Pythagoras' the-orem (see Chapter 21) since OPQ is a right-angledtriangle.yPQ xOxr yFigure 24.1Hence, r2 = (x2 + y2), from which r = x2 + y2From trigonometric ratios (see Chapter 21), tanθ =yxfrom which θ = tan−1 yxr = x2 + y2 and θ = tan−1 yxare the two formulaewe need to change from Cartesian to polar co-ordinates.The angle θ, which may be expressed in degrees or radi-ans, must always be measured from the positive x-axis;i.e., measured from the line OQ in Figure 24.1. It issuggested that when changing from Cartesian to polarco-ordinates a diagram should always be sketched.Problem 1. Change the Cartesian co-ordinates(3, 4) into polar co-ordinatesA diagram representing the point (3, 4) is shown inFigure 24.2.P43yxOrFigure 24.2From Pythagoras' theorem, r =√32 + 42 = 5 (notethat −5 has no meaning in this context).By trigonometric ratios, θ = tan−1 43= 53.13◦ or0.927rad.Note that 53.13◦ = 53.13 ×π180rad = 0.927rad.DOI: 10.1016/B978-1-85617-697-2.00024-7
Cartesian and polar co-ordinates 217Now try the following Practice ExercisePracticeExercise 95 Changing polar toCartesian co-ordinates (answers onpage 350)In problems 1 to 8, express the given polar co-ordinates as Cartesian co-ordinates, correct to 3decimal places.1. (5, 75◦) 2. (4.4, 1.12rad)3. (7, 140◦) 4. (3.6, 2.5rad)5. (10.8, 210◦) 6. (4, 4rad)7. (1.5, 300◦) 8. (6, 5.5rad)9. Figure 24.10 shows 5 equally spaced holeson an 80mm pitch circle diameter. Calculatetheir co-ordinates relative to axes Ox and Oyin (a) polar form, (b) Cartesian form.10. In Figure 24.10, calculate the shortest dis-tance between the centres of two adjacentholes.yxOFigure 24.1024.4 Use of Pol/Rec functions oncalculatorsAnother name for Cartesian co-ordinates is rectangularco-ordinates. Many scientific notation calculators havePol and Rec functions. 'Rec' is an abbreviation of 'rect-angular' (i.e. Cartesian) and 'Pol' is an abbreviation of'polar'. Check the operation manual for your particularcalculator to determine how to use these two func-tions. They make changing from Cartesian to polar co-ordinates, and vice-versa, so much quicker and easier.For example, with the Casio fx-83ES calculator, or sim-ilar, to change the Cartesian number (3, 4) into polarform, the following procedure is adopted.1. Press 'shift' 2. Press 'Pol' 3. Enter 34. Enter 'comma' (obtained by 'shift' then ) )5. Enter 4 6. Press ) 7. Press =The answer is r = 5,θ = 53.13◦Hence, (3, 4) in Cartesian form is the same as(5, 53.13◦) in polar form.If the angle is required in radians, then before repeatingthe above procedure press 'shift', 'mode' and then 4 tochange your calculator to radian mode.Similarly, to change the polar form number (7, 126◦)into Cartesian or rectangular form, adopt the followingprocedure.1. Press 'shift' 2. Press 'Rec'3. Enter 7 4. Enter 'comma'5. Enter 126 (assuming your calculator is in degreesmode)6. Press ) 7. Press =The answer is X = −4.11 and, scrolling across,Y = 5.66, correct to 2 decimal places.Hence, (7, 126◦) in polar form is the same as(−4.11, 5.66) in rectangular or Cartesian form.Now return to Practice Exercises 94 and 95 inthis chapter and use your calculator to determine theanswers, and see how much more quickly they may beobtained.
Revision Test 9 : Trigonometric waveforms and practical trigonometryThis assignment covers the material contained in Chapters 22–24. The marks available are shown in brackets at theend of each question.1. A sine wave is given by y = 8sin4x. State itspeakvalue and its period, in degrees. (2)2. A periodic function is given by y = 15tan2x.State its period in degrees. (2)3. The frequency of a sine wave is 800Hz. Calculatethe periodic time in milliseconds. (2)4. Calculate the frequency of a sine wave that has aperiodic time of 40μs. (2)5. Calculate the periodic time for a sine wave havinga frequency of 20kHz. (2)6. An alternating current completes 12 cycles in16ms. What is its frequency? (3)7. A sinusoidal voltage is given bye = 150sin(500πt − 0.25) volts. Determine the(a) amplitude,(b) frequency,(c) periodic time,(d) phase angle (stating whether it is leading orlagging 150sin500πt). (4)8. Determinetheacuteanglesin degrees,degreesandminutes, and radians.(a) sin−1 0.4721 (b) cos−1 0.8457(c) tan−1 1.3472 (9)9. Sketch the following curves, labelling relevantpoints.(a) y = 4cos(θ + 45◦) (b) y = 5sin(2t − 60◦)(8)10. The current in an alternating current cir-cuit at any time t seconds is given byi = 120sin(100πt + 0.274) amperes. Determine(a) the amplitude, frequency, periodic time andphase angle (with reference to 120sin100πt),(b) the value of current when t = 0,(c) the value of current when t = 6ms.Sketch one cycle of the oscillation. (16)11. A triangular plot of land ABC is shown inFigure RT9.1. Solve the triangle and determineits area. (10)ABC15m15.4m718Figure RT9.112. A car is travelling 20m above sea level. It thentravels 500m up a steady slope of 17◦. Determine,correct to the nearest metre, how high the car isnow above sea level. (3)13. Figure RT9.2 shows a roof truss PQR with rafterPQ = 3m. Calculate the length of(a) the roof rise PP ,(b) rafter PR,(c) the roof span QR.Find also (d) the cross-sectional area of the rooftruss. (11)PQ RP93m408 328Figure RT9.214. Solve triangle ABC given b = 10cm,c = 15cmand ∠A = 60◦. (10)15. Change the following Cartesian co-ordinates intopolar co-ordinates, correct to 2 decimal places, inboth degrees and in radians.(a) (−2.3,5.4) (b) (7.6,−9.2) (10)16. Change the following polar co-ordinates intoCartesian co-ordinates, correct to 3 decimalplaces.(a) (6.5,132◦) (b) (3,3rad) (6)
Chapter 25Areas of common shapes25.1 IntroductionArea is a measure of the size or extent of a plane surface.Area ismeasured in square units such as mm2, cm2 andm2. This chapter deals with finding the areas of commonshapes.In engineering it is often important to be able to cal-culate simple areas of various shapes. In everyday lifeits important to be able to measure area to, say, lay a car-pet, order sufficient paint for a decorating job or ordersufficient bricks for a new wall.On completing this chapter you will be able to recog-nize common shapes and be able to find the areas of rect-angles, squares, parallelograms, triangles, trapeziumsand circles.25.2 Common shapes25.2.1 PolygonsA polygon is a closed plane figure bounded by straightlines. A polygon which has3 sides is called a triangle – see Figure 25.1(a)4 sides is called a quadrilateral – see Figure 25.1(b)5 sides is called a pentagon – see Figure 25.1(c)6 sides is called a hexagon – see Figure 25.1(d)7 sides is called a heptagon – see Figure 25.1(e)8 sides is called an octagon – see Figure 25.1(f)25.2.2 QuadrilateralsThere are five types of quadrilateral, these being rect-angle, square, parallelogram, rhombus and trapezium.If the opposite corners of any quadrilateral are joinedby a straight line, two triangles are produced. Since thesum of the angles of a triangle is 180◦, the sum of theangles of a quadrilateral is 360◦.RectangleIn the rectangle ABCD shown in Figure 25.2,(a) all four angles are right angles,(b) the opposite sides are parallel and equal in length,and(c) diagonals AC and BD are equal in length and bisectone another.SquareIn the square PQRS shown in Figure 25.3,(a) all four angles are right angles,(b) the opposite sides are parallel,(c) all four sides are equal in length, and(d) diagonals PR and QS are equal in length and bisectone another at right angles.ParallelogramIn the parallelogram WXYZ shown in Figure 25.4,(a) opposite angles are equal,(b) opposite sides are parallel and equal in length, and(c) diagonals WY and XZ bisect one another.RhombusIn the rhombus ABCD shown in Figure 25.5,(a) opposite angles are equal,(b) opposite angles are bisected by a diagonal,(c) opposite sides are parallel,(d) all four sides are equal in length, and(e) diagonals AC and BD bisect one another at rightangles.DOI: 10.1016/B978-1-85617-697-2.00025-9
Areas of common shapes 225Total area of brickwork = 48 + 12= 60m2.Now try the following Practice ExercisePracticeExercise 97 Areas of commonshapes (answers on page 351)1. Name the types of quadrilateral shown inFigure25.18(i)to (iv)and determine for each(a) the area and (b) the perimeter.120 mm30 mm(iii)26 cm12 cm16 cm10 cm(iv)4 cm(i)5.94 cm3.5cm6 mm(ii)30mm38mmFigure 25.182. A rectangular plate is 85mm long and42mm wide. Find its area in square centi-metres.3. Arectangular field hasan areaof1.2hectaresand a length of 150m. If 1hectare =10000m2,find (a)thefield'swidth and (b)thelength of a diagonal.4. Find the area of a triangle whose base is8.5cm and perpendicular height is 6.4cm.5. A square has an area of 162cm2. Determinethe length of a diagonal.6. A rectangular picture has an area of 0.96m2.If one of the sides has a length of 800mm,calculate, in millimetres, the length of theother side.7. Determine the area of each of the angle ironsections shown in Figure 25.19.7cm1cm1cm(a) (b)2cm2cm30mm25mm8mm10mm50mm6mmFigure 25.198. Figure 25.20 shows a 4m wide path aroundthe outside of a 41m by 37m garden. Calcu-late the area of the path.414 37Figure 25.209. The area of a trapezium is 13.5cm2 and theperpendicular distance between its parallelsides is 3cm. If the length of one of the par-allel sides is 5.6cm, find the length of theother parallel side.10. Calculate the area of the steel plate shown inFigure 25.21.2560140Dimensionsin mm1002525Figure 25.21
226 Basic Engineering Mathematics11. Determine the area of an equilateral triangleof side 10.0cm.12. If paving slabs are produced in 250mm by250mm squares, determine the number ofslabs required to cover an area of 2m2.Here are some further worked problems on finding theareas of common shapes.Problem 11. Find the area of a circle having aradius of 5cmAreaof circle = πr2= π(5)2= 25π = 78.54cm2Problem 12. Find the area of a circle having adiameter of 15mmAreaof circle =πd24=π(15)24=225π4= 176.7mm2Problem 13. Find the area of a circle having acircumference of 70mmCircumference, c = 2πr,henceradius,r =c2π=702π=35πmmArea of circle = πr2= π35π2=352π= 389.9mm2or 3.899cm2Problem 14. Calculate the area of the sector of acircle having radius 6cm with angle subtended atcentre 50◦Area of sector =θ2360(πr2) =50360(π62)=50 × π × 36360= 15.71cm2Problem 15. Calculate the area of the sector of acircle having diameter 80mm with angle subtendedat centre 107◦42If diameter = 80mm then radius, r = 40mm, andarea of sector =107◦42360(π402) =1074260360(π402)=107.7360(π402)= 1504mm2or 15.04cm2Problem 16. A hollow shaft has an outsidediameter of 5.45cm and an inside diameter of2.25cm. Calculate the cross-sectional area of theshaftThe cross-sectional area of the shaft is shown by theshaded part in Figure 25.22 (often called an annulus).d 52.25 cmD 55.45 cmFigure 25.22Area of shaded part = area of large circle – area ofsmall circle=π D24−πd24=π4(D2− d2)=π4(5.452− 2.252)= 19.35cm2Now try the following Practice ExercisePracticeExercise 98 Areas of commonshapes (answers on page 351)1. A rectangular garden measures 40m by 15m.A 1m flower border is made round the twoshorter sides and one long side. A circularswimming pool of diameter 8m is constructed
Areas of common shapes 227in the middleof the garden. Find, correct to thenearest square metre, the area remaining.2. Determine the area of circles having (a) aradius of 4cm (b) a diameter of 30mm (c) acircumference of 200mm.3. An annulus has an outside diameter of 60mmand an inside diameter of 20mm. Determineits area.4. If the area of a circle is 320mm2, find (a) itsdiameter and (b) its circumference.5. Calculate the areas of the following sectors ofcircles.(a) radius 9cm, angle subtended at centre75◦.(b) diameter 35mm, angle subtended atcentre 48◦37 .6. Determine the shaded area of the templateshown in Figure 25.23.120mm90mm80mmradiusFigure 25.237. An archway consists of a rectangular openingtopped by a semi-circular arch, as shown inFigure 25.24. Determine the area of the open-ing if the width is 1m and the greatest heightis 2m.1 m2 mFigure 25.24Here are some further worked problems of commonshapes.Problem 17. Calculate the area of a regularoctagon if each side is 5cm and the width across theflats is 12cmAn octagon is an 8-sided polygon. If radii are drawnfrom the centre of the polygon to the vertices then 8equal triangles are produced, as shown in Figure 25.25.12cm5 mFigure 25.25Area of one triangle =12× base × height=12× 5 ×122= 15cm2Area of octagon = 8 × 15 = 120cm2Problem 18. Determine the area of a regularhexagon which has sides 8cm longA hexagon is a 6-sided polygon which may be dividedinto 6 equal triangles as shown in Figure 25.26. Theangle subtended at the centre of each triangle is 360◦ ÷6 = 60◦. The other two angles in the triangle add up to120◦ and are equal to each other. Hence, each of thetriangles is equilateral with each angle 60◦ and eachside 8cm.4cm8cm8cm608hFigure 25.26Area of one triangle =12× base × height=12× 8 × h
Areas of common shapes 229(a) its area in hectares (1 ha = 104 m2).(b) the length of fencing required, to the near-est metre, to completely enclose the plotof land.25.4 Areas of similar shapesFigure 25.29 shows two squares, one of which has sidesthree times as long as the other.3x3xxx(a) (b)Figure 25.29Area of Figure 25.29(a) = (x)(x) = x2Area of Figure 25.29(b) = (3x)(3x) = 9x2Hence, Figure 25.29(b) has an area (3)2; i.e., 9 times thearea of Figure 25.29(a).In summary, the areas of similar shapes are pro-portional to the squares of corresponding lineardimensions.Problem 20. A rectangular garage is shown on abuilding plan having dimensions 10mm by 20mm.If the plan is drawn to a scale of 1 to 250, determinethe true area of the garage in square metresArea of garage on the plan = 10mm × 20mm= 200mm2Since the areas of similar shapes are proportional to thesquares of corresponding dimensions,True area of garage = 200 × (250)2= 12.5 × 106mm2=12.5 × 106106m2since 1m2 = 106 mm2= 12.5m2Now try the following Practice ExercisePracticeExercise 100 Areas of similarshapes (answers on page 351)1. The area of a park on a map is 500mm2. Ifthe scale of the map is 1 to 40000, deter-mine the true area of the park in hectares(1hectare = 104 m2).2. A model of a boiler is made having an overallheight of 75mm corresponding to an overallheight of the actual boiler of 6m. If the area ofmetal required for the model is 12500mm2,determine, in square metres, the area of metalrequired for the actual boiler.3. The scale of an Ordnance Survey map is1:2500. A circular sports field has a diam-eter of 8cm on the map. Calculate its areain hectares, giving your answer correct to3 significant figures. (1 hectare = 104 m2.)
Chapter 26The circle26.1 IntroductionA circle is a plain figure enclosed by a curved line,everypoint on which is equidistant from a point within, calledthe centre.In Chapter25,worked problemson theareasofcirclesand sectors were demonstrated. In this chapter, proper-ties of circles are listed and arc lengths are calculated,together with more practical examples on the areas ofsectors of circles. Finally, the equation of a circle isexplained.26.2 Properties of circles(a) The distance from the centre to the curve is calledtheradius,r,ofthecircle(see OP in Figure26.1).CBQOPRAFigure 26.1(b) The boundary of a circle is called the circumfer-ence, c.(c) Any straight line passing through the centre andtouching the circumference at each end is calledthe diameter, d (see QR in Figure 26.1). Thus,d = 2r.(d) The ratiocircumferencediameteris a constant for any cir-cle. This constant is denoted by the Greek letter π(pronounced 'pie'), where π = 3.14159, correctto 5 decimal places (check with your calculator).Hence,cd= π or c = πd or c = 2πr.(e) A semicircle is one half of a whole circle.(f) A quadrant is one quarter of a whole circle.(g) A tangent to a circle is a straight linewhich meetsthecircleat onepoint only and doesnot cut thecir-cle when produced. AC in Figure 26.1 isa tangentto the circle since it touches the curve at point Bonly. If radius OB is drawn, angleABO is a rightangle.(h) The sector of a circle is the part of a circlebetween radii (for example, the portion OXY ofFigure 26.2 is a sector). If a sector is less than asemicircle it is called a minor sector; if greaterthan a semicircle it is called a major sector.XYTSROFigure 26.2(i) The chord of a circle is any straight line whichdivides the circle into two parts and is termi-nated at each end by the circumference. ST, inFigure 26.2, is a chord.(j) Segment is the name given to the parts into whicha circle is dividedby a chord. If thesegment is lessthan a semicircle it is called a minor segment(see shaded area in Figure 26.2). If the segmentDOI: 10.1016/B978-1-85617-697-2.00026-0
The circle 231is greater than a semicircle it is called a majorsegment (see the un-shaded area in Figure 26.2).(k) An arc is a portion of the circumference of a cir-cle. The distance SRT in Figure 26.2 is calleda minor arc and the distance SXYT is called amajor arc.(l) The angle at the centre of a circle, subtendedby an arc, is double the angle at the circumfer-ence subtended by the same arc. With referenceto Figure 26.3,Angle AOC = 2×angle ABCQAPCOBFigure 26.3(m) Theanglein asemicircleisaright angle(seeangleBQP in Figure 26.3).Problem 1. Find the circumference of a circle ofradius 12.0cmCircumference,c = 2 × π × radius = 2πr = 2π(12.0)= 75.40cmProblem 2. If the diameter of a circle is 75mm,find its circumferenceCircumference,c = π × diameter = πd = π(75)= 235.6mmProblem 3. Determine the radius of a circularpond if its perimeter is 112mPerimeter = circumference, c = 2πrHence, radius of pond, r =c2π=1122π= 17.83cmProblem 4. In Figure 26.4, AB is a tangent to thecircle at B. If the circle radius is 40mm andAB = 150 mm, calculate the length AOABrOFigure 26.4A tangent to a circle is at right angles to a radius drawnfromthepoint ofcontact; i.e.,ABO = 90◦.Hence,usingPythagoras' theorem,AO2= AB2+ OB2from which, AO = AB2 + OB2= 1502 + 402 = 155.2mmNow try the following Practice ExercisePracticeExercise 101 Properties of a circle(answers on page 351)1. Calculate the length of the circumference ofa circle of radius 7.2cm.2. If the diameter of a circle is 82.6mm, calcu-late the circumference of the circle.3. Determine the radius of a circle whose cir-cumference is 16.52cm.4. Find the diameter of a circle whose perimeteris 149.8cm.5. A crank mechanism is shown in Figure 26.5,where XY is a tangent to the circle at pointX. If the circle radius OX is 10cm and lengthOY is 40cm, determine the length of theconnecting rod XY.XYO 40 cmFigure 26.56. If the circumference of the earth is 40000kmat the equator, calculate its diameter.
The circle 2333. Convert to degrees:(a)7π6rad (b)4π9rad (c)7π12rad4. Convert to degrees and minutes:(a) 0.0125rad (b) 2.69rad(c) 7.241rad26.4 Arc length and area of circles andsectors26.4.1 Arc lengthFrom the definition of the radian in the previous sectionand Figure 26.7,arc length,s = rθ where θ is in radians26.4.2 Area of a circleFrom Chapter 25, for any circle, area = π × (radius)2i.e. area = πr2Since r =d2, area = πr2orπd2426.4.3 Area of a sectorArea of a sector =θ360(πr2) when θ is in degrees=θ2ππr2=12r2θ when θ is in radiansProblem 8. A hockey pitch has a semicircle ofradius 14.63m around each goal net. Find the areaenclosed by the semicircle, correct to the nearestsquare metreArea of a semicircle =12πr2When r = 14.63m, area =12π (14.63)2i.e. area of semicircle = 336m2Problem 9. Find the area of a circular metal platehaving a diameter of 35.0mm, correct to the nearestsquare millimetreArea of a circle = πr2=πd24When d = 35.0mm, area =π (35.0)24i.e. area of circular plate = 962mm2Problem 10. Find the area of a circle having acircumference of 60.0mmCircumference,c = 2πrfrom which radius,r =c2π=60.02π=30.0πArea of a circle = πr2i.e. area = π30.0π2=286.5mm2Problem 11. Find the length of the arc of a circleof radius 5.5cm when the angle subtended at thecentre is 1.20 radiansLength of arc, s = rθ, where θ is in radians.Hence, arc length, s = (5.5)(1.20) = 6.60cm.Problem 12. Determine the diameter andcircumference of a circle if an arc of length 4.75cmsubtends an angle of 0.91 radiansSince arc length, s = rθ thenradius, r =sθ=4.750.91= 5.22 cmDiameter = 2×radius = 2 × 5.22 = 10.44 cmCircumference, c = πd = π(10.44) = 32.80 cmProblem 13. If an angle of 125◦ is subtended byan arc of a circle of radius 8.4cm, find the length of(a) the minor arc and (b) the major arc, correct to3 significant figuresSince 180◦ = π rad then 1◦ =π180rad and125◦ = 125π180rad(a) Length of minor arc,s = rθ = (8.4)(125)π180= 18.3cm,correct to 3 significant figures
234 Basic Engineering Mathematics(b) Length of major arc = (circumference − minorarc) = 2π(8.4)−18.3 = 34.5cm, correct to 3significant figures.(Alternatively, major arc = rθ= 8.4(360 − 125)π180= 34.5cm.)Problem 14. Determine the angle, in degrees andminutes, subtended at the centre of a circle ofdiameter 42mm by an arc of length 36mm.Calculate also the area of the minor sector formedSince length of arc, s = rθ then θ =srRadius, r =diameter2=422= 21mm,hence θ =sr=3621= 1.7143 radians.1.7143rad = 1.7143 ×180π◦= 98.22◦ = 98◦13 =angle subtended at centre of circle.From equation (2),area of sector =12r2θ =12(21)2(1.7143)= 378mm2.Problem 15. A football stadium floodlight canspread its illumination over an angle of 45◦ to adistance of 55m. Determine the maximum area thatis floodlit.Floodlit area = area of sector =12r2θ=12(55)245 ×π180= 1188m2Problem 16. An automatic garden sprayerproduces spray to a distance of 1.8m and revolvesthrough an angle α which may be varied. If thedesired spray catchment area is to be 2.5m2, to whatshould angle α be set, correct to the nearest degree?Area of sector =12r2θ, hence 2.5 =12(1.8)2αfrom which, α =2.5 × 21.82= 1.5432 radians1.5432rad = 1.5432 ×180π◦= 88.42◦Hence, angle α = 88◦, correct to the nearest degree.Problem 17. The angle of a tapered groove ischecked using a 20mm diameter roller as shown inFigure 26.8. If the roller lies 2.12mm below the topof the groove, determine the value of angle θ2.12mm20mm30mmFigure 26.8In Figure 26.9, triangle ABC is right-angled at C (seeproperty (g) in Section 26.2).2.12mm210mmBAC30mmFigure 26.9Length BC = 10 mm (i.e. the radius of the circle), andAB = 30 − 10 − 2.12 = 17.88mm, from Figure 26.9.Hence, sinθ2=1017.88andθ2= sin−1 1017.88= 34◦and angle θ = 68◦.Now try the following Practice ExercisePracticeExercise 103 Arc length and areaof circles and sectors (answers on page 351)1. Calculate the area of a circle of radius 6.0cm,correct to the nearest square centimetre.2. The diameter of a circle is 55.0mm. Deter-mine its area, correct to the nearest squaremillimetre.
The circle 2353. The perimeter of a circle is 150mm. Find itsarea, correct to the nearest square millimetre.4. Find the area of the sector, correct to thenearest square millimetre, of a circle havinga radius of 35mm with angle subtended atcentre of 75◦.5. An annulus has an outside diameter of49.0mm and an inside diameter of 15.0mm.Find its area correct to 4 significant figures.6. Find the area, correct to the nearest squaremetre, of a 2m wide path surrounding acircular plot of land 200m in diameter.7. A rectangular park measures 50m by 40m.A 3m flower bed is made round the twolongersidesand oneshort side.Acircularfishpond of diameter 8.0m is constructed in thecentre of the park. It is planned to grass theremaining area. Find, correct to the nearestsquare metre, the area of grass.8. With reference to Figure26.10, determine (a)the perimeter and (b) the area.17cm28cmFigure 26.109. Find the area of the shaded portion ofFigure 26.11.10mFigure 26.1110. Find the length of an arc of a circle of radius8.32cm when the angle subtended at the cen-tre is 2.14 radians. Calculate also the area ofthe minor sector formed.11. If the angle subtended at the centre of a cir-cle of diameter 82mm is 1.46rad, find thelengthsof the (a) minorarc and (b) major arc.12. A pendulum of length 1.5m swings throughan angle of 10◦ in a single swing. Find, incentimetres, the length of the arc traced bythe pendulum bob.13. Determine the length of the radius and cir-cumference of a circle if an arc length of32.6cm subtends an angle of 3.76 radians.14. Determine the angle of lap, in degrees andminutes, if 180mm of a belt drive are incontact with a pulley of diameter 250mm.15. Determine the number of complete revo-lutions a motorcycle wheel will make intravelling 2km if the wheel's diameter is85.1cm.16. A floodlightat a sports ground spread its illu-mination over an angle of 40◦ to a distanceof 48m. Determine (a) the angle in radiansand (b) the maximum area that is floodlit.17. Find the area swept out in 50 minutes by theminute hand of a large floral clock if the handis 2m long.18. Determine(a)theshaded areain Figure26.12and (b)thepercentageofthewholesectorthatthe shaded area represents.50mm0.75 rad12mmFigure 26.12
Revision Test 10 : Areas and circlesThis assignment covers the material contained in Chapters 25 and 26. The marks available are shown in brackets atthe end of each question.1. A rectangular metal platehas an area of 9600cm2.If the length of the plate is 1.2m, calculate thewidth, in centimetres. (3)2. Calculate the cross-sectional area of the angle ironsection shown in Figure RT10.1, the dimensionsbeing in millimetres. (4)282751934Figure RT10.13. Find the area of the trapezium M N O P shown inFigure RT10.2 when a = 6.3cm,b = 11.7cm andh = 5.5cm. (3)M NOPhabFigure RT10.24. Find the area of the triangle DE F in FigureRT10.3, correct to 2 decimal places. (4)5. Arectangularpark measures150m by 70m.A2mflower border is constructed round the two longersides and one short side. A circular fish pond ofdiameter 15m is in the centre of the park and theremainder of the park is grass. Calculate, correctto the nearest square metre, the area of (a) thefish pond, (b) the flower borders and (c) the grass.(6)6. A swimming pool is 55mlong and 10mwide. Theperpendicular depth at the deep end is 5m and at8.75cm12.44cmFEDFigure RT10.3the shallow end is 1.5m, the slope from one endto the other being uniform. The inside of the poolneeds two coats of a protective paint before it isfilled with water. Determine how many litres ofpaint will be needed if 1 litre covers 10m2. (7)7. Find the area of an equilateral triangle of side20.0cm. (4)8. A steel template is of the shape shown inFigure RT10.4, the circular area being removed.Determine the area of the template, in squarecentimetres, correct to 1 decimal place. (8)30mm45mm130mm70mm70mm 150mm60mm30 mm50mmdia.Figure RT10.49. The area of a plot of land on a map is400mm2. If the scale of the map is 1 to 50000,
Revision Test 10 : Areas and circles 239determine the true area of the land in hectares(1 hectare = 104 m2). (4)10. Determine the shaded area in Figure RT10.5,correct to the nearest square centimetre. (3)20cm2 cmFigure RT10.511. Determine the diameter of a circle, correct tothe nearest millimetre, whose circumference is178.4cm. (2)12. Calculate the area of a circle of radius 6.84cm,correct to 1 decimal place. (2)13. The circumference of a circle is 250mm. Find itsarea, correct to the nearest square millimetre. (4)14. Find the area of the sector of a circle having aradius of 50.0mm, with angle subtended at centreof 120◦. (3)15. Determine the total area of the shape shown inFigure RT10.6, correct to 1 decimal place. (7)7.0 m10.0m6.0 mFigure RT10.616. The radius of a circular cricket ground is75m. Theboundary is painted with white paint and 1 tin ofpaint will paint a line 22.5m long. How many tinsof paint are needed? (3)17. Find the area of a 1.5m wide path surrounding acircular plot of land 100m in diameter. (3)18. A cyclometer shows 2530 revolutions in a dis-tance of 3.7km. Find the diameter of the wheelin centimetres, correct to 2 decimal places. (4)19. The minute hand of a wall clock is 10.5cm long.How far does the tip travel in the course of24 hours? (4)20. Convert(a) 125◦47 to radians.(b) 1.724 radians to degrees and minutes. (4)21. Calculate the length of metal strip needed tomake the clip shown in Figure RT10.7. (7)30mm rad15mm rad15mm rad70mm70mm75mmFigure RT10.722. A lorry has wheels of radius 50cm. Calculate thenumber of complete revolutions a wheel makes(correct to the nearest revolution) when travelling3 miles (assume 1mile = 1.6km). (4)23. The equation of a circle isx2 + y2 + 12x − 4y + 4 = 0. Determine(a) the diameter of the circle.(b) the co-ordinates of the centre of the circle.(7)
Chapter 27Volumes of common solids27.1 IntroductionThe volume of any solid is a measure of the space occu-pied by the solid. Volume is measured in cubic unitssuch as mm3, cm3 and m3.This chapter deals with finding volumes of commonsolids; in engineering it is often important to be able tocalculatevolumeorcapacity to estimate,say,theamountof liquid, such as water, oil or petrol, in different shapedcontainers.A prism is a solid with a constant cross-section andwith two ends parallel. The shape of the end is used todescribe the prism. For example, there are rectangularprisms (called cuboids), triangular prisms and circularprisms (called cylinders).On completing this chapter you will be able to cal-culate the volumes and surface areas of rectangular andother prisms, cylinders, pyramids, cones and spheres,together with frusta of pyramids and cones. Volumes ofsimilar shapes are also considered.27.2 Volumes and surface areas ofcommon shapes27.2.1 Cuboids or rectangular prismsA cuboid is a solid figure bounded by six rectangularfaces; all angles are right angles and opposite faces areequal. A typical cuboid is shown in Figure 27.1 withlength l, breadth b and height h.Volume of cuboid = l × b × handsurface area = 2bh + 2hl + 2lb = 2(bh + hl + lb)hblFigure 27.1A cube is a square prism. If all the sides of a cube are xthenVolume = x3and surface area = 6x2Problem 1. A cuboid has dimensions of 12cm by4cm by 3cm. Determine (a) its volume and (b) itstotal surface areaThe cuboid is similar to that in Figure 27.1, withl = 12cm,b = 4cm and h = 3cm.(a) Volume of cuboid= l × b × h = 12 × 4 × 3= 144 cm3(b) Surface area = 2(bh + hl +lb)= 2(4 × 3 + 3 × 12+ 12 × 4)= 2(12 + 36 + 48)= 2 × 96 = 192 cm2Problem 2. An oil tank is the shape of a cube,each edge being of length 1.5m. Determine (a) themaximum capacity of the tank in m3 and litres and(b) its total surface area ignoring input and outputorificesDOI: 10.1016/B978-1-85617-697-2.00027-2
242 Basic Engineering MathematicsOuter diameter, D = 25cm = 0.25m and inner diame-ter, d = 12cm = 0.12m.Area of cross-section of copper=π D24−πd24=π(0.25)24−π(0.12)24= 0.0491 − 0.0113 = 0.0378m2Hence, volume of copper= (cross-sectional area) × length of pipe= 0.0378 × 2.5 = 0.0945 m327.2.3 More prismsA right-angled triangular prism is shown in Figure 27.4with dimensions b,h and l.IbhFigure 27.4Volume =12bhlandsurface area = area of each end+ area of three sidesNotice that the volume is given by the area of the end(i.e. area of triangle = 12 bh) multiplied by the length l.In fact, the volume of any shaped prism is given by thearea of an end multiplied by the length.Problem 6. Determine the volume (in cm3) of theshape shown in Figure 27.512 mm16 mm40 mmFigure 27.5The solid shown in Figure 27.5 is a triangular prism.The volume V of any prism is given by V = Ah, whereA is the cross-sectional area and h is the perpendicularheight. Hence,volume =12× 16 × 12× 40 = 3840mm3= 3.840 cm3(since 1cm3 = 1000mm3)Problem 7. Calculate the volume of theright-angled triangular prism shown in Figure 27.6.Also, determine its total surface area6 cm40cmAB C8 cmFigure 27.6Volume of right-angled triangular prism=12bhl =12× 8 × 6 × 40i.e. volume = 960 cm3
Volumes of common solids 243Total surface area = area of each end + area of threesides.In triangle ABC, AC2= AB2+ BC2from which, AC = AB2 + BC2 = 62 + 82= 10cmHence, total surface area= 212bh + (AC × 40) + (BC × 40) + (AB × 40)= (8 × 6) + (10 × 40) + (8 × 40) + (6 × 40)= 48 + 400 + 320 + 240i.e. total surface area = 1008 cm2Problem 8. Calculate the volume and totalsurface area of the solid prism shown in Figure 27.715cm5cm5cm5cm4cm11cmFigure 27.7The solid shown in Figure 27.7 is a trapezoidal prism.Volume of prism = cross-sectional area × height=12(11 + 5)4 × 15 = 32 × 15= 480 cm3Surface area of prism= sum of two trapeziums + 4 rectangles= (2 × 32) + (5 × 15) + (11 × 15) + 2(5 × 15)= 64 + 75 + 165 + 150 = 454 cm2Now try the following Practice ExercisePracticeExercise 105 Volumes and surfaceareas of common shapes (answers onpage 351)1. Change a volume of 1200000cm3 to cubicmetres.2. Change a volume of 5000mm3 to cubiccentimetres.3. A metal cube has a surface area of 24cm2.Determine its volume.4. A rectangular block of wood has dimensionsof 40mm by 12mm by 8mm. Determine(a) its volume, in cubic millimetres(b) its total surface area in square millime-tres.5. Determine the capacity, in litres, ofa fish tankmeasuring 90cm by 60cm by 1.8m, given1litre = 1000cm3.6. A rectangular block of metal has dimensionsof 40mm by 25mm by 15mm. Determine itsvolume in cm3. Find also its mass if the metalhas a density of 9g/cm3.7. Determine the maximum capacity, in litres,of a fish tank measuring 50cm by 40cm by2.5m(1litre = 1000cm3).8. Determine how many cubic metres of con-crete are required for a 120m long path,150mm wide and 80mm deep.9. A cylinder has a diameter 30mm and height50mm. Calculate(a) its volume in cubic centimetres, correctto 1 decimal place(b) the total surface area in square centime-tres, correct to 1 decimal place.10. Find (a) the volume and (b) the total sur-face area of a right-angled triangular prismof length 80cm and whose triangular endhas a base of 12cm and perpendicularheight 5cm.11. A steel ingot whose volume is 2m2 is rolledout into a plate which is 30mm thick and1.80m wide. Calculate the length of the platein metres.
244 Basic Engineering Mathematics12. The volume of a cylinder is 75cm3. If itsheight is 9.0cm, find its radius.13. Calculate the volume of a metal tube whoseoutside diameter is 8cm and whose insidediameter is 6cm, if the length of the tube is4m.14. The volume of a cylinder is 400cm3. Ifits radius is 5.20cm, find its height. Alsodetermine its curved surface area.15. A cylinder is cast from a rectangular piece ofalloy 5cm by 7cm by 12cm. If the length ofthe cylinder is to be 60cm, find its diameter.16. Find the volume and the total surface areaof a regular hexagonal bar of metal of length3m if each side of the hexagon is 6cm.17. A block of lead 1.5m by 90cm by 750mmis hammered out to make a square sheet15mm thick. Determine the dimensions ofthe square sheet, correct to the nearest cen-timetre.18. How long will it take a tap dripping at a rateof 800mm3/s to fill a 3-litre can?19. A cylinder is cast from a rectangular pieceof alloy 5.20cm by 6.50cm by 19.33cm. Ifthe height of the cylinder is to be 52.0cm,determine its diameter, correct to the nearestcentimetre.20. How much concrete is required for the con-struction of the path shown in Figure 27.8, ifthe path is 12cm thick?2 m1.2m8.5 m2.4m3.1 mFigure 27.827.2.4 PyramidsVolume of any pyramid=13×area of base × perpendicular heightA square-based pyramid is shown in Figure 27.9 withbase dimension x by x and the perpendicular height ofthe pyramid h. For the square-base pyramid shown,volume =13x2hhxxFigure 27.9Problem 9. A square pyramid has aperpendicular height of 16cm. If a side of the baseis 6cm, determine the volume of a pyramidVolume of pyramid=13× area of base × perpendicular height=13× (6 × 6) × 16= 192 cm3Problem 10. Determine the volume and the totalsurface area of the square pyramid shown inFigure 27.10 if its perpendicular height is 12cm.Volume of pyramid=13(area of base) × perpendicular height=13(5 × 5) × 12= 100 cm3
Volumes of common solids 2473. A sphere has a diameter of 6cm. Determineits volume and surface area.4. If the volume of a sphere is 566cm3, find itsradius.5. A pyramid having a square base has a per-pendicular height of 25cm and a volume of75cm3. Determine, in centimetres, thelengthof each side of the base.6. A cone has a base diameter of 16mm and aperpendicular height of 40mm. Find its vol-ume correct to the nearest cubic millimetre.7. Determine (a) the volume and (b) the surfacearea of a sphere of radius 40mm.8. The volume of a sphere is 325cm3. Deter-mine its diameter.9. Given the radius of the earth is 6380km,calculate, in engineering notation(a) its surface area in km2.(b) its volume in km3.10. An ingot whose volume is 1.5m3 is to bemade into ball bearings whose radii are8.0cm. How many bearings will be producedfrom the ingot, assuming 5% wastage?27.3 Summary of volumes and surfaceareas of common solidsA summary of volumes and surface areas of regularsolids is shown in Table 27.1.Table 27.1 Volumes and surface areas of regularsolidsRectangular prism(or cuboid)hbl Volume = l × b × hSurface area = 2(bh + hl + lb)CylinderhrVolume = πr2hTotal surface area = 2πrh + 2πr2Triangular prismIbhVolume =12bhlSurface area = area of each end +area of three sidesPyramidhAVolume =13× A × hTotal surface area =sum of areas of trianglesforming sides + area of baseConehrlVolume =13πr2hCurved surface area = πrlTotal surface area = πrl + πr2SphererVolume =43πr3Surface area = 4πr227.4 More complex volumes andsurface areasHere are some worked problems involving more com-plex and composite solids.Problem 16. A wooden section is shown inFigure 27.14. Find (a) its volume in m3 and(b) its total surface area
Volumes of common solids 251Curved surface area of cylinder, Q = 2πrh= 2 × π × 3 × 8= 48π m2The slant height of the cone, l, is obtained by Pythago-ras' theorem on triangle ABC, i.e.l = 42 + 32 = 5Curved surface area of cone,R = πrl = π × 3 × 5 = 15π m2Total surface area of boiler= 18π + 48π + 15π= 81π = 254.5 m2Now try the following Practice ExercisePracticeExercise 107 More complexvolumes and surface areas (answers onpage 351)1. Find the total surface area of a hemisphere ofdiameter 50mm.2. Find (a) the volume and (b) the total surfacearea of a hemisphere of diameter 6cm.3. Determine the mass of a hemispherical cop-per container whose external and internalradii are 12cm and 10cm, assuming that1cm3 of copper weighs 8.9g.4. A metal plumb bob comprises a hemispheresurmounted by a cone. If the diameter of thehemisphere and cone are each 4cm and thetotal length is 5cm, find its total volume.5. A marquee is in the form of a cylinder sur-mounted by acone.Thetotal height is6mandthe cylindrical portion has a height of 3.5mwith a diameter of 15m. Calculate the surfacearea of material needed to make the marqueeassuming 12% of the material is wasted inthe process.6. Determine (a) the volume and (b) the totalsurface area of the following solids.(i) a cone of radius 8.0cm and perpendi-cular height 10cm.(ii) a sphere of diameter 7.0cm.(iii) a hemisphere of radius 3.0cm.(iv) a 2.5cm by 2.5cm square pyramid ofperpendicular height 5.0cm.(v) a 4.0cm by 6.0cm rectangular pyra-mid of perpendicular height 12.0cm.(vi) a 4.2cm by 4.2cm square pyramidwhose sloping edges are each 15.0cm(vii) a pyramid having an octagonal base ofside 5.0cm and perpendicular height20cm.7. A metal sphere weighing 24kg is melteddown and recast into a solid cone of baseradius 8.0cm. If the density of the metal is8000kg/m3 determine(a) the diameter of the metal sphere.(b) the perpendicular height of the cone,assuming that 15% of the metal is lostin the process.8. Find the volume of a regular hexagonal pyra-midiftheperpendicularheight is16.0cmandthe side of the base is 3.0cm.9. A buoy consists of a hemisphere surmountedby a cone. The diameter of the cone andhemisphere is 2.5m and the slant height ofthe cone is 4.0m. Determine the volume andsurface area of the buoy.10. A petrol container is in the form of a centralcylindrical portion 5.0m long with a hemi-spherical section surmounted on each end. Ifthe diameters of the hemisphere and cylinderare both 1.2m, determine the capacity of thetank in litres (1litre = 1000cm3).11. Figure 27.18 shows a metal rod section.Determine its volume and total surface area.1.00cmradius 1.00m2.50cmFigure 27.18
252 Basic Engineering Mathematics12. Find the volume (in cm3) of the die-castingshown in Figure 27.19. The dimensions arein millimetres.6030 rad2550100Figure 27.1913. The cross-section of part of a circular ven-tilation shaft is shown in Figure 27.20, endsAB and CD being open. Calculate(a) the volume of the air, correct tothe nearest litre, contained in thepart of the system shown, neglect-ing the sheet metal thickness (given1litre = 1000cm3).(b) the cross-sectional area of the sheetmetal used to make the system, insquare metres.(c) thecost ofthesheet metal ifthematerialcosts £11.50 per square metre, assum-ing that 25% extra metal is required dueto wastage.500mmAB2m1.5m1.5m800mmDCFigure 27.2027.5 Volumes and surface areas offrusta of pyramids and conesThe frustum of a pyramid or cone is the portionremain-ing when a part containing the vertex is cut off by aplane parallel to the base.The volume of a frustum of a pyramid or coneis givenby the volume of the whole pyramid or cone minus thevolume of the small pyramid or cone cut off.The surface area of the sides of a frustum of a pyra-mid or cone is given by the surface area of the wholepyramid or cone minus the surface area of the smallpyramid or cone cut off. This gives the lateral surfacearea of the frustum. If the total surface area of the frus-tum is required then the surface area of the two parallelends are added to the lateral surface area.There is an alternative method for finding the volumeand surface area of a frustum of a cone. With referenceto Figure 27.21,hRIrFigure 27.21Volume =13πh(R2+ Rr + r2)Curved surface area = πl(R + r)Total surface area = πl(R + r) + πr2+ πR2Problem 24. Determine the volume of a frustumof a cone if the diameter of the ends are 6.0cm and4.0cm and its perpendicular height is 3.6cm(i) Method 1A section through the vertex of a complete cone isshown in Figure 27.22.Using similar triangles,APDP=DRBRHence, AP2.0=3.61.0from which AP =(2.0)(3.6)1.0= 7.2cm
254 Basic Engineering MathematicsProblem 26. A storage hopper is in the shape of afrustum of a pyramid. Determine its volume if theends of the frustum are squares of sides 8.0m and4.6m, respectively, and the perpendicular heightbetween its ends is 3.6mThe frustum is shown shaded in Figure 27.23(a) as partof a complete pyramid. A section perpendicular to thebase through the vertex is shown in Figure 27.23(b).8.0m4.6cm8.0m(a)B2.3m1.7m 2.3m(b)4.0m2.3m 3.6mCAEDGFH4.6cmFigure 27.23By similar trianglesCGBG=BHAHFrom which, heightCG = BGBHAH=(2.3)(3.6)1.7= 4.87mHeight of complete pyramid = 3.6 + 4.87 = 8.47mVolume of large pyramid =13(8.0)2(8.47)= 180.69m3Volume of small pyramid cut off =13(4.6)2(4.87)= 34.35m3Hence, volume of storage hopper= 180.69 − 34.35= 146.3 m3Problem 27. Determine the lateral surface area ofthe storage hopper in Problem 26The lateral surface area of the storage hopper consistsof four equal trapeziums. From Figure 27.24,Area of trapezium PRSU =12(PR + SU)(QT )4.6m8.0mPUTSO8.0mQR4.6mFigure 27.24OT = 1.7m (same as AH in Figure 27.23(b) andOQ = 3.6m. By Pythagoras' theorem,QT = OQ2 + OT 2 = 3.62 + 1.72 = 3.98mArea of trapezium PRSU=12(4.6 + 8.0)(3.98) = 25.07m2Lateral surface area of hopper = 4(25.07)= 100.3 m2Problem 28. A lampshade is in the shape of afrustum of a cone. The vertical height of the shadeis 25.0cm and the diameters of the ends are 20.0cmand 10.0cm, respectively. Determine the area of thematerial needed to form the lampshade, correct to 3significant figuresThe curved surface area of a frustum of a cone= πl(R +r) from page 252. Since the diameters ofthe ends of the frustum are 20.0cm and 10.0cm, fromFigure 27.25,r5 5.0cmIh525.0cmR510.0cm5.0cmFigure 27.25
Volumes of common solids 255r = 5.0cm, R = 10.0cmand l = 25.02 + 5.02 = 25.50cmfrom Pythagoras' theorem.Hence, curved surface area= π(25.50)(10.0 + 5.0) = 1201.7cm2i.e., the area of material needed to form the lampshadeis 1200 cm2, correct to 3 significant figures.Problem 29. A cooling tower is in the form of acylinder surmounted by a frustum of a cone, asshown in Figure 27.26. Determine the volume of airspace in the tower if 40% of the space is used forpipes and other structures12.0m25.0m12.0m30.0mFigure 27.26Volume of cylindrical portion = πr2h= π25.022(12.0)= 5890m3Volume of frustum of cone =13πh(R2+ Rr +r2)where h = 30.0 − 12.0 = 18.0m,R = 25.0 ÷ 2 = 12.5mand r = 12.0 ÷ 2 = 6.0m.Hence, volume of frustum of cone=13π(18.0) (12.5)2+ (12.5)(6.0) + (6.0)2= 5038m3Total volume of cooling tower = 5890 + 5038= 10 928m3If 40% of space is occupied thenvolume of air space= 0.6 × 10928 = 6557 m3Now try the following Practice ExercisePracticeExercise 108 Volumes and surfaceareas of frusta of pyramidsand cones(answers on page 352)1. The radii of the faces of a frustum of a coneare 2.0cm and 4.0cm and the thickness of thefrustum is 5.0cm. Determine its volume andtotal surface area.2. A frustum of a pyramid has square ends,the squares having sides 9.0cm and 5.0cm,respectively. Calculate the volume and totalsurface area of the frustum if the perpendiculardistance between its ends is 8.0cm.3. A cooling tower is in the form of a frustum ofa cone. The base has a diameter of 32.0m, thetop has a diameter of 14.0m and the verticalheight is 24.0m. Calculate the volume of thetower and the curved surface area.4. A loudspeaker diaphragm is in the form of afrustum of a cone. If the end diameters are28.0cm and 6.00cm and the vertical distancebetween the ends is 30.0cm, find the area ofmaterial needed to cover the curved surface ofthe speaker.5. A rectangular prism of metal having dimen-sions 4.3cm by 7.2cm by 12.4cm is melteddown and recast into a frustum of a squarepyramid, 10% of the metal being lost in theprocess. If the ends of the frustum are squaresof side 3cm and 8cm respectively, find thethickness of the frustum.6. Determine the volume and total surface areaof a bucket consisting of an inverted frustumof a cone, of slant height 36.0cm and enddiameters 55.0cm and 35.0cm.7. A cylindrical tank of diameter 2.0m and per-pendicular height 3.0m is to be replaced bya tank of the same capacity but in the formof a frustum of a cone. If the diameters ofthe ends of the frustum are 1.0m and 2.0m,respectively, determine the vertical heightrequired.
256 Basic Engineering Mathematics27.6 Volumes of similar shapesFigure 27.27 shows two cubes, one of which has sidesthree times as long as those of the other.3xxxx3x3x(b)(a)Figure 27.27Volume of Figure 27.27(a) = (x)(x)(x) = x3Volume of Figure 27.27(b) = (3x)(3x)(3x) = 27x3Hence, Figure 27.27(b) has a volume (3)3, i.e. 27, timesthe volume of Figure 27.27(a).Summarizing, the volumes of similar bodies areproportional to the cubes of corresponding lineardimensions.Problem 30. A car has a mass of 1000kg.A model of the car is made to a scale of 1 to 50.Determine the mass of the model if the car and itsmodel are made of the same materialVolume of modelVolume of car=1503since the volume of similar bodies are proportional tothe cube of corresponding dimensions.Mass=density×volume and, since both car and modelare made of the same material,Mass of modelMass of car=1503Hence, mass of model= (mass of car)1503=1000503= 0.008 kg or 8 gNow try the following Practice ExercisePracticeExercise 109 Volumes of similarshapes (answers on page 352)1. The diameter of two spherical bearings are inthe ratio 2:5. What is the ratio of their vol-umes?2. An engineering component has a mass of400g. If each of its dimensions are reducedby 30%, determine its new mass.
Chapter 28Irregular areas and volumes,and mean values28.1 Areas of irregular figuresAreas of irregular plane surfaces may be approximatelydetermined by using(a) a planimeter,(b) the trapezoidal rule,(c) the mid-ordinate rule, or(d) Simpson's rule.Such methods may be used by, for example, engineersestimating areas ofindicatordiagrams ofsteam engines,surveyors estimating areas of plots of land or navalarchitects estimating areas of water planes or transversesections of ships.(a) A planimeter is an instrument for directly mea-suring small areas bounded by an irregular curve.There are many different kinds of planimeters butall operate in a similar way. A pointer on theplanimeter is used to trace around the boundaryof the shape. This induces a movement in anotherpart of the instrument and a reading of this is usedto establish the area of the shape.(b) Trapezoidal ruleTo determine the area PQRS in Figure 28.1,(i) Divide base PS into any number of equalintervals, each of width d (the greater thenumber of intervals, the greater the accu-racy).(ii) Accurately measure ordinates y1, y2, y3, etc.y1y2y3y4y5y6y7RQPd d d d d dSFigure 28.1(iii) Area PQRS= dy1 + y72+ y2 + y3 + y4 + y5 + y6 .In general, the trapezoidal rule statesArea =width ofinterval12first + lastordinate+⎛⎝sum ofremainingordinates⎞⎠⎤⎦(c) Mid-ordinate ruley1y2y3y4y5y6CBAd d d d d dDFigure 28.2To determine the area ABCD of Figure 28.2,DOI: 10.1016/B978-1-85617-697-2.00028-4
262 Basic Engineering MathematicsProblem 7. The power used in a manufacturingprocess during a 6 hour period is recorded atintervals of 1 hour as shown below.Time (h) 0 1 2 3 4 5 6Power (kW) 0 14 29 51 45 23 0Plot a graph of power against time and, by using themid-ordinate rule, determine (a) the area under thecurve and (b) the average value of the powerThe graph of power/time is shown in Figure 28.10.Graph of power/timePower(kW)50403020100 1 2 3Time (hours)4 5 621.57.0 42.0 49.5 37.0 10.0Figure 28.10(a) Thetimebaseisdivided into 6 equal intervals,eachof width 1 hour. Mid-ordinates are erected (shownby broken lines in Figure 28.10) and measured.The values are shown in Figure 28.10.Area under curve= (width of interval)(sum of mid-ordinates)= (1)[7.0 + 21.5 + 42.0 + 49.5 + 37.0 + 10.0]= 167kWh (i.e. a measure of electrical energy)(b) Average value of waveform =area under curvelength of base=167kWh6h= 27.83 kWAlternatively, average value=sum of mid-ordinatesnumber of mid-ordinatesProblem 8. Figure 28.11 shows a sinusoidaloutput voltage of a full-wave rectifier. Determine,using the mid-ordinate rule with 6 intervals, themean output voltage0 308608908 1808 2708 3608210Voltage(V)322Figure 28.11Onecycleoftheoutput voltageiscompleted inπ radiansor 180◦. The base is divided into 6 intervals, each ofwidth 30◦. The mid-ordinate of each interval will lie at15◦,45◦,75◦, etc.At 15◦ the height of the mid-ordinate is10sin 15◦ = 2.588V,At 45◦ the height of the mid-ordinate is10sin 45◦= 7.071V, and so on.The results are tabulated below.Mid-ordinate Height of mid-ordinate15◦10sin15◦= 2.588V45◦ 10sin45◦ = 7.071V75◦ 10sin75◦ = 9.659V105◦ 10sin105◦ = 9.659V135◦ 10sin135◦ = 7.071V165◦ 10sin165◦ = 2.588VSum of mid-ordinates = 38.636VMean or average value of output voltage=sum of mid-ordinatesnumber of mid-ordinates=38.6366= 6.439V(With a larger number of intervals a more accurateanswer may be obtained.)For a sine wave the actual mean value is0.637 × maximum value, which in this problemgives 6.37V.Problem 9. An indicator diagram for a steamengine is shown in Figure 28.12. The base line has
Irregular areas and volumes, and mean values 263been divided into 6 equally spaced intervals and thelengths of the 7 ordinates measured with the resultsshown in centimetres. Determine(a) the area of the indicator diagram usingSimpson's rule(b) the mean pressure in the cylinder given that1cm represents 100kPa.12.0cm3.6 4.0 3.5 2.9 2.2 1.7 1.6Figure 28.12(a) The width of each interval is12.06cm. UsingSimpson's rule,area =13(2.0)[(3.6 + 1.6) + 4(4.0 + 2.9 + 1.7)+ 2(3.5 + 2.2)]=23[5.2 + 34.4 + 11.4] = 34cm2(b) Mean height of ordinates =area of diagramlength of base=3412= 2.83cmSince 1cm represents 100kPa,mean pressure in the cylinder= 2.83cm × 100kPa/cm = 283kPaNow try the following Practice ExercisePracticeExercise 112 Mean or averagevalues of waveforms (answers on page 352)1. Determine the mean value of the periodicwaveforms shown in Figure 28.13 over a halfcycle.2. Find the average value of the periodic wave-forms shown in Figure 28.14 over one com-plete cycle.(a)0Current(A)210 2022t (ms)(b)0Voltage(V)1005 102100t (ms)(c)Current(A)0515 3025t (ms)Figure 28.13Voltage(mV)0102 4 6 8 10 t(ms)Current(A)052 4 6 8 10 t(ms)Figure 28.143. An alternating current hasthefollowing valuesat equal intervals of 5ms:Time (ms) 0 5 10 15 20 25 30Current (A) 0 0.9 2.6 4.9 5.8 3.5 0Plot a graph of current against time and esti-mate the area under the curve over the 30msperiod, using the mid-ordinate rule, and deter-mine its mean value.4. Determine, using an approximate method, theaverage value of a sine wave of maximumvalue 50V for (a) a half cycle (b) a completecycle.5. An indicator diagram of a steam engine is12cm long. Seven evenly spaced ordinates,including the end ordinates, are measured asfollows:5.90, 5.52, 4.22, 3.63, 3.32, 3.24 and 3.16cm.Determine the area of the diagram and themean pressure in the cylinder if 1cm repre-sents 90kPa.
Revision Test 11 : Volumes, irregular areas and volumes, and mean valuesThis assignment covers the material contained in Chapters 27 and 28. The marks available are shown in brackets atthe end of each question.1. A rectangular block of alloy has dimensions of60mm by 30mm by 12mm. Calculate the volumeof the alloy in cubic centimetres. (3)2. Determine how many cubic metres of concrete arerequired for a 120m long path, 400mm wide and10cm deep. (3)3. Find the volume of a cylinder of radius 5.6cmand height 15.5cm. Give the answer correct to thenearest cubic centimetre. (3)4. A garden roller is 0.35m wide and has a diame-ter of 0.20m. What area will it roll in making 40revolutions? (4)5. Find the volume of a cone of height 12.5cm andbase diameter 6.0cm, correct to 1 decimal place.(3)6. Find (a) the volume and (b) the total surfacearea of the right-angled triangular prism shown inFigure RT11.1. (9)9.70cm4.80cm11.6cmFigure RT11.17. A pyramid having a square base has a volumeof 86.4cm3. If the perpendicular height is 20cm,determine the length of each side of the base. (4)8. A copper pipe is 80m long. It has a bore of 80mmand an outside diameter of 100mm. Calculate, incubic metres, the volume of copper in the pipe. (4)9. Find (a) the volume and (b) the surface area of asphere of diameter 25mm. (4)10. A piece of alloy with dimensions 25mm by60mm by 1.60m is melted down and recast intoa cylinder whose diameter is 150mm. Assum-ing no wastage, calculate the height of thecylinder in centimetres, correct to 1 decimalplace. (4)11. Determine the volume (in cubic metres) and thetotal surface area (in square metres) of a solidmetal cone of base radius 0.5m and perpendicularheight 1.20m. Give answers correct to 2 decimalplaces. (6)12. A rectangular storage container has dimensions3.2m by 90cm by 60cm. Determine its volume in(a) m3 (b) cm3. (4)13. Calculate (a) the volume and (b) the total surfacearea of a 10cm by 15cm rectangular pyramid ofheight 20cm. (8)14. A water container is of the form of a central cylin-drical part 3.0m long and diameter 1.0m, with ahemispherical section surmounted at each end asshown in Figure RT11.2. Determine the maximumcapacity of the container, correct to the nearestlitre. (1 litre = 1000cm3.)3.0m1.0mFigure RT11.2(5)15. Find the total surface area of a bucket consist-ing of an inverted frustum of a cone of slantheight 35.0cm and end diameters 60.0cm and40.0cm. (4)16. A boat has a mass of 20000kg. A model of theboat is made to a scale of 1 to 80. If the model ismade of the same material as the boat, determinethe mass of the model (in grams). (3)
Revision Test 11 : Volumes, irregular areas and volumes, and mean values 26517. Plot a graph of y = 3x2+ 5 from x = 1 to x = 4.Estimate,correct to 2 decimal places,using 6 inter-vals, the area enclosed by the curve, the ordinatesx = 1 and x = 4, and the x-axis by(a) the trapezoidal rule(b) the mid-ordinate rule(c) Simpson's rule. (16)18. A circular cooling tower is 20m high. The insidediameter of the tower at different heights is givenin the following table.Height (m) 0 5.0 10.0 15.0 20.0Diameter (m) 16.0 13.3 10.7 8.6 8.0Determine the area corresponding to each diame-ter and hence estimate the capacity of the tower incubic metres. (7)19. A vehicle starts from rest and its velocity ismeasured every second for 6 seconds, with thefollowing results.Time t (s) 0 1 2 3 4 5 6Velocity v (m/s) 0 1.2 2.4 3.7 5.2 6.0 9.2Using Simpson's rule, calculate(a) the distance travelled in 6s (i.e. the area underthe v/t graph),(b) the average speed over this period. (6)
Chapter 29Vectors29.1 IntroductionThis chapter initially explains the difference betweenscalar and vector quantities and shows how a vector isdrawn and represented.Any object that is acted upon by an external forcewill respond to that force by moving in the line ofthe force. However, if two or more forces act simul-taneously, the result is more difficult to predict; theability to add two or more vectors then becomesimportant.This chapter thus shows how vectors are added andsubtracted, both by drawing and by calculation, and howfinding the resultant of two or more vectors has manyuses in engineering. (Resultant means the single vectorwhich would have the same effect as the individual vec-tors.) Relative velocities and vector i,j,k notation arealso briefly explained.29.2 Scalars and vectorsThe time taken to fill a water tank may be measured as,say, 50s. Similarly, the temperature in a room may bemeasured as, say, 16◦C or the mass of a bearing may bemeasured as, say, 3kg. Quantities such as time, temper-ature and mass are entirely defined by a numerical valueand are called scalars or scalar quantities.Not all quantities are like this. Some are definedby more than just size; some also have direction. Forexample, the velocity of a car may be 90km/h duewest, a force of 20N may act vertically downwards,or an acceleration of 10m/s2 may act at 50◦ to thehorizontal.Quantities such as velocity, force and acceleration,which have both a magnitude and a direction, arecalled vectors.Now try the following Practice ExercisePracticeExercise 113 Scalar and vectorquantities (answers on page 352)1. State the difference between scalar and vectorquantities.In problems 2 to 9, state whether the quantitiesgiven are scalar or vector.2. A temperature of 70◦C3. 5m3 volume4. A downward force of 20N5. 500J of work6. 30cm2 area7. A south-westerly wind of 10knots8. 50m distance9. An acceleration of 15m/s2 at 60◦ to thehorizontal29.3 Drawing a vectorA vector quantity can be represented graphically by aline, drawn so that(a) the length of the line denotes the magnitude of thequantity, and(b) the direction of the line denotes the direction inwhich the vector quantity acts.An arrow is used to denote the sense, or direction, of thevector.The arrow end of a vector is called the 'nose' and theother end the 'tail'. For example, a force of 9N actingDOI: 10.1016/B978-1-85617-697-2.00029-6
Vectors 267at 45◦to the horizontal is shown in Figure 29.1. Notethat an angle of +45◦ is drawn from the horizontal andmoves anticlockwise.9N0a45ЊFigure 29.1A velocity of 20m/s at −60◦ is shown in Figure 29.2.Note that an angle of −60◦ is drawn from the horizontaland moves clockwise.60Њ20m/s0bFigure 29.229.3.1 Representing a vectorThere are a number of ways of representing vectorquantities. These include(a) Using bold print.(b)−→AB where an arrow above two capital lettersdenotes the sense of direction, where A is thestarting point and B the end point of the vector.(c) AB or a; i.e., a line over the top of letter.(d) a; i.e., underlined letter.The force of 9N at 45◦ shown in Figure 29.1 may berepresented as0a or−→0a or 0aThe magnitude of the force is 0a.Similarly, the velocity of 20m/s at −60◦ shown inFigure 29.2 may be represented as0b or−→0b or 0bThe magnitude of the velocity is 0b.In this chapter a vector quantity is denoted by boldprint.29.4 Addition of vectors by drawingAdding two or more vectors by drawing assumes thata ruler, pencil and protractor are available. Resultsobtained by drawing are naturally not as accurate asthose obtained by calculation.(a) Nose-to-tail methodTwo force vectors, F1 and F2, are shown inFigure 29.3. When an object is subjected to morethan one force, the resultant of the forces is foundby the addition of vectors.F2F1Figure 29.3To add forces F1 and F2,(i) Force F1 is drawn to scale horizontally,shown as 0a in Figure 29.4.(ii) From the nose of F1, force F2 is drawn atangle θ to the horizontal, shown as ab.(iii) The resultant force is given by length 0b,which may be measured.This procedure is called the 'nose-to-tail' or'triangle' method.F2F1ab0Figure 29.4(b) Parallelogram methodTo add the two force vectors, F1 and F2 ofFigure 29.3,(i) A line cb is constructed which is parallel toand equal in length to 0a (see Figure 29.5).(ii) A line ab is constructed which is parallel toand equal in length to 0c.(iii) The resultant force is given by the diagonalof the parallelogram; i.e., length 0b.This procedure is called the 'parallelogram'method.
268 Basic Engineering Mathematics0cF1F2abFigure 29.5Problem 1. A force of 5N is inclined at an angleof 45◦ to a second force of 8N, both forces acting ata point. Find the magnitude of the resultant of thesetwo forces and the direction of the resultant withrespect to the 8N force by (a) the nose-to-tailmethod and (b) the parallelogram methodThe two forces are shown in Figure 29.6. (Although the8N force is shown horizontal, it could have been drawnin any direction.)45Њ5N8NFigure 29.6(a) Nose-to-tail method(i) The 8N force is drawn horizontally 8 unitslong, shown as 0a in Figure 29.7.(ii) From the nose of the 8N force, the 5N forceis drawn 5 units long at an angle of 45◦ to thehorizontal, shown as ab.(iii) The resultant force is given by length 0b andis measured as 12 N and angle θ is measuredas 17◦. 4585N8N a0bFigure 29.7(b) Parallelogram method(i) In Figure 29.8, a line is constructed whichis parallel to and equal in length to the 8Nforce.(ii) A line is constructed which is parallel to andequal in length to the 5N force.45Њ5N8Nb0Figure 29.8(iii) The resultant force is given by the diagonalof the parallelogram, i.e. length 0b, and ismeasured as 12 N and angle θ is measuredas 17◦.Thus, the resultant of the two force vectors inFigure 29.6 is 12 N at 17◦ to the 8 N force.Problem 2. Forces of 15 and 10N are at an angleof 90◦ to each other as shown in Figure 29.9. Find,by drawing, the magnitude of the resultant of thesetwo forces and the direction of the resultant withrespect to the 15N force15N10NFigure 29.9Using the nose-to-tail method,(i) The 15N force is drawn horizontally 15 unitslong, as shown in Figure 29.10.10NR15NFigure 29.10(ii) From the nose of the 15N force, the 10N forceis drawn 10 units long at an angle of 90◦ to thehorizontal as shown.
Vectors 269(iii) The resultant force is shown as R and is measuredas 18N and angle θ is measured as 34◦.Thus, the resultant of the two force vectors is 18 N at34◦ to the 15 N force.Problem 3. Velocities of 10m/s, 20m/s and15m/s act as shown in Figure 29.11. Determine, bydrawing, the magnitude of the resultant velocity andits direction relative to the horizontal15Њ32130Њ10m/s20m/s15m/sFigure 29.11When more than 2 vectors are being added the nose-to-tail method is used. The order in which the vectors areadded does not matter. In this case the order taken is ν1,then ν2, then ν3. However, if a different order is takenthe same result will occur.(i) ν1 is drawn 10 units long at an angle of 30◦ to thehorizontal, shown as 0a in Figure 29.12.b195Њ105Њ030Њ132raFigure 29.12(ii) From the nose of ν1, ν2 is drawn 20 units long atan angle of 90◦ to the horizontal, shown as ab.(iii) From the nose of ν2, ν3 is drawn 15 units long atan angle of 195◦ to the horizontal, shown as br.(iv) The resultant velocity is given by length 0r andis measured as 22m/s and the angle measured tothe horizontal is 105◦.Thus, the resultant of the three velocities is 22 m/s at105◦ to the horizontal.Worked Examples 1 to 3 have demonstrated how vec-tors are added to determine their resultant and theirdirection. However, drawing to scale is time-consumingand not highly accurate. The following sections demon-strate how to determine resultant vectors by calculationusing horizontal and vertical components and, wherepossible, by Pythagoras' theorem.29.5 Resolving vectors into horizontaland vertical componentsA force vector F is shown in Figure 29.13 at angle θto the horizontal. Such a vector can be resolved intotwo components such that the vector addition of thecomponents is equal to the original vector.FFigure 29.13The two components usually taken are a horizontalcomponent and a vertical component.Ifaright-angledtriangle is constructed as shown in Figure 29.14, 0a iscalled the horizontal component of F and ab is calledthe vertical component of F.0aFbFigure 29.14From trigonometry(see Chapter 21 and remember SOHCAH TOA),cosθ =0a0b, from which 0a = 0bcosθ = F cosθ
Vectors 271total horizontal component of the two forces,H = F1 cosθ1 + F2 cosθ2The vertical component of force F1 is F1 sinθ1 and thevertical component of force F2 is F2 sinθ2. The totalvertical component of the two forces,V = F1 sinθ1 + F2 sinθ2Since we have H and V , the resultant of F1 and F2is obtained by using the theorem of Pythagoras. FromFigure 29.19,0b2= H2+ V 2i.e. resultant = H2 + V 2 at an anglegiven by θ = tan−1 VHVbResultantaH0Figure 29.19Problem 7. A force of 5N is inclined at an angleof 45◦ to a second force of 8N, both forces acting ata point. Calculate the magnitude of the resultant ofthese two forces and the direction of the resultantwith respect to the 8N forceThe two forces are shown in Figure 29.20.4588N5NFigure 29.20The horizontal component of the 8N force is 8cos0◦and the horizontal component of the 5N force is5cos45◦. The total horizontal component of the twoforces,H = 8cos0◦+ 5cos45◦= 8 + 3.5355 = 11.5355The vertical component of the 8N force is 8sin0◦andthe vertical component of the 5N force is 5sin45◦. Thetotal vertical component of the two forces,V = 8sin0◦+ 5sin45◦= 0 + 3.5355 = 3.5355ResultantH ϭ11.5355 NV ϭ3.5355NFigure 29.21From Figure 29.21, magnitude of resultant vector= H2 + V 2= 11.53552 + 3.53552 = 12.07NThe direction of the resultant vector,θ = tan−1 VH= tan−1 3.535511.5355= tan−10.30648866... = 17.04◦Thus, the resultant of the two forces is a single vectorof 12.07 N at 17.04◦ to the 8 N vector.Problem 8. Forces of 15N and 10N are at anangle of 90◦ to each other as shown in Figure 29.22.Calculate the magnitude of the resultant of thesetwo forces and its direction with respect to the15N force10N15 NFigure 29.22The horizontal component of the 15N force is 15cos0◦and the horizontal component of the 10N force is10cos90◦. The total horizontal component of the twovelocities,H = 15cos0◦+ 10cos90◦= 15 + 0 = 15
272 Basic Engineering MathematicsThe vertical component of the 15N force is 15sin0◦andthe vertical component of the 10N force is 10sin90◦.The total vertical component of the two velocities,V = 15sin0◦+ 10sin90◦= 0 + 10 = 10Magnitude of resultant vector=√H2 + V 2 =√152 + 102 = 18.03NThe direction of the resultant vector,θ = tan−1 VH= tan−1 1015= 33.69◦Thus, the resultant of the two forces is a single vectorof 18.03 N at 33.69◦ to the 15 N vector.There is an alternative method of calculating the resul-tant vector in this case. If we used the triangle method,the diagram would be as shown in Figure 29.23.15 N10NRFigure 29.23Since a right-angled triangle results, we could usePythagoras' theorem without needing to go through theprocedure for horizontal and vertical components. Infact, the horizontal and vertical components are 15Nand 10N respectively.This is, of course, a special case. Pythagoras can onlybeusedwhenthere isanangleof90◦ betweenvectors.This is demonstrated in worked Problem 9.Problem 9. Calculate the magnitude anddirection of the resultant of the two accelerationvectors shown in Figure 29.24.15m/s228m/s2Figure 29.24The 15m/s2acceleration is drawn horizontally, shownas 0a in Figure 29.25.015a28bR␣ Figure 29.25From the nose of the 15m/s2 acceleration, the 28m/s2acceleration isdrawn at an angleof 90◦ to thehorizontal,shown as ab.The resultant acceleration, R, is given by length 0b.Since a right-angled triangle results, the theorem ofPythagoras may be used.0b = 152 + 282 = 31.76m/s2and α = tan−1 2815= 61.82◦Measuring from the horizontal,θ = 180◦ − 61.82◦ = 118.18◦Thus, the resultant of the two accelerations is a singlevector of 31.76 m/s2 at 118.18◦ to the horizontal.Problem 10. Velocities of 10m/s, 20m/s and15m/s act as shown in Figure 29.26. Calculate themagnitude of the resultant velocity and its directionrelative to the horizontal20 m/s10 m/s15m/s158308123Figure 29.26The horizontal component of the 10m/s velocity= 10cos30◦ = 8.660 m/s,
Vectors 273the horizontal component of the 20m/s velocity is20cos90◦ = 0 m/sand the horizontal component of the 15m/s velocity is15cos195◦ = −14.489m/s.The total horizontal component of the three velocities,H = 8.660 + 0 − 14.489 = −5.829m/sThe vertical component of the 10m/s velocity= 10sin30◦ = 5m/s,the vertical component of the 20m/s velocity is20sin90◦ = 20 m/sand the vertical component of the 15m/s velocity is15sin195◦= −3.882 m/s.The total vertical component of the three forces,V = 5 + 20 − 3.882 = 21.118m/s5.82921.118R␣ Figure 29.27From Figure 29.27, magnitude of resultant vector,R = H2 + V 2 = 5.8292 + 21.1182 = 21.91m/sThe direction of the resultant vector,α = tan−1 VH= tan−1 21.1185.829= 74.57◦Measuring from the horizontal,θ = 180◦ − 74.57◦ = 105.43◦.Thus, the resultant of the three velocities is a singlevector of 21.91 m/s at 105.43◦ to the horizontal.Now try the following Practice ExercisePracticeExercise 114 Addition of vectorsby calculation (answers on page 352)1. A force of 7N is inclined at an angle of 50◦ toa second force of 12N, both forces acting ata point. Calculate the magnitude of the resul-tant of the two forces and the direction of theresultant with respect to the 12N force.2. Velocities of 5m/s and 12m/s act at a pointat 90◦ to each other. Calculate the resultantvelocity and itsdirectionrelativeto the12m/svelocity.3. Calculate the magnitude and direction of theresultant of the two force vectors shown inFigure 29.28.10N13NFigure 29.284. Calculate the magnitude and direction of theresultant of the two force vectors shown inFigure 29.29.22N18NFigure 29.295. A displacement vector s1 is 30m at 0◦. Asecond displacement vector s2 is 12m at 90◦.Calculate the magnitude and direction of theresultant vector s1 + s26. Three forces of 5N, 8N and 13N act asshown in Figure 29.30. Calculate the mag-nitude and direction of the resultant force.5 N13N8 N708608Figure 29.30
276 Basic Engineering MathematicsR027.676.99␣Figure 29.38Thus, v2 − v1 − v3 = 28.54 units at 194.18◦This result is as expected, since v2 − v1 − v3= −(v1 − v2 + v3) and the vector 28.54 units at194.18◦ is minus times (i.e. is 180◦ out of phasewith) the vector 28.54 units at 14.18◦Now try the following Practice ExercisePracticeExercise 115 Vector subtraction(answers on page 352)1. Forces of F1 = 40 N at 45◦ and F2 = 30 N at125◦act at a point. Determine by drawing andby calculation (a) F1 + F2 (b) F1 − F22. Calculate the resultant of (a) v1 + v2 − v3(b) v3 − v2 + v1 when v1 = 15m/s at 85◦,v2 = 25m/s at 175◦ and v3 = 12 m/s at 235◦.29.8 Relative velocityFor relative velocity problems, some fixed datum pointneeds to be selected. This is often a fixed point on theearth's surface. In any vector equation, only the startand finish points affect the resultant vector of a system.Two different systems are shown in Figure 29.39, but,in each of the systems, the resultant vector is ad.a db(a)adbc(b)Figure 29.39The vector equation of the system shown inFigure 29.39(a) isad = ab + bdand that for the system shown in Figure 29.39(b) isad = ab + bc + cdThus, in vector equations of this form, only the first andlast letters, a and d, respectively, fix the magnitude anddirection of the resultant vector. This principle is usedin relative velocity problems.Problem 13. Two cars, P and Q, are travellingtowards the junction of two roads which are at rightangles to one another. Car P has a velocity of45km/h due east and car Q a velocity of 55km/hdue south. Calculate (a) the velocity of car Prelative to car Q and (b) the velocity of car Qrelative to car P(a) The directions of the cars are shown inFigure 29.40(a), which is called a space diagram.The velocity diagram is shown in Figure 29.40(b),in which pe is taken as the velocity of car P rela-tive to point e on the earth's surface. The velocityof P relative to Q is vector pq and the vector equa-tion is pq = pe + eq. Hence, the vector directionsare as shown, eq being in the opposite directionto qe.(a) (b) (c)QPENWS55 km/h45km/hpeqpeqFigure 29.40From the geometry of the vector triangle, the mag-nitude of pq =√452 + 552 = 71.06km/h and thedirection of pq = tan−1 5545= 50.71◦That is, the velocity of car P relative to car Q is71.06 km/h at 50.71◦(b) The velocity of car Q relative to car P is given bythe vector equation qp = qe + ep and the vectordiagram is as shown in Figure 29.40(c), having epopposite in direction to pe.
Chapter 30Methods of addingalternating waveforms30.1 Combining two periodicfunctionsThere are a number of instances in engineering and sci-ence where waveforms have to be combined and whereit is required to determine the single phasor (calledthe resultant) that could replace two or more separatephasors. Uses are found in electrical alternating cur-rent theory, in mechanical vibrations, in the addition offorces and with sound waves.There are a number of methods of determining theresultant waveform. These include(a) Drawing the waveforms and adding graphically.(b) Drawing the phasors and measuring the resultant.(c) Using the cosine and sine rules.(d) Using horizontal and vertical components.30.2 Plotting periodic functionsThis may be achieved by sketching the separate func-tions on the same axes and then adding (or subtracting)ordinates at regular intervals. This is demonstrated inthe following worked problems.Problem 1. Plot the graph of y1 = 3sin A fromA = 0◦ to A = 360◦. On the same axes ploty2 = 2cos A. By adding ordinates, plotyR = 3sin A + 2cos A and obtain a sinusoidalexpression for this resultant waveformy1 = 3sin A and y2 = 2cos A are shown plotted inFigure 30.1. Ordinates may be added at, say, 15◦intervals. For example,at 0◦, y1 + y2 = 0 + 2 = 2at 15◦, y1 + y2 = 0.78 + 1.93 = 2.71at 120◦, y1 + y2 = 2.60 + −1 = 1.6at 210◦, y1 + y2 = −1.50 − 1.73 = −3.23, andso on.yy15 3 sin Ay25 2 cos AyR 53.6 sin(A 1348)A023222133.621348908 1808 2708 3608Figure 30.1The resultant waveform, shown by the broken line,has the same period, i.e. 360◦, and thus the same fre-quency as the single phasors. The maximum value,or amplitude, of the resultant is 3.6. The resultantDOI: 10.1016/B978-1-85617-697-2.00030-2
280 Basic Engineering Mathematics2 angle t19Њ19Њi1 ϭ 20 sintiR ϭ 20 sin t ϩ10 sin(tϩ )90Њ 180Њ 270Њ 360ЊϪ30Ϫ20Ϫ10102026.5303223i2 ϭ 10 sin(tϩ )3Figure 30.43. Express 12sinωt + 5cosωt in the formAsin(ωt ± α) by drawing and measurement.30.3 Determining resultant phasorsby drawingThe resultant of two periodic functions may be foundfrom their relative positions when the time is zero.For example, if y1 = 4sinωt and y2 = 3sin(ωt − π/3)then each may be represented as phasors as shown inFigure 30.5, y1 being 4 units long and drawn horizon-tally and y2 being 3 unitslong,lagging y1 by π/3 radiansor60◦.To determinetheresultant of y1 + y2, y1 isdrawnhorizontally as shown in Figure 30.6 and y2 is joined tothe end of y1 at 60◦ to the horizontal. The resultantis given by yR. This is the same as the diagonal of aparallelogram that is shown completed in Figure 30.7.608 or /3 radsy15 4y25 3Figure 30.5Resultant yR, in Figures 30.6 and 30.7, may be deter-mined by drawing the phasors and their directions toscale and measuring using a ruler and protractor. In thisy15 4y253 608yRFigure 30.6y1ϭ 4y2ϭ 3yRFigure 30.7example, yR is measured as 6 units long and angle φ ismeasured as 25◦.25◦= 25 ×π180radians = 0.44 radHence, summarizing, by drawing,yR = y1 + y2 = 4sinωt + 3sin(ωt − π/3)= 6sin(ωt − 0.44).If the resultant phasor, yR = y1 − y2 is required theny2 is still 3 units long but is drawn in the oppositedirection, as shown in Figure 30.8.Problem 5. Two alternating currents are given byi1 = 20sinωt amperes and i2 = 10sin ωt +π3amperes. Determine i1 + i2 by drawing phasors
Methods of adding alternating waveforms 285(b) the frequency of the supply.7. If the supply to a circuit is v = 20sin628.3tvolts and the voltage drop across one ofthe components is v1 = 15sin(628.3t − 0.52)volts, calculate(a) the voltage drop across the remainder ofthe circuit, given by v − v1, in the formAsin(ωt ± α)(b) the supply frequency(c) the periodic time of the supply.8. The voltages across three components in aseries circuit when connected across an a.c.supply are v1 = 25sin 300πt +π6volts,v2 = 40sin 300πt −π4volts andv3 = 50sin 300πt +π3volts. Calculate(a) the supply voltage, in sinusoidal form, inthe form Asin(ωt ± α)(b) the frequency of the supply(c) the periodic time.
Revision Test 12 : Vectors and adding waveformsThis assignment covers the material contained in Chapters 29 and 30. The marks available are shown in brackets atthe end of each question.1. State the difference between scalar and vectorquantities. (2)2. State whether the following are scalar or vectorquantities.(a) A temperature of 50◦C.(b) 2m3 volume.(c) A downward force of 10N.(d) 400J of work.(e) 20cm2 area.(f) A south-easterly wind of 20 knots.(g) 40m distance.(h) An acceleration of 25m/s2 at 30◦ to thehorizontal. (8)3. A velocity vector of 16m/s acts at an angle of−40◦ to the horizontal. Calculate its horizontaland vertical components, correct to 3 significantfigures. (4)4. Calculate the resultant and direction of the dis-placement vectors shown in Figure RT12.1, cor-rect to 2 decimal places. (6)41 m26 mFigure RT12.15. Calculate the resultant and direction of the forcevectors shown in Figure RT12.2, correct to 2decimal places. (6)6. If acceleration a1 = 11m/s2at 70◦anda2 = 19m/s2 at −50◦, calculate the magnitudeand direction of a1 + a2, correct to 2 decimalplaces. (8)7. If velocity v1 = 36m/s at 52◦ and v2 = 17m/s at−15◦, calculate the magnitude and direction ofv1 − v2, correct to 2 decimal places. (8)5N7NFigure RT12.28. Forces of 10N, 16N and 20N act as shown inFigure RT12.3. Determine the magnitude of theresultant force and its direction relative to the 16Nforce(a) by scaled drawing.(b) by calculation. (13)F1510NF2 516NF3 520N458608Figure RT12.39. For the three forces shown in Figure RT12.3,calculate the resultant of F1 − F2 − F3 and itsdirection relative to force F2. (9)10. Two cars, A and B, are travelling towards cross-roads. A has a velocity of 60km/h due south andB a velocity of 75km/h due west. Calculate thevelocity of A relative to B. (6)
Chapter 31Presentation ofstatistical data31.1 Some statistical terminology31.1.1 Discrete and continuous dataData are obtained largely by two methods:(a) By counting – for example, the number of stampssold by a post office in equal periods of time.(b) By measurement – for example, the heights of agroup of people.When data are obtained by counting and only wholenumbers are possible, the data are called discrete. Mea-sured data can have any value within certain limits andare called continuous.Problem 1. Data are obtained on the topics givenbelow. State whether they are discrete or continuousdata.(a) The number of days on which rain falls in amonth for each month of the year.(b) The mileage travelled by each of a number ofsalesmen.(c) The time that each of a batch of similarbatteries lasts.(d) The amount of money spent by each of severalfamilies on food.(a) The number of days on which rain falls in a givenmonth must be an integer value and is obtained bycounting the number of days. Hence, these dataare discrete.(b) A salesman can travel any number of miles(and parts of a mile) between certain limits andthese data are measured. Hence, the data arecontinuous.(c) The time that a battery lasts is measured andcan have any value between certain limits. Hence,these data are continuous.(d) The amount of money spent on food can only beexpressed correct to the nearest pence, the amountbeing counted. Hence, these data are discrete.Now try the following Practice ExercisePracticeExercise 122 Discrete andcontinuous data (answers on page 353)In the followingproblems, state whether data relat-ing to the topics given are discrete or continuous.1. (a) The amount of petrol produced daily, foreach of 31 days, by a refinery.(b) The amount of coal produced daily byeach of 15 miners.(c) The number of bottles of milk delivereddaily by each of 20 milkmen.(d) Thesizeof10 samplesofrivetsproducedby a machine.2. (a) The number of people visiting an exhi-bition on each of 5 days.(b) The time taken by each of 12 athletes torun 100 metres.DOI: 10.1016/B978-1-85617-697-2.00031-4
Presentation of statistical data 289(c) The value of stamps sold in a day by eachof 20 post offices.(d) The number of defective items producedin each of 10 one-hour periods by amachine.31.1.2 Further statisticalterminologyA set is a group of data and an individual value withinthe set is called a member of the set. Thus, if themasses of five people are measured correct to the near-est 0.1kilogram and are found to be 53.1kg, 59.4kg,62.1kg, 77.8kg and 64.4kg then the set of masses inkilograms for these five people is{53.1,59.4,62.1,77.8,64.4}and one of the members of the set is 59.4A set containing all the members is called a pop-ulation. Some members selected at random from apopulation are called a sample. Thus, all car registrationnumbers form a population but the registration numbersof, say, 20 cars taken at random throughout the countryare a sample drawn from that population.The number of times that the value of a memberoccurs in a set is called the frequency of that mem-ber. Thus, in the set {2,3,4,5,4,2,4,7,9}, member 4has a frequency of three, member 2 has a frequency of2 and the other members have a frequency of one.The relative frequency with which any member of aset occurs is given by the ratiofrequency of membertotal frequency of all membersFor the set {2,3,5,4,7,5,6,2,8}, the relative fre-quency of member 5 is29. Often, relative frequency isexpressed as a percentage and the percentage relativefrequency is(relative frequency × 100)%31.2 Presentation of ungrouped dataUngrouped data can be presented diagrammatically inseveral ways and these include(a) pictograms, in which pictorial symbols are usedto represent quantities (see Problem 2),(b) horizontal bar charts, having data representedby equally spaced horizontal rectangles (see Prob-lem 3), and(c) vertical bar charts, in which data are repre-sented by equally spaced vertical rectangles (seeProblem 4).Trends in ungrouped data over equal periods of timecan be presented diagrammatically by a percent-age component bar chart. In such a chart, equallyspaced rectangles of any width, but whose height cor-responds to 100%, are constructed. The rectanglesare then subdivided into values corresponding to thepercentage relative frequencies of the members (seeProblem 5).A pie diagram is used to show diagrammatically theparts making up the whole. In a pie diagram, the area ofa circle represents the whole and the areas of the sectorsof the circle are made proportional to the parts whichmake up the whole (see Problem 6).Problem 2. The number of television setsrepaired in a workshop by a technician in sixone-month periods is as shown below. Present thesedata as a pictogramMonth January February MarchNumber repaired 11 6 15Month April May JuneNumber repaired 9 13 8Each symbol shown in Figure 31.1 represents two tele-vision sets repaired. Thus, in January, 512symbols areused to represent the 11 sets repaired; in February, 3symbols are used to represent the 6 sets repaired, andso on.JanuaryFebruaryMarchMonth Number of TV sets repaired ;2 setsAprilMayJuneFigure 31.1
290 Basic Engineering MathematicsProblem 3. The distance in miles travelled byfour salesmen in a week are as shown below.Salesman P Q R SDistance travelled (miles) 413 264 597 143Use a horizontal bar chart to represent these datadiagrammaticallyEqually spaced horizontal rectangles of any width, butwhose length is proportional to the distance travelled,are used. Thus, the length of the rectangle for sales-man P is proportional to 413 miles, and so on. Thehorizontal bar chart depicting these data is shown inFigure 31.2.0PQSalesmenRS100 200 300Distance travelled, miles400 500 600Figure 31.2Problem 4. The number of issues of tools ormaterials from a store in a factory is observed forseven one-hour periods in a day and the results ofthe survey are as follows.Period 1 2 3 4 5 6 7Number of issues 34 17 9 5 27 13 6Present these data on a vertical bar chart11020Numberofissues30402 3 4 5 6Periods7Figure 31.3In a vertical bar chart, equally spaced vertical rect-angles of any width, but whose height is proportionalto the quantity being represented, are used. Thus, theheight of the rectangle for period 1 is proportional to 34units, and so on. The vertical bar chart depicting thesedata is shown in Figure 31.3.Problem 5. The numbers of various types ofdwellings sold by a company annually over athree-year period are as shown below. Drawpercentage component bar charts to presentthese dataYear 1 Year 2 Year 34-roomed bungalows 24 17 75-roomed bungalows 38 71 1184-roomed houses 44 50 535-roomed houses 64 82 1476-roomed houses 30 30 25A table of percentage relative frequency values, correctto the nearest 1%, is the first requirement. Sincepercentage relative frequency=frequency of member × 100total frequencythen for 4-roomed bungalows in year 1percentage relative frequency=24 × 10024 + 38 + 44 + 64 + 30= 12%The percentage relative frequencies of the other typesof dwellings for each of the three years are similarlycalculated and theresultsareasshown in thetablebelow.Year 1 Year 2 Year 34-roomed bungalows 12% 7% 2%5-roomed bungalows 19% 28% 34%4-roomed houses 22% 20% 15%5-roomed houses 32% 33% 42%6-roomed houses 15% 12% 7%Thepercentagecomponent barchart isproduced by con-structing three equally spaced rectangles of any width,corresponding to the three years. The heightsof the rect-angles correspond to 100% relative frequency and are
Presentation of statistical data 291subdivided into the values in the table of percentagesshown above. A key is used (different types of shadingor different colour schemes) to indicate correspondingpercentage values in the rows of the table of percent-ages. The percentage component bar chart is shown inFigure 31.4.11020Percentagerelativefrequency304050607080901002 3YearKey6-roomed houses5-roomed houses4-roomed houses5-roomed bungalows4-roomed bungalowsFigure 31.4Problem 6. The retail price of a product costing£2 is made up as follows: materials 10p, labour20p, research and development 40p, overheads70p, profit 60p. Present these data on a pie diagramA circle of any radius is drawn. The area of the circlerepresents the whole, which in this case is £2. The circleis subdivided into sectors so that the areas of the sectorsare proportional to the parts; i.e., the parts which makeup the total retail price. For the area of a sector to beproportionalto a part, the angle at the centre of the circlemust be proportional to that part. The whole, £2 or 200p,corresponds to 360◦. Therefore,10p corresponds to 360×10200degrees, i.e. 18◦20p corresponds to 360×20200degrees, i.e. 36◦and so on, giving the angles at the centre of the circlefor the parts of the retail price as 18◦,36◦,72◦,126◦ and108◦, respectively.The pie diagram is shown in Figure 31.5.Problem 7.(a) Using the data given in Figure 31.2 only,calculate the amount of money paid to eachsalesman for travelling expenses if they arepaid an allowance of 37p per mile.1088Ip ; 1.881883687281268OverheadsProfitLabourResearch anddevelopmentMaterialsFigure 31.5(b) Using the data presented in Figure 31.4,comment on the housing trends over thethree-year period.(c) Determine the profit made by selling 700 unitsof the product shown in Figure 31.5(a) By measuring the length of rectangle P, themileage covered by salesman P is equivalent to413 miles. Hence salesman P receives a travellingallowance of£413 × 37100i.e. £152.81Similarly, for salesman Q, the miles travelled are264 and his allowance is£264 × 37100i.e. £97.68Salesman R travels 597 miles and he receives£597 × 37100i.e. £220.89Finally, salesman S receives£143 × 37100i.e. £52.91(b) An analysis of Figure 31.4 shows that 5-roomedbungalows and 5-roomed houses are becomingmorepopular,thegreatest changein thethreeyearsbeing a 15% increase in the sales of 5-roomedbungalows.(c) Since 1.8◦ corresponds to 1p and the profit occu-pies 108◦ of the pie diagram, the profit per unit is108 × 11.8i.e. 60pThe profit when selling 700 units of the prod-uct is£700 × 60100i.e. £420
292 Basic Engineering MathematicsNow try the following Practice ExercisePracticeExercise 123 Presentation ofungrouped data (answers on page 352)1. The number of vehicles passing a stationaryobserver on a road in six ten-minute intervalsis as shown. Draw a pictogram to representthese data.Period of time 1 2 3 4 5 6Number ofvehicles 35 44 62 68 49 412. The number of components produced by afactory in a week is as shown below.Day Mon Tues Wed Thur FriNumber ofcomponents 1580 2190 1840 2385 1280Show these data on a pictogram.3. For the data given in Problem 1 above, drawa horizontal bar chart.4. Present the data given in Problem 2 above ona horizontal bar chart.5. For the data given in Problem 1 above,construct a vertical bar chart.6. Depict the data given in Problem 2 above ona vertical bar chart.7. A factory produces three different types ofcomponents. The percentages of each ofthese components produced for three one-month periods are as shown below. Show thisinformation on percentage component barcharts and comment on the changing trendin the percentages of the types of componentproduced.Month 1 2 3Component P 20 35 40Component Q 45 40 35Component R 35 25 258. A company has five distribution centres andthe mass of goods in tonnes sent to eachcentre during four one-week periods is asshown.Week 1 2 3 4Centre A 147 160 174 158Centre B 54 63 77 69Centre C 283 251 237 211Centre D 97 104 117 144Centre E 224 218 203 194Use a percentage component bar chart to pre-sent these data and comment on any trends.9. The employees in a company can be splitinto the following categories: managerial 3,supervisory 9, craftsmen 21, semi-skilled 67,others 44. Show these data on a pie diagram.10. The way in which an apprentice spent histime over a one-month period is as follows:drawing office44 hours,production 64 hours,training 12 hours, at college 28 hours. Use apie diagram to depict this information.11. (a) With reference to Figure 31.5, determinethe amount spent on labourand materialsto produce 1650 units of the product.(b) If in year 2 of Figure 31.4 1% cor-responds to 2.5 dwellings, how manybungalows are sold in that year?12. (a) If the company sells 23500 units perannum of the product depicted inFigure 31.5, determine the cost of theiroverheads per annum.(b) If 1% of the dwellings represented inyear 1 of Figure 31.4 corresponds to2 dwellings, find the total number ofhouses sold in that year.31.3 Presentation of grouped dataWhen the number of members in a set is small, say ten orless, the data can be represented diagrammatically with-out further analysis, by means of pictograms, bar charts,percentage components bar charts or pie diagrams (asshown in Section 31.2).For sets having more than ten members, those mem-bers having similar values are grouped together in
Presentation of statistical data 293classes to form a frequency distribution. To assist inaccurately counting members in the various classes, atally diagram is used (see Problems 8 and 12).A frequency distribution is merely a table show-ing classes and their corresponding frequencies (seeProblems 8 and 12). The new set of values obtainedby forming a frequency distribution is called groupeddata. The terms used in connection with grouped dataare shown in Figure 31.6(a). The size or range of a classis given by the upper class boundary value minus thelower class boundary value and in Figure 31.6(b) is7.65−7.35; i.e., 0.30. The class interval for the classshown in Figure 31.6(b) is 7.4 to 7.6 and the classmid-point value is given by(upper class boundary value) + (lower class boundary value)2and in Figure 31.6(b) is7.65 + 7.352i.e. 7.5Class interval(a)(b)LowerclassboundaryUpperclassboundaryClassmid-point7.35to 7.3 7.4 to 7.6 7.7 to7.657.5Figure 31.6One of the principal ways of presenting grouped datadiagrammatically is to use a histogram, in which theareas of vertical, adjacent rectangles are made propor-tional to frequencies of the classes (see Problem 9).When class intervals are equal, the heights of the rect-angles of a histogram are equal to the frequencies of theclasses. For histograms having unequal class intervals,the area must be proportional to the frequency. Hence,if the class interval of class A is twice the class inter-val of class B, then for equal frequencies the heightof the rectangle representing A is half that of B (seeProblem 11).Anothermethod of presenting grouped data diagram-matically is to use a frequency polygon, which is thegraph produced by plottingfrequency against class mid-point values and joining the co-ordinates with straightlines (see Problem 12).A cumulative frequency distribution is a tableshowing the cumulative frequency for each value ofupper class boundary. The cumulative frequency for aparticular value of upper class boundary is obtained byadding the frequency of the class to the sum of the pre-vious frequencies. A cumulative frequency distributionis formed in Problem 13.The curve obtained by joining the co-ordinates ofcumulative frequency (vertically) against upper classboundary (horizontally) is called an ogive or a cumu-lative frequency distribution curve (see Problem 13).Problem 8. The data given below refer to the gainof each of a batch of 40 transistors, expressedcorrect to the nearest whole number. Form afrequency distribution for these data having sevenclasses81 83 87 74 76 89 82 8486 76 77 71 86 85 87 8884 81 80 81 73 89 82 7981 79 78 80 85 77 84 7883 79 80 83 82 79 80 77The range of the data is the value obtained by tak-ing the value of the smallest member from that of thelargest member. Inspection of the set of data shows thatrange = 89 − 71 = 18. The size of each class is givenapproximately by the range divided by the number ofclasses. Since 7 classes are required, the size of eachclass is 18÷ 7; that is, approximately 3. To achieveseven equal classes spanning a range of values from 71to 89, the class intervals are selected as 70–72, 73–75,and so on.To assist with accurately determining the number ineach class, a tally diagram is produced, as shown inTable 31.1(a). This is obtained by listing the classesin the left-hand column and then inspecting each of the40 members of the set in turn and allocating them tothe appropriate classes by putting '1's in the appropri-ate rows. Every fifth '1' allocated to a particular row isshown as an obliqueline crossing the four previous '1's,to help with final counting.A frequency distribution for the data is shown inTable 31.1(b) and lists classes and their correspond-ing frequencies, obtained from the tally diagram. (Classmid-point values are also shown in the table, since theyare used for constructing the histogram for these data(see Problem 9).)
294 Basic Engineering MathematicsTable 31.1(a)Class Tally70–72 173–75 1176–78 1111 1179–81 1111 1111 1182–84 1111 111185–87 1111 188–90 111Table 31.1(b)Class Class mid-point Frequency70–72 71 173–75 74 276–78 77 779–81 80 1282–84 83 985–87 86 688–90 89 3Problem 9. Construct a histogram for the datagiven in Table 31.1(b)The histogram is shown in Figure 31.7. The width ofthe rectangles corresponds to the upper class boundaryvalues minus the lower class boundary values and theheightsoftherectanglescorrespond to theclassfrequen-cies. The easiest way to draw a histogram is to mark the71426Frequency10812141674 77 80 83Class mid-point values8986Figure 31.7class mid-point values on the horizontal scale and drawthe rectangles symmetrically about the appropriateclassmid-point values and touching one another.Problem 10. The amount of money earnedweekly by 40 people working part-time in a factory,correct to the nearest £10, is shown below. Form afrequency distribution having 6 classes for thesedata80 90 70 110 90 160 110 80140 30 90 50 100 110 60 10080 90 110 80 100 90 120 70130 170 80 120 100 110 40 11050 100 110 90 100 70 110 80Inspection of the set given shows that the majority ofthe members of the set lie between £80 and £110 andthat there is a much smaller number of extreme val-ues ranging from £30 to £170. If equal class intervalsare selected, the frequency distribution obtained doesnot give as much information as one with unequal classintervals. Since the majority of the members lie between£80 and £100, the class intervals in this range areselected to be smaller than those outside of this range.There is no unique solution and one possible solution isshown in Table 31.2.Table 31.2Class Frequency20–40 250–70 680–90 12100–110 14120–140 4150–170 2Problem 11. Draw a histogram for the data givenin Table 31.2When dealing with unequal class intervals, the his-togram must be drawn so that the areas (and notthe heights) of the rectangles are proportional to the
296 Basic Engineering MathematicsTo assist with accurately determining the numberin each class, a tally diagram is produced as shownin Table 31.4. This is obtained by listing the classes inthe left-hand column and then inspecting each of the 50members of the set of data in turn and allocating it to theappropriate class by puttinga '1' in the appropriate row.Each fifth '1' allocated to a particular row is marked asan oblique line to help with final counting.Table 31.4Class Tally7.1 to 7.3 1117.4 to 7.6 11117.7 to 7.9 1111 11118.0 to 8.2 1111 1111 11118.3 to 8.5 1111 1111 18.6 to 8.8 1111 18.9 to 9.1 11A frequency distribution for the data is shown inTable 31.5 and lists classes and their corresponding fre-quencies. Class mid-points are also shown in this tablesince they are used when constructing the frequencypolygon and histogram.Table 31.5Class Class mid-point Frequency7.1 to 7.3 7.2 37.4 to 7.6 7.5 57.7 to 7.9 7.8 98.0 to 8.2 8.1 148.3 to 8.5 8.4 118.6 to 8.8 8.7 68.9 to 9.1 9.0 2A frequency polygon is shown in Figure 31.9,the co-ordinates corresponding to the class mid-point/frequency values given in Table 31.5. The co-ordinates are joined by straight lines and the polygonis 'anchored-down' at each end by joining to the nextclass mid-point value and zero frequency.A histogram is shown in Figure 31.10, the widthof a rectangle corresponding to (upper class boundary7.24206Frequency12108147.87.5 8.1Class mid-point values8.7 9.08.4Frequency polygonFigure 31.9value – lower class boundary value) and height corre-sponding to the class frequency. The easiest way to drawahistogramisto mark classmid-point valueson thehori-zontal scale and to draw the rectangles symmetricallyabout the appropriate class mid-point values and touch-ing one another. A histogram for the data given inTable 31.5 is shown in Figure 31.10.7.24206Frequency10812147.57.357.657.958.258.558.859.157.8 8.1 8.4Class mid-point values9.08.7HistogramFigure 31.10Problem 13. The frequency distribution for themasses in kilograms of 50 ingots is7.1 to 7.3 37.4 to 7.6 57.7 to 7.9 98.0 to 8.2 148.3 to 8.5 118.6 to 8.8 68.9 to 9.1 2Form a cumulative frequency distribution for thesedata and draw the corresponding ogive
Chapter 32Mean, median, mode andstandard deviation32.1 Measures of central tendencyA single value, which is representative of a set of values,may be used to give an indication of the general size ofthe members in a set, the word 'average' often beingused to indicate the single value. The statistical termused for 'average' is the 'arithmetic mean' or just the'mean'.Other measures of central tendency may be used andthese include the median and the modal values.32.2 Mean, median and mode fordiscrete data32.2.1 MeanThe arithmeticmean value isfound by adding togetherthe values of the members of a set and dividing by thenumber of members in the set. Thus, the mean of the setof numbers {4, 5, 6, 9} is4 + 5 + 6 + 94i.e. 6In general, the mean of the set {x1, x2, x3,...xn} isx =x1 + x2 + x3 + ···+ xnnwritten asxnwhere is the Greek letter 'sigma' and means 'the sumof ' and x (called x-bar) is used to signify a mean value.32.2.2 MedianThe median value often gives a better indication of thegeneral size of a set containing extreme values. The set{7, 5, 74, 10} has a mean value of 24, which is not reallyrepresentative ofany of the values of the members of theset. The median value is obtained by(a) ranking the set in ascending order of magnitude,and(b) selecting the value of the middle member for setscontaining an odd number of members or findingthe value of the mean of the two middle membersfor sets containing an even number of members.For example, the set {7, 5, 74, 10} is ranked as{5, 7, 10, 74} and, since it contains an even numberof members (four in this case), the mean of 7 and 10 istaken, giving a median value of 8.5. Similarly, the set{3, 81, 15, 7, 14} is ranked as {3, 7, 14, 15, 81} andthe median value is the value of the middle member,i.e. 14.32.2.3 ModeThe modal value, or mode, is the most commonlyoccurring value in a set. If two values occur withthe same frequency, the set is 'bi-modal'. The set{5, 6, 8, 2, 5, 4, 6, 5, 3} has a modal value of 5, since themember having a value of 5 occurs the most, i.e. threetimes.Problem 1. Determine the mean, median andmode for the set {2, 3, 7, 5, 5, 13, 1, 7, 4, 8, 3, 4, 3}DOI: 10.1016/B978-1-85617-697-2.00032-6
300 Basic Engineering MathematicsThe mean value is obtained by adding together thevalues of the members of the set and dividing by thenumber of members in the set. Thus,mean value,x=2 + 3 + 7 + 5 + 5 + 13 + 1 + 7 + 4 + 8 + 3 + 4 + 313=6513= 5To obtain the median value the set is ranked, that is,placed in ascending order of magnitude, and since theset contains an odd number of members the value of themiddle member is the median value. Ranking the setgives {1, 2, 3, 3, 3, 4, 4, 5, 5, 7, 7, 8, 13}. The middleterm is the seventh member; i.e., 4. Thus, the medianvalue is 4.The modal value is the value of the most commonlyoccurring member and is 3, which occurs three times,all other members only occurring once or twice.Problem 2. The following set of data refers to theamount of money in £s taken by a news vendor for6 days. Determine the mean, median and modalvalues of the set{27.90, 34.70, 54.40, 18.92, 47.60, 39.68}Mean value=27.90 + 34.70 + 54.40 + 18.92 + 47.60 + 39.686= £37.20The ranked set is{18.92, 27.90, 34.70, 39.68, 47.60, 54.40}.Since the set has an even number of members, the meanof the middle two members is taken to give the medianvalue; i.e.,median value =34.70 + 39.682= £37.19Since no two members have the same value, this set hasno mode.Now try the following Practice ExercisePracticeExercise 125 Mean, median andmode for discrete data (answers onpage 353)In problems 1 to 4, determine the mean, medianand modal values for the sets given.1. {3, 8, 10, 7, 5, 14, 2, 9, 8}2. {26, 31, 21, 29, 32, 26, 25, 28}3. {4.72, 4.71, 4.74, 4.73, 4.72, 4.71, 4.73, 4.72}4. {73.8, 126.4, 40.7, 141.7, 28.5, 237.4, 157.9}32.3 Mean, median and mode forgrouped dataThe mean value for a set of grouped data is found bydetermining thesumofthe(frequency× classmid-pointvalues) and dividing by the sum of the frequencies; i.e.,mean value,x =f1x1 + f2x2 + ···fnxnf1 + f2 + ···+ fn=( fx)fwhere f is the frequency of the class having a mid-pointvalue of x, and so on.Problem 3. The frequency distribution for thevalue of resistance in ohms of 48 resistors is asshown. Determine the mean value of resistance20.5–20.9 321.0–21.4 1021.5–21.9 1122.0–22.4 1322.5–22.9 923.0–23.4 2The class mid-point/frequency values are 20.7 3, 21.210, 21.7 11, 22.2 13, 22.7 9 and 23.2 2.For grouped data, the mean value is given byx =( f x)fwhere f is the class frequency and x is the class mid-point value. Hence mean value,(3 × 20.7) + (10 × 21.2) + (11 × 21.7)x =+(13 × 22.2) + (9 × 22.7) + (2 × 23.2)48=1052.148= 21.919...i.e. the mean value is 21.9 ohms, correct to 3 significantfigures.32.3.1 HistogramsThe mean, median and modal values for grouped datamay be determined from a histogram. In a histogram,
Mean, median, mode and standard deviation 301frequency values are represented vertically and variablevalueshorizontally.Themean valueisgiven by thevalueof the variable corresponding to a vertical line drawnthroughthe centroid of the histogram.The median valueis obtained by selecting a variable value such that thearea of the histogram to the left of a vertical line drawnthroughthe selected variable value is equal to the area ofthe histogram on the right of the line. The modal valueis the variable value obtained by dividing the width ofthe highest rectangle in the histogram in proportion tothe heights of the adjacent rectangles. The method ofdetermining the mean, median and modal values froma histogram is shown in Problem 4.Problem 4. The time taken in minutes toassemble a device is measured 50 times and theresults are as shown. Draw a histogram depictingthe data and hence determine the mean, median andmodal values of the distribution14.5–15.5 516.5–17.5 818.5–19.5 1620.5–21.5 1222.5–23.5 624.5–25.5 3The histogram is shown in Figure 32.1. The mean valuelies at the centroid of the histogram. With reference toany arbitrary axis, say YY shown at a time of 14 min-utes, the position of the horizontal value of the centroidcan be obtained from the relationship AM = (am),where A is the area of the histogram, M is the hor-izontal distance of the centroid from the axis YY, ais the area of a rectangle of the histogram and m isthe distance of the centroid of the rectangle from YY.The areas of the individual rectangles are shown circled14 15 16 17 18 19 20 21 22 23 24 25626 27426Frequency108121614Time in minutes1224321610EDAYY5.6ModeMedian MeanFCBFigure 32.1on the histogram giving a total area of 100 squareunits.The positions,m, of the centroids of the individualrectangles are 1,3,5,... units from YY. Thus100M = (10 × 1) + (16 × 3) + (32 × 5) + (24 × 7)+ (12 × 9) + (6 × 11)i.e. M =560100= 5.6 units from YYThus, the position of the mean with reference to thetime scale is 14 + 5.6, i.e. 19.6 minutes.The median is the value of time corresponding to avertical line dividing the total area of the histogram intotwo equal parts. The total area is 100 square units, hencethe vertical line must be drawn to give 50 units of area oneach side. To achieve this with reference to Figure 32.1,rectangle ABFE must be split so that 50 − (10 + 16)units of area lie on one side and 50− (24 + 12 + 6) unitsof area lie on the other. This shows that the area of ABFEis split so that 24 units of area lie to the left ofthe line and8 units of area lie to the right; i.e., the vertical line mustpass through 19.5 minutes. Thus, the median value ofthe distribution is 19.5 minutes.The mode is obtained by dividing the line AB, whichis the height of the highest rectangle, proportionally tothe heights of the adjacent rectangles. With referenceto Figure 32.1, this is achieved by joining AC and BDand drawing a vertical line through the point of inter-section of these two lines. This gives the mode of thedistribution, which is 19.3 minutes.Now try the following Practice ExercisePracticeExercise 126 Mean, median andmode for grouped data (answers onpage 354)1. 21 bricks have a mean mass of 24.2kg and29 similar bricks have a mass of 23.6kg.Determine the mean mass of the 50 bricks.2. The frequency distribution given below refersto the heights in centimetres of 100 people.Determine the mean value of the distribution,correct to the nearest millimetre.150–156 5157–163 18164–170 20171–177 27178–184 22185–191 8
302 Basic Engineering Mathematics3. The gain of 90 similar transistors is measuredand the results are as shown. By drawing ahistogramofthisfrequency distribution,deter-mine the mean, median and modal values ofthe distribution.83.5–85.5 686.5–88.5 3989.5–91.5 2792.5–94.5 1595.5–97.5 34. The diameters, in centimetres, of 60 holesbored in engine castings are measured andthe results are as shown. Draw a histogramdepicting these results and hence determinethe mean, median and modal values of thedistribution.2.011–2.014 72.016–2.019 162.021–2.024 232.026–2.029 92.031–2.034 532.4 Standard deviation32.4.1 Discrete dataThestandard deviation ofaset ofdatagivesan indicationof the amount of dispersion, or the scatter, of membersof the set from the measure of central tendency. Its valueis the root-mean-square value of the members of the setand for discrete data is obtained as follows.(i) Determine the measure of central tendency, usu-ally the mean value, (occasionally the median ormodal values are specified).(ii) Calculate the deviation of each member of the setfrom the mean, giving(x1 − x),(x2 − x),(x3 − x),...(iii) Determine the squares of these deviations; i.e.,(x1 − x)2,(x2 − x)2,(x3 − x)2,...(iv) Find the sum of the squares of the deviations, i.e.(x1 − x)2+ (x2 − x)2+ (x3 − x)2,...(v) Divide by the number of members in the set, n,giving(x1 − x)2+ (x2 − x)2+ x3 − x2+ ···n(vi) Determine the square root of (v).The standard deviation is indicated by σ (the Greekletter small 'sigma') and is written mathematically asstandard deviation,σ =(x − x)2nwhere x is a member of the set, x is the mean value ofthe set and n is the number of members in the set. Thevalue of standard deviation gives an indication of thedistance of the members of a set from the mean value.The set {1, 4, 7, 10, 13} has a mean value of 7 and astandard deviation of about 4.2. The set {5, 6, 7, 8, 9}also has a mean value of 7 but the standard deviation isabout 1.4. This shows that the members of the secondset are mainly much closer to the mean value than themembers of the first set. The method of determining thestandard deviation for a set of discrete data is shown inProblem 5.Problem 5. Determine the standard deviationfrom the mean of the set of numbers{5, 6, 8, 4, 10, 3}, correct to 4 significant figuresThe arithmetic mean, x =xn=5 + 6 + 8 + 4 + 10 + 36= 6Standard deviation, σ =(x − x)2nThe (x − x)2 values are (5 − 6)2,(6 − 6)2,(8 − 6)2,(4 − 6)2,(10 − 6)2and (3 − 6)2.The sum of the (x − x)2 values,i.e. (x − x)2, is 1 + 0 + 4 + 4 + 16 + 9 = 34and(x − x)2n=346= 5.·6since there are 6 members in the set.Hence, standard deviation,σ =(x − x)2n= 5.·6=2.380,correct to 4 significant figures.
304 Basic Engineering Mathematicsinto four equal parts. Thus, for the set {2, 3, 4, 5, 5, 7,9, 11, 13, 14, 17} there are 11 members and the valuesof the members dividing the set into four equal partsare 4, 7 and 13. These values are signified by Q1, Q2and Q3 and called the first, second and third quartilevalues, respectively. It can be seen that the second quar-tile value, Q2, is the value of the middle member andhence is the median value of the set.For grouped data the ogive may be used to determinethequartilevalues.In thiscase,pointsareselected on thevertical cumulative frequency values of the ogive, suchthat they divide the total value of cumulative frequencyinto four equal parts. Horizontal lines are drawn fromthese values to cut the ogive. The values of the variablecorresponding to these cutting points on the ogive givethe quartile values (see Problem 7).When a set contains a large number of members, theset can be split into ten parts, each containing an equalnumber of members. These ten parts are then calleddeciles.Forsetscontaining avery largenumberofmem-bers, the set may be split into one hundred parts, eachcontaining an equal number of members. One of theseparts is called a percentile.Problem 7. The frequency distribution givenbelow refers to the overtime worked by a group ofcraftsmen during each of 48 working weeks in ayear. Draw an ogive for these data and hencedetermine the quartile values.25–29 530–34 435–39 740–44 1145–49 1250–54 855–59 1The cumulative frequency distribution (i.e. upper classboundary/cumulative frequency values) is29.5 5, 34.5 9, 39.5 16, 44.5 27,49.5 39, 54.5 47, 59.5 48.The ogive is formed by plotting these values on a graph,as shown in Figure 32.2. The total frequency is dividedinto four equal parts, each having a range of 48 ÷ 4,i.e. 12. This gives cumulative frequency values of 0 to12 corresponding to the first quartile, 12 to 24 corre-sponding to the second quartile, 24 to 36 correspondingto the third quartile and 36 to 48 corresponding to thefourth quartile of the distribution; i.e., the distribution2510Cumulativefrequency403020504030 35Q1 Q2 Q345Upper class boundary values, hours55 6050Figure 32.2is divided into four equal parts. The quartile valuesare those of the variable corresponding to cumulativefrequency values of 12, 24 and 36, marked Q1, Q2 andQ3 in Figure 32.2. These values, correct to the nearesthour, are 37 hours, 43 hours and 48 hours, respec-tively. The Q2 value is also equal to the median valueof the distribution. One measure of the dispersion of adistribution is called the semi-interquartile range andis given by (Q3 − Q1) ÷ 2 and is (48 − 37) ÷ 2 in thiscase; i.e., 512hours.Problem 8. Determine the numbers contained inthe (a) 41st to 50th percentile group and (b) 8thdecile group of the following set of numbers.14 22 17 21 30 28 37 7 23 3224 17 20 22 27 19 26 21 15 29The set is ranked, giving7 14 15 17 17 19 20 21 21 2222 23 24 26 27 28 29 30 32 37(a) Thereare20 numbersin theset,hence thefirst 10%will be the two numbers 7 and 14, the second 10%will be 15 and 17, and so on. Thus, the 41st to 50thpercentile group will be the numbers 21 and 22.(b) The first decile group is obtained by splitting theranked set into 10 equal groups and selecting thefirst group; i.e., the numbers 7 and 14. The seconddecile group is the numbers 15 and 17, and so on.Thus, the 8th decile group contains the numbers27 and 28.
Mean, median, mode and standard deviation 305Now try the following Practice ExercisePracticeExercise 128 Quartiles, decilesand percentiles (answers on page 354)1. The number of working days lost due to acci-dents for each of 12 one-monthly periods areas shown. Determine the median and first andthird quartile values for this data.27 37 40 28 23 30 35 24 30 32 31 282. The number of faults occurring on a produc-tion line in a nine-week period are as shownbelow. Determine the median and quartilevalues for the data.30 27 25 24 27 37 31 27 353. Determine the quartile values and semi-interquartile range for the frequency distribu-tion given in problem 2 of Practice Exercise126, page 301.4. Determine the numbers contained in the 5thdecile group and in the 61st to 70th percentilegroups for the following set of numbers.40 46 28 32 37 42 50 31 48 4532 38 27 33 40 35 25 42 38 415. Determine the numbers in the 6th decile groupand in the 81st to 90th percentile group for thefollowing set of numbers.43 47 30 25 15 51 17 21 37 33 44 56 40 49 2236 44 33 17 35 58 51 35 44 40 31 41 55 50 16
Chapter 33Probability33.1 Introduction to probability33.1.1 ProbabilityThe probability of something happening is the likeli-hood or chance of it happening. Values of probabilityliebetween 0 and 1, where 0 represents an absolute impos-sibilityand 1 represents an absolute certainty. The prob-ability of an event happening usually lies somewherebetween these two extreme values and is expressedas either a proper or decimal fraction. Examples ofprobability arethat a length of copper wirehas zero resistance at 100◦C 0that a fair, six-sided dicewill stop with a 3 upwards16or 0.1667that a fair coin will landwith a head upwards12or 0.5that a length of copper wirehas some resistance at 100◦C 1If p is the probability of an event happening and q is theprobability of the same event not happening, then thetotal probability is p + q and is equal to unity, since itis an absolute certainty that the event either will or willnot occur; i.e., p + q = 1.Problem 1. Determine the probabilities ofselecting at random (a) a man and (b) a womanfrom a crowd containing 20 men and 33 women(a) The probability of selecting at random a man, p,is given by the rationumber of mennumber in crowdi.e. p =2020 + 33=2053or 0.3774(b) The probability of selecting at random a woman,q, is given by the rationumber of womennumber in crowdi.e. q =3320 + 33=3353or 0.6226(Check: the total probability should be equal to 1:p =2053and q =3353, thus the total probability,p + q =2053+3353= 1hence no obvious error has been made.)33.1.2 ExpectationThe expectation, E, of an event happening is definedin general terms as the product of the probability p ofan event happening and the number of attempts made,n; i.e., E = pn.Thus, since the probability of obtaining a 3 upwardswhen rolling a fair dice is 1/6, the expectation of gettinga 3 upwards on four throws of the dice is16× 4,i.e.23Thus expectation is the average occurrence of anevent.DOI: 10.1016/B978-1-85617-697-2.00033-8
Probability 307Problem 2. Find the expectation of obtaining a 4upwards with 3 throws of a fair diceExpectation is the average occurrence of an event and isdefined as the probability times the number of attempts.The probability, p, of obtaining a 4 upwards for onethrow of the dice is 1/6.If 3 attempts are made, n = 3 and the expectation, E, ispn, i.e.E =16× 3 =12or 0.5033.1.3 Dependent eventsA dependent event is one in which the probability ofan event happening affects the probability of anotherever happening. Let 5 transistors be taken at randomfrom a batch of 100 transistors for test purposes and theprobability of there being a defective transistor, p1, bedetermined. At some later time, let another 5 transistorsbe taken at random from the 95 remaining transistorsin the batch and the probability of there being a defec-tive transistor, p2, be determined. The value of p2 isdifferent from p1 since the batch size has effectivelyaltered from 100 to 95; i.e., probability p2 is depen-dent on probability p1. Since transistors are drawn andthen another 5 transistors are drawn without replacingthe first 5, the second random selection is said to bewithout replacement.33.1.4 Independent eventsAn independent event is one in which the probabilityof an event happening does not affect the probabilityof another event happening. If 5 transistors are taken atrandom from a batch oftransistorsand the probabilityofa defective transistor, p1, is determined and the processis repeated after the original 5 have been replaced inthe batch to give p2, then p1 is equal to p2. Since the5 transistors are replaced between draws, the secondselection is said to be with replacement.33.2 Laws of probability33.2.1 The addition law of probabilityThe addition law of probability is recognized by theword 'or' joining the probabilities.If pA is the probability of event A happening and pBis the probability of event B happening, the probabilityof event A or event B happening is given by pA + pB.Similarly, the probability of events A or B or C or...N happening is given bypA + pB + pC + ··· + pN33.2.2 The multiplicationlaw of probabilityThe multiplication law of probability is recognized bythe word 'and' joining the probabilities.If pA is the probability of event A happening and pB isthe probability of event B happening, the probability ofevent A and event B happening is given by pA × pB.Similarly, the probabilityof events A and B and C and...N happening is given bypA × pB × pC × ...× pNHere are some worked problems to demonstrate proba-bility.Problem 3. Calculate the probabilities ofselecting at random(a) the winning horse in a race in which 10 horsesare running and(b) the winning horses in both the first and secondraces if there are 10 horses in each race(a) Since only one of the ten horses can win, theprobability of selecting at random the winninghorse isnumber of winnersnumber of horses,i.e.110or 0.10(b) The probability of selecting the winning horse inthe first race is110The probability of selecting the winning horse inthe second race is110The probability of selecting the winning horsesin the first and second race is given by themultiplication law of probability; i.e.,probability =110×110=1100or 0.01Problem 4. The probability of a componentfailing in one year due to excessive temperature is1/20, that due to excessive vibration is 1/25 and thatdue to excessive humidity is 1/50. Determine theprobabilities that during a one-year period acomponent(a) fails due to excessive temperature andexcessive vibration,
308 Basic Engineering Mathematics(b) fails due to excessive vibration or excessivehumidity,(c) will not fail because of both excessivetemperature and excessive humidityLet pA be the probability of failure due to excessivetemperature, thenpA =120and pA =1920(where pA is the probability of not failing)Let pB be the probability of failure due to excessivevibration, thenpB =125and pB =2425Let pC be the probability of failure due to excessivehumidity, thenpC =150and pC =4950(a) The probability of a component failing due toexcessive temperature and excessive vibration isgiven bypA × pB =120×125=1500or 0.002(b) The probability of a component failing due toexcessive vibration or excessive humidity ispB + pC =125+150=350or 0.06(c) The probability that a component will not fail dueexcessive temperature and will not fail due toexcess humidity ispA × pC =1920×4950=9311000or 0.931Problem 5. A batch of 100 capacitors contains 73which are within the required tolerance values and17 which are below the required tolerance values,the remainder being above the required tolerancevalues. Determine the probabilities that, whenrandomly selecting a capacitor and then a secondcapacitor,(a) both are within the required tolerance valueswhen selecting with replacement,(b) the first one drawn is below and the secondone drawn is above the required tolerancevalue, when selection is without replacement(a) The probability of selecting a capacitor withinthe required tolerance values is 73/100. The firstcapacitor drawn is now replaced and a second oneis drawn from the batch of 100. The probabilityof this capacitor being within the required toler-ance values is also 73/100.Thus, the probabilityofselecting a capacitor within the required tolerancevalues for both the first and the second draw is73100×73100=532910000or 0.5329(b) The probability of obtaining a capacitor belowthe required tolerance values on the first drawis 17/100. There are now only 99 capacitorsleft in the batch, since the first capacitor is notreplaced. The probability of drawing a capacitorabove the required tolerance values on the seconddraw is 10/99, since there are (100 − 73 − 17),i.e. 10, capacitors above the required tolerancevalue. Thus, the probability of randomly select-ing a capacitor below the required tolerance valuesand subsequently randomly selecting a capacitorabove the tolerance values is17100×1099=1709900=17990or 0.0172Now try the following Practice ExercisePracticeExercise 129 Laws of probability(answers on page 354)1. In a batch of 45 lamps 10 are faulty. If onelamp is drawn at random, find the probabilityof it being (a) faulty (b) satisfactory.2. A box of fuses are all of the same shape andsize and comprises 23 2A fuses, 47 5A fusesand 69 13A fuses. Determine the probabilityof selecting at random (a) a 2A fuse (b) a 5Afuse (c) a 13A fuse.3. (a) Find the probability of having a 2upwards when throwing a fair 6-sideddice.(b) Find the probability of having a 5upwards when throwing a fair 6-sideddice.(c) Determine the probability of having a 2and then a 5 on two successive throws ofa fair 6-sided dice.
Probability 3094. Determine the probability that the total scoreis 8 when two like dice are thrown.5. The probability of event A happening is35and the probability of event B happening is23. Calculate the probabilities of(a) both A and B happening.(b) only event A happening, i.e. event Ahappening and event B not happening.(c) only event B happening.(d) either A, or B, or A and B happening.6. When testing 1000 soldered joints, 4 failedduring a vibration test and 5 failed due tohaving a high resistance. Determine the prob-ability of a joint failing due to(a) vibration.(b) high resistance.(c) vibration or high resistance.(d) vibration and high resistance.Here are some further worked problems on probability.Problem 6. A batch of 40 components contains 5which are defective. A component is drawn atrandom from the batch and tested and then a secondcomponent is drawn. Determine the probability thatneither of the components is defective when drawn(a) with replacement and (b) without replacement(a) With replacementThe probabilitythat the component selected on thefirst draw is satisfactory is 35/40 i.e. 7/8. The com-ponent is nowreplaced and a second draw is made.The probability that this component is also satis-factory is 7/8. Hence, the probability that both thefirst component drawn and the second componentdrawn are satisfactory is78×78=4964or 0.7656(b) Without replacementThe probability that the first component drawnis satisfactory is 7/8. There are now only 34satisfactory components left in the batch and thebatch number is 39. Hence, the probability ofdrawing a satisfactory component on the sec-ond draw is 34/39. Thus, the probability that thefirst component drawn and the second componentdrawn are satisfactory i.e., neither is defective is78×3439=238312or 0.7628Problem 7. A batch of 40 components contains 5which are defective. If a component is drawn atrandom from the batch and tested and then a secondcomponent is drawn at random, calculate theprobability of having one defective component, both(a) with replacement and (b) without replacementThe probability of having one defective component canbe achieved in two ways. If p is the probability of draw-ing a defective component and q is the probability ofdrawing a satisfactory component, then the probabilityof having one defective component is given by drawinga satisfactory component and then a defective compo-nent or by drawing a defective component and then asatisfactory one; i.e., by q × p + p × q.(a) With replacementp =540=18and q =3540=78Hence, the probability of having one defectivecomponent is18×78+78×18i.e.764+764=732or 0.2188(b) Without replacementp1 =18and q1 =78on the first of the two drawsThe batch number is now 39 for the second draw,thus,p2 =539and q2 =3539p1q2 + q1 p2 =18×3539+78×539=35 + 35312=70312or 0.2244
310 Basic Engineering MathematicsProblem 8. A box contains 74 brass washers,86 steel washers and 40 aluminium washers. Threewashers are drawn at random from the box withoutreplacement. Determine the probability that allthree are steel washersAssume, for clarity of explanation, that a washer isdrawn at random, then a second, then a third (althoughthis assumption does not affect the results obtained).The total number of washers is74 + 86 + 40, i.e. 200The probability of randomly selecting a steel washer onthe first draw is 86/200. There are now 85 steel washersin a batch of 199. The probability of randomly selectinga steel washer on the second draw is 85/199. There arenow 84 steel washers in a batch of 198. The probabilityof randomly selecting a steel washer on the third drawis 84/198. Hence, the probability of selecting a steelwasher on the first draw and the second draw and thethird draw is86200×85199×84198=6140407880400= 0.0779Problem 9. For the box of washers given inProblem 8 above, determine the probability thatthere are no aluminium washers drawn when threewashers are drawn at random from the box withoutreplacementThe probability of not drawing an aluminium washeron the first draw is 1 −40200i.e., 160/200. There arenow 199 washers in the batch of which 159 are not madeof aluminium. Hence, the probability of not drawingan aluminium washer on the second draw is 159/199.Similarly, the probability of not drawing an aluminiumwasher on the third draw is 158/198. Hence the proba-bility of not drawing an aluminium washer on the firstand second and third draws is160200×159199×158198=40195207880400= 0.5101Problem 10. For the box of washers in Problem 8above, find the probability that there are two brasswashers and either a steel or an aluminium washerwhen three are drawn at random, withoutreplacementTwo brass washers (A) and one steel washer (B) can beobtained in any of the following ways.1st draw 2nd draw 3rd drawA A BA B AB A ATwo brass washers and one aluminium washer (C) canalso be obtained in any of the following ways.1st draw 2nd draw 3rd drawA A CA C AC A AThus, there are six possible ways of achieving the com-binations specified. If A represents a brass washer,B a steel washer and C an aluminium washer, thecombinations and their probabilities are as shown.Draw ProbabilityFirst Second ThirdA A B74200×73199×86198= 0.0590A B A74200×86199×73198= 0.0590B A A86200×74199×73198= 0.0590A A C74200×73199×40198= 0.0274A C A74200×40199×73198= 0.0274C A A40200×74199×73198= 0.0274The probability of having the first combination or thesecond or the third, and so on, is given by the sum ofthe probabilities; i.e., by 3 × 0.0590 + 3 × 0.0274, i.e.0.2592
Probability 311Now try the following Practice ExercisePracticeExercise 130 Laws of probability(answers on page 354)1. The probabilitythat component A will operatesatisfactorily for 5 years is 0.8 and that B willoperate satisfactorily over that same period oftime is 0.75. Find the probabilities that in a 5year period(a) both components will operatesatisfactorily.(b) only component A will operatesatisfactorily.(c) only component B will operatesatisfactorily.2. In a particular street, 80% of the houses havelandline telephones. If two houses selected atrandom are visited, calculate the probabilitiesthat(a) they both have a telephone.(b) one has a telephone but the other doesnot.3. Veroboard pins are packed in packets of 20by a machine. In a thousand packets, 40 haveless than 20 pins. Find the probability that if 2packets are chosen at random, one will containless than 20 pins and the other will contain 20pins or more.4. A batch of 1kW fire elements contains 16which are within a power tolerance and 4which are not.If 3 elements are selected at ran-dom from the batch, calculate the probabilitiesthat(a) all three are within the power tolerance.(b) two are within but one is not within thepower tolerance.5. An amplifier is made up of three transis-tors, A, B and C. The probabilities of A, Bor C being defective are 1/20, 1/25 and1/50, respectively. Calculate the percentage ofamplifiers produced(a) which work satisfactorily.(b) which have just one defective transistor.6. Abox contains14 40Wlamps,28 60Wlampsand 58 25W lamps, all the lamps being of thesame shape and size. Three lamps are drawn atrandom from the box, first one, then a second,then a third. Determine the probabilities of(a) getting one25W,one40Wand one60Wlamp with replacement.(b) getting one25W,one40Wand one60Wlamp without replacement.(c) getting either one 25W and two 40Wor one 60W and two 40 Wlamps withreplacement.
Revision Test 13 : Statistics and probabilityThis assignment covers the material contained in Chapters 31–33. The marks available are shown in brackets at theend of each question.1. A company produces five products in the follow-ing proportions:Product A 24Product B 6Product C 15Product D 9Product E 18Draw (a) a horizontal bar chart and (b) a piediagram to represent these data visually. (9)2. State whether the data obtained on the followingtopics are likely to be discrete or continuous.(a) the number of books in a library.(b) the speed of a car.(c) the time to failure of a light bulb. (3)3. Draw a histogram, frequency polygon and ogivefor the data given below which refers to thediameter of 50 components produced by amachine.Class intervals Frequency1.30–1.32mm 41.33–1.35mm 71.36–1.38mm 101.39–1.41mm 121.42–1.44mm 81.45–1.47mm 51.48–1.50mm 4(16)4. Determine the mean, median and modal values forthe following lengths given in metres:28,20,44,30,32,30,28,34,26,28 (6)5. The length in millimetres of 100 bolts is as shownbelow.50–56 657–63 1664–70 2271–77 3078–84 1985–91 7Determine for the sample(a) the mean value.(b) the standard deviation, correct to 4 significantfigures. (10)6. The number of faulty components in a factory ina 12 week period is14 12 16 15 10 13 15 11 16 19 17 19Determine the median and the first and thirdquartile values. (7)7. Determine the probability of winning a prize ina lottery by buying 10 tickets when there are 10prizes and a total of 5000 tickets sold. (4)8. A sample of 50 resistors contains 44 which arewithin the required tolerance value, 4 which arebelow and the remainder which are above. Deter-mine the probability of selecting from the samplea resistor which is(a) below the required tolerance.(b) above the required tolerance.Now two resistors are selected at random fromthe sample. Determine the probability, correct to3 decimal places, that neither resistor is defectivewhen drawn(c) with replacement.(d) without replacement.(e) If a resistor is drawn at random from thebatch and tested and then a second resistor isdrawn from those left, calculate the probabil-ity of having one defective component whenselection is without replacement. (15)
324 Basic Engineering MathematicsNow try the following Practice ExercisePracticeExercise 138 Rates of change(answers on page 355)1. An alternating current, i amperes, is given byi = 10sin2πft, where f is the frequency inhertz and t is the time in seconds. Determinethe rate of change of current when t = 12 ms,given that f = 50 Hz.2. The luminous intensity, I candelas, of a lampis given by I = 8 × 10−4 V2, where V is thevoltage. Find(a) the rate of change of luminous intensitywith voltage when V = 100 volts.(b) the voltage at which the light is increas-ing at a rate of 0.5 candelas per volt.3. The voltage across the plates of a capacitor atany time t seconds is given by v = V e−t/CR,where V,C and R are constants. GivenV = 200V,C = 0.10μF and R = 2M , find(a) the initial rate of change of voltage.(b) the rate of change of voltage after 0.2s4. The pressure p of the atmosphere at height habove ground level is given by p = p0e−h/c,where p0 is the pressure at ground level and cis a constant. Determine the rate of change ofpressure with height when p0 = 1.013 × 105pascals and c = 6.05 × 104 at 1450 metres.
Chapter 35Introduction to integration35.1 The process of integrationThe process of integration reverses the process ofdifferentiation. In differentiation, if f (x) = 2x2 thenf (x) = 4x. Thus, the integral of 4x is 2x2; i.e., inte-gration is the process of moving from f (x) to f (x). Bysimilar reasoning, the integral of 2t is t2.Integration is a process of summation or adding partstogether and an elongated S, shown as , is used toreplace the words 'the integral of'. Hence, from above,4x = 2x2 and 2t is t2.In differentiation, the differential coefficientdydxindi-cates that a function of x is being differentiated withrespect to x, the dx indicating that it is 'with respectto x'.In integration the variable of integration is shownby adding d (the variable) after the function to beintegrated. Thus,4x dx means 'the integral of 4x with respect to x',and 2t dt means 'the integral of 2t with respect to t'As stated above, the differential coefficient of 2x2 is 4x,hence; 4x dx = 2x2. However, the differential coeffi-cient of 2x2 + 7 is also 4x. Hence, 4x dx could alsobe equal to 2x2 + 7. To allow for the possible presenceof a constant, whenever the process of integration isperformed a constant c is added to the result. Thus,4x dx = 2x2+ c and 2t dt = t2+ cc is called the arbitrary constant of integration.35.2 The general solution of integralsof the form axnThe general solution of integrals of the form axn dx,where a and n are constants and n = −1 is given byaxndx =axn+1n + 1+ cUsing this rule gives(i) 3x4dx =3x4+14 + 1+ c =35x5+ c(ii)49t3dt dx =49t3+13 + 1+ c =49t44+ c=19t4 + c(iii)2x2dx = 2x−2dx =2x−2+1−2 + 1+ c=2x−1−1+ c = −2x+ c(iv)√x dx = x12 dx =x12 +112 + 1+ c =x3232+ c=23√x3 + cEach of these results may be checked by differentia-tion.(a) The integral of a constant k is kx + c. Forexample,8dx = 8x + c and 5dt = 5t + cDOI: 10.1016/B978-1-85617-697-2.00035-1
Introduction to integration 331The shaded area in Figure 35.1 is given byshaded area =bay dx =baf(x) dxThus, determining the area under a curve by integrationmerely involves evaluating a definite integral, as shownin Section 35.4.There are several instances in engineering and sciencewhere the area beneath a curve needs to be accuratelydetermined. For example, the areas between the limitsof avelocity/time graph gives distance travelled,force/distance graph gives work done,voltage/current graph gives power, and so on.Should a curve drop below the x-axis then y(= f (x))becomes negative and f (x)dx is negative. Whendetermining such areas by integration, a negative signis placed before the integral. For the curve shown inFigure 35.2, the total shaded area is given by (area E+area F +area G).E0 FGya b c d xy ϭ f(x)Figure 35.2By integration,total shaded area =baf(x) dx −cbf(x) dx+dcf(x) dx(Note that this is not the same asdaf (x) dx)It is usually necessary to sketch a curve in order to checkwhether it crosses the x-axis.Problem 26. Determine the area enclosed byy = 2x + 3, the x-axis and ordinates x = 1 andx = 4y = 2x + 3 is a straight line graph as shown inFigure 35.3, in which the required area is shown shaded.1210864210 2 3 4 5 xy5 2x 13yFigure 35.3By integration,shaded area =41y dx =41(2x + 3)dx =2x22+ 3x41= [(16 + 12)−(1 + 3)] = 24 square units(This answer may be checked since the shaded areais a trapezium: area of trapezium =12(sum of paral-lel sides)(perpendiculardistance between parallel sides)=12(5 + 11)(3) = 24 square units.)Problem 27. The velocity v of a body t secondsafter a certain instant is given by v = 2t2 + 5 m/s.Find by integration how far it moves in the intervalfrom t = 0 to t = 4sSince 2t2 + 5 is a quadratic expression, the curvev = 2t2 + 5 is a parabola cutting the v-axis at v = 5,as shown in Figure 35.4.The distance travelled is given by the area under the v/tcurve (shown shaded in Figure 35.4). By integration,shaded area =40vdt =40(2t2+ 5)dt =2t33+ 5t40=2(43)3+ 5(4) − (0)i.e. distance travelled = 62.67 m
334 Basic Engineering MathematicsShaded area =π/30y dx=π/30sin2x dx = −12cos2xπ/30= −12cos2π3− −12cos0= −12−12− −12(1)=14+12=34or 0.75 square unitsNow try the following Practice ExercisePracticeExercise 141 Area under curves(answers on page 355)Unless otherwise stated all answers are in squareunits.1. Show by integration that the area of a rect-angle formed by the line y = 4, the ordinatesx = 1 and x = 6 and the x-axis is 20 squareunits.2. Show by integration that the area of the trian-gle formed by the line y = 2x, the ordinatesx = 0 and x = 4 and the x-axis is 16 squareunits.3. Sketch the curve y = 3x2 + 1 betweenx = −2 and x = 4. Determine by integrationthe area enclosed by the curve, the x-axis andordinates x = −1 and x = 3. Use an approx-imate method to find the area and compareyour result with that obtained by integration.4. The force F newtons acting on a body at adistance x metres from a fixed point is givenby F = 3x + 2x2. If work done =x2x1F dx,determine the work done when the bodymoves from the position where x1 = 1 m tothat when x2 = 3m.In problems 5 to 9, sketch graphs of thegiven equa-tions and then find the area enclosed between thecurves, the horizontal axis and the given ordinates.5. y = 5x; x = 1, x = 46. y = 2x2 − x + 1; x = −1, x = 27. y = 2sin2x; x = 0, x =π48. y = 5cos3t; t = 0,t =π69. y = (x − 1)(x − 3); x = 0, x = 310. The velocity v of a vehicle t seconds after acertain instant isgiven by v = 3t2+ 4 m/s.Determine how far it moves in the intervalfrom t = 1s to t = 5s.
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Resources
The following resources are available to you to help you find success in your learning of the world of mathematics.
Practice Tests - This is a collection of various practice tests to determine if you are sufficently prepared for a particular math course.
Advanced Writing - Great information on advanced writing and how it applies to mathematics.
Normal Probability - is a visualization program offering statistics students insights and computations for the relationship between z-values and areas under the standard normal curve. (Please note that this may not be accessible from the BYU-Idaho campus wireless network.)
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Find a Marana MathMy entire Controls and Stability course was involved with MATLAB. Attitude Dynamics was based in MATLAB. I'm an aerospace engineer, every single calculation that is too complex for a calculator, or requires modelling, is done in MATLAB
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The teacher's guide includes...
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Provide students with a college-prep Algebra II course that will allow them to easily progress onto even more difficult mathematical challenges. Saxon Algebra 2, 4th Edition prepares students for calculus through explicit embedded geometry instruction. Trigonometry concepts, statistics, and applications for other subjects such as physics and chemistry are also included. Incremental lessons include a Warm Up activity; New Concepts section that introduces new concepts through examples with sidebar hints and notes; and Lesson Practice questions with lesson reference numbers underneath the question number. Online connections are given throughout for additional help. Real-world applications and continual practice & review provide the time needed to master each concept, helping students to...
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Provide students with a college-prep math course that will give them the foundation they need to successfully move into higher levels of math. Saxon Algebra 1, 4th Edition covers all of the traditional first-year algebra topics while helping students build higher-order thinking skills, real-world application skills, reasoning, and an understanding of interconnecting math strands. Saxon Algebra 1 focuses on algebraic thinking through multiple representations, including verbal, numeric, symbolic, and graphical, while graphing calculator labs model mathematical situations. Incremental lessons include a Warm Up activity; New Concepts section that introduces new concepts through examples with sidebar hints and notes; and Lesson Practice questions with lesson reference numbers underneath the...
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Step-by-step, to-the-point lessons will help both students looking to incorporate supplemental teaching into their Saxon curriculum, as well as struggling students who need additional help. With one session per Saxon lesson, Art Reed presents the concepts taught in that chapter with clarifying examples. The actual problems from the textbook aren't used; students should read the text, watch the video, and work the workbook problems with a more fully-understood knowledge of the process. Set up with Mr. Reed using a whiteboard to illustrate examples in a classroom, students can easily pause or rewind if needed. Algebra 8/7, 2nd or 3rd Edition. DVDs come in plastic clamshell case
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Product Details
See What's Inside
Product Description
By Gwendolyn Lloyd, Sybilla Beckmann, Rose Mary Zbiek
Are sequences functions? Why can't the popular "vertical line test" be applied in some cases to determine if a relation is a function? How does the idea of rate of change connect with simpler ideas about proportionality as well as more advanced topics in calculus?
How much do you know… and how much do you need to know?
Helping your high school students develop a robust understanding of functions requires that you understand mathematics deeply. But what does that mean?
This book focuses on essential knowledge for teachers about functions. It is organized around five big ideas, supported by multiple smaller, interconnected ideas—essential understandings. Taking you beyond a simple introduction to functions, thisFocus on the ideas that you need to understand thoroughly to teach confidentlyCustomers Who Bought This Also Bought...
Connect the process of problem solving with the content of the Common Core. The first of a series, this book will help mathematics educators illuminate a crucial link between problem solving and the Common Core State Standards.
This book focuses on essential knowledge for teachers about proof and the process of proving. It is organized around five big ideas, supported by multiple smaller, interconnected ideas—essential understandings.
Middle school students consolidate their understanding of integers and rational numbers, increasing their facility with fractions, decimals, and percents and encountering proportionality. This book shows how students can explore these important ideas.
This book examines the study of geometry in the middle grades as a pivotal point in the mathematical learning of students and emphasizes the geometric thinking that can develop in grades 6–8 as a result of hands-on exploration.
The National Council of Teachers of Mathematics is the public voice of mathematics education, supporting teachers to ensure equitable mathematics learning of the highest quality for all students through vision, leadership, professional development, and research.
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Importance of maths not fully understood by students
Jun 25, 2014
Too many sixth form students do not have a realistic understanding of either the relevance of Mathematics and Statistics to their discipline or of the demands that will be put upon them in undergraduate study, according to a new report published today by the Higher Education Academy (HEA). The report examines the mathematical and statistical needs of students in undergraduate disciplines including Business and Management, Chemistry, Economics, Geography, Sociology and Psychology.
Professor Jeremy Hodgen, lead author of the report from the Department of Education & Professional Studies, said: 'Too few students in the UK study Mathematics after the age of 16, yet the study demonstrates that Mathematics matters across a range of subjects at university. The report recommends that prospective undergraduates are better informed of this when applying to higher education.'
Lack of confidence and anxiety about Mathematics and Statistics is also a problem for many students, making the transition into higher education particularly challenging. A number of recommendations are made within the report to address this problem, but overall it calls for better dialogue between the sectors so that pre-university students have a better understanding of what is expected of them and the higher education sector has a better understanding of what their undergraduates can do.
The report also draws attention to developments at pre-university level, where new 'Core Maths' courses are being designed to meet the needs of the many students (the report estimates at least 200,000 a year) who need Mathematics but for whom a full A-level would not be appropriate. It calls for higher education to become actively involved in and to influence this work.
Dr Mary McAlinden, Discipline Lead for Mathematics, Statistics and Operational Research at the HEA said: 'Many students are surprised at the amount of mathematical content in their undergraduate programmes and some struggle to cope with this content.
'This project, and the accompanying reports, seeks to promote greater understanding between the higher education and pre-university sectors so that students will arrive at university better prepared and better able to cope with the mathematical and statistical demands of their undergraduate studies.'
Dr Janet De Wilde, Head of STEM at the HEA said: 'This report demonstrates the importance that the HEA places on this topic. The recommendations it contains are valuable to the sector to help further the discussion between the secondary and tertiary sector to inform policy development and teaching practice to address the importance of mathematical and statistical skills.'
The solution to England's poor participation rate in post-16 maths education could lie in a new qualification that provides a clear and attractive alternative for students who don't currently go on to study ...
For many years, studies have shown that American students score significantly lower than students worldwide in mathematics achievement, ranking 25th among 34 countries. Now, researchers from the University of Missouri haveAll young people should continue to study maths at least until they are 18, even if they have already gained a good GCSE in the subject, the Sutton Trust said today, because the GCSE curriculum fails to give them the practical
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Core-Plus Mathematics, is a standards-based, four-year integrated series covering the same mathematics concepts students learn in the Algebra 1-Geometry-Algebra 2-Precalculus sequence.
Concepts from algebra, geometry, probability, and statistics are integrated, and the mathematics is developed using context-centered investigations.
Developed by the CORE-Plus Math Project at Western Michigan University with funding from the National Science Foundation (NSF), Core-Plus Mathematics is written for all students to be successful in mathematics.
Core-Plus Mathematics is the number one high school NSF/reform program and it is published by Glencoe/McGraw-Hill, the nation's number one secondary mathematics company.
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My daughter's school district gave her this book to study for a summer math test-out program. I have to agree with the other reviewers that it is a very poor choice. It is essentially a book of questions, some of which have longer introductions that try to serve as vignettes to teach the concepts. There is virtually no formal teaching of concepts. I have 6 years of engineering education and I had a hard time understanding the point of a lot of the questions.
My daughter needed the book for her math class in school so I looked for it on Amazon and paid half for it that I would have in a regular store. The book came exactly how it was described and I am very pleased with it!!! The book is in great condition, almost like new, no markings or torn pages. The covers and binder are in great shape, too! Great choice and great buy!!!
I bought this book for my nephew to study in the summer. When I got the book and went over it, I realized this book is not good. The reasons: 1) Each chapter has only 1 or two pages describing about the chapter then followed by many problems. The explanation is insufficient and there is no example. 2) There is no answer in the back for any of the problems.
Conclusion: This is a bad series of books and it is a waste of money. I have read many mathbooks and this is definitely the worst mathbook.
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This Book
* The basic approach and application of algebra to problem solving
* The number system (in a much broader way than you have known it from arithmetic)
* Monomials and polynomials; factoring algebraic expressions; how to handle algebraic fractions; exponents, roots, and radicals; linear and fractional equations
* Functions and graphs; quadratic equations; inequalities; ratio, proportion, and variation; how to solve word problems, and more
Authors Peter Selby and Steve Slavin emphasize practical algebra throughout by providing you with techniques for solving problems in a wide range of disciplines—from engineering, biology, chemistry, and the physical sciences, to psychology and even sociology and business administration. Step by step, Practical Algebra shows you how to solve algebraic problems in each of these areas, then allows you to tackle similar problems on your own, at your own pace. Self-tests are provided at the end of each chapter so you can measure your mastery.
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Meet the Author
PETER SELBY (deceased) was Director of Educational Technology at Man Factors Associates, a human factors engineering consulting firm. He is the author of two other self-teaching guides: Quick Algebra Review: A Self-Teaching Guide and Geometry and Trigonometry for Calculus: A Self-Teaching Guide, both published by Wiley.
STEVE SLAVIN, Ph.D., is Associate Professor of Economics at Union County College, Cranford, New Jersey. He has written over 300 newspaper and magazine articles, and is the author of four other books, including All the Math You ll Ever Need: A Self-Teaching Guide and Economics: A Self-Teaching Guide 18, 2002
Excellent book!
I'm a high school student and I was struggling with algebra for quite some time. I bought this book and in about four weeks I mastered the subject. With this book just about anyone can learn algebra.
3 out of 3 people found this review helpful.
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Anonymous
Posted February 1, 2007
Very good book....
I was worried that this book maybe outdated but I went ahead and bought it since it was only $17 something. I was amazed at how well it was written for the price. I have read textbooks that run in the hundreds and their not written as well as this book. This book not only is up to date but I find it very easy to read. You don't have to know anything about algebra to start reading this book. Because it explaines it all to you. If you would like to learn algebra then buy this book.
2 out of 2 people found this review helpful.
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Anonymous
Posted May 25, 2006
Out of date teaching method
I bought this book to get up to speed for College Algebra class. I am 55 years old and took my last Intermeadiate Algebra Class 12 years ago. This book uses old methods of teaching. It is black and white and does not do a very good job of holding your attention. The expainations are very poor. My advice is to buy a modern graphic rich text book. This book was a waste of my money.
1 out of 3 people found this review helpful.
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Anonymous
Posted June 18, 2006
Great review book
I used this book to review for a basic skills teacher certification test and aced my test. This book was very easy to follow. The practice problems in each chapter started easy and progressed to a more challenging level. Each practice question was referenced back to a specific section in the chapter making further review quick and easy.
1 out of 1 people found this review helpful.
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Anonymous
Posted July 19, 2005
preparing for IBEW aptitude test
I recently used this book as a resource in preparation for the IBEW aptitude test in NYC. As a music major in college, I took no math classes and even though I did well in high school math, this book proved to be an invaluable resource. I used it studying on my own outside of a classrom and found it very easy to understand, and even though I had only completed 50% of the book upon taking the test, I felt completely prepared for the algrbraic sections. This is an excellent book for anyone needing to brush up on their algebra.
1 out of 1 people found this review helpful.
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Anonymous
Posted June 26, 2003
Very nice
This is the best book in the world. I learned so fast with this book, its so easy!
1 out of 1 people found this review helpful.
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Anonymous
Posted October 9, 2004
great book
After being out of school for 20 odd years. i am going to go to college and needed to re-learn alegbra.. this book is easy to understand, and pick up quickly the methods taught. I highly recommend this book..
1 out of 1 people found this review helpful.
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Anonymous
Posted December 28, 2003
If You always wanted to learn Algebra but.....
I am 43 years old and had taken several algebra course in college when I was younger but failed them. Since then, I thought I was unteachable when It came to higher math until I came across this book. Quite simply, I cannot say enough praise for this book. It has opened up a new world for me which I have always wanted to learn. Without hesitation or reservation, I recomend this book to anyone who thought Algebra was beyond their reach. I did it, you can too!
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Anonymous
Posted August 8, 2003
Very good review text.
I used this book as a rush to catch up in preparation for taking a calculus course. It has been quite a few years since I last took Algebra. This material is laid out in a wonderful sequential order that as long as you can understand the previous chapters you will be fully prepared for the next. One thing I wish the book had, but it didn't was one more chapter introducing imaginary numbers. If you are doing this your first time trying to fully 'get' Algebra this is definately a good place for you to get started. Very good problem sets, and excellent explainations.
1 out of 1 people found this review helpful.
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Anonymous
Posted August 12, 2003
Nothing but compliments
I went through college without opening a math textbook. I'm going back to school for pre-vet stuff, and need to skip ahead in math as far as possible. I disliked algebra in school, I found it hard to understand and terribly boring. This book put a new spin on things. It is well written, almost error-free and makes this subject interesting (as much as algebra can be interesting) and easy. I strongly encourage anyone who needs to learn or re-learn algebra to get this book. It's the only one you'll need.
1 out of 1 people found this review helpful.
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Elementary Numerical Analysis
9780471433378
ISBN:
0471433373
Edition: 3 Pub Date: 2003 Publisher: Wiley
Summary: Offering a clear, precise, and accessible presentation, complete with MATLAB programs, this new Third Edition of Elementary Numerical Analysis gives students the support they need to master basic numerical analysis and scientific computing. Now updated and revised, this significant revision features reorganized and rewritten content, as well as some new additional examples and problems. The text introduces core areas... of numerical analysis and scientific computing along with basic themes of numerical analysis such as the approximation of problems by simpler methods, the construction of algorithms, iteration methods, error analysis, stability, asymptotic error formulas, and the effects of machine arithmetic.
Kendall Atkinson is the author of Elementary Numerical Analysis, published 2003 under ISBN 9780471433378 and 0471433373. Five hundred eighty six Elementary Numerical Analysis textbooks are available for sale on ValoreBooks.com, sixty six used from the cheapest price of $117.34, or buy new starting at $960471433373 0471433373
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Advanced Numeracy Studies
This course deepens students' understanding of the impact of the principles and practices of teaching and learning mathematics in primary schools. Students will explore the numeracy demands of all key learning areas in the primary curriculum. Students will learn how to interpret school-based and system-wide numeracy data in order to make informed decisions about student numeracy needs. With a focus on helping students to reason or "work mathematically", this course will place particular attention upon instructional strategies and educational technologies for teaching and assessing the foundational mathematical concepts of number and algebra, measurement and geometry, statistics and probability.
Available in 2014
Callaghan Campus
Semester 1
Previously offered in 2013, 2012, 2011, 2010, 2009
Objectives
Upon completion of this course, students will have the capability to:
1. Identify the numeracy demands of all key learning areas in the primary curriculum. 2. Identify the numeracy demands of everyday life. 3. Interpret school-based and system-wide numeracy data in order to make informed decisions about student numeracy needs. 4. Design instruction that assists students to work mathematically. 5. Select appropriate instructional strategies and educational technologies for teaching and assessing the foundational mathematical concepts of quantity, measurement, spatial representation, and generalisation.
Content
* The numeracy demands of all key learning areas in the primary curriculum. * Numeracy in everyday life. * School-based and system-wide numeracy data. * Working mathematically. * Instructional strategies and educational technologies for teaching and assessing the foundational mathematical concepts of number and algebra, measurement and geometry, statistics and probability.
Replacing Course(s)
Not Applicable
Transition
Not Applicable
Industrial Experience
0
Assumed Knowledge
Students must complete EDUC6166 and EDUC6739 prior to enrolling.
Modes of Delivery
Internal Mode
Teaching Methods
Lecture Tutorial
Assessment Items
Essays / Written Assignments
1,000 words equivalent, consisting of data analysis and the design, implementation and evaluation of an individual learning plan.
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Written in a readable, yet mathematically mature manner appropriate for college algebra level students, Coburn's Precalculus uses narrative, extensive examples, and a range of exercises to connect seemingly disparate mathematical topics into a cohesive wholePrecalculus, Fifth Edition, by Lial, Hornsby, Schneider, and Daniels, engages and supports students in the learning process by developing both the conceptual understanding and the analytical skills necessary for success in mathematics. With the Fifth Edition, the authors adapt to the new ways in which students are learning, as well as the ever-changing classroom environment.
Get a good grade in your precalculus course with Precalculus
Mike Sullivanís time-tested approach focuses students on the fundamental skills they need for the course: preparing for class, practicing with homework, and reviewing the concepts. In the Ninth Edition, Precalculus has evolved to meet todayís course needs, building on these hallmarks by integrating projects and other interactive learning tools for use in the classroom or online.
WithThis market-leading text continues to provide both students and instructors with sound, consistently structured explanations of the mathematical concepts. Designed for a one- or two-term course that prepares students to study calculus, the new Eighth Edition retains the features that have made PRECALCULUS a complete solution for both students and instructors: interesting applications, cutting-edge design, and innovative technology combined with an abundance of carefully written exercises.
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Evergreen ICSE Self-Study in Mathematics Class-10 is a comprehensive book for students of standard X studying in schools affiliated to the Council for the Indian School Certificate Examinations. The book comprises of chapters which covers the prescribed mathematics syllabus in detail. In addition, the book consists of previous years' solved papers and model papers for thorough revision and final practice. This book is essential for students studying in ICSE-affiliated schools.
About Evergreen Publications
Evergreen Publications was established in the year 1939, with a vision of developing and publishing quality books for children and other school books. Their sister concern include Nova Publications and Ved Parkash & Sons among others. Evergreen Publications' books conform to several state boards' syllabi and are thus widely read by students. Their products range from text books, children's books, activity books, Manuals, e-Books, corporate books, magazines, catalogues, diaries and other educational books. Some of the books published by Evergreen Publications are My Way Fun with Numbers, Different Strokes Cursive Writing, Candid New Trends in English Grammar and Composition, Evergreen Learning with Activity Mathematics and CBSE Self-Study in Social Science.
Goodreads reviews
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The field of binary Logics has two main areas of application, the Digital Design of Circuits and Propositional Logics. In both cases it is possible to teach the theoretical foundations and to do some exercises, but in both cases the examples that can be done in class and by hand are far away from examples that are relevant for practical problems. more...
Almost all of a pendulum clock's accuracy resides in its pendulum. If the pendulum is accurate, the clock will be accurate. In this book, the author describes how to make a more accurate pendulum or pendulum clock. All scientific aspects of the pendulum design and operation are explained in simple terms and backed up with experimental data. - ;The... more...
Many interesting problems in mathematical fluid dynamics involve the behavior of solutions of nonlinear systems of partial differential equations as certain parameters vanish or become infinite. This book introduces the problems involving singular limits based on the concept of weak or variational solutions. more...
At the 19th Annual Conference on Parallel Computational Fluid Dynamics held in Antalya, Turkey, in May 2007, the developments and implementations of large-scale and grid computing were presented. This book comprises of the invited and selected papers of this conference. more...
The Advances in Chemical Physics series provides the chemical physics and physical chemistry fields with a forum for critical, authoritative evaluations of advances in every area of the discipline. Filled with cutting-edge research reported in a cohesive manner not found elsewhere in the literature, each volume of the Advances in Chemical Physics... more...
Everyday math skills can be painlessly learned and easily mastered, transforming readers from a person who doesn't know the meaning of APR into someone who understands credit card rates. Ryan's guide is broken into sections which review basic arithmetic from fractions to percents. more...
One of the most striking features of mathematics is the fact that we are much more certain about the mathematical knowledge we have than about what mathematical knowledge is knowledge of. Are numbers, sets, functions and groups physical entities of some kind? Are they objectively existing objects in some non-physical, mathematical realm? Are they ideas... more...
This is the first publication to describe the evolution of fluid dynamics as a major field in modern science and engineering. It contains a description of the interaction between applied research and application, taking as its example the history of fluid mechanics in the 20th century. The focus lies on the work of Ludwig Prandtl, founder of the aerodynamic... more...
Here, the author, a researcher of outstanding experience in this field, summarizes and combines the recent results and findings on advanced two-phase flow modeling and numerical methods otherwise dispersed in various journals, while also providing explanations for numerical and modeling techniques previously not covered by other books. The resulting... more...
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On the Shoulders of Giants : New Approaches to Numeracy - 90 edition
ISBN13:978-0309084499 ISBN10: 0309084490 This edition has also been released as: ISBN13: 978-0309042345 ISBN10: 0309042348
Summary: What mathematics should be learned by today's young people as well as tomorrow's workforce? On the Shoulders of Giants is a vision of richness of mathematics expressed in essays on change, dimension, quantity, shape, and uncertainty, each of which illustrate fundamental strands for school mathematics. These essays expand on the idea of mathematics as the language and science of patterns, allowing us to realize the importance of providing hands-on experience and the d...show moreevelopment of a curriculum that will enable students to apply their knowledge to diverse numerical problemsA1Selections burlington, WI
0309084490 fast shipping excellent service
$3.003090844904.94 +$3.99 s/h
Good
invisibledog Salt Lake City, UT
0309084490 Unmarked text. In a Good jacket.
$5.99 +$3.99 s/h
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MyEmporium belvidere, IL
0309084490 Book is in good condition. Light stain at top of book
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Fall 2013 Basic Mathematics (3 units)
Section MTH-11-0420 Class Begins August 19
Course Description
This course will be a study of the fundamental concepts and processes of basic arithmetic with whole numbers, fractions and decimals, including the fundamentals of percents; converting between fractions, decimals, and percents; unit conversion; and applications.
Instructor: Jason Edington
Email: [email protected]
After the class has started, please use MyMathLab Private Messages for communication.
Estimated Time per Week: Students can expect to spend approximately 9 hours per week reading, working on research assignments, taking quizzes and participating in online class discussions. This class is not self-paced. Specific assignments and discussions are meant to be completed for the whole class at the same time.
Special Requirements: Log into ETUDES by the first day of class and begin follow the instructions. Be sure to go through the orientation and follow directions to MyMathLab.
Assignments & Tests: Assignments and tests are outlined on MyMathLab.
Additional Comments: The entire course will be conducted online through the MyMathLab course management system. Students are required to have Internet access, an active email account, the ability to use word processing, conduct Internet searches, attach files, send emails and private messages, and work independently.
New to Etudes: Here is an Online Orientation (Flash presentation opens in a new window) that will show you the basics of how to use Etudes. Here is a flash tutorial (Flash presentation opens in a new window) that demonstrates the log in protocol. Be sure to check System Requirements before getting started with Etudes. You need to do this on each computer you use while taking a class through Etudes.
Etudes Course: You will log into the Etudes classroom with the log-in information provided below.
Login ID
Password
First 2 letters of first name +
First 2 letters of last name +
Last 5 digits of Student COLLEAGUE ID
(Type using all lower case letters)
Example: Jose A. Garcia
Student ID: 1021945
Username = joga21945
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Maple has a rich collection of Clickable Math features for easy problem-solving and exploration, as well as tools that make it easy to create your own interactive Math Apps that you can share and grade through The Möbius Project. In this session, you will learn a few simple techniques that will allow you to use Clickable Math features to compose, visualize, and solve a wide variety of mathematical problems.
In this session, you will learn:
Simple techniques to solve a wide variety of mathematical problems using Maple's Clickable Math features
This is a free session and there are only a limited number of spaces available. To RSVP to attend, please fill out the form. Please note, this session requires that each attendee bring a laptop equipped with a copy of Maple 17. If you don't have a working copy of Maple 17 on your laptop, request a copy in the form below and a temporary download copy of Maple 17 will be emailed to you. At the workshop, all the files needed for the exercises will be provided.
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[they're forcing you to analyze math equations so you get used to analyzing so that when you're confronted with things you can't solve with calculators, you'll be able to do so and not feel immediately overwhelmed]
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Major Requirements
Coursework Requirements
Students majoring in mathematics must complete MATH 115 and one of 116/120 (or the equivalent) and at least eight units of 200-level and 300-level courses. These eight units must include 205, 206, 302, 305, and two additional 300-level courses. (Thus a student who places out of 115/116 and starts in 205 requires only eight courses.) At most two of 206, 210 and 215 may be counted towards the major. These courses must be completed for the mathematics major:
Math 115: Calculus I and Math 116: Calculus II, or the equivalent
Math 205: Multivariable Calculus
Math 206: Linear Algebra
Math 302: Elements of Analysis I
Math 305: Abstract Algebra
At least two elective 300-level courses not counting any of 350, 360, 370.
A student may count Math 215/Phys 215 towards her mathematics major. However, she may count at most two of the course 206, 210, and 215 toward the major. Credit for Math 216/Phys 216 satisfies the requirement that a math major take 205, but cannot be counted as one of the 200- or 300-level units required for the major.
Major Presentation Requirement
Majors are also required to present one classroom talk in either their junior or senior year. This requirement can be satisfied with a presentation in the student seminar, but it can also be fulfilled by giving a talk in one of the courses whose catalog description says"Majors can fulfill the major presentation requirement in this course." In addition, a limited number of students may be able to fulfill the presentation requirement in other courses, with permission of the instructor
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BEGIN:VCALENDAR
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DTSTART:20131020T053000Z
DTEND:20131020T060000Z
LOCATION:On Alabama Public Television Channel:Main
TRANSP: OPAQUE
SEQUENCE:0
UID:
DTSTAMP:20140922T105014Z
SUMMARY:GED Connection Formulas
DESCRIPTION:This episode of GED Connection focuses on formulas. Formulas are writt en in the language of algebra using letters to stand for numbers. Thes e letters are called variables. A list of formulas will be supplied to the GED test taker but it is important to understand what the formula s mean in order to put them to use. The formulas covered in this progr am are: the formula for the area of a rectangle, how to calculate dist ance, circumference of a circle, area of a circle, volume of an object , and areas of two separate shapes or odd shapes. An architect shows t he viewer how she uses algebraic formulas to calculate the volume of a basement in an ancient building; and a construction manager goes thro ugh a series of formulas as he calculates the volume of a cylindrical hole and the cost of the concrete to fill it
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In the past 50 years, discrete mathematics has developed as a far-reaching and popular language for modeling fundamental problems in computer science, biology, sociology, operations research, economics, engineering, etc. The same model may appear in different guises, or a variety of models may have enough similarities such that same ideas and techniquesMULTIPLY your chances of understanding DISCRETE MATHEMATICS If you're interested in learning the fundamentals of discrete mathematics but can't seem to get your brain to function, then here's your solution. Add this easy-to-follow guide to the equation and calculate how quickly you learn the essential concepts.
Written by award-winning math professor...
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Grosse Pointe Calculus where my student?s difficulties are coming from, give focused instruction on the concept, and give additional problem sets in the area to reinforce good habits.I majored in Engineering and minored in Math in college, and this required a strong foundation in Algebra. These concepts were ...The reading section has two main areas where students can improve in. The first is quick reading comprehension. Some students are naturally gifted with (or have developed) a natural ability to read quickly.
...They will understand and be able to solve one-step and two-step algebra equations and inequalities.
d. Students will be able to solve algebra problems by using elimination and substitution methodologies.
e. Students will be able to graph equations and inequalities, and read graphs to locate and identify key points/information.
f.
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Algebra 1
From the first day your students begin to learn the vocabulary of algebra until the day they take final exams and standardized tests, these programs ...Show synopsisFrom the first day your students begin to learn the vocabulary of algebra until the day they take final exams and standardized tests, these programs strengthen student understanding and provide the tools students need to succeed.Hide synopsis
Description:Good. 0078250838 Some visible wear, and minimal interior marks....Good. 0078250838 Some visible wear, and minimal interior marks. Unbeatable customer service, and we usually ship the same or next day. Over one million satisfied customers0078250835-5 Hardcover. May include moderately worn cover, writing,...Good. Hardcover. May include moderately worn cover, writing, markings or slight discoloration. SKU: 9780078250835-4 0078651131 FAST SHIPPING! ! Student Edition. No CD...Fair. 00786511Fair. 0078250838 Student Edition. Missing many pages. Heavy...Fair. 0078250838
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11,445calculus
| 10+ other
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A TS Calc document contains a series of equations and a list of variables and constants used by the equations. When the document is set up, the user can insert the input value and see the output generated by the TS Calc document using the equations to solve the problem.
TS Calc requires Mac OS X 10.8 or higher. It costs US$15
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Get the confidence and math skills you need to get started with calculus Are you preparing for calculus? This hands-on workbook helps you master basic pre-calculus concepts and practice the types of problems you'll encounter in the course. You'll get hundreds of valuable exercises, problem-solving shortcuts, plenty of workspace, and step-by-step... more...
The contributions that are included in this e-book have been selected from those presented at the first conference on 'Models and mathematical methods and their applications to biology and industry' held in La Roche sur Yon, France, in December 2007. The aim of the conference was to present mathematical and numerical methods for solving problems
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Tenafly Ge fundamental theorem of Calculus, relating derivatives and integrals, is demonstrated by analyzing the rate of change (derivative) of the area function (integral).
Techniques of differentiation are introduced, with applications to maximum/minimum problems and curve sketching. Integrals are a...
...Quadrilateral: any object with four sides
Parallelograms: a quadrilateral with two pairs of parallel sides. Has 2 congruent sides
Rhombus: A parallelogram with 4 congruent sides
Rectangle. A parallelogram which forms 4 right angles.
...Working with a student and creating a customized learning curriculum makes every new student a challenge and opportunity. SAT math seems daunting to many students at first, particularly after the first real test. It does not have to be
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Proposes a system of methodical digit construction based upon readily perceived rules of glyph design. In other words, the number value can be calculated by observation based on simple rules of digit design. This is for easier operations when using large number bases.
Mathematics V10 Home StudyHow can we restructure the calendar so that days no longer 'leap' around month to month and year to year, and so that seasons don't shift through the months of the calendar? This book answers that question.
There's an important distinction between a calendar and time. A calendar is a tool; durations of time are what the tool measures.
Learn Mathematics on your SmartphoneThe book asks why people might believe that numbers 'exist', rather than simply being a concept of our minds? In particular, why should we believe that numbers that consist of the sum of an infinite number of other numbers added together exist? This book presents a convincing argument against the independent 'existence' of such concepts.
The book covers new theorems and formulas by the author Adonia Yaniso in mathematics that cover subjects such as relationships between sides of a right-angled triangle, newly discovered relationships between reversed numbers, combinatorial analysis, Pythagorean triples.
"Algebra, Trigonometry, and Statistics" helps in explaining different theorems and formulas within the three branches of mathematics. Use this guide in helping one better understand the properties and rules within Algebra, Trigonometry, and Statistics.
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books.google.com - Clear, comprehensive introduction emphasizes graph imbedding but also covers thoroughly the connections between topological graph theory and other areas of mathematics. Discussion of imbeddings into surfaces is combined with a complete proof of the classification of closed surfaces. Authors explore the... Graph Theory
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...
1994 This is an ex-library book and may have the usual library/used-book markings inside. This book has soft covers. In fair condition, suitable as a study copy., 1350grams, ...ISBN: 9780750619554 technology; PART THREE introduces, and explains, the more advanced aspects that are required in certain areas of technology; PART FOUR deals with all aspects of handling data and elements of statistics.
A new textbook for beginners on technician engineering courses. Packed full of `tried and tested' examples, exercises and suggestions for exploratory work. Unlike many textbooks, the maths is related to examples in practical engineering.
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Meet the Author
Table of Contents
Part One - Plus and minus; Times and divide; Fractions; Letters for numbers; Lines, angles and shapes; Areas and volumes; Graphs; Using essential maths; Part two - Factors; Ratios and proportions; Indexes; Logs; Factors in algebra; Handling equations and formulae; More about triangles; Circles; More areas and volumes; Working with graphs; Using basic maths; Part Three - Simultaneous equations; Using graphic calculators; Polar graphs; Vectors; Complex numbers; Rates of change; Equations for summing; Planning projects with maths; Using spreadsheets; Analysing numbers; Using maths topics; Part Four - Collecting data; Displaying data; More about spreadsheets; Analysing data; Quality
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Arem's CONQUERING MATH ANXIETY workbook presents a comprehensive, multifaceted treatment approach to reduce math anxiety and math avoidance. The author offers tips on specific strategies, as well as relaxation exercises. The book's major focus is to encourage students to take action. Hands-on activities help readers explore both the underlying causes of their problem and viable solutions. Many activities are followed by illustrated examples completed by other students. The free accompanying CD contains recordings of powerful relaxation and visualization exercises for reducing math anxiety.
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PERIPHERALS; No-Frills Mathematics Instruction
By L. R. SHANNON Published: January 23, 1990
THERE is plenty of educational software, much of it so enhanced with color and pictures that learning seems like a game. Nothing is wrong with that, especially for younger children, but sometimes a more straightforward approach can be refreshing, especially for older students.
The ''Step by Step'' mathematics tutorials from Professor Weissman's Software consist of six no-frills but sophisticated programs that teach algebra.All the equipment needed is an I.B.M.-compatible computer with 384 kilobytes of internal memory and a color graphics adapter or Hercules card.
It has been a long time since I had to deal with signed numbers, exponents or order of operations, among the topics dealt with on the first disk, and I must have been absent the day that combining like terms was explained. A nice feature of the tutorial is that when a wrong answer is given, the program shows, step by step, how to arrive at the correct one, so no textbook is needed. As you get better, the randomly generated problems get harder. The course deals with areas too numerous to list here. Among them are linear equations, multiplying binomials, scientific notation, factoring trinomials, quadratic equations and radicals. The program keeps track of your progress.
There really is a Professor Weissman. He is Martin Weissman, who has taught mathematics at Essex County Community College in Newark for more than 20 years and has worked as a computer consultant. The nerve center of his company is his kitchen table in Staten Island. Well, Apple started in a garage.
The basic cost of the disks is $30 apiece in either the 3.5-inch or 5.25-inch size. There are special deals for students and educators. For details, write Professor Weissman at 246 Crafton Avenue, Staten Island, N.Y. 10314, or call (718) 698-5219.
Blockout is a game, and it doesn't pretend to be educational, but the skills required to master it are not unrelated to mathematics, particularly geometry. You must manipulate blocks falling into a pit so that they fill the layers at the bottom, rewarding the player with the best and swiftest sense of spatial relationships.
If that sounds like Tetris, or its sequel, Welltris, it's because it is similar to those two addictive exercises. They were created by a Soviet mathematician. Blockout also comes from behind the formerly iron curtain: it was developed by Aleksander Ustaszewski of Warsaw for California Dreams, 780 Montague Expressway, No. 403, San Jose, Calif. 95131.
After you have mastered the standard Blockout setup, almost 1,000 variations are available.
Blockout is available for I.B.M.-compatible computers, Amigas and the Macintosh. The suggested retail price is $39.95.
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Basic Technical Mathematics with Calculus - 8th edition
Summary: This tried-and-true text from the pioneer of the basic technical mathematics course now has Addison-Wesley's amazing math technologies MyMathLab and MathXL helping students to develop and maintain the math skills they will need in their technical careers.
Technical mathematics is a course pioneered by Allyn Washington, and the eighth edition of this text preserves the author's highly regarded approach to technical math, while enhancing the integration of te...show morechnology in the text. The book is intended for a two- or three-semester course and is taught primarily to students who plan to pursue technical fields. The primary strength of the text is the heavy integration of technical applications, which aids the student in pursuit of a technical career by showing the importance of a strong foundation in algebraic and trigonometric math.
Allyn Washington defined the technical math market when he wrote the first edition of Basic Technical Mathematics over forty years ago. His continued vision is to provide highly accurate mathematical concepts based on technical applications. The course is designed to allow the student to be simultaneously enrolled in allied technical areas, such as physics or electronics. The material in the text can be easily rearranged to fit the needs of both instructor and students. Above all, the author's vision of this book is to continue to enlighten today's students that an understanding of elementary math is critical in many aspects of life.
Special Caution and Note indicators identify and aid students on difficult topics throughout the text.
Flexibility of Material Coverage. An important and critical feature to the Washington approach to technical math is the flexibility of the table of contents. The chapters of the text are easily adapted to the specific needs of the students as well as the instructor. Certain sections or chapters may be omitted without loss of continuity, and chapters may be reorganized for a customized syllabus. Notes and suggestions on how to reorganize the material are contained in the Answer Book.
Graphing Calculator. The graphing calculator is integrated and emphasized throughout the text, though it is still not required for the course. This integration includes over 160 graphing calculator screens pictured in the text.
Design. The open design is an important aspect that continues in the eighth edition. The spacious layout allows for additional graphing calculator screen graphics in the margin, which helps students visualize the graphing technology.
Word Problems and Solutions. Throughout the text, approximately 120 examples present complete solutions to word problems. These examples are clearly indicated in the margin by the phrase "solving a word problem." In addition, the text includes over 900 word problems within the exercise sets.
Exercises and Figures. The eighth edition includes over 1500 new exercises. Over 200 new figures have been added to help students visualize applications and concepts.
Writing Exercises. The number of writing exercises has been increased. An icon highlights the writing exercises in the text. These exercises reinforce student understanding, as they require students to verbalize their answers.
Applications. Over 200 technical applications have been added to the eighth edition. A hallmark strength of the Washington text, the applications challenge students to apply the mathematics to technically-based problems, reinforcing the relevancy of mathematics in the technical fields.
Introduction to Functions. More About Functions. Rectangular Coordinates. The Graph of a Function. Graphs on the Graphing Calculator. Graphs of Functions Defined by Tables of Data.
4. The Trigonometric Functions.
Angles. Defining the Trigonometric Functions. Values of the Trigonometric Functions. The Right Triangle. Applications of Right Triangles.
5. Systems of Linear Equations; Determinants.
Linear Equations. Graphs of Linear Functions. Solving Systems of Two Linear Equations in Two Unknowns Graphically. Solving Systems of Two Linear Equations in Two Unknowns Algebraically. Solving Systems of Two Linear Equations in Two Unknowns by Determinants. Solving Systems of Three Linear Equations in Three Unknowns Algebraically. Solving Systems of Three Linear Equations in Three Unknowns by Determinants.
6. Factoring and Fractions.
Special Products. Factoring: Common Factor and Difference of Squares. Factoring Trinomials. The Sum and Differences of Cubes. Equivalent Fractions. Multiplication and Division of Fractions. Addition and Subtraction of Fractions. Equations Involving Fractions.
7. Quadratic Equations.
Quadratic Equations; Solution by Factoring. Completing the Square. The Quadratic Formula. The Graph of the Quadratic Function.
8. Trigonometric Functions of Any Angle.
Signs of the Trigonometric Functions. Trigonometric Functions of Any Angle. Radians. Applications of Radian Measure.
9. Vectors and Oblique Triangles.
Introduction to Vectors. Components of Vectors. Vector Addition by Components. Applications of Vectors. Oblique Triangles, the Law of Sines. The Law of Cosines.
Limits. The Slope of a Tangent to a Curve. The Derivative. The Derivative as an Instantaneous Rate of Change. Derivatives of Polynomials. Derivatives of Products and Quotients of Functions. The Derivative of a Power of a Function. Differentiation of Implicit Functions. Higher Derivatives.
Antiderivatives. The Indefinite Integral. The Area Under a Curve. The Definite Integral. Numerical Integration: The Trapezoidal Rule. Simpson's Rule.
26. Applications of Integration.
Applications of the Indefinite Integral. Areas by Integration. Volumes by Integration. Centroids. Moments of Inertia. Other Applications.
27. Differentiation of Transcendental Functions.
Derivatives of the Sine and Cosine Functions. Derivatives of the other Trigonometric Functions. Derivatives of the Inverse Trigonometric Functions. Applications. Derivative of the Logarithmic Function. Derivative of the Exponential Function. Applications.
28. Methods of Integration.
The General Power Formula. The Basic Logarithmic Form. The Exponential Form. Basic Trigonometric Forms. Other Trigonometric Forms. Inverse Trigonometric Forms. Integration by Parts. Integration by Trigonometric Substitution. Integration by Partial Fractions: Nonrepeated Linear Factors. Integration by Partial Fractions: Other Cases. Integration by Use of Tables.
29. Expansion of Functions in Series.
Infinite Series. Maclaurin Series. Certain Operations with Series. Computations by Use of Series Expansions. Taylor Series. Introduction to Fourier Series. More about Fourier Series
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About the book:
Although perspective, mathematics majors are spared repetition and provided with new insights, while other students benefit from the consequent simplicity of the proofs for many theorems. Among the topics covered in this accessible, carefully designed introduction are multiplicativity-divisibility, including the fundamental theorem of arithmetic, combinatorial and computational number theory, congruences, arithmetic functions, primitive roots and prime numbers. Later chapters offer lucid treatments of quadratic congruences, additivity (including partition theory) and geometric number theory. Of particular importance in this text is the author's emphasis on the value of numerical examples in number theory and the role of computers in obtaining such examples. Exercises provide opportunities for constructing numerical tables with or without a computer. Students can then derive conjectures from such numerical tables, after which relevant theorems will seem natural and well-motivated.., 1994 Usually ships within 1 - 2 business days, This brand new copy is waiting for you in our warehouse and should be with you within 10-11 working days via Air Mail.
Softcover, ISBN 0486682528 Publisher: Dover Pubn Inc, 1995 Paperback. Written by a distinguished mathematician and teacher, this undergraduate text uses a combinatorial approach to accommodate both math majors and liberal arts students. In addition to coveri.Shipping may be from multiple locations in the US or from the UK, depending on stock availability. 259 pages. 0.299
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Better World Books via United States
Softcover, ISBN 0486682528 Publisher: Dover Pubn Inc Acceptable, Usually ships in 1-2 business days, Used may contain ex-library markings, notes or highlighting, may no longer have its dust jacket if applicable
Softcover, ISBN 0486682528 Publisher: Dover Publications, 1994 Good. Former Library book. Shows some signs of wear, and may have some markings on the inside. Shipped to over one million happy customers. Your purchase benefits world literacy!
Softcover, ISBN 0486682528 Publisher: Dover Publications
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Survey of Mathematics with Applications, A (9th Edition)
9780321759665
ISBN:
0321759664
Edition: 9 Pub Date: 2012 Publisher: Addison Wesley
Summary: This textbook serves as a broad introduction to students who are looking for an overview of mathematics. It is designed in such a way that students will actually find the text accessible and be able to easily understand and most importantly enjoy the subject matter. Students will learn what purpose math has in our lives and how it affects how we live and how we relate to it. It is not heavy on pure math; its purpose ...is as an overview of mathematics that will enlighten students without an intense background in math. If you want to obtain this and other cheap math textbooks we have many available to buy or rent in great condition online.
Allen R. Angel is the author of Survey of Mathematics with Applications, A (9th Edition), published 2012 under ISBN 9780321759665 and 0321759664. Six hundred four Survey of Mathematics with Applications, A (9th Edition) textbooks are available for sale on ValoreBooks.com, two hundred forty five used from the cheapest price of $55.68, or buy new starting at $170.41 ***Warning***Text Only. Still in Shrink Wrap Annotated Instructor's Copy, 9th edition but No Supplementary Materials otherwise same as student with help added tips,and answers.Shipping from California.[lessTo me, the least helpful part was the first few pages of each section. It was almost as if trying too hard to explain something which would make easy things seem complicated. I deffinatly the practice questions were great practice for preparing for tests/quizzes and having some answers in the back of the book were helpful in checking myself.
The primary subject of this book was general college math and it was deffinatly effective. There is a lot of information in here that was very simple for me to understand and then there were other lessons that were much more complicated and requiread a lot more practice and hair pulling for me.
Honestly there is nothing I would change about this book. It worked great for the class I took. It explained the lesson in the beginning of the chapter then gave problems pertaining to the lesson. It was very helpful. Having the odd number answers in the back of the book helped a lot especially when you couldn't figure out how to do a problem. I would take the answer and answer the problem until I got that answer.
I learned how to do numerous things in this book, such as writing expressions, solving for X, rise over run and other problems. This book helped me a lot in my math skills. I haven't had math in a couple years, and this book helped refresh my math skills. Thanks to this book and the teacher I passed my HESI test for nursing with flying colors.
Adding because it help me learn the value of money and putting he decimals in the right place.It help me focus on adding nonstandard getting the right change back if you buy something in the near future.
I learn how to add, multiply, subtracting, and division in many ways the purpose of that is to help me build focus in this skill.Math is my favorite subject in this world I love it because without it we would not have paychecks ,banks or even stores who accept money.
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Overview
ALEKS (Assessment and LEarning in Knowledge Spaces) is an artificial intelligence-based system for individualized learning, available from McGraw-Hill over the World Wide Web. ALEKS delivers precise, qualitative diagnostic assessments of students' knowledge, guides them in the selection of appropriate new study material, and records their progress toward mastery of curricular goals in a robust classroom management system. ALEKS interacts with the student much as a skilled human tutor would, moving between explanation and practice as needed, correcting and analyzing errors, defining terms and changing topics on request. By sophisticated modeling of a student's "knowledge state" for a given subject matter, ALEKS can focus clearly on what the student is most ready to learn next. When a student focuses on exactly what they are ready to learn, they build learning momentum and success. ALEKS is both comprehensive and flexible. Although it can serve as a resource replacement for a traditional textbook and can be used as a full course solution in business statistics, ALEKS can be used effectively in tandem with any business statistics textbook as well. As a tool for individual assessment and self-study, ALEKS is designed to be used in parallel with the material covered in standard business statistics course―not in addition to it. By assessing each student and developing an individualized plan of self-study for each, ALEKS frees the instructor to focus on those areas of study that are central to business statistics, leaving it to ALEKS to attend to the computational study needs of individual
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Applied Mathematics
The subject of mathematics is related to almost all the other subjects. The advancement in the fields of engineering, science, economics statistics etc. are facilitated by use of mathematics. In other words, the application of mathematics helps in development and easier understanding of topics in other subjects. The branch of mathematics that is used for such a purpose is called as applied mathematics. It may be noted that in this branch of math, the important terms and constants used in the other topics also form as parts.
A car leaves an airport at 8 am and runs at an average velocity 45 mph. At 9 am another car leaves the same airport in the same direction and it has to meet the first car before noon. What should be the minimum average velocity with which the second car should run to achieve the necessity?
The situation described in the above physics problem is not very unusual. A person traveling in the first car might have left out something and his friend at the airport may like to reach that article. And by noon, the first person might reach his destination and may not be reachable thereafter. Let us see how applied mathematics helps us to solve. The velocity refers to the rate of change of distance with respect to time. Hence the distance traveled by the first car in the 4 hours (from 8 am to noon) is given by the mathematical equation d1 = v1* t1 = 45*4 = 180 miles, since we know that v1 = 45 mph and t1 = 4 hours. The concept of applied math is same for the second car but now the equation is d2 = v2* t2. But in this case the known values are d2 = 180 miles, since at the point of interception both cars must have traveled the same distance and out of necessity the maximum value of t2 can only be 3 hours (from 9 am to noon). Therefore, 180 = v2*3, which gives the solution as v2 = 60 mph.
Let us study another problem related to physics but which can be considered as an applied mathematics. Two wires of ½ in. diameter are anchored at a ceiling roof as shown in the diagram. These wires are riveted to a hook which is used to hold heavy weights. The wires make angles of 45o and 60o with the ceiling. The wires have an ultimate tensile strength of 16T per sq.in. What could be the maximum weight that can be loaded on the hook?
The concept of this problem is used in material lifting equipment. Ultimate tensile strength of 16T/ sq.in. means the wire can take a load of only 16T for a cross section area of 1 in. Since the diameter of the wires are ½ in. each wire can take only a load of 16(π/4)(1/2)2 = π tons ≈ 3.14 tons. Now mathematically we can draw a vector diagram and find the solution. The same is drawn below.
The wires on the left and right take the loads that are the projection of the main load W in the direction of wires. Let those components be P and Q respectively. As per vector algebra, P = (√2)W/2 and Q = (√3)W/2. Obviously the magnitude of Q is greater and therefore it must be equal to 3.14T. Hence W can be equal to a maximum of 3.14/0.866 = 3.63T approximately.
The concept of matrices is widely used in statistical fields. We will give an introduction to matrices in our next topic.
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Unless and until we apply our book knowledge into practical world we found it useless and uninteresting. If we are taught to learn things not just for reading books and solving mathematical problems but also by incorporating these ideas in to real life situations, the interest will develop.
I never analyzed this concept of applied mathematics, It seems really great. Can that be applied to trigonometric problem solving too. I think this has much of application on work and energy concept of physics where a stress and strain calculation for a particular object or a bridge needs to be calculated. Hope I am right!!
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11.1 Taylor Series
11.2 The Error in Taylor Series
11.3 Why the Error in Taylor Series Is Controlled by a Derivative
11.4 Power Series and Radius of Convergence
11.5 Manipulating Power Series
11.6 Complex Numbers
11.7 The Relation between the Exponential and the Trigonometric Functions
11.S Summary
12. Vectors.
12.1 The Algebra of Vectors
12.2 Projections
12.3 The Dot Product of Two Vectors
12.4 Lines and Planes
12.5 Determinants
12.6 The Cross Product of Two Vectors
12.7 More on Lines and Planes
12.S Summary
13. The Derivative of a Vector Function.
13.1 The Derivative of a Vector Function
13.2 Properties of the Derivative of a Vector Function
13.3 The Acceleration Vector
13.4 The Components of Acceleration
13.5 Newton's Law Implies Kepler's Laws
13.S Summary
14. Partial Derivatives.
14.1 Graphs
14.2 Quadratic Surfaces
14.3 Functions and Their Level Curves
14.4 Limits and Continuity
14.5 Partial Derivatives
14.6 The Chain Rule
14.7 Directional Derivatives and the Gradient
14.8 Normals and the Tangent Plane
14.9 Critical Points and Extrema
14.10 Lagrange Multipliers
14.11 The Chain Rule Revisited
14.S Summary
15. Definite Integrals over Plane and Solid Regions.
15.1 The Definite Integral of a Function over a Region in the Plane
15.2 Computing |R f(P) dA Using Rectangular Coordinates
15.3 Moments and Centers of Mass
15.4 Computing |R f(P) dA Using Polar Coordinates
15.5 The Definite Integral of a Function over a Region in Space
15.6 Computing |R f(P) dV Using Cylindrical Coordinates
15.7 Computing |R f(P) dV Using Spherical Coordinates
15.S Summary
16. Green's Theorem.
16.1 Vector and Scalar Fields
16.2 Line Integrals
16.3 Four Applications of Line Integrals
16.4 Green's Theorem
16.5 Applications of Green's Theorem
16.6 Conservative Vector Fields
16.S Summary
17. The Divergence Theorem and Stokes' Theorem.
17.1 Surface Integrals
17.2 The Divergence Theorem
17.3 Stokes' Theorem
17.4 Applications of Stokes' Theorem
17.S Summary
Appendices:
A. Real Numbers.
B. Graphs and Lines.
C. Topics in Algebra.
D. Exponents.
E. Mathematical Induction.
F. The Converse of a Statement.
G. Conic Sections.
H. Logarithms and Exponentials Defined through Calculus.
I. The Taylor Series for f(x,y).
J. Theory of Limits.
K. The Interchange of Limits.
L. The Jacobian.
M. Linear Differential Equations with Constant Coefficients.
Answers to Selected Odd-Numbered Problems and to Guide Quizzes
List of Symbols
Index
About the Authors
Sherman Stein, received his Ph.D. from Columbia University. After a one-year instructorship at Princeton University, he joined the faculty at the University of California, Davis, where he taught until 1993. His main mathematical interests are in algebra, combinatorics, and pedagogy. He has been the recipient of two MAA awards; the Lester R. Ford Award for Mathematical Exposition, and the Beckenbach Book Prize for Algebra and Tiling (with Sandor Szabo). He also received The Distinguished Teaching Award from the University of California, Davis, and an honorary Doctor of Humane Letters from Marietta College.
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Challenge Exercises for Introduction to Statistics
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Using the author's considerable experience of applying Mathcad to engineering problems, Essential Mathcad introduces the most powerful functions and features of the software and teaches how to apply these to create comprehensive calculations for any quantitative subject. The simple, step-by-step approach makes this book an ideal Mathcad text for professional engineers as well as engineering , science, and math students. Examples from a variety of fields demonstrate the power and utility of Mathcad's tools, while also demonstrating how other software, such as Excel spreadsheets, can be incorporated effectively. A companion CD-ROM contains a full non-expiring version of Mathcad (North America only). *Many more applied examples and exercises from a wide variety of engineering, science, and math fields * New: more thorough discussions of differential equations, 3D plotting, and curve fitting. * Full non-expiring version of Mathcad software included on CD-ROM (North America only) * A step-by-step approach enables easy learning for professionals and students alike
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The style and structure of Concepts in Abstract Algebra are designed to help students learn the core concepts and associated techniques in algebra deeply and well. Providing a fuller and richer account of material than time allows in a lecture, this text presents interesting examples of sufficient complexity so that students can see the concepts and results used in a nontrivial setting. Charles Lanski gives students the opportunity to practice by offering many exercises that require the use and synthesis of the techniques and results. Both readable and mathematically interesting, the text also helps students learn the art of constructing mathematical arguments. Overall, students discover how mathematics proceeds and how to use techniques that mathematicians actually employ.
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Find a Pembroke Park, FL CalculusPrealgebra concerns include the quick and accurate application of arithmetic to real numbers. Topics for Prealgebra include Solving 1 and 2 Step Linear equations; adding, subtracting, multiplying, and dividing integers; combining like terms; The Order of Operations. The most difficult area for
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Pacemaker® Basic Mathematics, Third Edition
Pacemaker® Basic Mathematics, Third Edition
Pacemaker Basic Math is a comprehensive program that provides a solid, well-balanced approach to teaching math content and building math skills in whole numbers, basic arithmetic operations, and mastery of simple geometry and algebra as it prepares students for the rigors of difficult standards and proficiency tests.
This program provides educators with tools to meet the needs of diverse classrooms, keep learning up-to-date and relevant, and create supportive learning environments for a range of learning styles. Correlated to the NCTM standards, the materials and techniques used in the program are accessible, predictable, age-appropriate, and relevant as it bridges the gap between varied abilities of students and the ladder to success in algebra.
Visual learners and struggling readers are supported with photographs, charts, graphs, and illustrations, and high-interest projects gear up students for lessons.
To view sample lessons and pages, click on Download a Brochure to the left. For ISBNs and prices, click on Program Components
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Product Description
Each lesson plan lists the primary concepts taught, learning objectives, materials needed (with page numbers referenced for the student materials), teaching tips, and assignments for the student to complete. The solutions-keys for the student book and worksheets found in the (sold-separately) "Tests & Resources" book are also included. General notes on preparing lessons and administering tests, a scope & sequence, lesson-to-worksheet correlation chart, and "appearance of concepts" list are also included. 399 pages, softcover.
I loved the elementary levels and was very happy to see the Pre-Algebra and Algebra 1 just when I needed them. I found them (teacher's guides) for both levels very difficult to use. The pages with answers are 1/3 the size found in the students books making it very hard to see. They didn't show how they came up with answers so I had to figure out making it very time consuming. I used both for my two oldest children and felt no different giving it two years-hoping I would grow accustomed to them. Hopefully they will revise and correct these issues. The curriculum is advanced but without a good teachers guide, it makes it more difficult
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This eBook introduces the student to number patterns and sequences, including odd and even numbers, square numbers, square roots, cube numbers, cube roots, factors, prime numbers, multiples, linear sequences, square number and cube number sequences, Fibonacci number sequences, triangular number sequences and sequences of the powers of 2, 3, 4, 5 and 10.
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Griffith ExcelPre-Algebra/Elementary Algebra
Pre-Algebra (23%). Questions in this content area are based on basic operations using whole numbers, decimals, fractions, and integers; place value; square roots and approximations; the concept of exponents; scientific notation; factors; ratio, proportion, and perce...
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Homer Glen Geometry algebra 2 typically ends with complex numbers, logarithmic and exponential functions. When I teach and tutor algebra 2, I stress mathematical reasoning and multiple approaches (algebraic, geometric and numeric). For example, many students are unaware that arithmetic on complex numbers can be done on the TI83/84 calculator. Teaching them this skill gives them a way to check answers.
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Precalculus Essentials -With 2 CDs - 2nd edition
Summary: This text takes the same approach as the regular Blitzer Precalculus version by deleting chapters. The text explores math the way it evolved: by describing real problems and how math explains them. It is interesting, lively (with applications you won't see in any other math book), and exceedingly clear. Blitzer's philosophy: present the full scope of mathematics, while always (1) engaging the student by opening their minds to learning (2) keeping the student engaged...show more on every page (3) explaining ideas directly, simply, and clearly. Students are strongly supported by a consistent pedagogical framework. A "See it, Hear it, Try it?" format consistently walks students through each and every example in just the same way that an instructor would teach this example in class. Blitzer liberally inserts voice balloons and annotations throughout the text helping clarify the more difficult concepts for students
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More About
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Overview
This book provides the mathematics skills required for a successful career in a technical or engineering technology discipline. It presents topics in an intuitive manner with technical applications integrated throughout whenever possible. The Third Edition has been carefully reviewed and special efforts have been taken to emphasize clarity and accuracy of presentation.
Editorial Reviews
Booknews
Discusses the mathematical skills that students need to pursue a career in a technical or engineering technology program. Subject areas include fundamental concepts of operations with numbers, the metric system, basic algebraic concepts, exponential and logarithmic functions, right-triangle trigonometry, complex numbers, analytic geometry, and the trigonometric functions, formulas, and identities. The authors provide 5,000 plus exercises, with an answer manual for even-numbered items, and append a review of essential geometry
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1 definition
by
Nick Bentzen
Calculus is how we know everything about everything, except sex, maybe.
It is also a name for the page and a half of indecipherable foreplay used in university physics textbooks beforethey give you the formula for something.
Engineering student: I forgot the formula for the volume of a sphere, but I was able to figure it out from the area of a circle using calculus.
Non-engineering student: I looked in the back of the textbook.
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English
Basic Complex Analysis skillfully combines a clear exposition of core theory with a rich variety of applications. Designed for undergraduates in mathematics, the physical sciences, and engineering who have completed two years of calculus and are taking complex analysis for the first time..
Basic Multivariable Calculus fills the need for a student-oriented text devoted exclusively to the third-semester course in multivariable calculus. In this text, the basic algebraic, analytic, and geometric concepts of multivariable and vector calculus are carefully explained, with an emphasis on developing the student's intuitive understanding and computational technique. A wealth of figures supports geometrical interpretation, while exercise sets, review sections, practice exams, and historical notes keep the students active in, and involved with, the mathematical ideas. All necessary linear algebra is developed within the text, and the material can be readily coordinated with computer laboratories. Basic Multivariable Calculus is the product of an extensive writing, revising, and class-testing collaboration by the authors of Calculus III (Springer-Verlag) and Vector Calculus (W.H. Freeman & Co.). Incorporating many features from these highly respected texts, it is both a synthesis of the authors' previous work and a new and original textbook.
Designed for courses in advanced calculus and introductory real analysis, Elementary Classical Analysis strikes a careful balance between pure and applied mathematics with an emphasis on specific techniques important to classical analysis without vector calculus or complex analysis. Intended for students of engineering and physical science as well as of pure mathematics.
This volume provides a detailed account of the theory of symplectic reduction by stages, along with numerous illustrations of the theory. It gives special emphasis to group extensions, including a detailed discussion of the Euclidean group, the oscillator group, the Bott-Virasoro group and other groups of matrices. The volume also provides ample background theory on symplectic reduction and cotangent bundle reduction.
More editions of Hamiltonian Reduction by Stages (Lecture Notes in Mathematics, Vol. 1913):More editions of Introduction to Mechanics and Symmetry: A Basic Exposition of Classical Mechanical Systems (Texts in Applied Mathematics):
A monograph on some of the ways geometry and analysis can be used in mathematical problems of physical interest. The roles of symmetry, bifurcation and Hamiltonian systems in diverse applications are explored.
More editions of Lectures on Geometric Methods in Mathematical Physics (CBMS-NSF Regional Conference Series in Applied Mathematics):
The use of geometric methods in classical mechanics has proven fruitful, with wide applications in physics and engineering. In this book, Professor Marsden concentrates on these geometric aspects, especially on symmetry techniques. The main points he covers are: the stability of relative equilibria, which is analyzed using the block diagonalization technique; geometric phases, studied using the reduction and reconstruction technique; and bifurcation of relative equilibria and chaos in mechanical systems. A unifying theme for these points is provided by reduction theory, the associated mechanical connection and techniques from dynamical systems. These methods can be applied to many control and stabilization situations, and this is illustrated using rigid bodies with internal rotors, and the use of geometric phases in mechanical systems. To illustrate the above ideas and the power of geometric arguments, the author studies a variety of specific systems, including the double spherical pendulum and the classical rotating water moleculeMore editions of Manifolds, Tensor Analysis and Applications (Global analysis, pure and applied): me chanics, electromagnetism, plasma dynamics and control thcory arc given in Chapter 8, using both invariant and index notation. The current edition of the book does not deal with Riemannian geometry in much detail, and it does not treat Lie groups, principal bundles, or Morse theory. Some of this is planned for a subsequent edition. Meanwhile, the authors will make available to interested readers supplementary chapters on Lie Groups and Differential Topology and invite comments on the book's contents and development. Throughout the text supplementary topics are given, marked with the symbols ~ and {l:;J. This device enables the reader to skip various topics without disturbing the main flow of the text. Some of these provide additional background material intended for completeness, to minimize the necessity of consulting too many outside references. We treat finite and infinite-dimensional manifolds simultaneously. This is partly for efficiency of exposition. Without advanced applications, using manifolds of mappings, the study of infinite-dimensional manifolds can be hard to motivateA presentation of some of the basic ideas of fluid mechanics in a mathematically attractive manner. The text illustrates the physical background and motivation for some constructions used in recent mathematical and numerical work on the Navier- Stokes equations and on hyperbolic systems, so as to interest students in this at once beautiful and difficult subject. This third edition incorporates a number of updates and revisions, while retaining the spirit and scope of the original book.
More editions of A Mathematical Introduction to Fluid Mechanics (Texts in Applied Mathematics):
This advanced-level study approaches mathematical foundations of three-dimensional elasticity using modern differential geometry and functional analysis. It is directed to mathematicians, engineers and physicists who wish to see this classical subject in a modern settingMore editions of Mathematical Foundations of Elasticity (Dover Civil and Mechanical Engineering):
This advanced-level study approaches mathematical foundations of three-dimensional elasticity using modern differential geometry and functional analysis. It is directed to mathematicians, engineers and physicists who wish to see this classical subject in a modern settingVector Calculus helps students foster computational skills and intuitive understanding. This edition offers revised coverage in several areas and a new section looking at applications to Differential Geometry, Physics and Forms of Life as well as a large number of new exercises and expansion of the book's signature Historical Notes.
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Thomas' Calculus - 11th edition
Summary: The new edition of Thomas is a return to what Thomas has always been: the book with the best exercises. For the 11th edition, the authors have added exercises cut in the 10th edition, as well as, going back to the classic 5th and 6th editions for additional exercises and examples.
The book's theme is that Calculus is about thinking; one cannot memorize it all. The exercises develop this theme as a pivot point between the lecture in class, and the understand...show moreing that comes with applying the ideas of Calculus.
In addition, the table of contents has been refined to match the standard syllabus. Many of the examples have been trimmed of distractions and rewritten with a clear focus on the main ideas. The authors have also excised extraneous information in general and have made the technology much more transparent.
The ambition of Thomas 11e is to teach the ideas of Calculus so that students will be able to apply them in new and novel ways, first in the exercises but ultimately in their careers. Every effort has been made to insure that all content in the new edition reinforces thinking and encourages deep understanding of the material99 +$3.99 s/h
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Maths Quest General Mathematics Preliminary Course
2nd edition
Maths Quest General Mathematics Preliminary Course by Rowland
Book Description
Maths Quest General Mathematics Preliminary Course Second edition is specifically designed for the General Mathematics Stage 6 Syllabus. This text provides comprehensive coverage of the five areas of study: Financial mathematics, Data analysis, Measurement, Probability and Algebraic modelling. This student textbook offers these new features: * graphics calculator tips throughout the text * a quick and easy way for students to identify formulae that will appear on the HSC examination formula sheet * A CD-ROM that contains the entire student textbook with links to: * interactive technology files; * SkillSHEETS, which assist students to revise and consolidate essential skills and concepts; *2 WorkSHEETS for each chapter, which assist students to further consolidate their understanding * Test Yourself multiple-choice questions.
The following award winning features continue to be offered in this edition: * full colour with photographs and graphics to support real-life applications * carefully graded exercises with many skill and application problems, including multiple-choice questions * cross-references to relevant worked examples matched to questions throughout the exercises * comprehensive chapter summaries and chapter review exercises with practice examination questions * a glossary of mathematical terms, simply defined * investigations, spreadsheet applications and more. The teacher edition contains everything in the student edition package and more: * answers printed in red next to most questions in each exercise * annotated syllabus information * detailed work programs The teacher edition CD-ROM contains 2 tests per chapter, complete with fully worked solutions, WorkSHEETS and their solutions, and syllabus advice - all in editable World format.
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A+ National Pre-traineeship Maths and Literacy for Retail by Andrew Spencer
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Pre-traineeship Maths and Literacy for Retail is a write-in workbook that helps to prepare students seeking to gain a Retail Traineeship. It combines practical, real-world scenarios and terminology specifically relevant to the Retail Industry, and provides students with the mathematical skills they need to confidently pursue a career in the Retail Trade. Mirroring the format of current apprenticeship entry assessments, Pre-traineeship Maths and Literacy for Retail includes hundreds of questions to improve students' potential of gaining a successful assessment outcome of 75-80% and above. This workbook will therefore help to increase students' eligibility to obtain a Retail Traineeship. Pre-traineeship Maths and Literacy for Retail also supports and consolidates concepts that students studying VET (Vocational Educational Training) may use, as a number of VCE VET programs are also approved pre-traineeships. This workbook is also a valuable resource for older students aiming to revisit basic literacy and maths in their preparation to re-enter the workforce at the apprenticeship level.
Buy A+ National Pre-traineeship Maths and Literacy for Retail book by Andrew Spencer from Australia's Online Bookstore, Boomerang Books.
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Math B (Barron's Regents Exams and Answers Math B)
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mat205
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This course introduces the concepts of finite mathematics, with a focus on real-world application. Students will explore linear functions and equations, linear programming, and the use and application of matrices. Mathematical applications of finance, statistics, and probability are also reviewed.
Logic, Probability, and Statistics, Part I
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Logic, Probability, and Statistics, Part II
Apply counting principles to solve problems.
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Paul Zeitz had written an enlightening and entertaining book about mathematical problem-solving. The target audience of the book are students with some mathematical background, for example gifted high-schools students.
Paul Zeitz won in 1974 the USA Mathematical Olympiad, and he has edited problems for mathematics contests for several decades. Many examples and exercises arise from mathematics competitions, but the book has a wider scope. Zeitz encourages the reader towards a wide curiosity about mathematical problems. The tone of the book is playful, avoiding too much seriousness although going quite deeply into mathematics.
Zeitz' book can be compared with the classic books about mathematics and problem-solving, for example "How to Solve It" by Polya and "What is Mathematics?" by Courant, Robbins, and Stewart.
A lot of the motivation for Zeitz book stems from Polya's classic. But Zeitz targets a somewhat wider audience, young people interested in problems and perhaps not mathematics as such. The essence of mathematics is not quite so firmly discussed, and perhaps it is better to encourage curiosity first and provide sermons later.
Other recent books about problem-solving are "How to Solve It: Modern Heuristics" by Michalewicz and Fogel and "The Nature of Mathematical Modeling" by Gershenfeld. These books are targeted towards large-scale applications, not so much basic mathematical problem-solving. Real-world examples of using mathematics are somewhat lacking in Zeitz's book, but the author has made a clear choice which has to be respected.
Fostering curiosity abot a wide range of mathematical topics and techniques is not a small accomplishment. Here Zeitz succeeds admirably. I hope many young students get their hands on this book, and start trying out the examples and exercises.
Even though my student years are behind, I got interested in many of the questions and started to think about them. Just as Zeitz recommended the reader to do.
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Discrete Mathematics, 6/E
LibraryThing Review
User Review - sloDavid - LibraryThing
Why is this book so tall? Good grief, it's like a notebook. I guess it's easier to fit into one's backpack though. Introduces all the basic concepts of the "science" of computer science. Logic ... Read full review
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Geometry Seeing, Doing, Understanding
9780716743613
ISBN:
0716743612
Edition: 3 Pub Date: 2003 Publisher: W H Freeman & Co
Summary: Jacobs innovative discussions, anecdotes, examples, and exercises to capture and hold students' interest. Although predominantly proof-based, more discovery based and informal material has been added to the text to help develop geometric intuition.
Jacobs, Harold R. is the author of Geometry Seeing, Doing, Understanding, published 2003 under ISBN 9780716743613 and 0716743612. One hundred forty Geometry Seein...g, Doing, Understanding textbooks are available for sale on ValoreBooks.com, thirty used from the cheapest price of $49.17, or buy new starting at $118.592003 (3rd ed. ) *** POOR COPIES *** An acceptable hardcover textbook with usual school stamps and labels. Heavy cover wear with cardboard exposed on corners and spine ends. Cracked hinges. Student scribbling inside covers and on page edges. Text pages are clean other than an occasional stray mark. Book only-no additional resources. Booksavers receives donated books and recycles them in a variety of ways. Proceeds benefit the work of Mennonite Central Committee (MCC) in the U.S. and
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Weekly Lesson Plan
Teacher: Click here to enter text. Content Area: Math Subject: Click here to enter text.
School: Jackson County High School Class Period: Choose an item. Grade (s) Date Click here to enter text.:
Student Friendly Higher Level Questions Procedures
Learning Targets (Include your formative assessment and instructional strategies
(I Can Statements…)
Monday
Tuesday
Wednesday
Thursday
Friday
IEP Modifications M T W Th F IEP Modifications M T W Th F
☐ Extended Time ☐ Scribe
☐ Highlight Information to be Learned ☐ Slow the Rate of Presentation
☐ Individualized Assistance ☐ Modified Grading
☐ Preferential Seating ☐ Oral Assessment
☐ Reading Assistance ☐ Use of Calculators
☐ Reduced Work ☐ Other Click here to enter text.
GT Modifications M T W Th F GT Modifications M T W Th F
☐ Additional Instruction and Assistance ☐ Research
☐ Enrichment Activities ☐ Other Click here to enter text.
Program Reviews
(Please indicate how this lesson address Art &
Humanities, Practical Living or Writing)
Kentucky Core Standards for Math—Algebra II
(Click in the box to indicate which standard you will be teaching. At the end of the standard, please indicate what day(s) you will be teaching that standard.)
POLYNOMIAL, RATIONAL AND RADICAL RELATIONS M T W Th F
☐ N.CN.1 Know there is a complex number i such that i2 = −1, and every complex number has the form a + bi with a and b real.
☐ N.CN.2 Use the relation i2 = –1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.
☐ N.CN.7 Solve quadratic equations with real coefficients that have complex solutions.
☐ N.CN.8 (+) Extend polynomial identities to the complex numbers. For example, rewrite x2 + 4 as (x + 2i)(x – 2i).
☐ N.CN.9 (+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials.
☐ A.SSE.1 Interpret expressions that represent a quantity in terms of its context.
☐ a. Interpret parts of an expression, such as terms, factors, and coefficients.
☐ b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1+r)n as the product of P and
a factor not depending on P.
☐ A.SSE.2 Use the structure of an expression to identify ways to rewrite it. For example, see x4 – y4 as (x2)2 – (y2)2, thus recognizing it as a difference
of squares that can be factored as (x2 – y2)(x2 + y2).
☐ A.SSE.4 Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems. For
example, calculate mortgage payments.
☐ A.APR.1 Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition,
subtraction, and multiplication; add, subtract, and multiply polynomials.
☐ A.APR.2 Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and
only if (x – a) is a factor of p(x).
☐ A.APR.3 Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function
defined by the polynomial.
☐ A.APR.4 Prove polynomial identities and use them to describe numerical relationships. For example, the polynomial identity (x2 + y2)2 = (x2 – y2)2 +
(2xy)2 can be used to generate Pythagorean triples.
☐ A.APR.5 (+) Know and apply the Binomial Theorem for the expansion of (x + y)n in powers of x and y for a positive integer n, where x and y are any
numbers, with coefficients determined for example by Pascal's Triangle.
☐ A.APR.6 Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are
polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer
algebra system.
☐ A.APR.7 (+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction,
multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.
☐ A.REI.2 Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.
☐ A.REI.11 Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the
equation f(x) = g(x); find the solutions approximately, e.g., using technology to graph the functions, make tables of values, or find successive
approximations. Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions.
☐ F.IF.7 Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more
complicated cases.
POLYNOMIAL, RATIONAL AND RADICAL RELATIONS M T W Th F
☐ a. Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior.
TRIGONOMETRIC FUNCTIONS M T W Th F
☐ F.TF.1 Understand radian measure of an angle as the length of the arc on the unit circle subtended by the angle.
☐ F.TF.2 Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian
measures of angles traversed counterclockwise around the unit circle.
☐ F.TF.5 Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline.
☐ F.TF.8 Prove the Pythagorean identity sin2(θ) + cos2(θ) = 1 and use it to find sin (θ), cos (θ), or tan (θ), given sin (θ), cos (θ), or tan (θ), and the
quadrant of the angle.
MODELING WITH FUNCTIONS M T W Th F
☐ A.CED.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic
functions, and simple rational and exponential functions.
☐ A.CED.2 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels
and scales.
☐ A.CED.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or
non-viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different
foods.
☐ A.CED.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm's law
V = IR to highlight resistance R.
☐ F.IF.4 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and
sketch graphs showing key features given a verbal description of the relationship. Key features include: intercepts; intervals where the function is
increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; end behavior; and periodicity.
☐ F.IF.5 Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. For example, if the function
h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the
function.
☐ F.IF.6 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the
rate of change from a graph.
☐ F.IF.7 Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more
complicated cases.
☐ b. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions.
☐ e. Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigonometric functions, showing period, midline, and
amplitude.
☐ F.IF.8 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.
☐ F.IF.9 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal
descriptions). For example, given a graph of one quadratic function and an algebraic expression for another, say which has the larger maximum.
☐ F.BF.1 Write a function that describes a relationship between two quantities.
MODELING WITH FUNCTIONS M T W Th F
☐ b. Combine standard function types using arithmetic operations.
For example, build a function that models the temperature of a cooling body by adding a constant function to a decaying exponential, and relate these
functions to the model.
☐ F.BF.3 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find
the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. Include recognizing
even and odd functions from their graphs and algebraic expressions for them.
☐ F.BF.4 Find inverse functions.
☐ a. Solve an equation of the form f(x) = c for a simple function f that has an inverse and write an expression for the inverse. For example, f(x) = 2 x3
or f(x) = (x+1)/(x-1) for x ≠ 1.
☐ F.LE.4 For exponential models, express as a logarithm the solution to a bct = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate
the logarithm using technology.
INFERENCES AND CONCLUSIONS FROM DATA M T W Th F
☐ S.ID.4 Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that
there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve.
☐ S.IC.1 Understand statistics as a process for making inferences about population parameters based on a random sample from that population.
☐ S.IC.2 Decide if a specified model is consistent with results from a given data-generating process, e.g., using simulation. For example, a model says a
spinning coin falls heads up with probability 0.5. Would a result of 5 tails in a row cause you to question the model?
☐ S.IC.3 Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization
relates to each.
☐ S.IC.4 Use data from a sample survey to estimate a population mean or proportion; develop a margin of error through the use of simulation models
for random sampling.
☐ S.IC.5 Use data from a randomized experiment to compare two treatments; use simulations to decide if differences between parameters are
significant.
☐ S.IC.6 Evaluate reports based on data.
☐ S.MD.6 (+) Use probabilities to make fair decisions (e.g., drawing by lots, using a random number generator).
☐ S.MD.7 (+) Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a
game
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MathForum.org - The math forum is the leading online resource for improving math learning, teaching, and communication since 1992.
MathWorld.wolfram.com - MathWorld is the web's most extensive mathematical resource, provided as a free service to the world's mathematics and internet communities as part of a commitment to education and educational outreach by Wolfram Research, makers of Mathematica.
Linear Algebra
Math.odu.edu/~bogacki/lat/ - This Linear Algebra Toolkit is composed of the modules. Each module is designed to help a linear algebra student learn and practice a basic linear algebra procedure, such as Gauss-Jordan reduction, calculating the determinant, or checking for linear independence.
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45,800
algebra 2
| 10+ other
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documents. Both sets of standards give emphasis on high quality mathematics or science education for every single student. These standards suggest the use of various types of instruction, including cooperative groups, open-ended questioning, extended inquiry, and problem solving in addition to improving...
National Venue and Date of reporting will follow. |
POWER POINT PRESENTATION:-
(a) For Students:
GROUPS | TOPIC | REMARKS |
VI-VIII | Any maths related topic from the syllabus of concern group | Two best entries to be sent to the regional office and again two best entries from each region to...
Math has many different forms ranging from your basic adding and subtracting to complex trigonometry. It is common for a student now days to take four to six math courses before they see their high school graduation. There are three main branches in math including: Algebra, Geometry and Analysis. Algebra...
Mathematics has a pervasive influence on our everyday lives, and contributes to the wealth of the country.
The importance of mathematics
The everyday use of arithmetic and the display of information by means of graphs, are an everyday commonplace. These are the elementary aspects of mathematics. Advanced...
of the impacts that both of these have on our daily lives. I have become fascinated by Pharmacy as a career because it brings together Chemistry and Maths and directly effects on the lives of people in the community. Pharmacists are now more actively involved with the patients and have become more accessible...
communicate to each other by using language. The communication is the most important part of our lives and if there were no it, the world would not be as developed as today. Only humans can speak and that is a trait which differs us from animals. But what exactly is language which we use everyday? It is...
Top Algebra Apps to improve your Maths
Introduction
Gone are the days when studying is about reading and learning from the book. It's more about learning through electronic gadgets and smartphone applications. As an evolved parent, many of them make use of the technologies that are developed to assist...
schools, as well as providing more flexibility for parents in deciding what schools their children will attend. Also, it encourages increased focus on math and reading. The purpose of "No Child Left Behind" is that state academic achievement standards will be met by all children so that they can reach...
no other one has experienced before. Technologies are now widely used especially when it comes into game industry. Games are the most popular product of information technology.
The researchers concentrate to those students who are computer addicts and think that Mathematics and Physics subjects are...
it is enough for students to know only how to perform mathematical procedures. They must have to make sense of mathematics so that they are able to use it in meaningful ways.
Background of the Study
Nowadays, many students, from elementary to college, are distracted by many factors and do not focus...
good neighbors who can think deeply and intelligently about issues of self and society, take care for and respect others, take care of their family needs, and contribute to the welfare of others" (Glickman 48). Is school necessary for developing this type of educated citizen? If not, how is it we measure...
music and its effect on studies and grades. 400 students were selected randomly from the different universities. As everybody in his life ever tried to use music during his studies so we wanted to know its effects. We used a questionnaire for that task. Different socio cultural peoples were used but main...
logical-mathematical side, according to Mr. Howard Gardner when he proposed the theory in the year 1983. He also mentioned that we should focus on the other types of intelligences to keep a balanced capability among the different types of the Multiple Intelligence theory.
A person's Multiple Intelligence...
Method of education, or the method itself.
Planes of development – Four distinct periods of growth, development, and learning that build on each other as children and youth progress through them: ages 0 – 6 (the period of the "absorbent mind"); 6 – 12 (the period of reasoning and abstraction); 12 –...
valid without variation for all cognizant beings. A posteriori judgments, on the other hand, are empirical and as such are necessarily synthetic. In the case of synthetic claims, the predicate is not contained in the subject, and are therefore ampliative and augment our knowledge. For example, the claimschools are run and owned by the LEA. They employ the staff, decide upon the admissions policy and own the building and surrounding land, which they may use to provide facilities for adult learning or childcare. This helps to develop strong links and encourage support within the local community. (L/O 1.2)
...
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Florida End Of Course Assessment: Algebra
Algebra
Biology
Geometry
Civics
US History
Florida - Algebra
Do you put the fun in functions, or is rational the last thing you feel when you encounter a rational expression? Was PEMDAS the name of your first dog, or is FOIL just something you wrap lovingly over last night's meatloaf? (You may think we've run out of math puns, but we're just getting warmed up.) Shmoop's guide to the Florida Algebra I End-of-Course Assessment gives you everything you'll need to navigate the variable-infested waters of functions, polynomials, quadratics, and more.
Try a sample:
Ready to get your math on?
At Shmoop, we heart that idea that Floridians can access high-quality education from absolutely anywhere. In a bus! On a train! In your room! On a plane!*
*We know, we know. We're basically poets.
But don't take our word for it—take us for a test spin. No license required.
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Algebra 1 is one of the most intimidating subjects for math students. The reason is that prior to this point students have been dealing with numbers such as "3" and "43" and now suddenly in Pre-Algebra And Algebra 1 they begin to deal with variables such as "x" and "y". This concept at first can seem a bit daunting with all of those letters running around!
The Math Video Tutor is named "Fractions Thru Algebra" because it starts off with the very basics - fractions - and from there slowly progresses into Pre-Algebra and Algebra 1. A great deal of time is spent on fractions and the concept of a negative number before the student is presented with Algebraic Expressions, which is where most Algebra 1 Textbooks begin. In this way a solid foundation is laid upon which Algebra 1 can be conquered.
The core philosophy of this 10 hour DVD course is that "Algebra Is Easy". The way that you make it easy for a student is to start off with the basics and gradually move on to the tougher material. In this way the student builds confidence in his or her skills and confidence is the key to success in Algebra.
The Math Video Tutor - Fractions Thru Algebra is a 10 hour course that will fully help a student master all of the core topics in Pre-Algebra and Algebra 1. After mastering the concepts in this DVD series the student will be ready and able to move and tackle Algebra 2, Trigonometry, and beyond.
How are the MathTutorDVD.com line of DVDs different? The answer is simple. Most math instruction involves a lengthy discussion of the theory behind the Math before instructing the student in how to solve problems. In the vast majority of the cases the student quickly gets bored and frustrated by the time he or she starts to solve the problems. This DVD, in contrast, teaches all of the concepts by working fully worked problems step-by-step, which is a much more engaging way to learn.
Exceptional value and affordability. MathTutorDVD.com believes in providing value for our customers. This is a 10 Hour DVD course. This entire DVD costs less than an hour of an in-home tutor, and you may rewatch the lessons as many times as needed to master the material!
What is our teaching style like? All topics on this DVD are taught by working example problems. There are no traditional lectures of background material that won't help you solve problems and improve your skills. We believe in teaching-by-doing and that is what you will receive by watching this DVD. Solving Equations is explained, for example, by working many problems in step-by-step detail. We begin with the easier problems and work our way up to the harder problems.
The student immediately gains confidence, does not get bored, and quickly feels like he or she can conquer the material. This method is extremely powerful and has proven itself time and again. Perhaps most importantly, problem solving skills are honed early on that will help with homework and taking exams even after watching the very first lesson.
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Mathematics is a hierarchical subject that continually builds upon what has gone before. The assimilation of earlier material is essential if it is going to be possible for students to be taught and learn new mathematical ideas. This indeed is the view adopted by the benchmarking group developing national standards for the teaching of undergraduate programmes in Mathematics, Statistics and Operational Research (MSOR) within the UK. The draft statement from this group makes a very important observation about the learning of mathematics:
The subjects included in MSOR are largely cumulative: what can be taught and learned depends very heavily and in considerable detail on previously learned material. This applies to MSOR very much more than to many other disciplines. An MSOR programme must be designed to follow a logical progression, with prerequisite knowledge always taken into account. Advanced areas of pure mathematics cannot be treated until corresponding elementary and intermediate areas have been covered. Development of application areas can often be done in parallel with other work, but it is always necessary to ensure that the required methods and techniques have been dealt with
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Precalculus is a mathematics course typically taught in secondary schools and in colleges. Precalculuscourses assume a prerequisite knowledge of concepts in intermediate algebra, and is generally designed to prepare students for the study of Calculus.
shitty wast of time that really fucking sucks if you take advanced. dont even waste your time trying to go for a challenge at this level of math. Take the dipshit level class and maybe you can pass with a d+
A course on outdated mathematics that is not necessary when a higher level of math has been reached. Its main purpose is to amplify the beauty of Calculus, where everything is so much easier thanks to the derivative.
Pre-Calculus Student: Hey, can you help me with my pre-calculus homework?
Me: No, I'm a mathematician.
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In the Classroom: Math for Artists
It may not occur at first to most people that art and math are
dependent upon each other, but Ivona Grzegorczyk says that in many
design-related fields, math knowledge becomes essential to fulfilling
the creative goal of the project.
That is why Grzegorczyk, the academic coordinator for the
mathematics and computer sciences programs, said mathematics for
artists is her favorite course to teach.
"This is mostly about designing repetitious designs. When you design
something like that, you're using abstract algebra. You have to think
it through before you draw it so that it will fit," she said.
The class is especially suited for students interested in
architecture, interior design and other people who will work with
repeated designs, such as tilers and quilters.
Although computers are used for the class, it is not a class on
using computers for creating. The computers are used to facilitate the
student's projects and students are first taught how to calculate their
projects without the computer.
Grzegorczyk said many of her art students taking the course believe
they are not good at math.
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Melanie Traxler gave some resources that are supposed to be useful to high
school students who are thinking about buying Mathematica. In doing so she
referred readers to:
"< for
downloadable notebooks that you and your son can look through. In
particular, I would recommend that you download the tutorial notebooks
entitled "Intersession" from this page. This group of notebooks includes
lessons on solving equations, defining functions, graphing, and more with
Mathematica. In order to view the notebooks, you may download a copy of
MathReader."
--------------------
I went to that web page.
1- I had a very hard time going to that web page without having my web
browser (Microsoft IE) crash right after I attempt to scroll down to the
list of notebooks. It seems the only way to make this work was to ensure the
web browser window is expanded to take up the whole screen. Ok, so this
looks like a Microsoft bug, but Wolfram Research should design the web page
to avoid this bug.
2 - A potential customer shouldn't be required to download and install Math
Reader to see examples of what Mathematica can do. One should be able to
access (via the above URL) an HTML version of the above notebooks. Hey
doesn't Mathematica have a feature to convert notebooks to web pages?
Wolfram Research could make the HTML versions with a few mouse clicks.
3. I down loaded the Introduction notebook, and it had no output cells! A
potential customer using MathReader can only read the description of what
would be displayed if the input cells were evaluated using Mathematica, but
the notebook doesn't make this clear. Next I down loaded the Basic graphing
notebook, and it had no output cells and no graphics. As a result the
potential customer using MathReader can only read about the graphics that
would be displayed if they had Mathematica.
--------------------
If the potential customer was patient enough to get that far they
experienced multiple crashes of their web browser. They went to the trouble
to down load and install MathReader. After all that they still can't see
examples of Mathematica output and graphics (at least not from the above
URL). How often will one of these potential customers to put "Mathematica
for Students" on their shopping list?
--------------------
Regards,
Ted Ersek
Down Load Mathematica tips, tricks from
| 677.169 | 1 |
...
More About
This Book
decisions, like using angles and parallel lines to crack a mysterious CSI case. Put geometry to work for you, and nail your class exams along the way.
We think your time is too valuable to waste struggling with new concepts. Using the latest research in cognitive science and learning theory to craft a multi-sensory learning experience, Head First 2D Geometry uses a visually-rich format designed for the way your brain works, not a text-heavy approach that puts you to sleep.
Meet the Author
Lindsey Fallow has spent the past decade exploring science and technology as a writer, software developer, and television personality. Following an undergraduate degree in Manufacturing Engineering, she fronted a science show for 8-12 year-olds on Disney, and went on to become a reporter & Associate Producer for Tomorrow's World (the BBC's #1 prime-time UK science and technology show) from 1998-2002. She's stood on the top of the Golden Gate bridge, fed sharks, filmed brain surgery, flown in military planes, and been bitten by a baby tiger, but is the most excited by far when her 13-year-old stepson 'gets' his homework.
Dawn Griffiths started life as a mathematician at a top UK university. She was awarded a First-Class Honours degree in Mathematics, and was offered a university scholarship to undertake a PhD studying particularly rare breeds of differential equations. She moved away from academia when she realized that people would stop talking to her at parties, and went on to pursue a career in software development instead. She currently combines IT consultancy with writing and mathematics.
When Dawn's not working on Head First books, you'll find her honing her Tai Chi skills, making bobbin lace or cooking nice meals. She hasn't yet mastered the art of doing all three at the same time.
She also enjoys traveling, and spending time with her lovely husband, Introduction to the subject
I wanted a quick and general review of simple geometry, and that's exactly what this is. It's lightweight enough to be a good refresher on the topic but doesn't include enough information to be of much use as a self-learning tool. What it does include is presented in a very clear and easily digested format. The topics are lean, but they are supposed to be with the new format of the Head First series. It promises to be a quick read with just the raw facts, and it delivers on that promise. I wouldn't expect to be a wiz in geometry after reading it, but you will have a cleaner grasp of the overall subject.
Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged.
| 677.169 | 1 |
Precalculus
Precalculus, Fifth Edition, by Lial, Hornsby, Schneider, and Daniels, engages and supports students in the learning process by developing both the ...Show synopsisPrecalculus, Fifth Edition, by Lial, Hornsby, Schneider, and Daniels, engages and supports students in the learning process by developing both the conceptual understanding and the analytical skills necessary for success in mathematics. With the Fifth Edition, the authors recognize that students are learning in new ways, and that the classroom is evolving. The Lial team is now offering a new suite of resources to support today's instructors and students. New co-author Callie Daniels has experience in all classroom types including traditional, hybrid and online courses, which has driven the new MyMathLab features. For example, MyNotes provide structure for student note-taking, and Interactive Chapter Summaries allow students to quiz themselves in interactive examples on key vocabulary, symbols and concepts. Daniels' experience, coupled with the long-time successful approach of the Lial series, has helped to more tightly integrate the text with online learning than ever before.Hide synopsis
Description:Fair. 0321783808 -used book-book appears to be recovered-has...Fair. 0321783808 -used book-book appears to be recovered-has some used book stickers-free tracking number with every order. book may have some writing or highlighting, or used book stickers on front or back
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Geometry Workbook For Dummies
9780471799405
ISBN:
0471799408
Pub Date: 2006 Publisher: Wiley & Sons, Incorporated, John
Summary: From proofs to polygons -- solve geometry problems with ease Got a grasp on the terms and concepts you need to know, but get lost halfway through a problem or worse yet, not know where to begin? No fear -- this hands-on guide focuses on helping you solve the many types of geometry problems you encounter in a focused, step-by-step manner. With just enough refresher explanations before each set of problems, you'll shar...pen your skills and improve your performance. You'll see how to work with proofs, theorems, triangles, circles, formulas, 3-D figures, and more! 100s of Problems! Step-by-step answer sets clearly identify where you went wrong (or right) with a problem Get the inside scoop on geometry shortcuts and strategies Know where to begin and how to solve the most common problems Use geometry in practical applications with confidence
Ryan, Mark is the author of Geometry Workbook For Dummies, published 2006 under ISBN 9780471799405 and 0471799408. One hundred eighty Geometry Workbook For Dummies textbooks are available for sale on ValoreBooks.com, sixty nine used from the cheapest price of $0.75, or buy new starting at $9.95
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Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for).
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books.google.com - Intended for the one- or two-semester course required of Education majors, Mathematics for Elementary School Teachers, 4/e, offers pre-service teachers a comprehensive mathematics course designed to foster concept development through examples, investigati... for Elementary School Teachers
Mathematics for Elementary School Teachers
Intended for the one- or two-semester course required of Education majors, Mathematics for Elementary School Teachers, 4/e, offers pre-service teachers a comprehensive mathematics course designed to foster concept development through examples, investigati
About the author (2007)
Tom Bassarear is a professor at Keene State College in New Hampshire. He received his BA from Claremont-McKenna College, his MA from Claremont Graduate School, and was awarded an Ed.D degree from the University of Massachusetts. Tom's complementary degrees in mathematics and educational psychology have strongly influenced his convictions about education--specifically, mathematics education. Before teaching at the college level, he taught both middle school and high school mathematics. Since arriving at Keene State College, Tom has spent many hours in elementary classrooms observing teachers and working with them in school and workshop settings, plus, he has taught 4th grade math every day for a semester at a local elementary school.
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45
Total Time: 7h 33m
Use: Watch Online & Download
Access Period: Unlimited
Created At: 07/29/2009
Last Updated At: 10/10/2011
In this 45-lesson series, we'll cover sequences and series. We'll start with sequences - what are they, how do you tell if they diverge or converge, what are the different types of sequences, etc. Then, we'll move on to series - what are they, what are the different types, how can one tell if a series converges or diverges, and how can one use various tests for convergence - e.g. the integral test, the direct comparison test, the limit comparison test, the root test, the ratio test, etc.
This series will also cover absolute versus conditional convergence and linear approximation. Once we master linear polynomial approximation, we'll move on to quadratic polynomial approximation and higher-degrees of polynomial approximation. This will lead us to lessons on Taylor Polynomials, Maclaurin Polynomials, the Taylor Series, and the Maclaurin series.
Last, we'll focus on the Power Series, its interval and radius of convergence, and how we can use it to find integrals of functions that we are otherwise unable to integrate.
Taught by Professor Edward Burger, this seriesexcellent job explaining!!! just excellent! the $2 i spent on this has probably been the most worthy investment ive ever made!
Below are the descriptions for each of the lessons included in the
series:
Calculus: The Determining the Squeeze and Absolute Value Theorems Monotonic and Bounded Sequences Summation of Geometric Telescoping Properties of Convergent nth-Term Test for Diver Defining p-Series Direct Invert Series in calculus Alternating Series Estimating the Sum of Absolute and Conditional Root Polynomial Approx - Elementary Functions
Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be Higher-Degree Approximations Maclaurin Remainder of a Taylor Polynomial
Taught by Professor Approximating the Value of a Function Taylor and Maclaurin his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
Calculus: New Convergence of Definition of Interval and Radius of include 1 2 Interval and Radius of Convergence: Pt 3 Differentiation-Integration professional Differentiation Integration Integrating Functions Using
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Homework Helpers Pre-calculus
9781564149404
1564149404
Summary: The book pays particular attention to strategies for understanding and constructing logical arguments and proofs. Common patterns that reappear in many proofs are highlighted. In addition, geometry problems that review arithmetic and algebraic skills are presented and fully explained. Homework Helpers: Pre-Calculus is a straightforward, understandable, and thorough review of the topics in a typical pre-calculus class..., including: - Linear functions - Polynomials - Rational functions - Exponential functions - Logarithmic functions - Systems of equations
Szecsei, Denise is the author of Homework Helpers Pre-calculus, published 2007 under ISBN 9781564149404 and 1564149404. Nine hundred five Homework Helpers Pre-calculus textbooks are available for sale on ValoreBooks.com, three hundred fifty seven used from the cheapest price of $2.99, or buy new starting at $10.85
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You are here
Analysis: A Gateway to Understanding Mathematics
Publisher:
World Scientific
Number of Pages:
301
Price:
64.00
ISBN:
9789814401388
This book is based on a first year course for students of Economics and Finance at University College of Dublin, and it was part of a programme of Business Studies that required 'sophisticated mathematics' during each year of undergraduate study.
Because mathematics courses for engineers, accountants, economists etc, are all too often of the handle-turning, 'ask-no-questions' variety, it's very refreshing to discover that Sean Dineen's philosophy is the complete antithesis of such educational short-sightedness. In fact, Dineen's approach is a true reflection of the attitude to mathematics shown by Laplace in his open lecture of January 20, 1795. For this book, it means that:
1. The vast majority of topics are set in an historical context.
2. Mathematics is seen as a way of knowing — not just as a body of knowledge.
3. Readers are led to explore the inter-connectedness of mathematical ideas.
4. The book will prepare students for ongoing mathematical study.
Basically, this book is suited to a course in one variable calculus and real analysis; but it would be wasteful to confine its use to students of business studies. For example, it is ideally suited for use on courses for future high school teachers, and it would lay good foundations for first year maths majors.
The first chapter centres upon an exploration of quadratic equations from the perspective of the formula that provides their roots. It examines the conditions under which this formula is applicable, and eventually considers analogous results for cubics. Because the approach is simultaneously heuristic and strongly historical, students will have begun to develop an expanded vision as to how, and why, mathematics is created.
This philosophy pervades the second chapter, which examines the role of diagrams and graphs in the mathematical thinking. Subsequent chapters then introduce the concepts of sets, functions and relations that are essential for the development of analytical ideas. Again, the presentation exemplifies a good combination of both horizontal and vertical thematic sequencing of ideas.
Real analysis begins (and continues) by avoiding the Weirerstass \(\varepsilon\)-\(\delta\) definitions. Convergence of real sequences is expressed with respect to lower and upper bounds, whilst continuity of a real valued function is defined in terms of its action upon a convergent sequence. Much of the material on differentiation and integration is standard, but the treatment is very much in the spirit of the aforementioned principles.
There's no way round the fact that real analysis (\(\varepsilon\)-\(\delta\) or not) remains a significant hurdle for incipient mathematicians — and this book is no easy option. Moreover, there are three challenging chapters on constructional approaches to number systems that (although they may be omitted on a first reading) reinforce the theoretical underpinnings of real analysis. For example, the chapter that defines real numbers in terms of convergent sequences of dyadic numbers would challenge any honours mathematics student.
One other attractive feature of this book is the nature of the exercise sets. The problems are truly instructional and a sufficient number of them are supplied with solutions. Overall, this book is highly recommended as a refreshingly different introduction to undergraduate mathematics.
Peter Ruane's career was centred upon primary and secondary mathematics education.
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How does high school maths compare to university maths?
I am in my final year of high school, I am doing the highest level of maths available but I don't find it difficult at all, so I was wondering whether or not I should cover some university level mathematics before doing bachelor of science (Advanced) next year.
For this BSc I am allowed to do two majors (1 major replaces electives) over 3 years, the available majors are physics, chemistry, maths, nanotech, medical science, computer science, biology erc... so I am thinking about doing physics and maths.
The topics I will cover by the end of high school (October 2013) are curve sketching, complex numbers, polynomials, advanced calculus, mechanics, conics, series and induction, inequalities, circle geo, combinations and permutations, binomial theorem, probability, trig, parametric equations and more... so which topics should I study during the 2013 December holidays before uni starts? I am thinking differential equations because we don't study those in HS, what else? By the way the university website doesn't say what topics are covered in first semester for physics and maths so I have no idea what to study?
University mathematics is vastly different from high school mathematics. Unless you went to a really good high school, your current high school experience in mathematics is probably memorizing formulas and rules and plug-n-chug exercises. This will end in university. Mathematics there is much more proof based. Instead of being able to calculate things, you will be expected to know why and how something works.
That you did well in high school is not a guarantee that you will be good in university math. The opposite is also true: I know people who were very bad in high school math, but who excelled in university.
If I were you, I would absolutely try to learn basic logics and proofs. Being able to prove things is so important in mathematics.
You can learn proofs either from a proof book or by working through an actual math course. If you want a proof book, then I don't think there is a better book than Velleman.
If you want to do some actual mathematics, then you might want to think of going through Spivak (this is a highly nontrivial book!!!) or you might want to consider this gem:
The trick to writing proofs is to make as much proofs as you can, and then to present the proofs you have written to somebody else (for example, on our homework forum). Then you should ask that other person to give as much criticism of the proof as he can. In the beginning, he will probably rip your entire proof to shreds. But you will see that you will get gradually better in time. This is really the only way of learning proofs.
Synchronised
#3
Dec5-12, 11:31 AM
P: 47micromass
#4
Dec5-12, 12:08 PM
Mentor
P: 18,346
How does high school maths compare to university maths?
Quote by Synchronised
You seem to have the wrong idea. The goal is not that you should learn famous proofs. The goal should be that you could invent and make a proof yourself. Reading proofs is good, but it won't help you as much as doing proofs yourself. It's a bit like watching others swim: you might learn from seeing their technique, but it won't teach you to swim.
So no, you shouldn't "learn proofs", you should learn how to prove!! That's a big difference. At university, they will often just give a statement and ask you to prove it yourself. In the beginning, they might give some hints or instructions, but that won't last very long.
You mention all these famous theorems. But you won't really see the significance of them until you know how to prove things yourself. If you read mathematics, then your focus should be on "how can I use this theory to prove things myself later?" and "Does this proof contain important ideas that might be helpful?" So you should always read mathematics with the goal of proving things yourself later on. Doing exercises is an essential part of this.
Synchronised
#5
Dec5-12, 12:24 PM
P: 47Klungo
#6
Dec5-12, 12:34 PM
P: 136
My way of looking at a introductory pure math course at the University level is much like building something completely from scratch.
You are given axioms (laws) that "govern" how the objects they describe are related.
From these axioms, you construct results or theorems by combining them. And from there, more and more abstract theorems and objects are manufactured until you construct an entire subfield of mathematics.
But it's a bit more complicated than that. Different math subfields collaborate for instance.
micromass
#7
Dec5-12, 12:42 PM
Mentor
P: 18,346
Quote by SynchronisedAll the textbooks I mentioned have exercises. In fact, I suggested the textbooks especially because they have good exercises.
pasmith
#8
Dec5-12, 12:54 PM
HW Helper
Thanks
P: 1,021
You may want to read this post by Tim Gowers, which although expressly aimed at Cambridge students is in fact generally applicable.
Synchronised
#9
Dec5-12, 12:55 PM
P: 47
I will buy the Daniel J. Velleman's book since it is the cheapest :) thank you again!
Alex Wiseman
#10
Dec5-12, 06:04 PM
P: 9
I'm just about to finish first year Linear Algebra tomorrow morning (that's my final), and all I can say is it truly does not compare to high-school maths at all. High-school was based on, "here's an equation, here's some value, solve for x", university level maths are more like, "here are some values, what can I do with them and why am I doing it? How can I take these values and apply them to things we encounter on a daily basis?".
As an example, in high-school I learned how to solve simple matrices, in university I've learned how these matrices essentially make up the way Googletm ranks pages in a search. Or how the 3x3 lights out game is mathematically defined.
There's not a whole lot you can do to prepare for university maths, seeing as it is like starting from scratch all over again. But as micromass suggested, learning to prove things on your own could be of benefit to you.
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Saxon Math programs are designed and structured for immediate, measurable and long-lasting results. By employing a proven method of incremental development and continual review strategies, each piece of supplementary curriculum provides time to practice, process and learn beyond mastery.
This Math Intermediate 3 Written Practice Workbook contains exercises designed to refresh students' memories, deepen understanding of concepts, shift gears between different types of problems, and see how different math topics are related. (Workbook reprints exercises from the text with space for the student to show their work).
Product:
Saxon Math Intermediate 3: Written Practice Workbook
Author:
Hake
Prepared by:
Saxon Publishers
Edition Description:
Student
Binding Type:
Paperback
Media Type:
Book
Minimum Age:
8
Maximum Age:
8
Minimum Grade:
3rd Grade
Maximum Grade:
3rd Grade
Number of Pages:
240
Weight:
0.89 pounds
Length:
10.9 inches
Width:
8.3 inches
Height:
0.41 inches
Publisher:
Saxon Publishers Written Practice Workbook.
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Elementary Algebra Expression Practice Book 2, Grades 4-5
Enhance essential elementary algebra skills with these twenty-one practice problems. Each problem has an expression with mixed operators to reinforce the concept of operator precedence and use of parentheses. Choose a problem from the problem list, and then confirm your answer by easily navigating the link to the complete instructive solution. Most appropriate for 4th and 5th grade students. More
Enhance essential elementary algebra skills with these twenty-one practice problems. Each problem has an expression with mixed operators to reinforce the concept of operator precedence and use of parentheses. Choose a problem to solve from the problem list, work the problem, and then confirm your answer by easily navigating the link to the complete instructive solution. Return to the practice problems to select another. Problems start with low difficulty and gradually increase to challenging. The second workbook in this subject series. Most appropriate for 4th and 5th grade students.
As a long time fan of fairy tales and the joyous enthusiasm of youth, Ned presents stories appropriate to children of all ages. Whether the story teaches and instructs the student, or merely entertains, the act of reading should always expand our comprehension of the world about us. With three university degrees, he continues to write for education as well as adventure.
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Numbers and Proofs' presents a gentle introduction to the notion of proof to give the reader an understanding of how to decipher others' proofs as well as construct their own. Useful methods of proof are illustrated in the context of studying problems concerning mainly numbers (real, rational, complex and integers). An indispensable guide to all students of mathematics. Each proof is preceded by a discussion which is intended to show the reader the kind of thoughts they might have before any attempt proof is made. Established proofs which the student is in a better position to follow then follow.
Presented in the author's entertaining and informal style, and written to reflect the changing profile of students entering universities, this book will prove essential reading for all seeking an introduction to the notion of proof as well as giving a definitive guide to the more common forms. Stressing the importance of backing up "truths" found through experimentation, with logically sound and watertight arguments, it provides an ideal bridge to more complex undergraduate maths.
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The material here consists of foundation knowledge of Boolean algebra. Boolean algebra is introduced here from a very...
see more
The material here consists of foundation knowledge of Boolean algebra. Boolean algebra is introduced here from a very elementary level. The notes are easy to follow and can be useful to students and enthsiasts of any levelDiscrete stochastic processes are essentially probabilistic systems that evolve in time via random changes occurring at...
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This course is an introduction to arithmetic geometry, a subject that lies at the intersection of algebraic geometry and...
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This course is an introduction to arithmetic geometry, a subject that lies at the intersection of algebraic geometry and number theory. Its primary motivation is the study of classical Diophantine problems from the modern perspective of algebraic geometry.
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Frames for Undergraduates
9780821842126
ISBN:
0821842129
Edition: 1 Pub Date: 2008 Publisher: American Mathematical Society
Summary: "Frames for Undergraduates is an undergraduate-level introduction to the theory of frames in a Hilbert space. This book can serve as a text for a special-topics course in frame theory, but it could also be used to teach a second semester of liner algebra, using frames as an application of the theoretical concepts. It can also provide a complete and helpful resource for students doing undergraduate research projects u...sing frames." "The early chapters contain the topics from linear algebra that students need to know in order to read the rest of the book. The later chapters are devoted to advanced topics, which allow students with more experience to study more intricate types of frames. Toward that end, a Student Presentation section gives detailed proofs of fairly technical results with the intention that a student could work out these proofs independently and prepare a presentation to a class or research group. The authors have also presented some stories in the Anecdotes section about how this material has motivated and influenced their students."--BOOK JACKET.
Han, Deguang is the author of Frames for Undergraduates, published 2008 under ISBN 9780821842126 and 0821842129. Sixteen Frames for Undergraduates textbooks are available for sale on ValoreBooks.com, nine used from the cheapest price of $76.31, or buy new starting at $45.00.[read moreBoonsboro, MDShipping:Standard, ExpeditedComments:Brand new. We distribute directly for the publisher. A frame in a finite-dimensional inner-produ... [more][ adaptability to existing conditions allows frames to be used in applied settings including signal processing, imaging, sampling, and cryptography. The study of frames, particularly in finite dimensions, begins with exactly the topics from an undergraduate linear algebra course. This makes the topics particularly accessible to undergraduate students, yet the theory contains deep unsolved problems. This book can be used as a resource for an REU or for a topics course about frames. It is also a suitable textbook for a second linear algebra course, using frames as a thematic example to demonstrate and explore the new material. The theory of frames is increasingly broad with widespread applications. "Frames for Undergraduates" introduces students to this vibrant and important area of mathematics.[less]
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This exercise shows a practical application of trigonometry in the aviation environment, where student pilots consider the relationship between altitude and distance to complete a landing. It requires a scientific calculator. This resource is from...(View More) PUMAS - Practical Uses of Math and Science - a collection of brief examples created by scientists and engineers showing how math and science topics taught in K-12 classes have real world applications.(View Less)
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More About
This Textbook
Overview
Now available in a new three-volume paperback edition, Morris Kline's monumental work presents the major creations in mathematics from its beginnings in Babylonia and Egypt through the first few decades of the twentieth century. Organized around the central ideas of mathematical thought, as well as the men responsible for them, this comprehensive history provides a broad panorama of the development of mathematics, displaying the unity behind the disconnected branches of the discipline today. Beginning with the origins of mathematics in Babylonia and Egypt, Volume One includes chapters on classical Greek and Alexandrian mathematics, Hindu and Arabic contributions, algebra in the sixteenth and seventeenth centuries, coordinate geometry, and the creation of calculus.
Together, these volumes offer a lucid and authoritative study which demonstrates how the character of mathematics as a whole has changed over
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Maths for science and technology
You're about to start a course in science and technology and you're wondering whether...
You're about to start a course in science and technology and you're wondering whether your level of maths is going to be enough to get you through. This unit will show you how to reflect on what you know, identify which skills you might need for your course, and help you to learn those skills using worked examples and activities.
Through a number of activities, you will be taught how to reflect on the maths you need for your course, in order to identify which areas you will need to concentrate.
Through instruction, worked examples and practice activities, you will gain an understanding of the following mathematical concepts:
indices;
equations and algebra;
units, significant figures and scientific notation;
basic trigonometry;
logarithms.
You will be provided with a list of references to further reading and sources of help, which can help you improve your maths skills.
Maths for science and technology
Introduction
How do you feel about the mathematics in your course materials? It is quite possible that you do not feel as confident as you would like. If you have not done any mathematics recently, you may feel that you need to refresh your memory or there may be a specific topic that you never really fully understood.
Other students have said:
When I'm faced with a triangle, I never know which trig ratio to use.
I'm alright until I get to the point where a question asks me to use logs to solve a problem, then my mind goes blank.
This unit is one in a series of Student Toolkits; there are others to help you with such things as note taking, essay and report writing, effective use of English, revision and examinations, and working with charts, graphs and tables.
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About this course
This distance learning course provides the information you will need to prepare for the AQA A-Level in Maths with Statistics. In this home study course, you will focus on four core topics of algebra, geometry, trigonometry and calculus, which make up two-thirds of the A-Level qualification. The remaining third is focused on the study of statistics, including estimation, probability and distributions. The course is optimized for students studying at home and includes full tutor support via email.
A-Level Maths with Statistics is a valuable complement to other A-Level courses with a statistical element, such as biology, sociology or psychology, and for those wishing to study these subjects at a higher level. A-Level Maths with Statistics is also applicable to many jobs and careers and is a well-respected qualification that can be used for career progression and further training whilst in employment.
Entry requirements
English reading and writing skills, and maths to at least GCSE grade C or equivalent are required. You will need to have general skills and knowledge base associated with a GCSE course or equivalent standard.
This specification is designed to:
develop the student's understanding of mathematics and mathematical processes in a way that promotes confidence and fosters enjoyment
develop abilities to reason logically and to recognise incorrect reasoning, to generalise and to construct mathematical proofs
extend their range of mathematical skills and techniques and use them in more difficult unstructured problems
use mathematics as an effective means of communication
acquire the skills needed to use technology such as calculators and computers effectively, to recognise when such use may be inappropriate and to be aware of limitations
develop an awareness of the relevance of mathematics to other fields of study, to the world of work and to society in general
On this course you will study six units:
AS Level
Unit 1 MPC1 Core 1
Unit 2 MPC2 Core 2
Unit 3 MS1B Statistics 1B
A2 Level
Unit 4 MPC3 Core 3
Unit 5 MPC4 Core 4
Unit 6 MS2B Statistics 2
Each unit has 1 written paper of 1 hour 30 minutes.
Course Content
AS Level
Unit 1 MPC1 Core 1
Co-ordinate Geometry
Quadratic functions
Differentiation
Integration
Unit 2 MPC2 Core 2
Algebra and Functions
Sequences and Series
Trigonometry
Exponentials and logarithms
Differentiation
Integration
Unit 3 MS1B Statistics 1B
Statistical Measures
Probability
Discrete Random Variables
Normal Distribution
Estimation
A2 Level
Unit 4 MPC3 Core 3
Algebra and Functions
Trigonometry
Exponentials and Logarithms
Differentiation
Integration
Numerical Methods
Unit 5 MPC4 Core 4
Algebra and Functions
Coordinate Geometry in the (x, y) plane
Sequences and Series
Trigonometry
Exponentials and Logarithms
Differentiation and Integration
Vectors
Unit 6 MS2B Statistics 2
Poisson distribution
Continuous random variables
The t-distribution
Hypothesis Testing
Chi-squared tests
AS +A2 = A Level in Maths with Statistics. Both AS and A2 level courses and examinations must be successfully completed to gain a full A Level.
AQA Specification 6360
The course comes to you as a paper-based pack
You will get full tutor support via email
You will receive feedback on your assignments from our experienced tutors
You will be given guidance through the Study Guide on the nuts and bolts of studying and submitting assignments
Postal assignments cannot be accepted without prior permission from the tutor
The course contains a number of assignments which your tutor will mark and give you valuable feedback on. We call these Tutor Marked Assignments (TMAs). You need only send the TMAs to your tutor for comment, not the self-assessment exercises which are also part of the course to help you gauge your progress.
Exams are taken at an AQA centre and we can provide an extensive list of centres for you. Please read our FAQs for further information
Our A Levels come with tutor support for 24 months.
You will have access to a tutor, via email, who will mark your work and guide you through the course to help you be ready for your examinations. In addition you will be supplied with a comprehensive Study Guide which will help you through the study and assessment processFollowing the popular AQA English Literature B specification, this home study course offers students the chance to study a variety of texts, including three of their own choice. This distance learning course aims to engage students in the study of literature, encouraging a multi-layered approach to the reading of literary texts. Focusing on the central place of narrative in the construction of texts, encouraging critical debate and fostering a recognition of the role of genre, this comprehensive home study course guides students through the study of A-Level English Literature, including full tutor support via email.
An understanding and appreciation of literature can bring an enhanced enjoyment of reading and a lifelong love of language to all students. This distance learning course is suitable for students wishing to increase their general knowledge of English literature or for those seeking to progress to the study English literature or related subjects at a higher levelThis home study course from UK Distance Learning & Publishing is designed to guide you through the AQA A-Level Geography specification. Focusing on issues and impacts, the course takes a modern approach to geography, allowing students to focus on contemporary issues within geographical study. The course covers such core topics as human and physical geography, and emphasizes topical issues where appropriate, such as human impacts and sustainability.
Guided by online support from your tutor, this distance learning course will allow you to take your study of geography to the next level. It is ideal for students wishing to study geography at university or for entry to related careers.
This online distance learning course in A-Level Pure Mathematics is designed to support students through their study of the AQA Pure Mathematics A-Level. The course offers a comprehensive guide to the study of pure mathematics, encouraging a sound understanding of algebra, trigonometry, calculus, logarithms, differentiation, mathematical reasoning and proofs.Including tutor support via email, this online home study course provides the information you will need to study this challenging, rigorous and rewarding discipline. Pure mathematics is a well-respected A-Level and is relevant to both employment and to higher level study in many subjects, including science, computing and engineering. Its combination of numeracy, logic and reasoning provide a solid basis of transferable skills that can aid progression in the workplac
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46,285elementary mathelementary math
| 3 other subjects
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not Theory, a lively exposition of the mathematics of knotting, will appeal to a diverse audience from the undergraduate seeking experience outside the traditional range of studies to mathematicians wanting a leisurely introduction to the subject. Graduate students beginning a program of advanced study will find a worthwhile overview, and the reader will need no training beyond linear algebra to understand the mathematics presented. The interplay between topology and algebra, known as algebraic topology, arises early in the book, when tools from linear algebra and from basic group theory are introduced to study the properties of knots, including one of mathematics' most beautiful topics, symmetry. The book closes with a discussion of high-dimensional knot theory and a presentation of some of the recent advances in the subject - the Conway, Jones and Kauffman polynomials. A supplementary section presents the fundamental group, which is a centerpiece of algebraic topology.
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'Get knotted ... ' Scouting for Boys
Book Description
Knot Theory, a lively exposition of the mathematics of knotting, will appeal to a diverse audience of mathematical readers, from undergraduates to professionals. The author introduces tools from linear algebra and basic group theory and uses these to study the properties of knots, high-dimensional knot theory and the Conway, Jones and Kauffman polynomials.
Most Helpful Customer Reviews
This book is an excellent introduction to knot theory for the serious, motivated undergraduate students, beginning graduate students,mathematicains in other disciplines, or mathematically oriented scientists who want to learn some knot theory. Prequisites are a bare minimum: some linear algebra and a course in modern algebra should suffice, though a first geometrically oriented topology course (e. g., a course out of Armstrong, or Guillemin/Pollack) would be helpful. Many different aspects of knot theory are touched on, including some of the polynomial invariants, knot groups, Alexander polynomial and related abelian invariants, as well as some of the more geometric invariants. This book would serve as a nice complement to C. Adams "Knot Book" in that Livingston covers fewer topics, but goes into more mathematical detail. Livingston also includes many excellent exercises. Were an undergraduate to request that I do a reading course in knot theory with him/her, this would be one of the two books I'd use (Adam's book would be the other). This book is intentionally written at a more elementary level than, say Kaufmann (On Knots), Rolfsen (Knots and Links), Lickorish (Introduction to Knot Theory) or Burde-Zieshcang (Knots), and would be a good "stepping stone" to these classics.
I really do enjoy this book - but picked it up as a means of teaching myself Knot Theory... as was the case with many of my text books in college, brevity (for the sake of publishing costs) makes some concepts more of a challenge to grasp. Overall, the illustrations are great, and if you do the exercizes, the material tends to flow more easliy. It seemed to me the book worked backwards a bit - first covering a subject, than introducing it comprehensively later on - not what I'm used to.
Keep in mind, I'm not a Mathematician, merely a graduate student of mathematics, who is interested in learning about this subject on my own.
Livingston does a good job on basic knot theory in this text. While Adams seems to jump around a bit in his book, Livingston keeps a nice flow to his work. The proofs require another text and a good background in algebra to understand, but the problems are wonderful for a deeper understanding of the material.
The perfect book for undergraduates interested in learning knot theory without the algebraic topology prerequisites. This book is great as an introduction, and develops as much of the material as possible without the use of homology.
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This is a free, online textbook that can be downloaded as a pdf. ISBN: 978-87-7681-624-7 - According to the website,...
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This is a free, online textbook that can be downloaded as a pdf. ISBN: 978-87-7681-624-7 - According to the website, "Mathematics is an exceptionally useful subject, having numerous applications in business, computing, engineering and medicine to name but a few. `Applied mathematics' refers to the study of the physical world using mathematics. This book approaches the subject from an oft-neglected historical perspective. A particular aim is to make accessible to students Newton's vision of a single system of law governing the falling of an apple and the orbital motion of the moon. The book and its associated volume of practice problems give an excellent introduction to applied mathematics.״
This is a free textbook from BookBoon.'Applied Statistics has come into existence as an outcome of an experiment and wide...
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This is a free textbook from BookBoon.'Applied Statistics has come into existence as an outcome of an experiment and wide experience of more than 40 years. Applied Statistics is intended to introduce the concepts, definitions, and terminology of the subject in an elementary presentation with minimum mathematical background which does not surpass college algebra. Applied Statistics should prepare the reader to make a good decision based on data. The material, contained in Applied Statistics, can be covered in a 15-week, 3-hours-per-week semester, with little adjustment as time allows.'
According to The Orange Grove, "Astronomy Notes is a web-book covering topics such as a brief overview of astronomy's place...
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According to The Orange Grove, "Astronomy Notes is a web-book covering topics such as a brief overview of astronomy's place in the scientific endeavor, the philosophy of science and the scientific method, astronomy that can be done without a telescope, a history of astronomy and science, Newton's law of gravity and applications to orbits, Einstein's Relativity theories, electromagnetic radiation, telescopes, all the objects of the solar system, solar system formation, determining properties of the stars, the Sun, fusion reactions, stellar structure, stellar evolution, the interstellar medium, the structure of the Milky Way galaxy, extra-galactic astronomy including active galaxies and quasars, cosmology, and extra-terrestrial life. The Astronomy Notes website also has pages giving angular momentum examples, a quick mathematics review, improving study skills, astronomy tables, and astronomy terms.״
This is a free, online textbook. According to the author, "This free online textbook (e-book in webspeak) is a one semester...
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This is a free, online textbook. According to the author, "This free online textbook (e-book in webspeak) is a one semester course in basic analysis. These were my lecture notes for teaching Math 444 at the University of Illinois at Urbana-Champaign (UIUC) in fall 2009. The course is a first course in mathematical analysis aimed at students who do not necessarily wish to continue a graduate study in mathematics. A Sample Darboux sums prerequisite for the course is a basic proof course. The course does not cover topics such as metric spaces, which a more advanced course would. It should be possible to use these notes for a beginning of a more advanced course, but further material should be added.״
״This book helps the student complete the transition from purely manipulative to rigorous mathematics. The clear exposition...
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״This book helps the student complete the transition from purely manipulative to rigorous mathematics. The clear exposition covers many topics that are assumed by later courses but are often not covered with any depth or organization: basic set theory, induction, quantifiers, functions and relations, equivalence relations, properties of the real numbers (including consequences of the completeness axiom), fields, and basic properties of n-dimensional Euclidean spaces. The many exercises and optional topics (isomorphism of complete ordered fields, construction of the real numbers through Dedekind cuts, introduction to normed linear spaces, etc.) allow the instructor to adapt this book to many environments and levels of students. Extensive hypertextual cross-references and hyperlinked indexes of terms and notation add truly interactive elements to the text.״
This is a free, open textbook that is part of the Connexions collection at Rice University. "This is a textbook for Basic...
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This is a free, open textbook that is part of the Connexions collection at Rice University. "This is a textbook for Basic Mathematics for community college students. The content in this book has been collected from three text books and modified, as specified by Robert Knight.״
This is a free online textbook offered by BookBoon.'Blast into Math! A fun and rigorous introduction to pure mathematics, is...
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This is a free online textbook offered byThis is a free textbook from BookBoon.'Blast into Math! A fun and rigorous introduction to pure mathematics, is suitable for...
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This is a free textbook fromAccording to Student PIRGS, "Book of Proof is an introduction to the language and methods of mathematical proofs. The text is...
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According to Student PIRGS, "Book of Proof is an introduction to the language and methods of mathematical proofs. The text is meant to bridge the computational courses that students typically encounter in their first years of college (such as calculus or differential equations) to more theoretical, proof-based courses such as topology, analysis and abstract algebra. Topics include sets, logic, counting, methods of conditional and non-conditional proof, disproof, induction, relations, functions and infinite cardinality.Although this book may be more meaningful to the student who has had some calculus, there is no prerequisite other than a measure of mathematical maturity. The text is an expansion and refinement of the author's lecture notes developed over ten years of teaching proof courses at Virginia Commonwealth University. The text is catered to the program at VCU to an extent, but the author kept the larger audience of undergraduate mathematics students in mind.
The emphasis in this free, online textbook is on problems - calculations and story problems. "The more problems you do, the...
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The emphasis in this free, online textbook is on problems - calculations and story problems. "The more problems you do, the better you will be at doing them, as patterns will start to emerge in both the problems and in successful approaches to them.״
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Algebra 2 - 03 edition
Summary: Applications with "Real" Data Since the graphics calculator is recommended, students experience excitement as they use "real" data in Algebra 2. Students investigate and extend relevant applications through engaging activities, examples, and exercises. Graphics Calculator Technology In Algebra 2, the graphics calculator is an integral tool for presenting, understanding, and reinforcing concepts. To assist student...show mores in using this tool, a detailed keystroke guide is provided for each example and activity at the end of each chapter. Functional Approach Algebra 2 examines functions through multiple representations, such as graphs, tables, and symbolic notation. Working with transformations (investigating how functions are related to each other and their parent functions) prepares students for advanced courses in mathematics by developing an extensive, workable knowledge of functions. ...show less
Acceptable Please Read This Description-Used-Acceptable. No apparent writing/highlighting. Written note on/near free endpaper. Binding in good condition. Some wear around edges. Some scuffs and scra...show moretches. --Please read our Seller Info and Policies prior to placing your order. Stock photo used for item picture. ...show less
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GoodwillTucson Tucson, AZ
Please Read This Description- Used - Acceptable. No apparent writing/highlighting. Written note on/near free endpaper. Binding in good condition. Some wear around edges. Some scuffs and scratches. --P...show morelease read our Seller Info and Policies prior to placing your order. Stock photo used for item picture. ...show less
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Yankee Clipper Books Windsor, CT
Sail the Seas of Value30660542-5-0
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northeastbooks FL Miami, FL
Hardcover Fair 00306605482.60 +$3.99 s/h
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northeastbooks FL Miami, FL
Hardcover Good 003066053.99 +$3.99 s/h
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Borgasorus Books, Inc. MO Wright City, MO
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Helping Hands Global Kansas city, MO
2003 Hardcover Good condition, Ex-library book with some writings and jacket is slightly torn and taped on corners. multiple copies available. Same day shipping. Thank you.
$9.25 +$3.99 s/h
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Helping Hands Global Kansas city, MO
2003 Hardcover Good condition, Ex-library book with some writings and the jacket is slightly torn and taped on corners. Multiple copies available
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...Algebraic math is a major stepping stone to multiple sciences and must be mastered to facilitate future academic progress in the sciences. As algebra skills consolidate, we move toward calculus (differential and integration calculus) which employs extremely powerful math skills. Pre-calculus is the bed rock of algebraic math upon which calculus stands.
| 677.169 | 1 |
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