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The Maths Handbook Author: Richard Elwes
ISBN: 9781782069454 Format: Paperback Number Of Pages: 298 Published: 5 June 2014 Country of Publication: GB Dimensions (cm): 19.8 x 12.9 x 1.8 Description: Covering all the basics including fractions, equations, primes, squares and square roots, geometry and fractals, Dr Richard Elwes will lead you gently towards a greater understanding of this fascinating subject. Even apparently daunting concepts are explained simply, with the assistance of useful diagrams, and with a refreshing lack of jargon.
So whether you're an adult or a student, whether you like Sudoku but hate doing sums, or whether you've always been daunted by numbers at work, school or in everyday life, you won't find a better way of overcoming your nervousness about numbers and learning to enjoy making the most of mathematics.
About the Author
Dr Richard Elwes is a writer, teacher and researcher in Mathematics and a Visiting Fellow at the University of Leeds. He contributes to New Scientist and Plus Magazine and publishes research on model theory. He is the author of Mathematics 1001 published by Quercus | 677.169 | 1 |
Students: Features and Benefits
Calculus is the principal gateway to pursuing education and careers in STEM (Science, Technology, Engineering, and Math) fields.
Many students struggle to learn Calculus and resort to memorization techniques. They never achieve a solid understanding of the fundamental concepts and topics. This can interfere with learning more advanced concepts, not only in Calculus, but in the student's chosen fields of study. Even if the student graduates, job performance and career advancement could be affected since solid problem solving skills were never mastered.
Using Gradarius, each student is encouraged to solve a specific problem in his or her own personal style. A student can enter every step of a solution in an unstructured manner. During each step, Gradarius provides the student with immediate feedback, explanations, and helpful hints to guide the student to solve the problem in a valid, effective and efficient fashion. Along the way, Gradarius teaches students to logically craft robust Calculus solutions in a structured manner using correct mathematical notation.
Using Gradarius, students can:
Solve Calculus problems online, in a step-by-step manner.
Experiment with various problem-solving procedures and techniques.
Receive immediate feedback and guidance during each step of solving a problem.
Learn Calculus topics and concepts quickly and correctly the first time. | 677.169 | 1 |
Methods of Integration - for AP Calculus BC
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PRODUCT DESCRIPTION
I created a review game, using Snakes and Ladders. A smartphone or mobile device with a QR-code reader is required. There are 48 integration problems to work on (if you don't like the game). The answers can be obtained by scanning the QR-code.
There are 4 categories of problems: Integration by Part, Trig Integrals, Trig Substitution, and Potpourri (which contains some trig integrals and partial fractions.
Since I also use this for my own classes, the links from the QR-codes will remain in place. I am in the process of creating more of these for both my Calculus class as well as my Algebra II class.
If you encounter any problems with the cards, please send me an email! MrsCarlsonCHS(at) | 677.169 | 1 |
Elementary Linear Algebra
Author:Stewart Venit - Wayne Bishop
ISBN 13:9780534951900
ISBN 10:534951902
Edition:4
Publisher:Brooks Cole
Publication Date:1995-11-10
Format:Hardcover
Pages:490
List Price:$261.95
 
 
This outstanding text starts off using vectors and the geometric approach, featuring a computational emphasis. The authors provide students with easy-to-read explanations, examples, proofs, and procedures. Elementary Linear Algebra can be used in both a matrix-oriented course, or a more traditionally structured course.
Booknews
New edition of a standard college textbook. Annotation c. Book News, Inc., Portland, OR (booknews.com) | 677.169 | 1 |
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QuickMath allows students to get instant solutions to all kinds of math problems, from algebra and equation solving right through to calculus and matrices. Universal Math Solver software will solve and explain step-by-step problems of any complexity from such areas of mathematics as: arithmetic; basic math; Table 1.1 Problems recently encountered and solved. What is a Formal Problem? There is a substantial amount of research literature on problem.
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[content block] Dr. Ken Shore's Classroom Problem Solver Stealing in the Classroom. Students steal for a variety of reasons. Some steal simply because they want an. Get help with any math problem with free math problem solver online. You can solve any problem in a jiffy and get your answer within minutes. Math problem solver free. Pentominos Puzzle Solver. THE APPLET ON THIS PAGE can solve pentominos puzzles. The applet appears as a button bellow. When you click this button, the program itself.
Free math problem solver answers your algebra homework questions with step-by-step explanations. Table 1.1 Problems recently encountered and solved. What is a Formal Problem? There is a substantial amount of research literature on problem. Basic Solver Selection. ode45 performs well with most ODE problems and should generally be your first choice of solver. However, ode23 and ode113 can be more. Free Math problem solver can be a tool or a person. There are a a number of websites that provide this facility. These sites have certain free tools like online.
Solve math problems online. Get free answers to math questions instantly with the help of a free online math problem solver and thus improve your math practice. Geometry is a fun and simple area of Math, which appeals to many students who are normally not too fond of Math. The combination of numbers and theorems with visual. Please use this form if you would like to have this math solver on your website, free of charge. Name: This free calculator will try to solve any math problem: algebra (equation solver, system of equations solver, least common multiple, greatest common
[content block] Dr. Ken Shore's Classroom Problem Solver Stealing in the Classroom. Students steal for a variety of reasons. Some steal simply because they want an. Free online math word problem solver has excellent tutors who solve problems and explain concepts you don't understand. Practice what you learn immediately with. | 677.169 | 1 |
Geology/School Project
I am a high school student doing a project in which I need to ask a professional about their career and how much math was integrated into their study. If you have time, I have some questions for you I would like to ask asap.
Sincerely,
Bella
ANSWER: I had quite a bit of math in college; probably more that I needed for a degree in Geology because I was an engineering major before taking a geology course. I had algebra (two semester), calculus (two semesters), statistics, and differential equations.
Much more important than what math I had in college is how much math did I use as a working geologistGood math skills are not just used in science and/or engineering but are really more about every day problem solving. Someone with good math skills can fit into almost any job (maybe except sales) and excel.
Let me know if you need more information, my phone number is (304) 586-0973
---------- FOLLOW-UP ----------
QUESTION: Thank you for your quick response!! If you wouldn't mind answering these twenty questions for me, it would be deeply appreciated. Some of these you have already responded to.
1. What sorts of math classes were required for you to take in college?
2. How many years of schooling were required for your degree?
3. What kinds of math do you use?
4. How often do you use math on a day-to-day basis?
5. How much of your degree is based on mathematics?
6. Was math one of your strong suits through school? Did it have to become one of your strong suits?
7. How difficult was it for you to obtain your degree?
8. How difficult was it for you to secure a job after graduating?
9. In your particular field, is your job in high demand?
10. How often do you use problem solving in your field?
11. If a high school student had particular interest in going into this field, what would you recommend for them to do in order to prepare for it?
12. What was the application process like at your college/university when applying for this degree?
13. Describe your usual day on-the-job.
14. Which is one time that you can recall that you messed up on your calculations? How much of a difference did it make to the problem?
15. Have you ever thought about going into a different math-based career? Why or why not?
16. How long had you known that you wanted to go into your field before you went to college?
17. Did you spend time in other fields before choosing your current one? If so, why did you change?
18. What was the last math class you took in high school?
19. Did you have an internship at all through college? If so, what was that experience like and how has that helped you?
20. Why should students be interested in your field?
Again, thank you for your time.
Sincerely,
Bella
Answer 1. Most schools require two semesters of algebra, two semesters of calculus and statistics.
2. My BS degree required 128 semester hours. This assumes 16 semester hours per semester to graduate in four years.
34. As a geologist I used algebra almost every day.
5. I would suspect that 1/4 - 1/3.
6. No, math was not one of my strong suits but science was and in order to work in science it requires a lot of math; so I learned what I needed to do the science.
7. It took me 12 years but I had several major changes and ended up with 180 semesters because of this but I also got married, had a child, and spent two years in the military (Vietnam) during this 12 years.
8. I worked on weekends and summers as a geologist for two years before getting my degree and then continued on with that same company after I graduated.
9. No, geology is not always in high demand. It depends on the world economy but it is very very interesting. I never really understood geology until I had worked in the field for at least five years.
10. Continuously, that is what geology is.
11. That certainly depends on the field but for sciences and engineering math is always good.
12. I first applied for acceptance into a college in the mid 1960's. Things are significantly different today then they were then. Basically all that I had to do was have a High School degree and take the ACT test and score reasonably well.
13. There were not usual days. I started out working as a consultant core drilling and field mapping mostly in the eastern US in coal. I opened an office for this consulting company in Golden, CO about five years after I started working. I realized, very quickly that my eastern sediment education was not good enough to compete in the Western US, as a consultant, so I went back to college at the Colorado School of Mines in Golden. After I completed my MS degree I went to work in Arizona for a uranium mining company where I identified deposits, drilled them and tried to understand the geology of uranium (which in my opinion is the most difficult geology in the world). When uranium fell out of favor in the US I moved on to run a gold mine in Durango, Colorado. When gold prices got so low that we had to shut down the mine I came back to the eastern US and worked as a mineral economist and expert wittiness in trails which I did for the last 20 - 25 years of my career. Recently I counted the states that I have worked in. I have done geology in 46 states. The states that I have not done geology in are: Hawaii, Delaware, Rhode Island, and Maine. I have also done geology in at least 20 countries.
14. I'm sure that there are times that I messed up a calculation but I don't recall any but a messed up calculation would make a tremendous difference. I did however miss seeing an analysis one time because it was on a core log rather than on a separate form. The company was sued and ended up paying about $250,00 for the mistake.
15. No, because there is nothing that I would like as much as geology. You get to spend at least 1/2 of your working time outside looking at rocks and traveling all over the world. I had an attorney at my office once and he went into my library (where most of my college books were along with numerous geology and mining books). He asked did you take all of these courses; mostly math and science, to which I said yes. He said you could have gone to medical school. I had never thought about that because I had been interested in geology since I was first introduced to it but he was right.
16. I didn't that is why it took me so long to get a degree. If I would have know that I wanted to be a geologist when I was in High School; I could have got a degree in half the time.
17. I served a four year apprenticeship as a brick mason; although I was going to college during these four years at nights. Brick masonry has served me well; no one wants to be a brick mason it is very hard work and requires being outside 95 percent of the time but it pays well and has got me through many rough times financially.
18. calculus
19. I suppose working on week ends and during the summer months for a consultant was an internship and it was overwhelmingly helpful.
20. Because it is so very interesting. I still love geology after working over 40 years as a geologist and it pays good, lots of travel, meeting interesting people and seeing interesting rocks and things. Geology is everywhere, you go, so if you understand geology (and believe me understanding geology takes a while) you will have fun where ever you are and whether being paid or not. And that is also why I like being an expert on all experts because I still enjoy geology and sometimes researching questions and promoting geology to young people like yourself. Best of luck on you project. Let me know if you ever need any more advice about geology.
Expertise
I am an economic geologist. An economic geologist does | 677.169 | 1 |
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The Algebra Word Problem Tutor DVD series teaches students how to set up algebra word problems and solve them. This episode, Number Problems, teaches students how to solve word problems that involve number relations, including how to set up the appropriate algebraic equation and solve for the unknown. Grades 5-9. 35 minutes on DVD. | 677.169 | 1 |
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Unformatted text preview: Math 330 Exam 2 Outline Exam 2 will take place on Monday March 2. No calculators will be allowed or be needed. Matrix Operations 20 pts This comes from section 2.1 and involves addition, scalar multiplication, matrix multiplication and various scenarios about these topics. (There will be 4 scenarios worth 5 points each). In addition to being able to do computations you should know when certain operations would be defined or undefined and how to find the missing matrices in calculations such as we did in Q2.1. Possible scenarios could include questions such as given this 2 2 matrix A , find a 2 2 matrix B such that AB negationslash = BA and make these calculations. Although we havent said anything about this in class (it is in the homework), be sure to know how to compute the transpose of a matrix. Inverse Matrices 20 points You will be asked to find the inverse of two matrices of different sizes or to show that these matrices are not invertible. Be sure to know the formula for the inverse of a 2 2 matrix as this can save you some time. A big warning here is that you need to be highly organized with your row operations. Dont underestimate the importance of a precise algorithm for doing this! Many students were not so organized (I think they underestimated the difficulty) on Q32 and this lead to errors and large amounts of unnecessary computations. See the homework and quiz solutions for examples. Here are my tips on the process: It is important to work from left to right in establishing pivots. If a column fails to have a pivot then you have easily demonstrated that the matrix is not invertible....
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This note was uploaded on 09/19/2009 for the course MATH 330 taught by Professor Johnson,j during the Spring '08 term at Nevada. | 677.169 | 1 |
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PRODUCT DESCRIPTION
This lesson is meant to be used as guided notes and assessments for an entire class. In the first section, students learn how to solve equations that involve the distributive property on one side or on both. In the second section, they learn how to solve equations that involve distributing and combining. There is helpful 5-step guide at the top of the notes.
There are assessments for both sections, which are based off of the guided | 677.169 | 1 |
Ready to learn math fundamentals but can't seem to get your brain to function? No problem! Add Pre-Algebra Demystified, Second Edition, to the equation and you'll solve your dilemma in no time.
Written in a step-by-step format, this practical guide begins by covering whole numbers, integers, fractions, decimals, and percents. You'll move on to expressions, equations, measurement, and graphing. Operations with monomials and polynomials are also discussed. Detailed examples, concise explanations, and worked problems make it easy to understand the material, and end-of-chapter quizzes and a final exam help reinforce learning.
This text-workbook gives students a firm foundation in medical terminology meanings and pronunciations. It provides a concise presentation of terms reinforced by practice exercises. It can be used in a short course or a review course of medical terminology or as a self-paced text. | 677.169 | 1 |
HARCOURT MATHEMATICS 12
Rich Learning Links Consultant Dan Schnabel York Region District School Board Technology Consultants Patrick Grew Limestone District School Board Darren Luoma Simcoe District School Board Robert Sherk Limestone District School Board
Contributors Bill Bisset (formerly) North York Board of Education Ray MacDonald Bluewater District School Board C. Gary Reid York Region District School Board
Acknowledgements
A special thanks to the reviewers listed below for their helpful observations and recommendations. Feedback from reviewers has been extremely valuable in creating a text that fulfills the requirements of both teachers and students in Ontario.
Jacquie Brodsky Mathematics Teacher Medway High School Thames Valley District School Board Robert Gleeson Mathematics Teacher Kincardine District Secondary School Bluewater District School Board Darren Luoma Mathematics Teacher Bear Creek Secondary School Simcoe County District School Board Marisa Palma Department Head of Mathematics St. Elizabeth Catholic High School York Catholic District School Board C. Gary Reid Head of Mathematics Sutton District High School York Region District School Board Susan Brooks Retired Mathematics High School Teacher Elmira District Secondary School Waterloo Region District School Board Patrick Grew Head of Mathematics Frontenac Secondary School Limestone District School Board Diane McCombs Mathematics and Music Teacher Regiopolis Notre Dame Catholic High School Algonquin and Lakeshore Catholic District School Board Mark Pankratz Head of Mathematics and Computer Science Brookfield High School Ottawa-Carleton District School Board Ann Rubin Retired Head of Mathematics Milliken Mills High School York Region District School Board Alexis Galvao Mathematics Department Head John Cabot Catholic Secondary School Dufferin-Peel Catholic District School Board Angelo Lillo Head of Mathematics Sir Winston Churchill Secondary School District School Board of Niagara Shelley McLean Mathematics Department Head/Teacher Holy Cross Catholic Secondary School Algonquin and Lakeshore Catholic District School Board Brian Poste Department Head (Mathematics, Computer Science, Business) North Hastings High School Hastings and Prince Edward District School Board David Rushby Head of Mathematics Martingrove Collegiate Institute Toronto District School Board
We gratefully acknowledge the following educators for participating in our mathematics discussion group meetings throughout Ontario.
Jeff Anderson Head of Mathematics Forest Heights Collegiate Institute Waterloo Region District School Board W. K. Dutton Mathematics Teacher Lester B. Pearson High School Halton District School Board Ken Billey Head of Mathematics Holy Names High School Windsor-Essex Catholic District School Board Lorenzo Ciapanna Head of Mathematics St. Jean de Brébeuf Catholic Secondary School Hamilton-Wentworth District School Board Chris Brady Mathematics Teacher Sherwood Secondary School Hamilton-Wentworth District School Board Wendy Fitzsimmons Mathematics Teacher Milton District Secondary School Halton District School Board
iii
Michele Goveia Head of Mathematics Father Henry Carr Toronto Catholic District School Board Garry Kiziak Head of Mathematics and Science Burlington Central High School Halton District School Board Darren Luoma Mathematics Teacher Bear Creek Secondary School Simcoe County District School Board Bob McRoberts Head of Mathematics Dr. G.W. Williams Secondary School York Region District School Board Mark Pankratz Head of Mathematics and Computer Science Brookfield High School Ottawa-Carleton District School Board John Santarelli Head of Mathematics Cathedral High School Hamilton-Wentworth Catholic District School Board Scott Taylor Head of Mathematics Bell High School Ottawa-Carleton District School Board Peter Wei Head of Mathematics North Toronto Collegiate Institute Toronto District School Board John Yakopich Head of Mathematics Sandwich Secondary School Greater Essex County District School Board
Patrick Grew Head of Mathematics Frontenac Secondary School Limestone District School Board Mike Lawson Head of Mathematics Father Michael Goetz Secondary School Dufferin-Peel Catholic District School Board Glenn McDermott Head of Mathematics Woodstock Collegiate Institute Thames Valley District School Board Chris Monk Head of Mathematics Marc Garneau Collegiate Toronto District School Board C. Gary Reid Head of Mathematics Sutton District High School York Region District School Board Jenny Stillman Head of Mathematics Central Secondary School Thames Valley District School Board Joan A. Tomiuk Mathematics Teacher Glebe Collegiate Institute Ottawa-Carleton District School Board Shelley Wilton Head of Mathematics Westminster Secondary School Thames Valley District School Board Laurie A. Zahnow Head of Mathematics Silverthorn Collegiate Institute Toronto District School Board
John C. Holden Mathematics Teacher Ridgemont High School Ottawa-Carleton District School Board Frank LoForte Head of Mathematics Riverdale Collegiate Institute Toronto District School Board Cheryl McQueen Head of Mathematics Central Elgin Collegiate Institute Thames Valley District School Board Peter O'Hara Mathematics Teacher Glendale High School Thames Valley District School Board David Rushby Head of Mathematics Martingrove Collegiate Institute Toronto District School Board Dwight Stead Head of Mathematics Cardinal Leger Secondary School Dufferin-Peel Catholic District School Board Jane Uloth Mathematics Teacher Lester B. Pearson High School Halton District School Board Beryl Wong Mathematics Teacher Holy Name of Mary Secondary School Dufferin-Peel Catholic District School Board
Using Geometry and Discrete Mathematics
A GUIDED TOUR OF YOUR TEXTBOOK
CHAPTER OPENER
You will be introduced to each chapter by reading about some real-life applications of the mathematical concepts that will be presented within the chapter. A colourful image will accompany this introduction.
Chapter 3
PROPERTIES OF CIRCLES
In Chapter 3, we will consider the closed plane curve of a circle. The circle is the most symmetrical shape, and very important in art and design. The properties of circles were fundamental to the ancient Greeks, and especially important were the formulas for circumference and area. These formulas, C 2 r and A r2, have been part of the study of mathematics throughout history. Why do we continue to study them? Aside from the practical value that they have for architects, designers, and engineers, these formulas introduce the number (pi). CHAPTER EXPECTATIONS In this chapter, you will
A list of skills corresponds to specific curriculum expectations addressed in the chapter. References point you to the section in which each expectation is addressed.
• •
A G U I D E D TO U R O F YO U R T E X T B O O K
ix
REVIEW OF PREREQUISITE SKILLS
Before proceeding to the new concepts introduced in each chapter, you can review the knowledge and skills you will need.
Review of Prerequisite Skills
A vector is a quantity, an inseparable part of which is a direction. Pause for a moment and think about physical quantities that have a direction. Force is an example. The force of gravity acts only downward, never sideways. Wind is another example. A wind from the north and a wind from the south have different physical consequences, even if the wind speeds are the same. Temperature, on the other hand, is not a vector quantity. Temperature does not go in any direction. Temperature is referred to as a scalar quantity. We need both scalar and vector quantities to model complex physical systems. Meteorologists, for example, need data on air temperature and wind velocity, among other things, to make weather forecasts. The object of this chapter is to introduce the mathematical properties of vectors and then show how vectors and scalars are used to describe many features of the physical world. In this chapter, we introduce the concept of a vector, a mathematical object representing a physical quantity that has both magnitude and direction. We will concentrate on geometric representations of vectors, so that most of our discussion will be of two-dimensional vectors. In later chapters we will introduce algebraic representations of vectors, which will be more easily extended to higher dimensions. Before we begin this chapter, we will review some basic facts of trigonometry.
T R I G O N O M E T R I C R AT I O S
B
In a right-angled triangle, as shown, sin tan
a c a b
cos
b c
c
a
Note: The ratios depend on which angle is angle is 90°.
and which
A b C
120 C H A P T E R 4
x
U S I N G G E O M E T RY A N D D I S C R E T E M AT H E M AT I C S
LESSONS
Lessons provide you with opportunities to explore concepts independently or to work with others.
Section 3.2 — Angles in a Circle
If an arc subtends an angle at the centre of a circle, we can determine the size of the angle.
EXAMPLE 1
In the diagram, the radius of the circle is 6 and the length of the arc is . Determine the measure of ∠AOB. Solution The circumference of the circle is C If ∠AOB xº, then
x 360 12 360 12
A
O B
2 r
12 .
x ∠AOB
30
30º
D 360 – α O α A C B
Worked examples with solutions help you build an understanding of a concept.
In this example, we have created a central angle. A central angle is an angle that has its vertex at the centre of a circle and stands on an arc (or chord) of the circle. In the diagram, two central angles are shown: ∠AOB, which is less than 180º, and reflex ∠AOB. The angle α stands on the minor arc ACB (or chord AB). In the same way, the angle 360 α stands on the major arc ADB. Each of the two angles is described as a central angle because each has its vertex at the centre of the circle and each stands on an arc of the circle. An angle at the circumference of a circle is an angle that has its vertex on the circumference of a circle and that stands on an arc (or chord) of a circle. ∠ADB and ∠ACB are examples of two such angles which stand on the minor arc AEB (or chord AB). ∠AEB is an example of an angle which stands on the major arc AFB (or chord AB).
Central Angles
Definitions and tips are easily found in highlighted boxes.
C
F
D
A E
B
Angles at the circumference
84
CHAPTER 3
EXERCISES
Exercises follow each lesson and are organized by level of difficulty. Questions allow you to master essential mathematical skills and attempt more challenging and thought-provoking problems.
Application
Exercise 3.1
Part A 1. Determine the length of the chord AB if OA ON 3. 5 and
A O N B
2. If AB
10 and OA
13, determine the length of ON.
O B N A
You are given multiple opportunities to practise concepts introduced in each lesson. There are many opportunities to use technical tools.
Knowledge/ Understanding
3. Calculate the distance between the parallel chords PQ and XY if PQ 6, XY 8, and the radius of the circle is 5.
P O X
Q
Y B O D
4. The two parallel chords AB and CD are a distance of 14 units apart. If AB has length 12 and the radius of the circle is 10, calculate the length of CD. Part B 5. Two circles with centres A and B have radii 5 and 8, respectively. The circles intersect at the points X and Y. If XY 8, determine the length of AB, the distance between the centres. 6. The distance between the parallel chords PQ and RS is 10. If PQ 8 and RS 12, determine the length of the radius of the circle.
7. A chord of a circle is 4 units away from the centre of the circle, which has a radius of 10. What is the length of the chord?
Communication
8. The point P is any point placed inside a circle. Is it always possible to draw a line through P that is bisected at the point P? Explain.
P
CHAPTER 3
A G U I D E D TO U R O F YO U R T E X T B O O K
xi
the art of secret messages. PQ and SR are parallel. QR and PS can both be shown to have a slope of 3 . or 1010.
V E C T O R S I N C O M M U N I C AT I O N
You have seen examples of geometric and algebraic vectors. respectively. BC. where they allow the light-bulb end to be moved without changing the direction in which the bulb points. Using strings of length 5 allows for the creation of 32 strings: 00000.
investigate
C H A P T E R 2 : VA R I G N O N PA R A L L E L O G R A M
Parallelograms are quadrilaterals with opposite sides that are parallel. In a room with light switches at both ends of the room. 7
A y P6 4 2 –6 –4 –2 –2 S –4 D –6 2 R 4 6 C Q x B
Discussion questions require you to explain how mathematical principles will be applied. A vector of length four might be 0010. 2). as shown.. We define addition as follows: a. Such vectors are composed of 0s and 1s in a row. and DA are found to be P (–1. the number 2 to represent B. the message This is clever can be transmitted as 101000100001001100110100110011000110110000101101100010110010. the resulting interior quadrilateral is a parallelogram. They are used in some jointed desk lamps... Thus. However.
DISCUSSION QUESTIONS
1. Thus. Using this system. Investigate and Inquire In geometry. They are a common shape because rectangles are a type of parallelogram. The second is due to electronic properties. scenes that require a moving camera can be made jitter-free by using a Steadicam. Satisfy yourself that there are 32 length-5 strings. Note that it is up to the receiver to create words from a string of letters and that. CD. 7 Similarly. Q(6. There is no "carrying" from column to column. There are two reasons for this. it usually has a special name. Hence. PQRS is a 2 parallelogram. In movies. surprisingly. –4). a parallelogram is a special quadrilateral with opposite sides that are parallel. they are used in some binocular mounts for amateur astronomers and some motorcycle suspension systems. 11111. if we connect the midpoints of the sides of the quadrilateral. there is no punctuation. What are some other special quadrilaterals? 2. –3). but they do arise as the solution to certain engineering and design problems. 0 0 0 1 0 1 1 1 0 1 1 0
b. They are not very secure if privacy is desired. The first is that adding five-digit strings will always give a five-digit result. Try this with four other points. easily overcome by defining an arithmetic of addition (and subtraction) which can effectively hide the message. . ●
28
CHAPTER 2
E X T E N D I N G A N D I N V E S T I G AT I N G
The Extending and Investigating feature at the end of most chapters presents applications of mathematics that enable you to explore links that connect different branches of mathematics.
E X T E N D I N G A N D I N V E S T I G AT I N G
199
xii
U S I N G G E O M E T RY A N D D I S C R E T E M AT H E M AT I C S
. 00011. if we connect four points on a plane. This provides the foundation of cryptography.
A similar calculation shows that the slope of SR is also 3 . and so on. 5). and D(–3. the lights are off if both switches are off (0 position) OR if both switches are on (1 position). are of great value in computer communication.RICH LEARNING LINK
The Rich Learning Link feature at the beginning of each chapter presents a real-world scenario and allows you the opportunity to apply your learning to real issues. –5) on a Cartesian plane. Using Geometer's Sketchpad® will allow you to move your four points independently. 7). –1). B(5. 00001. If a quadrilateral has special properties.. You are encouraged to think about and use prior knowledge in math. The midpoints of AB. This problem is. so they are also parallel. the addition works easily in electronic form. and reflect on your own life experiences to guide you through these investigations. because they correspond to current on (1) or current off (0). 3). mounted against the side of a wall). The first advantage gained is that this allows the simple transmission of messages. unless one of the unused strings is designated for the purpose. Such vectors. or 1111. R(2. the result is not likely to be a parallelogram. For example. Non-rectangular parallelograms are much less common. What are some other special parallelograms? 3. For similar reasons.1 string vector. The only surprise in the addition process is 1 1 0. A rectangle is a special type of parallelogram. Describe the possible motions of the parallelogram. The slope of PQ is x2 2
y y1 x1
2 6 5 ( 1) 3 . Using 26 of these we can define the alphabet as follows: A: 00001 G: 00111 M: 01101 B: 00010 H: 01000 N: 01110 C The weakness of this system is that messages such as this are easily intercepted. and S(–5. Consider a parallelogram frame that has one side held in position (e. you will note that this is a simple assignment of the number 1 in a five-digit display to represent A. a device which works because of the contribution of parallelograms. and so on. Pick four points A(–7.g. C(7. A third common type of vector is the 0 . You will have the opportunity to explore interesting applications of mathematics that will point toward research problems.
are much less obvious. One day in class Sunil was multiplying some numbers and made the following observation: 12 112 1112 11112 11 1112 1 121 12 321 1 234 321 123 454 321
Technology icons highlight opportunities for you to choose to use calculators. we can construct a triangle such as the one shown and fold vertex B over onto vertex C. 485
e
He concluded that he had found a very simple number pattern for squaring a number consisting only of 1s. if we want to show that an isosceles triangle has two A A equal angles. The Greek mathematicians who first endeavoured to establish proofs applying to all situations took a giant step forward in the development of mathematics. In the example just considered. where n is a positive integer. Is this statement true for all values of n? Does this expression ever generate a perfect square? We start by trying small values of n. We follow their lead in establishing the concept of proof. Consider the statement." The class checked and found that she was right. Other examples. What if BC is lengthened or shortened? Even if we construct hundreds of triangles. B every isosceles triangle imaginable? Consider the following example. never generates a perfect square. ∠B ∠C.
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APPENDIX P.TECHNOLOGY
Technology features are integrated throughout. The expression 1 1141n2. we can see that some patterns that appear to be true for a few terms are not necessarily true when extended. can we conclude that ∠B ∠C for B fold C C.1 — What Is Proof?
t chnology
The concept of proof lies at the very heart of mathematics. For example. we use careful and convincing reasoning to demonstrate the truth of a mathematical statement. graphing calculators. In this example. but that is only for this triangle.
Section 1. 1 W H AT I S P R O O F ?
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A G U I D E D TO U R O F YO U R T E X T B O O K
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. the pattern breaks down quickly. How many 1s did Jennifer use? From this example. When we construct a proof. we will learn how proofs are constructed and how convincing mathematical arguments can be presented. "This pattern breaks down. The class immediately jumped in to verify these calculations and was astonished when Jennifer said. however. Some icons have page references that direct you to the technology appendix. In this chapter.
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1 . Early mathematicians in Egypt "proved" their theories by considering a number of specific cases. and computers.
what are s and t?
2. 3i 4j ? b. both graphically and algebraically. Diagrams drawn free hand are sufficient. Note carefully that. Two forces F1 and F2 act on an object.
REVIEW EXERCISE
153
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U S I N G G E O M E T RY A N D D I S C R E T E M AT H E M AT I C S
. if a and b have opposite directions.
KEY CONCEPTS REVIEW
151
REVIEW EXERCISE
The chapter Review Exercise integrates the principles taught throughout the chapter.KEY CONCEPTS REVIEW
The principles taught are clearly restated in summary form at the end of the chapter. if u v
M
0. (a
Communication
b)u
au
bu
b. Once you have calculated answers. If i and j are perpendicular unit vectors. If tv c. find the tension in the string b. what is t? v. you have been introduced to the concept of a vector and have seen some applications of vectors. F1 b. This means that AC AB BC. then u
v
Knowledge/ Understanding
ˆ ˆ 5. a. Show that a b a b. you can travel directly along DB. ai bj ? 6.
D C
A
B
DIFFERENCES
Using the same diagram. and if they are physically reasonable. and then forwards along AB. and then along BC. Pay attention to and become familiar with details such as these. the order of the initial point D and the end point B are reversed. F1 54 N. and u is not parallel to v. b. What effect does this have on the tension in the line? Explain. Explain these properties of the zero vector: a. which of course is just the difference DB AB AD. You can refer to this summary when you are studying or doing homework. F2 34 N. If a horizontal force of 12 N pulls the mass to the side a. This translates into the equation DB AD AB. The supports A and B are not at the same level. Using vector diagrams. You will be able to draw and interpret vector diagrams and handle vector equations more quickly and correctly if you do. Two forces at an angle of 130º to each other act on an object. what is the magnitude of ˆ ˆ ˆ ˆ ˆ ˆ a. Which part of the line supporting the mass has the greater tension? Explain. Determine their magnitudes if the resultant has a magnitude of 480 N and makes an angle of 55º with one of the forces. Perhaps the most important mathematical skill to develop from this chapter is that of combining vectors through vector addition.
Key Concepts Review
In this chapter. or you can detour through B.
SUMS
Speaking informally. a. but observe how the detour point fits into the equation: it is the second letter of the first vector and the first letter of the second vector. and the detour point is the initial letter of the two vectors. v 0 v c. F2 21 N. on the right hand side of the equation. ask yourself if the calculated angles and magnitudes are consistent with your diagram. If v b. if you want to go from D to B.
Review Exercise
Communication
1. or you can detour through A. but try to make them realistic. find the angle the string makes with the vertical
Knowledge/ Understanding
8. if you want to go from A to C you can travel directly along the vector AC.
7. A 3-kg mass is hanging from the end of a string. travelling first backwards along AD. and the angle between them is 55º 45 N. Determine the magnitude of the resultant if a. It is not difficult to draw angles that are correct to within about 10º and to make lengths roughly proportional to the magnitudes of the vectors in a problem. 0v 0 b. If sv
t
v. what is t? tu. allowing you to practise and reinforce your understanding of the concepts and skills you have learned. (ab)u
a(bu)
A B
3. and the angle between them is 140º
9. show that a. 4. travelling first along AB. A mass M is hung on a line between two supports A and B. 24i 7j ? c.
3xn
1
7. there are no restrictions? b. 9} and any of the 26 letters selected from {a. The sequence x1. in what direction must the boat steer and how long will it take to cross? 8. How many such sequences are there if a. and
i 1
ai
60.n 1
9. Under what conditions is u
v
u
v?
2. A ferry boat crosses a river and arrives at a point on the opposite bank directly across from its starting point. 1. In the set of numbers U squares?
3. 3 … (2n 1. 5. b. n n 1 n 8. xn r 1) n. What is the product rule used in counting arguments?
10 20
1 20 . Concepts covered in the preceding chapters are further practised through additional exercises and problems. No letter or digit may be repeated. b. b) 4a
a
b
3. The boat can travel at 4 m/s and the current is 1. b. x3 and x4. 6. Determine the tension in each part of the cable. What passwords are in the subset A c. Find the magnitude of the resultant. 1 4. Simplify 3(4u
4.
A G U I D E D TO U R O F YO U R T E X T B O O K
xv
. Using mathematical induction. u b. 2. then accurately draw the following three vectors. A steel cable 14 m long is suspended between two fixed points 10 m apart horizontally. a.5 m/s. prove that 1 n 1C U M U L AT I V E R E V I E W
This feature appears at the end of chapters 3. Do not x
6. Find n(U). and c onto graph paper. In an arithmetic sequence a1. w a b
2 b 3
c
3c a 5c v) a 2u 3(u v). Illustrate in a diagram the vector property 4(a property called?
4b. Guess and prove a formula for xn.CHAPTER TEST
The chapter test allows you to measure your understanding and relate your results to the curriculum achievement charts. a. …. The cable supports a mass of 50 kg at a point 6 m from one end. z}.
Find
i 1
ai. a2. {1. A sequence of length 4 is used as a password. 5 8 1 3. how many are not perfect B?
2. and 12. What is this
5. the sequence starts and ends with a different digit? 4. Find n(A B). 4. …. …. Copy the three given vectors a. …. If the river is 650 m wide at the crossing point.
60 i=1
ai
20.
482 C U M M U L AT I V E R E V I E W C H A P T E R S 1 0 – 1 2
2. b. 8. 7. v c. Evaluate x2. 6. ….
Cumulative Review
CHAPTERS 10–12
1.
Chapter 4 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application Questions 2. The elements of the sequence can be any of the digits selected from {0. Let U be the set of all such passwords and A: the subset of passwords that start with a digit B: the subset of passwords that end with a digit a. A binary sequence of length 6 is formed using any number of 0s or 1s. Prove Pascal's identity r r r
n2 for all
1 . Forces of 15 N and 11 N act a point at 125º to each other. 1000}. 7
The achievement chart indicates how questions correlate to the achievement categories in Ontario's Mathematics Curriculum. x2. an.
1. … is defined by the recursion x1 n 2. the sequence must have exactly three 1s? c. Find the coefficient of x10 in the binomial expansion of 2x simplify your answer.
In these books. 1. you will
• • •
understand the principles of deductive proof.C. Euclid made such effective use of proof in geometry that the books became the standard in mathematics for over 2000 years.Chapter 1
INTRODUCTION TO PROOF
The British detective Sherlock Holmes was famous for solving crimes by using deductive logic. the Greek mathematician Euclid put together all the geometric knowledge of the day in a set of books called The Elements. a proof is a finite sequence of well-formed formulas. In systems of logic and mathematics. In this chapter. 1.4 prove some properties of plane figures using deduction. but became an additional tool for mathematicians to use. solved geometric problems by applying the newly developed skills in algebra. This new mathematics was called analytic geometry.3 prove some properties of plane figures algebraically. In 300 B. you will be introduced to the concept of proof and will learn how to construct a proof that presents a convincing mathematical argument.4
. economics. Section 1. It did not displace the geometry of Euclid. René Descartes. Section 1. Section 1.3. 1. medicine. you will use proof to establish the validity of your propositions. CHAPTER EXPECTATIONS In this chapter.. Our understanding of geometric proof has its roots in ancient history.1. a French mathematician. In the seventeenth century.2. or mathematics. Whether you work in law. politics.
Then prove that it works for any a and b. The numbers (3. 32 42 52. Just pick two numbers a and b with a b and set x 2ab. and are known as trivial solutions. which. Does x 3 y 3 z 3 have a non-trivial solution? 2. Investigate Fermat's Last Theorem is closely related to the Pythagorean Theorem.
DISCUSSION QUESTIONS
1. in 1988." The result has come to be known as Fermat's Last Theorem because it was the last of his conjectures to remain unresolved after his papers were published. Do you think the attention was justified? ●
2
CHAPTER 1
. y a2 b2. and z are non-zero integers. For example. "I have discovered a truly marvellous proof of this. y. There are an infinite number of Pythagorean triples. this margin is too small to contain. Similar to Fermat's Last Theorem is Leonhard Euler's conjecture that there are no non-trivial solutions to x 4 y 4 z 4 w 4. and z a2 b2. Naom Elkies found that 2 682 440 4 15 365 639 4 18 796 760 4 20 615 673 4
Since x2 y2 z2 has infinitely many solutions and x 4 y 4 z 4 w 4 has at least one solution. how could we be sure there were no solutions among the negative integers? 3. a lawyer and amateur mathematician. Then x2 y2 z2. This question remained unresolved for over 200 years until. Fermat claimed.D. however. and it is known that x2 y2 z2 has integer solutions. conjectured that if n is a natural number greater than 2.investigate
C H A P T E R 1 : F E R M AT ' S L A S T T H E O R E M
Around 1637. When Fermat's Last Theorem was finally proven.) Fermat wrote the statement in the margins of his Latin edition of a book called Arithmetica. it is hard to believe that x n y n z n has no solution. It became famous among mathematicians because for hundreds of years many great mathematicians attempted to prove it and achieved only partial results. its proof made headlines in newspapers around the world. 5) are called a Pythagorean triple. Why is there no need to consider negative integer solutions to x n y n z n? That is. the equation x n y n z n has no solutions where x. Pierre de Fermat. Try it. if we knew there were no solutions among the positive integers. written by the Greek mathematician Diophantus in the third century A. (Solutions where some of the integers are zero are possible but not interesting. 4.
We follow their lead in establishing the concept of proof. "This pattern breaks down. but that is only for this triangle. Early mathematicians in Egypt "proved" their theories by considering a number of specific cases.Section 1.
e
1 . we use careful and convincing reasoning to demonstrate the truth of a mathematical statement. the pattern breaks down quickly. When we construct a proof. B every isosceles triangle imaginable? Consider the following example. Other examples. In this chapter. can we conclude that ∠B ∠C for B fold C C. Is this statement true for all values of n? Does this expression ever generate a perfect square? We start by trying small values of n. In this example. we will learn how proofs are constructed and how convincing mathematical arguments can be presented. 485
The concept of proof lies at the very heart of mathematics. What if BC is lengthened or shortened? Even if we construct hundreds of triangles." The class checked and found that she was right. we can see that some patterns that appear to be true for a few terms are not necessarily true when extended. however. never generates a perfect square. For example. The expression 1 1141n2. are much less obvious. In the example just considered. we can construct a triangle such as the one shown and fold vertex B over onto vertex C. One day in class Sunil was multiplying some numbers and made the following observation: 12 112 1112 11112 11 1112 1 121 12 321 1 234 321 123 454 321
He concluded that he had found a very simple number pattern for squaring a number consisting only of 1s. if we want to show that an isosceles triangle has two A A equal angles. How many 1s did Jennifer use? From this example. ∠B ∠C. 1 W H AT I S P R O O F ?
3
. The class immediately jumped in to verify these calculations and was astonished when Jennifer said.1 — What Is Proof?
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APPENDIX P. The Greek mathematicians who first endeavoured to establish proofs applying to all situations took a giant step forward in the development of mathematics. where n is a positive integer. Consider the statement.
which illustrates that we must be careful about drawing conclusions based on calculations alone. By examining the location of infection and analyzing the data collected. When a proof is completed and there is agreement that a particular statement can be useful. A theorem is a proven statement that can be added to our problem-solving arsenal for use in proving subsequent statements. which is not a square 10 270. Our conclusion depends on the quality of the data we collect and the tests we use to test our hypothesis. we develop a chain of unshakeable facts in which the proof of any statement can be used in proving subsequent statements.1141(1)2 1141(2)2 1141(3)2 1141(4)2
1 1 1 1
1142. a medical doctor in London. (We will consider a very powerful form of proof called inductive proof later in this book. which is not a square
Can we conclude that this expression never generates a perfect square? It turns out that the expression is not a perfect square for integers from 1 through to 30 693 385 322 765 657 197 397 207. there is no dependence upon collected data. Mathematics depends on being able to draw conclusions based on rules of logic and a minimal number of assumptions that we agree are true at the outset. England. It takes only one case where the conclusion is incorrect (a counter example) to prove that a statement is wrong. although collected evidence can lead us to statements we can prove. the epidemic was controlled. It is a perfect square for the next integer. is called inductive reasoning. Deductive reasoning allows us to prove a
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. The water was obtained from the Thames River. Inductive reasoning rarely leads to statements of absolute certainty. it is important to explain our reasoning and to make sure that assumptions and definitions are clearly indicated to the person who is reading the proof. he concluded that the source of the epidemic was contaminated water. Theorems can be used to help prove other ideas and to draw conclusions about specific situations. In 1854. the statement is called a theorem. John Snow. We can use calculations or collected data to draw general conclusions. In other words. in which we draw general conclusions from collected evidence or data. Theorems are derived using deductive reasoning. This type of reasoning. In writing a proof. By shutting off the contaminated water. which is not a square (use your calculator to verify this) 4565. In mathematics. downstream from sewage outlets. the best we can normally say is that there is evidence either to support or deny the hypothesis posed. which is not a square 18 257. we also rely on definitions and other ideas that have already been proven to be true. was trying to establish the source of a cholera epidemic that killed large numbers of people.) After we collect and analyze data. Frequently.
and fourth are each greater by 5 than the one preceding. The expression n2 integer value of n.
Knowledge/ Understanding Thinking/Inquiry/ Problem Solving
2. determine whether or not the claim made is likely to be true. 3. say (11. 13). 6. 3. For those that appear to be true. you are given a mathematical statement. Inductive reasoning is a method of reasoning in which specific examples lead to a general conclusion. 14. we can observe that every triple has exactly one number that is a multiple of 3. Using inductive reasoning (that is. gives a composite number for all values of n. Test for n 1. Inductive reasoning can give us a hypothesis. 12. In every set of four positive integers such that the second. 1. exactly one of the three is divisible by 3. try to develop a deductive proof to support the claim. 2. which might then be proved using deductive reasoning. The expression n2 integer value of n. Part B 4.1
Part A In each of the following exercises. but it is not proof. Deductive reasoning is a method of reasoning that allows for a progression from the general to the particular. a. 5. 6. third.
Exercise 1. there is always one divisible by 4. 6. All integers ending in 5 create a number that when squared ends in 25. b. Test for the first ten positive integers ending in 5. 8. The expression f(n) n3 5n2 5n 6.statement to be true. 4. and 17 as first numbers in the set. This provides strong evidence for us to conclude inductively that every such triple contains exactly one multiple of 3. 1 W H AT I S P R O O F ?
5
. 16). 11. 7. 11. (One such set is 1. c. If a number of such triples are written (say two or three by everyone in the class). note that if we write triples of consecutive integers. As an example of inductive reasoning. Test using 5. testing specific cases). We will consider deductive proof in the other sections of this chapter. The expression n2 integer value of n. n n n 5 generates a prime number for every positive 11 generates a prime number for every positive 41 generates a prime number for every positive
1 . where n is a positive integer.
7. Test this claim for n 1. 6. 6. 15. Test this claim for n 5. 13. 4. 7. John Snow. 9. 6. If 3 is subtracted from the square of a positive integer n greater than 4. is often called the father of statistics.
Communication
9. It is claimed that the expression n2 – 79n 1601 generates a prime number for all integer values of n from 1 to 200.
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CHAPTER 1
. 2. the first line divides the plane into two regions. 8. 7. Discover more about his interesting work in the great cholera epidemic by searching on the London Cholera Epidemic on the Internet. 8. Write a computer program to determine whether or not a given number is prime and use it to test the claim. Test this claim for n 5. If 9 is subtracted from the square of an odd integer n greater than 3. Test this claim for n 4.Application
5. mentioned earlier. 8. Straight lines are drawn in a plane such that no two are parallel and no three meet in a common point. 9. For example. 3. creating one region in addition to the original one. the result is a composite number. the result is a number that is composite (has factors other than 1). If 9 is subtracted from the square of an even integer n greater than 2. 5. the result is a number that is divisible by 8. 12. 10. Part C
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10. It is claimed that the nth line creates n new regions.
suppose that the integer a when divided by 3 leaves a remainder of 2. then the next integer. Exactly one of the numbers is divisible by 3. we stated clearly the fact on which the proof was developed: when a positive integer is divided by 3. suppose that a leaves a remainder of 1 when divided by 3. We have considered all possibilities. The next example demonstrates these principles and provides a commentary to justify each step. Second. so the triple (a. Using the same argument as before. and each one convinces us that when we take three consecutive positive integers there is always exactly one of them divisible by 3. we considered all possibilities. there must be a remainder of 0. a 1.Section 1. A good proof is complete. or 2. There are a number of things to observe about this proof. so a 1 is divisible by 3. a 2. a 1. a 2) represents any set of three consecutive positive integers. If a is divided by 3. We conclude that for any set of three consecutive positive integers exactly one of them will be divisible by 3. leaving nothing to chance. First. a 1 and a 2 leave remainders of 1 and 2 on division by 3. Finally. 2 A N I N T R O D U C T I O N TO D E D U C T I V E P R O O F
7
. and the third integer. 1. then a 1 and a 2 are the next two integers. and very importantly. exactly one of the numbers is divisible by 3. will leave a remainder of 0 when divided by 3. no matter how large or small.1. or 2. the argument was presented completely in the proof. Exactly one of the numbers is divisible by 3. the result is completely general.
EXAMPLE 1
Prove that in any set of three consecutive positive integers exactly one of them is divisible by 3.
1 . but we always strive to be brief in its presentation. Third. so a 2 is divisible by 3. we said that when we are proving something to be true we must clearly explain the steps in our reasoning and state any assumptions that we are making. will leave a remainder of 2. will leave a remainder of 1. Next. a 1. the next largest integer. Further. If a is the smallest of the three consecutive integers. it leaves a remainder of 0. For any set of three consecutive positive integers.2 — An Introduction to Deductive Proof
In Section 1. Finally. we are guaranteed that exactly one of them is divisible by 3. If a leaves a remainder of 0. Again. then it is divisible by 3. The third integer. will leave a remainder of 0. 1. Proof If a is any positive integer. a 2.
the roots of the equation are equal. we have constructed proofs that are entirely general in nature.
In these examples. Note that the fact proven in Example 1 makes our task in Example 2 simple once we factor f(n).
EXAMPLE 3
If a. n. then f(n)
n3
n is always divisible by at
Solution Factoring the expression. that the numerical examples we considered in Section 1. n least 6. First.Mathematical proof enables us to construct a chain of proven facts. n 1 is a set of three consecutive integers. one of these numbers is divisible by 3. then n – 1. for a quadratic equation having equal roots. c form an arithmetic sequence. We used no specific numerical examples in our proof. In the next example. we use the result of Example 1 to prove a new fact. D (c – a)2 – 4(b – c)(a – b) [(a 2d) – a]2 – 4[(a d) – (a (2d)2 – 4( d)( d) 4d2 – 4d2 0 2d)][a – (a d)]
Since D
0.
EXAMPLE 2
Prove that if n is an integer. we obtain f(n) n3 – n n(n2 – 1) n(n – 1)(n 1)
If n is an integer and is 2 or greater. and no inductive reasoning.1 did give us some assurance that the result is true. then there is a constant difference between consecutive terms. prove that the equation (b – c)x2 (c a)x (a – b) 0 has equal roots. however.
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. say d. if a. they have a constant difference. if we have three consecutive integers. at least one of them is even (divisible by 2) by an argument similar to that of Example 1. Proof Two facts will help. and c form an arithmetic sequence.
2. Since a. Hence. c form an arithmetic sequence. Further. Then b a d and c b d or c a 2d. Then the product of the three numbers is divisible by 2 3 or 6. b. the discriminant is 0. b. Second. b. Note.
ending in 5 creates a number that when squared ends in 25. for example. n 2.
EXAMPLE 4
Prove that there is exactly one triple of numbers (n. and n 2 is divisible by 3. 2 A N I N T R O D U C T I O N TO D E D U C T I V E P R O O F
9
. which is not a prime. and hence is a composite number. Problems involving parity usually are posed for positive numbers only.2
Part A 1. 5. because one of n. Prove that if n is an odd positive integer. then n 2 and n 4 are even also. then n 4 (n 1) 3 is also. Any other triple (n.
2. n 2.
1 . Prove that if n is an even positive integer. n three are prime numbers. and n always divisible by 3. 2. so there is one triple (3. then n 2. one of the numbers is even and the other is odd. the purpose is to convince a reader that you have constructed a clear. explain your steps and always state the facts on which you base the proof. n 2. one of n. either they are both even or they are both odd. 4 is
Exercise 1. so if n is an even prime. that number is not a prime unless it is 3. If two numbers have different parity. n 4) with n odd must contain a number that is a multiple of 3. n 4 is divisible by 3. so there is only one such triple. Now. then one of the numbers n n 7 is divisible by 4. while 3 and 8 have different parity. n 3 because that is the smallest odd prime. n 1.Our next example uses the idea of parity. 7). then n3 – 4n is always divisible by 48. n 2 is 4. If one of the numbers n. When two integers have the same parity. that the numbers 3 and 11 or 2 and 6 have the same parity. n
4) in which all
Proof If n is an even number. We can say. Hence.
Knowledge/ Understanding
5 or
3. But the only even prime is 2. When writing a proof. Now. Remember. Prove that every positive integer. If n 1 is. airtight argument. then n 2 and n 4 are odd also. If n is odd. It is not possible for n to be even.
Prove that the square of an odd integer is always of the form 8k is an integer. Prove that if 25 is subtracted from the square of an odd integer greater than 5. the result is a composite number. the last two digits of 74 are 01. Let a1.
Thinking/Inquiry/ Problem Solving
7. 3k 2. 11. where k is an integer. the last two digits of 73 are 43. Prove that if 4 is subtracted from the square of an integer greater than 3.) 10. and a5 be any distinct positive integers.Application
4. a3. Prove that the last two digits of 7201 are 07. Show that there exists at least one subset of three of these integers whose sum is divisible by 3. (Use the fact that every integer can be written in one of the forms 3k. 3k 1. 9. Part B
1. Observe that the last two digits of 72 are 49. a4. Given that p and q are two consecutive odd prime integers.
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. 6. the resulting number is always divisible by 8. prove that their sum has three or more prime divisors (not necessarily distinct).
8. a2. Prove that there are no integer solutions to the equation 2x Part C 4y 5. Prove that n5 – 5n3
4n is divisible by 120 for all positive integers n
3. where k
Communication
5. and the last two digits of 75 are 07.
As we proceed. Using these facts we can establish new conclusions.
CB. and as we proceed we mark other facts as they are discovered. the base angles are equal. Isosceles Triangle Property In any isosceles triangle.
EXAMPLE 1
Triangle ABC has a right angle at B. AC is extended to D so that CD bisector of angle A meets BD at E. you will realize that these properties are not difficult to prove. we begin by assuming three basic facts (or properties) to be true.2. We emphasized the following points: 1. we are describing every triangle. we illustrated some important ideas about the construction of a proof. in the context of geometry. Since we require a starting point. 3 P R O O F I N G E O M E T RY
11
. They serve as a convenient starting point in our discussion of geometric proof. The most important characteristic of theorems is that they are general in nature. In geometric proofs. Angles in a Triangle The sum of the interior angles in a triangle is 180˚. The proof of a property is called a theorem. we can be certain that its three angles will always add to this constant sum. when we say. On it we indicate all given information. we do not have to verify their correctness each time we wish to use them. The sum of the interior angles in a triangle is 180˚.Section 1. Given any triangle. Now we'll continue our discussion of proof. 2. When we use theorems. Opposite Angles When two straight lines intersect. Prove that ∠AEB 45˚. which we call theorems. The
1 . A proof is based on assumptions and facts that we accept as true or that we have proven to be true. opposite angles are equal. a diagram is imperative. However.3 — Proof in Geometry
In Section 1. at this point they will be accepted as being true because they have been used in your past study of geometry and have become statements of fact that can be accepted as being true. For example.
Note that we use the symbol ∠CDB both as a name of the angle and to represent the measure of the angle. Then In ∆ABE. ∠CDB ∠CBD. Then ∠BAC 2x.
A convex polygon
A non-convex polygon
12
Proof Let P be any point inside the polygon. the sum of the interior angles is 180(n – 2)˚. we let ∠BAE and ∠EAD be x.
CHAPTER 1
A3 A2 P A1 An
A4 A5
A6
. Proof In ∆BCD.) There are n triangles. (2x) (y) (90 y) 180. This is supplemental information that may be of use to us as we proceed. or 180(n – 2)˚. CB CD. There are points inside a non-convex polygon that cannot be joined to all vertices by lines that lie completely inside the polygon. so the sum of the angles in the n triangles is 180n˚. so ∠CDB ∠CBD. Join P to each of the vertices. Since ∠A is bisected.
A xx
B
y E
C
y D
Since the sum of the angles in a triangle is 180˚ in ∆ABD. Because CB CD. each having an angle sum of 180˚. and they are each y. The sum of the n angles at P is 360˚. (This is why we specified a convex polygon. Comment First we clarify the term convex.Comment We first draw a diagram. a complete rotation. x y Then 45 Hence 90 90 2x 2y x y ∠AEB ∠AEB ∠AEB 90 45 180 180 45 45˚.
∠AEB
EXAMPLE 2
Prove that for any convex polygon of n sides. marking given information. Then the sum of the angles at the vertices is 180n˚ – 360˚. A convex polygon is a polygon in which each of the interior angles is less than 180˚.
the exterior angle is the angle between the extended and adjacent sides. In each of the triangles having one of the required angles. There are two exterior angles associated with each vertex because there are two sides that can be extended. It is easy to see that these two angles are equal.
B C
A
E
D
Proof At any vertex. the angle sum is 180˚. we dealt with a specific situation. Exterior Angles in a Convex Polygon Theorem The sum of the exterior angles of any convex polygon is 360˚. the sum of the interior angles is equal to 180(n – 2)˚. Regardless of the number of sides in a convex polygon.
EXAMPLE 4
For the star-shaped figure shown. C. Then the sum of the exterior angles is 180n˚ – (180n – 360)˚ 360˚. the sum of the exterior angles is 360˚. because they are formed by intersecting lines. every interior angle has an exterior angle associated with it. If a side of the polygon is extended. The sum of the n interior angles is 180(n – 2)˚ (180n – 360)˚. B. we count only one angle at each vertex. Proof Since vertically opposite angles formed by intersecting lines are equal. we can consider it to be a new theorem and remember it for future use. prove that the sum of the angles at A. while in the second we proved a property of convex polygons in general. and E is 180˚. Here we have another general result. 487
e
A D
E
1 . 3 P R O O F I N G E O M E T RY
13
. one that is rather surprising. the sum of the interior and exterior angles is 180˚. Angles in a Convex Polygon Theorem In any convex polygon with n sides (or vertices). In considering the exterior angles of a polygon.In the first example.
B C
t chnology
APPENDIX P. we indicate equal pairs in the diagram. Since this property has general applicability. D. the sum of all interior and exterior angles is 180n˚.
EXAMPLE 3
Prove that the sum of the exterior angles for any convex polygon is 360˚. If there are n vertices in the polygon. In a convex polygon.
x v
E
Exercise 1. Then Therefore x ∠A ∠B ∠A ∠C ∠B y ∠D ∠C z u v ∠E 720 ∠D ∠E 360 900 180
A B C x y y z z v u u D
The sum of the five angles is 180˚. the sum of the exterior angles is 360˚.)
14
CHAPTER 1
. P is the midpoint of the line segment LM. The number of degrees in one interior angle of a regular polygon is x˚. 4. Prove that if PL PK PM. ∠LKM 90˚. Prove 360 that a formula for the number of sides of the polygon is 180 – x . PQ PR. Given that all interior angles are equal. prove that every angle is 150˚. Part B
Knowledge/ Understanding
3.3
Part A 1. In ∆KLM. 2. In ∆ABC. we have tried to give you an idea of the nature of geometric proof and how to construct a proof. A convex polygon has 12 sides. (Remember that a regular polygon has equal sides and equal interior angles. The exercises give you an opportunity to develop the skill of writing simple proofs. Prove that the sum of the exterior angles at opposite vertices of any quadrilateral is equal to the sum of the interior angles at the other two vertices. PQ is extended to S so that QS Prove that ∠PRS 3(∠QSR). Prove that ∠BDC 135˚. In ∆PQR. In this section.
Application
5.
6.(∠A v x) (∠E u v) Then ∠A ∠B
(∠B x y) (∠C y z) 5(180) ∠C ∠D ∠E 2(x y z
(∠D u v)
z
u) 5(180) 900
For the interior polygon. The bisectors of ∠B and ∠C meet at D. ∠A is a right angle.
QR.
respectively. determine the sum of the angles at the tips of the star. the bisector of the interior angle at A and the bisector of the exterior angle at B intersect at P. b. a. prove that ∠ACB 108˚. For the following seven-pointed star. ∆ABC is obtuse-angled at C. The bisectors of the exterior angles at A and B meet BC and AC extended at D and E. If a star has n points. Prove that ∠APB 1 ∠C.7. find a formula for the sum of the angles at the tips of the star. 8. where n is an odd number.
1 . 2
Thinking/Inquiry/ Problem Solving
9. 3 P R O O F I N G E O M E T RY
15
. If AB AD BE. In ∆ABC.
4 P R O O F W I T H A N A LY T I C G E O M E T RY
17
. 2 2
1 .You know 1. y2) is x1 x2 y1 y2 . how to determine the equation of a circle given its radius and centre We can use these facts to consider the following examples. y2) M(x. y1) P(x. 2 . y) Q(x2. . that two lines are parallel if their slopes are equal 5. y1) and Q has coordinates (x2. how to determine the equation of a line given two points on the line 4.
EXAMPLE 2
Prove that the midpoint of the line segment connecting the points A(x1. how to determine the distance between two points 2. 2
The midpoint of the line segment connecting points (x1. y) A(x1. AM ∠APM ∠MAP MB ∠MQB ∠BMQ
0
(Right angle) (Lines parallel)
∆APM is congruent to ∆MQB. y1) x
Proof Let the midpoint of the line segment be M(x. In ∆APM and ∆MQB. y1) and B(x2. that two lines are perpendicular if their slopes are negative reciprocals or if the product of the slopes is –1 6. Draw the run and the rise from A to M and from M to B. y1) and (x2. Then Then or or AP MQ and PM QB x – x1 x2 – x and y – y1 2x x1 x2 and 2y y1 x
x1
y2 y2
y
x2 y1 y2 and y 2 2 x1 x2 y1 y2 The coordinates of M are . y).
y2
2
. how to determine the equation of a line given its slope and a point on the line 3. y2) has coordinates
x1 x2 y1 2
. y). Then P has coordinates (x.
y B(x2.
Now the slope of SP is
b e e –2 2 d c d –2 2 e –0 2
–0 2 b and the slope of RQ is c a a c –2 2
b
b c
The slopes are equal. 0) and one side along the x-axis. e). 0 . so SP is parallel to RQ. We must also be aware that geometric conditions will impose conditions on coordinates.
18
CHAPTER 1
C(b. B(c.
y
Proof Let the rhombus have coordinates O(0. b)
Q( 2 . PQRS is a parallelogram. c). 2 .
EXAMPLE 4
Prove that the diagonals of a rhombus bisect each other at right angles. A(a. Q c 2 a . o). o)
.) R ( 2 . e) v
d+c b+e
)
v
B(c.) Since OABC is a rhombus. C(b. b . Also. 0) simplify calculations. 0). the midpoints of OC.EXAMPLE 3
Prove that the lines joining consecutive midpoints of the sides of a convex quadrilateral form a parallelogram. 2
e Similarly. and CB are S d . This will a 0 O(0. 0)
P( 2 . OA OC. b). and C(d. and 2 2 P d 2 c . the slope of RS is
d–a d a – 2 2 b e b –2 2 e d c c a d a – 2 2
=
e
and the slope of QP is
Then RS is parallel to QP. BA. Then a2 b2 c2. we first calculate the coordinates for the midpoints of the four sides. 2 ) (Note that we have placed one vertex at (0.
y
Proof Let the coordinates for the quadrilateral be d e O(0. 2 C(d. 0). 0)
c+a b
Since we wish to show that the midpoints of the quadrilateral form a parallelogram. The midpoint of OA is R a . Since the opposite sides are parallel. b 2 e . c)
B(a + b. it was noted that convenient placing of a figure relative to the axes can simplify calculations. 0). respectively. (This makes CB OA. and B(a b. so c2 a2 – b2. c). A(a. o) A(a. In Example 3. This is important in the next example. 2 )
x A(a. S( 2 . c)
x 0 O(o.
)
1 . 8). Determine the equation of a line drawn from R that is perpendicular to PQ. 2 . Prove that the diagonals of a parallelogram bisect each other. 3). so the product of the slopes is a b c2 c . slope OB
c a b
But c2
c and slope AC b c a . 2). and Z(9.
b. 0). 1) lies on the median drawn from vertex Z. c. Determine the equation of the line drawn from vertex X to the midpoint of YZ.Now.
Knowledge/ Understanding
2. 2 . 3). Since these coordinates are the same. Prove that the two medians from parts a and b intersect at the point (4. What conclusion can be drawn about the medians of this triangle? 4. The midpoint of one diagonal is also the midpoint of the other. P and Q are the same point. 5) perpendicular to AC passes through the midpoint of AC. b–a b2 – a2 2 2 a2 – b2. 1. a.
1). 1). so the product is a2 – b2 1. Triangle ABC has vertices A( 1. and C(7. Prove that the line through B(5. Y( 2.4
Part A Always try to choose coordinates of points so as to simplify your work. (The line in a triangle from a vertex to the midpoint of the opposite side is called a median.
c c The midpoint of OB is P a 2 b . Prove that this triangle is isosceles. The midpoint of AC is Q a 2 b . d. and R(4. Verify that the point (4. The ∆XYZ has its vertices at X(5. 4 P R O O F W I T H A N A LY T I C G E O M E T RY
19
. The ∆PQR has its vertices at P( 6. a. It can sometimes be helpful to work a specific example before doing the general case. Q(0. b –a
and the diagonals are perpendicular. Determine the equation of the median drawn from vertex Y. Part B 3.
Exercise 1. (This line is called the altitude from R to PQ. 5). B(5. 4). a.) b. Then the diagonals of a rhombus bisect each other at right angles. e. 0).
In ∆ABC. Prove that the line segments joining the midpoints of any rectangle form a rhombus. 8. a2 b2 c2. Prove that the diagonals in a rectangle are equal. c. that is. b.
20
CHAPTER 1
. d. 2AD2. Determine the coordinates of the point of intersection of the two altitudes found in parts a and b. Show that the altitude from Q contains the point of intersection of the other two altitudes. a. What conclusion can be drawn about the altitudes of this triangle?
Application
5. Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and has a length equal to one half of it.
Application
7. 6. 10. In the interior of rectangle ABCD. Prove that the diagonals in a square are equal. 13. Prove that the altitudes of a triangle are concurrent. e. 9. Prove that the medians of a triangle are concurrent.) 11. Determine the equation of the altitude drawn from P to QR.b. Prove that PA2 PC2 PB2 PD2.
a c b
Part C
Thinking/Inquiry/ Problem Solving
12. Prove this result using the Pythagorean Theorem. prove that AB2 AC2 2BD2 (Let the length of BC be 2a units and use B or D as the origin. the square on the hypotenuse is equal to the sum of the squares on the other two sides. where AD is the median. a point P is chosen at random. Prove that the sum of the squares of the lengths of the sides of a parallelogram is equal to the sum of the squares of the lengths of the diagonals.
Pythagorean Theorem
If a triangle is right-angled.
Prove that if n is a natural number greater than 2. It is undoubtedly not the proof that Fermat himself claimed to have. mathematicians had proved the theorem for values of n up to four million. and z are non-zero rational numbers. When the corrected proof was published it was well over 100 pages long. over 350 years after Fermat first stated the theorem.
INDEPENDENT STUDY
What kinds of questions are studied by mathematicians working in the branch of mathematics known as number theory? What is the process by which a new mathematical result is accepted by the mathematical community? Research other famous mathematicians who contributed to Fermat's Last Theorem. the equation x n y n z n has no solutions where x. What are some other significant unresolved questions in mathematics? ●
RICH LEARNING LINK WRAP-UP
21
. Hint: Show that if there is a solution among the non-zero rational numbers then there is also a non-trivial integer solution (which we know is false). finally proved it for all natural numbers n. Show that if there are no non-trivial solutions to x 4 y 4 z 2.) 2. equivalently. Most proofs of Fermat's Last Theorem for the case n 4 focus on the equation x 4 y 4 z 2. y. In 1994. including a final year to fix an error pointed out after he first announced that he had the proof. (Or. Andrew Wiles. either by working out particular cases or by contributing ideas subsumed by Wiles' proof. By 1992. It took Wiles more than seven years to produce the proof. y. then there is a non-trivial solution to x 4 y 4 z 2. then there are no non-trivial solutions to x 4 y 4 z 4. if there is a non-trivial solution to x 4 y 4 z 4. See the section on indirect proof in the next chapter. and z are non-zero integers. most mathematicians working in number theory suspect that Fermat did not have a complete proof. Investigate and Apply 1. a Cambridge mathematician.wrap-up
C H A P T E R 1 : F E R M AT ' S L A S T T H E O R E M
investigate and apply
Fermat's Last Theorem states that if n is a natural number greater than 2. the equation x n y n z n has no solutions where x.
4. Prove. In the diagram. Julie writes the equation x2 2x 3 0. the sum always has at least three divisors. is D 22 – 4(1)(3) 8. She does another example. prove that the right bisectors of the sides meet at a common point. ABCD is a trapezoid with AB CD. R. 0). and C(b. and makes the same observation. such as (n – 1)x2 nx n 1 0. EB bisects angle ABC.
22
CHAPTER 1
. B(0. 7 1 All
1.
B C D
A E P D
A
B Q C
E
6. then this equation always has imaginary roots. c). b. 7. 5x2 6x 7 0. She observes that D.Chapter 1 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application Questions All 2. (Do not use the formula for the sum of the exterior angles of a convex hexagon. 4). the discriminant.) 3. Prove that when three consecutive even numbers are squared and the results are added. Given the two points A( 2. P is the midpoint of AD. using analytic methods. determine ∠E. 4) and B(6. S. a. 5. and Q lie in the same straight line. which implies that the given quadratic equation has imaginary roots. If ∠A is 58˚. that the points P. Prove that the sum of the exterior angles for a convex hexagon is 360˚. and Q is the midpoint of BC. Prove that if an equation has consecutive positive integers as its coefficients. what must be true about a point C(x. a). In ∆ABC. y) that is equidistant from A and B? 2. and EC bisects angle ACD. Be certain to explain the steps in your reasoning. The diagonals AC and BD have midpoints R and S. with vertices A(0. Verify that the equation 8x2 9x 10 0 has imaginary roots.
you will expand your knowledge of proofs to include theorems and plane figures. 2.1. you can put forward a theorem as something that can be proven from other propositions or formulas. extending it to include indirect proofs. 2. and other polygons. To develop our understanding of the properties of triangles. Section 2.Chapter 2
PLANE FIGURES AND PROOF
Now that you have learned more about the basic concept of proof. 2.3 prove some properties of plane figures using indirect methods.5.2. your theorem may be expressed as an equation formula. quadrilaterals.6 understand the principles of deductive proof.
CHAPTER EXPECTATIONS In this chapter. you will
• • •
prove some properties of plane figures using deduction. In mathematics.4
. As a rule or law. Section 2. as well.2. Section 2. 2. we continue to make use of the idea of proof in this chapter.
y x z
x
y
z
180º
Isosceles Triangle Property
In any isosceles triangle. constantly expanding our list of known facts. the base angles are equal.
x y x y
Pythagorean Theorem
If a triangle is right-angled.
THEOREMS ASSUMED IN CHAPTER 1 Angles in a Triangle
The sum of the angles in a triangle is 180º. we can prove new facts. accept them as true.Review of Prerequisite Skills
In Chapter 1. we discussed some of the basic aspects of geometrical and non-geometrical proof. opposite angles are equal.
a c b
24
CHAPTER 2
. we continue with proof in geometry. Using these theorems. In this chapter. and use them as building blocks for the development of further theorems. showing how we can take basic definitions and theorems. in the diagram. the square on the hypotenuse is equal to the sum of the squares on the other two sides.
a
b
a
b
Opposite Angles
When two lines intersect. a2 b2 c2.
c b
a
a
b
c
REVIEW OF PREREQUISITE SKILLS
25
.THEOREMS PROVED IN CHAPTER 1 Angles in a Convex Polygon
In a convex polygon of n sides. 2
b1 b2 h1 h2
A
1 bh 2 1 1
1 bh 2 2 2
Exterior Angle Property
An exterior angle of a triangle is equal to the sum of the two interior and opposite angles.
x3 x4
x2
x1
x5
x1
x2
x3
….
xn
180(n – 2)
Exterior Angles in a Convex Polygon
The sum of the exterior angles of a convex polygon is 360º. the sum of the interior angles is 180(n – 2)º.
a3 a4 a2 a1
a5
a1
a2
a3
…
an
360
THEOREMS ASSUMED IN CHAPTER 2 Area of a Triangle
The area of a triangle having base b and height h is A 1 bh.
What is the area of ∆ABC?
E 12 A C
2. Determine the length of AF.Parallel Line Properties
If two lines are parallel 1. alternate angles a and b are equal 2. DC
12 and AE
gm
10.
26
CHAPTER 2
.
gm
B
a. opposite angles are equal 3. the area is A bh
x B b1 b2 A w h 1 h2 y C D z
AB x A
CD and AD BC z and w y b1h1 b2h2
These relationships are the basis for the theorems that we will prove in this chapter. Determine the area of
ABCD.
Exercise
Part A 1. distance between the lines is constant
a b d h1 h2 c
Properties of a Parallelogram
In a parallelogram 1. In AE
gm
ABCD.
10
a. corresponding angles a and c are equal 3. the sum of co-interior angles a and d is 180º 4.
D
13 F 5 E C
b. opposite sides are equal 2. EC. Calculate the area of
ABCD.
D
b. lengths are marked as shown. In
gm
A B
ABCD.
A
D
x B
x
y y C E
t chnology
APPENDIX P. If ∠ABC 70º and ∠A 30º.3. Quadrilateral ABCD has BD 12 and AC 16.
A
D C
B
Part B 5. determine the measure of ∠BDC. Determine the length of the altitude from B to AC. AC ⊥ BD. In ∆ABC. 490
7. Determine the area of quadrilateral ABCD. The rhombus PQRS has diagonals of length d1 and d2 as shown. AC 10. and the altitude from C is 3. 2
S P
Q
d1
R d2
6.
A 17 8 D 6 B
C
4. Calculate the area of figure ABCD. Prove that the area of a rhombus is given by the formula A
d1d2 .
A
e
B
C
REVIEW OF PREREQUISITE SKILLS
27
. AB 12.
the resulting interior quadrilateral is a parallelogram. 5). it usually has a special name. 7). Describe the possible motions of the parallelogram. If a quadrilateral has special properties. as shown. QR and PS can both be shown to have a slope of 3 . a device which works because of the contribution of parallelograms. Thus. DISCUSSION QUESTIONS
y
y
y P6 A 4 2 –6 –4 –2 –2 S –4 D –6 2 R 4
B
Q x 6 C
1. They are used in some jointed desk lamps. For example. A rectangle is a special type of parallelogram. and S(–5. PQ and SR are parallel.
3 2 5 . Pick four points A(–7. Using Geometer's Sketchpad® will allow you to move your four points independently. ●
28
CHAPTER 2
. Non-rectangular parallelograms are much less common. Investigate and Inquire In geometry. but they do arise as the solution to certain engineering and design problems. Q(6. 7 Similarly. mounted against the side of a wall). where they allow the light-bulb end to be moved without changing the direction in which the bulb points. However. –5) on a Cartesian plane. Try this with four other points. if we connect the midpoints of the sides of the quadrilateral. Thus. The midpoints of AB. B(5. What are some other special quadrilaterals? 2. BC. CD. C(7. if we connect four points on a plane. 2). they are used in some binocular mounts for amateur astronomers and some motorcycle suspension systems. What are some other special parallelograms? 3.g. The slope of PQ is x2 x1 7 6 ( 1) 2 1 A similar calculation shows that the slope of SR is also 3 . the result is not likely to be a parallelogram. PQRS is a 2 parallelogram. Consider a parallelogram frame that has one side held in position (e. For similar reasons. –3). They are a common shape because rectangles are a type of parallelogram. In movies. –4). and DA are found to be P(–1. 3). scenes that require a moving camera can be made jitter-free by using a Steadicam.. respectively. a parallelogram is a special quadrilateral with opposite sides that are parallel. –1). R(2. so they are also parallel. and D(–3.investigate
C H A P T E R 2 : VA R I G N O N PA R A L L E L O G R A M
Parallelograms are quadrilaterals with opposite sides that are parallel.
we write ∆ABC ∆DEF.1 PROOFS USING CONGRUENT TRIANGLES
29
. we generate only one triangle. All triangles with the same three properties must be congruent. or if 2. (hypotenuse-side) the hypotenuse and one other side of one right triangle are respectively equal to the hypotenuse and one other side of a second right triangle. such as the lengths of corresponding line segments. The combinations that specify triangles are given by the following theorem. For example. We write ∆ABC ∆DFE.1 — Proofs Using Congruent Triangles
t chnology
APPENDIX P. Many other properties are equal because of the congruence. We use triangle congruence extensively in geometric proof. the measures of corresponding angles. are equal.
e
2. Thus measurable quantities. 491
Two geometric figures are congruent if they are exactly the same except for position. If we specify three properties (sides or angles) in the right combination. (side-angle-side) two sides and the contained angle of one triangle are respectively equal to two sides and the contained angle of the second triangle. we omit the proof. Without referring to the diagram. (side-side-side) three sides of one triangle are respectively equal to three sides of the second triangle. the order of the letters gives the correspondence of the figures. (angle-side-angle) two angles and a side of one triangle are respectively equal to two angles and the corresponding side of the second triangle. In this statement. we know from the statement that ∠B ∠F and AB DF.
A o D o
x B C E
x F
The two triangles shown above are congruent. or if 3. Since you have seen this result earlier. and the areas. mainly because triangles are easily defined geometric objects. That is. there is a way to match one figure with the other so that corresponding parts of the figures coincide.Section 2. or if 4. Triangle Congruence Theorem Two triangles are congruent if 1. If we wish to indicate that the triangles are equal in area but are not necessarily congruent. the altitude from A and the altitude from D are equal.
this was referred to as the ambiguous case. state the congruence. saying so does not give a third property. the third pair must be equal.
EXAMPLE 1
t chnology
APPENDIX P. The equal angles are neither right angles nor contained by the equal sides. as long as they are the same position in both triangles.
A 9 9 50° B C E 8 50° F 8 D
In writing a proof using congruent triangles. Note that if it is necessary. especially where the equality is not obvious 3. ∠ABC and ∠DEF have two pairs of equal sides and a pair of equal angles. 2. we will generally do this before identifying the triangles. implying the correspondence by the order of the vertices (it is customary to refer to the triangle congruence combination being used) Once we have established that the triangles are congruent. 3. the angles must be contained (where the pairs of sides meet) or right angles (the third property in the hypotenuse-side combination). Angle-angle-angle is not a triangle-congruence combination. For example. identify the triangles we are considering (it is a matter of preference whether we are careful about the correspondence of vertices at this stage) 2. but they are not congruent.
30
CHAPTER 2
e
. Recall from your study of trigonometry that there can be two triangles in some cases. giving reasons.Here are three additional observations: 1. we usually 1. we know that when two pairs of angles are equal. we can draw conclusions about components of the triangles that were not previously known to be equal. If there are two pairs of equal sides and one pair of equal angles in two triangles. to appeal to facts not relevant to the triangle congruence. 501
Prove that the median to the base of an isosceles triangle bisects the vertical angle. Since the total of the angles is fixed. in order to establish one of the three properties. The angle-side-angle combination can have the equal sides in any position relative to the pairs of equal angles. list the three pairs of equal components.
1 PROOFS USING CONGRUENT TRIANGLES
31
. Prove ∆ABC ∆DBE. the side-side-side property b. the hypotenuse-side property
B
A D
C
2. Thus we have an isosceles triangle ABC. ∆ABC ∆DBE
A
E B
(given) (vertically opposite angles) (given) (side-angle-side)
C
D
Exercise 2.
B
A
D
C
Notice that the same pair of congruent triangles show that ∠B ∠C.Solution On the diagram we indicate what is known at the beginning.
EXAMPLE 2
In the diagram. the triangle congruence combination would have been side-angle-side. state all properties that must be present for ∆ABC to be congruent with ∆ADC. AB
DB and CB
Proof In ∆ABC and ∆DBE. so BD DC. We have drawn the median AD. the side-angle-side property c. Use a. the Isosceles Triangle Property.1
Part A
Communication
1. EB. AB AC (given) BD CD (median) AD AD (common) Then ∆ABD ∆ACD (side-side-side) Therefore ∠BAD ∠CAD (triangles congruent) Then the median bisects the vertical angle. We need to prove that ∠ BAD ∠ CAD. with AB AC. BA BD ∠ABC ∠DBE BC BE Therefore. Had we used this instead of the common side AD. Proof In ∆ABD and ∆ACD. the angle-angle-side property d. For the diagram given.
Is the statement that follows true or false? Justify your answer. In the given diagram.
A
D
d. Find a pair of congruent triangles in each diagram. Given the statement ∆ABC ∆DEF. AB
AC. a. could you state enough reasons for ∆DBC to be congruent with ∆ECB? c.
O B C C B
AB and CD are diameters 5. Could you now prove ∆FDB ∆FEC?
B D F C E
3. Which angle in ∆DEF is equal to ∠ABC in ∆ABC? b. PT.
P
D
C Q S A T D R
c. b. a. In the following diagrams. Which side in ∆ABC is equal to DF in ∆DEF?
Knowledge/ Understanding
4. draw a diagram illustrating the triangles.
A
a. PQ PR and PS Prove that ∠QRS ∠RQT.
B A
D P
C
Part B 6.
Q
R
S
T
32
CHAPTER 2
.Communication
2. we have marked the equal angles and sides. If ∆ABE ∆ACD.
A B
b. The median of a triangle divides a triangle into two congruent triangles. In the given diagram. State one additional fact that would prove that ∆ABE ∆ACD. and write a proof explaining why the triangles are congruent.
If the diagonals PR and QS are equal. S is the midpoint of side QR of ∆PQR. Prove that QT RW. 13. In quadrilateral ABCD.
A x
e
B D x
C
E
F
2. Using ∆ABC and ∆DEF. write an informal proof showing that this statement is true. QT and RW are drawn perpendicular to PS or PS extended. Prove that AC bisects ∠BCD and BC DC.1 PROOFS USING CONGRUENT TRIANGLES
33
. Prove that if the opposite sides of a quadrilateral are equal then the diagonals bisect each other.
t chnology
APPENDIX P. In the given diagram. Prove that
S
P
Q
T
R
10. In ∆PQR. PQRS is a quadrilateral in which PQ SR. then prove that ∠PQR ∠SRQ. the diagonal AC bisects ∠DAB and AB AD. prove that AC DE.
B x x D
A
C
8. If AB BE. ABE and CBD are straight lines. 503
11.
∠RPT. and DE and BE are also perpendicular.
D A B E C
9. 12.Application
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e
APPENDIX P. PQ PR and ∠QPT ∠SQT ∠SRT. AC and AB are perpendicular. 502
7. We know that two triangles are congruent if two sides and the contained angle of one are respectively equal to two sides and the contained angle of a second triangle.
the triangle congruent situation angle-side-angle? c. a. Prove that AX AY. respectively.Thinking/Inquiry/ Problem Solving
14. the triangle congruent situation side-angle-side? b. X and Y are points on BD and CD.
X B
D
Y C
34
CHAPTER 2
. Discuss the conditions needed for two quadrilaterals to be congruent. the triangle congruent situation side-side-side? Part C 15. Prove that AP b. The diagram shows a rectangular solid in which A is the midpoint of SR. PB. so that BX CY. and B is the midpoint of UV.
U B A V P X S A R
Q
W
16. Prove that AX WB. Is there a possible statement about quadrilaterals that parallels a. A tetrahedron has four identical equilateral faces.
the resulting number p2 has exactly three divisors: 1. In mathematical terms. When we write p → q. The combination of the two simple statements. we read it as If p. 5. We use this type of logical inference frequently in everyday speech. Before considering further mathematical examples we examine the structure of conditional statements using examples from everyday life. When we apply it in mathematics we are attempting a similar kind of reasoning. If a prime number is squared then the resulting square has exactly three divisors. suppose we let p and q represent the following: p: My family pet has four legs. If I toss a stone into a pond then ripples are produced in the pond. then your knowledge will increase. In this section. The conclusion follows as a natural consequence of the premise that a stone is tossed into the pond. we are saying that the conclusion q follows as a result of p being true. This statement is not necessarily true. For example. If a triangle is drawn having exactly two equal sides then the triangle will have exactly two equal angles. Is this statement true? Our experience tells us that ripples are produced when a stone is tossed into a pond. we read the statement p → q. If we accept p as being true. or If you study diligently. Verify that this statement is true using the primes 3. as If my family pet has four legs then it is a dog. then q. We might hear someone say. q: Ripples are produced in the pond. This statement implies that if we choose a prime and square it the result is always a number having three divisors.2 — Conditional Statements
We often hear people make statements in which one action is a consequence of another. If we write p → q. but with a higher degree of precision. this statement is true. In this example. we are saying. If we square a prime number p. does the conclusion q follow absolutely? The truth of such statements depends on the premises that we accept. Consider the following statements: p: I throw a stone into a pond.Section 2. One example of a mathematical conditional statement is.
2 . p. together with If… then… creates a conditional statement. q: It is a dog. we examine the structure of conditional statements and their role in mathematical proof. Statements in which the first part implies the second part as a natural conclusion are called conditional statements. 2 C O N D I T I O N A L S TAT E M E N T S
35
. we are asking. As we know. and 7. If you tip that glass the water in it will run out. and p2. Consider the statement. The fact that my family pet has four legs does not guarantee that it is a dog. When considering the truth of q in the statement p → q.
We can do so using the theorems stated at the beginning of the chapter. Both parallelograms have opposite sides in the two parallel lines WX and YZ. QA RD. PQ RS.
(Parallelogram Area Property Theorem)
TQRD and gm BRDA have RD as a common base and lie between parallel lines RD and QA. if we can prove that a statement q is always true if a given statement p is known. BRDA. and gm BRDA. THEOREM Consider the following statements: p: Two parallelograms have equal bases and lie between the same parallel lines. Since p q and q r. then the conditional statement defines a theorem. we will always be able to make an immediate conclusion when we encounter such parallelograms.In mathematical applications. 505
36
CHAPTER 2
e
Prove that if a triangle and a parallelogram are on the same base and between the same parallels. their areas are equal. the height is the distance between the parallel lines. For each parallelogram. we know that such distances are equal. Prove that the area of gm PQRS Proof gm PQRS and gm TQRD have QR as a common base and lie between parallel lines QR and PD. Parallelogram Area Property Two parallelograms having equal bases (or the same base) and lying between two parallel lines have the same area.
(Parallelogram Area Property Theorem)
Notice that here we have used extended reasoning. Then area Then area
gm gm
TQRD PQRS
area area
gm gm
BRDA. then the area of the triangle equals 1 that of the parallelogram. then p r. Then area
gm gm
A P B Q R T S D
PQRS
area
gm
TQRD.
W A D E H X
Y
B C F G
Z
Proof Let the parallelograms be ABCD and EFGH with BC FG. PD QR. Since the parallelograms have equal bases and equal heights. This type of reasoning is used frequently in the construction of proofs.
EXAMPLE 1
In the given diagram. q: The parallelograms have the same area. Using the theorem of Parallel Line Properties.
EXAMPLE 2
t chnology
APPENDIX P. If we can show that the conditional statement p → q is always true. BR AD. 2
.
we have a conditional statement or an If … then… statement.
X
P R
Q
Proof Y Let the distance from Q to XY be h1 and the distance from P to XZ be h2.
W h Y
E
D
C
X
A B Since the distance between parallel lines is constant. (Parallel Line Properties Theorem)
Z
Since h is the height of ∆EAB. Also ∆XQY ∆XPZ But ∆PRY Then ∆XQY
1 (XY)h1 2
Z
∆PRY ∆QRZ ∆QRZ ∆XPZ
1 (XZ)h2 2
quad XPRQ quad XPRQ
X
h1
Q
But Then
XY h1
XZ h2
Y X
The distances are equal. Note that h1 is the altitude in ∆XQY and that h2 is the altitude in ∆XPZ. When we write the Z statement p → q.
2 .
P
h2
In this section we have introduced a number of proofs that involve simple conditional reasoning. XY XZ and P and Q are on the sides of ∆XYZ. 2
In the diagram given. represent this distance by h. 2 C O N D I T I O N A L S TAT E M E N T S
37
.Proof Let the triangle EAB and the parallelogram DABC have the common base AB and lie between the parallel lines WX and YZ. Prove that if the area of ∆PRY equals the area of ∆QRZ. area ∆EAB area
gm
1 (AB)h 2
ABCD
(AB)h
Thus area ∆EAB EXAMPLE 3
1 (area gm ABCD). We start with a premise p that we assume to be correct and attempt to prove that the condition q follows. then the distance from P to XZ is equal to the distance from Q to XY.
…. PQRS is a parallelogram.Exercise 2. 506
2. c. 2
P A Q
S
B
R
38
CHAPTER 2
. U. 2. and VU QP. a. and V are the midpoints of PQ. (v) If a five-sided figure has 3 angles each equal to 90º. then the resulting number always gives a remainder of 1 when divided by 4. then the remaining two angles must both be obtuse. (iv) If a parallelogram has an area of t square units. if the conclusion is incorrect. state an amendment that makes it correct. draw a second parallelogram having an area double that of parallelogram ABCD. (i) If an integer ends in a 0 or a 5. identify the premise and conclusion. Part B 5. T. Given a parallelogram PQRS. PR.
Knowledge/ Understanding
t chnology
APPENDIX P.2
Part A
Communication
1. If TU QR. then the resulting number will always have an even number of divisors. If A is the midpoint of PQ and B is the midpoint SR. prove that the area of 1 gm ASBQ area of gm PQRS. For each of the following statements. draw a rectangle equal in area to the parallelogram. 4. respectively. determine whether or not the conclusion follows from the premise. and QR. then the remainder is an integer in the set 0. TV PR. name three parallelograms having equal area. then the integer is divisible by 0 or 5. In ∆PQR. (iii) If a positive integer is divided by a positive integer n. then a parallelogram of 2t square units has a base that is twice as large as the first parallelogram. (ii) If an odd integer is squared.
P
T
U
e
Q
V
R
Knowledge/ Understanding
3. n 1. 1. b. (vi) If an integer is squared. Given a parallelogram ABCD.
then ∆AEF ∆FBD.
A E F C D B
t chnology
APPENDIX P. Prove that if AD and BE are medians in the given triangle. then the quadrilateral is a parallelogram. Prove that ∆ABX ∆ADX. 2
S P T Q
R
12.
A
E
B
D
C Q
9. ABCD is a quadrilateral whose area is bisected by the diagonal AC. 11. and ∆TRQ all have equal area. 2 C O N D I T I O N A L S TAT E M E N T S
39
. BE and CF are two medians of ∆ABC. A parallelogram ABCD has diagonal AC extended to X. A point O is on the side BC of gm ABCD. then the line bisects the area of the triangle.
P
T S R
10. 13. Prove that BD is bisected by AC. ∆TSR. PQRS is a parallelogram and T is any point inside the parallelogram. AD is the median of ∆ABC. intersecting at O. Prove that ∆PQT. Prove that if the diagonals of a quadrilateral divide it into four triangles of equal area. 2
2 . PQRS is a parallelogram. 15. Prove that ∆OFA ∆OEA. ∆PTS. Prove that ∆TSR ∆TQP
1 gm PQRS. Prove that ∆ABE
∆ACE.Application
6. 508
e
8. 7. 14. Prove that if a line in a triangle is a median. Prove that ∆AOB ∆DOC 1 gm ABCD.
AC is extended to E so that CE BD. 509
e
D
E
B
C
18.
A
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APPENDIX P. ABC is an isosceles triangle in which AB AC. describe a procedure for the construction of a triangle that is equal in area to the quadrilateral. AB AC and D is any point on AB. In ∆ABC. For a quadrilateral ABCD. BE and CD are drawn perpendicular to AC and AB respectively. DE cuts BC at K.
40
CHAPTER 2
.Part C
Thinking/Inquiry/ Problem Solving
16. 17. Prove that DE BC. Prove that DK KE.
"p implies q. If one side of a balance falls. The biconditional statement is." We sometimes write this as "p iff q. if the converse is true. The converse is If we stop the car." When we write p ↔ q we recognize that the truth of either of the statements depends upon the truth of the other. 510
The theorems we have accepted and the results we have proved from them are examples of conditional statements.Section 2. If one of two integers is even and the other is odd.3 — The Converse of a Conditional Statement
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APPENDIX P. we write p ↔ q. which is read. we write p → q. 2. The converse of this statement is.
2 . One side of a balance falls if and only if there is more weight on that side than on the other. statement 1. The sum of two integers is odd if and only if one of them is even and the other is odd. which is read." or "if p then q. Solution 1. state the converse b. The biconditional statement is. This statement is true for both legal and safety reasons. we can stop the car for a variety of reasons. Symbolically. This is certainly true. which we write as q → p. 3 T H E C O N V E R S E O F A C O N D I T I O N A L S TAT E M E N T
41
. This statement is not necessarily true. if p is the premise and q is the conclusion. determine whether the converse is a true statement c. This is certainly true.
EXAMPLE 1
e
For each of the following statements a. then one of the integers is even and one is odd. then one side of the balance falls. If the traffic light is red then we stop the car. then the sum of the integers is odd. then the light is red. 2." An idea closely related to a conditional statement p → q is that of its converse. If one statement (the premise) is true. If it is spring. This type of statement is said to be
biconditional. 3. We cannot conclude that the converse is true because the original statement is true. The converse of this statement is. If a statement is true (p → q) and its converse is also true (q → p). then the grass is green. Consider the statement. If there is more weight on one side of a balance than on the other. then another statement (the conclusion) is true. "p if and only if q (or q if and only if p). If the sum of two integers is odd. there is more weight on that side than on the other. restate the sentence as an …if and only if….
If the grass is green. If two angles are equal then they are vertically opposite angles.3.
A
B
C
In isosceles triangle ABC. If two angles are vertically opposite then they are equal. Solution First we prove that any point on the right bisector of a line segment is equidistant from the end points of the line segment. We can demonstrate that this statement is not true by constructing an example showing it to be false. This statement is certainly not true. q: The two angles are equal. then it is spring.
B
D
Part 1 q→p The second statement is. for these two statements? p: Two angles are vertically opposite. ∠COB 180º. Then p ↔ q is not true because p → q and q → p are not both true.
C
EXAMPLE 2
Since ∠COA Since ∠COA Therefore
∠COB ∠AOD ∠COB
180º. A point is on the right bisector of a given line segment if and only if it is equidistant from the ends of the segment. The converse of this statement is.
42
CHAPTER 2
. ∠ABC ∠ACB but these two angles are not vertically opposite each other. ∠AOD ∠AOD
180º 180º
∠COA ∠COA
O
A
The statement is true. Solution Part 1 p→q The first statement we will prove is. Is it true that p ↔ q.
THEOREM
Prove the biconditional statement.
∠PMB 90º
A
P
M
B
Then PM is the right bisector of AB. so P is a point on DE. Right Bisector Theorem A point is on the right bisector of a given line segment if and only if it is equidistant from the end points of the line segment. P is a point on FG. Join P to A and P to B. Let P be any point on XY. Proof We are given PA PB. and C that are not in a straight line. The point P is equidistant from A.
D
Solution If P is equidistant from A and B it must lie on the right bisector of AB. B. cutting AB at D. the midpoint of AB.Proof Let AB be any line segment and XY be its right bisector. PA PB (given) AM BM (constructed) PM PM (same line segment) Then ∆PAM ∆PBM (side-side-side) Then ∠PMA ∠PMA ∠PMB and. We combine the results of these two properties in the Right Bisector Theorem. and so PA PB PC. In triangles PAM and PBM. AD ∠PDA PD Then Then ∆PAD PA BD ∠PDB PD ∆PBD PB
(right bisector) (right angles) (same line) (side-angle-side)
A D Y B X P
Now we prove that if a point is equidistant from the endpoints of a line segment. Similarly.
EXAMPLE 3
Determine the position of a point P equidistant from three given points A.
A
F
P
C
G B E
2 . then it is on the right bisector of the line segment. the right bisector of BC. 3 T H E C O N V E R S E O F A C O N D I T I O N A L S TAT E M E N T
43
. We join P to M. and C. since their sum is 180º. B. In triangles PAD and PBD. the right bisector of AB.
THEOREM
A point is on the bisector of an angle if and only if it is equidistant from the arms of the angle.
44
CHAPTER 2
. Let BD be the bisector of ∠ABC and let P be any point on BD. it is equidistant from the arms of the angle.
B x x Y
A X P D C
From P draw perpendicular lines to meet the sides BA and BC at X and Y respectively. We will prove that PX PY. Let P be any point on line BD such that perpendiculars PX and PY are equal.
Angle Bisector Theorem A point is on the bisector of an angle if and only if it is equidistant from the sides of the angle. In ∆PBX and ∆PBY PX PY PB is common ∠PXB ∠PYB Then Then ∆PXB ∠PBX ∆PBY
(given)
B A X P D C
Y
90º (given)
(hypotenuse-side)
∠PBY and BP is the bisector of the angle. ∠PBX ∠PBY PB PB ∠PXB ∠PYB
(given) (same line) (construction)
90º
Therefore ∆PBX ∆PBY (angle-angle-side) Then PX PY and P is equidistant from the arms of the angle. In ∆PXB and ∆PYB. Part 2 If a point is equidistant from the angle arms. Proof Part 1 If a point is on the bisector of an angle. then it is on the bisector of the angle.
then it has three unequal angles. An integer is divisible by 2 if and only if it is even. b. the statement and its converse form a biconditional statement 3. 3 T H E C O N V E R S E O F A C O N D I T I O N A L S TAT E M E N T
45
. f. a. 3. 5. the converse is true c. If it rains.
A
3 km
6 km
B
4 km
C
2 . b. If a triangle has three unequal sides. Figures are congruent if and only if they are similar. Determine which of the following statements are true. d. State the converse of each of the following statements. b. then it is the weekend.
Communication
2. A four-sided figure is a parallelogram if and only if it is a rectangle. the statement is true b. a. then it is not divisible by 2. If all answers on a test are incorrect. B. a.Exercise 2. For each of the statements in Question 1. An integer is divisible by 5 if and only if it ends in a 5. Prove this converse. State the converse of the Isosceles Triangle Property Theorem. d. then it is a banana. g. A new school is being built so that it will be equidistant from three small towns A. or 7. f. then several errors have been made. e. A positive integer is prime if and only if it is odd. We now consider the property to be an if and only if statement. An animal is a cat if and only if it has four legs. determine an approximate location for the new school. c. then it is a square. and C. If the fruit is yellow. If an integer is a prime. If the distances between the towns are as shown in the diagram. If today is Saturday. c.
Application
4.3
Part A
Knowledge/ Understanding
1.
Thinking/Inquiry/ Problem Solving
5. e. then we will get wet. determine whether a. If a four-sided figure has four equal angles.
Draw a series of five circles that pass through two given points. C. 12. What can you say about the centres of all circles that pass through the two given points? Why?
46
CHAPTER 2
e
. a. Prove that the right bisectors of the sides of a triangle pass through a common point. Find the location of a point that is equidistant from the two intersecting lines and is also equidistant from the two given points.
River
8. Determine the approximate location of a pumping station if the pumping station is equidistant from the two roads. Explain why it is not likely that a circle would pass through the four points. Where should the gates be located? Provide justification for your answer.
A B
Thinking/Inquiry/ Problem Solving
9. Provide a justification for your answer. Suppose that you are given four points. Two sheds.
t chnology
APPENDIX P. Y and Z are two given points on the circumference of a circle. A. Find a point X also on the circumference such that ∆XYZ is isosceles. S1 and S2. and D. 513
11.
S1
S2
Part B
Thinking/Inquiry/ Problem Solving
7. are located in a circular compound as shown. 10. Where do the centres of these circles lie? b. A river crosses two roads as shown. B.6. Two gates are to be built so that each gate is equidistant from the two sheds.
Prove or disprove the following statement: The angles in a triangle are in arithmetic sequence if and only if one of the angles equals 60º. Suppose we are given ∆ABC as shown.
B A
C
14. Show how to draw a line parallel to BC that meets AB at D and AC at E so that DE DB EC.Part C 13. ax2 bx c 0 is the negative of
2 .
B
A
C
15. Find a point D in side AB that is equidistant from A and the midpoint of BC. 3 T H E C O N V E R S E O F A C O N D I T I O N A L S TAT E M E N T
47
. Prove that one of the roots of x3 another if and only if c ab. Suppose we are given ∆ABC as shown. 16.
EXAMPLE 1
Ten different teams are playing in a championship tournament in which each team plays every other team exactly once. In this example. Doctors. Auto mechanics. If we can eliminate them by showing them to be impossible. Certain neurological illnesses are diagnosed only by eliminating all other possibilities. If no team has played more than two games. decide on treatments to prescribe. using Sherlock's approach. by eliminating possibilities. this is not possible. In mathematics. must be the truth. whatever remains. Proof There are only two possibilities: either there is a team that has played at least three games or every team has played two games or less. It depends on a complete listing of all possible outcomes and the elimination of all but one. Since eleven games have been played. 513
Sherlock Holmes. We list all possible outcomes and examine those that seem incorrect. This line of reasoning is called Indirect Proof or Proof by Contradiction. Consider an example.
48
CHAPTER 2
e
. Our approach is to consider all possibilities and show that all but the one we wish to prove leads to a false conclusion or a contradiction. This method of indirect proof was first developed by Euclid in 300 B. or that one team has played more than three. we arrive at the conclusion that the only possible truth is the remaining possibility. Thus far in the tournament. Notice that we cannot say that two teams have played three games. deduce the source of a problem. the great fictional detective. eleven games have been played. These two possibilities cannot both be true.4 — Indirect Proof and Parallel Lines
t chnology
APPENDIX P. The only other possibility is that at least one team has played three games or more. This method of arriving at a conclusion is used frequently in real life by a variety of people. Prove that one team has played at least three games. we identified the two possibilities and showed that one of them led to a contradiction.Section 2. was famous for remarking that after we have eliminated the impossible. however improbable. then.C. by eliminating things that cannot be wrong. some things that we sense to be true but cannot prove directly are approached in the same way. the maximum number of games is 10 2 2 10.
THEOREM
Part 1 Prove that if a straight line crosses two straight lines in a plane so that the alternate angles are equal. with ∠CKB 0º. y and a 0. We conclude that x y a y
x y
a . a
This assumption leads to a contradiction. Proof Let EF and GH be two lines cut by the line AD at B and C so that ∠EBC ∠BCH. We can now prove these properties using indirect proof. a
We previously assumed certain properties of parallel lines. a
a . then the two straight lines are parallel. 4 I N D I R E C T P R O O F A N D PA R A L L E L L I N E S
49
.EXAMPLE 2
If x and y are different numbers and a Solution There are only two possibilities: either x y Since we wish to prove x y
x y
0. Either EF GH or EF GH. we start by assuming that x a y
x y
a . x a If x .
A E B E F α C α G D A B F K G C α H H
α
D
There are only two possibilities. ∠EBC is an exterior angle. then they must meet at some point K so that ∆BCK is formed. prove that x y
x y a or x a y x y
x y
a . If the lines are not parallel. x(y a) y(x a) y y a xa xa Then x But we are given that x Thus we conclude that x y xy xy ya ya y or a 0. We make the assumption that EF GH. x a is not true. Then But Then ∠EBC ∠EBC ∠EBC ∠BCK ∠BCK ∠BCK ∠CKB ∠CKB
(Exterior Angle Property Theorem)
2 . a a . In ∆KBC. We wish to prove that EF is parallel to GH.
List all the possibilities including the one that must be proved. then the alternate angles formed are equal. alternate angles are equal or 2. which is impossible. We wish to prove that A ∠EBC ∠BCH.
CKB
0°. and we can state the complete parallel line theorem. Show that each of them leads to an incorrect conclusion or a contradiction. corresponding angles are equal
Summary of Indirect Proof 1.
50
CHAPTER 2
. since Then EF GH. Remove the alternative that is to be proved and then consider each of the other possibilities.
Part 2 If a straight line cuts two parallel lines.
(alternative angles equal)
D
Then XY and EF are intersecting lines and are both parallel to GH.
G E X C H B
Y F
∠BCH and extend XB to Y. The Parallel Line Theorem Two straight lines are parallel if and only if 1. There are only two possibilities: ∠EBC ∠BCH or ∠EBC ∠BCH Assume that ∠EBC Draw ∠XBC Then XY GH. 3.
Using the alternate angle properties it is easy to prove similar results for corresponding angles. 2. Conclude that the one remaining possibility must be true. ∠BCH. The assumption that ∠EBC Then ∠EBC ∠BCH. Proof Let EF and GH be parallel lines cut by AD at B and C.The assumption that EF GH is not true. ∠BCH is false.
A. b. 6. In ∆ABC. with M being on the circle with centre P. PX and QY are radii in the given circles such that PX QY. It is often assumed that the right bisectors of two sides of a triangle meet. A quadratic equation ax2 bx c 0.
Communication
5.4
Part B 1. Prove that CD EF. b. the line cuts the circles at M and N. Prove that they do by using indirect proof. Describe a real-life situation in which you have used indirect reasoning. 4 I N D I R E C T P R O O F A N D PA R A L L E L L I N E S
51
. and c are not 0. The lines CD and EF are each parallel to the line AB. then ∆ABC has two equal angles. 8. Prove that PM QN. Prove that if the bisector of exterior ∠ACD is parallel to AB. Prove that the line whose equation is y with equation y x4 3x2 2x. Prove that a. Prove that X. 2x 1 does not intersect the curve
4.
A
B
C
2 . where a.Exercise 2. When X and Y are joined.
B A D
E
C
Communication
2. Use indirect reasoning to show that ∠DBE ∠DEB.
X Y P Q
Thinking/Inquiry/ Problem Solving
9.
B A
t chnology
APPENDIX P. and c cannot be consecutive terms of a geometric sequence. AB AC and DE AC. 514
e
C
D
Application
7. has real roots. and Y lie in a straight line. 3. The medians BD and CE in ∆ABC are produced to X and Y respectively so that BD DX and CE EY.
You are given an 8 8 checkerboard as shown in the diagram.
52
CHAPTER 2
. prove that x
1.Part C 10. is it possible to tile the new board with dominoes that look like ?
Knowledge/ Understanding
11. If x is a real number such that x9
7x
10. If we cut out two opposite white corners.1.
we can show that the areas of triangles having equal bases are proportional to their heights. 515
e
53 m 36 m
If he divides the 36 m side into parts of 9 m and 27 m and divides the field with a line parallel to the 53 m side.5 — Ratio and Proportion
Farmer Collins wishes to divide a rectangular field into two parts so that the ratio of the smaller part to the larger is 1:3. ∆ABC ∆AEF Then
1 h(BC) 2 1 h(EF) 2
BC EF
∆ABC : ∆DEF
BC : EF
When referring to the area of a triangle. he can accomplish his goal. How can he do this? How confident can he be that he is correct in the division?
t chnology
APPENDIX P. Proof Two triangles. we usually use the triangle name. The context makes it clear that we are referring to the area.
2 . have height h. We wish to calculate the ratio of their areas. ABC and DEF.
THEOREM
Triangles having equal heights have areas proportional to their bases.Section 2. By a similar approach. Can we justify this statement? Does the same thinking apply for a triangle or other geometric shapes? In this section we examine some of the mathematical properties that use ratios. 5 R AT I O A N D P R O P O RT I O N
53
. The area of The area of
A
∆ABC ∆AEF
1 h(BC) 2 1 h(EF) 2
D
h
h
B
C
E
F
Q
Comparing areas.
c a . D is a point on AB such that AD 1 and DB 2. We will prove that SQ Join SR and QT. What is the ratio ∆ADE : ∆ABC? Solution Join B to E and E to D. E is a point on AC such that AE 3 and EC 1. d b b c
1
d d
c d
1
3
3b b
c d c
3
3d d
THEOREM
A line in a triangle is parallel to a side of the triangle if and only if it divides the other sides in the same proportions. Since ∆ADE and ∆ABE have the same height. TR
54
CHAPTER 2
.Triangle Area Property If triangles have equal heights.
EXAMPLE 1
In ∆ABC. their areas are proportional to their bases and If triangles have equal bases. ∆ABE 3 ∆ABC (Triangle Area Property) 4
B
C
Substituting ∆ADE Then
∆ADE ∆ABC
1 3 ∆ABC 3 4 1 4
[
]
c d
1 ∆ABC 4
EXAMPLE 2
Prove that if a b Proof Since Then Also Then
a b a a b a b
c then a b b d
d
and a b 3b
c d
3d
.
∆ADE ∆ABE
A 1 D 2 3
or Since
∆ADE
1 3 1 ∆ABE 3
(Triangle Area Property)
E 1
∆ABE and ∆ABC have the same height for bases AE and AC.
PT . Proof Part 1 PS Let ST be a line in ∆PQR that is parallel to QR. and AB 3AD. their areas are proportional to their heights.
Therefore ∆STQ Then Therefore
∆PST ∆STQ PS SQ
∆STR
∆PST ∆STR PT TR
Note as an extension that we easily obtain PS 1 PT 1 TR SQ
PS SQ SQ PQ SQ PT TR PR TR
P X S T
TR
Part 2 Let PQR be a triangle and let S and T be points
PS in the sides so that SQ PT . and we do so by indirect proof. SQ PX Then PT TR XR PX XR
Q R
(Part 1 of theorem)
This means that the points T and X divide PR in the same ratio. which is impossible.
PS Because SX QR.
∆PST ∆STR PT TR
(Triangle Area Property)
Q R
Because ST QR. then draw SX QR meeting PR at X. In ∆ABC. ∆STQ and ∆STR have the same base ST and equal altitudes.
∆PST ∆STQ PS SQ
(Triangle Area Property)
S
P
T
Since ∆PTS and ∆STR have the same altitude with bases PT and TR. Triangle Proportion Property Theorem A line in a triangle is parallel to a side of the triangle if and only if it divides the other sides in the same proportion.Since ∆PST and ∆STQ have the same altitude with bases PS and SQ. Then the statement ST QR is not true. DE BC if and only if
AD DB AE . Therefore ST QR. We can prove that TR
ST QR. EC
B C
A
D
E
DE BC
2 . 5 R AT I O A N D P R O P O RT I O N
55
. If ST QR. Either ST QR or ST QR.
Prove that XY YZ
PQ . From K. any point in BC. prove that C is 2
P 1 B A C B Z A R
e
the midpoint of BZ. respectively.
13. Z. 518
11. and KF is drawn parallel to BD to meet CD at F. Y. If X is joined to Y.
58
CHAPTER 2
. In gm ABCD. and BY YC. prove that XY DC AB. three parallel lines are cut by lines at X. determine ∆BHF : rectangle ABCD. R. ABCD is a trapezium in which AB DC. KE is drawn parallel to BA to meet AC at E. If AP PB
AR 3 and RC 4 3 . In the diagram. X and Y are the midpoints of AD and BC. ABCD is a rectangle in which ED
1 and G is the 2
A I H
B
midpoint of DC.
E
F
D
G
C
10. If EF AB.AE 9. respectively.
X
Y
D
C
14. QR
X Y Z R Q P
t chnology
APPENDIX P. Q. ABC and BDC are two triangles on the same side of BC. Prove that EF AD. AX XD. Prove that BX and DY trisect AC. Part C 12. and P.
we create a new shape similar to the original. taken in order. are respectively equal to the angles of the other. the figures are not similar.
2.6 — Similar Figures
t chnology
APPENDIX P. PT BC QR
A P B Q E T
C
D
R
e
S
The converse of this statement is also true. and AQ P CD RS DE ST AE .
Thus. ∠B ∠Q.6 SIMILAR FIGURES
59
. 519
Similarity is one of the most important properties of plane figures. Two figures are similar if the angles of one. if polygons ABCDE and PQRST are similar.Section 2.
A 1 D 2 2 B 1 C H 1 E 2 F
G
We saw earlier that there are a number of conditions on two triangles that make the triangles congruent. ∠D ∠S and
B ∠E = ∠T. If we take a photograph of any shape and enlarge the photo. then ∠A ∠P. so the pentagons are not similar. Again. ∠C ∠R. and the corresponding sides are proportional. in the same order.
E I
D H
G C
A
J
F
B
The rectangle ABCD and the parallelogram EFGH have sides that are proportional (in this case equal) but the figures are not equiangular. There are three sets of conditions that make two triangles similar. We examine these in the Similar Triangle Theorem. The two pentagons ABCDE and JFGHI are equiangular but their side lengths are not in proportion.
and the angles contained by these sides are equal. ∠B
∠E.
A
∠D. or if 3. two pairs of sides are proportional. (Parallel Line Property) BC EF
AB Then DE AC DF AB Then DE AC DF BC EF AC DF BC EF
E F D (A)
(Triangle Proportion Property)
By repeating this process with ∠C on ∠F. To indicate that figures are similar. or if 2.
B D C
Construct ∆EFG equiangular to ∆BCA. they are equiangular.
B C
The triangles are similar. their sides are proportional. ∠B ∠E.
∠F. we write ∆ABC Part 2 AB Let ∆ABC and ∆DEF be such that DE We prove that ∠A Since ∠ABC ∆ABC
AB Then EG BC EF AC DF BC EF. ∠FEG and ∠ACB ∆GEF
AC FG AC DF BC EF
E
F
AB We are given that DE
60
CHAPTER 2
G
.
AB We prove that DE AC DF BC . Since ∠ABC ∠DEF and ∠ACB ∠DFE. Proof Part 1 Let ∆ABC and ∆DEF have ∠A ∠D.
∆DEF. Then AB falls along DE and B is on DE. EF
B
A
D
C E F
Translate the smaller triangle ∆ABC onto ∆DEF so that ∠A fits on ∠D.Similar Triangle Theorem Two triangles are similar if 1. Also AC falls along DF and C is on DF. and ∠C ∠F. and ∠C ∠EFG.
AB Since DE AC . DF
(Triangle Proportion Property)
BC EF Then ∠ABC Then ∆ABC ∠DEF and ∠ACB ∆DEF ∠DFE
(Parallel Lines Property) (equal angles)
We now have all conditions for similar triangles. Proof Let ∆ABC and ∆DEF be similar. and ∠EGF ∠ABC.
EXAMPLE 1
Prove that if two triangles are similar.Since BC is common to both proportions. AB falls along DE and AC falls along DF. and ∠BAC
∠EDF ∠BAC ∠EDF
Therefore ∠BCA Then
Part 3 Let ∆ABC and ∆DEF be triangles in AB AC . which ∠A ∠D and DE DF We will prove that the triangles are similar. ∠ABC ∆DEF.
D (A) A B C
∠FED.6 SIMILAR FIGURES
61
. EG DE FG FD EF FE Therefore ∆EFG ∆EFD (side-side-side) Then But ∠EFG ∠EFG ∆ABC ∠EFD. meeting BC at G. and ∠EGF ∠FED. and h2 be the altitude from D. ∠FEG ∠BCA. EF
AB EG AB C and AG DE F AC DF
Then EG DE and FG DF In ∆EFG and EFD. Let h1 be the altitude from A.
B D
t chnology
APPENDIX P. corresponding altitudes have the same ratio as a pair of corresponding sides. as shown.
B
C
E
F
Translate ∆ABC so that ∠A coincides with ∠D. ∠FEG ∠EFD. meeting EF at H. 522
e
A h2 h1
G
C
E
H
F
2.
We cannot determine the area of either triangle.
EXAMPLE 2
Prove that the areas of similar triangles are proportional to the squares on corresponding sides. can we make any statement about the ratio of their areas? Suppose. Proof Let ∆ABC and ∆DEF be two similar triangles and let h1 and h2 be their altitudes. but from Example 1 we can say that h1 is 2 also 1 . that ∆ABC
AB ∆DEF and DE BC EF AC DF 1 . 2
∆ Then ∆ABC DEF
1 (BC)h1 2 1 (EF)h2 2
D A h1 B C E h2 F
h
BC EF 1 2 1 4 1 2
h1 h2
Now we can prove the result in general. for example. 2
Can we determine the ratio of their areas? The answer is that we can. and using this we can determine the ratio of the triangle areas.
AB DE AC DF BC EF
A
t chnology
APPENDIX P. 523
e
h1
D C h2
h1 h2
B
G
∆ABC ∆DEF
1 (BC)h1 2 1 (EF)h2 2
E
H
F
62
CHAPTER 2
.In ∆ABG and ∆DEH. ∠ABG ∠DEH ∠AGB ∠DHE ∠BAG ∠EDH Then ∆ABG ∆DEH
AB Therefore DE AG DH
(similar triangles) (right angles) (angle sum) (Similar Triangle Theorem)
h1 h2
AC DF BC EF
But Then
∆ABC
h1 h2
AB DE
AB ∆DEF. so DE AG DF BC EF
If two figures are similar. Since ∆ABC ∆DEF.
two congruent pentagons
Knowledge/ Understanding Application
2. and AB
12. ∆ABC
60.
4. 50 m.
3. and DE
9. ∆ABC and ∆DEF are similar.
∆ABC ∆DEF 60 ∆DEF
AB 2 DE 12 2 9
4 2 3
16 9
16 ∆DEF Then ∆DEF
9
9 16
60 60 33.6 SIMILAR FIGURES
63
. DF determine the lengths of AC and BC. two hexagons that have equal angles c. In a map of the region containing the field.75
Exercise 2. two right-angled triangles. EF
5. the scale is 1:100. The sides of a triangular field are 40 m. two equilateral triangles g.6
Part A
Communication
1. If DE 2.BC EF BC EF ∆ Then ∆ABC DEF BC2 EF2
h1 h2
BC EF
EXAMPLE 3
If ∆ABC Solution
∆DEF. two circles f. which pairs are similar figures? a. In the following list. AB
12. and 65 m. two squares d. each having an acute angle of 30º h. two isosceles triangles e. What are the dimensions of the field in the map?
2. a quadrilateral and a rectangle b. determine ∆DEF.
AD DB 4. D and E are in AB and AC respectively such that DE BC. a point A is chosen. then BC is drawn parallel to QR. In ∆ABC. quadrilateral DBCE
Knowledge/ Understanding
7.2 m long.1 m long? Part B 5. A man 2 m tall casts a shadow 5. (Hint: Let the ratio between a pair of corresponding sides be k: 1. In quadrilateral PQRS. how do the side lengths of the two triangles compare? 8. S is any point on QR and S is joined to P. Prove each of the following: (i) (ii) (iii) AD2 AC2 AB2 (BD)(DC) (BC)(DC) (BC)(BD)
B D C A S P A B Q O C R D
t chnology
APPENDIX P. Determine each of the following: a. Prove that BC
2FD and
e
13.
6. How tall is a pole that casts a shadow 9.
Thinking/Inquiry/ Problem Solving
t chnology
APPENDIX P.Application
4. A line is drawn parallel
QS . ∆ABC b. b. From X. and PR in C. The medians BD and CE of ∆ABC intersect at F. Prove that if two triangles are equiangular. SR
64
CHAPTER 2
. name all pairs of similar triangles.
2 . prove that AC2 AB2 BC2. and CD is drawn parallel to RS.
t chnology
APPENDIX P. O is any point and is joined to each vertex. the bisectors of two corresponding angles have the same ratio as any pair of corresponding sides. PS in B. 526
12. a line is 3
15. 525
e
e
c. If BC XY. calculate the length of 3.) 10. If two similar triangles have areas in the ratio 4:49. 524
9. their perimeters have the same ratio as a pair of corresponding sides. the line AB is drawn parallel to PQ. a. Prove that AD PS. In OP. In the diagram. Prove that BF CF 2FE. meeting PQ in A. In ∆PQR. and ∆ADE 81. 11.
AB to QR. Using parts (ii) and (iii). Prove that if two triangles are equiangular. From A. A point X is chosen in side AB of ∆ABC such that AX XB drawn parallel to BC and meets AC at Y.
Conditional reasoning. you were introduced to indirect reasoning.
I M P O R TA N T C O N C E P T S
1. we also introduced properties of area and similarity of plane figures. In addition. biconditional. If two triangles have the same bases but different heights. we introduced you to different methods of proof through a discussion of plane figures. One of the main features of this chapter was the introduction of conditional. means showing q to be correct as a condition of p being true. there is no guarantee that p necessarily implies q. then their areas are proportional to their corresponding heights. EF
66
CHAPTER 2
. In addition. Triangles are similar under the following conditions: (i) corresponding angles are equal (ii) corresponding sides are proportional (iii) an angle of one triangle equals an angle in the second triangle. and indirect reasoning. 3. for instance. 2. If q → p is correct. then ∆ABC DEF AB 2 DE AC 2 DF BC 2 . In this type of reasoning. If ∆ABC
∆ ∆DEF. We noted that the two statements p → q and p ↔ q are not the same. Parallelograms having the same bases and between the same parallels have the same area. and sides about the angle are proportional 4.Key Concepts Review
In Chapter 2. you list all possible outcomes for a certain premise and then eliminate all possibilities except one. thereby implying that the remaining one must be correct. or p → q.
a rectangle and a parallelogram b. If DE BC and trapezoid DECB determine DE:BC. Write an expression for the perimeter of this rectangle in terms of A and L. Determine the following: a.Review Exercise
1. Which of the following pairs are similar figures? a.
D
E
F A C
REVIEW EXERCISE
67
. two rectangles c.
B
8
10 D
A
6
5. Determine the area of figure ABCD. two right-angled triangles f. The two diagonals DB and CA intersect at E. 3. In the trapezoid ABCD. two rectangles that measure 8 e. AE:EC b. two quadrilaterals each containing two angles of 100º and 120º 2. two congruent triangles d. Prove that the two triangles are similar. ∆ABC and ∆DEF are two isosceles triangles in which the vertical angles ∠ABC and ∠DEF are equal. ∆ABE:∆CDE
D
A E
B
C
B
6.
7 ∆ABC. 16
D B A
3 and 32
12
E C C
4. The area of a rectangle is A and it has a length of L. AB:DC 2:5.
BX DY
AX b. What is the area of the larger triangle? 12.
3. Determine ∆BED:∆ABC. XD AH HM
A
c. a.
6. while the hypotenuse of a similar triangle is 15. b. The legs of a right-angled triangle are 5 and 10. a line is drawn parallel to BC that meets AB and AC at E and F.2
10. a median is drawn from A to the point D on BC. Prove that the diagonals of the trapezoid trisect each other. AH d. respectively. AB 1 and DC 2.7. In the trapezoid ABCD. In ∆ABC. Determine ∆BED:∆ADC.
A B
13. BX and DY trisect AC 9. Prove a. CM
HM HM
D
e.
A
B 1 F
Q 1 S
2
R
D
C
2
D 1 E
3
C
11.2
3.6 y
13. Through any point P on AD. Determine the value of y in the diagram shown. ABCD is a parallelogram where X and Y are the midpoints of AD and BC. Prove that EP PF. In ∆ABC.
A
4 E 1 B 2 D 3 C B H X M C Y
8. respectively. Into what fractional parts do the dotted lines divide each of the following? a.
A E
B
c. D is a point on BC such that BD 2 and DC The point E is on DA such that DE 1 and EA 4.
D A
C
E
P
F
B
D
C
68
CHAPTER 2
.6 2.
P
b.
prove that the area of the trapezoid is 1 (a b)h.
A H D I E F C B
G
21. and point I is any point on the diagonal AC. A line is drawn through I such that EG is parallel to AD and BC. AG Prove AE .
A 4
C B
7
x
t chnology
APPENDIX P. Determine the value of x. In ∆ABC. Determine the ratio of the area of ∆ABF to quadrilateral EFCD. In the parallelogram ABCD. 16. find the area of ∆ABC. 527
15. the vertices A. BE. If the triangle and the square have equal areas. If a and b are the lengths of the two parallel sides of a trapezoid and h is the distance between them. Prove that the three altitudes of a triangle are equal if and only if the triangle is equilateral. AE:ED 1:2. B. 20. If ∆BDE 6.14. a line is drawn through I that is parallel to DC and AB. A straight line through A cuts BD at E. ABCD is a parallelogram. and C lie in a straight line. the medians AD. ABCD is a parallelogram.
B A E F G D C
e
18. Similarly. and meets DC extended at G. The base of a triangle is four times as long as a side of a square. and CF intersect at the point G. 2 17. and BC at F. and E divides BC in the ratio 3:4. 22. In ∆ABC. EF AF
A E D C F G B
REVIEW EXERCISE
69
. Prove that the six interior triangles in ∆ABC have equal areas. Prove that parallelogram EBFI parallelogram HIGD. In the adjacent squares shown. D divides AB in the ratio 1:2. find the ratio of the altitude of the triangle to a side of the square.
B
A F
E
D
C
19.
C(x3. 2. Prove Varignon's Parallelogram Theorem using the midpoint and slope formulas from analytic geometry. Investigate and Apply Geometer's Sketchpad® may be useful as you carry out some of these investigations. y3). Varignon accidentally came across a copy of Euclid's Elements and was inspired to a career in mathematics. and D(x4. y2). If the initial quadrilateral is a square. a rectangle. y1). b) Consider four points A(x1. Prove that the area of a Varignon Parallelogram is one half the area of the initial quadrilateral. The parallelogram formed by joining the midpoints of a quadrilateral is called a Varignon parallelogram.wrap-up
C H A P T E R 2 : VA R I G N O N PA R A L L E L O G R A M
investigate and apply
Varignon's Parallelogram Theorem states that the figure formed by joining the midpoints of any convex quadrilateral is a parallelogram. a rhombus. B(x2. or a square? If the initial quadrilateral is already a parallelogram. y4). or a parallelogram? 3. a trapezoid. The theorem is named after Pierre Varignon. does the interior Varignon Parallelogram have any special properties? What if the initial quadrilateral is a rectangle. Hint: Consider the initial quadrilateral as two triangles. alternately inwards and outwards. a French mathematician who lived from 1654 to 1722. 1. ●
70
CHAPTER 2
. c) Compare and contrast the proofs from parts a and b.
INDEPENDENT STUDY
What special characteristics must the initial quadrilateral have in order for its Varignon Parallelogram to be a rhombus. what special properties must it have in order to be similar to its Varignon Parallelogram? Prove that if equilateral triangles are drawn on the sides of a quadrilateral. Like so many others. a) Prove Varignon's Parallelogram Theorem using the methods of this chapter. He was the first to prove the theorem. Which is more convincing? Which is easier to understand? A third method of proof will be encountered in Chapter 6. their vertices will form a parallelogram.
with the right angle at B as shown. If the area of the trapezoid is 120.
B
A
D
C
c. What is the area of the rhombus? b.) a. ∠C 90º. The midpoint E of the median AD is joined to B. determine the distance between the parallel lines. Answer each of the following. The lengths of the parallel sides of a trapezoid 10 are 8 and 12 units. The sides of a right-angled triangle ABC are 10. 24. determine the area of ∆DEB. a. In ∆ABC. If P is any point inside the parallelogram. If AC 8 and BC 6. 3
1. Prove that the converse of this theorem is true. 2. Determine the length of MY. and 26. A line segment MY is drawn from M. Restate the original statement and its converse in "if and only if " form. c. D is a point on BC. State the converse of this theorem. the midpoint of AB. then AD is a median.
A 10
Y 26
M B 24 C
CHAPTER 2 TEST
71
. b. ABCD is a parallelogram in which DC 10 with a height of 8 units as shown. Consider the following statement: If ∆ADC ∆ADB. perpendicular to AC. In ∆ABC.
3. A rhombus has diagonals of length 12 and 20. calculate ∆APB ∆CPD.
A P B
8
D C d. respectively. (You do not need to provide proofs.Chapter 2 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application Questions all 7 1 2.
1 AD.
B
P D C
6. ∆ABC and ∆DBC have the same base BC and equal areas. AD is a median of ∆ABC.
A D G
(Hint: Select a point G on BD. PBCQ is a trapezoid in which PQ BC and PQ:BC 2:3. E is on AD so that AE determine ∆ABC. and assume that AG BC.)
B
C
72
CHAPTER 2
. Prove that the line joining A to D is parallel to BC. If ∆ABC 36. If ∆AEC 4
A
36. In parallelogram ABCD.4. Prove that ∆APD ∆ABP ∆DCP. calculate the area of trapezoid PBCQ. Prove the following statement using indirect reasoning. P is any point on BC. PC and QB intersect at A. 5.
B
P A
Q
C
7.
4. You are encouraged to develop other questions of interest for research. select any three non-collinear points A. a. Keep in mind that after you've made a conjecture. we will ask you to prove this same result where the point P is either in the interior or the exterior of the rectangle. In Chapter 9. Determine by experimentation the location of a point O that is equidistant from A. If three points are chosen that form an obtuseangled triangle. you were asked to prove that for any rectangle ABCD. Locate the position of point O. where P is an
P A B A B C O
interior point in the rectangle. c. Draw the circle which passes through A. B. and C. On a blank screen. PA2 PC2 PB2 PD2. B.C O M P U T E R I N V E S T I G AT I O N S
These investigations are provided for enrichment activity and as an introduction to investigation in mathematics. B. b. On a blank screen. B. Prove or disprove the conjecture that PA2 PC2 PB2 PD2. it must always be proved. and C. 2.
D A B D
C
P
C
E X T E N D I N G A N D I N V E S T I G AT I N G
73
. 1. and C. Verify that this result is correct using Geometer's Sketchpad®. They are designed to be used with Geometer's Sketchpad®. Move the point P to the outside of the rectangle and verify that this result is still true. select any three non-collinear points A. In Exercise 1. which is equidistant from A. what can we say about the location of a point equidistant from the three points? Is the same conclusion true if the three points form the vertices of an acute-angled triangle? a right-angled triangle? 3. and C. and C such that these points form an obtuse-angled triangle. ABCD is a parallelogram and P is any point in either the interior or exterior of the parallelogram. Verify that point O is at the point of intersection of the right bisectors of AB and BC using the Perpendicular Line function. B.
and very important in art and design. Why do we continue to study them? Aside from the practical value that they have for architects.Chapter 3
PROPERTIES OF CIRCLES
In Chapter 3. CHAPTER EXPECTATIONS In this chapter.3 prove some properties of plane figures algebraically. designers. 3. 3. These formulas. and especially important were the formulas for circumference and area. 3.3. Section 3. C 2 r and A r2.5
. Section 3. Section 3. The properties of circles were fundamental to the ancient Greeks.2. 3. 3. we will consider the closed plane curve of a circle. The circle is the most symmetrical shape.5 prove some properties of plane figures using indirect methods. and engineers.4. have been part of the study of mathematics throughout history.2. these formulas introduce the number (pi).1. you will
• • •
prove some properties of plane figures using deduction.4. 3.
The curved line enclosing the circle is the circumference. a diameter is a chord passing through the centre. A circle can also be considered as the surface area enclosed by the circumference.
60° r = 12
c. a. DFE is a minor arc. A semicircle is that part of a circle bounded by a diameter and an arc. and OA is the radius. DE is a chord.
r=6
b. Calculate the area of the circle and the area of the sector for each of the following.Review of Prerequisite Skills
D E F I N I T I O N S R E L AT I N G T O T H E C I R C L E
A circle is the locus of a point that moves so that it is always a constant distance from a fixed point.
A F D E
B
O
C
BC is a diameter. BGC is a semicircle.
80° r=5
76
CHAPTER 3
. A segment is that part of a circle bounded by a chord and an arc. The fixed point is the centre and the constant distance is the radius. and DGE is a major arc. O is the centre. In the diagram. a chord is a line segment connecting two points on the circumference. DFE and DGE are arcs. A sector is that part of a circle bounded by two radii and an arc. an arc is a portion of the circumference.
G
Segment
Sector Semicircle
Exercise
1.
100° r=5 100° r = 15
c. 7. Calculate the ratio of the areas of the two sectors in each of the following. each with a diameter of 1 cm.
r = 20
c.
120° r = 10
e. What is the value of n?
REVIEW OF PREREQUISITE SKILLS
77
.
180° r=7
f.d. in cm2.
50° r=8
2.
30° r=4 120° r=6
5. in cm. a. A set of n circles. a. 3. has a total area equal to that of a circle with radius 3 cm. The area of a given circle is 16 cm2.
80° r=6 40°
b. State a relationship between a sector area and the circle area in terms of the sector angle. Calculate the ratio of the areas of the two sectors for each of the following pairs of circles. What is the radius of the circle? 8. What is the circumference of this circle? 6. is equal to its area.
60° r=6 60° r=8
b.
40° r=5 80° r = 10
d. determine the area of the inscribed circle. a. What is the radius of this circle? b. The circumference of a circle.
100° r=9 70°
60°
4. If the area of the given square is 81 cm2.
Cartesian grid The perpendicular bisector of the segment PQ is the line of all points equidistant from P and Q. Q. investigators must look for relationships between the victims.investigate
CHAPTER 3: GEOGRAPHIC PROFILING
People who commit crimes sometimes seem difficult to catch. Q. using the distance formula. Given points P and Q as shown. mathematicians have developed a field called geographic profiling. Is there always a point equidistant from any three given points? What about four points. the line x 3 marks all points equidistant from Q and R. Q. Similarly.
DISCUSSION QUESTIONS
1. Instead.2). that C is equidistant from all three crime sites P. and the same distance from Q and R. Investigate Three crimes have occurred at locations y R Q indicated on the map by the points 5 A 4 P. Therefore the point C is equidistant from all three crime sites. that is. and R. there can be no short list of suspects. Verify. and R. We conclude that C marks the most likely location of the criminal's residence. This is the point C. To help police around the world. This means analyzing vast quantities of data. most of which is likely to be insignificant. and R. Is it reasonable to assume that the criminal's home is in a central position relative to the crime locations? ●
78
CHAPTER 3
. or more than four points? 3. It is the same distance from P and Q. B. How would you find the perpendicular bisector of PR? 2. or D most likely marks C B 2 the residence of the criminal? 1 D x We assume that the crimi1 2 3 4 5 nal's home is in a central P position relative to the crime City map with superimposed locations. Which of the points 3 A. the perpendicular bisector is the line y 2. When there is no clear relationship between the criminal and the victims. The intersection of these two lines is the point (3. Geographic profiling analyzes crime locations to find where the perpetrator is likely to live. C is the centre of a circle through the points P. C.
we reviewed basic properties of circles. Proof In ∆OAM and ∆OBM. 1 P R O P E RT I E S O F C H O R D S
79
. From O draw OM perpendicular to AB. Part 1 Let O be the centre of the circle and AB be a chord. OA OB OM is common AM BM Therefore Then But Then ∆OAM ∆OBM ∠OMA ∠OMB ∠OMA ∠OMB ∠OMA ∠OMB OM ⊥ AB
(Radii) (Given) (Side-side-side)
A M B O
180º 90º
Part 2 Let AB be a chord in the circle having centre O.Section 3. We prove that AM MB. 530
In the Review of Prerequisite Skills. We will prove that OM ⊥ AB.1 — Properties of Chords
t chnology
APPENDIX P. OA OB OM is common ∠OMA ∠OMB Therefore ∆OMA ∆OBM Then AM MB
(Radii)
e
O
90º
(Hypotenuse-side)
A M B
3 . Here we examine the properties of chords in a circle.
THEOREM
A line drawn from the centre of a circle to the midpoint of a chord is perpendicular to that chord if and only if it bisects the chord. Proof In ∆OAM and ∆OBM. From O draw OM where M is the midpoint of AB.
4.
EXAMPLE 1
A circle has a diameter of length 26. The perpendicular from the centre to a chord bisects the chord. 3. In ∆OMB and ∆OND. If a chord in the same circle has a length of 10. Proof From O draw OM and ON to the midpoints of the chords AB and CD. Join O to B and O to D. 2. OA 13. Since the diameter is 26. Part 1 Let AB and CD be equal chords in a circle whose centre is O.
B
D
M O
N
A
C
80
CHAPTER 3
. (Chord Right Bisector Property) Then AM2 52 MO2 MO2 MO2 MO OA2 (Pythagoras) 132 169 25 144 12 (MO 0)
M A 13 O
B
THEOREM
Two chords are of equal length if and only if they are the same distance from the centre of the circle. The line joining the centre to the midpoint of a chord is perpendicular to the chord.The Chord Right Bisector Property 1. If OM is perpendicular to AB. how far is the chord from the centre? Solution Let the circle centre be O and let the chord be AB. The right bisector of a chord passes through the centre. The centre of a circle is the intersection of the right bisectors of two non-parallel chords. OB OD (Radii) ∠OMB ∠OND (Chord Right Bisector) MB ND (Chord Right Bisector) ∆OMB ∆OND (Hypoteneuse-side) Then OM ON. ∠AMO 90º and AM MB 5. We will prove that the chords are equidistant from the centre.
P O X
Q
Y B O D
4. If XY 8. the distance between the centres. Two circles with centres A and B have radii 5 and 8. 6.1
Part A 1.
P
82
CHAPTER 3
. Part B 5. which has a radius of 10.Exercise 3. calculate the length of CD. Determine the length of the chord AB if OA ON 3. determine the length of the radius of the circle.
A 14 C X 5 A Y Q S P 10 R 8 B
7. If AB has length 12 and the radius of the circle is 10. respectively. and the radius of the circle is 5. The point P is any point placed inside a circle. The two parallel chords AB and CD are a distance of 14 units apart. Calculate the distance between the parallel chords PQ and XY if PQ 6. determine the length of AB. 5 and
A O N B
Application
2. The circles intersect at the points X and Y. XY 8. The distance between the parallel chords PQ and RS is 10. determine the length of ON. If PQ 8 and RS 12. What is the length of the chord?
Communication
8. A chord of a circle is 4 units away from the centre of the circle. If AB
10 and OA
13.
O B N A
Knowledge/ Understanding
3. Is it always possible to draw a line through P that is bisected at the point P? Explain.
1 P R O P E RT I E S O F C H O R D S
83
. Prove that when O is joined to X. Prove that a straight line drawn from the centre of a circle perpendicular to a chord will. ∠OXC ∠OXB. Prove that PX QY.
Thinking/Inquiry/ Problem Solving
10.9. where P and Q are the centres of the circles. if extended. Prove that XY is the right bisector of PQ.
A P B
T Q
3 . In the diagram. Two circles with centres X and Y intersect at P and Q.
X
Q P O
Y
11. When extended. Determine the length of AB if PQ 21 cm. PA 13 cm and QA 20 cm. PQ and XY are two equal non-parallel chords in a circle. A line is drawn through two concentric circles as shown. Part C 13. Prove that OQ OY. AB and CD are two chords of equal length that intersect at the point X in the circle with centre O.
P Q O Y X
12. bisect the arc cut off by this chord.
D A X C O B
14. they meet at a point O outside the circle.
∠ADB and ∠ACB are examples of two such angles which stand on the minor arc AEB (or chord AB).
Central Angles
C
F
D
A E
B
Angles at the circumference
84
CHAPTER 3
. then
x 360 12 360 12
A
O B
2 r
12 . two central angles are shown: ∠AOB. we have created a central angle. Determine the measure of ∠AOB. In the same way. A central angle is an angle that has its vertex at the centre of a circle and stands on an arc (or chord) of the circle. Each of the two angles is described as a central angle because each has its vertex at the centre of the circle and each stands on an arc of the circle. Solution The circumference of the circle is C If ∠AOB xº.
EXAMPLE 1
In the diagram. The angle α stands on the minor arc ACB (or chord AB). we can determine the size of the angle. the angle 360 α stands on the major arc ADB.Section 3. and reflex ∠AOB. In the diagram. the radius of the circle is 6 and the length of the arc is .
x ∠AOB
30
30º
D 360 – α O α A C B
In this example. which is less than 180º. An angle at the circumference of a circle is an angle that has its vertex on the circumference of a circle and that stands on an arc (or chord) of a circle.2 — Angles in a Circle
If an arc subtends an angle at the centre of a circle. ∠AEB is an example of an angle which stands on the major arc AFB (or chord AB).
Join CO and extend it to D. OC Then ∠OCA Then ∠DOA OA (Radii) ∠OAC (Isosceles triangle) 2∠OCA (Exterior angle)
A C x y O x 2x 2y D X Y D O A x x y C y B B y
Similarly.
3. let X be a point on minor arc AB. Join OA and OB. standing on arc AXB.THEOREM
The angle at the centre of a circle is twice an angle at the circumference of the circle that stands on the same arc.
Angle at the Circumference Property An angle at the centre of a circle is twice the angle at the circumference standing on the same arc. Proof Let A and B be points on a circle having centre O. In ∆OCA. Proof Let A and B be two points on a circle having centre O. in ∆OCB. Then and so ∠APB ∠AQB ∠APB
1 ∠AOB 2 1 ∠AOB. and let ∠APB and ∠AQB be angles in the same segment. and let ∠ACB be an angle at the circumference. This theorem leads to two immediate conclusions. 2
(Angle at the circumference) (Same)
P
Q O
A X
B
∠AQB. it can be shown that reflex ∠AOB 2∠ACB.
THEOREM
Angles in the same segment of a circle are equal. standing on arc AXB. ∠DOB 2∠OCB Then ∠DOA ∠DOB 2(∠OCA ∠OCB) ∠AOB 2∠ACB By choosing a point Y on the major arc AB.2 ANGLES IN A CIRCLE
85
.
Equal Angles in a Segment Property Angles in the same segment of a circle are equal.
In a similar fashion. ∠BED α
D
α
C E A D α α B
Since CD and AB are two intersecting straight lines. ∠PQR ∠PSR 50º. ∆DEB is isosceles.
P Q 100° O X S R 40° T
The sum of the interior angles of the quadrilateral QXST is 360º. determine the size of ∠QXS. let ∠DBE Since BD Then ∠DBE ED. ∠CEA ∠BED α
86
CHAPTER 3
.
EXAMPLE 3
CD and BA are two chords of a circle that intersect at E. If DB DE. Since ∠PQR and ∠RQT are supplementary.
A
C E
B
Proof Join A to C. ∠PST Then ∠QXS 360º (130 130º. Proof Let AOB be a diameter in a circle and let ∠APB be an angle on AB. Then But Then ∠AOB ∠AOB ∠APB 2∠APB (Angle at the circumference) 180º 90º
A P
O
B
Angle in a Semicircle Property The angle in a semicircle is a right angle. and for convenience. prove that ∆ACE is isosceles.
EXAMPLE 2
If O is the centre of a circle such that ∠POR 100º. Solution Since ∠POR 100º and is at the centre of the circle. ∠RQT 130º. and ∠PTR 40º.THEOREM
The angle in a semicircle is a right angle. 130 40)º 60º.
The diagonals of the quadrilateral meet at the point X. d. prove that AC CD.
87
. Prove that ∆DCX and ∆ABX are similar. for each of the following. If AO is the diameter of the smaller circle and the radius of the larger circle. the two circles are tangent at A.
135° O B 100° C B 30° O C A
b. The quadrilateral ABCD has its vertices on a circle.Also Then
∠ACD ∠CEA
∠DBA ∠ACE
α α
(Equal angles in a segment)
Therefore ∆ACE is isosceles. ∠OCA Then OC ⊥ AD Then AC CD
(Chord right bisector)
D C
90º
(Angle in a semicircle)
O
A
Exercise 3. Proof Join OC. with AC EXAMPLE 4
AE. c. Determine the measure of the indicated angle. a.2
Part A
Application
1.
B A X D C
3.
A B 120° X O D C 30° E
Determine ∠ADC and ∠AXB.2 ANGLES IN A CIRCLE
2.
A X C 20° D B 22° E
Determine ∠BAC.
A D
Determine ∠BAC.
Determine ∠BXD.
In the diagram.
A D
C
O B
6. Discuss other necessary conditions for the congruence of the triangles. D is a point on the circumference between A and B. State two equalities in the triangles. A square ABCD. Prove that if the parallelogram ABCD has each of its vertices on a circle then ABCD must be a rectangle. with side length 2. then ∠CED 90º. The length of chord AB is equal to the radius of the circle with centre O. The point E is on the arc AB as shown. We want to prove ∆ABC ∆ABD. Prove that if CD AB. AB is a chord of a circle.
A O E C D B
Knowledge/ Understanding
8. a.
P O
A
B
A
B
5. Prove that the Angle at the Circumference Property is true for circles with centre O where P is a point on the circumference as shown. c. Prove ∆ABC Part B ∆ABD. a. Determine the numerical value of AE2 BE2 CE2 DE2.
P O
b. Prove that ∠ADB 5∠ACB.Communication
3. 7. b. is inscribed in a circle. AB is the diameter of a circle with centre at O.
D E
C
A
B
4. Points D and C are chosen on the circumference of the circle so that AD BC.
E A B
D
C
88
CHAPTER 3
.
KAL is drawn perpendicular to AB and meets the circles at the points K and L. The bisector of the angle at A meets the circumference at D.2 ANGLES IN A CIRCLE
89
. Prove that AE2 BE2 CE2 DE2 = 2d2 where d is the length of the diagonal of the rectangle. Prove that AB bisects ∠PAQ. Prove that ∆ABC is isosceles. The circles meet at the midpoint of AC. From D. two circles are drawn. Two circles intersect at A and B. Prove that DE is a diameter of the circle. 11. a line is drawn perpendicular to the chord BC so that it meets the circumference at E. Part C
Thinking/Inquiry/ Problem Solving
A
B E
D
C
10.
A E
B
D
C
12. AXB and CXD are two perpendicular chords of a circle with centre O. For ∆ABC. The first circle has AB as its diameter and the second has BC as its diameter. The rectangle ABCD has its vertices on a circle as shown. 13. ABC has its vertices on a circle as shown. Prove that ∠AOD ∠BOC 180º.9.
3. The lengths KB and LB are extended to meet the circles at P and Q respectively. Point E is taken to be any point on the circumference of the circle.
Let DE and FG intersect at O. Then OA OB OB OC OC. In this section we discuss the conditions necessary.Section 3. Draw DE.3 — Cyclic Quadrilaterals
We can draw a circle that passes through the vertices of any triangle. for which ∠BOD is the central angle. In other words.
(Right bisector) (Same)
B G A D F O E C
Therefore OA
Then a circle with centre O and radius OA passes through A. The quadrilateral must meet special criteria. we usually use the term to refer to a set of four or more points. Hence ∠BAD 1 ∠BOD (Angles in a circle) 2 Likewise. Every triangle has a circumscribing circle. B. Draw FG. but this is not true for every quadrilateral. the right bisector of BC. Join BO and DO.
A cyclic quadrilateral
THEOREM
In a cyclic quadrilateral. the four points of a cyclic quadrilateral are concyclic. is a quadrilateral whose vertices lie on a circle.
EXAMPLE 1
Describe a method of obtaining a circle that passes through the vertices of a triangle. Solution Let ABC be any triangle. ∠BAD is an inscribed angle standing on the minor arc BD. Are there quadrilaterals for which a circle can be drawn that passes through the four vertices? Yes. In the diagram. there are. Because any three non-collinear points lie on a circle. A set of such points is also referred to as a cyclic set of points. This circle is the circumscribed circle of the triangle. the right bisector of AB. and C.
Concyclic points
are points that lie on a circle. opposite angles are supplementary. Proof ABCD is a quadrilateral with vertices on a circle with centre O. ∠BCD
1 (reflex ∠BOD) (Angles in a circle) 2
D
A O B
C
90
CHAPTER 3
.
Extend CD to meet the circle at E. Assuming D is outside the circle (Diagram 2) leads to a similar contradiction. Hence D is on the circle. the quadrilateral is cyclic. or even if both are semicircles. Assume D is inside the circle (Diagram 1). There is a circle passing through A. You will be asked to prove this in Exercise 3. 3 C Y C L I C Q UA D R I L AT E R A L S
91
. ∠EAD
(Exterior angle)
B C A B
A E D
C
Diagram 1
D E
Diagram 2 This is a contradiction. The exterior angle of a cyclic quadrilateral is equal to the interior angle at the opposite vertex. Proof Let ABCD be a quadrilateral in which ∠ABC ∠ADC 180º. A corollary worth noting is that if we extend one side of a cyclic quadrilateral. The converse of the theorem is also true.
A D E B C
Angles in a Cyclic Quadrilateral Property A quadrilateral is cyclic if and only if its opposite angles are equal. B. so ∠ABC ∠AEC 180º.∴∠BAD
∠BCD
1 (∠BOD 2 1 (360º) 2
reflex∠BOD)
180º Note that it does not matter which arc BD is minor and which is major.3. or on the circle. ∠AEC ∠AEC.
THEOREM
If opposite angles of a quadrilateral are supplementary. Then D cannot be inside the circle.
3 . the proof is the same. But ∠ABC So But ∠ADC ∠ADC ∠ADC ∠ADC 180º. ∠AEC. Now ABCE is a cyclic quadrilateral. The point D is inside the circle. or outside the circle. the exterior angle so formed is equal to the interior angle at the opposite vertex. so ABCD is cyclic. and C.
Draw a circle that passes through A. Other properties of a circle also lead to theorems that prove that points lie on a circle. We will prove that ABDC is a cyclic quadrilateral. ∠BCD
110º. and ∠BAC
30º. AB AC. The proofs are similar to the indirect proof given above. DECB is cyclic.EXAMPLE 2
In a cyclic quadrilateral ABCD. Prove that DECB is a cyclic quadrilateral. This circle either passes through D or it does not. Then ∠ACB ∠AEB
(Angles in a cyclic quadrilateral)
C
A
B
92
CHAPTER 3
. and C.
E D
Proof Let ABDC be a quadrilateral such that ∠ACB ∠ADB. Suppose that it does not. B.
180º 70º 40º 40º 110º 55º. and let the circle cut AD at E. Join EB. AB What is ∠ABC? Solution ∠BCD ∠BAD ∠BAD ∠DAC ∠DBC ∠ABD Since then ∠ADB ∠ABD ∠ABC
AD. 40º 95º
(Angles in a cyclic quadrilateral) (Subtraction) (Angles in a circle) (Angles in a triangle) (Isosceles triangle)
A
30° D B 110° C
55º
EXAMPLE 3
In ∆ABC. We give two of the more useful results below. D and E are on AB and AC respectively so that DE BC. the quadrilateral is cyclic. THEOREM If one side of a quadrilateral subtends equal angles at the two remaining vertices. Proof ∠ABC ∠AED Then ∠ABC ∠AED ∠ACB ∠ACB ∠AED ∠DEC ∠DEC 180º 180º
(Isosceles triangle) (Parallel lines)
D E C A
(Straight line)
B
Therefore ∠ABC
Since ∠DBC and ∠ABC are the same angle.
it is not possible that the circle meets AD extended. AD and BE meet at O. Proof In quadrilateral OECD. Its proof is similar to the one above. Then the circle must pass through D. D is on BC and E on AC so that AD ⊥ BC and BE ⊥ AC.
THEOREM
The circle having as its diameter the hypotenuse of a right-angled triangle passes through the third vertex of the triangle. since EB and DB meet at B. 3 C Y C L I C Q UA D R I L AT E R A L S
.
C
A
B
EXAMPLE 4
In the diagram at the right. Prove that ∠OCD ∠BAD. By a similar proof. ∠OEC Then OECD is cyclic Therefore ∠OCD ∠OED ∠ODC
180º (Right angles)
(Angles in a circle) 93
3 .But
∠ACB EB DB
∠ADB. Solution ∠DAE Then ∠ACB ∠ACD ∠BDC ∠BEC
A
70º
(Angles in
∆ABE)
D 30° B 20° x 80° E x C
60º (Angles in ∆ABC) 30º 100º (Exterior angles. Then the supposition is false. and ABDC is a cyclic quadrilateral. ∆ADC) 100º (Straight line)
(Cyclic quadrilateral property)
∴∠BCED is cyclic EXAMPLE 5
In ∆ABC. ∠ADB. Cyclic Quadrilateral Property A quadrilateral is cyclic if and only if one side subtends equal angles at the remaining vertices or opposite angles are supplementary. This theorem is the converse of the result that the angle in a semicircle is a right angle.
(Parallel lines property)
Therefore ∠AEB Then
But this is impossible. the angles are given. Show that BCED is cyclic.
and S are concyclic. Is it true that any square is cyclic? Any rectangle? Any parallelogram? Explain your answer. Q.
C
F 100°
2.3
Part A
Communication
1.
A 80° P B D C Q
94
CHAPTER 3
.
B 50° E 80°
D A A
Knowledge/ Understanding
3.
A
so EDBA is cyclic. In the cyclic quadrilateral ABCD. If ∠ABD 50º and ∠AED 80º. The diagonal AC intersects BD at E. ∠BAD ∠OCD ∠BED ∠BED ∠OCD
(Angles in a circle) (From above)
B
E O
Therefore ∠BAD
D
C
Exercise 3.
(Equal angles on chord AB). In the given diagram. BC is extended to F and ∠DCF 100º. R.
Application
5. Determine the size of ∠PBD in the given diagram.Also ∠AEB Then But
∠ADB(90º). determine each of the remaining angles in the diagram. what four points are concyclic?
D F B
E
C R 60° Q 110° 50° T P S
4. Prove that the points P.
as are its diagonals. . and Q are concyclic. In the quadrilateral ABCD. Part C 12.. In the given diagram. If OT OS. Prove that if a trapezoid has its vertices on a circle. Prove that ABCD is a cyclic quadrilateral. PR PQ. M is the mid-point of ST. 7. . Prove that if one side of a cyclic quadrilateral is extended. PQ and PR are two chords of a circle with centre O. OT is perpendicular to PQ and OS is perpendicular to PR. prove that the points T. R. Also. 3 C Y C L I C Q UA D R I L AT E R A L S
95
.
R
x
P
2x S
T
10. S. A chord ST of constant length slides around a semicircle with diameter AB.
3 . 13. DB DC and ∠DBC 2xº.
Thinking/Inquiry/ Problem Solving
11. Prove that the points P. Q. X7. and X8 are chosen on successive arcs. and S are concyclic.
Q
9. and TS is parallel to RQ where S is a point on PQ. ∠HX8A. Points X1. Prove that the angle SPM is constant for all positions of ST. PT PR. The point T is chosen on PR. . In isosceles ∆PQR. A cyclic regular octagon ABCDEFGH has its vertices on a circle. X2. AB AD and ∠ABD xº. Determine the numerical value of ∠AX1B ∠BX2C . .
D A B x° 2x° C
8. R.Part B 6.. any exterior angle equals the opposite interior angle of the quadrilateral. and P is the foot of the perpendicular from S to AB. Prove that ∠QTS ∠SRQ. then its base angles are equal.
remaining parallel to the original secant. it is said to touch the circle. If you have taken any calculus. Since they have no particular relation to the circle. skew lines are of no current interest. the line l1 is a secant. We can use geometry software in two ways to compare properties of secants as they approach tangency:
t chnology
e
1. This requires P to move as well. C is the point of contact. but we can use some of these concepts to investigate properties of tangents. which does not intersect the circle. is said to be skew to the circle. It is. allow the secant to move towards the edge of the circle. Thus a secant contains a chord of the circle.4 — Tangents to a Circle
A straight line can intersect a circle in two places. What happens to the radii to A and B? As B gets closer to A. you probably know that the above definition of a tangent is insufficient to describe tangents to more general curves. touching the circle at the point C. This time. what do you observe about the secant? Most of the properties of tangents can be demonstrated using one of these techniques.
96
CHAPTER 3
. We will not do a formal study of slopes and limits in this book. Now let the point B move so that it gets closer to A.
C
l3
The line l3 in the diagram. As the chord AB gets smaller. A secant to a circle is a line that intersects the circle in two points. containing the chord AB.Section 3. what do you observe about the secant? 2. however adequate for circles. can touch the circle. Create a circle and a secant PABQ as in the example above. There are no other possibilities. In the diagram. with P outside the circle and A and B on the circle (start with A and B some distance apart on the circle). Create a circle and a secant PAB. the line l2 is a tangent. In the diagram.
A
B
l1 l2
A tangent to a circle is a line that intersects the circle at only one point. or make no contact at all.
touching the circle at P. Using indirect proof we will prove that ∠OPA 90º. If ∠OPB 90º.DEMONSTRATION
Using a computer (or using ruler and pencil). ∆OAB is isosceles. so OQ It is not true that ∠OPA Then ∠OPA Part 2 90º. Either ∠OPA 90º or ∠OPA 90º. the other less than 90º. (Isosceles triangle) This is impossible because Q is outside the circle.
P
Q
B
A straight line drawn at right angles to a radius of a circle at the circumference is a tangent to the circle. make a diagram with radii OA and OB and the secant through A and B.
A O
Assume that ∠OPA
Then one of ∠OPA and ∠OPB is greater than 90º.
3 . then there is a point Q on AB such that ∠OPQ ∠OQP. Proof Let AB be a tangent to a circle with centre O. 90º. But then OP OQ. so ∠OAB ∠OBA What happens to ∠OAB as the secant moves towards tangency?
1 (180º 2
O
∠AOB). OP. 4 TA N G E N T S TO A C I R C L E
97
.
A B
t chnology
e
THEOREM
Part 1 A tangent to a circle is perpendicular to the radius at the point of contact. Part 3 A line drawn at right angles to a tangent at the point of contact passes through the centre of the circle. 90º.
THEOREM
If tangents are drawn to a circle from an external point. In ∆POA and ∆POB. 2. Prove that a circle with centre P and radius PQ will have BA as a tangent. two tangents can be drawn. and PQ ⊥ BC. 3. the segments to the points of contact are equal.
EXAMPLE 1
The line BX is a bisector of ∠ABC. a tangent is perpendicular to the radius at the point of contact. a perpendicular to a tangent at the point of contact passes through the centre. and PR is a radius. BA is tangent to the circle. P is a point on BX. We will prove that PA PB. Then PQ PR
(Angle bisector)
R P B C A
X
Therefore a circle with centre P and radius PQ passes through R. 1. (Tangent radius property)
Q
P
O
From a point outside a circle.Tangent Radius Property For a given circle. and let A and B be the contact points of the tangents. Since PR ⊥ AB. PO PO OA OB ∠OAP ∠OBP ∆POA ∆POB Then PA PB
(Same line) (Radii) (Tangent radius property) (Hypotenuse-side)
A P B O
X
Y
98
CHAPTER 3
. a line at right angles to a radius at the circumference is a tangent. Join PO. Proof Let PX and PY be tangents to a circle with centre O. Proof Let PR be perpendicular to BA.
a.
7 O 25 A y B O y 25°
b. Solution PA CA DB Now perimeter ∆PCD PB CE DE PC PC PA 30 15
P E C
D
B Y
A X
(Tangents from a point) (Tangents from a point) (Same) ED BD DP DP
CE CA PB
The perimeter is 30.Tangent from a Point Property Tangent segments from an external point to a circle are equal. PY is a tangent contacting the circle at B. In the diagram. the centre is marked as 0. 4 TA N G E N T S TO A C I R C L E
99
. the radius is 5 and tangent AB has length 12.
3 . If PA 15. PX is a tangent contacting the circle at A. In each circle.
AB is a tangent to the circle.
B O C A
2. Determine the value of the variable(s) in each of the following. and CD is a tangent contacting the circle at E. determine the perimeter of ∆PCD.
Exercise 3. Determine the length of tangent AC and the length of OA.
EXAMPLE 2
In the given diagram.4
Part A
Knowledge/ Understanding
1.
Prove that AB CD AD BC. Concentric circles have the same centre but different radii. Prove that in concentric circles. a. If AB 10. Will a tangent to the circle with centre O. Circles with centres O and P touch at X. also be tangent to the second circle?
O X P
b.c. 8. determine the perimeter of ∆ADF. The quadrilateral ABDC is circumscribed about a circle. as shown. If the smaller circle is placed inside the larger. 6. what condition is necessary in order that a circle with centre B and radius BC will have AX and AY as tangents?
A
X C B D Y
4. In the diagram. where X is on the line OP.
D A E F B A C D B O
C
100 C H A P T E R 3
. two chords of the larger circle that are tangent to the smaller circle are equal. will a tangent to one circle be a tangent to the other? Part B 5. E.
O y 60°
d. drawn at X. that is. AC. and DF are tangents to a circle at points B. 7. and C.
z
y
130° O
x
Communication
3. each side of the quadrilateral is tangent to the circle. Prove that tangents to a circle drawn at the ends of a diameter are parallel. AB.
as shown. each with a radius of 2.
3 . Lines AC and BD are tangent to the circle with centre P and intersect at D. Determine the length of BD. The line OP cuts the circles at A and B. are tangent to each other. Prove that M is the midpoint of AB. The tangent at X meets the direct common tangent AB at M. Find the radius of the circle. as shown. Part C
Thinking/Inquiry/ Problem Solving
12. 4 TA N G E N T S TO A C I R C L E
101
. PQ 16 cm and QR 30 cm. touching all sides. A circle is drawn inside the triangle. ∆PQR is right-angled at Q. 10. Tangents at two points P and Q on a circle with centre O meet at an external point X. Two circles with centres O and P touch externally at X. Two circles with centres O and P.
C E A O P
D
B
A M B P
O
X
11.9. Prove that ∠XPQ ∠XOP. (This is called an inscribed circle or incircle).
Then Also Then ∠RTS But ∠SRQ Therefore Therefore Similarly. ∠RXS ∠TSR ∠TRS ∠TRS ∠SRQ ∠SRQ ∠RYS ∠RTS 90º 90º 90º ∠RTS ∠RXS ∠SRP
(Angles on chord RS) (Angle in a semicircle) (Angles in a triangle) (Tangent Radius Property)
Tangent Chord Property The angle formed by a tangent and a chord is equal to the angle subtended by the chord in the segment on the other side of the chord. rather than a specific one? Is it still true that ∠RXS ∠SRQ? THEOREM
S X O
P
R
Q
The angle formed by a chord and a tangent is equal to the angle subtended by the chord in the segment on the other side of the chord.
P X S O Y R Q T
Join RO and extend the line to meet the circle at T. Proof Let PQ be tangent at R to the circle with centre O. In the diagram. We will prove that ∠RXS ∠SRQ and ∠RYS ∠SRP. For any point X on the circle. If this line is extended. Let RS be any chord. and by extending RO to S we have RS as a diameter. and let Y be any point in the minor arc. What happens if RS is any chord.Section 3.
102 C H A P T E R 3
. Now ∠RXS ∠SRQ.5 — More About Tangents
We have seen that a line perpendicular to a tangent passes through the centre of a circle. let X be any point in the major arc created by RS. we create ∠RXS 90º. it becomes a diameter of the circle. PQ is a tangent at R to the circle with centre O.
We prove that AE • EB CE • ED. Then ∠PQD ∠EQR ∠DEQ ∠DEQ ∠EDQ ∠EQR ∠EDQ
(Tangent Chord Property) (Same) (Parallel lines)
D E
Therefore ∠DEQ
P
Q
R
Then ∆DEQ is isosceles (equal angles). From these intersecting lines. they will certainly intersect outside the circle if they are extended. Proof Let AB and CD be chords intersecting at E in a circle. Proof Join DQ and QE.
AE CE ED EB
(Vertically opposite) (Angles on arc AD) (Angles equal)
C A B
E
D
AE • EB
CE • ED
3 . If not. 5 M O R E A B O U T TA N G E N T S
103
. ∠AEC ∠DEB ∠ACE ∠EBD Then ∆ACE Therefore or ∆DBE. we obtain a useful theorem. When two non-parallel chords are drawn in a circle. they may intersect inside the circle. the product of the two parts of one is equal to the product of the two parts of the other. In ∆ACE and ∆DBE.
THEOREM
If two chords intersect. DE is a chord of the circle parallel to PQR. Join AC and BD.EXAMPLE 1
PQR is tangent to a circle at Q. These extended chords are secants. Prove that ∆QED is isosceles.
two chords AB and CD intersect at X.A B D C
Intersecting Chords Property If two chords in a circle intersect. we have a tangent. then instead of PC • PD. ADQP is a cyclic quadrilateral. we have PT 2. This indicates the following corollary to the Secant Intersection Property. while P stays fixed.
DEMONSTRATION
Draw the diagram for secant intersection.5. P and Q are the midpoints of XC and XB respectively.
EXAMPLE 2
In a circle. then PA • PB
PC • PD. When C and D become the same point. Therefore. Corollary If a tangent PT is drawn from a point on a secant AB. Prove that ADQP is a cyclic quadrilateral.
P
Intersecting Secants Property If two secants AB and CD intersect at point P.5. then PA • PB = PT2. the product of the two parts of one is equal to the product of the two parts of the other.
You are asked to prove the Intersecting Secants Property in Exercise 3. Now let the secant PCD move so that. the chord CD gets shorter. You are asked to prove this corollary in Exercise 3.
104 C H A P T E R 3
. if we call the common point T. Proof Then or or Also Then BX • XA 1 BX • XA 2 QX • XA
PX QX
D A X Q P C
CX • XD (Intersecting chords) 1 CX • XD 2 PX • XD
XA XD
B
∠PXQ ∠PQX
∠AXD ∆AXD ∠ADX
(Opposite angles) (Similar Triangle Property)
Therefore ∆PXQ
or ∠PQA ∠ADP These angles are both subtended by AP.
you have studied many of the properties of circles.
P
A
O
B
T
Key Concepts Review
In this chapter. Prove that AQBC is a cyclic quadrilateral. AB is a diameter of a circle with centre O.
Application
9.6. You should be familiar with the following: 1. 2. 7. 5. A line through P meets the circles at A and B. and AC is a diameter. b. Two circles intersect at P and Q. Two circles intersect at P and Q. P is on BA extended. 8. 4. The tangents at A and B meet at C. Prove the corollary to the Intersecting Secants Property. and PT is tangent to the circle. circle chord properties and conditions for equal chords the equality of angles in the same segment properties of cyclic quadrilaterals the tangent radius property the tangent chord property
106 C H A P T E R 3
. 3. Prove that ∠XPY ∠XQY 180º. chords. Part C
Thinking/Inquiry/ Problem Solving
10. A common tangent touches the circles at X and Y. Prove that AB bisects ∠DAC. Use the corollary to the Intersecting Secants Property to prove that PT ⊥ OT. and angles in a circle. a. AB is a chord of a circle. Prove the Intersecting Secants Property. The tangent to the circle at B contains a point D such that AD ⊥ BD.
Determine the value of the indicated variables in diagrams a and b.
R 5 P x A D x C 50° E B
T
PT is a tangent to the circle. A series of triangles is drawn on the same side of a common base AB with each angle opposite AB of the same size. f.
O A 150° x B C
DE is a tangent to the circle at C.) a.
5 4 D B x A A 6 2 3 E x C
b.
6 D
B
C
c.
O C x B
A
∠AOB
76º
3.
A 45° x
E
B
y C
D
2.
A 100° 120° x D B y E C
b. Calculate the value of the unknown quantities in each of the following diagrams. What can be said about the set of third vertices?
108 C H A P T E R 3
.
Q
7
d. O is the centre of the circle.Review Exercise
1. e. (Where it is used. a.
5. How would you locate the centre of this circle? 11. C. Suppose you are given three points A. If the diagonals AC and BD intersect at the centre O. A point is taken on the bisector of a given angle. The chord CD has length 10. XY. Side BC is extended to point F. How would you draw a circle that touches two given straight lines if a. If the radius of the smaller circle is 8 and the radius of the larger circle is 13. The inscribed circle of ∆ABC touches its sides at D. prove that AC BD. Prove that a circle may be drawn with centre P that has the arms of the angle as tangents. E. the two given straight lines are parallel? b.
B A C D
A
12. and C and are told that they are on the arc of a circle. find the size of each of the remaining angles in the quadrilateral. and Q lie on the same straight line. You are given a cyclic quadrilateral ABCD with the diagonals AC and BD intersecting at E. determine the measure of the angles of ∆DEH.
P A
b. 7. If AB and CD are two parallel chords.
B
H 60° D
E 45° C
REVIEW EXERCISE
109
. between the centres of the two circles. and ∠ABD 60º. ABCD is a cyclic quadrilateral. ∠AED 85º. B. If ∠ABC 60º and ∠ACB 45º.
D
10. determine the distance. 6. a. If exterior angle BCF 100º.4. prove that ABCD is a rectangle. Two circles with centres P and Q touch a straight line and each other at the point C. the two given straight lines are not parallel? 8.
C Q P Q
A
C
B
B C X Y
9. Prove that P. and H.
determine the size of ∠OAB in terms of x. RP and SQ meet at X when extended. If ∠ATB x. A circle with centre P and radius 10 is tangent to the sides of an angle of 60º. If AD and BC are extended to meet at Q. A circle is drawn so that it has as its diameter one of the equal sides of an isosceles triangle. and one of them passes through O. then ∠APD x. Prove that OA bisects the angle between the common chord AB and the tangent to the circle at point A. In the diagram.13. 17. determine the length of OT. A tangent is drawn to the larger circle at A. which is the centre of the other circle. then ∠AQB 2x. AT is the tangent at A to a circle with centre O.
A O B A T
15. The line of centres intersects the circles at R and S as shown. Prove that PXQT is a rectangle.
110 C H A P T E R 3
A C O
B
. TA and TB are tangents to a circle with centre O and radius 10. In the given diagram. Two circles intersect at the points A and B.
B
O
T
16. A larger circle with centre Q is tangent to the sides of the angle and to the first circle. What is the radius of the larger circle?
O
Q
D C 3x A B x P
Q P R S
A
T
B
Q P
19. circles with centres A and B are tangent externally at T. How do we know that this circle passes through the middle of the base? 14. Find the value of x. If AB and DC are extended to meet at P. ABCD is a cyclic quadrilateral with ∠DAB 3x. If ∠AOB 120º. 18. PQ is a common tangent line.
respectively. Prove that ∠QRP 180º. Prove that FJ is perpendicular to AB. AB is the diameter of a semicircle. Prove that PB BQ. ∠B 50º. B. R.
F
C E A J
D B
REVIEW EXERCISE
111
. and ∠C 70º. AC and BD are extended to meet at F. ABCD is a cyclic quadrilateral. and the points Q. A line is drawn through A that meets one circle at P and the other at Q. C and D are any two points on its arc. Y. Circles are drawn around ∆QAB and ∆PBC. The bisectors of angles A. 21.
P A Q
B
23. and Z. A circle is drawn that passes through the vertices A. AB and DC are extended to meet at point P. These circles intersect again at R. Determine the size of each of the angles in ∆XYZ. Two circles have the same radius and intersect at the points A and B. From F. The lines CB and DA are extended to meet at Q.20. and C meet the circumference of the circle at X. a line through the intersection of AD and BC meets AB at J. 22. ∆ABC is a triangle in which ∠A 60º. B. and P are on the same straight line. and C.
RICH LEARNING LINK WRAP-UP
107
. Is ABDC a cyclic quadrilateral? Provide as many 4th St.
INDEPENDENT STUDY
In addition to the location of central points. Investigate and Apply The map shows a section of a city where four simi11th St.
2nd Ave. 4th Ave. 1st St. 3rd St.
9th Ave. D.wrap-up
CHAPTER 3: GEOGRAPHIC PROFILING
investigate and apply
Geographic profiling is a means of linking a series of crime sites to find where the criminal lives. A. He says. 3. 5th St. 8th St. Geographic profiling is designed to track these people down. Kim Rossmo. "Seventy-five percent of serial killers hunt in their own community . 2. each of which 6th St.. different arguments as you can to justify your 2nd St. 6th Ave. 10th Ave. where is its centre? 4. Speculate on the most likely location of the perpetrator's home and justify your conclusion with reasoning that includes mathematical reasoning. It was created by Dr.. what other geographic factors should be taken into account when using crime locations to predict a criminal's place of residence? Do you think that they can all be accounted for using circle geometry? The RCMP's Violent Crime Linkage Analysis System relates dozens of separate factors to find patterns among crimes. B C 7th St. D answer. 7th Ave. 1. Find the centres of four circles. who is also a former Detective Inspector with the Vancouver Police Department. 8th Ave. The crimes were 10th St." The software he has developed considers many geographic aspects of crime sites. 1st Ave. lar crimes have been committed. 5th Ave. B. passes through three of the crime locations. C. including the location of central points. 3rd Ave. A committed on four consecutive days in the order 9th St. Does the interior of any of the four circles contain all four points? If so. an adjunct professor at Simon Fraser University's School of Criminology. What are some other quantifiable aspects of random crimes? How is geographic profiling used in a court of law? Can it be used as proof of guilt or innocence? ●
11th Ave.
7 2 All
1. XY is tangent to the circle at Y. Determine the measure for each of the indicated values in the following diagrams.Chapter 3 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application
Questions All 5. AB is tangent to the circle at C.
x° 20° 50° A C B
x b. 6. only the answer is required.
__________
C b E 2 8 3 D B
A
b
112 C H A P T E R 3
__________
. a. Two chords AB and CD intersect at E.
__________
8 P 4 X a
Q
Y
a c. It is not necessary to show your work.
and AP is a chord. O is the centre of the circle. In the given diagram.
O B C D E
5.
A 120°
B y
x 100° D C E
x y 2. Explain your reasoning. and OM 8. Two circles intersect at A and B. Prove that their midpoints lie on a circle concentric with the given circle. and PQ is a tangent to both circles. If ∠BAC 50º. In a given circle. In the diagram. Prove that CD DE. a. OC 10. AB is diameter of a circle. OB 17.
__________ __________
B 50° C A
O
D
3. ABCD is a cyclic quadrilateral. several equal chords are drawn. it bisects PQ at the point R. b.
CHAPTER 3 TEST
113
. Prove that when BA is extended.d. which has B as centre. determine the size of ∠BCD.
P
Q A
B
6. Prove that the tangent at T cuts AP extended at right angles. The two given circles are concentric with centre O. AT is another chord bisecting ∠BAP.
A
C
M O
D
B
4. Determine the length of AC. 7. OC is a radius of the larger circle and a diameter of the smaller circle. Prove that all such chords are tangents to this concentric circle.
A well known theorem involving Cevians is Ceva's Theorem. Ceva's Theorem The cevians of ∆ABC. The first circle passes through vertex C and the points R and Q.) From each vertex. I. The second circle passes through B. and CR are concurrent if
BP and only if AR • CQ • PC RB QA
C
1. Using Geometer's Sketchpad®. select three points P.
A R Q C A
R
B
P
B
2. (In our diagram we have drawn two of the circles. AP. altitudes.C O M P U T E R I N V E S T I G AT I O N S
1.)
114 C H A P T E R 3
. AC. c. is the point of intersection of the bisectors of the angles and that the incircle has centre I and is tangent to the three sides of the triangles.
P Q
a. and R and the third through A. and angle bisectors. and Q. Draw a variety of different triangles and Cevians. and R. BQ. P. Draw any triangle ABC and on the three sides AB. P. Draw ∆ABC and construct the incentre and incircle. and BC. respectively. Construct three circles according to the following instructions. verify that the three Cevians concur.) b. and convince yourself that Ceva's Theorem is correct. (Note that Ceva's Theorem is written in if and only if form and must thus be verified in both directions. A Cevian is a line drawn from a vertex of a triangle to the opposite side of the triangle. Q. Some examples of Cevians are medians. (Recall that the incentre. draw a line to the point of contact on the opposite side. Prove that the observation you made in b is correct by using Ceva's Theorem and the properties of tangents drawn to circles.
(ii) Prove that this relationship is true using Ptolemy's Theorem. M to Q. and Q are concyclic points. show that if the point P is on the arc BC. call it M. Using Geometer's Sketchpad®. (Hint: Join P to M. This is Ptolemy's
B
C
D
E X T E N D I N G A N D I N V E S T I G AT I N G
115
. a. and M to R and then use properties of cyclic quadrilaterals to show that A. PB PC. Prove that M lies on the circle passing through A. P and Q. C. Using Geometer's Sketchpad®. We have drawn two of the circles and we have labelled the point M as the point of intersection. (i) Using Geometer's Sketchpad®. B. Y. and D on the circumference of the circle. M.) c. Draw an equilateral triangle ABC and construct a circle that passes through A. using Ptolemy's Theorem. (i) Find an analogous relationship for the hexagon to that found in b for the equilateral triangle. Draw a circle and locate any four points A. and C. P. construct all three circles and show that all three points have a point. show that (AB)(DC) Theorem. b. that this result is (AD)(BC) (AC)(BD).
A B Y P
A Q
M R
Z C
3. in common. b. PA correct. c. B. Label the centres of the three circles as X. and Z and prove that ∆XYZ ∆ABC. (ii) Prove. Draw a regular hexagon ABCDEF in a circle and place a point on the arc BC.a.
3. If the sum of the angles B and C remains constant. A rectangular area measures 2. A and B are 8 cm apart. Find the area of a.3
. quadrilateral AEFD 6. Write an algebraic expression for the area of this triangle.7 cm by 3. ∆EBF b. The vertex A of ∆ABC assumes various positions on the same side of BC. The medians BD and CE of ∆ABC intersect at F. meet outside the circle at a point O. 2x 2. 5. ABGH and ADEF are squares.5 cm on the map. 4. respectively. Prove that ∆AOD ∆BOC. Find two points. AB and CD are two perpendicular chords of a circle with centre O. while BC remains fixed in length and position.
E
F B
A
D
C
116 C U M U L AT I V E R E V I E W C H A P T E R S 1 . and 2x 3. AB and CD are two chords of a circle that. Prove that ∆CGE is isosceles. The areas of ∆DFC and ∆FBC are 8 and 16. What is its actual area? 2. Prove that ∠AOD ∠BOC 180º. 1 cm represents 1 km. ABCD is a parallelogram. On a map.
H G
8. when extended.Cumulative Review
CHAPTERS 1–3
1. what is the path traced out by the vertex A? 7. each of which is 5 cm from A and 6 cm from B. The lengths of the sides of a right-angled triangle are 2x 5.
Prove that ∆PDA ∆PCB.
P C A B
D
12.
E
C
F
A
D
B
C U M U L AT I V E R E V I E W C H A P T E R S 1 – 3
117
. In the given figure. Prove that AD2 (BD)(DC) if and only if ∠BAC 90º. Prove that AC BD. Prove that AT BC and AC bisects BT. two chords are drawn such that AB CD. Find the ratio of the area of the inscribed square DEFB to the area of the triangle ABC.
C D
15. A quadrilateral is a rhombus if and only if it is a parallelogram.9. SRNM is a parallelogram with AN SC and NT SB. 11. In ∆ABC. b.
B T S C R B A
14. ABC is a right-angled triangle with AB 2BC. Two lines L1 and L2 are parallel if and only if a line crossing L1 and L2 makes alternate angles equal. prove that these sides are parallel. Quadrilaterals have equal areas if and only if they are congruent. Determine which of the following statements are true. c.
B
A
D M A
C N
13. a. In the diagram. 10. In a circle. a line is drawn from A perpendicular to BC so that it meets BC at D. ∠PDC 45º. If a quadrilateral has its area bisected by the line joining the midpoints of a pair of opposite sides.
Prove that HE TN. Prove that (AX)(BY)(CZ) (BX)(YC)(ZA).
C
21.16. Prove that DBCE is a cyclic quadrilateral. PS intersects the circle at T. determine the measure of ∠ADC. are angles of 120º. ABCD is a square. If ∠DBA 30º. How many sides does the polygon have?
A
19. In the diagram. The others. QR is extended to S. Prove that AE BF if and only if BF is perpendicular to AE. Prove that QT is the bisector of ∠PQR.
Q R
T
S B
24. respectively.
B X Z
Y
C
18. ABC is an isosceles triangle in which AB AC. O is the centre of the circle. CD 4 and DA 6. Prove that ∆BYC quad AMYN. which are equal.3
A 6
2 x
3 C 4
D T H B A N E C
. A circle is inscribed in ∆ABC as shown.
B
D
E C B A D
20. the length of AC. ABCD is a cyclic quadrilateral with AB 2. making RS PR. In the diagram. 25. Two angles of a convex polygon are 100º and 140º.
D
F
C
E A A B
22. ∆ABC is an isosceles triangle in which ∠B ∠C 2∠A. E is a point on BC. Prove that BX and XA are consecutive sides of a regular pentagon. A 17. BC 3. TAN is the tangent at A to the circle shown. BX bisects ∠B and meets AC in X. BE and CH are altitudes of ∆ABC.
118 C U M U L AT I V E R E V I E W C H A P T E R S 1 . M and N are the midpoints of AB and AC.
B P
M Y
N
C
23. Determine the value of x. The line DE is a straight line parallel to BC.
2 model and solve problems involving velocity and force. Physicists use vectors in determining the thrust required to move a space shuttle in a certain direction. pilots need to know what effect a crosswind will have on the direction in which they intend to fly. You will learn more about vectors in this chapter.1 determine the components and projection of a geometric vector. Section 4. In order to navigate. and how vectors represent quantities possessing both magnitude and direction. 4.Chapter 4
VECTORS
Have you ever tried to swim across a river with a strong current? Have you sailed a boat. you will
• • • •
represent vectors as directed line segments.4
. or run into a head wind? If your answer is yes.
CHAPTER EXPECTATIONS In this chapter. vectors became an essential tool of navigators. engineers need to know what load a particular design will support. Section 4. then you have experienced the effect of vector quantities. Vectors were developed in the middle of the nineteenth century as mathematical tools for studying physics. Section 4.3. In the following century.1 perform mathematical operations on geometric vectors. and physicists. In order to build bridges. engineers. Section 4.
The force of gravity acts only downward. Meteorologists. never sideways. In this chapter.
T R I G O N O M E T R I C R AT I O S
B
In a right-angled triangle.
and which
A b C
120 C H A P T E R 4
. In later chapters we will introduce algebraic representations of vectors. which will be more easily extended to higher dimensions. we introduce the concept of a vector. Temperature is referred to as a scalar quantity. Pause for a moment and think about physical quantities that have a direction. need data on air temperature and wind velocity. We will concentrate on geometric representations of vectors. we will review some basic facts of trigonometry. as shown. so that most of our discussion will be of two-dimensional vectors. an inseparable part of which is a direction. a mathematical object representing a physical quantity that has both magnitude and direction. The object of this chapter is to introduce the mathematical properties of vectors and then show how vectors and scalars are used to describe many features of the physical world. is not a vector quantity. for example. even if the wind speeds are the same.Review of Prerequisite Skills
A vector is a quantity. sin tan
a c a b
cos
b c
c
a
Note: The ratios depend on which angle is angle is 90°. to make weather forecasts. Wind is another example. on the other hand. Force is an example. among other things. Before we begin this chapter. A wind from the north and a wind from the south have different physical consequences. We need both scalar and vector quantities to model complex physical systems. Temperature. Temperature does not go in any direction.
XY 6. and AC c. In ∆PQR. 4. An aircraft control tower T is tracking two planes at points A. YZ. ∠X 60º. and QR angles to the nearest degree. State the exact value of each of the following. Three ships are at points A. State the exact value
3. a. 3. 6. sin 60º d. Determine the measures of the
5. tan 45º 10. PR 7. tan 120º 2. and B. B. ∠B 90º.THE SINE LAW
a sin A b sin B c sin C
c A
b
B
a
C
THE COSINE LAW
a2 b2 c2 2bc cos A or cos A
b2 c2 2bc a2
Exercise
1. determine the distance between the planes. cos 60º e. Determine the values of XZ. What is the distance between B and C? 7 km. and C such that AB 2 km. In ∆XYZ. AC and ∠BAC 142º. A triangle ABC has AB of tan A.5 km from T.
70º. PQ 4. 5. 6 km from T. cos 30º 6. b.
REVIEW OF PREREQUISITE SKILLS
121
. sin 135º f. and ∠Y and ∠Z to two-decimal accuracy. If ∠ATB 70º.
Neuroscientists have asked what it is about the activity in a group of cells with overlapping tuning regions that specifies the actual direction of a stimulus? For example. Different cells are tuned to different directions. how is it that we can point accurately in the direction of a distant sound without seeing its source? One answer is that a cell responds more vigorously when the distance stimulus is in its direction. 1 From the diagram. 2 1 1 Solving. What direction would be represented by a north-east cell responding three times as vigorously as an east cell?
DISCUSSION QUESTIONS
east 1 unit
north 2 units
direction of stimulus
1. we see tan . what is the direction of the stimulus? We can use vector addition to find out. and the regions to which different cells are tuned overlap considerably. The third side is the actual direction of the stimulus. How many cells would be needed to represent all the directions in the plane? 2. Each cell responds only to stimuli from a specific direction.investigate
CHAPTER 4: VECTORS AND THE SUPERIOR COLLICULUS
Neuroscientists have found cells in a deep layer of a part of the brain called the superior colliculus.6º east of north. If the north cell responds twice as vigorously as the east cell. The tuning is broad. The direction is determined not by which cell fires most vigorously.6º. Investigate and Inquire The type of addition performed in the brain can be illustrated by a simple case involving only two brain cells. Why do you think the direction is not just taken to be the one corresponding to the cell that fires most vigorously? ●
122 C H A P T E R 4
. the stimulus is 26. but by a type of addition of the degrees to which the various cells have responded to the stimulus. Thus. 2 So 26. These cells are tuned to the directions of distant visual and auditory stimuli.6º. The side pointing north is twice as long as the side pointing east. we find tan 26. while the other responds to stimuli that are approximately east. Suppose that one of these cells responds to stimuli that are approximately north. The answer is found by forming a triangle with a side pointing east and a side pointing north.
Consider a typical winter weather report that you might hear on the nightly news: The temperature is presently –11 ºC. Quantities having both magnitude and direction are called vectors. a vector is represented by an arrow: .
4 . Force is a vector quantity. They are classified as follows: Quantities having magnitude only are called scalars. There is no problem with this interpretation. The direction in which the arrow points is the direction of the vector. The other quantity (the wind velocity) has a numerical value (its magnitude) and also a direction associated with it. in most situations we find it easier to use positive and negative numbers as scalars. sometimes it is a useful way to look at such quantities. However. the temperature could be thought of as having magnitude (11º) and direction (negative). 1 V E C TO R C O N C E P T S
123
.
There seems to be some overlap here. For example.Section 4. The force of gravity is measured when you step on a scale. For now we will restrict our discussion to vectors in two dimensions or to situations that can be expressed in two dimensions. A magnetic field is a vector quantity. One quantity (the temperature) is expressed as a single numerical value. Some examples of vector quantities are Force The force of gravity has a well defined magnitude and acts in a specific direction (down). it could be considered as a one-dimensional vector. Displacement When you walk from point A to point B. and restrict the term vectors to quantities that require (at least) two properties to define them. The length of the arrow is a positive real number and represents the magnitude of the vector. others are weak. These quantities are typical of the kinds encountered in science. Some magnets are strong. This weather report contains two different types of quantities. All cause a compass needle to swing around and point in a particular direction. with a wind from the northwest at 22 km/h.1 — Vector Concepts
Vectors are a part of everyone's common experience. you travel a certain distance in a certain direction. in that sense.
Magnetic Field
In a diagram. Our definitions and conclusions are easily extended to three dimensions (or more). Displacement is a vector quantity.
Often it is necessary to explicitly state the initial point and the end point of a vector. Then. v are also vectors x. The magnitude of a vector is expressed by placing the vector symbol in absolute value brackets. then changing buses and riding a further 3 km north. a. two capital letters are used. N Solution Suppose you represent a 1-km distance by a 1-cm line segment. Its magnitude is AB. Represent these displacements on school a vector diagram. u. D Two vectors are parallel if their directions are either the same or opposite. a 2-cm arrow pointing left represents the first leg of the bus trip. such as AB and CD. v are vectors
u v W S bus stop home E
u.EXAMPLE 1
A student travels to school by bus. A 3-cm arrow pointing up represents the second leg. Scalar quantities are written as usual. v AB is the vector that starts at point A and ends at point B. v are the magnitudes of the vectors u. first riding 2 km west. b are scalars u. Two vectors are equal if and only if their magnitudes and their directions are the same.
A B
Certain other terms are used in connection with vectors. Give examples of vectors which are a. When two vectors are opposite. The total trip is represented by a diagram combining these vectors. parallel but having different magnitudes
F
A
B C
124 C H A P T E R 4
E
D
.
EXAMPLE 2
ABCDEF is a regular hexagon. Then. y. Some texts use boldface letters for vectors. Two vectors are opposite if they have the same magnitude but point in B C opposite directions. A one is the negative of the other: AB CD. equal b. The magnitude of a vector is a positive scalar. The notation used to describe vector quantities is as follows: The algebraic symbol used in this text for a vector is a letter with an arrow on top. Such vectors are referred to as point-to-point vectors.
but ED
b.
v u
ED EB CB DC CB.6º. The angle between OB and OC is 45º. that is.
c.
3 . starting at the same point. The angle between two vectors is the angle ( 180º) formed when the vectors are placed tail to tail. ∠EOC 6
O
6
C
26. The diagonal of the square bisects ∠AOC. equal in magnitude but opposite in direction d. FE d. There is no special symbol for the direction of a vector. To specify the direction of a vector. FA EB. OB and OC b. ED e. tan ∠EOC OE and OC is 26. E is the midpoint of BC.4º. DC There are other possible answers. OB and OE
6
B
E
Solution a. AB c. a. equal in magnitude but not parallel e. but FE DC. Using trigonometry. different in both magnitude and direction Solution a. but FA
One way to determine the angle between two vectors is to examine geometrical relationships and use trigonometry. so the angle between 26. b.6º. we state the angle it makes with another vector or with some given direction such as a horizontal or vertical axis or a compass direction. A Find the angle between the following vectors. OE and OC c. The angle between OB and OE is the difference 45º
4 .
EXAMPLE 3
OABC is a square with sides measuring 6 units.c.6º 18. 1 V E C TO R C O N C E P T S
125
. FB.
Express b and c each as a scalar multiple of a. Unit Vectors 1. b. It is true in general that two vectors u and v are parallel if and only if u kv. and c each in terms of the unit vector a. When a vector and a unit vector are denoted by the same letter.
v
EXAMPLE 4
Examine the vectors in the diagram. then AB 2AM. so the three vectors are parallel. each vector lies on the hypotenuse of a right-angled triangle with sides in the ratio 1:2. and since the directions of AB and AM are the same.
1 a. you should understand v to be a unit vector having the same direction as v. Express a. On the grid.
5a and c
126 C H A P T E R 4
ˆ b.
Thus. a. The unit vector in the direction of a is a ˆ ˆ b 5 5a. Such vectors are called unit vectors. Then a 5
52
102
5 5. b. and c can be found using the Pythagorean Theorem. for example v and ˆ ˆ v . ˆ b.When two vectors are parallel. M is the midpoint of the line segment AB.
a b c
Solution a. Since M is the midpoint.
ˆ 5a. A unit vector is denoted by a carat (ˆ) placed over the symbol. The magnitudes of a. one of the vectors can be expressed in terms of the other using scalar-multiplication. Suppose. Such vectors are valuable even though their direction is undefined. A particularly useful type of vector is a vector with magnitude 1. Any vector can be expressed as a scalar multiple of a unit vector. a and c Therefore b 12 22 32 62 5. The zero vector is denoted by 0. A unit vector in the direction of any vector v can be found by dividing v by its magnitude v: 1 ˆ v v 2. and c 3 5a.
. we write the vector equations AB 2AM or AM
1 2 AB or BM 1 2 AB. Any vector v can be expressed as the product of its magnitude v and ˆ a unit vector v in the direction of v ˆ v vv Another useful type of vector has magnitude 0. for example. multiplication of a vector by a scalar k results in a new vector parallel to the original one but with a different magnitude. b 3 5 3a.
the force exerted by a chain hoist carrying a load of 200 kg
4 . the density of a lead weigh q. the force of gravity generator k. the temperature of a swimming pool
3.1
Part A
Communication Communication
1. N and NE
Knowledge/ Understanding
b. the area of a parallelogram l. the friction on an ice surface b. the acceleration of a drag racer c. the speed of light g. Which of these physical quantities is a vector and which is a scalar? a. S and W
6. the volume of a box
i. Draw a vector to represent a. E and SW
c. the momentum of a curling stone n. in opposite directions
5. For each part of Example 2. In your own words. Draw vectors that represent the velocities of the two cars if they are going a. the pressure of the atmosphere r. 2. 1 V E C TO R C O N C E P T S
127
. the energy produced by an electric j.) b. What is the angle between the following directions? a. the velocity of a fishing boat travelling at 8 knots on a heading of S75ºW (A knot is a speed of one nautical mile per hour. One car travelling 75 km/h passes another going 50 km/h. the magnetic field of the earth p. the frequency of a musical note f. the velocity of a wave at a beach e. explain the difference between a scalar and a vector. the speedometer reading in an automobile m. state a second answer. Part B 4. the position of a city intersection 7 blocks east and 3 blocks south of your present position c. the age of a child h. in the same direction b. the mass of the moon d. the time on a kitchen clock o. the displacement of a crate that moves 6 m up a conveyor belt inclined at an angle of 18º d.Exercise 4.
Express each direction as an angle measured counter-clockwise from a unit vector in the positive x direction. 11. 1 EA 4 c. The aircraft are travelling at velocities of 450 kph N.
14. and 175 kph N20ºE.
2 GC 3
A
e. In a diagram. 5 DF 2 d. Determine the displacement vector for each leg of the trip. Radar in the control tower of an airport shows aircraft at directions of N50ºE. it turns to a heading of N 80º E and continues for another 2 hours before returning to base. a. E. The points A. Determine the direction of its velocity. ABCD is a rhombus.
y
a c d
b
x
e
12. respectively. find two vectors u and v in this diagram (expressed as point-to-point vectors) such that a. C. 550 kph N70ºW. Prove that two vectors u and v are parallel if and only if u
128 C H A P T E R 4
. N70ºW. travelling at a speed of 240 km/h. and G are equally spaced along a line. respectively. u c. After travelling for one hour and fifteen minutes.
Knowledge/ Understanding
8. Part C 13. a. A search and rescue aircraft. (v 0)? kv. b. D. F. and 12 km. an aircraft rises 100 m for every 520 m of horizontal motion. During takeoff. u v 2v b. u d.
2AD
D
Application
9. and distances of 5. draw small vectors to represent their velocities. starts out at a heading of N 20º W. draw vectors showing the positon of the three aircraft in relation to the tower. u v
1 v 2
B
P C
10. Find the total distance the aircraft travelled and how long it took. Determine the magnitude and the direction of each of the vectors in the given diagram. For each of the following. For what values of k is (k 2)v 4v. b. Name a vector which is equal to a. 8.Application
7. At the positon of each aircraft in part a. and S20ºW. 3BD b. B.
If the vectors are added in the opposite order. some people will walk along the sidewalk from A to B and then from B to C. The sum u v. Others may follow a shortcut through the park directly from A to C. Whichever route is followed. Triangle Law of Vector Addition To find the sum of two vectors u and v using the triangle law of vector addition. vectors can be represented geometrically by a directed line segment. for example. both get from A to C.
A Vector Diagram B
D
C
A
B
C
This model for vector addition is valid for all vectors.
v u+v u
v
4 . the result is the same. This demonstrates that vectors satisfy the commutative law of addition: u v v u.2 — Vector Laws
In many applications of vectors to physical problems. the displacement is the same. a parallelogram is formed. Therefore AB BC AC. This route is described by the displacement vector AC. Suppose the rectangle ABCD is a park at the corner of an intersection. is the combined effect of two or more forces acting on an object? How does wind velocity affect the velocity of an aircraft? To determine what the sum of two vectors is. draw the two vectors head to tail. we must find the combined effect or sum of two or more vectors. To get from A to C. They follow a route described by the sum of two displacement vectors: AB BC.Section 4. because. or resultant. let us look first for a geometrical answer. The order in which we add the vectors is unimportant. What. is the vector from the tail of the first to the head of the second. 2 V E C TO R L A W S
129
.
u v v+u u
u+v v
u
By combining the two triangles of the triangle law in one diagram. in general.
These two laws of addition are equivalent. Solution Adding a to b first.
u 30° v
130 C H A P T E R 4
. Complete the parallelogram with these vectors as sides. draw the two vectors tail to tail.
EXAMPLE 2
Find the magnitude and direction of the sum of two vectors u and v.
EXAMPLE 1
Given the three vectors a.Parallelogram Law of Vector Addition To find the sum of two vectors using the parallelogram law of vector addition. The sum u v is the diagonal of the parallelogram from the point where the tails are joined to the point where the heads meet. The method we use depends on which is the most convenient for the problem at hand. we obtain
c b a+b a a+b (a + b) + c
b
c
a
b
Adding b to c first. if their magnitudes are 5 and 8 units. start by drawing vector diagrams such as those on page 129. Complete the parallelogram and draw the resultant. When you set out to solve a problem involving vectors. b. It means that we can omit the brackets and write simply a b c. and c. respectively. b c. a (b c). and the angle between them is 30º. Solution Make a vector diagram showing the two vectors with an angle of 30º between them. we obtain
b b+c c a b b+c a + (b + c) c
This example illustrates that vectors satisfy the associative law of addition: a (b c) (a b) c. sketch the sums a and (a b) c.
6 2(5)(8) cos 150º
The direction of u v is expressed as an angle measured relative to one of the given vectors. we express the difference in terms of a sum.4º with v. say v. v is 12. Observe that the angle between the vectors is 5 30° 150° not an angle in this triangle. geometrically
AD
b. u Then u v2 v 52 82 158.4º Therefore. algebraically
A B
Solution a. so AB AD DB
D
C
AD´.
D C
EXAMPLE 3
In parallelogram ABCD. 2 V E C TO R L A W S
131
. (Why?) Use the angle of 150º and the cosine law to find the magnitude of the sum.6 units. AB
AD
AB AB DA DB
( AD) DA AB
D´
C´
(DA is the opposite of AD) (Rearrange the order of the terms)
4 . The difference of two equal vectors a a is the zero vector. The angle between the 8 vectors is equal to an exterior angle of the triangle. It can be found using the sine law. use the opposite of b and add it to a. Draw AD´ opposite to AD. Hence a b is equivalent to a ( b). and it makes an angle of
To subtract two vectors a and b.
sin 5 sin 150º 12.6 5 sin 150º 12. To find the vector a b.The resultant is the third side of a triangle with sides 5 u+v and 8. Using the parallelogram law.28 12. find the difference AB a. The zero vector has zero magnitude. draw the sum AB which is AC´ in the diagram.
A B
b. This is in the diagram. But AC´ DB. Its direction is indeterminate. the magnitude of u approximately 11.6
sin
11. denoted by 0.
. The lengths of the sides of the triangle are the magnitudes of the vectors. When does equality hold?
Solution Make a diagram of two vectors u and v. Therefore u
132 C H A P T E R 4
u+v u
v
v
u
v. The three vectors form a triangle. The following example illustrates this. There is no triangle if it is greater. We can demonstrate the validity of these relations by using vector diagrams. From the diagram. There are other basic vector relations that are universally true. the side u v must be less than the sum of the other two sides u v.
EXAMPLE 4
Show that u
v
u
v.In the parallelogram formed by two vectors u and v
D C u+v v u–v A u B
• the sum u v is the vector created by the diagonal from the tail of the two vectors u v AC • the difference u second diagonal u v DB v is the vector created by the
Properties of Vector Addition •a b b a Commutative Law • (a b) c a (b c) Associative Law Properties of Scalar Multiplication • (mn)a m(na) Associative Law • m(a b) ma mb Distributive Laws • (m n)a ma na Properties of the Zero Vector: 0 •a 0 a Each vector a has a negative ( a) such that • a ( a) 0 These laws state that you may add vectors in any order you like and that you may expand and factor expressions in the usual way. and their sum u v.
three vectors. For each of the following. 8a 3b
4 . Find the sum of the vectors u and v if a.
D u v C
B
Knowledge/ Understanding
2. F. state the name of a vector equal to u to u v. D. and G. v 21 and
is the angle between them. Express the vector BE as a.u
u+v u v
v
u
v. u 12. ˆ ˆ 6. What single vector is equivalent to each of these sums? a. the sum of two vectors. and u v u v. u
5. a. calculate ˆ ˆ ˆ ˆ a. and four vectors b. the difference of two vectors in two different ways 3. 3. 2 V E C TO R L A W S
133
. Triangle Inequality For vectors u and v. E. A tour boat travels 25 km due east and then 15 km S50ºE. are arranged in order from left to right on a single straight line. B. the triangle collapses to a single line.
Exercise 4. PT GE QT CE SR SQ
Knowledge/ Understanding Application
4. C.
D C
c. 3a 5b b. PT c. and 115º
70º b.2
Part A
Communication
1.
D v v A B A u B A u C
v and equal
b. If a and b are unit vectors that make an angle of 60º with each other. v 10. EA Part B TS CB SQ DB AD b. AC d. Seven points A. Represent these displacements in a vector diagram. then calculate the resultant displacement.When u and v have the same direction.
find x and y in terms of a and b. By considering the angles between the vectors. x v) ku kv. and state its magnitude and direction. 24. Let a a. 3x k. y 23. 3b 4c
b. for any scalar k and any vectors u and v. b 9y) i j
b. u v u v
8. Vectors a and b have magnitudes 2 and 3. u v u v b. If a
3j
3k. a. Illustrate for k
0 that k(u
15. find the vector 5a 2b. ABCDEF is a regular hexagon with sides of unit length. a 12. Show how to find the vectors themselves.7. respectively. k(u – v) ku kv. What conditions must be satisfied by the vectors u and v for the following to be true? a. a. and x y 30. a 5x
2y and b
4y. x
y 2 x x 2 y
b. If the angle between them is 50º.
134 C H A P T E R 4
. Simplify the following expressions using the properties of vector operations. u v u v c. 10. 8(3x
Application
5y 5y) 2i b 3x c
x
6y 4(6x k. 2x d. If x 11. have the zero vector as a resultant? 9. and 25. Part C 17.
x 2
y
x 2
y
14. 4x c. Check each identity algebraically. and illustrate with the use of a diagram. find x y. Find c.
13. The sum and the difference of two vectors u and v are given. respectively. Under what conditions will three vectors having magnitudes of 7.
u+v u–v A B C D F E
b and a
b are
19. 16. Find the magnitude and the direction of AB AC AD AE AF. show that a perpendicular when a b. 18. and c 2b – 3c 2i
4(x 6y
y) 4(2y x) 6x
11. Show geometrically that.
What is the magnitude of a face diagonal? A body diagonal? 21. the other body diagonals of the cube
D F A B G H
E
C
b.Thinking/Inquiry/ Problem Solving
ˆ ˆ ˆ 20. AC. BE. FG. a diagonal of the front face of the cube c. and AD that lie along adjacent edges of the cube in the given diagram. a. the other diagonals of the front. j . Represent by i . a body diagonal of the cube d. Prove that for any vectors u and v. the three vectors AB.
4 . j . and k. top and right faces of the cube
e. and k. u v2 u v2 2(u2 v2). Express each of the following vectors in ˆ ˆ ˆ terms of i . 2 V E C TO R L A W S
135
.
so it accelerates into the sky. Newton's First Law of Motion An object will remain in a state of equilibrium (which is a state of rest or a state of uniform motion) unless it is compelled to change that state by the action of an outside force. it will rise into the air.8 m/s2 as they fall. The engines provide thrust. which is abbreviated as N. and the buoyant forces and the force of gravity balance. gravity causes objects to accelerate at a rate of approximately 9.Section 4.3 — Force as a Vector
A force on any object causes that object to undergo an acceleration. it can float for days. You can feel a force pushing you back into your seat whenever the car you are riding in accelerates from a stop light. It is attracted by the force of gravity like everything else but upward forces are greater. the force propelling the aircraft forward. The magnitude of the gravitational force is the product of an
136 C H A P T E R 4
. The thrust is counterbalanced by a drag force coming from air resistance. When you release a helium-filled balloon. but that does not mean that the force that set the car in motion has ceased to exist. Eventually it reaches an altitude where the atmosphere is less dense. In this state of equilibrium.
lift drag aircraft weight thrust
The magnitude of a force is measured in newtons. It was Newton who first clarified these concepts and formulated the law that bears his name. At the earth's surface. A steady speed is an example of a state of equilibrium in which the net force is zero. a force which counterbalances the force of gravity and keeps the plane aloft. The air rushing past the wings produces lift. The outside force mentioned in Newton's First Law refers to an unbalanced force. Solution An aircraft flying at a constant velocity is in a state of equilibrium.
EXAMPLE 1
Describe the forces acting on an aircraft flying at constant velocity. Instead that force is now balanced by other forces such as air resistance and road friction. as weather balloons often do. You no longer feel any force once the car has reached a steady speed.
We refer to the quantietiesh and v as the horizontal and vertical components of the original force. a 1-kg object weighs approximately 9. Now consider that this force is the resultant of a horizontal force h and a vertical force v. what horizontal force is he exerting on the toboggan? Solution First draw a diagram showing the force and its direction. Since forces are vectors.
EXAMPLE 2
Jake is towing his friend on a toboggan. It is important to know the net effect of all these forces. In other words. h and v are perpendicular. 3 F O R C E A S A V E C TO R
137
. Sometimes a force acts on an object at an angle. the ropes pull one to each side. If Jake is pulling with a force of 70 N.
60 N 20° 20° 60 N
140° r
60
N
4 .
EXAMPLE 3
Jake and Maria are towing their friends on a toboggan. This single force is the resultant of all the forces. The gravitational force on a 1-kg object at the earth's surface is approximately 9.4
70
N
v
25° h
So the horizontal force is about 63. using a rope which makes an angle of 25º with the ground.8 N. Solution Make a diagram showing the forces. Since they are walking side by side. with the original 70 N force as the resultant. Find the force pulling the toboggan forward.object's mass and this acceleration. It is generally the case that several forces act on an object at once. because an object's state of motion is determined by this net force. the single force that has the same effect as all the forces acting together can be found by vector addition.8 N. so that only part of the force is affecting the motion of the object. we show the resultant r. the diagonal of the parallelogram. By completing the parallelogram. Each is exerting a horizontal force of 60 N.4 N. Now h 70 cos 25º 63. they each make an angle of 20º with the line of motion. We show this by forming a triangle.
r2 602 602 r 112. what should the magnitude of the force exerted by Jake at an angle of 20º be if the toboggan is to move straight forward without turning? According to the sine law. the direction of the resultant will be a little to the right of the axis of the toboggan. which keeps the towing force from accelerating the toboggan. the toboggan will travel in a circle. If these conditions remain unchanged. We could have solved this question by finding the component of each force along the direction of travel and adding the results.
60 N 30° 60 N
r
Maria
EXAMPLE 5
In Example 2. This is because there is a frictional force that is equal and opposite to the towing force. the toboggan is (probably) travelling at a constant speed. the angles made with the direction of travel would not have been equal. if Maria pulls with a force of 60 N at an angle of 30º.
EXAMPLE 4
In Example 2. This means that the toboggan will not travel forward in a straight line but will veer continually to the right.
sin 30º = sin 20º 60 F
Jake ? 20° 30° 60 N Maria F 20° 30° R 130° 60
F
88 N
138 C H A P T E R 4
. It exactly counterbalances the resultant. In Example 3.8
2(60)(60) cos 140º
The towing force is about 113 N. 1. In Example 2. The force that is equal in magnitude but opposite in direction to the resultant is called the equilibrant. 2. If the forces had not been equal. the force of friction is the equilibrant. indicating that there is no unbalanced force on it. what if Maria starts pulling at an angle of 30º instead of 20º? As the diagram shows.
draw the tension vectors parallel to the corresponding lines in the position diagram.Jake must pull with a force of 88 N. In the diagrams. with the guy wire at the steeper angle having the greater tension. In making the force diagram. the resultant will be greater than before:
sin 130º R sin 20º 60
R
134 N
As in Example 2 and the subsequent discussion. 3 F O R C E A S A V E C TO R
139
. Since Jake is pulling harder than before. Solution First draw the position diagram showing where the forces act.
24°
850 N
850 N T1 T2
T2 66° 58°
58°
66°
Position Diagram
T1 58° Force Diagram
Since all three angles in the force triangle are known. In this problem.
T1 sin 24º
850 sin 124º 850 sin 24º sin 124º
and and
T2 sin 32º
850 . find the tension in each of the wires. which is the equilibrant. the resultant of the two tensions must be 850 N to counterbalance the buoyant force of the balloon. and then how these angles are used to determine the angles inside the force triangle. and the two guy wires make angles of 58º and 66º with the horizontal. sin 124º 850 sin 32º sin 124º
Therefore T1
T2
417 N
543 N
The tensions in the guy wires are approximately 417 N and 543 N. If the buoyant force on the balloon is 850 N. make it a practice with force problems to look for ways to justify your numerical results and make them physically meaningful.
EXAMPLE 6
A large promotional balloon is tethered to the top of a building by two guy wires attached at points 20 m apart. observe step by step how the angles in the position diagram are first translated into the force diagram.
4 . the magnitudes of the tension vectors T1 and T2 can be calculated using the sine law.
by vector addition. whose sum. 72. however.9 N.8 N and Fv 90 sin(36)º 52. and 40 N act on it? Solution An object will be in a state of equilibrium when the resultant of all the forces acting on it is zero.8 N. when you push a lawn mower. an object cannot be in a state of equilibrium with the three given forces acting on it. For example. Solution a. you exert a force along the handle. is the original force. but in this case the magnitudes of the forces are such that 10 20 40. By the triangle inequality theorem. but the mower does not move into the ground along the line of the force. a. The components are the magnitudes of forces acting horizontally and vertically. which makes an angle of 36º with the ground. Often.EXAMPLE 7
Is it possible for an object to be in a state of equilibrium when forces of 10 N.
force
motion
vertical component
force
horizontal component
EXAMPLE 8
A lawn mower is pushed with a force of 90 N directed along the handle. b. Therefore. we assumed that an object is free to move in the direction of the force acting on it. The horizontal component of the force. how much of the force that you exert actually contributes to the motion? To answer this question. 52. Determine the horizontal and vertical components of the force on the mower. The vertical component of the force. we must resolve the force into horizontal and vertical components. the sum of any two sides must be greater than the third. This means that the three given force vectors must form a triangle. moves the lawnmower forward across the grass. Describe the physical consequences of each component of the pushing force.
140 C H A P T E R 4
. is in the same direction (down) as the force of gravity. The force diagram is a right triangle. 20 N. In the discussion of forces in the previous examples. The components are Fh 90 cos(36º) 72.9 N
Fv
Fh 36°
b. So. It moves horizontally. that is not the case.
3
Part A
Communication
1. forces of 7 N east and 12 N north c. a. Solution The force of gravity on the trunk is (20 kg) (9. 3 F O R C E A S A V E C TO R
141
. forces of 7 N east and 12 N west b. a. 200 N acting at an angle of 30º to the horizontal b.
Fp
15° 196 N position diagram
Fn 15° 196 N force diagram
The parallel component points down the slope of the ramp. 200 N. The magnitude of the force of friction is proportional to this component. forces of 6 N southeast and 8 N northwest
Knowledge/ Understanding
4 .8 m/s2) The parallel and perpendicular components are Fp 196 sin 15º and 51 N Fn 196 cos 15º 189 N
196 N acting down. It is opposed by the force of friction acting up the slope.EXAMPLE 9
A 20-kg trunk is resting on a ramp inclined at an angle of 15º. The perpendicular component presses the trunk against the ramp. Name some common household objects on which the force of gravity is approximately 2 N. Calculate the components of the force of gravity on the trunk that are parallel and perpendicular to the ramp. Find the resultant of each pair of forces acting on an object. Describe the physical consequences of each. 75 N acting at an angle of 51º to the vertical d. What is your weight in newtons? 2. 160 N acting at an angle of 71º to the horizontal c. It tends to cause the trunk to slide down the slope. forces of 6 N southwest and 8 N northwest d. 36 N acting vertically 3. 20 N. Find the horizontal and vertical components of each of the following forces.
Exercise 4.
A mass of 10 kg is suspended from a ceiling by two cords that make angles of 30º and 45º with the ceiling. 5 N c. 13 N
Knowledge/ Understanding
12.
142 C H A P T E R 4
. Two forces F1 and F2 act at right angles to each other. Find the magnitude of the resultant of the four forces shown in the given diagram. calculate the angle between the lines of action of the 5 N and 7 N forces 13. Find the magnitude and direction of the equilibrant of each of the following systems of forces. What is the tension in each rope? 14. 9. 7 N. 12 N. Find the magnitude and the direction (to the nearest degree) of the resultant of each of the following systems of forces. Express the magnitude and direction of F1 F2 in terms of F1 and F2. 14 N d. Two forces of equal magnitude act at 60º to each other.
10 N
9N 7N 5N
5. A man weighing 70 kg lies in a hammock whose ropes make angles of 20º and 25º with the horizontal. 7 N. forces of 3 N and 8 N acting at an angle of 60º to each other b. show how the forces must be arranged b. 10. Find the tension in each of the cords.Part B 4. 5 N. a. A force of 375 N pulls the wire down at a point 15 m from one end of the wire. State the tension in each part of the wire. 2 N. Which of the following sets of forces acting on an object could produce equilibrium? a. If their resultant has a magnitude of 30 N. 11. 27 N. Three forces of 5 N. forces of 15 N and 8 N acting at an angle of 130º to each other 7. 6. 13 N. and 8 N are applied to an object. If the object is in a state of equilibrium a. A steel wire 40 m long is suspended between two fixed points 20 m apart. 26 N. Is it easier to pull yourself up doing chin-ups when your hands are 60 cm apart or 120 cm apart? Explain your answer. 5 N. forces of 32 N and 48 N acting at an angle of 90º to each other b. 13 N b. forces of 16 N and 10 N acting at an angle of 10º to each other
Communication
8. a. find the magnitude of the equal forces.
what is the horizontal force moving the log? 18. find the components of the rotor force parallel to and perpendicular to the ground b. the force of friction that acts parallel to and up the ramp must have a magnitude of at least how many newtons? 22. If the handle of the roller makes an angle of 42º with the ground. A child with a mass of 35 kg is sitting on a swing attached to a tree branch by a rope 5 m in length. 3 F O R C E A S A V E C TO R
143
. v. find the tension in the wire and the compression in the steel brace. If the force of gravity on the sign is 850 N. what horizontal component of the force is causing the roller to move?
4 . The child is pulled back 1. What is the tension in the rope?
Application
20. An advertising sign is supported by a horizontal steel brace extending at right angles from the side of a building. What horizontal force will hold the child in this position? b.5 m measured horizontally. A tractor is towing a log using a cable inclined at an angle of 15º to the horizontal. A piece of luggage is on a conveyer belt that is inclined at an angle of 28º. a. If the helicopter flies with the rotor revolving about an axis tilted at an angle of 8º to the vertical a. explain the physical effect of each of these components 19. 16. explain the physical effect on the helicopter of each component of the rotor force 21. determine the components of the force of gravity parallel to and perpendicular to the conveyer belt b.15. If the tension in the cable is 1470 N. The main rotor of a helicopter produces a force of 55 kN. and by a wire attached to the building above the brace at an angle of 25º.
65° w = 12 y
25°
u =5 40°
x
v =9
17. and w. In order to keep a 250-kg crate from sliding down a ramp inclined at 25º. If the luggage has a mass of 20 kg a. A lawn roller with a mass of 50 kg is being pulled with a force of 320 N. Find the x-and y-components of each of the vectors u.
The smaller tug is 10º off the port bow and the larger tug is 20º off the starboard bow. Suppose that a weight of 400 N is suspended from a 200-cm length of string. If the magnitudes of these forces are 6 N. Three forces. respectively.
120 cm 100 cm 100 cm
400 N
b.)
10 N 15 N 6N
24. find the magnitude and direction of the resultant. Braided cotton string will break when the tension exceeds 300 N. each of which is perpendicular to the other two.Part C 23. when the weight is hung at the centre of the string. In what direction will the ship move? 25. a. Show that the string will support the weight. (State the angles that the resultant makes with the two larger forces. 15 N. and 10 N. act on an object. The larger tug pulls twice as hard as the smaller tug. Will the string break if the weight is 80 cm from one end of the string?
144 C H A P T E R 4
. Two tugs are towing a ship. the upper ends of which are tied to a horizontal rod at points 120 cm apart.
the resultant velocity determined from the addition of two velocity vectors is nevertheless a meaningful. When the direction of motion as well as its magnitude is important.
5 km/h 2 km/h v 400 m x
x 2
400 5
x
160
4 . Velocity vectors can be added. speed is a quantity having magnitude only. When you walk forward in the aisle of an aircraft in flight. the speed of a moving object is calculated by dividing the distance travelled by the travel time.8º
Therefore.
EXAMPLE 1
A canoeist who can paddle at a speed of 5 km/h in still water wishes to cross a river 400 m wide that has a current of 2 km/h. In any case. so it is classified as a scalar. The displacement triangle is similar to the vector triangle. 4 V E L O C I T Y A S A V E C TO R
145
. Speed is the magnitude of a velocity.4 km/h along a line at an angle of about 22º. So even though its heading is straight across the current. From the vector diagram. making your total velocity 502 km/hr. speed is defined more carefully as the rate of change of distance with time. determine the resultant velocity. the 2-km/hr velocity of your walk adds to the 500-km/hr velocity of the plane. Find the point on the opposite bank where the canoe touches. the correct term to use is velocity. Velocity is a vector quantity. v2 v (5)2 29 (2)2 and tan
2 5
400 m
x
current
Vector Diagram 5 km/h 2 km/h
5. In advanced work.4 — Velocity as a Vector
In elementary problems. When two velocities are not in the same direction. If he steers the canoe in a direction perpendicular to the current. physical quantity. Solution As the canoe moves through the water.4 km/h
21. its actual direction of motion is along a line angling downstream determined by the sum of the velocity vectors. the canoeist crosses the river at a speed of 5. it is carried sideways by the current.Section 4.
the canoeist must head upstream at an angle of about 24º.
146 C H A P T E R 4
. depending on the strength and direction of the wind. His crossing speed will be about 4. it can be represented by a vector whose direction is west 25º south. v2 v (5)2 21 (2)2 and sin ( )
2 5
Vector Diagram
5 km /h
2 km/h
4.
EXAMPLE 2
Suppose the canoeist of Example 1 had wished to travel straight across the river. it is the vector sum. which is perpendicular to the river bank. This time.2 min It takes the canoeist approximately 5. We could also find x using the angle . The time it takes to cross the river is calculated from t
river width crossing speed 0.087 h or 5.
400 m current
Solution In order to travel directly across the river. the canoeist must steer the canoe slightly upstream. This may be different in both magnitude and direction from the plane's ground speed. Determine the resultant ground velocity of the plane. Solution Since the wind is blowing from 25º north of east.4 21
(where the width is 0.6º
Therefore. Wind affects a plane's speed and direction much the same way that current affects a boat's. not the heading of the canoe. Determine the direction he must head and the time it will take him to cross the river.He touches the opposite bank at a point 160 m downstream from the point directly opposite his starting point. The airspeed of a plane is the plane's speed relative to the mass of air it is flying in.4 km) (we avoid using rounded values if possible)
0. From the vector diagram.6 km/h. but we must be careful not to round off in the process. This wind will blow the plane off its course. to travel straight across the river.2 minutes to cross the river.6 km/h
23.
EXAMPLE 3
An airplane heading northwest at 500 km/h encounters a wind of 120 km/h from 25º north of east.
110° 45° 65° 45° 25° A
A key step in solving problems such as that in Example 3 is to find an angle in the triangle formed by the vectors. two sides and the included angle are known. As they approach. Then draw the resultant velocity using the parallelogram law of vector addition.2º
The resultant velocity has direction 33º north of west and a magnitude of 553 km/h. 4 V E L O C I T Y A S A V E C TO R
147
.7
2(500)(120) cos 110º
500
v +w 110° A 120 O
Store this answer in your calculator memory. Then. so the magnitude of the resultant velocity can be calculated using the cosine law. ∠AOB can be calculated from the sine law. Vectors are needed to describe situations where two objects are moving relative to one another. Here is a helpful hint: identify which angle is formed by vectors whose directions are given. their relative velocity is zero.
sin ∠AOB 500
sin 110º v w
(use the value of v + w calculated above)
∠AOB ∠WOB
58. ∠COA 45º 25º 70º. On a set of directional axes.changing both its ground speed and its heading. in ∆OAB.
Plane Heading N B v v +w W 45° Wind direction S E W A w S O E C N
In parallelogram OCBA. and draw small axes at the vertex of that angle. draw the two velocity vectors. When astronauts want to dock the space shuttle with the international space station. they must match the velocities of the two craft. Let v be the airspeed of the plane and w be the wind speed. even though the two spacecraft are orbiting the earth at thousands of miles per hour. When they finally dock. Next.
4 . The diagram shows this alternate way to calculate that ∠OAB 110º in Example 3.
B
v w2 v w
5002 1202 552.2º 25º 33. so ∠OAB 110º. astronauts on each spacecraft can picture themselves to be stationary and the other craft to be moving.2º 58.
watching the two vehicles pass by.Relative velocity is the difference of two velocities. vrel2 vrel (96)2 (110)2 190. After the truck turns. a. when he perceives himself to be stationary.)
The angle of the relative velocity vector can be calculated from the sine law.4 km/h 2(96)(110) cos 135º
v truck v truck 135° v car v rel v car
(Store this in your calculator. It is what an observer measures. respectively.
sin 96 sin 135º 190.9º
148 C H A P T E R 4
. according to the driver of the car. Now what is the velocity of the truck relative to the car? Solution The vector diagram shows the velocity vectors of the car and the truck. These velocities are relative to someone standing by the side of the road. The truck turns onto a side road and heads northwest at the same speed. When two objects A and B have velocities vA and vB. b. the angle between the car and the truck velocities is 135º. vrel vtruck vcar ( 96) (110) 206 km/h or 206 km/h west
v truck v rel v truck
v car
v car
This is the velocity that the truck appears to have.4
20. the velocity of B relative to A is vrel vB vA
EXAMPLE 4
A car travelling east at 110 km/h passes a truck going in the opposite direction at 96 km/h. The magnitude of the sum is found using the cosine law. What is the velocity of the truck relative to the car? b. Since the car is going east. let its velocity be vcar 110. The principle that all velocities are relative was originally formulated by Einstein and became a cornerstone of his Theory of Relativity. Then the truck's velocity is vtruck 96.
1 km/h b. S 30º E d. Find a. respectively. the velocity of the taxi relative to the bus d. A streetcar. its velocity is 190 km/h in a direction W 21º N relative to the car. 4 V E L O C I T Y A S A V E C TO R
149
. and 50 km/h. How far downstream from the marina will the boat reach the other bank? b.
Exercise 4. a bus. It remains the same as long as the two vehicles continue to travel in the same directions without any changes in their velocities. A marina lies directly across the river on the opposite bank. The streetcar and the taxi are travelling north.
4 . S 80º W c.After the truck turns. A light plane is travelling at 175 km/h on a heading of N8º E in a 40-km/h wind from N80º E. How long will it take? 5.4
Part A
Communication
1. A plane is heading due east. How far will the plane travel in 3 h? b. and a taxi are travelling along a city street at speeds of 35. the bus is travelling south. a. An airplane is headed north with a constant velocity of 450 km/h. a. the velocity of the bus relative to the streetcar Part B 4. N 80º E
Knowledge/ Understanding
2. A motor boat that has a speed of 20 km/h in still water heads out from one bank perpendicular to the current. Will its ground speed be greater than or less than its airspeed. 42. the velocity of the streetcar relative to the bus c. Determine the plane's ground velocity. A man can swim 2 km/h in still water. A river is 2 km wide and flows at 6 km/h. Note that the relative velocity of the two vehicles does not depend on their position. and will its flight path be north or south of east when the wind is from a. N b. Find at what angle to the bank he must head if he wishes to swim directly across a river flowing at a speed of a. 4 km/h
Knowledge/ Understanding
3. The plane encounters a west wind blowing at 100 km/h. What is the direction of the plane?
Application
6. the velocity of the streetcar relative to the taxi b.
The time for the round trip when there is no wind. The time for the round trip when there is a constant wind blowing from Vancouver to Toronto. Determine the velocity of the truck relative to the car.Application
7.
150 C H A P T E R 4
. A car is 260 m north and a truck 170 m west of an intersection.5 m/s on a ship travelling north at 12 m/s. Determine which time is shorter. 8. If the destroyer has a top speed of 30 knots. and the truck from the west at 50 km/h. 13. A plane is steering east at a speed of 240 km/h. Calculate the horizontal and vertical components of its velocity at this moment. What is the speed of the sailor relative to the ocean floor? 14. Determine its velocity relative to the ground when there is a 2 m/s current from 40º east of north. An airplane flies from to Toronto to Vancouver and back. a. A pilot wishes to fly to an airfield S20º E of his present position. If the average airspeed of the plane is 520 km/h and the wind is from N80º E at 46 km/h. a jet raises its nose to an angle of 18º with the horizontal and begins to lift off the ground. what will the plane's ground speed be? 11. A sailor climbs a mast at 0. a. Upon reaching a speed of 215 km/h on the runway. b. b. What is the ground speed of the plane if the wind is from the northwest at 65 km/h? What is the plane's actual direction? 9. the car from the north at 80 km/h. in what direction should the pilot steer? b. while the current flows east at 3 m/s. a. at what heading should it travel to intercept the submarine? Part C 12. A boat heads 15º west of north with a water speed of 3 m/s. A destroyer detects a submarine 8 nautical miles due east travelling northeast at 20 knots. They are both approaching the intersection. What is the physical interpretation of each of these components of the jet's velocity? 10.
and if they are physically reasonable. if you want to go from A to C you can travel directly along the vector AC. or you can detour through B. on the right hand side of the equation. and then along BC. you can travel directly along DB. Pay attention to and become familiar with details such as these. You will be able to draw and interpret vector diagrams and handle vector equations more quickly and correctly if you do. It is not difficult to draw angles that are correct to within about 10º and to make lengths roughly proportional to the magnitudes of the vectors in a problem. Perhaps the most important mathematical skill to develop from this chapter is that of combining vectors through vector addition.
KEY CONCEPTS REVIEW
151
. and the detour point is the initial letter of the two vectors. travelling first backwards along AD. This means that AC AB BC. Once you have calculated answers. but observe how the detour point fits into the equation: it is the second letter of the first vector and the first letter of the second vector.
D C
A
B
DIFFERENCES
Using the same diagram. and then forwards along AB. This translates into the equation DB AD AB. Note carefully that. you have been introduced to the concept of a vector and have seen some applications of vectors. if you want to go from D to B. or you can detour through A. both graphically and algebraically. travelling first along AB. the order of the initial point D and the end point B are reversed. Diagrams drawn free hand are sufficient.Key Concepts Review
In this chapter. but try to make them realistic.
SUMS
Speaking informally. ask yourself if the calculated angles and magnitudes are consistent with your diagram. which of course is just the difference DB AB AD.
Using vector diagrams. and the angle between them is 55º 45 N. ai bj ? 6.
7. what is the magnitude of ˆ ˆ ˆ ˆ ˆ ˆ a.
REVIEW EXERCISE
153
. Two forces F1 and F2 act on an object. if u v
M
0. If v b. (a
Communication
b)u
au
bu
b. 24i 7j ? c. The supports A and B are not at the same level. F1 54 N. v 0 v c. A 3-kg mass is hanging from the end of a string. 4. If sv
t
v. A mass M is hung on a line between two supports A and B. what are s and t?
2. Explain these properties of the zero vector: a. and the angle between them is 140º
9. a. If a horizontal force of 12 N pulls the mass to the side a. find the angle the string makes with the vertical
Knowledge/ Understanding
8. show that a. Which part of the line supporting the mass has the greater tension? Explain. what is t? tu. if a and b have opposite directions. If i and j are perpendicular unit vectors. F2 21 N. what is t? v. Show that a b a b. 3i 4j ? b. F1 b. What effect does this have on the tension in the line? Explain. Two forces at an angle of 130º to each other act on an object. If tv c. F2 34 N. a. Determine their magnitudes if the resultant has a magnitude of 480 N and makes an angle of 55º with one of the forces. 0v 0 b.Review Exercise
Communication
1. and u is not parallel to v. then u
v
Knowledge/ Understanding
ˆ ˆ 5. b. find the tension in the string b. (ab)u
a(bu)
A B
3. Determine the magnitude of the resultant if a.
A pilot wishes to reach an airport 350 km from his present position at a heading of N 60º E. Forces of 5 N. There is a 80 km/h wind from the northeast. respectively. The pilot of an airplane that flies at 800 km/h wishes to travel to a city 800 km due east. all lying in the same plane. and then on to Nassau before returning to Miami. Calculate the wind speed. find a. What is the length of time required to cross the river if its width is 150 m? 15. Find the magnitude and direction of the resultant. and 12 N. what the ground speed of the plane will be c. what heading the pilot should steer b. How far downstream does she travel in 10 s? b. on a heading of E 14º N. Determine the displacement vector from Miami to Nassau. a. and the plane has an airspeed of 450 km/h. 2 N. Once a week the ship travels directly from Miami to Nassau. If au bv equal? 0 and u and v have different directions. 14. 12. Florida. The distance from Freeport to Nassau is 217 km on a heading of E 50º S. act on an object.
Application
11. She heads downstream at 30º to the current. which is flowing at 4 m/s.10. What should the plane's heading be? b. Find the tension in each string. An airplane heads due south with an air speed of 480 km/h. A 10-kg mass is supported by two strings of length 5 m and 7 m attached to two points in the ceiling 10 m apart. The 5 N and 2 N forces lie on opposite sides of the 12 N force at angles of 40º and 20º. The distance from Miami to Freeport is 173 km on a heading of E 20º N. Twice a week. Under what conditions
Application
19. How long will the trip take?
Thinking/Inquiry/ Problem Solving
13. a cruise ship carries vacationers from Miami. when its radar detects a tanker ahead at a distance of 9 nautical miles travelling with a relative velocity of 19 knots. how many minutes it will take for the plane to reach its destination
Thinking/Inquiry/ Problem Solving
16. Measurements made from the ground indicate that the plane's ground speed is 528 km/h at 15º east of south. What is the actual velocity of the tanker? 17. what must a and b v u v. Show geometrically that u does equality hold?
154 C H A P T E R 4
. If the wind is from S 25º E with a speed of 73 km/h. A camp counsellor leaves a dock paddling a canoe at 3 m/s. to Freeport in the Bahamas. a. A coast guard cutter is steering west at 12 knots. 18.
1997). is responding twice as vigorously as an east cell? 2. in turn. Consider an ensemble of 36 cells. Randy Gallistel. Michalels Gazzaniga. Each cell responds only to stimuli located within a cone of directions. One cell will represent 10º east of north. the next will represent 20º east of north. The vigour of a cell's response can be regarded as specifying the magnitude of a vector in the direction the cell represents.wrap-up
investigate and apply
CHAPTER 4: VECTORS AND THE SUPERIOR COLLICULUS
Brain cells in the superior colliculus are tuned to the directions of distant visual and auditory stimuli. a professor in the Department of Psychology at UCLA. whose research focus is in the cognitive neurosciences. representing directions evenly distributed around a circle. Ed. 3. Dr. which. a) Which cells will respond to a stimulus whose direction is north-east? b) A response pattern is a description of the relative proportions of the vigour of the various cells' responses. and so on." Investigate and Apply 1. Give two possible response patterns for the cells found in part a. How do you think the brain deals with the fact that several different response patterns can represent the same direction?
INDEPENDENT STUDY
Investigate the field of neuroscience. What other things can be represented in the brain using resultant vectors formed from cells representing individual vectors? What are some other questions to which neuroscientists are seeking answers? What role does mathematics play in the search for answers to these questions? ●
152 C H A P T E R 4
. with one cell representing north. A cell always responds to some extent whenever a stimulus is within 20º of the cell's direction. has suggested that these neurological resultant vectors are "the first new idea about how the nervous system represents the value of a variable since the beginning of the [twentieth] century (from Conservations in the Cognitive Neurosciences. MA: Bradford Books/MIT Press. The resultant vector formed by summing the vectors represented by the individual cells points in the direction of the stimulus. What direction would be represented by a north cell responding three times as vigorously as a north-east cell.
b.Chapter 4 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application Questions 2. a. A steel cable 14 m long is suspended between two fixed points 10 m apart horizontally. The cable supports a mass of 50 kg at a point 6 m from one end. Determine the tension in each part of the cable. v c. Simplify 3(4u
4. w a b
2 b 3
c
3c a 5c v) a 2u 3(u v). What is this
5. and c onto graph paper. 6.5 m/s. then accurately draw the following three vectors. Forces of 15 N and 11 N act a point at 125º to each other. b) 4a
a
b
3. A ferry boat crosses a river and arrives at a point on the opposite bank directly across from its starting point. Copy the three given vectors a. u b. 5 8 1 3. The boat can travel at 4 m/s and the current is 1. Illustrate in a diagram the vector property 4(a property called?
4b. 7
1. Find the magnitude of the resultant. 4. If the river is 650 m wide at the crossing point. Under what conditions is u
v
u
v?
2. 6. 7. in what direction must the boat steer and how long will it take to cross? 8.
we can use the so-called taxi-cab distance given by t(P. and so on. it makes sense to talk about planar figures such as triangles. from the given point (0. we normally define distance in Euclidean terms. Q) (x1 x2)2 (y1 y2)2
If we ask for the locus of all points that are a constant distance. For example. For example. the rules or assumptions we make are chosen to match our experience with the world we live in. y2) is d(P. 0). When we solve geometrical problems. then the distance between two points P(x1. One way to create a whole new geometry is to change the way we measure distance.N O N . y1) and Q(x2. Q) x1 x2 y1 y2
y
Q
y2 – y1
P
x2 – x1
R
x
156 C H A P T E R 4
. When we represent points and figures in terms of coordinates on the Cartesian plane.E U C L I D E A N G E O M E T RY
The word geometry comes from the Greek words for earth and measure. since locally the earth looks flat. we get the circle with equation x2 y2 1. say 1. circles. But what happens if we change the rules? For example.
y) is any point on the locus. The locus is plotted below.The taxi-cab distance between P and Q is the sum of the lengths PR and RQ. the right bisector of the line segment joining the two points.
y
(0. then the equation of the locus is x 0 y 0 1 or x y 1. With this definition of distance. In this case. it is easiest to break the problem into four cases depending on x and y being positive or negative. –1)
You can investigate many other locus problems in this new geometry. we get a straight line.
E X T E N D I N G A N D I N V E S T I G AT I N G
157
. find the set of points that are equidistant from (0. If we use Euclidean distance. Q) d(p. The graph can be produced by a graphing calculator or by hand. 0) (1. we can ask the same locus question. Q) for any pair of points P and Q. 1)
x (–1. Note that t(P. and turns out to be a square. 0)
(0. For example. What is the set of all points a taxicab distance of 1 from the origin? If P(x. 1). The reason for the colourful name is that it is the actual distance driven if a cab is restricted to a rectangular grid of streets. 0) and (1. The following diagram shows what happens with taxi-cab distance.
In this geometry. The line segment joining (0. try the following with taxi-cab distance: 1. 0) to (1. What happens to its length?
158 C H A P T E R 4
. 0) and (1. all points are equidistant from (0.
1. Find an equilateral triangle with taxi-cab side length 1. the right bisector is the line. Another example is to look at geometry on the surface of a sphere. 3. for x x 0. 2). 0)
For 0 x 1. as with Euclidean distance. straight lines (the shortest path between two points) become arcs of circles. 0) is rotated about the origin. 1). y
0 and
There are many other ways to generate non-Euclidean geometries. 1)
x (0. Are all angles equal? 2. y 1. For fun. However. 0) and (1.y
(1. Sketch the locus of all points that are equidistant from (0.
thanks to Descartes and analytic geometry.1 determine the intersection of a line and a plane in three-dimensional space.4.5 perform mathematical operations on Cartesian vectors. We will see the real power of vectors in this chapter. you will
• • • • •
determine equations of lines in two. Section 5. But what about a quantity that has more than three dimensions? In such cases.4. a line could be represented by a picture or by an equation. Section 5.Chapter 5
ALGEBRAIC VECTORS A N D A P P L I C AT I O N S
For quantities that have both magnitude and direction. Section 5. CHAPTER EXPECTATIONS In this chapter. A vector model allows you to add. 5. subtract. For example. the directed line segment or arrow is an excellent introductory method. The development of the vector model was made possible because. 5.and three-dimensional space. when we will use them to solve problems in the third dimension and beyond. many geometric ideas already had an algebraic counterpart.2 determine and interpret dot and cross products of geometric vectors.3. 5. Section 5. and multiply by a scalar vector. 5. Section 5. an algebraic vector model is required.3. 5.1.5
.1 represent Cartesian vectors. We can also use this model to multiply one vector by another vector.
5. Because there are three identical atoms surrounding the nitrogen atom. then the O hydrogen atoms are located symmetrically at –6 –4 –2 about (7.investigate
CHAPTER 5: MOLECULAR BOND ANGLES
Atoms bond together to form the molecules that make up the substances around us. the three are evenly spaced around a circle. For example. Formaldehyde (H2CO) is a trigonal planar molecule with O the carbon in the centre. The bond angle for each of the three bonds is. 5. the O-C-H bond angle or the H-C-H Formaldehyde bond angle? 3. It consists of four atoms in a plane: three oxygen atoms surrounding and individually bonding to a single nitrogen atom. six hydrogen atoms.5º. Can three-atom molecules always be studied in a plane? Can four-atom molecules always be studied in a plane? What about molecules with more than four atoms? ●
160 C H A P T E R 5
.88 = 180 – 2 tan–1 = 104. what is one way to assign planar coordinates to the atoms? 2. Ethyl alcohol and dimethyl ether are both formed from two carbon atoms. The –4 bond angle formed at the oxygen atom is –6 5. The properties of enzymes. the angle formed where two hydrogen atoms link to an oxygen atom to form water (H2O) is about 104. and one oxygen atom (C2H6O).5º. If the distance between the nitrogen atom and each oxygen atom in NO3– is 1. Which is likely to be smaller. Investigate A water molecule can be studied in a Cartesian plane.22 10 –10 metres. DISCUSSION QUESTIONS 1. If we allow each unit on the H 6 plane to represent 10 –11 metres and place 4 2 the oxygen atom at the origin. One aspect of molecular geometry that interests chemists is called the bond angle.59.88) and (–7. The bond between the carbon C and the oxygen is shorter than the bond between the H H carbon and either one of the hydrogen atoms. The geometry of molecules is a factor in determining many of the chemical properties of these substances. It is the angle between two bonds in a molecule. 360 3 = 120°.59. 7.59 Can you explain why this calculation is correct?
y
H
x 2 4 6
Nitrogen trioxide (NO3–) is an example of a trigonal planar molecule. depend upon precise fits between molecules with specific shapes.88). but they have very different chemical and physical attributes. protein molecules that speed up biochemical reactions. therefore.
Its endpoint will fall on some point P with coordinates (a. 1 A L G E B R A I C V E C TO R S
. or origin. A line is one-dimensional. The direction
b
u
P(a. which are its coordinates. A plane has two dimensions. and a is called the coordinate of P.
y
u
Let u be a vector in the plane. b) using the Pythagorean Theorem. we establish principles that allow the use of algebraic methods in the study of vectors.Section 5. b) x a 161
O
5 . Next.1 — Algebraic Vectors
In this chapter.
O origin u P a
The absolute value a is the magnitude of u. Move the vector until its initial point is at the origin. y But the same process leads to an algebraic representation P(a. and let the magnitude of a represent the distance from the origin to P. Any point P in the plane is identified by an ordered pair of real numbers (a. The coordinate a contains everything you need to describe the vector u. and every real number corresponds to one and only one point on the line. The axes are oriented so that a counter-clockwise rotation about the origin carries the positive x-axis into the positive y-axis. A line is a geometrical object. Its endpoint will fall on some point P with coordinate a. when Descartes introduced the concept of a coordinate system. We have now established the connection between the coordinates of a point and a geometrical vector on a line. b). This amounts to an algebraic representation of a geometrical vector. How? Let the sign of a indicate which side of the origin P is on. b) b of a vector in a plane. The result is known as the real number line. It is the first step in the development of algebraic methods to handle vector problems. How is a coordinate system O P for a line constructed? First. The correspondence between points on the line and real numbers is complete in this sense: each point on the line has a different real number as its coordinate. Now let u be a vector on this line. associate with each origin a point P on the line a real number a. choose an arbitrary point on the line as a reference point. The application of algebra to problems in geometry first became possible in 1637. b). Move u until its initial point is at the origin. and the sign of a tells you its direction. The magnitude of u can be determined from (a. The Cartesian coordinate system for a plane is constructed from two real number lines— x the x-axis and the y-axis—placed at right angles in the O a plane.
b) is referred to as an algebraic vector. Nothing else is needed. The context of a problem will tell you whether (a. Solution The point P( 3. Another notation commonly used to describe algebraic vectors in a plane employs unit vectors.and y-components of the vector. Therefore. Its direction makes an angle of approximately 113º with the positive x-axis. 0 360º. 7). 1). Define the vectors ˆ ˆ i (1. b) can be interpreted in two different ways: it can represent either a point with coordinates a and b. b).of u can be expressed in terms of the angle
between u and the positive x-axis.
We can observe that. The ordered pair (a. The formula above gives two values of . These are unit vectors that point in the direction of the positive x-axis and positive y-axis. It is important to remember that the ordered pair (a. the ordered pair (a. The magnitude and direction of OP are P y calculated as follows: OP2 OP ( 3)2 58 (7)2 tan
7 3
113º
x –3 O
Thus. express it in ordered pair notation. The position vector of P is OP ( 3.
EXAMPLE 1
The position vector of a point P is the vector OP from the origin to the point. b) lies. b) represents a point or a vector. the magnitude and direction of u are determined entirely by the coordinates of P. The actual value depends on the quadrant in which P(a. 7). respectively.
y
j i
x
162 C H A P T E R 5
. or a vector with components a and b. b) is a valid representation of the vector u. just as in the case of a line. where its magnitude u and direction are given by the equations u a2 b2 and tan
1 b
a
with measured counter-clockwise from the positive x-axis to the line of the vector. and determine its magnitude and direction. 7) is in the second quadrant. Any vector u in a plane can be written as an ordered pair (a. 0) and j (0. the magnitude of OP is 58. The values of a and b are the x. Draw the position vector of the point P( 3.
A plane in space that contains two of the coordinate axes is known as a coordinate plane.and y-axes. Some point in space is chosen as the origin. –2)
A coordinate system for three-dimensional space is formed in much the same way as a coordinate system for a two-dimensional plane. c). mathematicians use a right-handed coordinate system. 3)
y
R(0. is called the xy-plane.As you can see in the diagram. the position vector of point P(a. A point such as ( 4. b) or u OP ai bj
EXAMPLE 2
Express the position vector of each of the points shown in the diagram as an ordered pair and in unit vector notation. 7)
Solution OP (6. and j
y bj
j
P(a. Any algebraic vector can be written in either form: ˆ ˆ u OP (a. As a rule. can ˆ be expressed as the vector sum of scalar multiples of i ˆ. 1). The plane containing the x. 1 A L G E B R A I C V E C TO R S
163
. which has a y-coordinate of 0. 3) ˆ ˆ 3i 3j
OR
(0. If you could grasp the z-axis of a right-handed system with your right hand. There are two different ways to choose the positive directions of the axes. Through the origin. and the z-axis. ˆ 6i
x
2) ˆ 2j
OQ
( 3. three mutually perpendicular number lines are drawn. b). and thus any vector u in the plane. the y-axis. b) x ai
O i
Ordered pair notation and unit vector notation are equivalent. pointing your thumb in the direction of the positive z-axis. b. called the x-axis. Each point in space corresponds to an ordered triple of real numbers (a. 0.
Q(–3.
z
y O x
z
xz-plane
yz-plane
xy-plane
y
x
5 . your fingers should curl from the positive x-axis toward the positive y-axis. lies in the xz-plane. which are its coordinates on the three axes. 7) ˆ 7j
P(6. for instance. The other two coordinate planes are named similarly. A left-handed system would have the positive y-axis oriented in the opposite direction.
sketch the position vector OP in three dimensions. OP 3i 5j 4 k Solution a. c) P(a. –7. ˆ ˆ ˆ a. b. b. therefore. Any vector u in three-dimensional space can be written as an ordered triple. 2)
z
b.
P(–5. Its tip will then fall on some point P with coordinates (a. and then c units in the z direction. c). and k (0. b. u OP ai bj ck. Be sure each move is made along a line parallel to the corresponding axis. 0. 5. ˆ ˆ j (0.
z O 5 y
2 –7 O
–5 y x
3 –4 P(3. represents an algebraic vector in three dimensions. 0). then b units in the y direction. 1). 0. 1. b. Its magnitude is given by u a b c EXAMPLE 3 Locate the point P. c) a O b x x c y (a. P( 5. and calculate its magnitude. 0) O OP P (0. b. Alternatively. c). 0) y z
Just as in two dimensions. 0).
z (0. u OP (a. this ˆ ˆ ˆ ˆ vector could be expressed in terms of unit vectors i . Drawing a rectangular box will help you to see the three-dimensional aspect of such diagrams. j . 2) b. 7. from which its magnitude and direction can be determined. ˆ ˆ ˆ or in terms of unit vectors. –4)
x
OP
( 5)2 78
( 7)2
(2)2
OP
(3)2 50
(5)2
( 4)2
164 C H A P T E R 5
. b. The ordered triple (a.To plot a point P(a. c) in space. 0. move a units from the origin in the x direction. 0. any vector in space can be placed with its initial point at the origin. where i (1. c). 2 2 2. and k.
β.
34. γ 180°. the components a. cos β u b . y-. 1 A L G E B R A I C V E C TO R S
165
. Consequently. which is exactly (cos α. u
x α z
c γ
u β y
The other direction numbers and angles are related in the same way. b. b. β. called direction angles. The direction angles of a vector (a. u u u
It follows from this that the direction cosines. We can see the right triangle that relates u. the direction cosines are the components of a unit vector. and the direction angle γ. 5. u
Note that if you divide a vector (a. From any two of them you can find the third. cos β. In the given diagram. the direction angles are all acute angles. b . 34 3 . and z-axes. where 0° α . and cos γ u c . you create a unit vector with components cos2 α Thus.
EXAMPLE 4
Find the direction cosines and the direction angles of the vector u Solution The magnitude of u is (0)2 (5)2 ( 3)2 The direction cosines and angles are therefore cos α cos β cos γ
0 . from which it follows that cos γ
c .In two dimensions. c . In this context. In three dimensions. we can describe the direction of a vector by a single angle. we use three angles. respectively. b. where cos α
a .
z
3). and c of the vector u are referred to as direction numbers. and hence the direction angles.
121° 31° x
y
α β γ
90º 31º 121º
5 . and γ that the vector makes with the positive x-. 34 5 . 34
(0. c) are the angles α. cos2 β cos2 γ 1. β and γ .
The direction cosines of a vector are the cosines of the direction angles α.
a . are not all independent. cos γ). c) by its magnitude u. the direction number c.
In Example 1. these expressions must be equal. ˆ and j have different directions. or i . and k in three dimensions. The uniqueness of algebraic vectors leads to a fundamental statement about the equality of algebraic vectors. equivalently. A proof of uniqueness for vectors in three dimensions is more complicated and will be explored in Chapter 6. we establish the uniqueness of the algebraic representation of a vector in terms of these basis vectors. which means that the two representations of the vector u are not different after all.or three-dimensional spaces in which vectors exist.and y-components is unique.
170 C H A P T E R 5
. The only possible way the ˆ ˆ equation can be valid is if the coefficients of i and j are zero. They form a basis for the two. that is. which have been chosen to play this special role. ˆ ˆ ˆ a1i b1 j = a2i b2 j Some rearrangement produces the equations ˆ ˆ ˆ ˆ b1 j b2 j a1i – a2i (a1 ˆ a2)i ( b1 ˆ b2 )j ˆ ˆ The last equation states that a scalar multiple of i equals a scalar multiple of j . The unit vectors i multiplication by a scalar can make the vectors equal. and no ˆ But this cannot be true. all vectors can be expressed in terms of the unit vectors ˆ ˆ ˆ ˆ ˆ i and j in two dimensions. in ˆ ˆ ˆ terms of ordered pairs or triplets. Vectors such as i . and k.
EXAMPLE 1
Prove that the representation of a two-dimensional algebraic vector in terms of its x. or. are termed basis vectors.Section 5.2 — Operations with Algebraic Vectors
As we saw in Section 5. we begin by assuming that the vector u can be written in terms of components in two different ways: ˆ ˆ ˆ ˆ u a1i b1 j and u a2i b2 j Since they represent the same vector. Solution Using the method of proof by contradiction. j . j . Two algebraic vectors are equal if and only if their respective Cartesian components are equal. a1 a2 and b1 b2.1.
a. I(1. and a mass
174 C H A P T E R 5
. Find a point not on the y-axis that is equidistant from the points (2. 1. 9). A mass of 1 unit at (1. 1). 2. B(4. 1). 3). 0). 7). …. b. 16. 4) d. and (0. 3 . OA2. 0). OAn . for the points A(2. Find the point on the y-axis that is equidistant from the points (2. 1. OAn is the point with position vector OC
OA1 OA2 n … OAn
. 7). C( 2. 1) 1). a. The centroid of the n points with position vectors OA1. 0). …. …. K(0. A4(4. 2 b. a. 1). a mass of 3 units at ( 2. A mass of 2 units at (0. 0. 0). A2(1. 10). 4.Part C 15. 2). 0. 4. of 7 units at (1. Find the distance from A to the centroid of the triangle. and C(1. 4). 1). and (0. A1(3. Calculate the centre of mass in each case. A(1. 1)
17. is the point with position vector OG
m1OA1 m2OA2 … mnOAn . 1). 1) 19. 1). OA2. 3). K(0. 0). 1. 0. 2)
Find the centroid of each of the following sets of points. mn at the points with position vectors OA1. 0). J(0. and a mass of 1 unit at (11. 1. 18. Find the length of the median AM in the triangle ABC. 3. respectively. C(0. a. c. 2) to Y(1. m2. m1 m2 … mn
In some kinds of problems. J(0. 0. The centre of mass of the masses m1. 1) 1. b.
Thinking/Inquiry/ Problem Solving
1. 3. 0). a mass of 3 units at (4. A3(7. 0. B(3. for the purposes of calculation. a collection of masses can be replaced by a single large mass M m1 m2 … mn located at the centre of mass. Find the components of the unit vector with direction opposite to that of the vector from X(7. b. 0. 4. I(1. a mass of 5 units at ( 1. 2).
u Solution a. The dot product of two vectors is one of these combinations.
Proof The condition that u • v 0 is sufficient.
b. u 24. Other. because if then or u•v 0 0 uv cos cos
0. (since the vectors are non-zero)
5 . u • v uv cos 60º (7)(12)(0. more sophisticated combinations of vectors are required. u • v
EXAMPLE 2
Prove that two non-zero vectors u and v are perpendicular.5) 42 c. 60º c. u • v uv cos 34º (24)(9)(0. The dot product of two vectors u and v is u•v where uv cos . the dot product is also called the scalar product. where angle between the vectors. Nothing else is needed to guarantee that the vectors are perpendicular. u 7. 531
Certain applications of vectors in physics and geometry cannot be handled by the operatons of vector addition and scalar multiplication alone.1 b. a. if and only if u • v
0.Section 5. the dot product of two vectors is a scalar. For this reason. v 9. 3 T H E D OT P R O D U C T O F T W O V E C TO R S
175
.3 — The Dot Product of Two Vectors
t chnology
APPENDIX P. v 12.
Since the quantity uv cos on the right is the product of three scalars. v 34º uv cos 56 (20)(3) 30 3
3 2
e
is the
5 6
3. 20. is the angle between the two vectors.
EXAMPLE 1
Find the dot product of u and v in each of the following cases.8290) 179.
b. this does not necessarily mean that b equals c. Find the angles between the body diagonals of a cube. (This is known as the Cauchy-Schwarz Inequality. by drawing a geometrical diagram 22. A body diagonal of a cube is a line through the centre joining opposite vertices. y z 0. which satisfy the equation u v u v. if x 2. Calculate the value of 3. If a and b are perpendicular.
Thinking/Inquiry/ Problem Solving
25. and a b c. What is the usual name of this result? Part C
Communication
21. a. and a b. Find a unit vector that is parallel to the xy-plane and perpendicular to the ˆ ˆ ˆ vector 4i 3j k.
24. When does equality hold? Express this inequality in terms of components for vectors in two dimensions and for vectors in three dimensions. a. 23.
b. If a and b are not perpendicular. and z 4. y. b. Three vectors x. show that a2 What is the usual name of this result?
b2
a
b2. express c2 in terms of a and b. How is the figure defined by u and v related to the figure defined by a and b of part a? 26.20. If the dot product of a and b is equal to the dot product of a and c. Show why this is so a. Use the dot product to show that two vectors.)
180 C H A P T E R 5
. Under what conditions is (a b) • (a b) 0? Give a geometrical interpretation of the vectors a. a b. by making an algebraic argument b. Prove that a • b ab. must be perpendicular. and z satisfy x y x • y y • z z • x.
The cross product of a and b is chosen to be the vector of this form that has k 1. for some constant k. Let us find all the vectors v (x. a1x a2y a3z 0 b1x b2y b3z 0
a b
b a
1 2
We solve these two equations for x. Hence. b2. a3b2) and z k(a1b2 – a2b1).4 — The Cross Product of Two Vectors
We have already defined the dot product of two vectors. The cross product is useful in many geometric and physical problems in three-dimensional space. Multiply equation 1 by b3. These vectors satisfy both a • v 0 and b • v 0. The cross product of two vectors a and b is a vector that is perpendicular to both a and b. which gives a scalar quantity. and equation 2 by a3 to obtain 3 a1b3x a2b3y a3b3z 0 a3b1x a3b2y a3b3z 0
4
Now eliminate z by subtracting equation 3 from equation 4 to obtain (a3b1
x a2b3 a3b2
a1b3)x
(a3b2
y a1b3
z
a2b3)y
0
This is equivalent to
a3b1
Using a similar procedure. Hence. this cross product is defined only in three-dimensional space. 4 T H E C R O S S P R O D U C T O F T W O V E C TO R S
181
. In this section we introduce a new product called the cross product or vector product. y. y. a3) and b (b1. a2. a1b2 a2b1). and z. we eliminate x from the original equations to obtain
y a3b1 a1b3
a1b2 a2b1
Let y Then x
k(a3b1 k(a2b3
a1b3). it is used to help define torque and angular velocity in statics and dynamics. and it is also used in electromagnetic theory. z) that are perpendicular to both a and b.
Then the vector v perpendicular to both a and b is of the form v (x. a3b1 a1b3.
5 . z) k(a2b3 a3b2. We will use it to find vectors perpendicular to two given vectors. y. b3) be two vectors in three-dimensional space.Section 5. Let a (a1.
3. 18). there is an easy check that should always be done after calculating any cross product. for a k R. 2) • (33. 2(4) 1(7). In our example. If v a b. 7) (3(7) 2( 6). This is one of the most useful properties of the cross product. 2) and (4.
Hence. 1( 6) 3(4)) (33. 6. (33. one vector perpendicular to (1. 18) Hence. 1. 6. From the definition of the cross product.
The definition of cross product is motivated by the mechanical act of turning a
182 C H A P T E R 5
. b2. It can be expressed as follows: a2 a3 a3 a1 a1 a2 . 3. (4.The Cross Product or Vector Product of a b (b1. a b b2 b3 b3 b1 b1 b2
(
)
bc
where a b c d
ad
We showed above that any vector perpendicular to both the vectors a and b can be written as k(a b).
Solution The cross product will be one such vector.
18) is perpendicular to both (1.
Hint: It is very easy to make errors in calculating a cross product. . (1. 3. 1. then every vector perpendicular to both a and b is of the form k(a b). 7) is (33. a2. a3) and a2b1)
This is a rather complicated expression to remember. 1. 1. 7) • (33. a3b1 a1b3. Finding a Vector Perpendicular to Two Vectors If a and b are two non-collinear vectors in three-dimensional space. 6. 2) (4. 2) and (4.
EXAMPLE 1
Find a vector perpendicular to both (1. 1. you can always check that a • v = 0 and b • v 0.
6. a1b2
(a1. 2) and (4. 7). 7). b3) is the vector a b (a2b3 a3b2. However. 3. (1. 3. 18) 18) 1(33) 0 4(33) 0 3(1) 6(1) 2( 18) 7( 18) 6.
This gives a definition for the magnitude of the cross-product vector. b2. is the angle between the vectors. Hence. a2. a3b1 a1b3. a2b1) a2b1)2
We will prove that a Let a Then a
(a1. If. b.
5 .
a b
There are two vectors perpendicular to a and b with the same magnitude but opposite in direction. and a b form a right-handed system. a bolt with a right-hand thread is turned clockwise.bolt with a wrench. we have the following geometric description of the cross product. b. for instance. and a b form a right-handed system. a process which involves vectors pointing in three different directions. it moves down along the axis of rotation in a direction perpendicular to both the wrench handle and the turning force.
b a
The cross product of the vectors a and b in three-dimensional space is the vector whose magnitude is ab sin and whose direction is perpendicular to a and b. 4 T H E C R O S S P R O D U C T O F T W O V E C TO R S
183
. 0 b = ab sin . a3) and b (b1. and so a
b
ab sin . such that a.
axis of rotation
force
direction of travel
The magnitude of the cross product of two vectors a and b is a where b ab sin 180º. a1b2 b2 (a2b3 a3b2)2 (a3b1 (a2b2)2 a32)(b12 (a • b)2 (ab cos )2 cos2 ) a1b3)2 (a1b2
By adding and subtracting (a1b1)2 rewritten as a b2 (a12 a22 a2b2 a2b2 a2b2 (1 a2b2 sin2 Since 0 180º. b3) so that a b (a2b3 a3b2. The choice of direction of the cross product is such that a. sin
(a3b3)2 to the right side this can be b22 b32) (a1b1 a2b2 a3b3)2
0.
0. Use the cross product to find a vector perpendicular to each of the following pairs of vectors. a c) b. and w • (au
bv) are all zero. 6) 2) d. 2. (a c) b) (b • c) c) c) (b (b • c) c. 0) and (0. Find u v for each of the following pairs of vectors. a j. because it is a scalar quantity. or meaningless. i j k. For example. 4) c. you may form the dot product or the cross product of this vector with a third vector. (a • b) e. a. explain why w • u. (a • b)
Knowledge/ Understanding
4.
185
b. (a i. 1). (Why?) The quantity (u v) w is a vector and is called the triple vector product. Both of these quantities arise in the application of vectors to physical and geometrical problems. Notice that the first property means that the cross product is not commutative. (1. a • (b • c)
k. Find a unit vector perpendicular to a (4.
v =5 68° u = 12
b. (0. scalars. The quantity (u v) • w is known as the triple scalar product of three vectors.These properties can be checked by using the definition of the cross product. 1) and (6.
Exercise 5. (a f. Some of their properties are investigated in the exercises. 3. 2. 4. a. (a h.
5 . State whether the following expressions are vectors. but j i k.
u = 18 120° v = 25
c. w • v.4
Part A
Communication Knowledge/ Understanding
1. 4 T H E C R O S S P R O D U C T O F T W O V E C TO R S
. Check your answer using the dot product. 0)
1. (a • b)
l.
12° v =3 u =5
Communication
3. 3) and (1. Part B 5. 5) and ( 4. (a b) • c b) b) c c c (b • c) (b b ) • (b
g. 0. because u (v • w) is meaningless. v
2. 1) and b (2. If w
u
v. 9. Since the result of a cross product is a vector. State whether u is directed into or out of the page. 0. Brackets are required in this expression to specify the order of operations. (2. a • (b d. 3. The brackets are not really needed to specify the order of operations. a. (4.
and c (4. 3) and ( 2. b ( 1. 4. z). k j i. This means that. Find two vectors perpendicular to both (3.
1). 1). and k as ordered triples and show that ˆ ˆ ˆ ˆ ˆ ˆ a. b
(2. Find another vector v for which a
c. 0. b e. a d. By choosing u v. Express the unit vectors i . a. b.
ˆ ˆ ˆ 7. If the cross product of a and b is equal to the cross product of a and c. 1. Part C
Communication
14. find y and z.
b. v. Show why this is so a.
Application
w. if u and v are collinear. 2). b = (a • a)(b • b) (a • b)2. v b. c f. Explain why there are infinitely many vectors v for which a
186 C H A P T E R 5
. a. y. 3). (b c•a c) a c. 0.
6. calculate the following triple scalar and triple vector products. i j k. and (a b) a
12. and a v b. (c v) a•b a) b
11.
Thinking/Inquiry Problem Solving
15. Prove that a
10. show that a. this does not necessarily mean that b equals c. u v v v u for any vectors u and v. by making an algebraic argument. Given a (2. If a
(1. a are mutually perpendicular. show that u (v w) (u in general. Using components.
9. 8. the cross product is not associative. 1. b. 3. by drawing a geometrical diagram. 1. (a b•c b) c b. j . u b. Given two non-collinear vectors a and b.
v
b. b. Prove that the triple scalar product of the vectors u.6.
Thinking/Inquiry Problem Solving
13. show that a. v
( 3. and w has the property that u • (v w) (u v) • w. 5). Carry out the proof by expressing both sides of the equation in terms of components of the vectors. 0).
Therefore. in certain circumstances.5 — Applications of Dot and Cross Products
In this section. There is no special symbol for a projection.
5 .
3) onto v
(1. a projection is formed by dropping a perpendicular from each of the points of an object onto a line or plane. Projections Mathematically. The shadow of an object. the projection of u onto v is the vector ON.
EXAMPLE 1
Find the projection of u
(5.Section 5. the direction of ON is the same as the direction of v when is acute. we will apply the dot product and the cross product to problems in geometry and physics.
ucos v v uvcos v u • v v
A u O A u N B
The magnitude of ON is given by ucos
N
O
v
B
As you can see from the given diagrams. 4. In this text. The sign of cos in the dot product takes care of both possibilities. The projection of one vector onto another can be pictured as follows. and opposite to v when is obtuse. 5 A P P L I C AT I O N S O F D OT A N D C R O S S P R O D U C T S
187
. The projection of u onto v is Proj(u onto v) Its magnitude is
(u • v)v) v2 u • v v
. we use the notation ON Proj(u onto v). is a physical example of a projection. where u OA and v OB. In the given diagram. . 5). 6.
Proj(u onto v)
Area of a Parallelogram
u h The area of a parallelogram is the product of its base and its height: A bh. 3). 4. 1) (0.
188 C H A P T E R 5
A
. 2) 3. R(7. 2 2
b a c h b c
A parallelepiped is a box-like solid.
The area of a parallelogram having u and v as sides is A u v. 4. 3.
EXAMPLE 2
Find the area of the triangle with vertices P(7. PR Cross product Magnitude ( 5. 3 3 3
14
Therefore. and
Solution Start by finding the vectors that form two sides of this triangle. which you will recognize to be the magnitude of the cross product of the two vectors u and v that make up the sides of the parallelogram.Solution First. The base of the parallelogram v in the given diagram is v and its height is equal to usin . we calculate u • v and v2. 1) (0. 2. b. PQ PR (2. 6. the opposite faces of which are parallel and congruent parallelograms. . 2) ( 5)2 ( 4)2
4. 1. and c. Its edges are three non-coplanar vectors a. Its area is therefore A vusin . The area of the triangle is half the area of the parallelogram having these vectors as sides.
5). 2) (2)2 ( 5. 45 4.
6). 2)
The area of the triangle is therefore Volume of a Parallelepiped
45 3 5 or .
1. u•v v2 (5. 5) 52 42
(u • v)v v2 14(1. 3. 5) 42 1 4 5 . 1. 12 42 3) • (1. Q(9. Vectors PQ = (2.
which is in the direction of b c: h Proj(a onto b
a • (b c) b c
c)
The volume is therefore V Ah b c
a • (b c) b c
a • (b
c)
In other words. because the force displaces the object from a higher to a lower position. like that of a cylinder. The work done by a force is defined as the dot product F•d F d cos where F is the force acting on an object d is the displacement caused by the force and is the angle between the force and displacement vectors. and t is a constant. Since the volume is a constant. you are doing no work.The volume V of a parallelepiped. this result illustrates an important property of the triple scalar product. independent of which face is chosen as the base. the word work has a much narrower meaning: work is done whenever a force acting on an object causes a displacement of the object from one position to another. the volume of the parallelepiped is the magnitude of the triple scalar product of the three vectors that make up its edges. The area of the base is the area of the parallelogram determined by the vectors b and c: A b c
The height is the magnitude of the projection of a onto the normal to the base. which is measured along a line perpendicular to the base. if you do not move. the word work is applied to any form of activity that requires physical exertion or mental effort. Work In everyday life. work is done by the force of gravity when an object falls. If a • b c t. While it might seem like hard work to hold a heavy object. The unit of work is a joule (J). For instance. In physics. then b • c a c • a b t. 5 A P P L I C AT I O N S O F D OT A N D C R O S S P R O D U C T S
189
. and a • c b c • b a b • a c t. W Work is a scalar quantity.
5 . is the area of the base A times the height h.
Then Fa
245 cos 72º cos 22º
72° Fg
Fa
22° 18°
81. The maximum turning effect occurs when the force is perpendicular to the lever arm.) Solution a. The distance along the shaft of the pedal from the axis of rotation to the point at which the force is applied is known as the lever arm. The applied force acting up the ramp is Fa cos 22º. (Ignore friction. Determine the work done by the force of gravity as the box slides to the bottom of the ramp.7 The force must exceed 81. for example. This turning effect of a force is called torque. b.EXAMPLE 3
a. the force causes an angular rather than a linear displacement. acting at an angle of 40º to the horizontal.7 N. The force exerted by a cyclist on a bicycle pedal. required to slide the box back up the ramp. a force causes an object to turn. that is. Torque Sometimes. A 25-kg box is located 8 m up a ramp inclined at an angle of 18º to the horizontal. The force of gravity is Fg (25 kg)(9. instead of causing a change in position.
axis of rotation applied force
lever arm
190 C H A P T E R 5
.8 m/s2) 245 N. The angle between the displacement down the ramp and the force of gravity is the difference 90º 18º 72º. The gravitational force acting down the ramp is 245 cos 72º. the work done by gravity would be W Fg • d (245)(8) cos 72º 606 J b. Determine the minimum force. turns the pedal about an axis. Therefore.
the magnitude of the torque is T rF sin (0. is 100º.The torque caused by a force is defined as the cross product T r F ˆ rF sin n
where F is the applied force r is the vector determined by the lever arm acting from the axis of rotation is the angle between the force and the lever arm ˆ and n is a unit vector perpendicular to both r and F Torque is a vector quantity.16)(115)sin 100º 18. It is measured in units of newton metres (N-m). In this case.
15° F
r 25°
Solution As with any problem involving forces. if the shaft of the pedal is 16 cm long. Therefore. that means placing the vectors tail to tail and determining the angle between them. as you can see. This angle. the first step is to change the picture showing where the forces act into a vector diagram. as determined by the righthand rule. 5 A P P L I C AT I O N S O F D OT A N D C R O S S P R O D U C T S
191
.1 N-m
r 15° F r 25° picture 15° F vector diagram 25° 75°
The direction of the torque vector is into the page.
5 .
EXAMPLE 4
Find the torque produced by a cyclist exerting a force of 115 N on a pedal in the position shown in the diagram.
a. 1) to Q(5. At the top of the ramp. A 35-kg trunk is dragged 10 m up a ramp inclined at an angle of 12º to the horizontal by a force of 90 N applied at an angle of 20º to the ramp. d 2i
16. What is the maximum torque that can be exerted by a 50-N force on this wrench and how can it be achieved? Part C 19. How much work is done by gravity in causing a 30-kg rock to tumble 40 m down a slope at an angle of 52º to the vertical? 12. For each of the following.
Application
18. acting in the direction of the vector (1. Find the total work done. 5) to B(3. 5 A P P L I C AT I O N S O F D OT A N D C R O S S P R O D U C T S
193
. Under what circumstances is a. What is the torque on a bolt at the other end of the wrench? b. 1). Proj(u onto v) Proj(v onto u)? Proj(v onto u)?
5 . moves an object from P( 2. How much work is done against gravity by a workman carrying a 8-kg sheet of plywood up a 3-m ramp inclined at an angle of 20º to the horizontal? 14. A pedicab is pulled a distance of 300 m by a force of 110 N applied at an angle of 6º to the roadway. 17. 1. d ˆ 3i ˆ 10j ˆ 8j ˆ 4k (800. 6). ˆ ˆ ˆ a. which moves an object from A(2. d (20. F ˆ 4i ˆ j. Find the work done by a 30-N force acting in the direction of the vector ( 2. 5). 15. find the work done by a force F that causes a displacement d. 13. How much work is done in sliding a refrigerator 1. d 5i 6j b. 1. If a 10-N force. Proj(u onto v) b. 2). F 2i .10. 600). calculate the work done. The distance is in metres. the trunk is dragged horizontally another 15 m by the same force. Calculate the work done. F d. F c. The distance is in metres. 50) ˆ ˆ ˆ ˆ 12i 5j 6k. A 50-N force is applied to the end of a 20-cm wrench and makes an angle of 30º with the handle of the wrench. 1.5 m across a kitchen floor against a frictional force of 150 N?
Application
11.
REVIEW EXERCISE
195
. For any vectors a. onto each of the coordinate planes 14. 0). 3.c. 3. 5. 1. and ( 1. Find the coordinates of the four vertices. a. 0. 1. 1. onto each of the coordinate axes b. 1). 13. Find the volume of the tetrahedron with vertices (1. 6). one vertex at (0. (a b) b) c lies in the plane of a and b c (a • c)b – (b • c)a 4. 2). ( 7. (a b. b and c. (3. Find the projection of u (17. How far is the centroid from each vertex? 16. c. and one vertex. Determine the fourth vertex needed to complete a rectangle. 8)
a. with a positive x-coordinate. A regular tetrahedron has one vertex at the origin. on the xy-plane. Calculate the perimeter of triangle ABC. d. show that a. Use the cross product to find the area of the triangle whose vertices all lie in the xy-plane at coordinates A( 7.
17. 6. b. B(3. 0). 0) and C(2. 0). Find the coordinates of the centroid of the tetrahedron. 8). 15.
The carbon atom is located at the centre of the regular tetrahedron.
H
C H H Methane
H
One way to define a regular tetrahedron in three-dimensional space is to connect the four vertices at (0. The resulting shape for the four hydrogen atoms is called a regular tetrahedron. GeH4. Use dot product methods to find the angle formed between any two vectors extending from the centre of the regular tetrahedron to two of its vertices. and they are distributed evenly in three dimensions to be as far apart as possible.wrap-up
CHAPTER 5: MOLECULAR BOND ANGLES
investigate and apply
The geometry of a molecule is one factor in determining its properties. They theorized that bonds tend to keep as far apart as possible. and it is equidistant from each vertex. Other regular tetrahedral molecules include SiH4. 1). 4. 0. 1. 1. 2. but they all have the hydrogen atoms evenly distributed. Find the centre of the given regular tetrahedron. involved in the formation of chemical bonds. 3. Investigate and Apply Methane (CH4) consists of four hydrogen atoms bonded to a single carbon atom. therefore. Powell. and any other regular tetrahedral molecule?
INDEPENDENT STUDY
What are the bond angles in tetrahedral molecules such as CH3Cl. M. and (0. and SnH4. (1. The hydrogen atoms are all 1. 0. Why is your answer to question 4 the bond angle in CH4. 1). The term valence electrons refers to those electrons that are most weakly bound to an atom and are. (1. 5. whose shapes are not regular tetrahedrons? Explain the differences. 1. 0). Bond angles are one quantitative aspect of molecular geometry. V. Draw a three-dimensional coordinate system and then draw the vertices and edges of the given regular tetrahedron. SnH4. SiH4. Sidgwick and H. GeH4. and BrO3F. They are different sizes.095 10 11 metres from the carbon atom. What other methods do chemists use to determine bond angles? What are other quantitative aspects of molecules that chemists measure and use? ●
196 C H A P T E R 5
. A theory about the relationship between valence electrons and angles of chemical bonds was proposed in 1939 by N. 0). CH3Br. Hint: Its three coordinates are all equal. Verify that the four points listed are all equidistant from each other.
a. 5)
c. u 2. the position vector OP of the point P(3. Determine the magnitudes of the projections in a. a. (u ˆ k. 3). u v 0 v ˆ 6i 3v uv ˆ 3j b. a unit vector perpendicular to both u and v
3. 1. the area
5. A box is dragged 16 m across a level floor by a 75-N force at an angle of 35º to the floor. 2. In what direction should the force act to produce the maximum torque? (Draw a diagram. B(2. 4. Determine the total work done by the force. (u uv v) • u ˆ 3i 0 ˆ 4j c. 4u c. 6. u • v d. find v v) 0 u 0
ˆ 2k and v b. 3). What can you conclude about the vectors u and v if a. A force of 50 N acts at the end of a wrench 18 cm long. ii. y-. A parallelogram ABCD has vertices A( 1. the projection of OP onto the xy-plane b. 1. the projection of OP onto the z-axis iii. D( 3. and 2. u • v e. parts ii and iii. 3 7 1 4–6
1. It is then dragged by the same force 8 m up a ramp inclined at an angle of 20º to the floor. Draw x-. u f. the coordinates of C b. and z-axes and make a sketch of i. the angle at A 1). Given u a.Chapter 5 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application
Questions 2. Determine a. u • v
d.)
CHAPTER 5 TEST
197
.
What is the maximum torque? (State both magnitude and direction. At what angle will the force produce half the maximum torque? Indicate this angle on your diagram.b. 7. Use the dot product to find an expression for the cosine of the acute angle between the diagonals of a rectangle with sides a and b.
198 C H A P T E R 5
.) c.
because they correspond to current on (1) or current off (0). Hence. In a room with light switches at both ends of the room. the number 2 to represent B. Using strings of length 5 allows for the creation of 32 strings: 00000. and so on. you will note that this is a simple assignment of the number 1 in a five-digit display to represent A. and so on. unless one of the unused strings is designated for the purpose. surprisingly. The only surprise in the addition process is 1 1 0. Satisfy yourself that there are 32 length-5 strings. A vector of length four might be 0010. the addition works easily in electronic form..1 string vector.. Such vectors. The first is that adding five-digit strings will always give a five-digit result. This provides the foundation of cryptography. 0 0 0 1 0 1 1 1 0 1 1 0
b. The first advantage gained is that this allows the simple transmission of messages. 11111. This problem is. are of great value in computer communication. or 1010. They are not very secure if privacy is desired. We define addition as follows: a. Such vectors are composed of 0s and 1s in a row.V E C T O R S I N C O M M U N I C AT I O N
You have seen examples of geometric and algebraic vectors. the message This is clever can be transmitted as 101000100001001100110100110011000110110000101101100010110010. the art of secret messages. 00001. . The second is due to electronic properties. or 1111. There is no "carrying" from column to column. there is no punctuation.. Note that it is up to the receiver to create words from a string of letters and that. A third common type of vector is the 0 . The weakness of this system is that messages such as this are easily intercepted. Using this system. Using 26 of these we can define the alphabet as follows: A: 00001 G: 00111 M: 01101 B: 00010 H: 01000 N: 01110 C
E X T E N D I N G A N D I N V E S T I G AT I N G
199
. the lights are off if both switches are off (0 position) OR if both switches are on (1 position). easily overcome by defining an arithmetic of addition (and subtraction) which can effectively hide the message. 00011. There are two reasons for this.
The person knowing the key will be able to decrypt the message.Using this definition. by using vectors of length 6. A person receiving the message and knowing the key merely reverses the process. For interest. 0 1 0 0 0. 1 1. Discussion You can increase the complexity of the deciphering.
200 C H A P T E R 5
. we illustrate addition: 10010 01101 11011 01001 01001 00100 In the first example. messages can be encrypted and interceptors cannot decode the transmitted message. or 8. as follows: 10000111000100110101 11101111011110111101 01101000011010001000 Note that subtraction is identical with addition. 1
Key
The message is retrieved! You can easily create messages in code. 10010 (R) now looks as though it is 01001 (I). In the next chapter you will see how seven-digit vectors constructed using algebraic vector properties can be used to build codes that can be corrected when errors occur. 2. 7. using the key 11101 for each letter. the word MATH becomes: Key 01101000011010001000 Add 11101111011110111101 10000111000100110101
(Send this)
An interceptor would translate this to P?IU and be confused. By adding a five-digit key. Discuss the effect of doing so. and a person intercepting the message will have a difficult time in determining the true message. work with two friends and try the following: 1. It will be a real challenge for the remaining person to do so. Encrypt a message and give it to your friends. 0 1 1 0. For example. 3. Choose a key known to two but not the third.
6.4
.2. we will answer these questions as well as look at the use of vectors as another tool in the world of proofs. 6. Section 6. Section 6. but suppose we move to fourdimensional space or beyond. Section 6. Section 6. you will
• • • • • •
perform mathematical operations on geometric vectors. parallel to the first but with a different magnitude. 6. what does "lie in the same plane" mean in four-dimensional space? In this chapter. CHAPTER EXPECTATIONS In this chapter.or threedimensional space.1.3 represent Cartesian vectors.1 determine the components and projection of a geometric vector.3 perform mathematical operations on Cartesian vectors.Chapter 6
LINEAR C O M B I N AT I O N S
We have seen that if a vector v is multiplied by a scalar k.2. Section 6.and three-dimensional space. again.4 prove some properties of plane figures using vector methods. Section 6. Does the concept parallel mean anything? In two.or threedimensional space we can add two non-parallel vectors a and b and form a new vector c a b. This new vector c lies in the same plane as a and b but.3 determine equations of lines in two. the result is a new vector kv. This property is maintained in two.
It is possible to turn things around and write r
A
or t
d . what are the implications? This chapter begins by examining the dependence and independence of vectors and ends with applications of vectors to geometrical proofs. we are taking one quantity. In some way it is not as fundamental as the other quantity.
202 C H A P T E R 6
. The derived quantity is referred to as the dependent variable. to be a derived quantity. In making statements such as these.Review of Prerequisite Skills
One important use of mathematics is to describe a relation between two quantities. making the radius v
or the time the dependent quantity. and we write formulas such as A r2 or d vt. such as the area or the distance. We say the area of a circle depends on its radius. because its value depends on the value of the other variable. but the underlying idea of dependence is still present. or the distance travelled at a constant speed depends on the time. The way a formula is written is a matter of preference or circumstance. How does this basic idea of dependence apply to vectors? Under what circumstances does one vector depend on another? If vectors are independent.
which has the property that 0 v v. such that v ( v) 0. 4. 5. (iii) the set of all ordered triples of real numbers (x. Any two vectors can be added. mathematicians have learned much about them by studying what they all have in common. Scalar multiplication distributes over vector addition: a(u v) au av. (ii) the set of all ordered pairs of real numbers (x. including vectors. 3. Investigate A vector space is a set of objects. (v) the set of all polynomials. as do a number of other mathematical objects. How can addition and scalar multiplication be defined so that the set of all functions is a vector space? ●
RICH LEARNING LINK
203
. called vectors. y). 8. If a vector is multiplied by a real number. Verify that the following sets are all vector spaces: (i) the set of all vectors drawn in a plane. V. There is a zero element 0. Is exponentiation an associative operation on the set of real numbers? 2.investigate
C H A P T E R 6 : V E C T O R S PA C E S
Vectors can be added to each other and they can be multiplied by real numbers. 6. y. The set of all real numbers is itself a vector space. Addition is commutative: u v v u. Scalar multiplication distributes over scalar addition: (a b)v av bv. 9. In 1888. Polynomials also have these two properties. the result is in V. and the result will always be in V. Every vector v has an opposite v. Every vector is unchanged when multiplied by 1. the Italian mathematician Guiseppe Peano presented the first definition of what are now called vector spaces. Is the set of all complex numbers a vector space? What about the set of all integers? 3. (iv) the set of all polynomials with degree less than or equal to two. Calculate 2(34) and (23)4. Addition is associative: (u v) w u (v w). 10. 2. Vector spaces provide an abstract system for studying the common properties of many different mathematical objects. for which two operations are defined: addition and scalar multiplication. Instead of studying all these various objects separately. These operations must have the following ten properties: 1. 7. Scalar multiplication is associative: a(bv) (ab)v.
DISCUSSION QUESTIONS
1. z).
that one way to write an ˆ ˆ ˆ algebraic vector was as a linear combination of the basis vectors i . where a and b are scalar quantities. and determine a numerical value for the scalar a only x when x and u are collinear. when can a vector x be expressed in the form x au bv …? How are the coefficients of such an expression to be found? What does it mean if forming such a linear combination proves to be impossible? We begin the investigation of these questions with the simplest case. The equation x au implies that the u two vectors x and u are parallel. express x as a scalar multiple of u. and w is written au bv cw.
EXAMPLE 1
If possible. x a. ˆ ˆ b. v.
Two vectors x and u are collinear if and only if it is possible to find a non-zero scalar a such that x au. x au cannot be written as a scalar multiple of u. Under what circumstances is it possible to express a given vector x in the form au? The answer is that we write x au. The vector au is a linear combination of the single vector u. Let a be a scalar. If x ˆ 8j ) ˆ Then (4i ˆ ˆ 4i 8j
204 C H A P T E R 6
au ˆ a(6i ˆ 6ai
ˆ 12j ) ˆ 12aj
. If x and u are not collinear. (5. and k: ˆ bj ck.Section 6. These expressions are linear because they consist only of the sum of scalar multiples of vectors and nothing else. Recall.1 — Linear Combinations of Vectors
The expression au bv. j . ˆ ˆ ai This section is concerned with the reverse problem. is a linear combinaof the vectors u and v. A linear combination of three vectors u. Under what circumstances can a given vector be expressed as a linear combination of other vectors? That is. 6)
Solution a.
8. 3) 4. it was by means of such expressions that sets of vectors were combined to form new vectors. x 4i 8j u 6i 12j u
(10.
tion
In the preceding chapters.
and v are coplanar. a(5. Proceed as in a. Then 4 a 6a
2 3
and
8 a
2 3
12a
The value for a in the two cases is consistent. and bv form a triangle. and x. If x lies between u and v in the same plane. x? Under what circumstances is it possible to v x express a given vector x in terms of two vectors u u and v? The answer is that you can write x au bv and determine numerical values for the scalars a and b only when the three vectors x. 6a). au.
v 30° 20° u
x
6 . Solution You are to find scalar multiples of u and v. 4.
Equating corresponding components. (5a. Consider now a more complicated situation. x 8. If x does not lie in the plane of u and v. 1 L I N E A R C O M B I N AT I O N S O F V E C TO R S
205
. v. the sum of which equals x. If then (10. Let x au bv. there is no scalar multiple of u which equals x. and x 24. then the coefficients of ˆ ˆ i and j on the left must be the same as the corresponding coefficients on the right. 6) 4a. u. making an angle of 20º with u and 30º with v. have magnitudes u 10. (10. These two vectors are non-collinear. Thus x 2 u and x and u are collinear. and v are coplanar if and only if it is possible to find non-zero scalars a and b such that x au bv.If the vector on the left is equal to the vector on the right. 3) au.
EXAMPLE 2
Three vectors u. 3 b. The equation x au bv implies that the three vectors x. this triangle cannot exist. u. v 15. express x as a linear combination of u and v. 3) 8. where a and b are coefficients to be determined. 10 a 5a 2 and 8 a 4a 2 and 3 a 6a
1 2
Since the scalar a must be the same for all components of the vectors. Three vectors x.
the vectors are not coplanar. If they are. 4) b(6. If we cannot find values for a and b.
bv 30° x 20° au
20° au
x 30° 130° bv
From
au sin 30º x sin 130º
24 sin 30º 10 sin 130º
and from
bv sin 20º x sin 130º
24 sin 20º 15 sin 130º
a
b
1. as expected. Write x au bv. The sides of this parallelogram are the required vectors au and bv.
EXAMPLE 3
Determine whether or not the three vectors u (3. and x (7.566u 0. so b will be less than one. Instead. v (6. If we can find values of a and b. the vectors are coplanar. 8). On the other hand. so you should expect a to be greater than one. bv is shorter than v. and x is expressed as a linear combination of u and v. Note that a is greater than one and b is less than one. We have (7. a 4b.714
Therefore. 4) are coplanar. which add to x. but we would be left with the problem of expressing x as a linear combination of u and v. x 1. 8) 6b. 4a 8b)
Equating components. express x as a linear combination of u and v. we proceed as follows. 4. 4) a(3. Solution We could calculate the triple scalar product. the vectors are coplanar. Now draw the triangle for the addition of the vectors. 1. In this particular case. 3. we get the system of equations 3a 6b 7 a 4b 3 4a 8b 4
206 C H A P T E R 6
. (3a 1. parallel to u and v. au is longer than u. If it is zero. 4).714v.566
0. 4. 3.Make a parallelogram by drawing lines from the tip of x in the vector diagram.
Is it always possible to express x as a linear combination of u and v? Explain. we find a 5 . b. a. when x does not lie in the plane of u and v.1
Part A
Communication
1. Write each of the following vectors as a linear combination of i and j . 4) and v ( 2. What information does the cross product of u and x give about the collinearity of u and x?
Communication
Knowledge/ Understanding
ˆ ˆ 4. the position vector of the point A(8. and x are any non-zero vectors in the xy-plane. The vectors u. say the first and second. 3a 6b 3a 12b 6b b 7 9 2
1 3
so
Substituting into either of the equations. a. v.We solve any two of these equations. 5)? Justify your answer. Can every vector in the xy-plane be written as a linear combination of u (1. and x
Exercise 6. as required Therefore the vectors are coplanar. 3
4. 3. a vector directed at an angle of 45º with a magnitude of
d. 3 We now see if these values satisfy the third equation 4 5 3 8 1 3
20 3 8 3 5 u 3 1 v. Write the vector ( 567. the vector p ( 4. 2. Explain why it is impossible to express x as a linear combination of u and v. 1 L I N E A R C O M B I N AT I O N S O F V E C TO R S
207
.
6 . a vector directed at an angle of 150º with a magnitude of 6 5. 669) in terms of u and v. 5) 3) 2 b.
c.
not both zero. We then say that u and v are linearly dependent vectors. On the other hand.
u u = 2v u – 2v = 0 m = kn
v
u
–2v
m
n
Parallel vectors
Non-parallel vectors
We can see that whenever two vectors u and v are parallel.2 LINEAR DEPENDENCE AND INDEPENDENCE
209
. Three vectors are linearly independent if they are not linearly dependent. unless a b 0. such that au bv cw 0. Two vectors are linearly independent if they are not linearly dependent. suppose u and v are two vectors such that u 2v. and c. Geometrically. Let us first consider just two vectors. Linear Dependence of Three Vectors Three vectors u. and w are linearly dependent if and only if there are scalars a . Their importance is both theoretical and practical.
Linear Dependence of Two Vectors Two vectors u and v are called linearly dependent if and only if there are scalars a and b. where a and b are not both zero. consider two vectors m and n that are not parallel. For example. there is a relationship au bv 0. This statement can be rewritten as u 2v 0. Two vectors are linearly dependent if they are collinear or parallel. such that au bv 0. These vectors are called linearly independent vectors. Three vectors are linearly dependent if they are coplanar.
6.2 — Linear Dependence and Independence
The concepts of linear dependence and independence are fundamental in vector algebra. v. b. this means that multiplying v by the scalar 2 and adding it to u brings us back to the zero vector 0. not all zero.Section 6. There is no possible way to combine multiples of these vectors so that am bn 0.
1) or 2( 1. eliminating a 2a b 4c 0 2a 14b 6c 0 15b 10c 0 2 ∴b c 3 To avoid fractions. which make the linear combination au bv cw equal to the zero vector. it is not possible to find unique values for a. 7) 5u and eliminating b 14a 7b 28c 0 a 7b 3c 0 15a 25c 0 ∴ a 5c 3 2 and a 5. and w are linearly dependent. we choose c Consequently.
210 C H A P T E R 6
. 3) are linearly dependent by showing that it is possible to find coefficients a. not all zero. b. b. 2a b a 7b 4c 3c 0 0
Since there are three variables but only two equations. 2
and
w
5 u 3
2 v 3
EXAMPLE 2
Prove that three non-collinear vectors u. c such that au a(2.
3. 7). b. 7) (2a b 4c. b. v 5
0. a bv cw c( 4. Express each vector as a linear combination of the other two. then
5 u 2 3 w. 1). b. b. and w ( 4. v. such that the linear combination au bv cw is equal to the zero vector. 3) 7b 3c) 0 0 0
The components of the vector on the left must be zero. and c. 1) b( 1. Demonstrate that the three vectors u (2. and c. 3) 2v 3w
Thus there are values of a. If the vectors are linearly dependent. whereupon b 0 0
3( 4. and w are linearly dependent if and only if they are coplanar. If 5u u
2 v 5
2v
3w
3 w. Solution a. then there are values a. not all zero. and c. Then the three vectors u.EXAMPLE 1
a. 5(2. v ( 1. v.
and w are coplanar.Solution First. c. for then. v. and c 0. w
a u c b v c
Thus. making the three vectors coplanar. Assume that c is one of the non-zero scalars. If the two vectors are non-collinear. suppose that u. the only values of a and b that can make the equation true are a 0 and b 0. d. equal the zero vector if we set all the coefficients a. if making coefficients equal zero is the only way the linear combination can equal the zero vector. Such vectors are said to be linearly independent. For instance. then the vectors cannot be dependent. x. v. Under these circumstances. we can always write an equation of the form au bv 0 for any pair of vectors u and v. We can make a similar observation about a linear combination of three vectors. It is then possible to divide by c and solve for w. and c that can make the equation true are a 0. w. of course. w is expressed as a linear combination of u and v. and w are linearly dependent. can be written as a linear combination of u. the only values of a. b. au bv w 0. suppose that u. b. We can always write an equation of the form ax bu cv 0. … equal to zero. so w must lie in the plane of u and v. b. A linear combination of vectors will. however. But if the three vectors are not coplanar. Hence.2 LINEAR DEPENDENCE AND INDEPENDENCE
211
. 0u 0v 0.
A set of vectors u. c. so that au bv cw 0 for some scalars a. b. We can always form a linear combination of vectors. c. and no two are collinear. and 1 are not all zero. the three vectors must be linearly dependent. not all zero. Conversely. Because the coefficients a. and v as w au bv. b 0. … is linearly independent if the only linear combination au bv cw dx … that produces the zero vector is the one in which the coefficients a. b. v. However. … are all zero. then it is impossible to express one of the vectors in terms of the other two. Then w. for example.
6.
let ˆ a(4i ˆ 4ai (4a ˆ 8j ) ˆ 8aj ˆ 6b)i au bv ˆ ˆ b(6i 3j ) ˆ 3bj ˆ 6bi ˆ (8a 3b)j 0 0 0 0
ˆ ˆ Since i and j are linearly independent vectors. ˆ ˆ ai bj 0 Either a 0 or a 0.
Solution For scalars a and b. It follows ˆ ˆ ˆ that 0i bj 0. the basis vectors for a plane. Thus. Consequently. Suppose that a ˆ 0. i must be a scalar multiple of j .
EXAMPLE 4
Show that u
ˆ 4i
ˆ 8j and v
ˆ 6i
ˆ 3j are linearly independent.
Then u and v are linearly independent. are linearly independent. i must equal the zero vector. Both of these statements are false. Solution (3 (3 s)u – 5u ( 2 s)u s)u tv tv 4sv (t 4s)v 5u 0 0 4sv
212 C H A P T E R 6
. determine the values of s and t.EXAMPLE 3
ˆ ˆ Prove that i and j . a
ˆ ˆ If b 0. so a must be zero.
EXAMPLE 5
Given that the vectors u and v are linearly independent and (3 s)u tv 5u – 4sv. the vectors i and j must be linearly independent. which means that b 0. both coefficients are equal to zero. ˆ If b 0. since j 0. Solution ˆ ˆ Start by forming a linear combination of i and j and setting it equal to zero. the assumption a 0 is impossible. Therefore. 4a 6b 8a 3b a b 0 0 0
Thus. the values a 0 and b 0 are the only values of a and b for which the linear combination ˆ ˆ ˆ ˆ ai bj 0. Then i
b ˆ j.
the vectors i and j are not the only vectors that could be used as basis vectors for a plane.2 LINEAR DEPENDENCE AND INDEPENDENCE
213
. Therefore. su 5)u 7v (t 5u 3)v tv 0
6. a. what can be said about w
a. Given three vectors in a plane. a. b. Every vector in the plane can be expressed as a linear ˆ ˆ combination of these basis vectors. they can be designated as basis vectors for space.2
Part A 1. then the three vectors are non-coplanar. Three vectors are linearly dependent if they are coplanar. (s d. Are three vectors lying in a plane always linearly dependent? Explain. Any pair of linearly independent vectors can be designated as basis vectors for a plane. su c.Since u and v are given to be linearly independent. if u and v are linearly dependent?
Communication
2. if u and v are linearly independent? b. they are linearly independent. Can any set of three non-collinear vectors be used as a basis for space? Explain. On the other hand. As a consequence. if the triple scalar product of three vectors is not zero.
Exercise 6. find the values of s and t for each of the following equations. They are just a particularly simple and convenient choice. under what circumstances is it impossible to express one of them as a linear combination of the other two? c. (s 2tv 2)u 0 (s t 3)v b. in terms of which every three-dimensional vector can be expressed. since (u v) is perpendicular to the plane of u and v. Thus. Any two non-collinear vectors define a plane. it must lie in the plane of u and v. If u and v are linearly independent vectors.
Knowledge/ Understanding
3. the coefficients of u and v must be zero. Three vectors are coplanar if (u v) • w 0. Therefore. and if w is perpendicular to this cross product. 2 s s 0 2 and t 4s t 0 8
Two vectors are linearly dependent if they are collinear. Given that w
Communication
au
bv.
1). 10)
11. 1. and k 8. 10. 3) relative to the basis. Determine whether the following sets of vectors form bases for three-dimensional space. 0. prove that au and bv are linearly independent vectors. (k 2)u (k (k 2)v 2)v 0 0 b. ru b. a. and w are linearly independent vectors. ˆ ˆ 7. if possible. for each of the following equations. v1 (3. v1 (1. s. determine the coordinates of v (1. a. 1. 1). 1). j . and a (a b) can be used as a basis for vectors in space. 5). basis for vectors in space. (1. Show that the vectors ( 1. determine the coordinates of v (8. v2 (1. 1. (k2 – 4)u Part B 6. v1 c. 1. v2 (2. 1. Given that the vectors u and v are linearly independent. Show that the representation of a three-dimensional vector in terms of i . 2. 1) ( 4. v3 (1. 1). a. v. 2). v2 (3. Given that u and v are linearly independent vectors and a and b are non-zero scalars. 0). v3 (2. v2 (1. ˆ is unique. 1) and (1. determine the value of k. 1. 0). If a set forms a basis. 3.4. 1). Show that the vectors a. 1) 5)
214 C H A P T E R 6
. 7) relative to this base. v1 ( 1. Determine whether the following sets of vectors form bases for two-dimensional space. a b. If u. 1. where a and b are any non-collinear vectors. 5) b. (1. and t for each of the following equations. 3. (6k d. ku 4)u 3v 0 (8 12k)v 0 c. v3 (3. a. 0. If a set forms a basis. v1 b. (r
Application
(2s s
1)v 5)u (r
(r s
s
t)w 1)v (r
0 st)w 0
5. v2 (6. find the values of r. 1. 1) can be used as a
Thinking/Inquiry/ Problem Solving
9.
and w are mutually perpendicular. b. are the vectors u v. and w u linearly dependent or linearly independent? 15. if the vectors (1 s)u 2 v and 3u sv are parallel.2 LINEAR DEPENDENCE AND INDEPENDENCE
215
. If (3u 4v) and (6u 2v) are linearly independent vectors. d is any three-dimensional vector. 3
6. If u. The vectors u and v are linearly independent. v w. show that u and v must be linearly independent also. 14. Find s. linearly independent vectors.
Thinking/Inquiry/ Problem Solving
13.Part C 12. v. The vectors a. and c is unique. b. and c are basis vectors for space. and d ka lb mc. Show that this representation of d in terms of a.
The derivation of this midpoint formula is valid for vectors in both two and three dimensions.Section 6. The position vectors of the points A. 6)]
4)
7.
The position vector of the midpoint is a linear combination of the position vectors of the endpoints of the line segment.
6). the three points are said to be collinear. If a third point lies on this line. Solution The position vectors of points A and B are OA and OB (8. (8.
216 C H A P T E R 6
. OA. This means that the vectors AM and MB are equal. such as Q or R in the given diagram. A.
1 [(10. 1. lie on an extension of the segment outside the interval AB. OM The midpoint M is (9. respectively. and P are collinear. When P lies between A and B on the line segment. a formula for the position vector of the midpoint can be found. 1. 5)
4)
(9. The third point P is called a division point of the segment. then they are said to divide the segment externally.3 — Division of a Line Segment
Two points determine a straight line. two of the points. B. 5).
P divides AB A internally Q and R divide AB Q A externally P B B R
The midpoint of a line segment is an example of an internal division point. 6). B. When points. From this equality. The midpoint M of a segment AB lies between A and B and divides the segment exactly in half. are taken as the end points of a line segment.
EXAMPLE 1
Find the midpoint of the line segment from A(10. 3. and OM.
7. and M relative to some origin O are. OB. 1. Therefore. 3.
4) to B(8. say A and B. 2
7. When three points. Since Then OM AM OA 2OM OM MB OB OA
1 OA 2
A
M
B
O
OM OB
1 OB 2
A M B
Therefore. then P is said to divide the segment internally.
(10.
P divides AB in the ratio 3: 1 or
AP PB 3 1
3 1 A B P
You can distinguish an external from an internal division by a negative sign in the ratio or fraction. such as that shown in this division-point diagram. Here. AP PB
B b. the segments AP and PB are both positive. such that AP and QB 3 units. P divide AB? b. so A divides BP in the ratio 7: 3 6 8 .3 DIVISION OF A LINE SEGMENT
217
.
c. B divide PQ?
d. In what ratio does a.
EXAMPLE 2
Points P and Q lie on line segment AB. and in the opposite direction as negative. if then P divides AB in the ratio 1:3 or
AP PB 1 3
1 A P 3 B
or
AM MB
1 1
the points must be arranged on the line as in this division-point diagram:
In reading or making a division-point diagram. AA P PB c. treat segments going in the same direction as the given segment as positive. so P divides AB in the ratio 6:8 or 3:4 8 14 . for instance. In the case of an external division of a line segment. BQ
6 .There are several equivalent ways to describe the division point of a line segment. The midpoint. can be spoken of as a ratio or written as a fraction. In the diagram above. QA AB
11:14
6. M divides AB in the ratio 1:1 In a similar manner. A divide BP?
6 units. A divide QB?
Solution Draw a division-point diagram containing the given information:
6 A P 5 Q 3 B
a. so B divides PQ in the ratio 8: 3 3 11 so A divides QB in the ratio 14
d. PQ
5 units. The negative sign is more conveniently placed on the smaller term of the ratio. P divides BA in the ratio 1:3 since BP has direction opposite to BA.
Express OP as a linear
5
2:5.
4
QS SR
3 4
3 S Q 5 S R R 1 P
SR RP
5 1
The numbers in the ratios are not the actual lengths of the segments. In what ratio does Q divide SP?
3:4. Think of them as the number of equal parts into which the segment has been divided. Q. SR must be divided into the same number of parts. and therefore that
SQ QP
15 9
so Q divides SP in the ratio 15:9 or 5:3. Therefore. Solution P divides AB in the ratio Therefore. To compare the ratios. R divides
Solution Write the fractions and draw the division-point diagrams for the two given ratios. multiply the first ratio by 5 and the second by 4. 5
2 A P
B
P
B
P
2(OB 2OB 5OA
5 OA 3
OP) 2OP 2OB
2 OB 3
O O
218 C H A P T E R 6
.
EXAMPLE 4
The point P divides the line segment AB in the ratio combination of OA and OB. and then arrange all the points on a single line. The points S and R are common to both segments. AP 5(OP 5OP OA) 5OA 3OP OP
2 PB 5
A
2:5. so AP PB
2 .EXAMPLE 3
The points P. R. S divides QR in the ratio SP in the ratio 5:1. and S are collinear.
20 15 S Q 5 R 4 P
This makes it clear that QR must contain 5 parts.
EXAMPLE 5
Prove that if OP kOA lOB.
6. B. and P are collinear if and only if OP
b a b
OA
a a b
OB. AP b(OP OA) bOP bOA (a b)OP OP
a PB b
P
a(OB OP) aOB aOP bOA aOB b OA a a b OB a b
A
O
Division-Point Theorem Points A.The formula for the position vector of the division point found in Example 4 can a be generalized. coplanar.3 DIVISION OF A LINE SEGMENT
219
. and P are collinear. PB b
B
Therefore. B. This proves the if part of the division-point theorem. and constitutes a proof of the only if part of the theorem. The derivation of the division-point formula shows it is necessary for collinearity.
This theorem shows that when three points are collinear. the vectors AP and AB are parallel. and P are collinear. this is sufficient to guarantee that
B A
(1 l)OA OA lOA l(OB lAB OA)
lOB lOB
kOA
P lOB
O
Therefore. their position vectors are linearly dependent and. k 1 l
l
1. then AP . B. and so A. If P divides AB in some ratio a:b. hence. and k the points A. Solution Since Then So OP k l OP OA AP 1.
a. 2. If the point P divides AB in the ratio 1:2 and the point Q divides AB in the ratio 1:2. 1:2 2:3 d. C. Determine a. the ratio in which A divides BD d. Points A. point D divides EF in the ratio Part B 4. and D are located on a line as shown in the given diagram. If T divides AB in the ratio 2: 1. point K divides MN in the ratio 5:8
220 C H A P T E R 6
. in what ratio does A divide QB? b. point A divides BC in the ratio 2:1 b. Can a line segment be divided in the ratio 1: 1? Explain. the ratio in which D divides AB e. the ratio in which B divides CD
A 4 B 3 C 3 D
Knowledge/ Understanding
3. point Q divides PR in the ratio e. point U divides ST in the ratio 3: 1 c.Exercise 6. a. Draw a division-point diagram for each of the following statements. in what ratio does B divide PA? 5. in what ratio does B divide QP? c.3
Part A
Communication Knowledge/ Understanding
1. the ratio in which B divides AD c. in what ratio does Q divide AP? d. the ratio in which C divides AD b. B. in what ratio does P divide QA? e. prove from first principles that OT 2OB OA.
Using vectors. For instance. Position vectors.
EXAMPLE 2
Two of the opposite sides of a quadrilateral are parallel and equal in length. One approach is to use point-to-point vectors. Remember that there are several things to do before you can actually start a proof. Sometimes one solution may be more direct and easier than another. which is not usually part of the figure. we gain insight into the mathematical principles involved and increase our confidence in the results. To do this. on the other hand. Euclidean proofs are the usual way to establish the properties of geometrical figures. point from some outside origin. Point-to-point vectors usually lie in the plane of a figure and join one point of the figure to another.
222 C H A P T E R 6
. prove that the other two opposite sides are also parallel and equal in length.4 — Vector Proofs in Geometry
We need to be able to solve a problem in several different ways. Using position vectors. so your first job is to express what is given and what is to be proved in the form of vector formulas or equations. The use of vectors is an alternate way to accomplish the same result.Section 6. There are two distinct approaches that can be taken when we use vectors to do proofs. so are the other two opposite sides. prove that the other two opposite sides are also parallel and equal in length. Using point-to-point vectors. we write AB DC. if two of the opposite sides of a quadrilateral are parallel and equal. If we try different methods of solution. The two methods are illustrated in Examples 1 and 2 below. what is to be proved can be written AD Then AD AB DC BD BC BD BD DC
B A
BC.
EXAMPLE 1
Two of the opposite sides of a quadrilateral are parallel and equal in length. to points in the figure. The other is to use position vectors. Solution Let ABCD be a quadrilateral in which AB CD and AB CD. Likewise.
D
C
Therefore. you will need a suitably labelled diagram. the proposition to be proved is usually expressed in words.
AB Since OB OD AB OA OA AD
DC is given. and AD DC OC OC BC
BC is to be proved. and OE 2 1 OA 2 1 OC 2
B A
E G
D
C
Let Then
OG
kOA kOA kOA
lOD. respectively. Let O be an origin that is not in the plane of the quadrilateral.
EXAMPLE 3
Prove that the medians of a triangle intersect at a point that divides each median in the ratio 2:1. If O is a point not in the plane of the triangle. 4 V E C TO R P R O O F S I N G E O M E T RY
223
. Sometimes a proof using position vectors requires the use of the division-point formula and the concept of linear independence. An example of that kind of proof is shown next.Solution Let ABCD be a quadrilateral having AB CD and AB CD. D and E are the midpoints of BC and AC.
OD OB
The conclusion is the same as that of Example 1. (k l( 1 OB 2
1 lOB 2
l
1)
1 OC) 2 1 lOC 2
Similarly
OG
mOB mOB
n OA 2
nOE. (m n( 1 OA 2 mOB
n
1 OC) 2 n OC 2
1)
These two expressions for OG must be equal. Solution In ∆ABC. Therefore. kOA or (k
l OB 2 n )OA 2 l (2 l OC 2 n OA 2 l (2
mOB
n )OC 2
n OC 2
m)OB
0
6 .
O B A C D
As in Example 1. then OD
1 OB 2 1 OC.
and Since Then k
l 2 n 2 n 2 n 2 l 0. l
1 OB 3
2m n 2m
1 . Therefore.Since the vertices of the triangle A. l 1.
C A O B
OC • OC
224 C H A P T E R 6
. and G divides AD in the ratio 2:1. If the dot product of the two vectors is zero. and C are not collinear. or k
1 3
k l OG
l 2 l 2
l
2 3
2 OD. so k
1 OA 3
2k 1. then m
2 OE. B. the position vectors OA. ∠ACB is a right angle. since OA and OC are radii.n 3 2 3
m OG
n
1. So the point of intersection G divides each of the medians in the ratio 2:1. Another type of problem asks for a proof that two line segments are perpendicular. OB. the result is the same. CA • CB (OA OC) • (OB OC) OA and OB are both radii and OB OA. they are linearly independent vectors. 3
If you repeat this work using AD. and OC are not coplanar. Solution Let O be the centre of a circle with diameter AB. Then CA • CB (OA OC) • ( OA OC) OA • OA OA • OC OC • OA OA2 OC2 0. Therefore. Draw angle ∠ACB in the semicircle. This linear combination can only equal 0 if each of the coefficients separately equals zero: k Since Since Now Then Then Similarly.
EXAMPLE 4
Prove that an angle inscribed in a semicircle is a right angle. This angle is the angle between the vectors CA and CB. then ∠C is a right angle. and G divides BE in the ratio 2:1. for instance. l 0. Such problems are handled by showing that the dot product of the corresponding vectors is zero. G is called the centroid of the triangle. n 0. 2
m 2k n 2k
l 0. 3
m
n 2
0. and the third median CF. 2
n 2
0
0.
for instance. Prove that if the diagonals of a quadrilateral bisect each other. 4 V E C TO R P R O O F S I N G E O M E T RY
225
.
A
B
D
C
BC • CD
0
CD.EXAMPLE 5
If the diagonals of a parallelogram are perpendicular. using point-to-point vectors b. If side BC of ∆ABC is trisected by points P and Q. show that AB AC AP AQ a. Part B 4. BC a rhombus. prove that OD OE OF OA OB OC. AB DC. using point-to-point vectors b. using position vectors
3. so AC • BD 0 (AB BC) • (BC CD) 0 ( CD BC) • (BC CD) 0 CD • BC CD • CD BC • BC CD2 BC2 0 Therefore. the quadrilateral is a parallelogram. using position vectors
Knowledge/ Understanding
2. Then opposite sides are equal vectors. and F are the midpoints of the sides of the triangle ABC. The diagonals are perpendicular. Prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length a. Let ABCD be a parallelogram. If D. E.4
Part A
Communication
1.
6 . Solution Draw and label a diagram. so adjacent sides are equal and the figure must be
Exercise 6. prove that the parallelogram is a rhombus.
OB • OD b. c. prove that AG
Knowledge/ Understanding
7. a.
Application
9. Prove that the diagonals of a parallelogram bisect each other. Explain why the results of part b prove that the three altitudes of a triangle intersect at a common point. and AP ⊥ BC. Let ABCD be a rectangle. BP ⊥ AC. Prove that OG
1 (OA 3
OB
OC). prove that it intersects the chord at its midpoint.) 13. OA2 OC2 OB2 OD2
226 C H A P T E R 6
. Show that CP ⊥ AB. AB show that AK
n n 1
nDC. Show that CP XA XB. If a line through the centre of a circle is perpendicular to a chord. b.5. in what ratio does G divide AD? c. OA • OC Part C 14. meeting at the point K. ∆ABC is inscribed in a circle with centre X. If G is the centroid of ∆ABC. (P is known as the orthocentre of the triangle. AC and BE. Define a point P by its position vector XP XA XB XC. Use the type of proof shown in Example 3. If the diagonals BD and AC meet at K. Prove that the sum of the squares of the diagonals of any parallelogram is equal to the sum of the squares of the four sides. Show that the midpoint of the hypotenuse of a right-angled triangle is equidistant from the vertices.
6. 8. 10. a. 11. Determine the ratios in which K divides AC and BE.
1 1
AD
n
AB
Application
12. in what ratio does D divide BC? b. In the trapezoid ABCD. BG CG 0. Prove that a. If G is the centroid of ∆ABC and AD is one of its medians. A regular hexagon ABCDEF has two of its diagonals.
17. In a triangle ABC. The point F divides AC in the ratio 2:3. DC is extended to E so that DE:EC 3: 2. 16. The line AE meets BC at F. a. AP a. Given the tetrahedron MNPQ with MN ⊥ PQ and MP ⊥ NQ.Thinking/Inquiry/ Problem Solving
15.
6 . AQ and BP BQ. b. In the quadrilateral APBQ. Determine the ratio of the area of the quadrilateral CEDF to the area of the triangle ABC. Determine the ratios in which F divides BC and F divides AE. The two line segments BF and AE intersect at D. 4 V E C TO R P R O O F S I N G E O M E T RY
227
. Prove that AB bisects PQ. the point E is selected on BC so that BE:EC 1:2. b. Find the ratios in which D divides AE and BF. In the parallelogram ABCD.
Thinking/Inquiry/ Problem Solving
18. Prove that AB is perpendicular to PQ. prove MQ ⊥ NP.
Three vectors in a two-dimensional plane and four vectors in three-dimensional space are always linearly dependent. If the linear combination au bv cw … is equal to 0 only if the coefficients a. or • if the cross product u v 0. and P are collinear and their corresponding position vectors are coplanar only if the coefficients m n 1. and w are linearly dependent • if they are coplanar • if u av bw where a and b are not both zero. the three points A. … is linearly independent. You should understand what the implications are when a set of vectors is found to be linearly independent or not. w. b. and OP b b a b
OA
a a b
OB. they must be linearly independent. The question of linear independence is usually approached indirectly by asking if a set of vectors is linearly dependent. v. then the set of vectors u. or • if the triple scalar product u v • w 0. Two non-zero vectors u and v are linearly dependent • if they are collinear • if u kv with k 0.
228 C H A P T E R 6
. Three non-collinear vectors u. If the vectors are not linearly dependent. In the linear combination OP mOA nOB. if you pay attention to the form of the equations and the positions of the letters representing the individual points. AP PB
a . v. You should be able to express the division of a line segment AB by a point P in three equivalent ways and to convert readily from one to the other.Key Concepts Review
The fundamental concept in this chapter is that of linear independence. That consists of trying to express one of the vectors in the set in terms of the others.
This is not difficult. P divides AB in the ratio a:b. c. B. The division of a line segment is also connected to the linear independence of vectors. … are each zero.
one approach may be more difficult than the other. using point-to-point vectors that lie in the plane of the figure 2. there is no way to predict this. However. To learn to carry out proofs successfully. and unfortunately. Expect to work through and write out the logic of a proof several times until you get it right. there is no substitute for doing many problems.The concept of linear independence and the properties of division points both play a role in vector proofs of geometrical propositions. Follow the examples. Two approaches have been illustrated: 1. try another.
KEY CONCEPTS REVIEW
229
. using position vectors from some origin to points in the figure It should be possible to prove a proposition using either approach. If you can make no progress using one method. Persevere. Start with the simpler proofs.
15. Determine the ratios in which R divides CQ and BP. In the parallelogram ABCD. Prove that AB • (QD b. 14.)
REVIEW EXERCISE
231
. E divides AB in the ratio 1:4 and F divides BC in the ratio 3:1. CA. In what ratio does K divide DE. Prove that the medians to the equal sides of an isosceles triangle are equal. E. Let R be the point of intersection of CQ and BP. Explain why these results prove that the perpendicular bisectors of the sides of a triangle meet at a common point. the points D. The perpendicular at E to AC meets the perpendicular at F to AB at the point Q.9. 0. (Q is called the circumcentre of the triangle. Prove that the median to the base of an isosceles triangle is perpendicular to the base. 10. and AB.
c. In ∆ABC. The point P divides the sides AC of the triangle ABC in the ratio 3:4 and Q divides AB in the ratio 1:6. Prove that the sum of the squares of the diagonals of a quadrilateral is equal to twice the sum of the squares of the line segments joining the midpoints of the opposite sides. a. Prove that a line that passes through the centre of a circle and the midpoint of a chord is perpendicular to the chord. and in what ratio does K divide AF? 11. respectively. Use parts a and b to prove that CB • QD
d. and F are the midpoints of sides BC. Prove that AC • (QD
1 AC) 2 1 AB) 2
0. 13. 0. The line segments DE and AF meet at K. 12.
INDEPENDENT STUDY
Investigate: How is the dimension of a vector space defined? What are the dimensions of the vector spaces we have encountered so far? Can vector spaces be infinitely dimensional? Investigate: Is the set of all matrices of a fixed size a vector space? If it is. (0. Give two examples of bases of the set of all ordered triples of real numbers. It is called the standard basis of this vector space. b) Let n be a fixed whole number. 3. –1)} is also a basis of the set of all ordered pairs. What is the smallest set of coins you would need in order to be able to make any amount of change less than one dollar? 2. is called a basis of V. 3). or you might have the wrong coins (seven dimes will not work). You might not have enough coins (you will need at least seven). no less) of the right type (linearly independent) so that every other vector in V can be written as a linear combination of the vectors in B. 0) and (0. How many different bases could a particular vector space have? What do they all have in common? 5. the same problems can occur. what is an example of a basis? ●
232 C H A P T E R 6
. there are two ways that you might be unsuccessful. 0). (1. When you try to express a vector as a linear combination of a set of vectors. In a particular vector space V.wrap-up
C H A P T E R 6 : V E C T O R S PA C E S
investigate and apply
When you try to make exactly 49 cents from the coins in your pocket. 1)} is a basis of the set of all ordered pairs.
I N V E S T I G AT E A N D A P P LY
1. a set of vectors B that has just the right number of vectors (no more. your set may not have enough vectors or it may have unsuitable vectors. 4. b) Show that {(2. So {(1. 1). a) Show that every ordered pair of real numbers can be written as a linear combination of (1. Find a basis of the vector space of all polynomials of degree less than or equal to n. a) Find a basis of the vector space of all polynomials with degree less than or equal to two.
Three vectors u. Express OR as a linear combination of OP and OQ. 2 5. v. b. 12. making an angle of 35º with v and 20º with w. a geometric example
a
b
4. b. and w are linearly independent.Chapter 6 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application
Questions 3.
c
b. Explain this concept using a. Draw a division-point diagram showing the relative positions of the four points. b. P divides QR in the ratio 10: 3. Determine values of r and s where c ra sb. an algebraic example 2. a. 4 7. 6
1. 8 1. Express u as a linear combination of v and w. The vectors u. F divides AP in the ratio 13: 8 and F divides PG in the ratio 4: 3. Express OP as a linear combination of OQ and OR. and 18 respectively. v. 5. In what ratio does P divide AG?
CHAPTER 6 TEST
233
. 3. a. u lies between v and w. Copy the three vectors shown in the given diagram onto graph paper and draw c accurately as a linear combination of a and b. and w are coplanar and have magnitudes 5. a.
8). 7. ABCD is a quadrilateral. Express OP as a linear combination of OA and OB. D lies on AB and E lies on AC such that DE kBC. P. Form two point-to-point vectors out of the three points A( 4. 8. b. Prove that AD kAB and AE kAC. 8. B( 1. and S are the midpoints of its sides. Prove using vectors that PQRS is a parallelogram. a.
234 C H A P T E R 6
. using the fact that AB and AC are linearly independent. In ∆ABC. 2). 2. R. 4. 2) and P(1. Q.6. Demonstrate that the two vectors are collinear.
By transposing the last four columns in B (that is. The vectors used for transmitting are formed as follows. It does not correct if there are two or more errors. b3. While this seems like an impossible task. For example. you saw that 0 . Since we have no access to the original message we need to be able to determine whether or not the message received is unchanged from the one that was sent. The message passes through electronic interference on its way and may be distorted. Suppose that a satellite sends electronic messages using this system from outer space to a receiver on earth. we have the following vectors: v1:1101000 v2:1010100 v3:0110010 v4:1110001
E X T E N D I N G A N D I N V E S T I G AT I N G
0 1 0
1 1 0
235
. 0 0 1 0 0 1
(Because 1 + 0 = 1 and 1 + 1 = 0)
To create the vectors. 1 0 0 We obtain 1001101 b1 B 0101011 b2 0010111 b3 This is a set B of three vectors. so a vector 0001000 is received as 0001001.1 strings can be used to create vectors for communication. The following system is the basis for the system originally used by NASA and provides a one-error correcting code. changing rows to columns) and then appending the 4 4 matrix with 1 in the diagonal and 0 elsewhere. it can be done if we are clever in constructing the vectors we use. that allows us to determine the validity of a vector received and to correct it if exactly one error occurs. we have (1001101) • (1101011) 1 1. we wish to correct it if it is changed.ERROR–CORRECTING CODES
In the last chapter. b1. we use the matrix 100 010 001 together with all possible linear combinations of the vertical vectors making up the matrix. b2. It uses the same addition as the earlier example but also employs the dot product of vectors. Further. Hence for vectors.
vectors. all with the same property. b3 . How can we tell whether it was originally sent in this form and. You should verify that this is so. or four together) we can create a total of 15 vectors. This is the location of the error. that their dot product with each of b1. Now suppose that v6 v1 v3 (1011010) is sent from space but in passing through a field of lightning is changed to (1010010). b3 is 0. the vector received is corrected to that which was sent. combinations of vectors. but we have shown how to create a system using the properties of matrices. you see that the clever use of number properties together with the algebra of vectors allows for the creation of impressive technological applications. In more advanced applications. b2. With this system we have only 15 vectors. By adding v1 v2. the approach described here is used in systems requiring the correction of errors when it is impossible to obtain original messages for comparison. we create v5 0111100. b3 is 0. In this example. if it wasn't. b2. Where is the error? The vertical vector 1 1 matches the vector in the array B in column 4. how it has been changed? We simply take the dot product of the vector received with each of b1. three. and dot product. b2. as follows: (1010010) (1010010) (1010010) (1001101) (0101011) (0010111) 1 1 0
Because we do not obtain 0 in each dot product there is an error.These vectors have the property that their dot product with each of the vectors b1. and by taking all possible combinations of them (two.
236 C H A P T E R 6
. By changing the 0 0 in the fourth position to 1.
CHAPTER EXPECTATIONS In this chapter. is a superior tool. There are wellestablished formulas for finding the slope or direction of a line in two-dimensional space. Both Euclidean and analytic methods are used for solving problems in three-dimensional space as well. and hyperbolas.Chapter 7
LINES IN A PLANE
When we solve geometric problems in twodimensional space.4 solve problems involving intersections of lines and planes. 7.4
.1. you will
• •
determine equations of lines in two. For solving problems involving curves such as parabolas. but how do you express direction in three-dimensional space? There are also formulas for lines in twodimensional space. but are there corresponding formulas for lines in three-dimensional space? In this chapter. however. 7. we will use vectors to develop these formulas and to solve problems involving points and lines in two and three dimensions. the analytic geometry of Descartes. ellipses. but vector methods are more powerful than either the Euclidean or the analytic method. 7.2.and threedimensional space. using the language of algebra. Euclid's methods are usually sufficient for problems involving polygons and circles.3. Section 7. Section 7.
A line segment has a finite length.
238 C H A P T E R 7
. but vectors are used to describe lines. Such lines are said to be coincident. A vector defines a direction with a clear distinction between forward and backward. We will also develop the principal concepts needed to solve problems about the intersections of and distances between straight lines in both two and three dimensions.
Lines Lines are bi-directional. This equation is not suitable for describing the equation of a line in space.
The equation of a straight line in a plane in the form y mx b is familiar from earlier mathematics courses. Two lines are the same when they have the same direction and same location. A line defines a direction. but there is nothing to distinguish forward from backward. The opposite sides of a parallelogram are two different line segments. vectors are used to investigate the geometry of straight lines and Euclidean planes in two and three dimensions. A line is infinite in extent in both directions. Such vectors are said to be equal. a new form of the equation of a line based on vectors is developed. Vectors Vectors are unidirectional. Lines and line segments have a definite location. In this chapter.Review of Prerequisite Skills
In this chapter and the next. Lines are not vectors. Vectors have a finite magnitude. Two vectors are the same when they have the same direction and the same magnitude. Their similarities and differences are presented in the following table. one that can be extended from two to three dimensions. The opposite sides of a parallelogram are described by the same vector.
A vector has no fixed location.
How has mathematics been used in artistic practices? ●
RICH LEARNING LINK
239
. y) ( . For example. 2). Here. Sometimes mathematicians produce interesting. As t increases. at time t.
y 2 2 t=4 t=1 x 2 4 6 8 2 4 6 8 x 2 4 6 8 10 x y 2 t = 10 y
This figure is called a trochoid. These will be defined more precisely in Chapter 8.investigate
C H A P T E R 7 : E Q U AT I O N S O F L I N E S
Some forms of mathematics use creativity and imagination in a way that is similar to artistic creation. (Note: when working with parametric curves. The result is shown below. if t radians. y). the point moves through the plane. Investigate and Inquire One way to create interesting images is through the use of parametric equations. even beautiful. to use your imagination? Why does math sometimes seem unlike art? 3. How can parametric equations be used to describe a curve in three dimensions? What about four or more dimensions? 2. images. t 0. Try drawing this on a graphing calculator. When has mathematics required you to be creative. then (x.
DISCUSSION QUESTIONS
1. as it moves through the plane. each value of t gives a point (x.) Many parametric equations can be interpreted as the position of a particle. we will evaluate trigonometric functions using radians. y 1 cos(t). but one example of a set of parametric equations is x t 2 sin(t). It is not uncommon to hear mathematicians refer to theorems as elegant or even beautiful.
Section 7.1 — Parametric and Vector Equations of a Line in a Plane
Imagine you are travelling at a constant speed along a perfectly straight highway that runs south and east from point A toward point B. Unlike the familiar formula y mx b. since you are travelling to the south. The highway from A to B is a straight line. y) on the highway at any time t. y0). when you describe a relation between two variables in an indirect manner using a third variable.
240 C H A P T E R 7
. but how? What are the equations which relate x and y to t?
A(x0. vy). your position. As you travel. which will be negative in this example. In this vector. It is important to realize that the equations derived above represent a new and different way to describe this straight line. where vx is the eastward component of your velocity. The distance that you have travelled east is the difference x x0 between your present position at point P and your starting position at point A. In mathematics. changes from moment to moment. Therefore In like manner x y x0 y0 vxt vyt
This pair of equations gives your position P(x. P. t. which will be positive in this example. vx is the eastward (x) component.and y-coordinates of your position depend on t. has passed since leaving point A. and t is the length of time you have been travelling: x x0 vxt. vy is the northward (y) component. here each of the coordinates x and y is expressed separately in terms of a third variable. which expresses y as a function of x. The x. after starting from A(x0. y0) v
north
east
P(x. depending on how much time. y)
B
Your velocity v is a vector (vx. Equations that show how the two variables depend on that parameter are called parametric
equations. t. Consider first your motion toward the east. the third variable is called a parameter. This distance is equal to vxt.
b) has slope b . is an example of a direction vector. a vector from one point to another on the line. and each point on the line is characterized by a unique value of t. at this point in time. 0). it 5
y (a. y0) (0. 102 0 85t. any vector d (a. The parameter t. then t takes 6 of an hour. a
7 . with (x0.The parametric equations of a straight line in a plane have the form x y x0 y0 at bt
where (x. y) is the position vector of any point on the line (x0. In general. 78. b) is a direction vector for the line and t R is the parameter. 1 PA R A M E T R I C A N D V E C TO R E Q UAT I O N S O F A L I N E I N A P L A N E
241
. is a real number that can take on any value. for you to reach a point on the highway that is 5 102 km east of Regina.
6 . A line with direction vector (a. Using the x-component of the velocity. you are Then y 0 ( 65) 6 or y 5 at a position on the highway that is 78 km south of Regina. This could be. The diagram below shows how the direction vector for a line is related to its slope. How far south of Regina are you when you are at a position 102 km east of Regina? Solution The parametric equations are x x0 85t. Saskatchewan. Suppose you travel on this highway with a constant velocity (expressed in component form. provided a 0. b) a x b
Any vector that is parallel to a line may be used as a direction vector for the line. which represents the travel time above. each value of t corresponds to a particular point on the line. we indicate that the units are in kilometres and hours. which in Example 1 was parallel to the highway. b) parallel to a line may be used as a direction vector for the line. Just as each point in time corresponds to a position on the highway. or 72 minutes. 65) in the example. Therefore. By choosing the vector (85. y y0 65t. where east and north are positive) v (85. Consequently.
EXAMPLE 1
Highway 33 from Regina to Stoughton. 65) km/h. y0) is the position vector of some particular point on the line (a. is an almost straight line. for example. The velocity vector.
The simplest such vector is (0. t R
(5. A line with a slope of
5 has a rise of 3
5 and a run of 3. This vector or any scalar multiple of it such as (2. 4) and D(7. Then.EXAMPLE 2
State a direction vector for a. Does the point (1. x 5 2(3) 11 y 2 6(3) 16 Therefore. c. 6). y0) of the line are x y 5 2 2t 6t. To determine if (1. Solution a. 2 4) (4. When t 3. a line that has slope
5 3
c. write the equation of the line in the form y mx b. 8) lies on the line.
2) with direction vector (2. What point on the line corresponds to the parameter value t c. 16) on the line corresponds to the parameter value t = 3. a vertical line passing through the point ( 6. The vector CD has components (7 3. State the parametric equations of this line. even though the slope of a vertical line does not exist.
a. try to find its parameter value. The point the line goes through is irrelevant. Find the y-intercept and the slope of the line. The vector (3.
2) and (a. A vector parallel to a vertical line has a horizontal component of zero. Substitute (1. 1). 5) Solution a. The parametric equations
b. b)
(2. So. b. 2) b.
EXAMPLE 3
A line passes through the point (5. b. 6). the point (11. a direction vector does. 2). 8) for (x. 8) lie on this line? 3?
d. 1) would be a suitable direction vector.
242 C H A P T E R 7
. c. It is given that (x0.
5)
is parallel to this line and would be a suitable direction vector. the line that passes through the points C(3. y) and solve for t.
Using the y-intercept 2 the equation of the line is y 3x 17. 1) and Q(7.
5). for the line. b)
The vector equation of a straight line in a plane has the form r (x0. y) or (x.
Let us now look at the parametric equations of a line from a vector viewpoint.
Solution The vector PQ from one point to the other on the line may be used as a direction vector. y) can be reinterpreted as the position vector of the point P(x. d OQ OP (7. 8) does not lie on the line. 6). y) is the position vector of any point on the line. y y0 bt becomes (x. b) where r (x. 1 PA R A M E T R I C A N D V E C TO R E Q UAT I O N S O F A L I N E I N A P L A N E
243
. y0 bt) (x0. a vector equation of the line is r (4. y0) t(a. (x0.
5 2
so t so y
17 17. 6)
Then. 1) t(3.
EXAMPLE 4
State a vector equation of the line passing through the points P(4.1 t
5 2
2t
8 t
2 1
6t
There is no single parameter value that satisfies both equations. x x0 at. set x 0 y 5 2 2t 6
5 2
0 and find the values of t and then y. 1) (3. 2). Consequently. y) (x0 at. 5) s(1. we can combine the two parametric equations into one vector equation. and t R. the parametric equations of a line are equations about the x. y0) is the position vector of some particular point on the line. 5) (4. t R. Therefore. Therefore.and y-components of vectors. b) is a direction vector for the line.
7 . y0) t(a. To find the y-intercept. y). the slope is 6 or 3. d. s R. (a. the point (1. d. Recall that the ordered pair (x.
Since the direction vector is (2. 6). so another vector equation of this line is r (7. You could also have used a shorter direction vector and the other point.
P0P is a scalar multiple of some direction vector (a. Since any vector parallel to the line will do as a direction vector. Consequently. (You may check in the same way that ( 9. b) for the line. The point (3. the point (3.
P0(x0 . 3). 1) b. y) x 0
OP0 is the position vector (x0. 10) from the second line lies on the first line with s 6. we check to see whether a point on one of the lines satisfies the vector equation of the other. 4) is on the first line.
To decide if the lines are coincident. or or OP OP0 td (x. yet still represent the same line. the vector equation of a line has an unusual feature.As Example 4 shows. to determine whether or not two different vector equations in fact represent two different lines. and the lines are coincident. 10) t( 6. It is important. The diagram shows that OP is the sum of the vectors OP0 to the line and P0P along the line: OP OP0 P0P. 4) from the first line does lie on the second line.
EXAMPLE 5
Are the lines represented by the following vector equations coincident? That is. 3)
Solution Check the direction vectors first. r (3. 10) t( 6.)
y
To summarize using vector language. and any point on the line can serve as the particular point required in the equation. d2 ( 6. 4) s(2. d1 Since d2 (2. y0) d P(x. then (3. r ( 9. b)
244 C H A P T E R 7
. 4) ( 9. Then 3 t 9 2 6t and 4 t 10 2 3t
Since the same parameter value is obtained from each equation. do these equations represent the same straight line? a. then. a. y0) of a particular point on the line. the vector equation of a line is a formula that gives the position vector OP of any point on the line. 1) b. 3)
3d1. y0) r r0 td t(a. two vector equations may look entirely different. y) (x0. If it is also on the second line. the direction vectors of the two lines are parallel.
a line parallel to r e. For each of the following. the line through (6. 0). r (4. 4) and ( 2. P(5. a.Exercise 7. State parametric equations a.
x
y
x
5. x 3 8t.1
Part A
Communication
1. 4) b. State a parametric equation and a vector equation for each of the following lines. a. 7) t(4. 5)
Knowledge/ Understanding
Knowledge/ Understanding
4.
y
b. State a direction vector for each of the following lines. y 4 6) t b. for the x-axis b. y 4t b. In each case. 3) 9 3t. d (1. d. a line parallel to x c. 1). 0) t(0. 1 PA R A M E T R I C A N D V E C TO R E Q UAT I O N S O F A L I N E I N A P L A N E
. a. for a line parallel to but not coincident with the x-axis
245
7 . a. the line y 3x 6 (1. P(1. What is a direction vector? What is a parameter? What role do these quantities play in the equation of a line? 2. r ( 3. d (4. 4) t(4. x y 6 1 2t 5t b. a. Graph the following lines. find the parametric equations of the line that passes through the point P with direction vector d. State the coordinates of two points on each of the following lines. 3)
7. a vertical line 3. find two points on the line different from P. a horizontal line f. 3)
6.
1 3 4 t 1. For each of the following lines. 4) and r t( 3. 3)
1). r 1 3t. 0º α 180º. P( 2. y
1 2
3
2 t 3
3
t 1. 1) t(5. For each of the following. 4). 5) and r ( 5. 1) contain the point (1. r
1 3 1 . Find a vector equation of the line that passes through the point (4. 10)
13. 7). (
4) 3. If possible. a. y 7 4t and x 2 (2.and y-axes. For each of the following lines. You are driving from point A(24. d
2 . 6) u(3. r c. Find the points where each of the following lines intersects the x. 6) t(3. (i) r (2. 0) (2. 0). P(0. 65) km/h. r
Application
6. 1) b. 5 2
Thinking/Inquiry/ Problem Solving
10.6 3
( 2. 96) on a map grid toward point B with a velocity defined by d (85.
Thinking/Inquiry/ Problem Solving
(3. a. P(1. 14. 1)
( 5. d
b. s(4.Part B 8. y (2. The angle α. Graph the line. 4) and r
11. Show that both lines r (3. that a line makes with the positive x-axis is called the angle of inclination of the line. 5) t(3. r c. 3)
9. a. x b. determine which pairs of lines are parallel and which are perpendicular. Find the angle of inclination of each of the following lines. a. 3s 4) 3) (1. d 4 d. 3 . 12. x b. P 2. 8) t(3. state a direction vector with integer components. 9) t (2. Find the acute angle of intersection of these lines to the nearest degree. Prove that the tangent of the angle of inclination is equal to the slope of the line. y s(3. 3)
1
7t t(1. name a point on the line with integer coordinates. find the vector equation that passes through the point P with direction vector d. d c. 5) and is perpendicular to the line r (1. 3 1 . a. r c. 0) 4s.
246 C H A P T E R 7
. 7). 7) t( 3. 4) (ii) r (6. x b. 2
2t. 7) (1.
Application
15.
8) and Q(17. B( 5. c. b. State the parametric equations of the highway line. 3) with direction vector (2. d. What are the coordinates of your position P at that time?
Thinking/Inquiry/ Problem Solving
16 a. 22) using parametric equations with suitable restrictions on the parameter. show that the equation of a line can be written in the form a 0 (provided neither a nor b is zero). When t 2. Part C 17. b.a. 2)
b. 8) to Q(17. Find the vector equations of the two lines that bisect the angles between the lines r1 r2 (5. a. b. For what values of t does the point R with position vector r lie between points P and Q on the line? c. By eliminating the parameter t from the parametric equations of a line.
x x y b y0
Thinking/Inquiry/ Problem Solving
18. y 3 5t (ii) r (0. 5) . 2) (5.
7 . Describe the line segment from P(5. For what values of t does the point R with position vector r lie closer to Q than P? 19. This is known as the symmetric equation of a line. Find a symmetric equation for each of the following lines. b. 4) t(4. 4). 2) t( 3. How long have you been travelling when you reach a point P 102 km east of where you started at point A? c. a. 22) are points on the line that passes through A(7. Find a symmetric equation for the line through the points A(7. 1) 2) and c. Are the two lines that bisect the angles made by the intersecting lines always perpendicular? Explain. draw a vector diagram that shows where point R with position vector r lies on the line relative to points P and Q. 6) u(11. 1 PA R A M E T R I C A N D V E C TO R E Q UAT I O N S O F A L I N E I N A P L A N E
247
. Show that the line that passes through P and Q has the vector equation r (1 t)p tq. Suppose p and q are the position vectors of points P and Q in the plane. (i) x 5 8t. Sketch all four lines. a. Show that P(5.
Then.2 — The Scalar Equation of a Line in a Plane
Another way to form the equation of a line is to use a vector that is perpendicular to the line rather than one that is parallel to the line. therefore. Write the equation of the line in the form Ax By C 0. n2) 0 2n1 5n2 0 One of the many ways this equation can be satisfied is by choosing n1 5 and n2 2 . 1) with normal (5. 2 b. a direction vector is defined by P0P (x 2. 2) is a normal to the line with direction vector (2. y 1) 5(x 2) 2(y 1) 5x 10 2y 2 5x 2y 8 0 0 0 0
P0(–2. The dot product of a normal vector and a direction vector is always zero because they are perpendicular. Any vector that is perpendicular to a line is called a normal vector or simply a normal to the line. 1). 2) x
This vector is perpendicular to the normal. (5. The dot product of these vectors must be zero. (2. This is the key to the use of normal vectors in two dimensions. y 1). The dot product of (2. The slope of the given line is
line is 1 . A vector normal to the line is. 2) • (x 2. 5). 5) and any normal vector (n1. 1)
This is the equation of the line through ( 2.Section 7. y) 2x (2. which passes through the point ( 2. EXAMPLE 1
Find a normal to the line a. 2). (x. The direction vector is (2.
P(x. 5).
248 C H A P T E R 7
. (2. n2) must be zero. y) y n = (5. y b. y) on the line. The slope of a line perpendicular to the given
Solution a. 5). 1). Solution For a point P(x. 5) • (n1. t R 2. 5 3) t(2. (5.
EXAMPLE 2
Find the equation of the straight line with normal (5. 2).
2) we can use (3. it is a relatively straightforward task to find the distance from a point to the line. following equation. y0) to any other point P(x. measured along the normal through Q.
EXAMPLE 3
Find the scalar equation of the straight line with normal ( 6. 4) 2(3. The vector P0P along a line from a fixed point P0(x0. y) must be perpendicular to the normal n (A. where C
Ax0
By0. Then n • P0P 0 (A.
y Q
2) as a normal to the line. The
Since the point ( 3. B).and y-terms. (Why?)
N P0 x
7 . its coordinates must satisfy the C C 0 5 2y 5 0. B) is a normal to the line. This distance is shorter than the distance from Q to any other point P0 on the line. we can see that the components of the normal end up as the coefficients of the x. B) • (x x0. The following derivation demonstrates that this will always be the case. y y0) 0 A(x x0) B(y y0) 0 Ax By ( Ax0 By0) 0 Ax By C 0. 2 T H E S C A L A R E Q UAT I O N O F A L I N E I N A P L A N E
249
. 7). equation must be of the form 3x 2y C 0 7) lies on the line.
The scalar or Cartesian equation of a straight line in a plane has the form Ax By C 0 where the vector (A. 3( 3) 2( 7)
The equation of the line is 3x
When the equation of a line is expressed in scalar form. The shortest distance from the point Q to the line l is QN.In the equation found in Example 2. 4) that passes through the point ( 3. Solution Since ( 6.
1)
P0(3. 2) gives P0Q (2. n (7. 6). y0) is on the line.
Solution In the diagram. y0) be a point on the line. Then QN is the magnitude of the projection of P0Q onto the normal to the line. y1
y0) • (A.EXAMPLE 4
Find the distance from the point Q(5. the distance. 6) • (7. 2) x
2 2 The distance from the point Q(5. so QN proj(P0Q onto n)
y Q(5. 1). 8) to the line 7x y 23 0 is 2 2 units. B) B2 Ax0 By0 B2
A2 Ax1 By1 A2
Since P0(x0. so Ax0 or C By0 Ax0 C By0 0
Ax1 By1 C . 8) to the line 7x
y
23
0. it satisfies the equation of the line. Also. we can find a simple formula for the distance from a point Q(x1. y1) to a line with scalar equation Ax By C 0. By working through the steps of the solution to Example 4 in general terms. 8) N n
7 1 14 6 50
2 2
(2. Choosing P0 to be (3. is d proj(P0Q onto n)
P0Q • n n (x1 x0. Letting P0(x0. y1) to the line Ax is given by the formula d
Ax1 By1 A2 B2 C
By
C
0
250 C H A P T E R 7
. A2 B2
Then the distance is d
The distance from the point (x1. the required distance is QN. where P0 is any point on the line. d. where N is the point where the normal through Q meets the line.
is parallel to 2x 3y 8 2y 3) 2) 0 12 0 b. State a normal of the line that is a. 4x c. find a normal vector. 7) t(8. 1 . Explain why there is one and only one scalar equation of a given line. P0 1 . Find the scalar equation of the line through (8. y 6y 3x 14 10
5. 7)
7.
7 . 2. 1) b. n 3 3 ( 4. Prove that both ( b. 2) that is parallel to the line x 4 5t. b) for all a and b. 0) ( 1. parallel to 2x d.2
Part A
Communication
1. and a point on each line. 3x d. 2 . y 11 3t by first finding the symmetric equation of this line. perpendicular to 2x b. y 2 6) and 4 9t (2. 7) (1. (x.
a) are perpendicular to (a. a. 6) t(2. Find the Cartesian equation of each of the following lines. n
4. For each of the following. r (4. and then simplifying it. For each of the following. y 18t. n (2. x d.
6. y) c. has a normal vector (3. a. P0(4. n 2 d. find the scalar equation of the line that passes through the point P0 and has normal vector n. 2) b. a. 5) c. 2) t(4. Find the scalar equation of the line that passes through (2. 0 5) 2) t(4. a) and (b. d. is perpendicular to 3x c. 1)
c. has a direction vector (2. P0(3. a. parallel to r 4y (2. P0 1 . 2 T H E S C A L A R E Q UAT I O N O F A L I N E I N A P L A N E
251
.
8. 5 0
Communication
Knowledge/ Understanding
3. a direction vector. 2). perpendicular to r 4y 5 (2. x 3 2t. whereas there are many different parametric and vector equations for the line.Exercise 7. 3). x 3y 5 12 0 b.
Using diagrams. of inclination of 120º.
Application
b.
252 C H A P T E R 7
. a. draw a normal to the line through A. 3x c. b. 1 5 6 b.
4x
6y
9
0
10. 2).1.) b. a.
20 3
14.
d. 5)
d. parametric. Prove that the shortest distance from a point to a line is the distance measured along the perpendicular from the point to the line. Through point B(8. Show that N is a point on the circle defined by AN • BN = 0 b. b. (4. n is a normal to a line and OP is the position vector of a point P(x. find the distance from Q(3. Describe the relationship between this circle and the points A and B. a. a. (See Exercise 7. (4. Find the distance from each of the following points to the line 6x 3y 10 0. Find the scalar equation of the line through the point (6. y) on the line. 16. Question 14. 11.
x 2 3 y 7 4
2) to each line. Find vector.
5. show that the line goes through the origin when n • OP 0. y). and symmetric equations of the following lines. a. a. 10). Find the angle of inclination of 2x 4y 9 0. Prove that two lines in a plane are parallel if and only if their normals are parallel. 7) Part C 13. Prove that two lines in a plane are perpendicular if and only if their normals are perpendicular. Prove that the line goes through the origin if and only if n • OP 0. For each of the following. Draw any line through point A(2. meeting it at the point N(x. 4) with an angle c. (0. a.
Thinking/Inquiry/ Problem Solving
15. x
5
12.
8)
c.Part B 9. 5x
Thinking/Inquiry/ Problem Solving Knowledge/ Understanding
3y
15
0
b. 6 7) 0 t 1. r 2y ( 3. Show that the equation of a line that has an angle of inclination α can be expressed in the form x sin α y cos α C 0.
z0) t(a. z)
where r is the position vector of any point on the line r0 is the position vector of some particular point on the line d is a direction vector for the line and t R. 1) 3) or. PQ OQ OP (2. 6) 4. we must introduce a z-coordinate for points and a z-component for vectors.
EXAMPLE 1
Determine a direction vector for a. therefore.
The numbers a. y.S PA C E
253
. the line that passes through the points P(6. 8. 1. (0. except that there is no scalar equation of a line in space because a line in space does not have a unique normal. b.
b. better still. (2. ( 2. 5) (6. a line perpendicular to the xz-plane Solution a. 5)
4. direction vector for this line. better.Section 7. 2. b. 1) and Q(2. A vector perpendicular to the xz-plane is parallel to the y-axis. 2. The equations are otherwise very similar. 3) could be used as a
This vector or. and c.3 — Equations of a Line in 3-Space
In generalizing the equations for a line from a two-dimensional plane to a threedimensional space. ( 4. 3 E Q UAT I O N S O F A L I N E I N 3 . b. The vector equation of a straight line in space has the form OP or or r r0 OP0 td (x0. 0). A suitable direction vector is. 8. y0. are known as direction numbers of the line.
7 . 4. c) td
(x. which are the components of the direction vector.
so the point Q does lie on the line. and 1) 8 t t(1. Does the point Q( 3. 13) lie on this line?
1) and
Solution a. Then or 3 t 8.
x
z P0(x0 .t
y b
y0
. and c are direction numbers for the line. parameter t such that ( 3. 8.The position vector OP to a general point P(x. P0P is some scalar multiple of a direction vector. Solving each of the parametric equations for the parameter t gives t
x a x0
.
1)
t(1. and a. or c is zero. A vector equation of the line is r b. b. z0) r0 O d r P(x.
254 C H A P T E R 7
. z) y
EXAMPLE 2
a. 0. 0. b. has direction numbers (1. 2. 2. 2. and t
z c
z0
provided that none of a. 0. and t R. 3). The parametric equations of a straight line in space have the form x y z x0 y0 z0 at bt ct
where (x0. y0.
13) lies on the line only if there is a value of the (1. y0. b. y. z0) on the line. 1 4 t 13)
(1. These expressions give an alternate form for equations of a straight line in space.
When each component of the vector equation is written out separately. z) on the line can be expressed as OP OP0 P0P. The point Q( 3. y0 . OP0 is the position vector of a particular point P0(x0. the resulting equations are the parametric equations of a straight line in space. 3). 3) 2t 4 and 13 t 1 4 3t
This vector equation is satisfied by t
4. Find a vector equation of the line that passes through the point P(1. y. 8. z0) are the coordinates of some particular point on the line.
and c are direction numbers for the line with a. 8) t(1. In cases like this. a vector equation of the line is r (2.S PA C E
255
. 2. Solution Since AB (3. 3 E Q UAT I O N S O F A L I N E I N 3 . 2). then (1.)
EXAMPLE 4
Write a vector equation for the line Solution Rewriting the equations. 0. 14). 2) is a direction vector for the line. and c 0. Using A as the fixed point. parametric. and a. 2. 0. z0) are the coordinates of some particular point on the line. there is no symmetric equation.
y
( 2) 1
z 1
(0)
Then by inspection. b. and symmetric equations of the line that passes through the points A(2. 1)
z 3 1
EXAMPLE 5
y 4 Do the equations x 2 5 5 represent the same line?
and x 41
y
11 10
z
4 6
7 .
x (0) 1
x
y
2
z. y0. Then the corresponding parametric equations are x 2 t y 2 z 8 2t and the corresponding symmetric equations are
x 1 2 z 8 . 1. 6). 2. b.
EXAMPLE 3
Find vector. 0) t ( 1. 0. a vector equation of the line is r (0.The symmetric equations of a straight line in space have the form
x a x0 y b y0 z c z0
where (x0. 2.y 2
2
There is no symmetric expression for y because the corresponding direction number b 0. when y does not change with t. you must still state its value. (If two direction numbers are 0. 8) and B(5.
1.
5)
b. For each of the following. if possible. 0. (11) 5 4 3. 6)
3. Parametric equations are x 0.Solution The direction vector of the second line ( 4. 3. d ( 1. 5). 10. a. 5. and symmetric equations of the y-axis. 2) 4
4 3
t(3. d c. Find a direction vector for a line a.3
Part A
Communication
1. if possible. parallel to x (7. P0(1. Give the coordinates of two points on each of the following lines. (0. A vector ˆ equation for the y-axis is r (0. r b. z
x
4. but no scalar equation? 2. parametric. z 0. through (0. Why does a line in space have a vector equation and a parametric equation.
Knowledge/ Understanding
(1. 0). 4. 0. 3) and (7. parallel to r c. therefore the lines are parallel and distinct. so the lines are parallel. 3). 1. ( 4) 1 1 3 The fractions are not equal. 4. They are coincident if the point ( 1. 11.
( 1) 2 5
EXAMPLE 6
Find vector. y
y 4 5
5t. 0) or simply r tj .
1 1
1) 5 4t
2t. 2. It has no symmetric equation because two of the direction numbers are zero.
y 2
Knowledge/ Understanding
9. a. x c. and. 4. d b. find vector. 6) is 2 times the direction vector of the first line (2. y t. parametric. 2
z
1. 3. P0(0. 1)
256 C H A P T E R 7
. 6. 4) on the second line satisfies the equation of the first line. 0). 3)
3
z 4
t( 4. 1. P0(2. 6). 2) 1) ( 1. 0. Solution ˆ The y-axis goes through the origin and has direction j (0.
Exercise 7. symmetric equations of the line that passes through P0 and has direction vector d.
2) and is perpendicular to both of the lines
x 4
y
10 6
z 3
2
and x 3 5
y 2
5
z 4
5
. 0. 1. Find the symmetric equations of the line that passes through the point ( 6.
10. Find symmetric equations for the line through the origin that is parallel to the line through the points (4. 2. y 3 t. 3.S PA C E
257
. 4. y 0. 3) and r (2. or neither of these. a. 4. y 5. 5). Describe the set of lines in space that have one direction number equal to zero. b. b. x
Thinking/Inquiry/ Problem Solving
t. r b. a. Part C 12.
( 2. z 1 2 t. 1) s(3. 2) 3. 6. x b. 0. Find parametric equations for the line that passes through the point (0. 6. 5) for even integer values 2t. 4) 1. 0) s(3. 6) and r s(6.
7. x c. 4. 1) and ( 2. z 1 t?
Application
6. 2. y
2. 2)
3) lies on the line. 2. 0. Sketch the lines.
8. z
1 1 2 t t t. 0. r (1. 1) t(2.
1. find the values of a and b.Application
5. 1) t(2. 5. 3. 5) 3. 3)
t(3. determine whether they represent the same line. parallel lines. 9. 2) b. 0) and r ( 3. Describe the set of lines in space that have two direction numbers equal to zero. 2) Q( 2. 4. a.
Thinking/Inquiry/ Problem Solving
1. For each of the following pairs of equations. Describe in words the lines having the following parametric equations. 2) t( 9. r c. Part B
S(6. 1) and the midpoint of the line segment from (2. Which of the following points lies on the line x P(2. List the points on the line r of t from 6 to 6. ( 5. 4. 3) (2. 3). 3 E Q UAT I O N S O F A L I N E I N 3 . a.
7 . If the point (a. 1. (1. 1) R(4. z
11. 2) to (4.
c. 3) to the line ( 2. 1.
258 C H A P T E R 7
. 0. 5. Describe the line segment from A to B using parametric equations with suitable restrictions on the parameter. Find the distance between the parallel lines r and r (2. Show that the points A( 9. 1. 4). 2) u(7. a. Find an equation of the line through the point (4. 6). 2. 1. 1) t(7. 3. 2) and has direction numbers (3. b. r (3. Prove that the distance from a point Q in space to a line through a point P with direction vector d is equal to
PQ d d
. 5) that meets the line
x 3 11
y
8 1
z 1
4
at right angles. 3. Find the distance from the point Q(1.
2.
15. 16) and B(6. 14) lie on the line that passes through (0. 2). 1.
Thinking/Inquiry/ Problem Solving
14. 4)
b. 0) t(1. 2. 3. a.13.
4 THE INTERSECTION OF TWO LINES
259
. thereby having an infinite number of common points. intersecting at a single point. you can find their point of intersection by the familiar method of elimination.
EXAMPLE 2
Find the intersection of the lines r r (18. 2) ( 5. they can cross.
EXAMPLE 1
Find the intersection of the lines and Solution Solving. intersecting at no points. 1)
7.Section 7.
l1 l2 l2 y l1 y l2 l1 y
x
x
x
parallel
intersecting
coincident
When the equations of two lines are expressed in scalar form. the point of intersection is (3.4 — The Intersection of Two Lines
What are the possible ways that two lines in a plane can intersect? They can be parallel (and distinct). 8). 2) t(3. 2x 2x 30 0 26 0 56 0 y 8 Substitute in the second equation. 4) s(2. x 2(8) 13 x 0 3 3y 4y 7y 2x x 3y 2y 30 13 0 0
Therefore. or they can be coincident.
Equation 2 (–2) Equation 3 4t 4t 3s 2s 5s 1 14 15 s 0 0 0 3
Substituting. or coincident. 8). Nevertheless. y. because they lie in different planes. 18 2 3t 2t 5 4 s 5. They just pass by each other like the vapour trails left by two aircraft flying at different altitudes. line 1 x y 18 2 3t 2t line 2 x y 5 4 s 2s
Equating the expressions for x and y.Solution First write the parametric equations of the lines. lines in space can be parallel. the coordinates of the intersection point are (3. they do not intersect. and z gives 1 1 2t 1 3s 7 s 2s or 3t 4t 2t 2s 0 3s 1 0 s 7 0
Solve for s and t using the second and third equations. 4t Verify that t 2 and s 1 0 t 2 3 satisfy the first equation. s
Substituting these into line 1 or into line 2. But there is also a new possibility: they can be skew. 4 and t 2s or 3t 2s 23 2t s 6 0 0
Solving. Like lines in a plane. Skew lines are not parallel. 3(3)
260 C H A P T E R 7
. intersecting at a point.
EXAMPLE 3
Find the intersection of line 1
{
3t 4t
x y z
1 1 2t 4t
3t
and
line 2
{
x y z
1 3s 7
2s s
Solution Equating the expressions for x.
It is now time to place the subject of this section. 0) t(1. so the lines either intersect or are skew. Therefore. 1)
Solution The direction vectors are not parallel. and z gives 2 1 t t t 3 2s 3s 1 s or t t t x y z 3 2s 3s 1 s
2s 1 0 3s 1 0 s 1 0
Solving the first and second equations. The point of intersection is (5. y. which is the point determined by t 2 on line 1. t s 1 (1) 2 0 (0) 1
The values of t and s do not satisfy the third equation. the lines have no point of intersection. 1) (3. 9. 4). 1. into a more general context. 1.
3t
2s
EXAMPLE 4
Find the intersection of line 1 line 2 r r (2. 1) s(2. (–1) Equation 1 Equation 2 t t 2s 3s 1 1 5s s 0 0 0 0.3(2) 2(3) 6 6 0 Therefore. and s 3 on line 2. so t
1
Finally. 0.
7. check to see if these values of t and s satisfy the third equation. They are skew lines. The scalar equation of a line is an example of a linear
equation.4 THE INTERSECTION OF TWO LINES
261
. The parametric equations are line 1 x 2 t line 2 y 1 t z t Equating the expressions for x. 3. intersections of lines. the two lines intersect at a unique point.
. Problems requiring the solution of a linear system arise in disciplines such as engineering. particularly if all coefficients are non-zero. assuming that no two customers make identical purchases.
262 C H A P T E R 7
.. and k are constants.. the third buys seven different items worth $52. or imagine the task of solving for 1 000 000 forces acting on the beams in a large building.. . From this we can construct 50 equations in 50 variables. A linear system is a set of two or more linear equations and may involve thousands of equations. You can get a reasonable picture of a real situation from the following examples. Otherwise the system is inconsistent. Fortunately. economics. Solving a linear system when the number of equations is large is an extremely challenging problem. physics. in real life many of the coefficients are zero. Try to imagine the amount of work required to solve ten equations with ten variables.. Solving such a system involves computer applications and clever algorithms. are variables and the ai. and people skilled in analyzing such systems are greatly in demand. A system involving a large number of equations and having many coefficients equal to 0 is referred to as a sparse system. From these equations we can determine the cost of each item. Suppose that a grocery store that stocks 50 different items has 50 customers. The first buys six different items worth $20. the focus has been on the geometrical interpretation of a linear system and its solutions.A linear equation is an equation of the form a1x1 a2x2 a3x3 . Now picture the situation if there are 50 000 items in the store. In this section. and biology. the second buys ten different items worth $37. It is true that many of the coefficients are 0. The study of linear systems is a highly developed area.. The intersection problems considered in this section are elementary examples of linear systems. A system of linear equations may have i) no solution ii) a unique solution iii) an infinite number of solutions A linear system is said to be consistent if it has at least one solution. and so on. . k where the xi.
2) with the sphere x2 y2 z2 6. 4. b. 1. 4) u( 1. 13. 6. 1) and (x. Find the points of intersection of the common perpendicular with each of the lines (x. a. Find the points of intersection of the line r (0. Find the equation of the line through the point ( 5. Determine the point of intersection. 8. 0) t(2. y-. Find their point of intersection. 1) r ( 2. 1. 19. 15.
t(3. 2. a. 2). 7. 2) u( 2. 11. 0) s(1. 3. 10.
Application
24
7t. Determine the point N at which the normal through the origin intersects the line Ax By C 0 in the xy-plane. 2. 2). The common perpendicular of two skew lines with direction vectors d1 and d2 is the line that intersects both the skew lines and has direction vector n d1 d2. 1) and
b. Find the magnitude of the position vector ON of point N. 16.
8. 4) and r (14.
Thinking/Inquiry/ Problem Solving
14. z) (0. 2. Show that the lines r (4. 4) and r (1.5.
264 C H A P T E R 7
. If they exist. y.
Communication
Application
12. 1. Is the segment of the line between the intersection points a diameter of the sphere? 13. 2. 2) at 90º. 2) that intersects the line at r (7. 1. 3. 5. Find the point at which the normal through the point (3. 3). 10x 4y 101 0 intersects the line. y. 3) t(1. 19) t(1. and z-intercepts of the line x y 4 t. Find a vector equation for the line through the origin that intersects both of the lines r (2. find the x-. What are the possible ways that three lines in space can intersect? Describe them all with diagrams. 3) intersect at right angles and find the point of intersection. 0) u(1. 7. 2. 2. z) ( 2. Part C
4) to the line
Communication
9. 1) t(4. z 20 5t. What are the possible ways that three lines in a plane can intersect? Describe them all with diagrams. 1. Find a vector equation for the line perpendicular to both of the given lines that passes through their point of intersection. 8) t(1. 5. Consider the lines r (1.
z t and x
PQ • n where n Proj(PQ onto n) or n
OP d1 (0. higher dimensional spaces.4 THE INTERSECTION OF TWO LINES
265
. c) of the direction vector and the coordinates (x0. 0) 5. z
z0 c
r0
td
x0
x0
y0
z
z0
ct
Ax
By
C
0 (in two dimensions only)
Notice the position in each equation of the components (a. r r0 td. b. z) x
x a
(x0. Try visualizing lines in three-dimensional space. x (0. The distance between the skew lines r Find the distance between the lines a. 6. This equation also describes a line in more abstract.
7. To master this material. b. 1. Work at converting from one form to another by inspection. perhaps using the lines along the corners of a room as coordinate axes. 1. 5. y
td1 and r d2. 6) 4 t(2. z
OQ
sd2 is
1) and r
s( 1. the vector equation the parametric equations the symmetric equations the scalar equation (x. y0. t. 2s. The form of the vector equation of a line. y. y0. learn the various forms of the equation of a line. is the same whether the line lies in a plane or in a three-dimensional space. remembering to move parallel to the axes when you plot coordinates of points or components of direction vectors. where the vectors have more than three components. y 2. z0) at. r b. c) or r bt. Practice sketching graphs of lines in two and three dimensions. y
y b y0
t(a.Thinking/Inquiry/ Problem Solving
16. 2) 3 s
Key Concepts Review
This chapter has illustrated how the algebraic description of straight lines can be formulated in terms of vectors. z0) of a point on the line.
0. that passes through ( 9. r (4. Find a vector equation of the line
a. 3) 3. 5) and is parallel to the y-axis 3. 8) with slope b. that passes through the points (3. c. 4. 0) t(0. 6)
3 and direction vector (1. Is it possible for this line to intersect just one coordinate plane? exactly two? all three? none at all? 2. that has an x-intercept of c. 0)
x 4 5
3) and ( 3. Find parametric equations of the line a. that is parallel to the line x 31 origin b. Is it possible for this line to intersect just one coordinate axis? exactly two? all three? none at all? b. that is parallel to (0. 2) b. that passes through the point (6. 5) c. 1) t(3. 0) and (0. that passes through the point ( 5. that is perpendicular to the line 2x point (0. b. Consider any line in space that does not pass through the origin.
266 C H A P T E R 7
. 2)
2 3
3) and is parallel to the line 5y 6 0 and passes through the
2) and is perpendicular to the line 2)
c. 9) and ( 4. Find a vector equation of the line a.
6 5
2)
7 and a z-intercept of 4
y 2 2
z
and passes through the point
5. r (4. 2. through the points (4. that has a z-intercept of
y 2 2
z
3 and passes through the
4. 6. that passes through the points (2. that passes through (3. a.Review Exercise
1. Find parametric equations of the line a.
Find the Cartesian equation of the line a. 0. 1) and (x. 2). if any c. if any b. Find the direction cosines and the direction angles (to the nearest degree) of the direction vectors of the following lines. z)
u(2. 5) 0
t( 3.
2. coincident. At what points does the line x 2 4 planes? 10. 1. 4. z t(4. 4) and C(3. 3) and is perpendicular to the line x 2 t. find the intercepts with the coordinate axes. 4) u. find the Cartesian equation of the line r b. y. y-. 0) and is parallel to the line passing through B( 2. Determine if the following pairs of lines are parallel and distinct. b. find the intersections with the coordinate planes. 0)
REVIEW EXERCISE
267
. find a vector equation of the line 5x c. n) is a point on l. 2. 2)
1. perpendicular. 3x 4y 5 0 2) and is parallel to the line
b.6.z t and x t( 1.
4 1
( 1. y
2
t. x 1 c. y. m. y 3 2t c. y
y
2u 2t. find a vector equation of the line y 11. 1) and r 3 . x 5 3 b. In the xy-plane. 1)
2 4
2t.
meet the coordinate
a. 8. 4. that passes through the origin and is perpendicular to the line x 4y 1 0 7. (x. a.
x 2
(2. 7. z-coordinate system 12. r
y 2 6
z 1 1
8t.
a. r b. z
z
1 and x
1
6 1
d. a.
2)
9. 8. or none of these. that passes through the point ( 1. Find the parametric equations of the line l that passes through the point A(6. y
y
u(6. 0)
4
4t
( 7. z)
(1.
1 3
2) 6 ( 3. find m and n. a. If ( 4. graph the line in an x-. 0. Given the line r (12. that passes through the point ( 7. 3) 10
1 2
t( 1. y 4t. 3) 1
1
t( 3. 0. 1). x c. 4) 2y
3 x 4
(2.
Find the distance from the point (1. Find a
6. Determine the coordinates of P2. b.
270 C H A P T E R 7
. Find any two of the three intersections of the line x 6 2 the coordinate planes.
5. Show that l1 and l2 are skew lines (that is. 4 5. 3) and x 15 8
y
120º and γ
45º. 1. 4. 0. 2. 1) t( 3. A line through the origin has direction angles β vector equation for the line. 4). its parametric equations d. 3. 2) and (3. 3)
8 2
z 6 3 y 1 1 z be two 3
7. Let l1:x 8 t. the point on l1 determined by t 2. its scalar equation
2. Determine the point of intersection of the two lines (x. State the coordinates of P1. y) (0. Find the scalar equation of the line which is perpendicular to the line 2x 3y 18 0 and has the same y-intercept as the line (x. its symmetric equation b. A line goes through the points (9. 6
1. z) ( 2. c. 3) to the line x
y 3 4
z 2 with 3
y
z
2. neither parallel nor intersecting). t(5. its vector equation c. z lines in three-dimensional space. the point on l2 such that P1 P2 is perpendicular to l2. and graph the line. y. 4).Chapter 7 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application
Questions 1. y 3 2t.
3t and l2: x 2 1
a. 7 3 2. Determine a.
f ( f(f (x0))). x2 0..0 and x0 0.32. 0).2).2). x3 0.2 0.2 0...2. We can define a sequence by specifying x0 and then each subsequent term by xn f (xn 1). The points on the curve have coordinates (xn. Note that the sequence can be written as x0.6 0. .8 0. f(0.. .4
y=x 0.2.491602. The next point is ( f(0. You can easily check that the r graph of this function is symmetric about x 1 with a maximum value of 4 ..CHAOS
The simple quadratic function f (x) rx(1 x) where r is a specified constant can be used to demonstrate some of the most interesting ideas in modern mathematics.2))and so on. n 1. . f (0. The line y also shown on the plot. 1... The next point is (0. We are interested in the 2 function when 0 x 1 and 0 r 4 so that 0 f (x) 1. 2.2)). We then move horizontally to the line y x to get the point (f(0. All terms are between 0 and 1. If we start with r 2.4352.4 0.6
y = 2x(1 – x) 0.2)).2.. f (x0). xn 1). We can trace the development of the sequence on a plot of the function shown below..8 1
We start at the point (0.
E X T E N D I N G A N D I N V E S T I G AT I N G
271
. . n 0. It is easy to calculate the terms of this sequence on a spreadsheet. x is
0.2
0 0 0.. f (f (x0)). we get x1 2 0. f(f(0.
799
If you construct the corresponding plot. then the sequence will converge to the same value — this is easy to see from the graph and to check with a spreadsheet calculation. Write a spreadsheet program to try some values of x0 and r.5. a very small change in the value of r can lead to a complete change of behaviour of the sequence. let's see what happens if we change the value of the multiplier to and carry out the same calculations with x0 0. 32 .2 1 0.513 and 0.799 5 0. In this case.800 3 0. The sequence is called chaotic.799. this abrupt change is fascinating. there are some values for which the sequence is chaotic and some for which there are regular oscillations. there is no apparent pattern in the sequence for many starting points and there are no points of convergence. the sequence oscillates between these two values.The key question is to determine what happens as n gets large. you will see that there are now two points of convergence at 0. … points of convergence.57.2. if we look at larger values of r. Suddenly.5. If we start with any value 0 x0 1. we find that there are four points of convergence for most starting points. For mathematicians used to continuous behaviour. In fact by slowly increasing r. Try generating this sequence on a spreadsheet with x0 0. This does not seem very interesting. we can get 8. a very active branch of modern mathematics.513 4 0. but some. Even stranger. The first few terms of the sequence are
n xn 0 0.513 6 0. produce a single point of convergence. The big surprise occurs about r 3. we see that the terms of the sequence will approach 0. such as x0 16 . Different starting values lead to different sequences. However. Can you figure out why? If we further increase r to 3. Most starting 5 values produce the same limiting behaviour. 16. Can you produce the plot of the sequence as described above? In the chaotic case.512 2 0. The study of this sequence and its generalizations is called chaos theory.2.
272 C H A P T E R 7
. When n is large.
Chapter 8
E Q U AT I O N S O F PLANES
The concepts of point and line are as fundamental to geometry in three-dimensional space as they are in two-dimensional space. In three-dimensional space, however, we have another fundamental issue to consider, that of the plane. Is there such a thing as the equation or equations of a plane? Does a plane have a direction in space? In this chapter, we will look at the relationships between a point and a plane, a line and a plane, two planes, and even three planes. CHAPTER EXPECTATIONS In this chapter, you will
• • • •
determine the vector, parametric, and scalar equations of planes, Section 8.1, 8.2 determine the intersection of a line and a plane in three-dimensional space, Section 8.3 determine the intersection of two or three planes, Section 8.4 solve systems of linear equations involving up to three unknowns using row reduction of matrices, with and without the aid of technology, Section 8.4 interpret row reduction of matrices as the creation of a new linear system equivalent to the original, Section 8.4 interpret linear equations in two and three unknowns, Section 8.5
• •
Review of Prerequisite Skills
Two points or one vector (a direction vector) and one point determine a line, a one-dimensional object. A third point not on that line opens the door to a second dimension. Thus, three non-collinear points or two non-collinear vectors and a point determine a plane, a two-dimensional object. Similarly, four non-coplanar points or three non-coplanar vectors determine a three-dimensional space. These concepts can be generalized to higher dimensional spaces, which despite their abstract nature have a surprising number of applications in the physical sciences, engineering, and economics. The equation of a two-dimensional plane in a three-dimensional space has several forms. These are developed in the first part of this chapter in much the same way as those of a straight line. Recall in particular the vector equation the parametric equations r x y z Ax r0 x0 y0 z0 By td at bt ct C 0
the scalar equation of a line in two-dimensional space
Equations of lines and planes are essential parts of computer systems used by engineers and architects for computer-assisted design. The remainder of this chapter is concerned with how lines can intersect with planes and planes can intersect with other planes. Intersection problems are geometrical models of linear systems. Therefore, this chapter includes an introduction to systematic methods for solving linear systems using matrices.
274 C H A P T E R 8
investigate
C H A P T E R 8 : S U N E L E VAT I O N
Astronomers have always sought methods for determining and predicting the positions of objects in the sky. One reason for doing this has been to test and improve upon our models of the solar system. In this way, astronomers have already learned, among other things, that the earth revolves around the sun once every 365.25 solar days in an elliptical orbit at a distance varying between 147.1 and 152.1 million kilometres. The earth rotates on its axis once every 23 hours, 56 minutes, and 4 seconds. This is known as a sidereal day. The axis of rotation is tilted 23.45º from perpendicular to the plane of the earth's orbit. We shall investigate, here and in the wrap-up at the end of the chapter, how to determine the angle of elevation of the sun at any given time of any given day at any given place on the surface of the earth. Investigate and Inquire The angle of elevation of the sun is the angle between the vector from the earth to the sun and the plane tangent to the surface of the earth. Planes will be studied in this chapter. Here we shall set out some groundwork for our calculations. We will make the following assumptions to simplify our calculations: the earth is a perfect sphere that orbits the sun every 365 days in a perfect circle whose radius is 150 million kilometres. We shall let d be the number of solar days past December 21, the date of the winter solstice in the northern hemisphere. We will assume that at noon on December 21, the north pole is pointed away from the sun as much as possible. We let h be the number of hours (positive or negative) from noon. Noon here means the time when the sun is highest, regardless of the standardized time-zone time. It is the time halfway between sunrise and sunset. Let be the latitude of the observer. If we place the sun at the origin of a three-dimensional Cartesian coordinate system, we can parameterize the earth's orbit in the xy-plane as x 150 sin 360d , y 150 sin 360d (See the Rich Learning Link on 365 365 parametric curves in Chapter 7.) The vector s, from the earth to the sun, will be 150 sin 360d , 150 cos 360d , 0 . Why is it ( x, y, 0) and s ( x, y, 0) 365 365 not (x, y, 0)?
DISCUSSION QUESTIONS
1. What is the angle of elevation of the sun at sunrise and sunset? How might we interpret a negative angle of elevation? 2. Why is there a difference between the length of the solar day (the 24 hours between successive noons) and the sidereal day? ●
RICH LEARNING LINK
275
Section 8.1 — The Vector Equation of a Plane in Space
The vector equation of a plane gives the position vector OP of any point P(x, y, z) in the plane. It is constructed in the same way as the vector equation of a line. First, write the position vector OP as the sum of two vectors: OP0, the vector from the origin to some particular point P0(x0, y0, z0) in the plane, and P0P, the vector from the particular point P0 to the general point P. OP OP0 P0P
z Po sa P tb x O y z Po
P y
x
O
Now, choose two non-collinear vectors in the plane as basis vectors for the plane. Call them a and b. These two vectors are known as direction vectors for the plane. Express the point-to-point vector P0P as a linear combination of a and b. We write P0P Therefore, OP sa OP0 r0 sa tb sa tb
or, letting r and r0 stand for the position vectors OP and OP0, respectively, r tb
The vector equation of a plane has the form r where and r0 sa tb a and b are direction vectors for the plane, r0 is the position vector of a particular point in the plane s, t R.
The coefficients s and t in the vector equation of a plane are parameters. There are two parameters because a plane is two-dimensional. The parametric equations of the plane are equations for the components of r.
276 C H A P T E R 8
The parametric equations of a plane have the form x x0 sa1 tb1 y y0 sa2 tb2 z z0 sa3 tb3 where (a1, a2, a3) and (b1, b2, b3) are components of the direction vectors a and b for the plane, (x0, y0, z0) are components of the position vector of a specific point in the plane, s, t R.
and
In reality, a plane is a flat surface that extends infinitely in all directions. In the diagrams on page 276, we have depicted a plane using a parallelogram. This gives a three-dimensional perspective to the diagrams and suggests that the plane may be oriented at some angle to the coordinate axes. Although not true graphs, such diagrams are adequate for analyzing most problems about lines and planes in three dimensions.
EXAMPLE 1
Find vector and parametric equations of the plane that contains the three points A(1, 0, 3), B(2, 3, 1), and C(3, 5, 3). Solution The point-to-point vectors AB and AC both lie in the plane. They are AB AC (1, 3, 4) (2, 5, 0)
Since these vectors are non-collinear, they can serve as direction vectors for the plane. Taking point A as the given point, r0 OA (1, 0, 3). Therefore, a vector equation of the plane is r (1, 0, 3) s(1, 3, 4) t(2, 5, 0)
The parametric equations can be written down by inspection. x 1 s 2t y 3s 5t z 3 4s It should be clear that the vector and parametric equations of a plane are not unique. In Example 1, if BA and BC had been chosen as direction vectors and point B as the given point, then the vector equation would have been r (2, 3, 1) s( 1, 3, 4) t(1, 8, 4)
8 . 1 T H E V E C TO R E Q UAT I O N O F A P L A N E I N S PA C E
277
When two equations look entirely different, how do you decide if they represent the same plane? This question will be addressed in the next section.
Solving the first two equations gives p 4, q 2. But these values of p and q do not satisfy the third equation. Therefore, the point does not lie in the plane. You can also see that these values of p and q produce z 8 for the z-coordinate of the point, not z 3 as they should.
Solution We take (2, 4, 1) from l1 as the position vector r0 of a given point on the plane and (3, 1, 1) as a, one of the direction vectors. For the second direction vector, use the point-to-point vector between the given points on the two lines, (1, 4, 4) (2, 4, 1) ( 1, 0, 3). A vector equation of the plane is thus r (2, 4, 1) t(3, 1, 1) s( 1, 0, 3).
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Exercise 8.1
Part A
Communication
1. Why does the vector equation of a plane have two parameters while the vector equation of a line has only one? 2. a. State two direction vectors for the xz-coordinate plane. b. What do all direction vectors for the xz-coordinate plane have in common?
d. the plane containing the two parallel lines r (3, 2, 2) t( 9, 6, and r (1, 6, 6) s(6, 4, 4) e. the plane containing the three points (2, 6, 5), ( 3, 1, 4), and (6, 2, 2) 9. Determine the vector equation of each of the following planes. a. the plane parallel to the yz-plane containing the point (6, 4, 2) b. the plane containing the origin and the points (3, 3, 3) and (8, c. the plane containing the x-axis and the point ( 1, 10. a. Explain why the three points (2, 3, do not determine a plane. b. Explain why the line r (4, 9, do not determine a plane. 3) 1), (8, 5, t(1, 4, 7) 1,
6)
1)
5), and ( 1, 2, 1) 7, 5)
4, 2) and the point (8,
280 C H A P T E R 8
Application
11. Find vector and parametric equations of the plane that contains the line x 7 t, y 2t, z 7 t and that does not intersect the z-axis. 12. Demonstrate that a plane with a vector equation of the form r (a, b, c) s(d, e, f ) t(a, b, c) passes through the origin. Part C
Thinking/Inquiry/ Problem Solving
13. a. The vectors a, b, and c are the position vectors of three points A, B, and C. Show that r pa sb tc, where p s t 1 is an equation of the plane containing these three points. b. What region of the plane is determined by the equation, when the parameters s and t are restricted to the values 0 s 1, and 0 t 1? (Hint: replace p with (1 s t).) 14. a. The equation r r0 td is a vector equation of a line and q is the position vector of a point Q not on the line. Show that r kr0 lq td, where k l 1 is an equation of the plane containing the line and the point. b. What region of the plane is determined by the equation, when the parameter k is restricted to 0 k 1? (Hint: replace l by (1 k).)
8 . 1 T H E V E C TO R E Q UAT I O N O F A P L A N E I N S PA C E
281
Section 8.2 — The Scalar Equation of a Plane in Space
Any vector that is perpendicular to a plane is a normal vector or simply a normal to the plane. You can find the normal to a plane by finding the cross product of the two direction vectors of the plane. Since every vector in the plane can be represented as a linear combination of the direction vectors, the normal is perpendicular to every vector in the plane.
z n b a O x y
EXAMPLE 1
a. Find a normal to the plane with vector equation r (3, 0, 2) s(2, 0, 1) t(6, 2, 0). b. Show that the normal is perpendicular to every vector in the plane. Solution a. The two direction vectors of the plane are (2, 0, 1) and (6, 2, 0). The cross product of the direction vectors is (2, 0, 1) (6, 2, 0) Thus, a normal to the plane is (2, 6, 4) or (1, 3, 2).
Since the dot product is zero, the two vectors must be perpendicular. This result is independent of the values of p and q. You can use the fact that the normal to a plane is perpendicular to every vector in the plane to derive the scalar equation of a plane. Let P(x, y, z) be any point in a plane with normal (A, B, C), and let P0(x0, y0, z0) be some particular point in the plane. The vector P0P must lie in the plane because its endpoints do. Therefore, it must be perpendicular to the normal (A, B, C), and their dot product must be zero. (A, B, C) • P0P (A, B, C) • (x x 0, y y0, z z 0) A(x x 0) B(y y0) C(z z 0) Ax By Cz ( Ax 0 By0 Cz 0) 0 0 0 0
282 C H A P T E R 8
The quantity in brackets is a constant because the components of the normal and the coordinates of the given point have particular numerical values. Letting D ( Ax0 By0 Cz0) the result is Ax By Cz D 0
The scalar or Cartesian equation of a plane in space is Ax By Cz D 0 where (A, B, C) is a vector normal to the plane.
Unlike the vector equation, the scalar equation of a plane is unique. For instance, the equations x 2y 3z 4 0 and 2x 4y 6z 8 0 represent the same plane, since one equation is a multiple of the other.
The scalar equation of this plane is x 3y 2z 7 0, so the two vector equations represent the same plane, or the planes represented by the two vector equations are coincident. The distance from a point to a plane in three dimensions is calculated in much the same way as the distance from a point to a line in two dimensions. It is measured along the normal to the plane. If Q is some point not in the plane and P0 is any point in the plane, then the distance QN from Q to the plane is the projection of P0Q onto the normal n. EXAMPLE 3 Find the distance from the point Q(1, 3,
z n N O x Po y Q
A general formula can be derived by following the same steps. If P0 Q is the vector from some point P0 on the plane Ax By Cz D 0 to a point Q(x1, y1, z1) off the plane, then the distance d from Q to the plane is the projection of P0Q onto the normal (A, B, C). d Proj(P0Q onto n)
P0Q • n n
(x1
A(x1 Ax1
284 C H A P T E R 8
x0, y1
x0) By1
A2
y0, z1 z0) • (A, B, C) B2 C2
z0) Cz0)
B(y1 y0) C(z1 A2 B2 C2 Cz1 A2 ( Ax0 By0 B2 C2
Since P0 is a point in the plane, it satisfies the equation of the plane, so Ax0 By0 Cz0 D 0 or D Ax0 By0 Cz0. Substituting this into the above equation gives the following result. The distance from the point (x1, y1, z1) to the plane Ax By Cz D 0 is given by the formula d
Ax1 By1 Cz1 D A2 B2 C2
Note the structure. The numerator uses the equation of the plane, with the coordinates of the point off the plane substituted for x, y, and z. The denominator is the magnitude of the normal. In the special case when the point Q(x1, y1, z1) is the origin, the distance to the plane Ax By Cz D 0 is d
A2 D B2 C2
EXAMPLE 4
What is the distance between the planes 2x 4x 2y 4z 9 0?
y
2z
3
0 and
Solution The planes are parallel, since n2 (4, 2, 4) is a multiple of n1 (2, 1, 2). The distance between the planes is the distance from a point in the first plane to the second plane. The point (0, 3, 0) is on the first plane. Then d
4(0) 2(3) 4(0) 9 (4)2 ( 2)2 ( 4)2 15 36 5 2
Exercise 8.2
Part A
Knowledge/ Understanding
1. For each of the following, find the scalar equation of the plane that passes through the point P0 and has normal n.
8 . 2 T H E S C A L A R E Q UAT I O N O F A P L A N E I N S PA C E
where a. b. and C. c2
Thinking/Inquiry/ Problem Solving
Thinking/Inquiry/ Problem Solving
20. the y-intercept. B. and C in the scalar equation of a plane are the components of a unit normal. b. Find a formula for the scalar equation of a plane in terms of a. Show that the scalar equation of the plane through A. The vectors a. and c. and the z-intercept of a plane. the plane 2x 3y kz 0 rotates about a line through the origin in the xy-coordinate plane. and d is the distance from the origin to the plane. and c are the x-intercept. and C can be expressed in the form (r a) • (a b b c c a) 0.
Thinking/Inquiry/ Problem Solving Thinking/Inquiry/ Problem Solving
3). 18. B.
17. 2).
288 C H A P T E R 8
. b. 1. b. respectively. what is a geometrical interpretation for the constant D? 19. B. 6. and the z-intercept of a plane. Find the scalar equation of the plane through the points A(8. 1). show that
1 d2 1 a2 1 b2 1 .16. Show that as k varies. a. b. B(5. 4. respectively. If a. When the coefficients A. Find parametric equations for this line. and C( 4. and c are the x-intercept. the y-intercept. respectively. and c are the position vectors of three points A.
6 3t. 4(1 2t) 4 2( 6 3t) ( 5 8t 12 6t 5 2t) 19 2t 19 4t 8 t 0 0 0 2
Therefore. these coordinates satisfy the equation of the plane. z
4
t and the plane
8.
EXAMPLE 2
Find the intersection of the line x x 4y 2z 4 0.
1). 0. y 6 3t.3 THE INTERSECTION OF A LINE AND A PLANE
289
. for some particular value of t. the coordinates of a point on the given line are (x. y.
2t. It can cut through the plane. y
1
t. 5 2t). intersecting it at no points. intersecting it at one point. Substituting. the point on the line with parameter t intersects the plane. Its coordinates are x y z 1 6 5 2(2) 5 3(2) 0 2(2) 1
2 is the point at which the line
The point of intersection of the line and the plane is (5.Section 8.3 — The Intersection of a Line and a Plane
What are the possible ways that a line and a plane in three dimensions can intersect? The line can be parallel to the plane. in which case every point on the line is a point of intersection. z) (1 2t. Solution In terms of t. This point will lie on the plane if. It can lie in the plane.
z z z
O x
y x
O line intersects the plane
y x
O line lies in the plane
y
line is parallel to the plane
EXAMPLE 1
Find the intersection of the line with parametric equations x 1 2t. z 5 2t and the plane whose scalar equation is 4x 2y z 19 0.
1. 1). is perpendicular to the normal to the plane. 0. 0) on the line is a point in the plane. ( 4 3t) 2(0) 3(t) 4 3t 3t 4 4 0t 0 0 0
In this case. 0. the equation is satisfied for all values of t. 4. every point on the line is an intersection point. 1) • (1. 1 t. This means that the line must be parallel to the plane. These intersections are the key to making sketches of planes in three dimensions. (2t) 4(1 t) 2( 4 2t 4 4t 8 t) 2t 4 4 0t 0 0 8
There is no value of t which satisfies this equation. m (2. y-. or a plane may be parallel to or contain an axis. y
0. m•n (2.
EXAMPLE 4
Determine the x-. 2.
8y
8z
24
0. 2) 2 4 2 0
EXAMPLE 3
Find the intersection of the line x x 2y 3z 4 0. 3). z
t and the plane
Solution Substitute the point ( 4 3t.
4
3t. must be perpendicular to the normal to the plane. y-. 2). A plane may intersect an axis at a point. (1. Therefore. 1). 4 t) into the equation of the plane. The intersection of the line and the plane is the entire line itself. Its direction vector. The x-. 1. (1. and z-axes are lines in space. so there is no point at which the line intersects the plane. You can confirm this conclusion by checking that the particular point ( 4. Indeed. and the line lies in the plane. and that the direction vector of the line.
290 C H A P T E R 8
. 4. t) into the equation of the plane to find the parameter value of the point of intersection. The intersections of a plane with these special lines are of particular importance.Solution We find the parameter value of the point of intersection by substituting the point (2t. and z-intercepts of the plane 3x Make a sketch of the plane. (3. 0.
(3. has no component in the z direction. Plot the intercepts.and z-intercepts are 3 and 3. 2. Clearly. 0) x z
(0. 0. 0. the y.
(–8. The flat region between the parallel lines is a representation of the plane in three dimensions. the x. draw lines parallel to the z-axis. which extends infinitely in the directions shown by the orientation of the triangle. Therefore. Make a sketch of the plane. 0). 0) x
z
The x-intercept of this plane is the point 8. and sketch the plane as a triangular surface. Now.3 THE INTERSECTION OF A LINE AND A PLANE
291
.Solution To find the x-intercept. through the intercepts. 0) y
As observed. join them with straight line segments. and there is no z-intercept. 0. 0. 3)
x
(0. but it also intersects a coordinate plane in a line. The vertical lines through the intercepts are the lines in which the plane intersects the xz-plane and the yz-plane. 3. Keep in mind that the plane extends infinitely up and down and left and right. plot these on the coordinate axes. This figure is a three-dimensional representation of the plane. The line joining the intercepts is the line in which the plane intersects the xy-plane. knowing how to find these intersection lines would help us make the sketch of a plane. Likewise. there is a simple way to find the equations of these lines. 0)
y
Note that the sides of the triangle formed by the line segments joining the intercepts are segments of the lines in which the plane intersects each of the three coordinate planes. Then. in the directions shown by the orientation of the shaded area.
EXAMPLE 5
Find the intersections of the plane 3x axes.
(6. 3x 8(0) 8(0) 3x 24 24 x 0 0 8
z
y (a.
2y
18
0 with the three coordinate
Solution The normal to this plane. the plane must be parallel to the z-axis. a plane not only intersects a coordinate axis in a point.
8.and y-intercepts are 6 and 9. Fortunately. By inspection. set y and z equal to zero. respectively. 0) (0. 9.
You can reach the same conclusion by observing that every point (0. x 4 3 t.3
Part A
Knowledge/ Understanding
1. z 6t. The normal to this plane is (5. 0). For each of the following. 0.
EXAMPLE 6
Sketch the plane 5x
2y
0. From this set of planes.
Solution Since D 0. so setting z equal to zero does not change the equation).
z
The set of planes with this property is illustrated in the given diagram. In Example 4. is the plane where the z-coordinate of every point is zero. The scalar equation of the xy-plane is z 0. this plane is parallel to the z-axis. Sketch the plane as a parallelogram. a. for example. z) on the z-axis satisfies the equation of the plane. for instance. the plane intersects the xy-coordinate plane in the line 3x 8y 8(0) 24 0 or 3x 8y 24 0. the plane intersects the xy-coordinate plane in the line 3x 2y 18 0 (there is no variable z in the equation of this plane. find the intersection of the line and the plane. But if the plane is parallel to the z-axis and contains the origin. the point (0. 2. we choose the one which intercepts the xy-plane along the line with equation 5x 2y 0. with the intersection line and the z-axis as sides. we are singling out those points in the plane that lie in the xy-coordinate plane. x b. By setting z equal to zero in the equation of a plane. z 2
1 2
t and 2x 3t and 3x
y
6z 4y
10 7z 7
0 0
292 C H A P T E R 8
. In Example 5.and y-intercepts are both zero.The xy-coordinate plane. 0) satisfies the equation of the plane. it must contain the entire z-axis. So this plane contains the origin. and by setting z 0 we obtain the equation. so as with Example 5. This parallelogram-shaped region represents a section of the plane 5x 2y 0 in three dimensions. y 4t. These are exactly the points on the intercept line.
y x z ?
y x (5x – 2y = 0)
Exercise 8. Consequently the x. y 6 2 2t. 0.
t(3. y
4
2t. 4. a. State whether it is possible for the lines and planes described below to intersect in one point. the yz-plane
b.
iii) the xz-plane 1) meet c. a. 7)
t( 7. In what line does this plane intersect the i) the xy-plane 5.
2. 2) intersects the plane. 1) t(3. For each of the following planes. and z-intercepts and make a three-dimensional sketch. a. or in no points. a line parallel to the y-axis and a plane parallel to the y-axis c. the z-axis?
b. 6. 1. 14. r e. a.
7
2t and 2x y 0 5
3y 2z
4z 6
7 0
0
(2. 10. Find the point of intersection of the plane 3x 2y 7z 31 0 with the line that passes through the origin and is perpendicular to the plane. 2y 5z 18 0
Application
9. the xy-plane
Communication
ii) the yz-plane (6. 3)
s( 2. the x-axis? Part B 4. 7. Does the line r ( 2. 3)
1. r
5
t. 2x y 3y z 4z 8 12 0 0 b. 7z 6
1) lie in the plane 1) lie in the plane 0 intersect c. find the x-. Find the point at which the normal to the plane 4x through the point (6. 1) and 3x 1) and z
t(0. in an infinite number of points. In what point does the plane r intersect i) the x-axis 2y
t(3. the xz-plane
6. 8. 5) 3x 4y z 25 0? b. x d. 4x 2y y z 2z 5 16 0 0
8. 12x c. 2) 3x 4y z 17 0? 3. 2. 2. a line perpendicular to the z-axis and a plane parallel to the z-axis
Application
7. y-. x d. 1) (5. 2. z t( 1. Where does the plane 3x a.c.
4. 6. Where does the line r a.
3)
ii) the y-axis
iii) the z-axis
b. a line parallel to the x-axis and a plane perpendicular to the x-axis b. 1.3 THE INTERSECTION OF A LINE AND A PLANE
293
. Does the line r (4. 4. the y-axis?
(6.
x d. y
3 2z
0 0
c. 3x Part C 11. none
x of which is zero. For what values of k will the line x 3 k x 4y 5z 5 0 a. A plane has an x-intercept of a. y-. in a single point? b. For each of the following planes. in an infinite number of points? c. a y-intercept of b. and z-intercepts. if they exist.10. and a z-intercept of c. x y
1 z
0 0
y 2
4
z 1
6
intersect the plane
12. in no points?
Thinking/Inquiry/ Problem Solving
y z
4 6
0 0
b. find the x-. Then make a three-dimensional sketch of the plane. a. and the intersections with the coordinate planes. Show that the equation of the plane is a
y b
z c
1.
294 C H A P T E R 8
. x e. 2y f.
4). hence not intersecting. They can intersect in a line.
z z z
e
O x planes are parallel
y x
O planes are coincident
y x
O planes intersect in a line
y
If the normals are collinear.4 THE INTERSECTION OF TWO PLANES
295
. These are not collinear. To find its equation. But there are no constraints on z. the planes must intersect in a line.
EXAMPLE 1
Find the intersection of the two planes 2x 2x y 4z 7 0. t R 1 6t
Parametric equations of the line of intersection of the two planes are x 4 t. intersecting at every point. y 1 6t. 2.Section 8. we solve the equations. If the normals are not collinear. z 2t. 532
What are the possible ways two planes can intersect? They can be parallel and distinct. 2x (1 6t) 8t 7 x 0 4 t z y 2t. so the planes intersect in a line. 1. They can be coincident.4 — The Intersection of Two Planes
t chnology
APPENDIX P. 5) and (2.
8.
2y
5z
10
0 and
Solution The equations of the two planes constitute a linear system of two equations with three variables. Let Then Substituting in equation two. The normals of the two planes are (2. Subtracting we obtain 2x 2x Then 2y y 3y 5z 10 4z 7 9z 3 y 0 0 0 1
3z
The value of y depends on the value of z. the planes are parallel and distinct or coincident.
The matrix method of solving the system of Example 1 starts with the augmented matrix and proceeds by performing arithmetic operations on its rows. It is called the coefficient matrix of the system. add it to the first row. for instance. column 2 equal to zero. which is equivalent to eliminating x in the corresponding equation. and then replace the second row with this difference.
In order to make the element in row 1. This step is the counterpart of subtracting the equations in Example 1. Each coefficient is an element of the matrix. For our purposes. The row and column position of each matrix element indicates the equation and the term to which the coefficient belongs. One of them makes use of matrices.The solution of systems of linear equations is such an important topic that several different methods to handle this problem have evolved. y. We write
This is equivalent to the removal of the factor subtraction in Example 1. The vertical bar in the matrix shows where the equal signs in the system are located. the linear system dealt with in Example 1. R1 R2 2 0 2 5 3 9 10 3
Observe how this operation on the rows of the matrix is expressed in symbolic form: R1 and R2 stand for the two rows. we indicate where the result is to be placed. a matrix is a rectangular array of numbers made to facilitate the solution of a linear system. and z in the two equations 2 2 5 2x 2y 5z 10 you can form the matrix 2 1 4 2x y 4z 7 This is a 2 3 matrix – it has two rows and three columns. Consider. From the coefficients of x. we multiply the second row by 2. The constant terms of the equations (which are here written to the right of the equal signs) can be included by adding another column to the coefficient matrix.
296 C H A P T E R 8
. These operations on the matrix have made the element in the lower left corner equal to zero. The next step is to divide each of the elements of the second row by R2 ( 3) 2 0 2 1 5 3 10 1 3 from the result of the 3. By placing R1 R2 beside row 2. and replace the first row with this sum. 2 2 2 1 5 4 10 7
This matrix is called the augmented matrix of the system. The first operation is to subtract the elements of the second row from those of the first.
Any row can be replaced by the sum (or difference) of that row and a multiple of another row. Lastly. Such operations on the rows of the matrix are legitimate. because they match similar operations that could be done on the corresponding equations. x and y are both functions of z. The matrix method of solving a system of linear equations. Row Operations 1. 3. The actual operations performed on the rows will depend on what the coefficients are.
t chnology
e
Using the matrix methods described above. Any row can be multiplied (or divided) by a non-zero constant. is referred to as Gauss-Jordan elimination. 2. y 1 6t. illustrated above. A 2 4 matrix of the form * * * * * * * * can be written down directly from the original equations of the linear system to be solved and then changed into reduced row-echelon form 1 0 * * 0 1 * * This form is only one step removed from the solution of the system.
8. divide the elements of the first row by 2. the solution of a linear system can be systematized so that it can be programmed on a calculator or computer. z 2t as before.4 THE INTERSECTION OF TWO PLANES
297
. The permissible operations that can be performed on the rows of a matrix arise from the algebraic operations that can be performed on the equations of the corresponding linear system. setting z 2t. The two equations corresponding to this matrix are x y
1 z 2
4 1
or
x y 1
4 3z
1 z 2
3z
Here. R1 2 1 0 0 1
1 2
3
4 1
At this point the matrix has served its purpose. So. Any two rows can be interchanged.We write 2R2 R1 2 0 0 1 1 3 8 1
This is equivalent to eliminating y in the first equation. but there are no restrictions on z. the equations of the line of intersection are x 4 t.
This makes it possible to find solutions to systems with many equations and variables.
EXAMPLE 3
Find the intersection of the two planes x 2x 8y 6z 11 0. which would be difficult. such as those in economics or statistics. to work out by hand. z t. try using it to work through the example above before continuing.
0 and
Solution The equations of the two planes form the linear system 4x 7y 33z 8x 5y 3z The augmented matrix of this linear system is 4 8 7 5 33 3
17 7 17 7
The solution is achieved by starting with the augmented matrix and carrying out the following row operations to change the matrix into reduced row-echelon form.
t chnology
e
The box on page 299 shows how to use a calculator to solve a system of linear equations. They are x 1 4t. Solution The augmented matrix of this system is
4y
3z
6
0 and
1 4 2 8
3 6
6 11
298 C H A P T E R 8
. If you have a calculator that can perform matrix operations. y 3 7t.
EXAMPLE 2
Find the intersection of the two planes 4x 7y 33z 17 8x 5y 3z 7 0 using Gauss-Jordan elimination. if not impossible. 2R1 R2 R1 R1 9 7R2 4 R2 4 7 0 9 4 7 0 1 33 63 33 7 17 27 17 3 4 3 1 3
4 0 16 0 1 7 1 0 0 1 4 7
The final matrix corresponds to the equations x y 4z 7z 1 3 or x y 1 3 4z 7z
Parametric equations of the line of intersection result when z is set equal to t.
To complete and execute the instruction. 4.
8. press 2 ENTER and 4 ENTER . To put the matrix in reduced row-echelon form. Since (2. press 4 ENTER . To select which matrix to reduce. The second row of this matrix corresponds to the equation 0z 1. select matrix A press 2nd . 6. (1. To define the matrix. To enter its elements. 3) and (2. The result is
1 0 0 1
4 7
1 3
Now write the corresponding equations and complete the solution. so the planes are parallel and distinct or coincident. for instance. then cursor down to B:rref( press 2nd and press ENTER ). select MATH. 11) 2(1. To set its dimensions.. We could say that the normals to the planes. To solve the linear system of Example 1. They must be parallel. 2. select matrix A press 2nd . select NAMES. If an elementary row operation makes all the elements of a row zero. the planes are distinct. etc. 6). MATRIX . QUIT Then press 2nd to return to the home screen. and the planes do not intersect. start with the augmented matrix
4 8
7 5
33 3
17 7
and follow the following steps (the instructions are for a TI-83 Plus calculator).The first operation is to put a zero in the lower left corner of the matrix 2R1 R2 1 4 0 0 3 0 6 1
There is no need to go further.4 THE INTERSECTION OF TWO PLANES
299
. and press ENTER . Hence. 3. press ) and press ENTER . 6). are collinear. but there is no value of z for which this equation is true. CALCULATOR APPLICATION Some calculators can put a matrix into reduced row-echelon form and thereby help you to find the solution to a linear system. there is no solution. then 7 ENTER . 8. for all eight elements. MATRIX . 4. 1. MATRIX . 8. this indicates that one equation is a multiple of the other and the planes are coincident. select EDIT. and press ENTER .
no intersection
3 planes are coincident.. The remaining examples in this section illustrate this method of solving a linear system. that is. the other crossing. no common intersection
two planes are coincident. intersection: a line
304 C H A P T E R 8
. continue by writing the corresponding equations.As you can see from Example 2. it is usually simpler to do Gaussian elimination. Using a calculator with matrix functions makes the work faster and easier (see the box on page 307). When working without a calculator. intersection: a plane
When only two of the normals of the planes are parallel. consider the normals of the three planes. When the normals of all three are parallel. the possibilities are
two planes are parallel and distinct. it is time to return to the question that started this section: What are the possible ways three planes can intersect? To answer this question. the other parallel. * * * * 0 * * * 0 0 * * which in Example 2 is x
1 0 0
3y y
3 2 1 5 0 11
2z 5z 11z
9 21 44
9 21 44
Then. it can be a complicated and lengthy process to work out a problem by hand using Gauss-Jordan elimination. no intersection
2 planes are coincident. the possibilities are
3 planes are parallel and distinct. putting the augmented matrix in row-echelon form.
and finally finish the problem by doing the substitutions as in Example 1. This consists of using matrix methods to get just the three zeros in the lower left corner. Try solving the problem in Example 2 using a calculator. Now that you have the tools to solve systems of three linear equations. the other crossing.
z) x y z the scalar equation Ax x0 y0 z0 By (x 0. a2. b3)
As with lines. whether are they consistent. Draw graphs. Finally. the vector methods used to find the equations of a line have been extended to planes. In a summary statement. diagrams. z0) sa1 sa2 sa3 Cz tb1 tb2 tb3 D 0 s(a1.Key Concepts Review
In this chapter. b2. y0. Look at the equations or numerical values of your answers and ask if they answer the question asked. express the solution in words. Make a connection between the algebraic equations and the geometrical position and orientation of a line or plane in space. by inspection when possible. it is essential to memorize these equations and to learn to convert quickly. a3) t(b1. y. try to invest your solutions to problems with meaning. from one form to another. and whether they meet your expectations. or sketches to increase your ability to visualize intersections.
310 C H A P T E R 8
. The resulting equations of a plane are the vector equation the parametric equations (x.
This is the angle of elevation of the sun. Justify this and then calculate n. The earth's axis is tilted 23. s2. and . 23 x z Mar. 21 y
We want to find the angle between s and the observer's plane of tangency to the earth. c) Why is the angle of elevation of the sun. As previously noted. 365 150cos 360d . s1. Use negative values of for southern latitudes. c cos b sin ). c). 3. If n2 (a. 0 . b) Calculate 90º . n. and let be the latitude of the observer. Pick specific values of d.wrap-up
C H A P T E R 8 : S U N E L E VAT I O N
investigate and apply
Like so many astronomers before us and throughout history. Calculate s. a) Given that s (s1.
How can we find the positions of the stars and the other planets? ●
RICH LEARNING LINK WRAP-UP
311
. the vector from the earth to the sun is s 150sin 360d . Now to find n we start by assuming the earth's axis is not tilted. b. What does it mean if the angle of elevation is negative? (In practice. 1. we will need the normal. b cos c sin . and (perhaps the current date and time and your current latitude ). through calculations. Investigate and Apply Let d be the number of days past December 21.45º away from the z-axis in the direction of the y-axis. h. h. 24 24 Why is this the correct normal for a person on the equator? (0. Why?)
INDEPENDENT STUDY
Develop a general formula for
in terms of d. 365
Sept. and not ? 4. a) Let be the angle between s and n. 21
June 21
Dec. of this plane of tangency. 0. let n1 s cos 360h ( s2. 0)sin 360h . n1 sin ). 0). the angle of elevation of the sun for any given time of any given day at any given place on the surface of the earth. the angle between a line and a plane will always be between 0º and 90º. we shall determine. then n (a. What does n2 represent? b) Let n2 n1cos 2. To do this. Let h be the number of hours (positive or negative) from noon.
2) t(4. 0. Find the scalar equation for the plane a.
312 C H A P T E R 8
. 5)? Explain. that contains the two intersecting lines r (3. Find vector and parametric equations of the plane a. that contains the line r r (3. 3) and is parallel to the y-axis c. For what value of k. 2) s(4. that passes through the point ( 1. b. 1)
b. 1. 2) 1. that passes through the point (0. 1) and r (3. 0. 4. 9) and has normal n (1. 0. 1. Can a plane be parallel to the yz-coordinate plane and contain the point ( 4. Find the scalar equation of the plane that contains the parallel and distinct lines x 5. perpendicular?
y 2 5
z 1 3
5. 1. 3. 2)
3. 1. 7. that contains the point (1. a. if any. that passes through the points (3. 0. 0. r (2. that contains the points (6. 3. 2). 1. y 3 t. 0. 2. 1. 6) and is parallel to the plane y d. that passes through the point (1. y 3. 2. ( 4. y 4 3
z and x 2
ky
z
6
0 and
b. 4) t(2. 0) s(5. 2) c. 1. and is parallel to the z-axis 4. 1. 0) and ( 2. 1) and the line 3
e. that has intercepts x 2. 2). 0) and is parallel to the line
f. 0) 3. z 4 2t e. 1. 3). 0. .Review Exercise
1. and z 4
y 4 z 5
x d. and ( 1. 7) 2. that passes through the points (1. 5) z 5 b. 0) and (3. 1. will the planes 3x 6x (1 k)y 2z 9 0 be a. parallel? 1. 0) and the line x (2. 2) and is parallel to the plane t(0.
s(0. 0. Can a plane be perpendicular to the x-axis and contain the line x z. y 0? Explain. that contains the point (3.
6. 2. At what point does this normal intersect the plane? 14. Find the distance between a. Show that the line x 2x y 2z 5 0. 3. Graph the following planes in an xyz-coordinate system: a. a y-intercept of 2.
0 2z 17 5 0
b.
15. z 6y 0 s( 12. 2) to the plane having an x-intercept of 1. 8) and the line r do not determine a plane. 2. 8. 5z 10 z 5
( 4. 1) t (2. y 4z (4. 3. 0) 4t. find the scalar equation of the plane. 2x c. 1) and the plane 3x 2y z 10 (1. 21. 5y z 20 0 meets the coordinate axes. 8) 1 4z t(8. 7. 3 24 5. A line that passes through the origin intersects a plane at the point (1. Explain why the point (2. 11. 2). 3. 3) ˆ ˆ ˆ ˆ and is parallel to the vectors 6 k and i 2 j 3k . 9. 2. 2. 3z y 9 z 0 5 3 0 b. the point (3. 13. 3y d. the planes x 7) and the plane 6y t (1. and a z-intercept of 3. For what values of k will the planes 2x 3x 9y 6z k 0 a. 3. 9. Find a vector equation of the plane that contains the origin and the point (2. the point (7. intersect in a plane?
REVIEW EXERCISE
313
. Find the scalar equation of the plane that contains the intersecting lines
x 2 1
y 3
1
z
1 x 1 and 1 1
y 5
1
z 4
1
. Find the distance from the point (1. 7. Find the scalar equation of the plane that passes through the point (1. 2) 2y 1) and the plane y 0 and 2x 4y 10z
12. the line r d. A normal to the plane 4x 2y 5z 9 0 passes through the origin. not intersect?
b. 8. Determine where the plane 4x and graph the plane. 8)
16. intersect in a line?
c. If the line is perpendicular to the plane. r 3t. c. 3)
10. 2) and is perpendicular to the plane x 2y z 3 0. 7.
5t lies in the plane 3 0 and
17.
x 4 2
y 1 2 z 1
3. 1. 2)
a. Solve the following system of equations and give a geometrical interpretation of the result. 0) t(2. Find the scalar equation of the plane containing the line x point (2. two planes. is parallel to the plane. or intersects the plane at a point. two planes. For each of the following. a. intersects the x-axis at what point? b. intersects the xz-coordinate plane in what line? 5. x 3t. 5) s(4. if their normals satisfy n1 • n2 b. 4. 0. three planes. 3 4. 6
1. x x 7y 2x y 2y 4z z z 3 13 4 0 0 0
CHAPTER 8 TEST
315
. 4). z 0 and the
6. Describe with diagrams all the ways that three planes can intersect in one or more common points.Chapter 8 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application Questions 2. 5. if their normals satisfy n1 c. The plane r (0. What can you conclude about the intersection of a. z 10t b. if their normals satisfy n1 n2 0? 0? 0? n2 • n3
2. 5 7 1. 0. state whether each line lies in the plane 4x y z 10 0. Give your reasons. y. y 5 2t.
7.
b. Find the distance from the point P(10. 10) to the plane 3x 2y z 14 0. Is P on the same side of the plane as the origin? Give evidence to support your answer. a.
316 C H A P T E R 8
. 10. Find the distance from the origin to the plane 3x
2y
z
14
0. c.
Use the key strokes MATH → → → PRB 5 to get to the function randInt. When we divide 37 by 8. or 7. Then enter 0.. and repeat the process. How does this work? One way to produce a sequence of random digits from the set {0. if the algorithm is well selected. 5. Of course. reach into the container and choose a ball. To see how the calculator generates a sequence of random numbers.. Then. 0. This is the advantage of simulation. Your calculator has a function that can generate random numbers. the numbers generated by the calculator are far from random. Spreadsheet programs such as EXCEL have a function mod(x. airlines often sell more seats than exist on a plane because they know that some ticket holders may not show up. 2. However. replace the ball in the container..G E N E R AT I N G R A N D O M N U M B E R S
Random numbers are used for computer simulations of processes that can be modelled using probability. For example. 4.8) ENTER to get the 8 random digits 9. if you try this you will get a different answer! How does the calculator produce this sequence since there is no one inside to shake up a container of balls? The answer is that the calculator uses an algorithm that is completely deterministic. The numbers produced will be exactly the same every time if you start with the same initial conditions. Of course. Note the digit on the ball you selected. the remainder is 5. we get a remainder of 0.. inclusive. Here is a sequence of random digits produced by a TI-83.9. we must look (perhaps suprisingly) at how division works. Using random numbers as part of a model. 1. the remainder r is an integer between 0 and m 1. We use a fancy notation x rmod(m) and say that x is congruent to rmodulo m". 2. without looking. 3. 9. if we divide any integer x by the integer m. or random digits within a specified range. . 1. That is. 1. usually between 0 and 1. 0 5. . then the airline will have to provide costly incentives to convince some of the extra passengers to wait for the next flight. the airline can simulate different seat-selling strategies without ever trying one in practice. 2. the sequence shares many properties with a sequence of random numbers and.. Hence. For example. You might get a sequence such as 7. m) that returns the remainder when x is divided by m. 8. 37 4 8 5. If we divide any integer by 8. Generally. 1. 2. 5. 0. 9} is to put 10 identical balls numbered 0 to 9 into a container and shake it vigorously.
E X T E N D I N G A N D I N V E S T I G AT I N G
317
. 37 5 mod(8) and 63 7 mod(8). This is not true with the container of balls. the numbers produced are good enough for practical purposes. if too many tickets are sold. 2.
2.. This looks better since the period is now 12. 2. investigate different values of a to determine the longest possible period. n 1. 1. which then repeats. the function rand returns a rational number between 0 and 1. For amusement. 8. Since the remainder xn xn is always less than m. 5. 6.. 4. 5. Can
you guess the answer for 2e for any integer value of e? 3.. a 427419669081.. we get the sequence 8. 2. What are the possible periods for various choices of a?
On your calculator. n 1. m and a are selected so that the sequence has a very large period and other good properties. one version of Waterloo MAPLE uses a generator with m 1012 11. Could the period be longer than 12 for any choice of a? In practice. 11. 1. For example.
318 C H A P T E R 8
. 1. n 1. 1. Explain why you must get a periodic sequence with this generator (try specific values for a and m first). you can try the following. If we substitute n 1. with seed x0 1. For m 23. where x0 is a specified number called the seed. 3. . . We say that the sequence has period 4. 1. which produces a sequence with period 1012 12 (do not try to check this by hand!). the number displayed is m . 12. 12. for example m 17. 8.. For example.. 9. consider the generator xn 8xn 1mod(13). 2. 12. 5. If we change a in the generator to 2 so that the equation is xn 2xn 1mod(13). The seed can be set by the user or determined in some other way (e. This sequence does not look very random since it repeats itself every four terms. 3.. we get the sequence 1. . 25. 8.One algorithm for generating a sequence of random numbers is a mathematical equation of the form xn axn 1mod(m). 12. 24. Suppose m is a prime. 2.g. . 2. 12. 5. 7. from the clock inside the calculator). 10....
3. Prove that (u v) u is perpendicular to v. For what value of k is BQR a straight line? 7. 0). Prove that W. and Z are coplanar. and w are coplanar. Show that the cross product of two unit vectors is not generally a unit vector. 0).Cumulative Review
CHAPTERS 4–8
1. 4. for k a real number. 6. P1P4 c. are consecutive vertices of a regular polygon with 12 sides. 2l). let W be the midpoint of AB. X. P2 x and P1P3 y.. P3. B(0. P3P7
8.. Write the vector (0. P is the midpoint of BC. 2k). 2l. P2. P2P3 b. v. 1). R is on AC such that AR kAC. express the following vectors in terms of x and y: a. 2. If P1. For the four points A(2k. P12. 8) as a linear combination of the vectors (2. 4) and ( 2. 0).
C U M U L AT I V E R E V I E W C H A P T E R S 4 – 8
319
. Let a. In ∆ABC. and C( 2. . 0. B(0. and c be linearly independent vectors in space. C(0. Find the cosine of ∠ABC. Y. and D(2l. Y the midpoint of CD. X the midpoint of BC. 2k. b. 5. P1. and Z the midpoint of DA. The points A(2. Q is the point that divides AP internally in the ratio 5:2. and let u v w 3a 2a a 2b 4c 3b kc c
Determine k so that u. 4). 0. 1) define a triangle in the plane.
3). y. and is parallel to the line x 3 2
y 3 1
z . b. 3) s(1. y 3 2t. y. Find the equation (in the form Ax By Cz D 0) of the plane that is tangent to the sphere at (2.
a. z)
(3. Let M be the point of intersection of the diagonals AC and BD. 11. Determine a point A on the line with equation (x. Prove that M is the midpoint of PQ. 13. 6. z) (3. Q( 1. Prove that the diagonals of a parallelogram bisect each other. 0) lies on the plane that passes through the three points P(1. 0). x b. 3). Find the point of intersection of the lines. 1). c.
1. 1). 1
15. x y 4x 4y z y 3z 2 2z 7 0 7 0 0 1 t. Consider the two lines with equations
x 1 8
y 3
4
z 1
2
and (x. 12. y. so that AB is parallel to m
(2. z t with
320 C U M U L AT I V E R E V I E W C H A P T E R S 4 – 8
. 10. 5). 1. and R( 5. 14. 0.
1.
1). 1. 2. 3)
t(4. 4. 2.9. Draw a quadrilateral ABCD with opposite sides AB and DC parallel. 2). The equation (x 1)2 (y 2)2 (z 3)2 9 defines a sphere in threedimensional space. Determine the equation in the form Ax By Cz D 0 of the plane that passes through the point P(6. 4. has z-intercept 4. Determine the intersection of the line x each of the following planes: a.
16. 3) t( 1. Prove that the bisector of the apex angle of an isosceles triangle is perpendicular to the base. 1. z) ( 3. 1. Through M draw a line parallel to AB that intersects AD in P and BC in Q. 5). a point at one end of a diameter of the sphere. 3. and a point B on the line (x. Show that the lines are perpendicular. 17. Determine whether the point O(0.
Determine all values of x. 20. 1) intersects the plane 3x 5y 2z 0 at point Q. 2. 22. 3. 4). 21. 1) where s and t are real numbers. 1) L2: (x. (2. 2. x 2x 3x 2y 5y 6y 3z 4z z 1 1 3
C U M U L AT I V E R E V I E W C H A P T E R S 4 – 8
321
. Determine the parametric equations of the line of intersection of the two planes 3x y 4z 6 0 and x 2y z 5 0. z) (2. 2. 0. 0. Consider two lines L1: (x. A plane passes through the points (2. The line through a point P(a. 25. 3) s(a. 2. y. 0. 3. The line through P with direction vector ( 3. 4). Determine the distance between the xy-intercept and the xz-intercept. Determine the components of a vector of length 44 that lies on the line of intersection of the planes with equations 3x 4y 9z 0 and 2y 9z 0. Determine the coordinates of Q. 1). Find the point of intersection of the plane and the line. 0) in the plane with equation 3x y z 12.18. b. b. z) (3. At what points does the line of intersection intersect the three coordinate planes? c. 2). 1) and (1. a. 19. For what choice of a is the distance between Q and R equal to 3? 24. y. y. 0) t(1. 2. Find a relationship between a and b (independent of s and t) that ensures that L1 and L2 intersect. 1) intersects the plane at point R. and (2. Find the point on the xy-plane that lies on the line of intersection of the planes with equations 4x 2y z 7 and x 2y 3z 3. 2. 23. a) with direction vector ( 1. and z satisfying the following system of equations. The point Q is the reflection of P( 7. A line passes through the points (3. 1.
determine the solution set and give a geometric interpretation.
322 C U M U L AT I V E R E V I E W C H A P T E R S 4 – 8
. k is a real number. 2x y y kx z z kz k 0 0 1
a. For what value(s) of k does the system i) have no solution? ii) have exactly one solution? iii) have an infinite number of solutions? b. In the following system of equations. For part a iii.26.
"if it was so. working backwards. Section 9." continued Tweedledee. and if it were so. but as it isn't it ain't. intuition. Section 9. you will
• • • • •
prove some properties of plane figures algebraically. Section 9. The ability to consider a problem from many angles is as valuable for business professionals as for researchers in academic environments.3 understand the relationship between formal proof and the use of dynamic geometry software. In this chapter.4
. you will consider the value of different approaches to solving a problem. and trial and error. That's logic. In mathematical problem solving.4 use technology in testing conjectures.Chapter 9
PROOF USING DIFFERENT APPROACHES
"Contrariwise." (Lewis Carroll) Mathematics is one of the best ways to teach logic and proof. it would be. Section 9. it might be. you can consider problems using deduction.2 solve problems by combining a variety of problem-solving strategies. CHAPTER EXPECTATIONS In this chapter.3 generate multiple solutions to the same problem.1. Section 9. 9.
y1) and (x2. y)
2x2
. 2. y2) is y y y y1 x2 x1 (x x1). k).
2
. and parallel lines. and their use in the equations of lines.
P(x1.
• The coordinates of the midpoint of the line segment joining (x1. similar triangles. y1) and (x2. y2) so that AP 2PB.Review of Prerequisite Skills
1. m1m2
x1 x2 y1 y2
1. y2) is d (x2 x1)2 (y2 y1)2.
2 1
• If two lines are perpendicular and have slopes m1 and m2. Recall the properties of congruent triangles. 3. y1) to the straight line whose equation is Ax By C 0 is d
Ax1 By1 A2 B2 C
. 0) and is (x h)2 (y k)2 r2 if the centre is (h. y1) with slope m is y y1 m(x x1). Recall the properties of vectors and combinations of vectors. Recall the properties of circles. chords. 4. y) divides the line segment connecting A(x1.
. • The distance between the points (x1. y1) d
• The perpendicular distance from P(x1. • The equation of the straight line through (x1.
324 C H A P T E R 9
. and tangents. y1) and B(x2. then (x.
x1 2y2 3 y1 3
• If P(x. • The equation of the straight line through (x1. y2) are
2
. y1) and (x2.
• The equation of a circle having radius r is x2 y2 r2 if the centre is (0.
AC:CB 2:3
4. 5) to B(5. The point (x. isosceles. 5). 7). 4) and radius 4? c. centre ( 1. 0). centre (3. (2. A circle has its centre on the x-axis and a chord that connects ( 2. 0). 4)? 7. 4) as it is from the origin. Determine all possible values of x. (1. The line segment joining A( 1. (1. 8. 2). C(8. 0) and radius 3? b. What are the coordinates of E and F. 4) c. 5. 8) and (1. 7). the midpoints of AC and BD? What are the coordinates of G. 2)? 10. 4) is twice as far from the point ( 9. 11). (7. 1). 4). 3) is divided internally by the point C. A triangle has vertices (1. 7). (4. ( 3. 1) and (10. congruent b. AC:CB 3:1 b. ( 1. similar 2. d. ( 1. 7) and (10. A quadrilateral has vertices A(0. What are the coordinates of the centre? 6. 3). or right-angled. List conditions under which two triangles are a. Determine whether the triangles defined by the following sets of points are equilateral. (1. a. Determine the coordinates of C if a. (0. (2. 2) b.
3. centre (0. 0). the midpoint of EF?
REVIEW OF PREREQUISITE SKILLS
325
. 5). 5). What are the coordinates of the midpoints of the sides? 9. 0). What is the equation of the median from the point (2. (7. 3) in a triangle if the other vertices are (5. Determine the coordinates of the point on the y-axis that is equidistant from the points (5.Exercise
1. B(6. (0. and D(3. and ( 3. 1) 1) 2). 2) and passing through (7. 6). What is the equation of a circle having a.
This network has a total length of 1 3 2.investigate
CHAPTER 9: STEINER NETWORKS
Consider the problem of providing a fibre-optic network to three or more sites on a university campus so that any two sites are connected by some fibre-optic path. Besides length. If ∆ABC is an acute triangle. Can you think of other applications in which a shortest-length network might be sought? ●
326 C H A P T E R 9
. The H-pattern network has total length of 3 units. Verify this calculation. The shortest network. B. The Steiner network for a rectangle ABCD is very similar to that of the square. Applying the Pythagorean theorem. then there is a point F inside the triangle. A shorter network is D C D C D C formed by the diagonals of the Steiner H-Pattern Diagonals Network square. production facility layouts. we find that this network has length 2 2 2.
DISCUSSION QUESTIONS
A B
E
F
D Steiner Network
C
1. Points E and F are located such that EF passes lengthwise through the centre of the rectangle and the angles ∠AED and ∠BFC are both 120º. and C. for which AF BF CF is less than for any other point. It is the centre of the Steiner network for the three points A. who lived from 1796 to 1863. the network must be as short as possible.83 units. To minimize cost and maximize communication speed. Applications of Steiner networks arise in designing transportation networks. and computer microchip design. How many Steiner networks does a square have? How many Steiner networks does a rectangle have? 2. a Steiner network. How can this be done? Such a shortest-length network is called a Steiner network. Will a Steiner network connecting five points always be longer than a network connecting four points? 3. what other concerns arise when designing communications or transportation networks? 4. is formed by locating two points E and F such that EF passes through the centre of the square and the angles ∠AED and ∠BFC are both 120º.73 units. Investigate A B A B A B The diagram shows a square ABCD with side length one unit E F and three possible networks. called the Fermat point. after the Swiss geometer Jacob Steiner.
one such as y 3x2 4x defines a parabola. we expand our ability to develop the equations of figures having specific conditions. An equation such as 3x 2y 6 0 defines a straight line.1 — Using Analytic Methods
An equation involving variables x and y represents a restriction of the xy-plane to points whose coordinates satisfy the equation. one such as x2 y2 9 defines a circle with centre at the origin and radius 3.Section 9. The radius is r AC [ 5 ( 2)]2 (1 5)2
A(–5.
EXAMPLE 1
Determine an equation for a circle such that the endpoints of its diameter are determined by the points A( 5. 1 U S I N G A N A LY T I C M E T H O D S
4y
327
. 5). 9 2 1 C( 2. Since the radius and/or the coordinates of the centre might be fractions. and so on.
EXAMPLE 2
Find the radius and centre of the circle defined by 3x2 3y2 10x 12y 13 0. In this section. This answer is given in standard form. The equation of this circle is (x 2)2 (y 5)2 25. x2 Rearranging. Solution Divide the given equation by 3. 9)
C(–2. the coordinates of the centre are at C 1 2 5. 1) and B(1. 9). 1) and B(1. we get x2 4x 4 y2 10y 25 25 x2 y2 4x 10y 4 0
This equation is in general form. y2 x2
10 x 3 10 x 3
4y y2
13 3
0
13 3
9 . we could multiply the equation by a number to eliminate fractions. If we expand the squares. Solution Since the end points of the diameter are A( 5. 5)
9 16 5. 9). 1) x
y
B(1.
so that instead of 13 on the right. 3) meets a line l at the point Q(2. adding the same numbers on the right as on the left. 5)
P(–1. then the line from the tangent contact point to the centre has slope 1 .
13 25 3 9 5 2 100 x– 3 (y 2)2 9 5 The centre is 3 . 2 and the radius is 10 . What is the equation of line l? Solution The slope of PQ is 2 5 ( 31) Then the slope of l is 3 . 5
2 . 4 1 is 4
P O
y2
17. and the tangent through P
x2 x
The point Q is (4. Replacing y by x2
1 2 x 16 17 2 x 16 1 x. and P is ( 4. 3 3 (x 2
y
Q(2. and no circle would exist. 1).
y
x Q
We determine the coordinates of P and Q by solving this equation with the circle equation.
EXAMPLE 3
A perpendicular from the point P( 1. 2 The equation of l is y or 3x 2y 16 0. the final result would have a 9 negative value for r2. we obtain 4
17 17 16 4 4(x 4). the equation of the diameter with slope y
1 x. Then. 1). 5). The tangent through Q has equation y 1 has equation y 1 4(x 4). 3)
l x
2)
EXAMPLE 4
Determine the equations of all tangents having slope 4 to the circle x2 Solution If the tangents have slope 4. 3 we got a negative number larger than 25 4. 4 since these lines are perpendicular.Complete the squares.
328 C H A P T E R 9
. 3
x2
10 x 3
25 9
y2
4y
4
4
If the constant had been a large positive number.
14. 8.6. Find the length of the tangent from (5. Find the length of the common chord of the two circles whose equations are x2 y2 4 and x2 y2 6x 2 0. 4) and which has its centre on the line with equation y 2x 4. For the circle x2 y2 16. 12. Find the equation of the circle that passes through the points (8. 2) and ( 2. Determine the shortest possible distance between a point on the circle x2 y2 9 and a point on the circle x2 y2 12x 6y 41 0. 10. Find the length of PQ.
330 C H A P T E R 9
. 7. Show that the point A(1. Find the coordinates of the other end of the diameter through A. determine the length of the longest possible chord passing through P(1. 7) to the circle with equation x2 y2 6x 2y 6 0. 5) is on the circle with equation x2 y2 4x 2y 20 0. 2). The point R(3. 4) is on a circle with its centre at the origin. 9. determine the equation of the circle. The tangent through R intersects the x-axis at the point P and the y-axis at the point Q. Part C 13. Find the coordinates of the intercepts of the circle whose equation is x2 y2 6x 2y 0. A circle lies in the third quadrant and touches both coordinate axes. 2) to the circle is 5. If the length of the tangent from the point A( 4.
Thinking/Inquiry/ Problem Solving
11.
In Chapter 1. the diagonals bisect each other. c). The vertex C in the diagram cannot be defined using the variable a. prove that the diagonals of a parallelogram bisect each other. c)
x O(0. We could have avoided fractions in the midpoint coordinates by letting the coordinates be A(2a. First.
y C x O A B
EXAMPLE 2
Using analytic methods. ensure that the location of figures used is to your advantage. we must ensure that OA BC and OC AB. 0). 2c). 2 c midpoint of AC is a 2 b . Second. Solution By making one vertex the origin and a second vertex a point on the x-axis. C(2b. we have
c midpoint of OB is a 2 b . Solution Using the parallelogram defined in Example 1. Then. If we are to make the best possible use of analytic methods. Let one vertex be O(0. 0). since CB a. determine the usefulness of analytic methods. We let C have coordinates (b. you saw how geometric facts can be proven using deductive thinking.
EXAMPLE 1
Describe an efficient way of defining a parallelogram in analytic terms. c).Section 9. c) B(a+b. By considering a few examples we can see how this can be done. We are free to define coordinates as long as we use properties of the figures. 2c).
EXAMPLE 3
Use analytic methods to prove that the medians of a triangle are concurrent and divide each other in a ratio 2:1 from a vertex to a midpoint. 2 P R O O F U S I N G A N A LY T I C G E O M E T RY
331
. and B(2a 2b.2 — Proof Using Analytic Geometry
In earlier chapters. In the examples above. two steps are necessary. we require only one variable. the coordinates of B are (a b. The strength of the analytic method is that we can choose coordinates for convenience. 0)
Since these are the same. we looked briefly at using analytic methods. 2
y C(b.
9 . The parallelogram is now defined using a minimum number of variables. 0) and a second vertex be A(a. 0) A(a.
Then the diameter on the x-axis has end points whose coordinates are A( r. We will prove that ∠ACB 90º. c). 3
a) If R divides CD in the ratio 2:1. 3 . 0). then R has coordinates 2(b 3 2b 2a 2c . Solution Let the circle have equation x2 y2 r 2. Q. prove that the angle in a semicircle is a right angle. so p2 Then m1m2
r2 p2 p2 r2
q2
r2 or q2
r2
p2
1 90º
Then AC ⊥ BC and ∠ACB
332 C H A P T E R 9
. 2c)
F
E
x B(2b. the midpoint of AC. Then D. If P divides AE in the ratio 2:1. and F. and C(0. 3 . 0)
If Q divides BF in the ratio 2:1.
EXAMPLE 4
Using analytic methods. 0). Therefore.
2c 3
or
But P. has coordinates (b. then Q has coordinates 2( a) 2b . 0). 3
A(–2a. 2c). 0)
• p r p r But C is on the circle. the midpoint of AB. 0) y C(p. 3 0 0
. 0).Solution Let any triangle have coordinates A( 2a. The slope of AC is m1 The slope of BC is m2 Then m1m2
q q q p q p r p2 q2 r2 r
A(–r. 2c or 3 3 2b 2a 2c . B(2b. 2c . q)
x B(r. has coordinates (b a. then P has coordinates 2b 3 2a . and R are the same point. the midpoint of BC. the medians intersect at the same point and divide each other in a ratio 2:1. E. q) be any other point on the circumference. 0) O D y C(0. 0) and B(r. has coordinates ( a. c). Let C(p.
9 . y1) is a point that subtends equal tangents to two circles C1 0 and C2 0.
Application
5. by writing C1: x2 y2 r2. Prove that the lines joining the midpoints of opposite sides of a quadrilateral bisect each other. Show that if P(x1. Part B 3. then P lies on the line whose equation is C1 C2 0. Prove that PC2 PD2 is independent of the position of CD. Prove that the diagonals of a rectangle are equal. 1. 7. A convenient way of expressing equations briefly is to give them a name. 9. y1) to a circle with equation (x h)2 (y k)2 r2 is l (x1 h)2 (y1 k)2 r2. Prove that the length l of the tangent from an external point P(x1. 8. Hence. Prove that if the diagonals of a parallelogram are equal. we can describe the circle x2 y2 r2 0 as C1 0.2
Part A Use analytic methods to solve the following problems. Prove that the midpoints of successive sides of a quadrilateral are the vertices of a parallelogram. CD is any chord of the circle parallel to AB. 6. 4. 2 P R O O F U S I N G A N A LY T I C G E O M E T RY
333
.Exercise 9. Prove that the line connecting the midpoints of two sides of a triangle is parallel to the third side and equal to one-half of it.
Knowledge/ Understanding
2. Prove that the midpoint of the hypotenuse of a right-angled triangle is equidistant from the three vertices of the triangle. the parallelogram is a rectangle. P is any point on the diameter AB of a circle.
y1)
X(x. Solution You have already seen a development of a formula using vectors. and B. There is value in trying different methods on a problem.
EXAMPLE 1
Find a formula for the coordinates of the point N(x. y) m
B(x2.Section 9. the best method is the one that leads you to a solution. n
334 C H A P T E R 9
. y2)
Y(x2. Here we use analytic methods. It is sometimes clear that one method is easier than others. N. y) by drawing lines parallel to the axes through A. y2) in the ratio m:n. we strengthen our understanding of concepts and increase our confidence that we can find a solution. In the given diagram.3 — Different Techniques of Proof
Many problems in most branches of mathematics can be solved by a variety of approaches. you proved a number of things that you had done earlier by other methods. y1) and Y(x2. However. y1)
x
∆BYN
AN NB m n
x x1 x2 x
y m and y n 2
y1 y
m n
nx (m x Similarly y
nx1 n)x
mx2 m my2 m
mx2 mx2
nx1 n ny1 n mx2 m
mx nx1
The coordinates of N are
nx1 my2 . By doing so. m n
ny1 . as shown. ∠NAX ∠BNY ∠NXA ∠BYN Then Therefore ∆NXA
AX NY XN YB
(parallel lines) (right angles)
y n N(x. y1) and B(x2. y)
A(x1. we determine points X(x. y) that divides the line segment joining A(x1. In ∆NXA and ∆BYN. In the last section. as you will see in the following examples.
9. so 1 3m k 0 4
2
m
k
1
0 1
4 and k = 3 7 7 4 AF and so AE:EF 7 3 DB and so DE:EB 7
Subtracting. b)
E x D(0. AE DE
4:3 3:4
Solution 3 Let the coordinates of the parallelogram be as in the given diagram. 0) C(4c. 0) F(3c. The equation of AF is y 0 a b 3c (x 3c) The equation of DB is y
b b a 4c
x
b 4c
y
For the point E. but to gain experience. There is no best approach. a 3c (x 3c) a (a 4c)(x 3c) (a 3c)x (a 4c a 3c)x 7cx x Then y (a (a 4c)3c 4c)3c
x
A(a. so DE:EB Also AE
4 7
3:4
a (a 3c
3 7
3(a
b
3 b 7
2
3c)2
3(a 7
b2
4c) 2 3 2 b 7
and EF
(3c 4:3
a)2
b2
Then AE:EF
Note that in this example the analytic approach is more difficult algebraically and the deductive method is shortest. we must do a number of problems. 0)
3 (a 4c) 7 b 3 (a a 4c 7 3 b 7 4c) 2 7
4c)
Then the coordinates of E and B are in the ratio of 3:7.Now AE
m or 34 b m or 34
DE ma kb
a kb ( m ka k a 1)a 0
Since a and b are sides of a parallelogram. 7 m 4 m Thus. b) B(a + 4c.3 DIFFERENT TECHNIQUES OF PROOF
337
. they are linearly independent. Experience allows us to choose the most likely approach. although some approaches may be better than others.
Part B 5. Draw a quadrilateral ABCD. The Theorem of Apollonius states that in ∆ABC. In a trapezium. E is the midpoint of BC. By letting ∠AMB α and ∠AMC 180 α. Previously. Prove that DE GH. Prove this theorem using analytic methods. A town K is 12 km from a straight railroad. A a. then attempt the questions using the methods you have agreed upon. For some questions. a. then PA2 PC2 PB2 PD2. How far apart are the stations? 4. pairs of students can agree to share approaches and then discuss their results with classmates. On each of the four sides of the quadrilateral select a point (not the midpoint) so that when these four points are joined in succession.
338 C H A P T E R 9
B
M
C
. Prove that this result is also true if the point P is not contained in the plane of rectangle ABCD but is above the rectangle. Part C 7. then AB2 AC2 2AM2 2MC2. we proved that if P is a point in the interior of rectangle ABCD. Prove that a line through the intersection of the diagonals and parallel to the base divides the nonparallel sides in the same ratio. AE and BD intersect at F. if M is the midpoint of BC. Prove that the line joining points that divide sides AB and AC in ∆ABC in the ratio 3:1 is parallel to BC and equal to 3 BC. Part A 1. D bisects AB.Exercise 9. you should discuss different approaches with classmates.
Application
3. Square ABCD has sides of length 2.
Thinking/Inquiry/ Problem Solving
A P B
D
C
8. What is the ratio of ∆BFE:∆FAD?
B D G
A
E FH C
6. Two stations on the railroad are 20 km and 13 km from K. What is the height from F of ∆BFE? b. use the cosine law in ∆AMB and derive the result. E bisects AC. b. Verify that your selection of points is correct by the use of analytic and deductive methods. G bisects FB and H bisects FC.3
In this exercise. 4 2. a parallelogram is formed. the ratio of the parallel sides is 5:3. In the given diagram.
What is the locus of their midpoints? In this section. like the path of a point that moves such that it is always a constant distance from a fixed point (a circle). If AB is extended to C so that AB BC. Chords are drawn from a fixed point on the circumference of a circle. What is its locus? 2. What is the locus of the set of points? 5. A is fixed.
EXAMPLE 1
A triangle ABC has a base BC triangle has an area of 40. or the path traced out by a point that moves according to a stated geometric condition. and B is allowed to move along the circumference.4 — Locus
Locus is a fundamental concept in mathematics. use technology in the study.4 LOCUS
339
.
8. What is the locus of an airplane flying at constant height from the equator to the north pole? 3. what is the locus of C as B moves? 7. A set of points has the property that every point is equidistant from two fixed intersecting lines.
t chnology
e
The purpose of this investigation is to examine the path of a moving point in differing situations. A point moves so that it is 3 cm above AB. Where possible. Describe the locus of the vertex A so that the
9. we examine some methods of identifying loci (the plural of locus). or the set of points that are equidistant from two fixed points (the right-bisector of a line segment). A quarter is rolled around the circumference of a nickel. and in some examples we employ it. AB is a chord in a circle. AB is a fixed horizontal straight-line segment. From earlier work you are familiar with examples of locus.
A locus is a set of points that satisfy a given condition. technology can be helpful. What is the locus of the midpoint of the stick? 6. always in contact with it.
INVESTIGATION
1. In problems of this type. What is the locus of the centre of the quarter? 4. A stick moves so that one end is in contact with the wall and the other end is in contact with the floor.Section 9.
The locus of A is a pair of parallel lines. if you prefer.
A P C P O P P C C C P A B B
t chnology
e
The locus of P appears to be a circle. Since the distance between parallel lines is constant by definition. Then OA is the diameter of a circle for all positions of P (angle in a semicircle).
B(–1. the midpoint of AB. Solution 1 ∠OPA 90º (P is the midpoint of AC). 5)
Thus. from our definition because A. (x (y (x 3)2 (y 7)2 x2 2x 1 y2 10y 25 C(x. is 10. we use Geometer's Sketchpad to draw the locus. This is only a strong hint. and C must form the vertices of a triangle. Determine the locus of P. C In order to see what the locus of P will look like. chords of the circle are drawn to a moving point C on the circle. Determine the locus of vertex C. then A must lie on a line parallel to BC. 7) D(1. 7) and B( 1. tell us what we are aiming for. the midpoint of AC. and 10 away from BC.
C(x. or BC extended. 6).
EXAMPLE 3
A circle with a radius r units has diameter AB and its centre at O. You can also draw a number of chords manually.
A 10
A
l1
B 10
8
C
A
A
l2
EXAMPLE 2
The points A(3. that is. B.
340 C H A P T E R 9
O
. however. it has radius 2 and its centre at the midpoint of AO. We must exclude point D(1. 5) are the vertices of the base of an isosceles triangle ABC. the height of ∆ABC must be 10.Solution Since the area of ∆ABC is 40. distance 10 away from it. The locus of A consists of all those points such that the distance from A to BC. y). 6)
CA. Since ∆ABC is isosceles. If we had said C is equidistant from A and B. The locus of P is a r circle with AO as diameter. y) x2 6x 9 y2 14y 49 2x y 8 0 The locus of vertex C is the line having equation 2x y 8 0. From A. then D would be included. and gives us confidence that we are proceeding properly. then CB 1)2 5)2
A(3. y)
Solution Let the coordinates of C be (x. it does not constitute a proof. one on each side of BC. It does.
Solution Let the intersecting lines be x 0 (the y-axis) and y mx. ∠BPC is constant for every position of A. so (AB)2 16. P lies on a circle segment on BC as chord (angles on a chord).2 tells us that N is (1. What is the locus of P? Solution ∠ABC ∠ACB Then ∠PBC But ∠BPC Then ∠BPC ∠BAC (angle sum) 90º
A
EXAMPLE 5
180º
∠PCB ∠PBC 90º
1 ∠BAC 2 ∠PCB 180º (angle sum) 1 ∠BAC 2
y y
P x x C
Since ∠BAC is constant. Triangle ABC is variable such that BC is fixed and ∠BAC is constant.
342 C H A P T E R 9
. y) are given by x b and 2 y a 2 mb . a) and the coordinates of B by (b. We represent the coordinates of A by (0. a)
y = mx B(b. Then b2 (mb a)2 16 4x2 (2mx 2y 2mx)2 16 4x2 4(2mx y)2 16 x2 4m2x2 4mxy y2 4 (4m2 1)x2 4mxy y2 4 The equation of the locus is (4m2 1)x2 – 4mxy y2
y P A(0. Let A be a point on x 0 and B be a point on y mx such that AB 4. which divides XY in the ratio 1:2. 0). If P is the midpoint of AB. Substituting N into our equation provides a simple way to check the accuracy of the work. with their bisectors meeting at P. Determine the equation of the locus of the midpoint of the line segment. Then BC subtends a constant B angle at P. The locus of P is a circle segment with BC as chord. Our formula from Section 9. Therefore. mb) x
4.We know that the locus contains the point N.
EXAMPLE 6
The ends of a line segment of length 4 move along two intersecting lines. then its coordinates (x. Then b 2x and y a 22mx or a 2y 2mx Now AB 4. mb). For any position of A. ∠ABC and ∠ACB are bisected.
(Use a unit square and a constant k. a circle with centre (0. find the locus of P (the intersection of the line through A and Y and the line through B and X). A and B are fixed points on a circle and XY is a diameter.4
Part A 1.
Exercise 9. Part B 2. XA and YB extended meet at P. Determine.) 4.
9. If A and B are fixed points on a given circle not collinear with centre 0 of the circle and if XY is a variable diameter. respectively. two equal tangents to a circle are drawn. Determine the locus of the set of points such that these equal tangents are always perpendicular. 0) and radius 2. Find an equation for the locus and show that it represents a circle. so that the lines are perpendicular.
Y B
O X
P
A
3. Part C 5. determine the equation for the locus of P where P divides AB in a 2:1 ratio.Note that if m 0. If AB has a length of 6 units. with proof. A and B are points on the lines y x and y 2x. From an external point. the equation becomes x2 y2 4. As XY revolves around the centre. the locus of P. A point moves so that the sum of the squares of the lengths of the perpendiculars from it to the four sides of a square is constant. the point P moves.4 LOCUS
343
.
vector.Key Concepts Review
CHAPTER 9
In a problem where you wish to use analytic geometry. setting the problem up so that you can simplify necessary algebra is of great importance. Remember that it is not necessarily true that the easiest approach for one person is also easiest for you. In locus problems. place one side of a figure on an axis • choose coordinates for vertices to simplify your work as much as possible • seek a solution that minimizes your work In problems where you can consider different approaches. discuss different approaches with classmates.
344 C H A P T E R 9
. When possible. and try to determine which will yield the easiest approach for you. consider how you might approach the question using deductive. keep the following ideas closely in mind: • position figures so as to minimize the number of variables needed. and develop different approaches to a given problem yourself. or analytic methods.
Repeat the steps of Question 2 with equilateral triangles on the edges AB and CD. Draw the network AG. 8). or in cases when only two perpendicular directions are permissible (if the network must follow the grid of city streets. B( 2. e. How do you find the Fermat point of an acute triangle? What are some of its other properties? What are some other geometric problems that Jacob Steiner considered? ●
RICH LEARNING LINK WRAP-UP
345
. Steiner networks are the topic of a significant amount of contemporary mathematical research because of their application to data communications and computer microchip design. H G d. Some of this research considers Steiner networks in three dimensions. What are ∠AGD and ∠BHC? 3. Construct equilateral triangles on the edges AD F and BC pointing away from ABCD. 0). for example). 0). or circumstances in which the shortest total length is not necessarily the most cost effective. Investigate and Apply We will find a Steiner network for the points A(0. a. DG. but it will be much easier to use Geometer's Sketchpad (using the GRID feature). Draw the line segment connecting E and F. Draw the quadrilateral ABCD in a Cartesian coordinate system. GH. Which one is it?
INDEPENDENT STUDY
The Fermat point of an acute triangle is the point for which the sum of the distances to the vertices is as small as possible. Draw the circumcircles of the two equilateral C triangles. and HC. This can be done by using analytic geometry. C(7. Label the two new vertices E and F. It should look a bit like the diagram shown. 5). Either this E network or the one produced in Question 2 is the Steiner network. D A Find AG DG GH HB HC. HBH. and D(6. . c. 1. Find the points G and H where the two B circumcircles intersect EF.wrap-up
CHAPTER 9: STEINER NETWORKS
investigate and apply
A Steiner network for a given set of points is a set of line segments joining the points in such a way that the total length of all the line segments is minimal. b. What is the sum of the lengths of the diagonals? 2.
P is a point in the interior of the rectangle as shown. 0) 2. 4) and B(0. Determine the coordinates of the point C where C divides AB in a 3:1 ratio. Determine the equation of the locus of a point P that moves so that the line joining P to A(2.and y-axes b. 2) always equals 4. If PA2 PC2 85 and we are told that the lengths PB and PD are integers. the point P divides DC in a 5:1 ratio. The point P is such that the difference between the squares of its distances from the two points A( 3. AE:EP b.
A P D C B D A B
E P C
7. 6) c. given that the pathway is 1 m in width?
346 C H A P T E R 9
. the set of points equidistant from A(0. Find the equation of its locus. 1). What is the area of the pathway. In the rectangle ABCD. Ian has a square garden that measures 10 m 10 m. the set of points 1 unit away from O(0. The line segment AB has end points A( 3. 7) and B(4. 3. y) such that P is always equidistant from the line x 3 and the point X(3. In the parallelogram ABCD. 6. DE:EB 5.Review Exercise
1. 9. Determine a. 5) and B(9. He constructs a pathway around the square garden so that the pathway is always 1 m away from the garden. the set of points equidistant from the x. The line joining A to P and the diagonal BD intersect at E. 16). Describe the following loci algebraically. Find an equation for the locus of points P(x. Find the equation of a locus such that its distance from the origin is numerically twice its distance from the line y 6. a. 5) always has an inclination of 45º. 4. 8. determine the possible values for PB and PD.
Determine the coordinates of the point that is equidistant from the points A(2. Show that the locus of P is a circle with centre at D. 1). Show that each of the following circles passes through the centre of the other. B(2.10. 5). a. 18. In the circle with equation x2 y2 6x 8y 24. 15. (Let the origin be at the centre of the parallelogram and the diagonal AC lie on the x-axis. 0) and the point A(2. 3). Identify this locus. Find the coordinates of the centre and the length of the radius in a. 2) is always 2 greater than the slope of the line joining P to the point (0. A point moves so that the sum of the squares of its distances from the vertices of a triangle is constant. Show that the locus of the vertex of a right-angled triangle whose hypotenuse is the line joining the points A(3. y) and the point (3. In parallelogram ABCD. Show that the set of points that bisects chords drawn through an end of the horizontal diameter of the circle represented by x2 y2 a2 is a circle.)
REVIEW EXERCISE
347
. 4). 13. a chord is bisected by A(5. 17. AC is the shorter diagonal and a point P moves so that AP2 CP2 BP2. 5) and B(7. and C( 6. b. 1). 14. x2 y2 4x 18y 60 0 and x2 y2 2x 10y 1 0. Find the equation of its locus. Find the equation of the circle that has its centre on the y-axis and passes through O(0. 11. 12. 7). The slope of the line joining P(x. Determine the equation of the chord. 11) is a circle. 16.
17) in a 1:4 ratio. The coordinates of a point A that divides the line segment joining A( 3. 5.
348 C H A P T E R 9
. 6. Find the equation of the locus of the centre of a circle that passes through the point (0. The rod is allowed to move so that the ends slide along the axes. Determine each of the following. 5) and B(1.Chapter 9 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application Questions all 6. a. 0) and cuts off a length 2 k from the line with equation x c. An equilateral triangle has a side length of 2 units. and C(5. 3)
3. The locus of a point P that moves so that P is the midpoint of the line segment joining the origin to any point on x 4. Determine the equation of the circle passing through A. 4. the point that divides AB in a 1:4 ratio. 2) 2. 4). b. a. 8). Describe each of the following loci in one sentence. x2 y2 2x 6y 3 0 b. the set of points equidistant from the lines at x c. 4). B(1. B. (x 1)2 (y 2)2 (z 3)2 9 3 and x 5 b. 7) and B(2. a. 7. 7 1. Determine the coordinates of a point that is equidistant from the points A(1. the set of points equidistant from A( 3. Determine the equation of the locus of P and show that it is a circle. Find the locus of P. Describe the following loci in algebraic terms. the set of points 5 units away from A( 3. 2 all
1. A and B are the ends of a rod of length 10 units with A on OX and B on OY. OX and OY represent two straight rulers placed at right angles to each other. and C. A point P moves so that the sum of the squares of the distances from P to the three vertices is 11.
Use a search engine to find a picture and a description of a CMM on the Web.C O O R D I N AT E M E A S U R I N G M A C H I N E S
Modern cars are made by mass production. The CMM determines the coordinates of the yellow dot and the corresponding coordinates of the point on the opposite side of the rod. To measure geometrical properties. For the engine to run properly.
E X T E N D I N G A N D I N V E S T I G AT I N G
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. the distance from the centre of the hole on the left (the larger end) and the centre of the hole on the right (the small end) can vary by no more than a few microns (1 micron 10 6 metres). y. This means that the dimensions of each rod must be very tightly controlled.
1
3
2
4
Thousands of rods are manufactured every day. to measure the thickness of the rod at one of the yellow dots on the illustration. The computer then calculates the x-. the rods must be interchangeable. For example. For example. often called a CMM. A computer-controlled arm with a stylus on the end can be moved so that the stylus touches the rod at any point. One necessary criterion for mass production to work is the ability to manufacture particular components so that they are all essentially the same.and z-coordinates of the point on the part relative to a fixed origin. Thousands of components are assembled to get a finished vehicle. Then the standard geometric formula is used to find the distance between the two points. When the engine is assembled. A V6 engine requires one rod for each of its six cylinders. the rod can be mounted in a clamp. a part is placed on the bed of the CMM. How can this distance be measured to such a high degree of precision? One common tool for measuring dimensions is a coordinate measuring machine. a rod (more fully called a connecting rod) is part of the engine.
suppose we wanted to find the coordinates of the centre of the hole at the larger end. Instruct the computer to carry out the following calculations: 1. There are many interesting mathematical problems associated with this algorithm. This algorithm is relatively easy to describe and implement. The problem is that the shape of the hole is not a perfect circle! In fact. we do not know the exact shape. Finally. we can then determine the critical distance between the two centres. if we repeat the measurement. C on the edge of the circle at the larger end. the coordinates of three points A. If we use the same procedure at the small end. the algorithm is repeated for each of the ten subsets of three points. We know that these three points form a triangle on a plane. B. They know all about points. What can go wrong? There is a long list. Find the coordinates of the midpoints P and Q of AB and AC.For a harder problem. C. There is no place to touch! Now we need to use some mathematical thinking.
2. They exploit this knowledge to measure. more precisely. through Q perpendicular to the line AC. planes. Where is the best place to put the five points? Would it be better to use six or seven or more points? How much variation will there be in the coordinates of the centre if the algorithm is repeated on the same rod? People who run CMMs are experts at three-dimensional geometry. The centre of the circle is the point of intersection of the right bisectors of the edges of the triangle. Find the point of intersection of the lines l1 and l2. because it is not possible for the CMM to move the stylus to exactly the same three points on the edge of the circle. cylinders. First. You might think that this would not matter since we know that the centre of the circle is the same for any three points on the edge.
5. the coordinates of several points (say five) on the edge of the circle are measured.
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. the average coordinates of the ten centres are determined. It is a good thing that we have computers to do all this work. Find the equation of the line l2 in through P perpendicular to the line AB. Find the equation of the line l1 in 4. Find the equation of the plane through A. B. lines. Then. and other geometric shapes. One algorithm is to first determine the coordinates of three of the four yellow dots or. To deal with this uncertainty. critical features of manufactured parts. For example. the above algorithm is modified. 3. This gives us the coordinates of the centre of the circle. we will not get the same answer. often indirectly. which will vary somewhat from one rod to the next.
2.5
. 10. Today. Section 10. counting's the best thing to do… The Count from Sesame Street had the right idea. you will
• • •
express the answers to permutation and combination problems. counting is marvelous.3. the herdsmen would know if any animals were missing. When evening came and the herds returned. Section 10. you will be introduced to the mathematics of counting or combinatorics. Ancient herdsmen counted their herds by matching each animal with a stone.Chapter 10
INTRODUCTION TO COUNTING
Counting is wonderful.1 solve problems using counting principles.4 solve problems involving permutations and combinations. the Census of Canada counts people so that resources may be allocated fairly. The properties of combinations are used in statistics and probability. even though they did not use a number system. Section 10. Counting is the oldest mathematical operation.
CHAPTER EXPECTATIONS In this chapter. In this chapter. 10. 10.1.
The following activity will help you understand the usefulness of what you will learn in the next two chapters.. Your chance of winning the grand prize depends directly on the number of possible combinations. F. . since the numbers are obviously very large. . Make a list of all possible ordered pairs of squares. B. we must be careful not to miss any possibilities or to count any combination more than once. we must be very organized. You will have to invent a way to record all of the possibilities so that you do not miss any.
A B
B A
Note that in any pair there are two different letters and that
is a different pair than the first. In other words. the balls are numbered from 1 to 49 and you win a very large prize if you hold a ticket that matches the set of six numbers drawn.
ACTIVITY
With a partner. One example is shown below.. Use the squares as a physical model to answer the following questions. how can we count the number of combinations of six different balls that are drawn from the drum in the 6/49 lottery? In this lottery. cut out six small squares of paper (approximately 5 cm 5 cm) and label them A. two. However. you will realize that there is much more to counting than pointing at objects and saying one. In trying to answer these questions. The mathematics of counting is called combinatorics.Review of Prerequisite Skills
Counting is a fundamental mathematical operation. three. We also might be interested in knowing how many ways the balls can be selected so that at least three of them match the number on our ticket. …. For example. 1. How many pairs do you have in your list? How many pairs in your list have A in the first position?
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. with a moment's reflection.
Repeat Question 1 using an ordered triple of squares. you will want to find a convenient way of writing it. To provide some organizational tools. we use the notation and properties of sets to develop some rules and strategies for counting the number of elements in a set.2. we will look at ways to count lists of objects. In the following sections. we start this chapter with some ideas from set theory. Sets are mathematical objects that describe how groups of things or elements can be organized in a formal way. we use sets to specify the elements that we are trying to count. Since the list is long. As a result of the activity. The squares A and B can be used to make two different ordered pairs. Make a list of all sets of two squares that can be formed from the six letters. List all sets of three squares that can be formed from your six squares.
REVIEW OF PREREQUISITE SKILLS
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. How many sets are in the list? How many of the sets contain square A? Can you explain the relationship between the answers here compared to those in Question 1? 4. you may have discovered some of the fundamental rules that we will use to solve such counting problems. Here. 3. How many sets can you form? How does this number compare to the answer in Question 2? Can you explain the relationship? In the next two chapters. but only one set. A set of two squares is not ordered.
the prize is divided equally among all the winners. If nobody wins.00) = $200.00. Otherwise. you lose. you expect to win $200. Many times there is no winning ticket.00 (2 $50. The probability of winning $200. Numbers less than 31 are also chosen more frequently than numbers larger than 31. Players choose six numbers and hope to match the six numbers selected on draw night.00 $200.00.00 once and $50. by playing over a long period of time. Lottery statistics show that people favour lower numbers.00 is 11 and the probability 00 1 of winning $50. per ticket. Sometimes there are four or five winning tickets. the holders of one Canadian lottery ticket collected $22. Investigate The expected value of a lottery ticket is the amount of money you can expect to lose.00 every time you play this lottery. If more than one ticket matches the six numbers drawn. One of the most popular lotteries in Canada is Lotto 6/49. the prize amount keeps growing. Why do you think this is the case? What does this suggest about strategies for picking lottery numbers? ●
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. Consider the following simple lottery: You buy ticket #17 for $5.00 twice. The other half are people's own choices.
DISCUSSION QUESTIONS
1. you win a $200. and you will find that the expected loss per play is $2. Does this strategy reduce their expected loss? 2.If the number drawn is 16 or 18. Approximately one-half of Lotto 6/49 numbers are quick picks (i. especially numbers between 1 and 12. Your total lossess are $500. A number between 1 and 100 is drawn at random. and they are not evenly distributed among all 49 possible numbers.00 prize. In other words. With only a small chance of winning. you lose on average $2. are lotteries worth playing? One way to answer this is through the concept of expected value.investigate
C H A P T E R 1 0 : L O T T E R I E S A N D E X P E C T E D VA L U E S
In March 1998.e.00 is 50 .00 prize. if you play this game about 100 times at a cost of $500. Some people only play Lotto 6/49 when the prize is very large.5 million.00. If the number drawn is 17.. you win a $50. The possibility of winning such a large sum of money entices millions of people to try their luck at lotteries. randomly selected). Divide this by the 100 times that you played. Hence.00.
The sets {1. We name the set using a capital letter and enclose the elements or members of the set in braces {}. 3.
The collection of numbers A {1. 29} are the same. 2. B. 2. The next example is more complicated. The set of all cells in the specified region is U {A1. 15. we specify the rule by the pattern of the elements listed. 22. and 3 and columns A. 15. and the rows by numbers 1. 2. N2L 3G1. We often use a partial listing such as this.
10. 22. B2. …. 22.1 SETS
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. The columns are designated by letters A. 49}. EXAMPLE 2 Postal codes are used to help sort mail automatically. we look at some basic properties of sets. C3. . An optical character reader scans the postal code on each letter to help direct the letter to a particular location. B1. 8. 5. D1. we do not need to write out the rule that defines U because U is small enough that we can list all of its elements. 8. A Canadian postal code is a sequence of length six. C1. the set of integers from 1 to 49 can be written B {1. 29} is not a set.. EXAMPLE 1 A spreadsheet program uses cells that are labelled by their column and row position. Write down the set of all cells that are in rows 1. since 29 appears twice. Therefore. 2. B3. Solution We can label a cell uniquely using its column letter and row number so A1 is the cell in the first column and first row. 8. Here are some examples. the 12 elements of the set U are cells. the elements of a set are unique. D2. so that we can improve our understanding of what the elements are. Second. For example. A3. There are two further considerations when we are specifying a set. B. {1.. 29. no two can be the same. C2. 3. D. 29. 8. the order in which we list the elements does not matter. C. for example. If there is no possibility of misunderstanding. . 10} is the set A with elements that are integers from 1 to 10. C.. 9.Section 10. It is identified by a rule for deciding whether or not a particular element is in the set.1 — Sets
In this section. A set is a collection of elements. D3}
Here. . First. 15.. 1. 36} and {36. 7. 6. 4. A2. In Example 1. even if the rule is not obvious and must be stated separately..
1. U {A0A 0A0. A0A 0A1. . We can describe the eight possible outcomes by the universal set U. Specify the set of possible postal codes. and the second. We will consistently name the universal set U. If the universal set is the integers from 1 to 10.The first. 9. 9} is a subset of U. 3. the sequence HTH indicates that the first coin came up heads. and sixth terms are digits from 0 to 9. and fifth terms are upper-case letters. Note also that C {1. we will learn how to count the number of postal codes.. and the third heads. This set is called the universal set. TTH. It is important to distinguish between the elements of a set and the number of elements. 4. It is helpful to specify the universal set to be sure that we understand the objects that we are counting. Each coin can come up either heads (H) or tails (T).. 3. Define the subsets corresponding to the following statements. I. HTT. U. O. THH. Another important idea is a subset of the universal set.. 7. for example. Z9Z 9Z9}
Each element in the set is a postal code. To make the idea of a universal set and its subsets clearer and also to demonstrate some games that we can play with subsets. That is. For this example. and W are left out because they look too much like other digits or letters to the character reader. HTH. 5. 2. U {HHH. and state the number of elements in each. 2} is not a subset of U. 8. We can imply the rule by writing the postal codes in alphabetic/numeric order. Solution The collection of all postal codes is a set U (a very large one). 5. E: the second coin has come up heads b. fourth. F: exactly one coin has come up heads c. Not all letters are used. we use the notation n(A). third. A subset A of U is a set whose elements are all elements of U. 7. HHT.
EXAMPLE 3
Suppose three coins are tossed. In the next sections. G: two out of the three coins have come up the same
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. THT. The set of all postal codes in Example 2 is the universal set if we are interested in counting postal codes with various properties. 10}. n(A) 5 and n(C) 2. Q. consider the next example. we usually start with the set of all the possible elements of the type being counted. If we wish to refer only to the number of elements. we list the terms that are letters in alphabetic order and the other terms in numerical order. The letters D. B {0. 3} is a subset of U and also a subset of A. 6. U {1. TTT}
where. F. then A {1. a. the second tails. In counting problems.
and n(G) 6. 12. HHT. …. THT. All the sequences in E have an H as the second term. 9. THH. We display a subset by enclosing the appropriate elements within a closed curve. 100} is the subset of integers in U greater than 10.
and C is the subset of integers in U that do not contain the digit 1. 100}. 21. 81. 5.1 SETS
357
. …. 91. Hence. That is.Solution a. …. THT. HTH. given in Example 3. C is the subset of U consisting of all integers from 1 to 100 that contain at least one digit 1. 10} {1. 2. U {1. 100}
Solution a. 31. so F {HTT. 11. THH. 99} is the subset of integers in U that are not perfect squares. {2. 3. E {HHH. Here we show a Venn diagram for subset E. 100} {1. we have A B {11. THT}. c. We can display the universal set and selected subsets in a picture called a Venn diagram. HTT}. b.
10. and n(F) 3. a. TTH. and n(E) 4. 19. The sequences in G have two Hs and one T or one H and two Ts. We assume the most logical rule to describe a set when not all the elements are listed. The answers to the following example might be different for different people. 12. A Venn diagram is useful for showing the relationships among subsets. For instance. …. TTH}. c. 3. A is the subset of U consisting of all integers from 1 to 10. A with a bar over the top. We usually denote the complement of A by A. The sequences in F are made up of one H and two Ts.
U HTT TTH TTT HTH HHH HHT THT THH
E
In Example 3. b. B c. …. that is. then the complement of A is the subset of U containing all the elements of U not in A. Hence. (Notice how many elements had to be listed to make this rule obvious!) If A is a subset of U. the object was to specify subsets based on a verbal description. A b.
EXAMPLE 4
Consider the universal set U of integers from 1 to 100. 10. Sometimes we need to play the game in reverse. Describe the following subsets of U in a simple sentence. 2. C {1. B is the subset of U consisting of all perfect squares up to 100. …. G {HHT. …. 4. in Example 4.
100
In Example 5.. 2. so it follows that n(A) n(U) n(A) 1000 200 800. 12.
EXAMPLE 5
How many integers from 1 to 1000 are not divisible by 5? Solution Let U {1. 15. List the elements of U and A.By definition. 10. 1000} {1 5. …. There are 800 integers from 1 to 1000 not divisible by 5. 200 5}
We can see that n(A) 200. it is difficult to count the elements in A but easy to count those in A. A Venn diagram demonstrates the rule clearly.
U A 1. n(A) n(A) n(U).. This simple rule is surprisingly useful. …. …. Suppose that U is the set of all two-digit integers and A is the subset of U of integers containing the digit 7.. 3..
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. .1
Part A 1. 3.. 1000} be the universal set so that n(U) 1000.
Exercise 10. You will also have to define U and find n(U). 2. If it looks difficult to count the number of elements in a subset directly. try instead to count the elements of the complement. 10 A 11.. n(A) n(U) n(A)
If it is difficult to count the elements in A. those in A and those in A. If A is the subset of integers not divisible by 5. 3 5. A {5. then the complement of A is the set of integers divisible by 5. All the elements in U are divided into two distinct groups. or more usefully. . When we are counting the number of elements. 13. 2 5. we can determine this number indirectly by finding n(A) and n(U). the complement of a subset A contains all the elements of U not in A.
List all the elements in the universal set U. What is n(U)? c. A student is allowed to sign out two books at one time. 4. Consider the universal set U of binary sequences of length 3. List the elements in the following subsets of U. e. List the set of all possible ways that two books can be signed out. E: the sequence has exactly one 1 F: the sequence has at least one 1 d. describe P and Q in words. Indicate the elements in the following subsets of U. …. A school library has five different calculus books. 3. a.Application
2. A binary sequence is a sequence of 0s and 1s. 10. 75.1 SETS
359
.
Knowledge/ Understanding
5. If V is the subset of vowels. If F is the subset of all elements in U that are not in F. and U on a Venn diagram. most of which are not real words. 15. describe F in words. 54. Show A. Describe the complements A and B in words. e. For P and Q given in b. Consider the set of integers U R {5. List the elements in the following subsets of U. List all the elements in U. list the set U of all possible meals. {53. b. A restaurant serves four main courses and three desserts. Verify directly that n(F) 7. Part B
Knowledge/ Understanding
6. b. 2. If a meal consists of a main course and a dessert. A: all words beginning with the letter a B: all words ending with t d. P: all integers that are greater than 70 Q: all integers that end in the digit 7 c. …. Let U be the set of all such words. The three letters of the word cat can be re-arranged to form other three-letter words. 100}. R
a. ….
{1. b. 76}. Describe the following subsets in words. What is n(U)? c. list the elements of V and its complement and find n(V) and n(V ).
10. S n(F ) n(U). 100}. 3. B. 55. The universal set U is made up of the letters of the alphabet. a.
and Y: the integers from 1 to 100 that are perfect squares. Two subsets of the set U {1. Let U be the set of all subsets of size 2 with elements selected from the six letters a. Is Y the complement of X? Explain.8. In words. 3. Note that 1 is not a prime number. 100} are X: the integers from 1 to 100 that are prime. Let U be the universal set of all possible outcomes. In a simple lottery. 1000} be the set of positive integers less than or equal to 1000. describe S and its complement. b. Balls are not replaced once they have been selected. 20. b. Find the following subsets of U. 2. 3. 2. H: the subset of integers not ending in 9
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. A: the ball labelled 4 is selected first B: the ball labelled 4 is selected
Communication
12. b. Do X and Y have any elements in common? Explain. Repeat Question 11 if the balls are replaced on each draw. What are n(S) and n(S )?
Communication
9. Suppose U1 is the set of all possible outcomes. 2. 1000}. …. …. W: each subset contains the letter a V: each subset contains at least one of a or f d. F: the subset of perfect squares c. Let U {1. a. a. e. 10. Write out all of the elements of U. Explain why n(U1) n(U). 1000} with subset S {10. f. c. three balls are selected to form a three-digit number from a set of nine balls numbered from 1 to 9. Suppose U is the set {1. …. d. Determine the number of elements of the following subsets of U. c. Show V and V on a Venn diagram. …. a.
Application
11. Develop an appropriate notation and rule to describe the elements of U. a. E: the subset of integers divisible by 7 b. b. 30. Find the size of the following subsets of U. G: the subset of integers not divisible by 3 d. 13. Find n(U).
15.1 SETS
361
.Part C
Thinking/Inquiry/ Problem Solving
14. V {(x. and C and the corresponding envelopes a. and V. b. V. B. 0 y 1}. y):x y Draw a Venn diagram that shows U. Suppose U is the set of points in the plane defined by U {(x. List the elements of the subset that corresponds to no letter being sent to the correct person. Suppose we label the letters A.
10. and c. Three letters that include the name of the recipients are folded and put into three envelopes.
1} is a subset of U. y):0 x 1. Construct a notation for the universal set to describe all the possible assignments of letters to envelopes.
and U in a Venn diagram. 20}. The total area covered by A and B represents all the elements in U that are found in either A or B or both. suppose the universal set is the positive integers from 1 to 20. 20}
U 2 8 4 6 10 20 1 3 7 9 11 13 15 17 19 5 B
The subset A {2. U {1.Section 10. so these two subsets are disjoint. 6. let C {5. …. 18. 4. those divisible by 2). 15. The intersection is represented by the shaded area on the Venn diagram.
In words. We call these subsets
disjoint. The general case is displayed in the Venn diagram. …. it appears only once in the union A B. 20} corresponds to the even integers in U (i. 6.2 — Combining Subsets
In this section. A B {10.
12 14 16 18 A
The fact that A and B overlap in the picture indicates that they have one or more elements in common.
Remember the fundamental rule that we do not repeat elements in a set. In this case. The elements 10 and 20 are found in both subsets. we look at two operations for combining subsets that will later be helpful in solving counting problems. 14. The subsets A and C have no common elements. Even though 10 is found in both A and B.e.
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U 3 1 7 9 11 13 17 19 A
5 15
C
6 2 4 14 10 12 0 8 2 16 18
. 10. 8 10.
U A A B B A B
In some cases. 2. 5. 20}. A B is the subset of U of integers divisible by either 2 or 5. This subset is called the union of A and B and is denoted by A B. A and C have no overlapping area. 4. We can display A. 15} be the subset of odd integers divisible by 5.
In the above example. Note that this Venn diagram has been simplified by omitting the individual elements. 12. A B contains all positive integers less than or equal to 20 that are divisible by both 2 and 5. The subset corresponding to the common elements of A and B is called the intersection of A and B and is denoted by A B. On the Venn diagram. and the subset B {5. 16. 15.. In words. B. two subsets have no elements in common. To demonstrate these operations. 20} corresponds to the integers in U divisible by 5. Here A B {2.
1101. 1100} and W V is the subset of binary sequences that have at least three 1s or start with 11. C. 0111. and V be the subset of sequences that include three or more 1s. B. so the complement of W is not the subset of binary sequences ending in 11. The Venn diagram is shown to the right.2 COMBINING SUBSETS
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. and C are three subsets. B. 1110. We have W {1111. intersection. R. The subset S is all arrangements in which B is at the right end of the shelf. b. Four books labelled A. 0010. c. 1100} and V {1111.EXAMPLE 1
Consider the set U of binary sequences of length 4. B. These definitions can be extended to any number of subsets. The complement of W contains all of the sequences in U that are not in W. The universal set U is the set of all possible arrangements. 1101} and W V is the subset of binary sequences with at least three 1s starting with 11. 0001. 0111}. Is this answer correct? Explain. Similarly. and C. 1011. Find the elements in W c. List the elements of U. We can build many subsets by using the union. W V {1111.
U V 1111 1110 1101 1100 W 1011 1010 1001 1000 0111 0110 0101 0100 0011 0010 0001 0000
d. R S and R S. Show U. S. For example. if A. For example. W V {1111. b. 1110. or C. ….
d. and complement. U {0000. a student explained that the complement of W is the subset of sequences that end with 11. 1011. 1101. then A B C is the subset containing all those elements that are common to A. Find the elements in W V and describe this subset in words.
Exercise 10. W. On a test. Solution a. and V on a Venn diagram.
10. and D are placed on a shelf. 1111}
Let W be the subset of binary sequences starting with 11. 1110. The subset R is all arrangements in which A is to the left of B.2
Part A 1. 1101. B. 1110. V and describe this subset in words. 0000 is in W. A B C is the subset containing all elements found in at least one of A. a.
No three points fall on a line. 9. Draw a Venn diagram to display U.
d. a. c.
E D A B C
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. B and describe this subset in words. For any two subsets A and B. where E is the subset of sequences that start with 1 and F is the subset of sequences that have at least two 1s. Let P be the subset of pizzas with pepperoni and M be the subset of those with mushrooms. and D. 8. Explain in words why n(A B) n(A) n(B). If A is the subset of choices with a blue shirt. C. A. Let U be the universal set of all such sequences.Application
2. n(A
Communication
7. find n(E F). Construct the set U of all possible triangles that can be formed using these five points. labelled A. A be the subset of such integers divisible by 3. and grey. and n(A B). Do not include the individual elements of U. are shown on the diagram. The universal set U is the set of all possible selections you can make. q are arranged in a sequence. D and describe this subset in words. Note that triangle ABC and triangle BCA are the same. can n(A B) n(A) n(B)? Explain. p. b. q. C be the sequences in U starting with p. Determine A B and describe this subset in words. You decide to buy a pair of jeans and a shirt. If U is the set of all binary sequences of length 4. find n(A) and n(A). b. and D be those sequences ending with p. A pizza special can be ordered with any three of nine toppings. Find a subset of U that is disjoint with C. Are A and B disjoint? e. Determine A d. 3. Suppose the 5 letters p. Let U be the set of positive integers from 1 to 16. Suppose A and B are two disjoint subsets of U. a. B. black. Find C c. Part B
Communication
4. Five points in the plane. C. B). A store stocks jeans and shirts in three colours: blue. Are P and M disjoint? Explain. 5. Find C D and describe this subset in words. D. E. and B. n(B). a. Write all the elements of U. and B be the subset of such integers greater than 10.
Thinking/Inquiry/ Problem Solving
6. p. Determine n(A).
the complement of A B. n(F). List the elements of F G and F G. On the plot. describe the subset A b. B the subset with 2 in the second place. C and three boys D. Draw a set of axes with a dot at each of the points in P.b. Consider the set P of points in the Cartesian plane defined by (i. describe the subset A B in terms of the elements of A and B. show that U can be written as the union of three subsets corresponding to a committee with two girls. and a committee with two boys. 4 are arranged to make a four-digit number. where c takes
Thinking/Inquiry/ Problem Solving
14. The letters of the word tree are scrambled. b. c. j 4. Find A
Application
B
C
D. F. show the subsets corresponding to i on values 2 to 8. a. a. The universal set U is the set of all possible arrangements. where i and j are integers. By listing the elements in each subset. 3. We want to select a committee of two people from six candidates. C the subset with 3 in the third place. Let U be the set of all such committees. 13. find n(E). where U is the universal set of all of the possible arrangements? Explain. In words. Is any pair of these subsets disjoint? d.
11. List the elements of the subset F of triangles that have the line segment AB as one side and the subset G of triangles that contain C as a vertex. n(E F) and n(E F).2 COMBINING SUBSETS
365
.
10. and D the subset with 4 in the fourth place. Let A be the subset of those numbers that start with 1. 12. Does A B C D U. The digits 1. a. Does their union include all the elements of P? Part C j c.
10. B. a committee with one boy and one girl. 2. c. 1 i. In words. c. If E is the subset of arrangements in which the two es are side by side and F is the subset of arrangements in which the t comes before the r. B. three girls named A. Is any pair of these subsets disjoint? Explain. E. j). b. Suppose A and B are two subsets of a set U.
B. Explain why A B A B. describe the subset A d. after Augustus De Morgan (1806–1871). A and B are two subsets of a set U as shown on the Venn diagram. the first Professor of Mathematics at University College. 15. and their complements.c.
e. In words.
A
B
Venn Diagram
(The results in Questions 14 and 15 are known as De Morgan's Rules.
U
16.)
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. Show the result in d by illustrating each set on a Venn diagram. London. Use the results from Question 14 to show that A B A B. Show that U can be written as the union of four pair-wise disjoint subsets defined in terms of A. B.
RYRB. Then we have E {RYRB. BYRR. RRYB. Let U be the set of all possible arrangements. n(E F) 1 We observe that n(E F) n(E) n(F) n(E F). Here.
10. BRRY. U
Sum Rule If E and F are two subsets. n(E F). {RRYB. YBRR. we need to determine the number of possible outcomes. we look at the relationship between n(E). Since we have counted the elements in E F twice. Many of the problems we encounter require only that we determine the number of elements in a set. RRBY represents the arrangement with two red blocks first. BYRR. we combined two subsets using the union and intersection operations.3 — The Sum Rule
In the previous section. we first count the number in E plus the number in F. BRYR. and finally the yellow block. BYRR. one is yellow. We call this result the sum rule. For this reason we will now focus on the number of elements in sets. YRBR. YBRR} E F {RRBY. and n(E F). RYBR} E F {BYRR} and n(E) 3. RYBR.
EXAMPLE 1
Four wooden blocks are arranged in a row. for any two subsets E and F of some set U. n(E F) 8. Two of the blocks are red. We can demonstrate this relationship with a simple example. YRRB. RYBR. RRYB. YRRB.3 THE SUM RULE
367
. RBRY. RRBY. BRRY. BYRR} F {RRBY. and one is blue. n(F). This rule is true in general and simply states that to count the number of elements that are in one or both of E or F. For example. if we wish to know our chances of winning a lottery. the number of elements found in one or both of E or F is n(E F) n(E) n(F) n(E F). RBYR. YBRR} where. but we do not need a listing of them all. YRRB. RYRB. Let E be the subset of U corresponding to the second block in the arrangement being yellow and F the subset corresponding to all arrangements in which the two red blocks are side by side. then the blue block. n(F) 6. we must subtract the number of elements that appear in both E and F. BRRY. for example.Section 10.
9. In total. 6. 96} { 1 6. 3 3. we get n(E F) n(E) n(F) F) n(E 6. Similarly.
368 C H A P T E R 1 0
. Then we can state the sum rule for disjoint sets. the number of elements found in E or F is n(E F) n(E) n(F). …. the subsets E and F are disjoint. …. 50 2} {3. …. the number of elements that are in E but not in F is shown as n(E) – n(E F). 6.A Venn diagram can help to demonstrate the sum rule. 99} {1 3. 3 F. Of course. Hence E F {6.
EXAMPLE 2
What percent of the integers from 1 to 100 are divisible by either or both of 2 or 3? Solution U {1. n(E Applying the sum rule.
Sum Rule for Disjoint Subsets If E and F are disjoint. F) 50 33 16 67
67 Hence. …. …. they have no common elements. remembering that for disjoint sets n(E F) 0. 2 3. Let E be the subset of integers that are divisible by 2 and F the subset of integers divisible by 3. this subset contains those that are divisible by 6. 33 3}
The subset E F is all integers in U that are divisible by both 2 and 3. 4. 2 2. 100 or 67% of the integers from 1 to 100 are divisible by 2 or 3. 6. 18. the number of elements in F but not in E is n(F) – n(E F). 2 The integers divisible by 2 or 3 are all those in E We find n(E) 50. Then E F {2. In many counting problems. 3 2. 100} is the set of integers from 1 to 100. In the diagram. 16 6}. the number n(E F)] n(E F)
We use the sum rule to count the number of elements in the union of two subsets. n(F) 33. That is. so that n(E F) 0. ….
n(U) n(E F) n(E)–n(E F) n(E) n(F)–n(E F) n(F)
The number of elements in the overlapping area is n(E of elements in E or F is n(E F) [n(E) n(E n(E) n(F) F)] n(E [n(F) F)
F). as illustrated below. we can always use the first rule.
16 by counting the elements directly. …. That is. 12. 2. 100} {1 2.
g) where 1 r. three boys and three girls. 100} is the set of integers from 1 to 100.EXAMPLE 3
We want to form a committee of two people. Considering the complement. The set U of possible outcomes is given by all ordered pairs of the form (r. If G is the subset of committees with two girls. the pair (3. Find n(A B). 3.
Application
10. Similarly. determine the number of arrangements that have at least two trees of the same type side by side. How many committees can we form so that both members are the same sex? Solution Let the three girls be denoted by a. Then n(G) 3.3 THE SUM RULE
369
. 2. are rolled. Two dice. g 6. The possible number of committees is six.3
Part A
Knowledge/ Understanding
1. The subset G B contains all committees with both members the same sex. b} is the committee of a and b. and c. so we have n(G B) n(G) n(B) 6. for example. Note that G and B are disjoint. Consider the following subsets of U. 2. For example. {a. 4) indicates that the red die came up 3 and the green die came up 4. A: both dice have the same value B: the sum of the values is 7 a. List all the pairs in these two subsets. if B is the subset of committees with only boys. {b. Suppose that U {1. c}. There are six people available. B. …. n(B) 3. then G {{a. b. If A is the subset of perfect squares and B is the subset of integers divisible by 4. Six trees—three cedars and three pines—can be planted in a row in 20 different ways. one red and one green. {a.
Exercise 10. 3. b}. verify that n(A B) n(A) n(B) n(A B) by directly counting the number of integers in each subset. b. c}}
where. List all pairs that are in A c.
That is U {abcd. 2. divisible by 7
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.. What is n(U)? 10. 2. 1. b. and B the subset of arrangements that have b in the second position. 17. She concludes that there are 19 such two-digit numbers. How many numbers from 1 to 1000 are a. A student was asked to count the number of two-digit positive integers that start or end with 7. 37 take Business 101. A first-year calculus class has 90 students. 0. Of the 90 students.Part B
Knowledge/ Understanding
4. Consider the set U of arrangements of the four letters a. a.e. 70 to 79) and 9 that end with 7 (i. Explain why n(U) n(E 0) n(E1) n(E2) n(E 3). abdc. verify the sum rule for the subset A B. find n(V W). The Venn diagram shows the subsets A and B of the set U and gives the number of elements in each non-overlapping area.) 8.
17 5 A A 3 8 B B
Thinking/Inquiry/ Problem Solving
12. She decides that there are 10 such numbers that start with 7 (i. 3 is B). 11. Describe the complement of V W in words and.e. i the subset of such sequences with exactly i 1s. ….
Let A be the subset of arrangements that start with the letter a. 6. n(B). hence. …. What is the maximum number of days in a non-leap year that can fall on the weekend? 13. ….. If 19 of the girls do not take Business 101. c. Let U be the set of two-digit integers. n(A b. Find n(A). 30} are not divisible by 3? (Hint: First look at the subset of U corresponding to integers that are divisible by 3. V is the subset of such integers that contain at least one 5 and W is the subset of integers that contain at least one 6. dcba}. 3. Prove that n(A B) n(A) n(B) for any subsets A and B. how many of the boys do? 9. divisible by 5 b. Explain why this answer is not correct. 7.
Communication
5. of whom 42 are girls. By listing the elements. 97). 27. d. How many integers in the set U {1. Suppose U is the set of binary sequences of length 3 and Ei .
17. C. 1 i. 16. E2. Three subsets of U are A: sequences that start with 1 B: sequences that end in 1 C: sequences that contain exactly two 1s. Use the sum rule for three events to find the number of integers between 1 and 1000 that are divisible by 2 or 3 or 5. (This development is known as the Principle of Inclusion and Exclusion. the sum rule for three events. A.)
10. By listing the sequences. 1111}. B C. A b. j. 1000. find the number of elements in the following subsets.c. … n. Show this result. on a Venn diagram. divisible by neither 5 nor 7 e. Consider the set U of all binary sequences of length 4. n(C) – n(A B) – n(A C) B. …. Verify that n(A B C) n(A) n(B) – n(B C) n(A B C). State a formula for n(E1 E2 … En) in terms of n(Ei). Use De Morgan's Rules (see Questions 14 and 15 in Section 10. …. Show this result on a Venn diagram. Suppose we have n subsets E1. divisible by 5 but not divisible by 7 14. 18. n(Ei Ej). n(Ei Ej Ek ). En. that is.2) to show that n(A B) n(U) n(A B). How many integers between 1 and 1000 are divisible by 7 or 13? Part C 15. l. Suppose A and B are two subsets of a set U. U {0000. a.3 THE SUM RULE
371
. A C. …. A B C
c. divisible by both 5 and 7 d. B.
one of a.Section 10. for example.4 — The Product Rule
Suppose Eric has five shirts and three sweaters. bC. d. The box on the left is the shirt choice and can be filled in five ways. Each way of filling the boxes will correspond to a unique element of U. The ordered pair bC corresponds to shirt b and sweater C. the most useful and powerful counting tool. eB. Once we have chosen the shirt. To answer the question posed. B. We count the elements of U constructively. then together the two tasks can be done in p q ways. eC}
Product Rule If the first of the two tasks can be done in p ways and. eC} represents all the possible shirt-sweater combinations. dA. cB. aB. e
Then. each way corresponding to a different choice of shirt. cC. dB.
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. b. We label the shirts a. e and the sweaters A. for each of these ways. d. Since there are three choices of sweater for each choice of shirt.
First. first the shirt and then the sweater. b. dC. We solve this problem with the product rule. we look at how many ways we can build an element of U by filling in the two boxes shown to the right. bB. fill this box in three ways. for example. U {aA. It will be used in almost every counting problem. How many different shirt-sweater combinations can he make? He doesn't worry about whether the colours match. the second task can be done in q ways. there are then three ways to fill the second box corresponding to the choice of sweater. we want to find n(U) where the universal set U {aA. there are 5 3 15 different combinations. aC. We will always list the shirt first to avoid confusion. eA. C. aC. he does know that sweaters can be worn only over shirts. aB. fill this box in five ways. cA. …. for each of these five ways. bA. B. A. however. c. That is. and C
Note that we build the combination in order. The set U has 15 elements listed below so that you can confirm the logic of the argument. c.
S. we simply use the product rule repeatedly. Now let's choose the second letter of the acronym. choose Y for the second letter. there were only two tasks: choose the shirt and then the sweater.
1
2
3
The subscript on each box indicates which letter in the acronym it represents. The product rule was used twice in the above example. there are more than two tasks.
EXAMPLE 1
Suppose we have three copies of each of the letters of the alphabet and we want to make a three-letter acronym such as IBM (which stands for International Business Machines). there are 26 choices for filling the third box. for example.4 THE PRODUCT RULE
373
. We find n(U) by looking at how we construct the acronyms. The set of all such acronyms is U {AAA. To build an acronym. …. there are 26 26 676 ways to construct the first two letters of the acronym. Finally. we place an X in the first box.In the shirt and sweater example. there are 17 576 three-letter acronyms. including X and Y. can be used. ZZZ}
U has a large number of elements and we will need a good strategy to find n(U). For these applications.
10. we complete the acronym by choosing the third letter. For each possible choice for the first letter. In most applications. AAB. The acronym IBM is different from BMI (an American performing rights organization that represents more than 140 000 U. We can write out a general form if there is a series of tasks to perform in sequence. the first box can be filled with any one of the 26 letters. there are 26 choices for the second box. publishers). Note that we can construct any three-letter acronym by placing one of the 26 letters in each of the boxes.
X
1 26 choices 2 3
X
1
Y
2 3
676 choices
X
1 676
Y
2 26 choices
Z
3
There are 676 26 ways to construct the three-letter acronym. even though both acronyms use the same three letters. songwriters and composers and over 60 000 U. Using the product rule. Suppose for example. For every possible choice for the first two letters. The order of the letters in the acronym is important. so any letter. We might.S. Hence.
abdc. Solution We have U {abcd. U is the set of all such words. then the entire sequence of tasks can be done in p q r …ways. To count the bytes. d with no repetition. Here we have eight boxes to fill. and C {abcd. 00000010. we look at how we can construct them. Each such sequence is called a byte. using 0 or 1 for each term in the sequence. We can make counting problems more interesting by looking at subsets of the universal set. A typical byte is 01001110 and the set of all bytes is U {00000000. for each of these ways. for each of these ways. we can construct a word by filling the four boxes as shown.
Each box can be filled in two ways and the conditions for the product rule apply. b. …}. and n(C). 00000001. …. B {bcda. Then determine n(U). B is the set of all words in U with last term a. c. the third task can be completed in r ways. n(B). and so on. List two elements of each set.
EXAMPLE 2
A computer codes information in a binary sequence of length 8. and C is the set of all words in U with third term b or c. bdca. …}. …} To count the elements of U. corresponding to the eight terms of the sequence. How many different bytes can be formed? Solution It is always a good idea to identify the objects being counted. the second task can be completed in q ways and.Generalized Product Rule If the first of a number of tasks can be done in p ways and. so there are 2 2 2 … 2 28
binary sequences of length 8. acbd.
EXAMPLE 3
A word is formed by arranging the four letters a.
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. 11111111}.
In the product rule. then two ways to select the second. Applying the product rule. we have n(C) 2 3 2 1 12. the first letter can be selected in three ways. there are three ways to select the second letter. There are then 4 3 ways to choose the first two letters. there is one way to select the third letter. using the product rule. since the number of possible letters would vary depending on whether we had already selected b. 2. there are 4 3 2 1 24 words in U. To count the words in B. The next example shows what can go wrong if the condition does not apply. There is only one choice. which is either b or c. the fourth letter can be selected in one way.4 THE PRODUCT RULE
375
. if we had started with the first letter (four choices) and then selected the second (three choices).
EXAMPLE 4
The digits 1. we can again count the ways of filling the four boxes. For any one of these choices. we would have a problem with the third letter. we start with the third letter of the word. 4. for each of these selections there is only one remaining letter and. This time we start with the fourth letter in the word. there are three ways to select the first letter. To find n(C). which must be an a. stated once again. Finally. or both. For each of these. For each of these choices. Hence. we see that we can construct sequences by starting with different terms. then together the two tasks can be done in a b ways.We can select any one of the four letters for the first letter of the word. the third letter can be selected in two ways from the remaining unused letters. for each of these ways. For each of these choices. A very good strategy is to start with the terms that are most restricted. you must always check that this condition is satisfied. In the above example. 5 can be arranged to form five-digit numbers. To count the elements of C. only one way to choose the fourth letter. Hence n(B) 1 3 2 1 6. the phrase and for each of these ways cannot be ignored. For this choice. How many of these numbers are even?
10. There are two choices. the second task can be done in b ways. then the second in two ways. 3. For each of these choices. When using the product rule in its simple or general form. or c. hence. Product Rule If the first task can be done in a ways and.
2. This helps to develop a good notation and to clarify any necessary restrictions. There are five choices for the first term. There are two possibilities. However. if we use 2 or 4 for the first term.
5 choices
The last term must be even. We will discuss these situations in the next section. so we consider it next. Then E {12354. Then the second term can be selected in three ways. List some of the sequences you are counting by defining the universal set and appropriate subsets. 3. 2. 3.
5 choices
1 or 2 choices depending on our choice for Box 1
Rethinking the solution. …}. For each of these.Solution A five-digit number is a sequence of length 5. Suppose we start with the first. three for position 2. Once the first digit is chosen in five ways. we fill the final position first and have two choices. Let E be all those five-digit numbers that are even. You can avoid mistakes by writing a brief explanation of the counting process. Start by constructing the most restricted terms. and one for position 4. 1. Sometimes it is not possible to do this. There are five boxes to fill. there are two choices for the last term. Make sure that the condition for each of these ways is met. and the fourth in one way. There are five choices. 14532. If that term is 1. the third in two ways. it is not true that for each of these ways there are two ways to select a digit for the fifth place. two for position 3. there is only one possibility for the last term. 3. The first term can be any one of 1.
When you are using the product rule to count sequences and arrangements. there are four choices for position 1. This is the wrong answer because our analysis is faulty. or 5. 4. there are 5 3 2 1 2 60 elements in E. or 5.
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. either 2 or 4. Then n(E) 4 3 2 1 2 48. Hence.
The solution to a clue in a crossword puzzle is an anagram (a rearrangement) of the word alerting. O. How many possible seven-digit telephone numbers are there within a given area code such that the first digit is 3. Using the product rule. U.4 THE PRODUCT RULE
377
.Exercise 10. four main courses. Based on the clues already solved. I. b. how many different meals can you order? Explain. 4. 2. F.4
Part A
Knowledge/ Understanding
1. How many arrangements are possible if the two tallest people are at opposite ends of the row? 7. Four different calculus books are arranged on a shelf. or 6? Part B 6. Explain why there are 24 different arrangements possible. one main course. you know that the second letter is i and the last letter is e. and W are not used? 5. as shown. and one dessert. How many possibilities are there?
s q u a c i r c l e r e
Application
10. Q. A computer program is available to look at all the possible arrangements of the remaining letters. write a clear explanation of how to find n(U). A restaurant menu has three appetizers. and three desserts. 5. a.
Communication
3. If you decide to order one appetizer. A four-digit PIN number can be represented as a sequence with four terms. A Canadian postal code has the form XxX xXx where X is an uppercase letter and x is a digit from 0 to 9. Six people arrange themselves in a row for a photograph. If each term can be any digit from 1 to 9. How many postal codes are there if the letters D. list three different elements in the set U of all possible four-digit PIN numbers.
e. Is this correct? Explain. Find n(A B). f can be rearranged to form a number of words or sequences of length 6 (no repeated letters are allowed). S. there are 3 of cat. anchovies. 3 3 27 possible arrangements of the letters
b. green pepper. onions. d. Hence. A three-letter acronym is a sequence of length 3 with terms that are letters of the alphabet. a.Communication
8. extra cheese. Find n(U) and the number of such words that do not begin with a. D. b. Find the size of each of these sets. each with four possible answers A. List two elements in each of U. Q. The second letter can be any one of the three letters. for example.99. Suppose that 30 000 students enter the contest. pineapple. The letters of the word cat can be rearranged to form six different words. How many different answer sheets are possible? 11. The third letter can be any one of the three letters.
Communication
9. c. you get a large pizza with any three toppings chosen from mushrooms. pepperoni. In words. A student uses the product rule to determine that there are 9 8 7 504 different pizzas possible.
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. C. c. A mathematics contest has seven questions. T. Let U be the set of all such sequences. describe the set of acronyms A
f. or T. B. or T. or olives. R. b. sausage. Consider the following subsets. C: acronyms made up of three different letters a. B. d. A. A.
Thinking/Inquiry/ Problem Solving
13. C. A pizza restaurant has a special deal: for only $6. Find n(A e. Find the number of three-letter acronyms that use one letter at least twice. Write a correct argument using the product rule. a student may choose not to answer any particular question. A: acronyms that start with a vowel B: acronyms using only the letters from the set P. 10. C. Let U be the set of all such words. There are ten questions on a true/false test. What is wrong with the following argument? a. The letters a. Is it possible that every answer sheet is different? Explain. Students attempt all questions. C. A. for example. or T. The first letter can be any one of the three letters. B. b. Show how an answer sheet can be represented by a sequence of length 10. C. for example. B). A.
Knowledge/ Understanding
12. As well.
We can always write a positive integer as the product of prime factors. b. and any symbol can be used repeatedly. where a
1. 9999}. How many passwords can be formed? b. 1.4 THE PRODUCT RULE
379
. regardless of position. 15. Every integer divisor of 12 can then be written in the form 2a 3 b where 0 a 2. 12 223. How many of these integers have no repeated digits? d. Use the same method to count the divisors of 144. For example. "repeated" means that the same digit can be used more than once. What fraction of the divisors is even?
10. a. a. What fraction of these divisors is even? Part C
Thinking/Inquiry/ Problem Solving
34
52. These symbols can be either upper. A computer password must have eight symbols.
19. How many passwords can be formed that have at least one 9? 17. Use the product rule to explain why there are 9000 such integers. How many divisors of 144 are odd? 18.14. How many passwords can be formed that start and end with a digit? c. An integer n can be factored as n a. b. b
1. How many of these integers have repeated digits? In these problems and elsewhere. 1001. a. 1002. Show that every divisor is equivalent to a sequence of length 2 where the first term is 0. How many of these integers end in 7 or 8? c. Repeat Question 13 assuming the word formed has only four letters selected from the given six. The integer 64 800 can be factored as 25 a. How many such sequences can be formed? c. c
1. How many divisors does 64 800 have? b. or 2 and the second term is 0 or 1. 16. How many divisors of 12 are there? d. e. ….or lower-case letters or digits from 0 to 9. 0 b 1. How many passwords can be formed with no repeated symbols? d. Consider the set of four-digit integers {1000.
b. How many divisors does n have?
2a3b5 c.
Using a calculator or spreadsheet program. a. In part b. each symbol can be repeated up to r times. Repeat Question 22 to find an expression for the fraction of sequences with two or more birthdays the same. each symbol can be repeated up to three times? 21.20. 2
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. b. we can create a sequence of length 5 to represent their five birthdays. The first term gives the first person's birthday. each symbol can be used at most once? b. The days of the year (not including leap years) can be labelled 1 to 365. Suppose there are n people in a room. a. If we have five people in a room. 22. and so on. What percent (to two decimals) of the sequences have five distinct birthdays? c. Suppose we have m symbols available. find the smallest value of n so that the fraction in part a exceeds 1 . How many sequences of length 3 can be formed if a. How many sequences are possible? b. Repeat Question 20 assuming the sequence is of length r. What percent (to two decimals) of the sequences have two or more birthdays the same?
t chnology
e
23.
we divide A into disjoint subsets and count the number of elements in each. In the list of numbers 1.5 THE USE OF CASES
381
. 2. inclusive. we can count the number of digits easily. we have discovered several useful tools such as the sum and product rules for solving counting problems. Case single digit numbers 1. we divide a problem into smaller (and hopefully easier) sub-problems and solve each of these separately. …. …. Note that the two cases are completely separate. In this section. 11. In the language of sets. 2.Section 10. how many years started with 1? Solution Let A be the subset of years that start with 1. to find the number of elements in a set A. how many digits are there in total? We consider two cases. 99. The Venn diagram is shown at the right. 99 Total Number of digits 90 9 2 189 180
For each case. there is no overlap. 9 two digit numbers 10.
EXAMPLE 1
From year 1 to year 2000.5 — The Use of Cases
To this point. …. consider the following question. For a simple example. A1: all years with a single digit starting with a 1 A2: all years with two digits starting with a 1 A3: all years with three digits starting with a 1 A4: all years with four digits starting with a 1 Every pair of these subsets is disjoint and A A1 A2 A3 A4. the challenge is to identify the disjoint subsets where we can easily count the number of elements.
U A 1 A2 A3 A4
10. we look at the use of cases. We partition A into four mutually disjoint subsets of years. In any problem. another simple but handy strategy. 3. 3. This is simply a special case of the sum rule applied to two or more disjoint sets. When we use cases. Then we combine the results to get the answer to our original problem.
we have n(A) n(A1) n(A2) n(A3) n(A4).. i. and n(A4) 1000 (1000 years starting with 1.e. we count the number of years in each of the simpler cases. years 10 to 19). to see that the number of sequences containing 4 is the same as the number of sequences containing 2 and so on. For example. it is easy to find n(A) here or. Again filling boxes (in each case place the 4 first in one way). Let A be the subset of U in which each sequence contains a 4. Then n(E1 E2 … Ek) n(E1) n(E2 ) … n(Ek).
EXAMPLE 2
A sequence of length 3 is formed from the digits 1.. and A3 the subset in which 4 is the third term.e. n(A3) 100 (100 years starting with 1. exactly 1 must contain 4. Extended Sum Rule The subsets E1. E2.e. E1 E2 … Ek is the subset that contains all elements in any one of E1. Now A1. …. and n(A3) 8 7 1. 124. 3
n (A)
EXAMPLE 3
User identifications for a local e-mail system must be formed from upper-case letters and can be five to eight letters long. A2. n(A2) 8 1 7.e. …} be the set of all sequences of length 3 formed from the 9 digits. 3. i. How many such identifications are possible?
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. alternatively. Filling three boxes and using the product rule.
168 1 . Hence. i. The fraction of sequences containing 1 is n (U) 504 3 There are other solutions. inclusive. …. n(A2) 10 (10 years starting with 1. 2.. years 1000 to 1999). Since each sequence contains three out of nine digits. 9 with no repetition allowed. We have n(A1) 1 (1 year starting with 1. Ek. What fraction of these sequences contains the digit 4? Solution Let U {123. starting with the digit 1. The formal extension to the sum rule is shown below. years 100 to 199). E2. i. year 1).Using a simple extension of the sum rule.. Next. n(A) n(A1) n(A2) n(A3) n(A4) 1111. and A3 are pairwise disjoint and A A1 A2 A3. There are 1111 years between 1 and 2000. Since the subsets are disjoint. n(A) n(A1) n(A2) n(A3) 168. especially after you see the answer. we have n(A1) 1 8 7. Ek of a universal set U are pairwise disjoint if every pair of subsets is disjoint. A2 be the subset in which 4 is the second term. …. we see that U has 9 8 7 504 elements. In this case. Let A1 be the subset of A in which the 4 is the first term.
7. n(A) 49 56 49 0 154. n(A4) 0. Clearly. A2 be the subset ending in 50. respectively.5 THE USE OF CASES
383
. n(B7) 267. 1. …}. there are then seven choices for the second digit. …}. By defining appropriate disjoint cases based on the number of days in a month. 2. n(A1) 7 7 1 1 49.
Exercise 10. A3 be the subset ending in 75. Then n(A) n(A1) n(A2) n(A3) n(A4). B7 {AAAAAA. For A2 there are eight choices for the first digit and seven choices for the second digit. AAAAAAAB.2 1011. That is. we have n(U) 265 266 267 268 2. since we cannot use 0. For A1 there are seven choices for the first digit. B8 {AAAAAAAA. Filling five boxes with 26 symbols. To find n(U). Let A1 be the subset of A ending in 25. AAAAAB. 2. B8 to contain the identifications of length 5.…. We have B5 {AAAAA. so n(A2) 8 7 1 1 56. B6. …}.5
Part A
Knowledge/ Understanding
1. Sometimes a problem that seems like a straight forward application of the product rule might be counted using cases because there are two restrictions. n(B6) 266. 6. since repetition of a digit is not allowed. look at four cases in which the length of the sequence is fixed. Similarly. and A4 be the subset ending in 00.Solution Let U be the set of possible identifications. each of which affects the other. Filling eight boxes with 26 symbols. AAAAAB.
EXAMPLE 4
A four-digit number is formed using the digits 0. or 5. 9 without repetition. n(B5) 26 26 26 26 26 265 B6 {AAAAAA. Now using the extended sum rule.
10. Hence. …}. How many such numbers are divisible by 25? Solution Let A be the set of four-digit numbers divisible by 25. Then. count the number of days in the year with a date of either 30 or 31. since we cannot use the one chosen for the first digit but can now use 0. define B5. AAAAB. and 8. n(A3) 7 7 1 1 49. Filling six boxes with 26 symbols. n(B8) 268. B7. These subsets are pairwise disjoint and U B5 B6 B7 B8. Filling seven boxes with 26 symbols.
…. o. How many sequences contain an o?
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. then using three letters (without repetition) from the chosen word. 1000}? 10. C. How many such words can be created? b. What is the total number of binary sequences with length no greater than 5?
Knowledge/ Understanding
4. How many sequences start with a vowel? d. How many binary sequences of length 6.
Thinking/Inquiry/ Problem Solving
7. This means that the number of rows times the number of columns must be 10. and y. mouse. for example. The number of 5s in an integer is the highest power of 5 that divides evenly into the integer. i.
Thinking/Inquiry/ Problem Solving
9. and 2. How many sequences are there? b. 2. Repeat Question 5 assuming repetition of the symbols is allowed. four. How many of these words end with a? c. B. a. and 25 are 1. 3. How many such sequences start with A or end with F? 11. Six letters. How many ways can the numbers be stored in the spreadsheet? 8. and d can be used to make words of length 1 to 4 if no repetition is allowed. the number of 5s in 10. How many words with three. b. A. 7. A sequence of length 3 is formed by first selecting one of the three words cat.Application
2. or five letters can be formed if each word must contain at least one of a. The numbers 1 to 10 are stored in a spreadsheet in a rectangular array. How many of these words contain the letter a? (Hint: Look at the complement. How many positive integers between 1 and 2000 inclusively have distinct digits? What fraction of these integers is odd? 12. and F. are arranged into a sequence of length 6. 12. e. E. or 8 begin with 0? Part B 5. c.) 6. a. The symbols a. How many sequences end in the letter s? c. D. What is the total number of 5s in the set of integers {1. respectively. 0. How many such passwords can be constructed that contain exactly one letter and no repeated digits? 3. or goldfish. A special password is a sequence of length 6 that uses the digits 0 to 9 and the letters a to z (lower case only). so. u.
9. How many of the sequences contain at least one A if a. Find n(B). Prove that the number of binary sequences with length less than k is 2 less than the number of binary sequences of length k. Find n(A). A is the subset of U corresponding to sequences with unique digits. 9 without repetition.5 THE USE OF CASES
385
. repetition of symbols is allowed? b. 1.13. Let A be the subset of U in which the terms of the sequence form an arithmetic progression. U is the set of all sequences of length 3 that can be formed using the digits 1. 16. no repetition of symbols is allowed? 15. B is the subset of U corresponding to sequences that contain at least one 0. Find n(U). Part C 14. b. 2. Find n(A). 3.
10. Consider the set U of all such sequences with length at most 10. 2. A sequence of length 3 is formed from r symbols that include the letter A. c. a. …. …. A sequence has terms selected from the digits 0.
Sometimes this is obvious (e. n(C). and n(B C). We hope that it will be easier to determine the size of the complement and the universal set and then apply the rule of the complement. you are learning to communicate in a way that can be applied to all sorts of problems. When you study mathematics. not just specified mathematical ones.
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. and use the sum rule to combine the results. Solving counting problems is a useful way to practise these skills. you might label the people A. Another good strategy for difficult problems is to use cases. To find n(A) we can use the sum rule if we can first find n(B). F and denote a committee by a set such as {A. then you should consider counting the objects by filling in boxes and using the product rule. If the problem as posed seems difficult.. remember to write complete solutions. In other problems. determine the size of each subset as a smaller and possibly easier problem.Key Concepts Review
This chapter introduced you to some tools and strategies for solving counting problems. Always start with the terms that are most restricted. It is not enough to just get the correct answer. we can sometimes write the set A in the form A B C. As a last resort. Once you are given a counting problem. For example. It is important that you be able to explain your solution to others. you will have to invent a notation to carry out this task. divide A into disjoint subsets. if the problem is to count the number of committees of three that can be formed from six people. with B and C not disjoint. C. write down a few of the objects you are counting. If the objects you are counting are sequences or arrangements. B. Note that you need to define a universal set in order to specify the complement. E. B. D. C}.g. all the arrangements of the letters of the word dog). When you work through the Review Exercise. how should you proceed? First. That is. then specify a set A so that the counting problem is to find n(A) and consider the complement of the set.
00 Lotto 6/49 ticket purchased in the year 2000. You do not need to match all six numbers to win some money playing Lotto 6/49.
1st Prize: Match all 6 draw numbers 2nd Prize: Match any 5 draw number and the bonus number 749 $93 300.75 3rd Prize: Match any 5 draw numbers 4th Prize: Match any 4 draw numbers 5th Prize: Match any 3 draw numbers
Number of winners: Average amount won:
114 $2 532 204. four.99
31 324 $1 798. Use the results from question 1 and the average amounts won to determine the expected loss of a $1. Investigate and Apply Lotto 6/49 draws involve randomly selecting six numbers and one bonus number from the numbers 1 through 49. how many different tickets could win each of the prizes? 2. The expected loss of a Lotto 6/49 ticket is lessened by the possibility of winning on five. In the year 2000. After the six numbers and bonus number have been drawn.80
N/A $10
1.wrap-up
C H A P T E R 1 0 : L O T T E R I E S A N D E X P E C T E D VA L U E S
investigate and apply
People who buy tickets for lotteries offering multi-million-dollar prizes are attracted to the possibility of changing their lives. or even three numbers.53
1 640 264 $65. Has the expected loss of a Lotto 6/49 ticket changed since 2000? Lotteries have been called a regressive form of taxation. many people need the occasional rewards associated with smaller prize amounts. Psychological theories suggest that in order to stay attracted to lotteries. Approximately how many people do you think won fifth prize in the year 2000?
INDEPENDENT STUDY
Research recent Lotto 6/49 payouts. 3. Investigate this statement. Lotto 6/49 paid out prizes as follows. ●
RICH LEARNING LINK WRAP-UP
387
.
describe the subsets G d. We have looked at four basic tools for solving counting problems. F and G F. . a. third. Find n(U).. b. Let U be the set of all such sequences and let
388 C H A P T E R 1 0
. Q. O. B. W are not used) and the second. Let U be the set of all postal codes. c. 2.. Calculate n(A B). b. c. and sixth terms are digits from 1 to 9. and n(F). Canadian postal codes are a sequence of length 6. In words. n(A). F. The letters of the word goldfinch can be rearranged into a large number of sequences of length 9.
4. and fifth terms are upper-case letters (D. A. Find n(U). cases 2. fourth. A sequence of length 3 can be formed from the digits 0. List two elements in each of U. Write a brief description of each tool. I.Review Exercise
1. the rule of the complement b. n(C). Are G and F disjoint? Explain. U. the product rule d. Let U be the set of all such sequences and let G: all sequences in U that start with gold F: all sequences in U that end with finch a. A: all codes in U starting with N B: all codes in U ending with 8 C: all codes in U that use the letter A D: all codes in U that use the letter N exactly once a. n(G). C and D. Consider the following subsets. 9 without repetition. Find n(G F) and n(G F). 1. The first. n(B).
3.. and n(D). the sum rule c. Use an example to demonstrate its use.
A PIN number for a photocopier can be any sequence of three or four digits. How many such PIN numbers are there? b. (If you are wondering why we have included so many binary string problems.) 6. it is because they are of fundamental interest in computer science. Local telephone numbers (within an area code) are a sequence of seven digits. which does not start or end with 1. A. he had originally listed some examples of binary sequences. How many integers between 1 and 1000 include the digit 7? 8. similarly. Being well trained. a. A student is trying to count the fraction of binary sequences of length 6 that start or end with 1. It would be friendly to give him a correct solution too. How many such numbers contain at least one 2?
REVIEW EXERCISE
389
. n(A). How many such numbers start with 2? c. b. all the sequences start or end with 1. and C. n(B). He then concludes that 2 1 the fraction of sequences that start or end with 1 must be 1 1. List two sequences in the intersection of A and B and then find n(A B). Help your poor fellow counter and show him the error in his thinking. exactly 1 of the sequences start with 1 2 and. He cannot remember the product rule but cleverly deduces that since the first term is either 0 or 1. B. In other 2 2 words. and n(C) c. exactly 1 of the sequences end with 1.A: all such sequences starting with an even digit B: all such sequences ending with an even digit C: all such sequences containing an even digit a. The first digit in an area code is selected from 2 to 9. Find n(U). How many usable numbers are there? What fraction of the numbers end in 99? 7. List two sequences in each of U. one of which was 011110. so that he knew his answer was wrong.
5.
F. The set U is all six-letter words that can be formed by rearranging the letters of euclid. …. {5. …. b. 50} {5 1. 50} that are exactly divisible by 2 or 5. Explain the error in the student's argument. A student is asked to count the number of integers in the set U {1. describe the complement of the subset A b. The set of binary sequences of length 5 is U {00000..3. B. He then concludes that there are 25 10 35 integers in U divisible by 2 or 10.Chapter 10 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application Questions all 8. 2 25} has 25 elements all divisible by 2. 4.
4. 9 2. 2. D. 4. C. How many ways can the three letters selected from A.995}.…. 3. Find n(A B). 5 2. …. 00001.50} {2 1. 6 all
1. a. E. 5. What is n(A)? 2. 5 10} has 10 elements all divisible by 10.….10. He first notes that the subset A {2.2. What is the product rule?
3. …. describe the subset A b. and the subset B {5. 11111}. G (with no repeats allowed) be arranged in a row if A or B must come first? Explain your reasoning. In words.999} is the set of positive integers less than 1000. Let A be the subset of such words that end id and B the subset of such words that end ic.
a.10.…. a. How many of these sequences start and end with the same digit? 6.
390 C H A P T E R 1 0
. U
{1. 2 2.…. In words. B and A B. Provide a correct solution to the question.15.
B. The lines on the given diagram represent connections between five locations labeled A.7. How many have at least one a? 8. D. How many have unique letters? c. 2. How many of these numbers are both even and greater than 5000? 10. and E. How many different ways can you go from A to E if you cannot pass through the same location twice?
B A
C
E
D
9. …. A four-digit number is formed by selecting digits from the set {1. C. a. 9} with no repetition allowed. How many such passwords are there? b. A password for a small computer system is a sequence of any four letters from the alphabet with repeated letters allowed. 3. How many integers between 1 and 1000 do not contain a digit 7?
CHAPTER 10 TEST
391
.
all these sequences are equally likely to occur. is incredibly small.. More importantly. and so on. We can represent the set of all possible birthdays by a sequence of length 20 with terms selected from the set of digits {1. It is also true that more babies are born in some periods of the year than others. Suppose that there are 20 students in your class.
That is. This is a very different question. To date. even a person who bought a ticket at every opportunity.05. With a class of 20 students. with 20 students in a class. Hence. Among your classmates. The chance of finding a few children with the same rare condition at a particular site is very small. Based on our assumptions. However. As far as we know. The chance that any particular person wins twice. we sometimes notice clusters where a number of children living near a power station or some high voltage lines all develop the same unusual disease. since they hardly ever reach the age of 25 in any case. However.. the sequence 332. The so-called birthday problem is a good example. Probability (or chance) is a tool that allows us to see if what appears to be unusual is really so. if you look at all the opportunities for clusters across the large number of such sites then the probability of one such cluster occurring just by chance is much larger. 20) 36520
two or more have the same birthday is more than 40 percent. the chance that someone wins twice is much. the answer is just over 0. the probability that two or more people have their birthday on the same day is 1
P(365. Is this a coincidence or does the observed pattern indicate that something about the power station or high voltage lines is causing the disease? One way to examine such events is to use the mathematical theory of probability. it can make the decision to investigate further much more difficult. there have been about 1800 runs of the 6/49.. given the large number of people who play every time.. The chance of finding a match of some two birthdays is surprisingly large. the chance of finding a match to your birthday is small. 346 such sequences. 23. For example. 365}. To simplify the calculations. 243 indicates that the first student's birthday is on day 332. 2. This is just like the birthday problem. With a little thought. we omit babies born on February 29 in leap years. this has not happened yet! We can use similar arguments to show that a cluster of cases of an unusual disease is not as rare by chance as you might expect. What appeared unusual is in fact quite likely to happen. . 124. There are P(365. We are interested in knowing about the subset of sequences in which two (or more) of the birthdays are the same. so winning is very unlikely. If about 2 million people bought one ticket on all 1800 plays. Suppose we assume that every student in a class is equally likely to be born on any one of the 365 days in the year. you should be able to show that for a class of 20 students. We consider the complementary subset of sequences in which every term is different. the second's is on day 23.
0.. You might ask how likely it is that someone in the class has the same birthday as you. there are 36520 possible sequences. The fact that we can explain the occurrence of a cluster of cases by chance does not rule out the possibility that there is a connection between the power plant and the disease.. but we ignore this point to keep matters simple. the chance of picking the 6 correct balls is 1 or approximately 49 6 1 in 14 million. so it is in fact quite unlikely for you to find someone who matches your birthday. the chance that
392 C H A P T E R 1 0
.. . then the chance of a double winner is about 3 percent.COINCIDENCE AND CHANCE
Life is full of strange coincidences — two people in your class have the same birthday or a newspaper reports that the same lucky person has won a lottery twice. much larger.. We can perform the same type of calculations for lottery winners. 20) 365 364 .411. However. For the 6/49 lottery.
Section 11. 11. 11.1. Section 11. how many vehicles can now be accommodated for licensing? Problems such as this one. you will
• • • • • •
solve problems using counting principles. Section 11. 11.5 evaluate expressions involving factorial notation. and computer passwords require careful counting techniques. 11. you will expand your combinatorics knowledge to derive general formulas that can be applied to a wide variety of problems. Section 11. In this chapter.Chapter 11
COUNTING METHODS
How many cars.2. 11.5 solve problems by combining a variety of problem-solving strategies.5 explain solutions to counting problems. Section 11.4. 11.3. 11.3. and trailers are there in Ontario? If each license plate is of the form XXXnnn. you learned how to develop rules and strategies for solving certain counting problems.5 solve problems involving permutations and combinations.4. In Chapter 10.1. Section 11. trucks.
CHAPTER EXPECTATIONS In this chapter. 11. as well as the design of social insurance numbers. do we have enough plates available? Since the new plates are in the XXXXnnn format. postal codes. health card numbers.2.5 express the answers to permutation and combination problems. where X is any one of the 26 letters of the alphabet and n is any digit from 0 to 9.5
.
In this chapter.12. One problem with this system is that governments may be formed by parties that do not have the support of the majority of Canadians. There are 359 eligible voters in seven riding voting for one of three political parties. Do you think mathematical methods can determine a fairer system for allocating seats? ●
394 C H A P T E R 1 1
. Eligible voters choose from a slate of candidates in their riding. which party gets which particular seats? Regional concerns may raise the importance of this question. in the 2000 Federal election. Typical methods would award four seats to Blue and three seats to Green. what criteria should be used for selecting the different ways of allocating them? 2.6% of the votes. Proportional representation would award seats in the ratio 177:160:22. the Liberals won 40% of the popular vote but gained 57% of the seats in Parliament.43. When the number of seats earned by each party is determined. For seven seats. Investigate Consider the simplified.3% of the votes.45. 0. Make a list of all the possible ways of assigning four seats to Blue and three seats to Green. election results tabulated below. Which of these 35 is the best way to allocate the seats? We will also see just how big the problem of seat allocation can become. There are many voting systems used throughout the world. Dealing with these decimals is one of the issues generated by proportional representation. each with its own strengths and weaknesses. One of these is a system of proportional representation. Blue would win 6 out of 7 seats with 49. albeit extreme. we will see that there are 35 different ways to allocate the seats. For example.1% of the votes. 3. Green would win no seats even though it had about 44. (Should Red get the seat in riding G?) The next question is.investigate
CHAPTER 11: VOTING SYSTEMS
In Canada. Provincial governments may be elected similarly. The political party with the most seats forms the government.
DISCUSSION QUESTIONS
1.
Ridings Parties Blue Green Red A 29 28 1 B 24 22 2 Blue C 25 23 2 D 18 17 1 E 39 38 3 Blue F 34 31 2 Blue G 8 1 11 Red 177 160 22 Total Votes
Winner Blue
Blue Blue
Using the current system in Canada. the unrounded allocation is 3. governments are elected by the people. The candidate with the most votes in a riding wins a seat in the House of Commons. and Red would win one seat despite earning only 6.
We started with n possibilities. Following the pattern. d. the first term can be chosen in n ways. By the rth term. Consider this question: if there are n different symbols available. gfed} We can count the number of sequences by using the product rule. there are 7 6 5 4 840 elements in U. For each of these ways. so we have left n (r 1) n r 1 possibilities. the third in n 2 ways. we looked at each problem and example from first principles. The hard part here is to see how many ways there are for selecting the final term. the second term can be selected in n 1 ways. In general. the second term can be selected in six ways. To get the answer. we review how to count such sequences. Note that r n. No matter which letter we choose. how many sequences of length r can be formed if no symbol can be used more than once? In Example 1. Suppose we have n symbols and we want to count the sequences of length r if we can use each symbol at most once.
11. if r 2. Hence. If r n. In this chapter.1 — Counting Sequences With Distinct Elements
In Chapter 10. The third can then be selected in five ways and the fourth in four ways. one for each term in the sequence.
EXAMPLE 1
How many sequences of length 4 can be formed using the seven letters a. then the answer is 0 because in building the sequence we run out of symbols before we finish the construction. …. f. abce. Look at it another way. Since we can use a letter only once. There are as many factors in the product as there are terms in the sequence. we derive general formulas that can be applied to a wide variety of problems. then the last (second) term can be selected in n 1 n 2 1 ways. we find the product of four factors. there are six letters left. then the last (fourth) term can be selected in n 3 n 4 1 ways. which is the rth term. the total number of sequences is n (n 1) … (n r 1). In doing so. b. Hence. Some typical elements in U are U {abcd. we developed rules and strategies for solving certain counting problems. For example. If r 4. g if no symbol can be used more than once? Solution Let U be the set of all such sequences.Section 11. We use exactly the same approach for the general problem. The first term can be selected in seven ways.1 COUNTING SEQUENCES WITH DISTINCT ELEMENTS
395
. the rth term can be selected in n r 1 ways. e. we have used r 1 of these. c. and so on.
First. In this instance. Hit ENTER (produces the value of n!) Calculating P(n. e. r). it may use the symbol nPr. Enter n 2. In this case. Your calculator may also have a function for P(n. The number of sequences of length r that can be formed using n different symbols. consider.We denote this quantity by P(n.333 10157. Go to MATH menu 3. r) refers to permutation. r) is the product of all of the integers from 1 to n. n! increases rapidly and soon is larger than the display of a calculator. Enter 2 (gives the P(n. r) gives the correct answer of 0. which is the product of exactly r factors. r) n (n 1) … (n r 1)
A sequence formed from symbols is sometimes called a permutation of the symbols. r) in terms of the factorial notation.g. The formula for P(n. Enter n 2.321 1081. r))
t chnology
e
Another special case occurs for r n. To see how this works. one of the factors in the product is 0. The righthand side is the product of the first four factors in 9! We can multiply top and bottom by the remaining five factors without changing the value of P(9.
396 C H A P T E R 1 1
. where each symbol can be used at most once. the first factor in P(n. so the formula for P(n. Enter 4 (gives the factorial function) 5. We call this product n factorial and use the notation n! n (n 1) … 3 2 1. Go to MATH menu 3.
Calculating n! on a TI-83 1. r) is n and the last is 1. in other words. The formula for P(n. your calculator gives 10! 3 628 800 and 60! 8. Scroll across to PRB sub-menu 4. r). Hit ENTER (produces the value of P(n. so this function is useful only for small values of n if exact values are desired. Here we are counting sequences of length n that can be built from n different symbols using each symbol once and only once. r) has some special cases. look what happens if r n. 4) 9 8 7 6. The P in P(n. Many calculators have a function to calculate n! For example. for example P(9. is P(n. Scroll across to PRB sub-menu 4.. It is sometimes convenient to express P(n. Enter r 6. r) on a TI-83 1. 4). 100! 9. As n gets larger. Some calculators will go higher. r) operator) 5. the length of the sequence is greater than the number of symbols.
3) n(n 1)(n 2) is much easier to work with than P(n. We now use the formula for P(n.
EXAMPLE 2
A five-digit integer is formed from the digits 1. If the integer begins with 3. Hence. In this instance. P(n. There are P(9. How many integers can be formed? b. P(n. P(9. For example. r) to solve some problems. 4) . Other strategies and tools. 9. are also still important. many of the formulas we develop will work in special cases if we define 0! = 1. such as the sum rule. 3) (n n! 3)! . Using a calculator. We can define 0! = 1 so that the formula in 0! the box gives the correct answer. How many of these integers contain the digit 6? Solution a. this calculation can be done directly. 2) 100 99 9900. which is correct but not very useful.. 5) 9! 15 120 integers. The number of integers is the number of sequences of length 5 formed from 9 symbols using each at most once. We know that the number of sequences of length n that can be formed using n symbols is P(n. r) n n
n! (n r)!
(n (n
1) 1)
(n (n
2) 2)
… …
(n
(n
r
r
1)
1)
(n (n r) r) (n (n r r 1) 1) … … 1 1
Be careful not to abuse this result.1 COUNTING SEQUENCES WITH DISTINCT ELEMENTS
t chnology
e
397
. The original definition can also be useful in algebraic situations. 3. with no digit used more than once. The fraction of integers beginning with 3 is P(8. or 1 . n) = n! . the formula tells us that P(100. . n) = n!. Using the formula given in the box above. 4) 1680 such integers. One final point about this notation. 2) 100! . 1 80 There are P(8.P(9. 5) 9
11. What fraction of these integers begin with 3? c. 116120 of the integers begin 9 5 with 3. a. 2. P(n. 4)
9
9 9! 5!
8
8
7
7 5 6 4
6
5 3 4 2 3 1 2 1
In general terms.1111. This definition is one of convenience. it is much 98! better to go back to the original form P(100. . the first position is fixed and the remaining four positions form a sequence of length 4 chosen from the remaining eight digits.. 4! b.
r). How many such words are there if a. o. 5). then the 6 can be in any of five positions. As there are P(n. r) P(n 1. r) P(n 1. In this case. the first letter is a vowel.… are available. Equating the results gives a relationship among the formulas we have developed. r) sequences in total. with the remaining four digits forming a sequence of length 4. There are two ways to count the number of sequences of length r that contain a particular symbol. the number of sequences containing a is P(n. Equating the two results gives the required result P(n.
Note that we can always use m!
EXAMPLE 3
A ten-letter word is formed from the 26 letters of the alphabet. Proof 2 Now we have an algebraic proof. r 1). we have P(n. 4) 8400. we consider the complement—those sequences that do not contain a. We determine the number of integers in two different ways. r)
n! (n n(n (n r)! 1)! r)! (n (n 1 (n (n 1)! r)!
multiply the second fraction by
n n r r
to make the denominators the same
r)(n 1)! r)(n r 1)! 1)! r)!
since n (n r 1) n 1 (r r 1) 1
[(n r [(n rP(n
(n
(n 1)
r)] (n (n
1)! (r 1)]!
1. can be constructed in P(n 1. Starting from the left side. The number of integers is 5 P(8. there are no restrictions? b. the a can be placed in any one of r places and then the rest of the sequence. (a. i. or u)? c. there are r P (n 1. r 1). using the product rule.c. b. r 1) ways. the number of five-digit integers that do not contain 6 is P(8. r) P(n 1. 5) 8400. Alternatively. r 1) sequences that contain a. an arrangement of n 1 symbols in r 1 positions. Proof 1 Suppose n different symbols a. r) P(n 1. First. so the number that contains 6 is P(9. either the first or the last letter is a vowel?
398 C H A P T E R 1 1
. r) r P(n 1. Alternately. No letter may be used more than once. say a. e. Since we can now use only n 1 symbols to build the sequences. If the integer contains a 6. r m
1) (m – 1)! when it is convenient. r) r P(n 1. Hence. there are P(n 1. we can use the same method to establish the general relationship P(n. Part c of Example 2 demonstrates an important principle. r) sequences that do not contain a. 5) – P(8.
The remaining nine terms can be formed in P(25. the number of words is the number of sequences of length 10 formed from 26 symbols using each at most once. b. Typical examples of words in E F are {abcdefghio. Each of the following questions involves forming sequences using the letters of the name Euclid at most once.820 1012.1 COUNTING SEQUENCES WITH DISTINCT ELEMENTS
399
. 2).Solution a. Using a 16! spreadsheet program. 31! 28! 1 is an integer. do not calculate.
Exercise 11. so we have n(E F) 5 4 P(24. 6) b. 9) 5 4 P(24.9275 1013 words. To calculate n(E F). where n 7! c. the number of sequences of length 4
11. a. This question is more difficult. r) for the appropriate choices of n and r. 9) 5 P(25. P(8. Simplify P(n. Using a calculator. 10) 26! 1. 8) ways. If there are no restrictions. d. then the first term can be chosen in five ways. we can evaluate P(26. the number of sequences of length 6 b. Evaluate the following expressions. 2) h. 10) as exactly 19 275 223 968 000. Hence. P(8. From part b. 8). the number of words that start or end with a vowel. 2)/P(n Part B
3. n(F) n(E). 8! f. 7! e. 9). If the first letter is a vowel. there are five choices (one of the vowels) for the first letter and then four choices for the tenth letter. We want to find n(E F). Express the answer in terms of P(n. Using the sum rule. By symmetry. The middle eight letters can then be arranged in P(24. … }. we have n(E F) 5 P(25. c. a. 9) words starting with a vowel.1
Part A
Knowledge/ Understanding
1. abcdefghij. The remaining nine terms are a sequence of length 9 formed from 25 symbols with no repeats. n(E F) 6. there are P(26. E and F are not disjoint. we know that n(E) 5 P(25. 8! 7! g. This is the number of words that begin or end with a vowel. 11! 9! 1. 8). 9) ways. Explain in each case whether or not using a calculator would be effective. Let E be the set of words with the first letter a vowel and F the set with the last letter a vowel. Hence there are 5 P (25. 79! 76!
2.
c. the number of sequences of length 5 that end with d d. the number of sequences of length 6 that start with a vowel e. the number of sequences of length 4 that start and end with a vowel
Communication Knowledge/ Understanding
4. Which is larger, P(10, 5) or P(10, 6)? Explain. 5. A six-digit integer is formed by selecting digits from the set {1, 2, …, 9} without replacement. What fraction of the possible numbers a. start with 6? b. are even? c. start with an odd digit? d. start or end with an odd digit? e. contain the digit 9? f. contain both digits 8 and 9? g. contain the digit 8 or 9 or both? h. are less than 460 000? (Hint: Consider cases.) 6. Twenty-four students arrange themselves in three rows of eight for a class picture. How many different arrangements are possible if the eight tallest are in the back row?
Application
7. A sorting algorithm puts lists of numbers into increasing order. To check the algorithm, a programmer prepares all possible lists of the integers 1, 2, …, 100. Of these lists, how many begin with an integer less than 10 and end with an integer greater than 90? Leave your answer in terms of P(n, r). 8. If the digits 0, 1, 2, …, 9 are formed into a sequence of length 6 with each digit used at most once, how many of the sequences start with 0 or end with 9? 9. Twelve different books—six mathematics text books and six novels—are arranged on a shelf. How many arrangements are there in each of the following? Leave your answers in terms of n! for appropriate n. a. the six novels are to the left of the six mathematics books b. the novels and the mathematics books alternate c. a novel is on the left end of the shelf d. a novel is on the left end of the shelf and a math book is on the right end e. a novel is on the left end of the shelf or a math book is on the right end 10. Sequences of length 4 are formed from n symbols, each used at most once. Using the P(n, r) notation, count the number of sequences in the following sets.
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U: all possible sequences A: all sequences that start with a particular symbol α B: all sequences that have two particular symbols α and β side by side in the order αβ C: all sequences that have two particular symbols α and β side by side in either order 11. A sequence of length r is formed from n symbols using each symbol at most once. Using the P(n, r) notation, repeat the calculations in question 10.
Thinking/Inquiry/ Problem Solving
n, r 12. Show that P(P(n, r) 1) is an integer for r an integer, 1
r
n.
13. A set of 40 cards consists of cards numbered 1, 2, 3,...,10 in red, yellow, green, and blue. Suppose the set is shuffled and the first five cards are set down in a sequence. Find the number of elements in the following sets. U: all possible sequences A: all sequences with cards all the same colour C: all sequences with two or more cards having the same number B: all sequences that start with 2 and end with 8 14. The numbers 1, 2, …, 144 are arranged in a square 12 12 array. What fraction of all possible arrangements have the perfect squares in increasing order on the main diagonal? 15. Ten children's blocks have the letters A, B, ..., J on one face. How many of the different arrangements of the blocks do not have the blocks with A and B adjacent?
Thinking/Inquiry/ Problem Solving
16. What is the largest power of 10 that divides 20!? Part C 17. By counting in two different ways the number of sequences of length r that can be formed from n symbols, prove that P(n, r) n P(n 1, r 1). Hint: Consider building the sequence by selecting the first element and then all the others. 18. Using algebraic manipulation, prove that P(n, r) 19. Find two proofs that P(n, r) 1 k r 1. P(n, k) P(n k, r n P(n 1, r 1).
k) for any integer
20. Find the largest value of k so that 10k divides evenly into 100!.
11.1 COUNTING SEQUENCES WITH DISTINCT ELEMENTS
401
Section 11.2 — Sequences With Unlimited Repeated Values
In this section, we look at the following question: If there are n symbols available, how many sequences of length r can be formed, if each symbol can be used as often as we like? If we change the question slightly so that each symbol can be used only once then, from the previous section, we know that the answer is P(n, r) n (n 1) … (n r 1). To see what difference allowing repeats makes, reconsider the example from Section 11.1.
EXAMPLE 1
Let T be the set of sequences of length 4 formed from the seven letters {a,b,c,d,e,f,g}, where we can use a letter as often as we like. Find n(T). Solution We can count the elements of T using the product rule. The first term can be chosen in seven ways, the second term can be selected in seven ways, the third in seven ways and the fourth in seven ways. Then n(T) 7 7 7 7 74 2401. There are four factors in the answer, one for each term in the sequence. For the general question, we proceed in the same way as in the example. To build a sequence of length r, the first term can be chosen in n ways. For each of these ways, the second term can be selected in n ways, the third in n ways, and so on until the rth term, which can also be selected in n ways. Hence, there are n n … n nr sequences of length r that can be formed using n symbols as often as we like. If there are n symbols available, the number of sequences of length r that can be formed if each symbol can be used as often as we like is n n … n nr.
EXAMPLE 2
The set U of binary sequences of length r has elements such as 01001…0, a string of r 0s and 1s. In computing language, a binary sequence is called a bit string, so U is the set of bit strings of length r. Let A be the subset of these strings that has at least one 0 bit. Let B be the subset of strings that start with 0 and end with 1. Find n(U), n(A), and n(B). Solution Here we have two symbols with unlimited repeats in a string of length r, so n 2 and n(U) 2r. To find n(A), consider the complement, the subset of strings with
402 C H A P T E R 1 1
no 0s. This set has exactly one element 111…1, so n(A) n(U) n(A) 2r 1 A typical element of B is 0100…01. The first and last terms of these elements are fixed by the definition of B. The r 2 middle terms of the elements of B are all the bit strings of length r 2. Hence, n(B) 2r 2.
EXAMPLE 3
A standard die has its faces numbered 1, 2, 3, 4, 5, 6. Suppose that six standard dice, coloured red, yellow, blue, white, green, and orange, respectively, are rolled simultaneously. In what fraction of the possible outcomes will six different values occur? Solution Each die comes up with an outcome 1 to 6. To relate the question to sequences, we can let the first term of the sequence be the outcome of the red die, the second term the outcome of the yellow die, and so on. Thus the possible outcomes of rolling six dice can be described by a sequence of length 6 made using the digits 1 to 6. For example, the sequence 222333 corresponds to the red, yellow, and blue dice coming up 2 and the remaining three dice coming up 3. The total number of possible outcomes is the number of such sequences, which is 66. One example in which all six different values occur is the sequence 654321. The outcomes in which all six values occur correspond to sequences of length 6 in which each digit from 1 to 6 is used once. From the previous section we know there are 6! such 6! 5 sequences. The fraction of sequences with distinct outcomes is 66 . 324
EXAMPLE 4
We can indicate the day of the year by a number between 1 and 365 (ignoring leap year). Suppose there are n people in a room. Of all the possible arrangements of birthdays, what fraction has two or more people with their birthday on the same day? For what values of n does this fraction exceed 0.5? Solution We can proceed by finding what fraction of the possible cases have all the people in the room with birthdays on different days. This corresponds to a sequence of length n, using 365 possible symbols in which the terms are all different. There are P(365, n) such sequences. The number of possible cases is found by counting the same sequences with repetition allowed. There are P(365, n) such sequences. The fraction of sequences corresponding to all people having their birthdays on different days is t(n)
P(365, n) 365n 365 364 … (365 365n n 1)
The fractions we just calculated can be interpreted as probabilities. For example, if you are in a group of 23 people (perhaps one of your classes) and everyone in the group is equally likely to have a birthday on any day of the year (no twins), the probability that there are two people in the group who have their birthday on the same day is about 1 . If there are more people, the probability is greater, since 2 t(n) is decreasing. In a group of 60, the probability is greater than 0.99.
Exercise 11.2
Part B
Application
1. In a plan for North American telephone numbers, each number is a sequence of 10 numbers of the form xyy xyy yyyy, where 2 x 9 and 0 y 9. How many different telephone numbers can be formed? 2. A sequence of length 7 is formed from the digits 0, 1, …, 9. Each digit can be used as often as you like. What fraction of these sequences a. begins with 1? b. begins and ends with 1? c. uses only even digits? d. begins and ends with an even digit? e. does not contain a 0?
Knowledge/ Understanding
3. A sequence of length 12 is formed using n different symbols including A. Each symbol can be used repeatedly. How many of these sequences a. begin with A? b. begin with AA? c. include at least one A? 4. A sequence of length 10 is formed using the letters a, b, c with unlimited repetition. How many of these sequences use only two symbols?
404 C H A P T E R 1 1
Knowledge/ Understanding
5. Let U be the set of bit strings (binary sequences) of length r of these strings a. begin with 1? b. begin and end with 1? c. begin or end with 1?
2. How many
6. How many binary sequences of length 12 start with 1 or end with 0? 7. Eight plain and eight blue tiles are available to cover the rectangular table top that is shown. How many different patterns can be made if the spaces labelled x must have the same type of tile, the spaces labelled y must have the same type of tile, and the tiles on x and y must be different?
Application
x y
x y
8. In a series of licence plates, the first three symbols are any of the 26 letters in the alphabet and the last three are any of the 10 digits from 0 to 9. a. How many license plates can be formed? b. How many plates can be formed with all symbols different? c. How many plates can be formed in which at least one symbol is repeated? d. How many plates can be formed in which at least one of the digits and at least one of the letters are repeated? 9. In a programming language, variable names are sequences of length 1 or 2 that use lower-case letters, upper-case letters, or digits. The name must start with a letter. The second symbol, if used, can be any letter or digit. How many different variable names can be constructed? 10. My password is a seven-symbol sequence formed from upper- and lower-case letters and the digits from 0 to 9. Yesterday, I forgot the last three symbols in my password. It takes me 20 seconds to try to log onto my computer. Approximately how many hours will it take for me to check all the possible passwords?
Thinking/Inquiry/ Problem Solving
11. A voice-mail system has 1253 users. A password to open a mailbox is a sequence of length r formed from the digits 0 to 9. To ensure confidentiality, there should be at least 1000 possible passwords for every user. What is the smallest possible value of r? 12. In a simple lottery there are 100 tickets numbered 1 to 100. You hold ticket number 1. Three tickets are drawn one after the other. After each draw, the ticket is replaced. In how many ways can the tickets be selected so that you win at least one prize?
1 1 . 2 S E Q U E N C E S W I T H U N L I M I T E D R E P E AT E D VA L U E S
405
13. A coin is flipped eight times to produce a sequence of heads and tails. What fraction of these sequences have the same result on the first and last flip? 14. Four switches each have three positions, up, middle, and down. How many different ways can the switches be arranged?
Thinking/Inquiry/ Problem Solving
15. A sequence of bits such as 101001 represents a six-digit binary number, one that is expressed in terms of powers of 2. For example, 101001 1 25 0 24 1 23 0 22 0 21 1 20. We can convert the binary number to a decimal by expanding the powers of 2 in decimal form so that 1 25 0 24 1 23 0 22 0 21 1 20 32 8 1 41. Conversely, we can represent any decimal number uniquely as a sequence of bits by expanding in powers of 2. For example, 27 1 1 16 24 1 1 8 23 0 0 4 22 1 1 2 21 1 1 1 20
a. What is the largest decimal number that can be represented by a sequence of length 6? b. What decimal numbers can be represented by bit sequences of length 6 that end with 1? c. What decimal numbers can be represented by bit sequences of length 6 that start and end with 1? 16. Suppose we want to represent all decimal numbers less than or equal to 1000 as bit sequences of length r, as in Question 15. How large must r be? 17. A subset is formed from the integers 1, 2, 3, …, 9. a. How many possible subsets can be formed? b. How many of these subsets contain 1? c. How many of these subsets contain 1 or 2?
Application
18. Six balls are drawn consecutively from a barrel that has 49 numbered balls labelled 1, 2, …, 49. After each draw, the ball is replaced. What percentage of the possible sequences have six different ball numbers? 19. A sequence of length r is formed using n symbols with unlimited repetition. What fraction of these sequences have all terms different? Part C 20. Suppose n symbols are available to construct a sequence of length r 2. One of the symbols is A. Show that the fraction of all sequences that contain exactly one A is greater if each symbol can be used at most once than if there is unlimited repetition.
406 C H A P T E R 1 1
21. A sequence is a palindrome if it looks the same read from either end. For example, the word level is a palindrome. Suppose that a sequence of length r is built from n distinct symbols with unlimited repetition. How many of these sequences are palindromes? 22. Suppose n 2 symbols are available to construct a sequence of length less than or equal to r where each symbol can be used an unlimited number of times. Show that the total number of sequences with length less than r is smaller than the number of sequences of length r. 23. A function from a set S to a set T assigns exactly one element of T to each element of S. For example, if S {a, b, c, d} and T {0, 1}, then one function f from S to T is f(a) 0, f(b) 1, f(c) 1, f(d) 0. In general, how many different functions can we construct from a set S to a set T? 24. In attempting to count the number of sets of any size that can be formed from the six letters A, B, C, D, E, F, we can define a sequence of length 6 using the symbols I (in) and O (out) to indicate whether a specific letter is included. Then IOIOOI corresponds to the set {A, C, F} and OOOOOO corresponds to the empty set . a. Is there a one-to-one correspondence between the possible sets and the sequences of length 6? b. Determine the number of sets that can be formed from the six letters.
Section 11.3 — Counting Subsets
Many problems involve counting subsets. For example, the results of the 6/49 lottery depend only on the balls drawn, not their order. That is, the results depend only on which subset of six balls is chosen. We can use the methods that we have developed for counting sequences to count the number of subsets. In this section, we look at systematic ways to count subsets. Remember that the fundamental difference between a subset and a sequence is that the order of the terms in a sequence is important. For example, suppose a set of six letters {a, b, c, d, e, f } is available. Two different sequences of length 3 are abc and bca. Note they are made up of the same three letters. However, the subset {b, c, a} is the same as the subset {a, b, c}. Two distinct subsets are {a, b, c} and {b, c, d}. Two distinct subsets must have at least one element that is different.
EXAMPLE 1
How many subsets of size 3 can be selected from the six letters {a, b, c, d, e, f}? Solution Let x represent the unknown number of subsets. Our strategy is to use two different methods to count the number of sequences of length 3 that can be formed using the six letters at most once. First, we count the number of sequences directly, so that the answer is P(6, 3). Alternatively, we can select a particular subset. Since there are x subsets, we can do this in x ways. Suppose we choose {a, b, c}. The three elements of the subset can be used to generate 3! sequences of length 3. In this case, the sequences are abc, acb, bac, bca, cab, cba. If we choose another subset, say {a, b, d}, then these three letters can be used to generate another set of 3! sequences that are all different from those listed above because they include a term d. In other words, each of the x subsets can be used to generate 3! different sequences. Using the product rule, there are a total of x 3! sequences that can be generated in this way. But the terms of every sequence of length 3 make up exactly one subset of size 3. Hence, we have counted all of the sequences of length 3. It follows that x 3! P(6, 3) or, solving, x
P(6, 3) 3! 6 5 4 3!
20 There are 20 subsets of size 3 that can be selected from {a, b, c, d, e, f}. The key to this result is that we have counted the number of possible sequences in two ways and noted that the two answers must be equal. The same strategy works in general. Suppose we have a set of n elements and we want the number of
408 C H A P T E R 1 1
subsets of size r. Again, let x represent this unknown number. As in Example 1, we use two different methods to count the number of sequences of length r that can be formed from the n elements, using each at most once. Using the direct method, there are P(n, r) such sequences. If we have x possible subsets of size r, then each of these will generate r! different sequences and so there are x r! sequences in total. Equating the two counts gives x r! P(n, r) and solving for the unknown number of subsets, we have x
P(n, r) r! n(n 1)…(n r! r 1)
A common but rather curious notation for the number of subsets of size r selected n . We read this expression as n choose r. We can also use the from n elements is r alternative form of P(n, r) in the numerator. Recall that P(n, r) (n n! r)! , so we n P(n, r) n! can write n choose r as . r! (n r)!r! r The number of possible subsets of size r that can be selected from a set of n different elements is n n (n 1) … (n r 1) r! r n n! or (n r)!r! r n can also be written C(n, r) or nCr. Here the letter C stands for r combinations, just as P used earlier stands for permutations. Your calculator probn . Note also from the final expression in the box that ably has a function for r The expression n r n n r
n , with the best choice depending r on the relative sizes of n and r. We can use the calculator function directly, or evaluate the first expression in the box by calculator or by hand. n n Alternatively, we can first replace by and then proceed. r n r There are thus several choices for evaluating
Note that the number of factors in the numerator and denominator is the same in the fundamental definition. This is a useful check to avoid silly mistakes. c. Using the last method, 100 100 100 99 98 97 4 3 2 1 96 4
3 921 225
Note also that if the numerical answer is not required, it is acceptable to n leave the answer in the form . To complete the notation, note that there is one r n n! . subset of size n that can be selected from n elements. That is, 1 0!n! n Since we have defined 0! 1, the formula makes sense. It will also be useful to n 1. define 0 EXAMPLE 3 Eight people—six students and two teachers—are available to serve on a committee. How many different committees of size 4 can be formed if a. b. c. d. there are no restrictions? Bob, one of the students, must be on the committee? the committee has exactly one teacher? at least one teacher must be on the committee?
410 C H A P T E R 1 1
This answer is wrong because we know that there are only 70 committees altogether and some of them have no teachers at all. There are 4 ber of committees with at least one teacher is 70 15 55. there are 40 such committees. we had to select a teacher in the first step. start by choosing 2 ways. where a. This problem is more difficult. Using the product rule. c. To form such a committee. Suppose we had used the following argument. when we separated the selection into two steps. An alternative approach is to use the complement. one teacher from the two available in 1 6 we can then choose the three students in ways. Thus there are 70 such comremaining seven people in 3 1 3 mittees. There are other people from the seven available in 3! 3 35 committees that include Bob. Since we know there is at least one teacher on the committee.Solution a. Be careful! It is very easy to make a mistake. For each of these selections. c}. we can select the 2 ways. A committee is a subset of size 4 selected from the eight available people. What went wrong? The error occurred because. Case 1: The committee contains exactly one teacher. a. There are then 40 15 55 committees that 2 2 contain at least one teacher. d. the number of committees in this 2 6 case is 15. You should number of committees with exactly one teacher is 1 3 verify this calculation. Case 2: The committee contains exactly two teachers.3 COUNTING SUBSETS
411
. Here we build the possible committees in two steps. There are 8 8! 70 such committees. The committees that do not contain at least one teacher are made up 6 15 such committees. One approach is to consider two disjoint cases. so the numentirely of students. 3) 35 ways. From part c. and we could select the second teacher in the second step. Every committee containing both teachers has been counted
11. b. b. 4!4! 4 b. A typical committee that includes Bob is {Bob. since we know that there are 70 committees in total. Constructing the possible committees in two steps as above. Now pick the other three committee members from the the teacher in 1 7 2 7 ways. First. c represent three people other than Bob. the 3 2 6 40. we select the three 7 P(7.
… Remember that it may be easier to count the complement. E. Whenever you see a problem like this—one containing the key words at least or at most— divide the problem into cases so that each case corresponds to exactly . do not contain either E or F? Part B
Thinking/Inquiry/ Problem Solving
5.
600 3
3. B. 3 10 2. F. A subset of size 3 is formed by selecting three letters from the set {A. How many subsets can be selected? b.3
Part A
Knowledge/ Understanding
1. 2. and once with the selection reversed.twice: once with teacher A selected separately and teacher B as one of the three other members. …. so our answer was too large by 15. D. contain exactly one vowel? c. How many of the subsets contain two even and three odd numbers? d. G}. How many of the subsets contain the number 10? f. contain the letter A? b. Suppose a sequence of length 4 is formed by choosing four digits from the set {0. a. How many of the subsets contain 9 or 10? 4. 6 a. Evaluate the following. 10}. What fraction of the possible subsets a.
Exercise 11. 1. a. How many of the subsets contain at least two even numbers? e. How many such sequences can be formed if no term in the sequence can be repeated?
412 C H A P T E R 1 1
.
60 3
c. There are 15 committees containing both teachers. How many of these subsets contain only numbers less than or equal to 7? c. Evaluate 3 11 4
b. 9}. …. C. A subset of five numbers is chosen from the set {1.
Y4. A set of 12 distinct wooden blocks has three that are red. B5. the committee has three women and two men? c. When are the two quanr
9. Ron and Enzo refuse to serve on the same committee? 11. Explain without any calculation why P(n. The blocks are labeled R1. How many of these subsets contain only even-numbered balls? d. Y1. and four that are yellow. n(A). How many subsets have two blocks of different colours? 10. c. R3. A: all subsets of two red blocks B: all subsets with one red and one yellow block C: all subsets with two blocks the same colour a. how many different sequences of length 4 can be formed? 10 P(10. express your answers in terms of r of n and r. where the letter matches the colour. R2.3 COUNTING SUBSETS
413
. ….. By counting the sequences in two ways. For the n for various choices following questions.b. n(C). C. …. 2. …. explain why 4! 4 6. In a lottery. 49. a. five that are blue. How many possible subsets of six balls are there? b. Consider the following subsets of U. the set of all possible subsets of two blocks. How many of these subsets contain ball 49? c. b. the committee must contain at least one man and one woman? d. B. How many of these subsets contain three even-numbered and three oddnumbered balls?
11. r) tities equal? n . How many committees of size 5 can be selected from 11 people—five men and six women—if a. Find n(U). 4) . How many ways can two girls and two boys be selected from a class of 12 girls and 10 boys?
Knowledge/ Understanding Communication
7. n(B). six balls are selected from 49 balls numbered 1. For any subset of size 4 selected from the above set of digits. B1. …. In how many ways can a subset of six letters be selected from {A. there are no restrictions? b. Z} so that both A and Z are included? 8.
what fraction of the possible samples contain at least one defective item? 14. 3). …. b. How many of the samples have no defects? c. 2. A box of 100 electronic components contains three that are defective. How many of the samples have exactly two defects? e. She reasons that she can select the two red blocks in ways 2 4 ways. a. R2. Part C 16. 2. If a sample of five components is tested. List all subsets that have at least two red blocks and count them directly. …. C. R3. a. How many of the samples have exactly one defect? d. 100) has three that are defective (labels 1.Application
12. A subset of three blocks is selected from a population of six blocks. If there are 300 students in each grade from 9 to 12. where the first three blocks are red. of which three are red. A sample of five bulbs is selected. How many different samples can be selected? b. A carton of 100 light bulbs (labelled 1. B. To learn how students feel about a proposed dress code. A student is trying to count the number of subsets with at least 3 two red blocks. Explain why the student got the wrong answer. How many samples can be selected that have 60 grade 12 students?
Application
13. the stuand then the remaining block in 1 3 4 12 subsets with at least two red dent calculates that there are 2 1 blocks. how many samples can be selected that have 15 students in each grade? c. a. A. How many sequences of length 4 can be constructed using the digits {1. the principal decides to survey a sample of 60 students from the school population of 1200 students. How many different samples are possible? b. 2. How many of the samples have exactly three defects? f. Using the product rule. Label the blocks R1. 9} if two of the terms are even and two are odd?
Communication
15. Explain how the results of parts a to e show that 100 97 3 97 3 97 3 97 3 5 5 0 4 1 3 2 2 3
414 C H A P T E R 1 1
.
How many of these solutions have x 3? y? y z 11.17. a. a. How many of these solutions have x c. How many different solutions are possible? 19. Show that any solution to x y z 11 with x same as a corresponding solution to x1 y1 z1 y1 0. z 0. y w 0 is the 0. Eleven stars are arranged in a row as shown. Consider the equation x y z 11 where x.
a. w are integers with x –2. We can create a group by placing two vertical bars in the spaces between the stars. y. z1 0. y z w 21?
20. How many ways can the three groups of stars be formed? b. w are positive integers. y. Each group must have at least one star. How many of these diagonals pass through the centre? 18. a. How many solutions are there to the equation x b. For example. a middle group. and the right group has four stars.
11. and a right group. *********** The object of the game is to divide the row into three groups—a left group. b. A diagonal of a regular n-gon is a line joining two vertices and lying inside the figure. Suppose that x.
21 where x. the middle has five stars. y –1. Be sure to do Question 18 before you attempt this one. * *|*****|**** the left group has two stars. in the diagram below. z 14 with x1 0. y. Suppose you want to count the solutions to x x. y 0. z.3 COUNTING SUBSETS
415
. Find the number of solutions to x y z
11 with x
21. How many diagonals are there? b. 0. z are positive integers. where we allow 0. z to be non-negative integers. z 0. y. Find the number of solutions to the equation x y z z. w 1.
we count sequences in which some of the elements are the same. the answers are P(7. the same specified the positions for the 0s. For example.
Let's look at the example 0001110100 in more detail. we looked at counting two types of sequences. there 4 10 210 bit strings of length 10 with six 0s and four 1s.Section 11. hence. In this section. 0001110100 is a bit string of length 10 with six 0s and four 1s. In doing this example we showed that each bit string of length 10 with
. We can create many new questions by varying how often the letters can be used in the sequence.4 — Counting Sequences With Repeated Elements
In earlier sections. The 10 order of selection does not matter. the sequence is completely determined. our answer would then be 6 answer. Hence building the sequence corresponds to selecting the four boxes from ten that will contain 1. Consider filling the following ten boxes. c. In Section 11. We could have are 4 10 .
0
0
0
1
1
1
0
1
0
0
416 C H A P T E R 1 1
Note that the four 1s are found in positions {4.
EXAMPLE 1
A bit string is a sequence in which each term is 0 or 1. b. g if i) no letter may be repeated? ii) a letter can be repeated as often as we like? Using the product rule. How many bit strings of length 10 can we make using six 0s and four 1s? Solution Since we are counting sequences. 6. 8}. This time we use the methods that we have developed for counting subsets to count sequences with repeated terms. You might wonder what happens if some of the letters can be used two or three times. f.3. e. we counted the number of possible subsets by counting sequences in two different ways. d. 4) and 74. respectively. a good strategy is to look at how many ways we can build the sequence. 5. Once we specify the positions for the 1s. We considered questions such as this one: How many four-letter words can be formed from the seven letters a. There are such selections and.
4 we can then select the three positions for the bs from the remaining five positions 5 ways. we used only two symbols to form the sequence. For each of these ways. 2. where n1 n2 … nk n is n n1 n n2 n1 … nk nk
n! n1! n2! … nk!
1 1 . 4 C O U N T I N G S E Q U E N C E S W I T H R E P E AT E D E L E M E N T S
417
. we look at a more general question. Hence. we can select the two positions for the cs from the remainin 3 2 9 5 2 ways. ….
9 ways. there is a one-to-one correspondence between the set of bit strings of length 10 having four 0s and the set of subsets of size 4 selected from {1. 10}. We can select the four positions for the as in We can rewrite this answer in a more memorable form 9 5 2 9! 5! 2! 5!4! 2!3! 0!2! 4 3 2
9! 4!3!2!
The two examples can be generalized as follows. n2 of the second type. 2. three bs. It then follows that these two sets have the same number of elements. The number of binary sequences of length n with r 1s and (n n n! r (n – r)!r! r) 0s is
The number of sequences of length n that can be formed using k symbols with n1 of the first type. each of these subsets corresponds to exactly one bit string. . and two cs? Solution We count these sequences by looking at how many ways we can construct them.four 0s corresponds to one subset of size 4 selected from the set of boxes {1. Formally. and nk of the kth type. Conversely. Finally.. 10}. In Example 1.
EXAMPLE 2
How many sequences of length 9 can be formed using four as. …. there are 1260 sequences ing two positions in 2 4 3 2 of length 9 that can be formed... In Example 2.
there are two es. There are n(P) 2!2!2!2!!2!2! such 2 sequences. What is the probability of getting two 1s. etc. two 2s. 2. b. …}. To count the sequences in which 1!1!1!1! the two es appear side by side. Thus. c? Solution Let the set of all such sequences be U {aabb. is P {112233445566. then there are 4! 12 different 2! sequences. Each element of P is a sequence of 1 length 12 with two 1s. two 2s. The probability that the required sequence will occur is n(P) 12!/(2!)6 0. aabc. …. where each term represents the outcome on a die of a particular colour. b. Note that we can leave out 2! all the 1! factors for convenience. and so on. Now we look at some applications. and one c. two ss. 6}. For example. a. and so on. 4. two ss and six other distinct ones including E.
EXAMPLE 3
How many sequences of length 9 can be made from the letters of the name Descartes? Of these sequences. Thus we have n(U) 612. The subset of sequences P with two 1s. 3. and with elements selected from {1.Each of these formulas can be developed using the methods shown in Examples 1 and 2. The set of all outcomes U is the set of sequences of length 12 using any of the numbers from 1 to 6 as often as desired. 121233445566. c. Now there are eight symbols to arrange. and two 6s? Solution Any possible outcome can be represented by a sequence of length 12. we put the two es together as a single symbol E ee.0034. …}. 112233445566 corresponds to the first two dice coming up 1. 5. and five other distinct letters. one b. the third and fourth dice coming up 2. Thus. how many have the two es side by side? Solution In the nine letters. Consider these two cases:
418 C H A P T E R 1 1
. For example. We know how to count the sequences once we know which letters are to be used.
EXAMPLE 4
Suppose 12 different coloured dice are rolled. the number of such sequences is 8! . the number of sequences is 2!2!1!9! 90 720. if there are two as. two 2s. n(U) 612
EXAMPLE 5
How many sequences of length 4 can be made using the letters a.
b. Hence in Case 1. Combining the two results. How many such sequences can be made? 2.4
Part A
Knowledge/ Understanding
1. 2.. we have 36 18 54 sequences in U.
a. there are 3 ways to select the letter which appears twice. two of each type (e. Hence. 1. How many of these anagrams start with the letter s? c. 4 C O U N T I N G S E Q U E N C E S W I T H R E P E AT E D E L E M E N T S
419
. 3 For Case 1. 2. For each of these choices. aabc) Case 2: Two different letters. Such an arrangement is called an anagram.g. How many of the anagrams have the two is side by side? e. two of one type and two others distinct (e. How many sequences can be formed? Part B
Knowledge/ Understanding
3.Case 1: Three different letters. For 1 each of these choices there are 12 different sequences as shown above. the letters can be arranged in 24! ! 6 !2 ways. aabb) These cases correspond to disjoint subsets A and B with the property that A B U so that n(U) n(A) n(B). 3. 3 3 ways to select the two letters. The letters of the name Mississauga are rearranged. there are 2 appears twice. A binary sequence of length 5 is formed using 0s and 1s by filling the five boxes shown below. How many of the anagrams start and end with s? d. each of which For Case 2. A sequence of length 5 is formed from the digits 1.
Exercise 11. Case 2 contains 18 sequences. List three such sequences. How many of the anagrams start with a vowel?
1 1 .g. How many possible anagrams can be formed? b. there are 36 different sequences. a..
the four treatments appear once in a random order. What fraction of sequences of eight flips of a coin. Given the four groups. how many configurations have exactly four switches turned on? c. Each of seven switches has two positions—off (O) and on (I). How many of the subsets contain only digits less than 7? c. including the empty set? b. Count all of the arrangements of the letters of the word Descartes that end with s. Of these. a. not necessarily in the same order. 9. what fraction have exactly three 1s in the first five terms? 11. group 1 is the four heaviest members of the group. 0 and 1. Of all such strings with exactly six 1s. How many subsets can be created. Suppose we want to create subsets of the ten digits {0. How many of the configurations have exactly four switches turned on including switch 1? 10. Alternately. How many arrangements can be made if there are two bulbs of six different colours available? 5. group 2 is the next set by weight. a. b. How many of the subsets contain 0 or 9?
Application
12. Each treatment is used four times. D. how many different ways can the treatments be assigned? (This is called a randomized block design. How many different ways can the treatments be ordered? This is called a completely randomized design. 8. For example. A bit string of length 10 is formed using two symbols.)
420 C H A P T E R 1 1
.4. and so on. How many of the configurations have at least two switches turned on? d. results in exactly four Hs? 6. 9}. a. C. How many binary sequences of length 5 have two or more 1s? 7. How many arrangements are possible if each row must have the same composition of trees. 1. there are four treatments labelled A. Within each group. 2. each giving H or T. the subjects can be divided into groups of four people on a specific characteristic. B. …. How many different ways can the switches be configured? b. In a statistically designed experiment. A string of lights has 12 sockets. which are applied to 16 subjects in a random order. and two spruce—are planted in two parallel rows of five trees. four cedars. so that every subject receives exactly one treatment. Ten trees—four pines.
) Part C 21. Explain why 0 1 10 d. r 10?
16. a. two 4s.
10 17. b. Verify this result by direct calculation. two 2s. In a binary sequence of length 10 made from 0s and 1s. Without evaluating the expression. What fraction of all the possible outcomes have a. a. How many are there for the word Mississippi? 14. c. How many bit strings of length 8 have at least one pair of consecutive 1s? 20. c. b. c. Explain what type of sequence is represented by 2!3!! . Consider arrangements of the symbols a. a. 3)?
1 1 . explain why 20!50! 15!15! tive integer. Be sure to explain your reasoning. 5!
must be a posib. c. Generalize the result in Question 10 to sequences of length n. 0) and moves at each step one unit to the right or one unit upwards. b. a. how many sequences have exactly r 1s where 0 10 10 10 … 210.Thinking/Inquiry/ Problem Solving
13. How many of the anagrams of the word Mississauga are palindromes? b. Consider a bit string of length 8 having five 0s and three 1s. How many of these strings have at least two consecutive 1s? 19. How many random walks end at the point (5. at least one repeated value? b. 4 C O U N T I N G S E Q U E N C E S W I T H R E P E AT E D E L E M E N T S
421
. a. Six different dice are rolled. and two 6s? c. Is (n
(3n)! 1)!n!(n 1)!
always a positive integer for n
1?
18. What fraction of these sequences contain a pair of consecutive cs? (Hint: Start building the sequence by arranging the as and bs.)
Communication
15. a. A random walk on the xy-plane starts at the point (0. how many sequences are possible? b. c. three odd values and three even values? (Hint: Count how many ways such a sequence can be constructed.
after one step. at least one A comes before the first B? c. Show that each possible path can be represented as a binary sequence. c. 0).22. the possible positions are (1. it moves one unit to the right and either one unit up or one unit down. At each step. a. 2). What are the possible end positions? b. at least one C occurs before the first A. 4) and (12. Consider the situation after 12 steps. A random walk starts at the point (0. How many arrangements of six As. four Bs. 0). 1) or (1. how many pass through the point (10. 0) or (2. 0)?
422 C H A P T E R 1 1
. A point on the plane starts at (0. How many of the paths in part b pass through both the points (8. 0) and at each step moves to the right one unit or upwards one unit. 10)? d. b. That is. all the As come before the first B? b. and three Cs can be formed if a. How many paths end at the point (12. using E to represent a unit move to the right and V to represent a unit move vertically. and at least one A occurs before the first B? 24. 12)? c. 1) and after two steps (2. 2) or (2. Show an equivalence between random paths and binary sequences. a. How many paths are possible that end at the point (20. 8)? 23. Of these paths. Note that there are two different paths to reach (2.
n1 alike of the first type. Counting problems have several peculiarities: • They all involve words.
1 1 . r) n(n 1)…(n r 1) (n n! r)! if each symbol can be used at most once b. nr if each symbol can be used as often as we like 5.Section 11. The number of ways of arranging n distinct symbols in a sequence of length is a.5 — A Strategy for Counting Problems
You now have the tools and strategies to solve a wide variety of counting problems. 2. then we can perform the two tasks together in m n ways. • They are easy to generalize. we can turn a relatively straightforward problem into one that is much more difficult. often in several different directions. The number of ways to arrange n symbols. Without a good strategy. (n r)!r! r! r 6. • With a small change. n(A). We can exploit this property in reverse by looking at particular simplifications of a problem to make sure that we understand what is being asked and also to get an idea of how to carry out the counting. is n !n n! ! . 3. where n1 n2 … nk n. …. The product rule: If we can perform a first task in m ways and a second task in n ways. which means that we have to translate the problem into a mathematical notation before we can get started.
The Tools 1. The number of ways to select a subset with r elements from a set of n elements is n n! P(n. In this section. This becomes n(A B) n(A) n(B) if A and B are disjoint. r) . We cannot use this rule
4. P(n. we review the tools and give you some ideas on how to approach a counting problem when you try to solve it. This means that it is difficult to judge whether a problem is hard or easy when you first read it. The sum rule: If A and B are two subsets of a set U. n2 alike of the second. you can be completely fooled. !…n
1 2 k
The Strategy You have done enough of the exercises and problems in Chapters 10 and 11 to realize that there is no single approach that we can use to solve every counting problem. then n(A B) n(A) n(B) n(A B). 5 A S T R AT E G Y F O R C O U N T I N G P R O B L E M S
423
. nk alike of the kth type. The rule of the complement: n(A) n(U) without defining the universal set U.
we look at two examples. • Decide if the objects you are dealing with are sequences or subsets.
424 C H A P T E R 1 1
. you can compare special cases. • Write down some specific examples. Step 1: Understand the problem • Invent a notation to describe the objects that you want to count.e. Then try to find a formal counting approach that agrees with your direct count. go back to Step 2. Not every step applies to every problem.To demonstrate the strategy. No letter may be used more than once. Different approaches can lead to different expressions. if it involves ns and rs). • Watch out for double counting or incomplete counting (especially with cases). • Use a constructive approach—the number of objects is equal to the number of ways that we can build them. • Write an explanation that will be clear to someone else. try a simpler version with specific small values and list (and then directly count) all possible objects. • If your approach does not seem to be working. • If the problem is general. • If it is not obvious how to count the number of ways of constructing the objects directly. consider cases or the complement. check the answer by looking at particular small cases where the count can be done directly. If these expressions lead to numeric values. it is easy for you to see whether your answer agrees with that of a classmate.. • Start the construction by satisfying the restrictions first. How many of the words contain the letter z?
EXAMPLE 2
Suppose we have n symbols to arrange in a row. Step 3: Implement your approach • Use the tools we have developed. Does the order of the elements matter? Step 2: Decide on an approach • If the problem is general (i. For answers involving symbols. How many arrangements can we make that start or end with the special symbol? These two problems have been chosen because it is difficult to write down their answers without some careful thought. We attack such problems using the following steps.
EXAMPLE 1
The letters of the alphabet are used to form a word of length k. Don't rely on formulae alone. but the steps give a useful guideline. We are given that r of the symbols are identical (here called special) and that the remaining n r symbols are all distinct.
Now we can choose the symbols A. C. ABCDAA. B. D where A is designated to be special. k). We want to count the number of ways that we can select the three letters. we can arrange them to form 3! 6 different words. we notice that once we have the three letters selected. 5 A S T R AT E G Y F O R C O U N T I N G P R O B L E M S
425
. No letter may be used more than once. all of which have distinct letters and include the letter z. k 1 Another approach to Example 1 is to use the rule of the complement to count the total number of k letter words. k). and the number of such words that do not include z. for k 3. we obtain 2 25 3! . As the examples show. We do not want to choose so small a problem that we lose the essential difficulties. We want to count how many arrangements we can make using these six symbols with A at the beginning or end.
EXAMPLE 1
The letters of the alphabet are used to form a word of length k. P(26. r 3. we can show that the two answers are algebraically the same. k). that appears quite different from the first answer. say n 6. How many arrangements can we make that start or end with the special symbol? Solution Step 1: Let's start with a specific problem. Looking at the examples. Hence there are 25 k! words of length k that contain z. BCADAA. say k 3. We want to count three-letter words such as zab or abz. Some typical examples are AABCAD. we can arrange the k letters into k! words. With some effort. we can choose the other two 25 letters in ways.Now consider using this strategy for the examples. A. order matters here—we want to count zab and abz as different words. Since z must be included. using the product rule. The third example raises an important aspect of the problem — are we counting arrangements that can start and end with the special symbol? The answer depends
1 1 . These examples are difficult problems chosen to show the power of the strategy. Since z is included. This approach gives an answer. How many of the words contain the letter z? Solution We start with a special case.
EXAMPLE 2
Suppose we have n symbols to arrange in a row. P(26. For each of k 1 these selections. We apply the same method to the general case. to make sure that we understand the problem. We are given that r of the symbols are identical (here called special) and that the remaining n r symbols are all distinct. k) P(25. A. Hence. P(25. we 2 25 can select the other k 1 letters for the word in ways.
we might consider two cases: Case 1: arrangements that start with A Case 2: arrangements that end with A One problem is that the cases are not disjoint. so n(U) 3!16! !1! 120. That is. D). the first position can be filled in n r ways. n(T). A. The universal set U contains all sequences of length 6 made from A." Does it mean to include those arrangements that both begin and end with A or not? There is no correct answer here—the language of the question is too vague. D. A. Step 2: Based on Step 1. we can now attack the general problem. the number of arrangements that begin or end with the special symbol is
426 C H A P T E R 1 1
. Let's agree that we will include arrangements that begin and end with A. Let's try this approach.on how we interpret the word "or. The middle n 2 positions can then be filled with an arrangement of n 2 symbols. of which r are the same in (n r! 2)! ways. three As and something different. and the last box in two ways (using only B. To count the number of arrangements that neither start r! nor end with the special symbol. if we let S be all the arrangements in Case 1 and T all the arrangements in Case 2. Using the product rule. the number of arrangements that begin or end with A is 120 24 96. The total number of arrangements is then 3 2 34! ! 24. !1 Using the rule of the complement. and the last in n r 1 ways. B. then we want to find n(S T). Some arrangements in Case 1 are also in case 2. then n(U) n! . Step 3: We will start with the particular case n 6. using the rule of the complement. !1
To count the number of arrangements that start and end with a symbol other than A. C. which means that we would have to find n(S). To use this approach. the set of all arrangements that do not begin or end with A. r 3 to make sure that our approach is feasible. we also need to count the universal set of all arrangements of the symbols without restriction. This is not necessary but is often helpful for building up courage to attack the general situation. If U is the set of all arrangements of n symbols with r that are alike and all the rest are neither. Finally. In set notation. we can fill the first box in three ways. Since our approach works well. C. There are 34! ! different arrangements for the !1 middle four boxes. The middle four boxes are filled with an arrangement of four letters. there are (n r) (n r 1)
(n 2)! r!
possible
arrangements. An alternate approach is to consider the complement. and n(S T).
5
Part B 1. One final worry is whether we have covered all the possible situations. 2. n. We have solved two difficult problems with tools and strategies. 2. 2. The following problems are chosen to give you practice solving counting problems. 3. A subset of k integers is selected from the set 1. This is the correct answer since we place the special symbol in two ways at the beginning or end and then the other symbols in (n – 1)! ways. 2. How many of these words contain the letter z if we allow letters to be repeated as often as we like? See Example 1. you should verify that we get 96 when n 6. To make sure that there are no errors. In how many of these subsets are the selected integers consecutive? 3. …. How many of these subsets have largest element L where 1 L n? 4. r n. For example. You can also check that we get the correct answer in the other extreme case. if r 1 so that there are no repeated symbols. Be brave and you will succeed!
Exercise 11. 2n + 1}. 100. r 3. …. The letters of the alphabet are used to form a word of length k. we get (n 0! 2)! (2n 2) 2 (n 1)!. A subset of five numbers is selected from the integers 1. Some of these are also difficult. …. 5 A S T R AT E G Y F O R C O U N T I N G P R O B L E M S
427
. does our formula give the correct answer? Substituting in the above expression.n! r! n (n
(n
(n
r)
(n
(n
2)!
r
1) (n (n r 1)
(n r!
2)!
1) r! 2)! 2)!
r)
(n r
r
1) 1)]
(n r!
2)!
r! (n r! (n (r 2)! 1)!
[n(n r (2n
1) (2n r
r)(n 1)
There is not much merit in this algebra since the final answer is not very illuminating. The unsimplified version is likely just as useful. Consider the set of numbers {1. How many subsets of size o e can be formed with o odd numbers and e even numbers?
1 1 .
Suppose you have n symbols of which r are identical (called special as in Example 2). A binary sequence of length n has exactly k 1s. each symbol may be used up to r times? 6. An arrangement of length r is constructed from a set of n distinct symbols that include the letter A. A sequence of length r is formed using the integers 1. How many of these sequences have at least one 2 before a 1? 11.5. no symbol may be used more than once? b. How many arrangements of these symbols can you make if the arrangement must start with a special symbol? 7. 2. How many of these arrangements contain A if a. b Bs and c Cs if no two of the As are consecutive?
428 C H A P T E R 1 1
. 1000. …. 2. How many arrangements can be formed from a As. at most once. How many such sequences can you make if the sequence begins or ends with 1? Part C 8 How many sequences of fixed length n can be formed by arranging the integers 1. and the remaining n r are distinct. two 2s. and so on to two ns. n so that 1 and n are separated by exactly one term? 9. In how many of these sequences does 1 occur before 2? 10. …. A sequence of length 2n is formed using two 1s.
r) or nPr) P(n. number of sequences is n n n! n ! 1 2!… k n or nCr r Counting Subsets Subsets of r elements from a set of n distinct elements n P(n. with unlimited repetition allowed. Next you must decide what restrictions have been placed on the objects you are counting and whether you can deal with these restrictions and be left with a standard question.5 gave some hints on problem solving. As was pointed out. Arrangements of r elements from a set of n distinct elements (P(n. it also occurs when counting subsets. n) n. We do not need basic concepts such as the product rule to solve every question. Arrangements of r elements from a set of n elements. if dealing with the restriction imposes order on the object. the terms permutation and arrangement are also used. so that n1 n2 … nk n. r) n(n 1)…(n r 1)
n! (n r)!
If r If r
n. You may have to separate the objects into cases. n2 of type 2. If it does. The first thing to decide is whether order matters. This happens when the way you deal with the restriction affects the number of ways to complete the objects. we get P(n. number of sequences nr c. Arrangements of n symbols of k types. you are counting subsets or combinations..Key Concepts Review
Most counting problems are variations of standard questions or can be converted into such questions. with n1 symbols of type 1. We can define some expressions for common situations. If it does not. r) n! r! (n r)!r! r n r n r n 0 n n n r n 1 … r 1 n r n n 1 1 2n
KEY CONCEPTS REVIEW
429
. you are counting sequences. r) 0
n!
b. etc. if the elements are distinct. you need to remember the following formulas: Counting Sequences a. Section 11. P(n.
the first or last digit is 9? e. A password for a voice-mail system is a sequence of five digits selected from {0. the sum of the first two terms is 7? g. the first digit cannot be 0? c. How many passwords can be formed if a. 1. all the digits are distinct? g. the sequence starts with 7? c. There are three green. the first and last digit are 9? d. at least two green wires are used? f. 9}. exactly three different colours are used?
REVIEW EXERCISE
431
. 2. exactly two different colours are used? g. there are no restrictions? b. three black. 1. …. the first two terms are both less than 3? e. the first and last symbol are the same? f. there are no restrictions? b. How many subsets of four wires can be selected if a. only green and red wires are used? d. there are no restrictions? b. If no repetition is allowed. at least one of the first two terms is less than 7? 2. at least one green wire is used? e. the digit 9 must be included? 3. the sequence starts with a digit less than 7? d.Review Exercise
1. and three white wires. the first two terms are both less than 7? f. three red. 2. 9} with unlimited repetition allowed. at least two different digits must be used? h. A sequence of length 6 is formed from the digits {0. A cable contains 12 wires that are colour coded. there is exactly one wire of each colour? c. how many of these sequences can be formed if a. ….
and 3 choices for the fourth. the two spruces cannot be planted side by side? f. why P(100. three pines.
7. Hence there are 240 sequences that start or end with a vowel. How many different arrangements of the ten trees can be planted if a. To count the number of sequences that start or end with a vowel. and two spruces available. a spruce is planted at each end of the row? d. there are no restrictions? b. no two cedars can be side by side? 6. the string starts or ends with 0? d.4. A bit string of length 8 is a binary sequence made using 0s and 1s. To count the number of sequences that contain at least one vowel. A landscaper plans to plant a row of ten trees. b. so there are 2 5 4 3 120 sequences that start with a vowel. one in the first half and one in the second half? h. consider two cases Case 1: sequence starts with a vowel Case 2: sequence ends with a vowel In Case 1. which occur consecutively? g. the string starts and ends with 0? c. How many such strings can be constructed if a. The remaining three letters can be chosen in 3
432 C H A P T E R 1 1
. the trees at the ends of the row are the same species? e. there are no restrictions? b. using words only. the string has at least 2 0s? 5. 10) 100 10 10!. the row starts or ends with a spruce tree? g. start with 2 choices for the last letter and so on. e. 5 choices for the second. c. A sequence of length 4 is formed using the letters {a. the string has exactly 2 0s. a. Explain what is wrong with the following arguments and provide a correct solution with a clear explanation. Explain. f} without repetition. the string has exactly 2 0s? e. select 5 the vowel in 2 ways. 4 choices for the third. so there are 120 sequences that end with a vowel. In Case 2. there are 2 choices for the first letter. all trees from each species must be planted side by side? c. d. the string has exactly two 0s. b. the string has exactly 2 0s and starts with 10? f. There are five cedars.
The five letters {a. A and B occurring before C?
REVIEW EXERCISE
433
. neither A nor B? 10. how many words are between adcbe and dacbe? c. A sequence of length r is formed from a set with n elements that include A.ways. A subset of size r is formed from a set with n elements that include A and B. B. how many words come before ceadb? b. what is the 61st word in the list? 9. e}are arranged to form a word. a. b. Then the four selected letters can be arranged in 4! ways so there are 5 4! 480 sequences that contain at least one vowel. If all possible words are arranged in a list in alphabetic order. A occurring before B? b. and C. d. c. 2 3 8. A or B? c. How many of the possible subsets contain a. A occurring before B occurring before C? c. How many of the sequences have a. both A and B? b.
5 seats in the centre.wrap-up
CHAPTER 11: VOTING SYSTEMS
investigate and apply
Voting systems in Canada and around the world vary. and 12 seats in the west. and 25 in the west. that used proportional representation.) 3. Canada's system gives voice to disparate regions but can result in governments that do not have a majority of the popular vote. 60 seats for the central region. 8 seats in the centre. That is. past or present. Party C wins 2 seats in the east. Each has its own strengths and weaknesses. no election procedure can always fairly decide the outcome of an election that involves three or more candidates. How many ways are there to assign these seats?
INDEPENDENT STUDY
Research a recent federal or provincial election and determine the number of seats each party would have if proportional representation seating had been used. and 40 seats for the west. Party B wins 10 seats in the east. but if regions are to be heard. After an election. How did they deal with determining who gets what seat? What did Kenneth Arrow mean by fair choice? ●
430 C H A P T E R 1 1
. How many ways can 12 seats in a small parliament be assigned to three political parties if the votes are in the proportions 8 to 3 to 1? 2. allocating the seats can be a problem. Party A wins 8 seats in the east. Proportional representation gives seats in parliament in direct proportion to the popular vote. 47 in the centre. mathematical economist Kenneth Arrow proved that there is no consistent method of making a fair choice among three or more alternatives. In 1951. Investigate and Apply 1. How many ways can 120 seats in a parliamentary house be assigned to three political parties if the votes are in the proportions 80 to 30 to 10? (Most scientific calculators will not be able to answer this—try to find computer software that can. How many ways could these seats have been allocated? How does the proportional representation seating compare to the actual number of seats awarded to each party? Research a voting system. and 3 seats in the west. Proportional representation is used in each of these regions. A 120-seat parliamentary house represents three regions: 20 seats for the east.
2. A committee of four people is to be formed from six men and six women. four each with possible answers A. there are no restrictions? b. How many different sets of ten questions can be formed if the set has exactly two questions with each possible answer? 6. or E. r) is the number of sequences of length r that can be formed n P(n. 3)
10 . Evaluate P(10. r) using n symbols with no repetition allowed. How many of these words have the two Ts separated by at least one other letter?
434 C H A P T E R 1 1
. A five-digit number is formed using the digits from the set {1. 8
1. D. Bob and Mary will not serve on the committee together? You may leave your answer in unsimplified form.Chapter 11 Test
Achievement Category Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application Questions all 7-9 2-4 5. start and end with an even digit? c. How many of these numbers a. 9} with no repetition. You are given a collection of 20 multiple choice questions. Given that P(n. 5. C. 3
2. B. start with an even digit? b. have exactly three even digits? 3. prove that r! r 4. …. both Bob and Mary must be on the committee? c. Be sure to explain your reasoning. All the letters of the word Toronto are used to form words of length 7. How many committees can be formed if a.
Binary sequences of length n are formed using 0s and 1s. The universal set U has N elements. Digits and letters may be repeated. 7. or 8 characters is formed using the 26 letters of the alphabet and the ten digits selected from the set {0. A password with 6. How many passwords can be formed if a password must contain at least one letter and at least one digit? 9. Show that the r fraction of all subsets of size r that contain A is N . 1. 9}. including the letter A. How many of these sequences start or end with 1? 8.7.
CHAPTER 11 TEST
435
. …. 2.
we represent the entries in the matrix as *. Next. where the subscript indicates the number of equations we are solving. such as 3 1 2 3 2 1 1 7 3 2 2 1
A computer can use the same algorithm to solve ten linear equations in ten unknowns or. 1) entry is 1 or a 0 appears that requires an interchange of rows. subtractions. there are algorithms that sort lists or search strings of characters. in general. we examine one way to assess algorithms. n linear equations in n unknowns. We have used a total of 20 operations to reduce the matrix to the form 1 * * * 0 * * * 0 * * *
436 C H A P T E R 1 1
. To assess the algorithm. we count the number of additions. A Web search engine uses a complex algorithm to quickly search a huge text string in order to identify sites that you are seeking. written in matrix form. Here. You have learned the Gauss-Jordan algorithm for solving systems of three linear equations. * * * * * * * * * * * * We do not worry about special cases when. let the number of operations be a3. This requires another eight operations. multiplications. The time taken to solve the system is closely related to the total number of operations required. We repeat this step for the third row. For the above example. the (1. and divisions it takes to solve the equations. 1) entry and subtract from the second row. giving another eight operations.ASSESSING ALGORITHMS
An algorithm is a recipe that a computer uses to solve a specific class of problems. Computer scientists are interested in studying the properties of algorithms to see how efficient and fast they are. for example. Since we only care about how many operations are needed and not the actual solution. For example. The first step of the algorithm is to divide every entry in the first row by the first element. we multiply each element in the first row by the (2. This requires four divisions.
This requires two multiplications and two subtractions. and. we substitute the solutions of these two equations into the equation corresponding to the first row and solve. n 1 2 3 4 5 6 7 8 9 10 an 1 12 36 77 139 226 342 491 677 904
E X T E N D I N G A N D I N V E S T I G AT I N G
437
. A computer can use this algorithm to solve a system of n equations. n 2. to find the first unknown. so we have a2 12. a3 36. simplifying. You can use the same argument to show that a2 a1 11. Finally. that is. and we require a2 operations to solve these equations. so we have a3 a2 24. You can use the same counting method to show that an (n 1) 2(n 1)(n 1) 2(n 1) an 1 or. We can use this formula to find an recursively for any value of n. we use the formula and a3 to find a4 and then repeat.The last two rows correspond to two equations in two unknowns. a1 1. obviously. The table gives the number of operations required for solving up to ten equations in ten unknowns. an 2n2 3n 3 an 1.
one in which the number of operations required is proportional to n2 as n gets large.
438 C H A P T E R 1 1
. The answer here is no. in certain cases. nn 3
4n3
15n2 6
7n
6
. when n is large and many of the coefficients are 0 (this is called a sparse system of equations). However.To assess the algorithm more generally. a more efficient algorithm can be found. Note that an is
2 for large values of n. we can show that an
a approximately proportional to n3 as n gets large. 3
You might wonder if it is possible to find a better algorithm that requires substantially fewer operations. for example. that is.
and medicine. CHAPTER EXPECTATIONS In this chapter.1 solve problems using counting principles. Section 12. will it always drop directly to the ground? A scientist would answer this question by repeating the experiment many times in carefully controlled circumstances before drawing a conclusion about the consequence of a continuous or connected series of actions. and diagrams will lead to important and practical applications in finance. Section 12.1.Chapter 12
SEQUENCES
If you release a tennis ball from your hand. Section 12.3 prove the binomial theorem. a set of numbers can be arranged according to some rule or sequence. computer science.4 prove relationships between coefficients in Pascal's triangle. Section 12.2
•
. In mathematics. 12. Section 12.2 prove formulas for the sums of series.4 determine terms in the expansion of a binomial. Section 12. Section 12. A closer look at sequences of numbers and sequences of functions.4
• •
use sigma notation.3 understand mathematical induction. values of r coefficients. statements. you will
• • • • • •
solve problems involving permutations and combinations. Section 12.4 describe the connections between Pascal's n and values for binomial triangle. Section 12.
…. 28. For f(n) 7 . 18. 1.
440 C H A P T E R 1 2
. In a geometric sequence having first term g1 a and constant ratio r. 3n 1. tn 4 (n 1)3 3n 1. …. …. The sum of the first n terms of an arithmetic sequence is Sn a (a d) (a 2d) … (a (n 1)d)
n [2a 2
(n
1)d]
For the arithmetic sequence 2. Every linear function defines an arithmetic sequence if the variable has the set of positive integers as its domain. 7. we obtain the geometric sequence 2. 2n 1. In this sequence. . …. we obtain the sequence 7. If a 4 and d 3. g3 ar2. 5]
555
GEOMETRIC SEQUENCES
In a geometric sequence. g4 ar3. 5. In an arithmetic sequence having first term a1 a and constant difference d. the ratio of consecutive terms is constant. a4 a 3d. An exponential function of the form f(n) a • bn 1 defines a geometric sequence if the variable has the set of positive integers as its domain. …. the value of a is 2. the difference between consecutive terms is a constant. 4 (n 1)3. successive terms are g2 ar. 2. Here we review some ideas about arithmetic and geometric sequences. we obtain the arithmetic sequence 4. we obtain the sequence 4. a3 a 2d. …. d is 5. 56. 6. 7. …. 3n 7.
ARITHMETIC SEQUENCES
In an arithmetic sequence. and an a (n 1)d. 2n 1.Review of Prerequisite Skills
This chapter deals with sequences. 5n and the sum of the first 15 terms is S15 2 7 12 … 72
15 [4 14 2 15 [74] 2
3. If a 2 and r 3. successive terms are a2 a d. …. 54. …. 10. For f(n) 3n 7. 2 . 12. …. 7 . gn ar n 1.. …. 14.
find an expression for the general term bn and show that b1. Consider the sequence tn 2 . identify them as arithmetic. In an arithmetic sequence. … is an arithmetic sequence with first term a and common difference d and bn a2n 1. …. b2. Find the tenth term and the sum of the first ten terms. n 1. In a geometric sequence. Evaluate the first five terms of the sequence. n a. …. Is this
8. 3. the first term is 2 and the fifth term is 32. a2. tn n ( 2)n n ( 2)n n( 2)n 7n 5 3 . geometric. 7. 5. n 1. tn f. … also form an arithmetic sequence. an. 4. bn. tn d. the sum of the first 12 terms is
1)
Exercise
1. n
1. n 1 has both a and r positive. or something else.The sum of the first n terms of a geometric sequence is Sn a ar ar2 … arn
1
a rr
n
1 1
For the geometric sequence with a 12 S12 7 33 11
7 12 (3 2
7 and r
3. tn e. n 1?
REVIEW OF PREREQUISITE SKILLS
441
. Find the sum of the first ten terms. Consider the new sequence defined by an sequence arithmetic? tn
1
n
tn. find the sum of the first twenty terms. 6. What can you say about the sequence with terms an log(gn). If a1. n 1. Can a sequence be both arithmetic and geometric? Explain. If the sum of the first five terms of an arithmetic sequence is 30 and the sum of the first ten terms is 10. tn c. For the following sequences. the second term is 7 and the fifth term is 16. In all cases. tn b. Is it arithmetic? b. a. Suppose a geometric sequence with general term gn arn 1. 2n 1
2.
tn tn 1 tn 2. published in 1202. He introduced the sequence in his book Liber Abaci. named in honour of Leonardo da Pisa. What are fractals and where are they used? ●
442 C H A P T E R 1 2
. There are parallel ideas in sociology (cultural evolution) and in philosophy (Hegel's dialectic method). 34.investigate
CHAPTER 12: RECURSIVE SEQUENCES
All living things receive their DNA from the generation that came before them. 2. 5.. Changes to the DNA as it is passed along are fundamental to the evolution of species. Fibonacci numbers arise in several other contexts. n 1 n 2.. recursively defined sequences are used to solve complicated mathematical equations indirectly when no direct method exists. . Fractals are widely used mathematical objects that can be constructed by repeating a process in which each step adds greater complexity to the results from the previous step. Calculate the first ten terms of this sequence tn 1 t 1 . 3. The numbers themselves are called Fibonacci numbers. 5. 3.
DISCUSSION QUESTIONS
1. It can also be approximated using the recursive sequence t1 1.. 13. 8. 21.. . Investigate Possibly the most famous recursively defined sequence in mathematics is the Fibonacci sequence. They answer questions like "How many sequences of 0s and 1s of length n 1 have no 0. Among other things.. Recursively defined sequences are numerical sequences in which each term is found from the previous term (or terms) through some specific process. The first few terms of this sequence are 1.. 4.
2
This ratio occurs in nature and is used in art and architecture because it is considered to be aesthetically pleasing. etc. who lived around the beginning of the 13th century and whose nickname was Fibonacci. . When is a direct formula for the nth term in a sequence preferable to a recursive formula? 2.0 sub-sequence?" and "How many subsets of the integers from 1 to n 1 contain no consecutive integers?" 1 5 The ratios of the Fibonacci numbers approach the Golden Ratio . There are also parallels in mathematics.. t2 2. as the solution to the following problem: How many pairs of rabbits will be produced in a year if every month each pair of adult rabbits gives birth to a new pair that becomes mature in the following month? The answer is the twelfth term in the sequence defined by t1 1. n 3.
and diagrams. we start with the first term. The terms of a sequence may be functions. such as those used to sort and merge lists. In this chapter. D1. are based on sequences. … where S1: the sum of the first natural number is 1 S2: the sum of the first two natural numbers is 3 Sn: the sum of the first n natural numbers is n(n 2 1) We know that every term in this sequence of statements is true. repeat the removal
12. a sequence has the form t1. and tn is the nth or general term.
D1
D2
D3
To produce this sequence of diagrams. In medicine. Sn. such as S1. For example. sequences are used to model the growth and decline of epidemics of infectious diseases in a population. statements. which are the subject of modern mathematical research. t2.1 SEQUENCES
443
.Section 12. are defined as a sequence of operations. t2 is the second term. In computer science. ….1 — Sequences
We define a sequence of numbers by its terms. such as t1(x) 1 x. we use a method of proof called mathematical induction to investigate other sequences of statements to see if they are true. In Section 3 of this chapter. leads to Sierpinski's triangle. Financial calculations. an equilateral triangle. such as monthly mortgage payments. This leaves three smaller equilateral triangles as shown. join the midpoints of the three sides of the initial triangle and remove the equilateral triangle in the centre. …. …. That is.
t chnology
e
We can also construct sequences of diagrams. … or statements. amazing computer-constructed diagrams. D2. many algorithms. one term for each positive integer. we look at sequences of numbers and also sequences of functions. t2(x) (1 x)2. Enter fractal in a search engine and you will soon locate some beautiful examples. tn(x) (1 x)n. To get the second term. S2. To produce the third term. tn. Sequences are of practical importance. a simple fractal. the following sequence of diagrams. … where t1 is the first term. D3. One use of these sequences is to construct fractals.
Again. tn(x) (1 x)tn 1(x). what happens to the terms of a geometric sequence as n gets large? If you have taken calculus.
a. tn tn 1 d. n 2. the general arithmetic sequence b. Using the above examples. Recursion as a way to define sequences is very important in computer science and other areas of application. tn tn 1 d for all n 2. t n r for all n 2. That is. The above sequence of diagrams is defined recursively. The first term of the general arithmetic sequence is t1 a and the difference between two consecutive terms is d. the sequence of diagrams converges to Sierpinski's triangle. In recursive definitions of a sequence. Instead. we produce the nth diagram by using the removal process on all the equilateral triangles in the (n 1)st diagram. That is. The first term of the geometric sequence is t1 a and the ratio of two t consecutive terms is r. …. n 2.
D6 As n gets large.
EXAMPLE 1
Provide a recursive definition for a. Here is what happens after n 6. n 2. we rearrange the
n 1
equation to get the recursive definition of the geometric sequence t1 tn r tn 1. Rearranging this equation gives the recursive definition of the arithmetic sequence t1 a. In general. For any fixed x. For example.process on all the remaining triangles in the second diagram. we do not give a formula for finding the nth term directly. the general geometric sequence c. we specify the first term and then define each term by a process applied to the preceding terms. Applying the result from above we have t1(x) 1 x.
c. the sequence is a geometric sequence. tn(x). … where tn(x)
(1
x)n
Solution a. the sequence of functions t1(x). b. t2(x). There are many Web sites that will show you dynamically how this sequence evolves. then you know this is the limit of tn as
444 C H A P T E R 1 2
. we can ask many mathematical questions about sequences.
n 10. evaluate p18. n
2.1
Part A
Knowledge/ Understanding
1. Show that the sequence pn.1 SEQUENCES
445
. …. A family is saving to pay for their child's university education. 1. t1 b. 18 satisfies the recursion p10 2000. On each subsequent birthday up to and including the 18th. the family puts $2000 into a savings account that pays 5% annual interest. The following sequences are defined recursively. gn(x) x. especially as n gets large. Let p10.. There are many ways to do this. In an arithmetic sequence. the third term is 12 and the 15th term is a. p11. Show that if the second and third terms are also equal.. 11 n 18. another $2000 dollars is added to the account.. A geometric and arithmetic sequence have a common first term. Evaluate the first five terms. 11th.
12. How many terms in the sequence are positive? Part B
Application
3. n 3 2 2 48. …. hn(x)
xgn 1(x)
2hn 1(x).
1. n 1. show that the sum of any 20 consecutive terms is 20 times the average of the first and last term in the sum. f2 2 1. fn 3tn 1. then all terms are equal. b. On the child's tenth birthday. we could ask about the area and the number of triangles in the nth diagram. a.n → . Find the formula for the general term of the sequence.
Exercise 12. a. g1(x) d. We can also generate interesting questions by creating new sequences from a given sequence. 11. Using a calculator.05pn 1 2000. 5. h1(x) 1. 18th birthday. b. 4. p18 be the amount in the account after the payment is made on the 10th. For the sequence of diagrams. pn 1. n fn 1
1
2 fn 2. For any arithmetic sequence.. tn 1. f1 c.
At the end of each month. A sequence has general term gn tn tn 1. … is defined with first term a and the sum of any two consecutive terms a constant s. Create a new sequence according to the instructions given and determine whether each new sequence is geometric. n 2. Prove that every point in the sequence lies on a straight line. The monthly interest rate is 0. A sequence of points in the plane P1. n 2. The first row is 1 to 10. where tn 3 2n 1. vn. P2. Use this recursive definition and a spreadsheet program or calculator to determine how many months it takes before the loan is paid.
Communication
9.
Communication
12. In that this is a geometric sequence. n 2. Explain why pn 4937. Show that p2 b. Note that p1 5000. At the start of the nth month. n 1. c. Sketch the position of the first four points in the sequence. a2. b. ….
Thinking/Inquiry/ Problem Solving
10. Pn. v2. … is defined recursively so that v1 2 and vn 3 vn 1. Suppose that the sequence v1.
c. …. 2. what must be the value of g1? 8. …. An anti-arithmetic sequence a1. n 1. Consider the geometric sequence with general term tn 3 2n 1. a. Find an expression for rk. the second 11 to 20. 11. 0) and Pn 1 (1. The nth term of the new sequence is tn tn 1. n 1.50 1. Find the general term of the sequence. n 2. The nth term of the new sequence is the square of the nth term of the given sequence. … is defined recursively with P1 given by (1. an. a. and so on.Knowledge/ Understanding
6. the remaining debt owed on a student loan of $5000 is pn. Verify that vn 2 3n 1 satisfies this recursive definition for all n 1.
446 C H A P T E R 1 2
. n 2 that appear in the kth row of the array. … are written in a spreadsheet with ten columns. 7. 3. The first term of the new sequence is 3. an an 1 3. The terms of the new sequence are the odd-numbered terms of the given sequence. 2) Pn. the student makes a payment of $100. a.0075pn
1
100 for n
2. The numbers 1. Let rk be the number of terms of the arithmetic sequence a1 4. b.75%.
n 2. let an be the number of equilateral triangles in the nth diagram.5. Suppose x1 0. Find an expression for fn(x) and show that it satisfies the recursion. c. a2. x1 1. a. sn. s2. s2. …. Use a calculator or spreadsheet to evaluate x2. x3. ln. … is constant.
12. evaluate xn for all n
2
b. …. x2. If x1
t chnology
e
0. … alternates in value? 16. In the sequence of Sierpinski triangles. b. with x1 between 0 and 1. and a4. …. Sketch a graph of y g(x) and the line y x on the same set of axes. n 2.Part C 13. b. A sequence of lines l1. xn. s3 and s4. What can you con-
14.
l4
.n
l2
l3
15. …. Let s1. a. The first four lines are shown on the diagram. x15. fn(x1). a. Consider the function f (x) 2x(1 x) and define a sequence x1. The first term f1(x) g(x) 1 x x and fn(x) g(fn 1(x)). 2 satisfies the recursive definition in b. Are there any values x1 so that the sequence f1(x1). …. c. f2(x1). fn(x0). …. n
2
2. Evaluate a1. and xn f(xn 1). a3. … recursively. d. … is drawn in the plane so that no two are parallel and no three intersect in a common point. … be a sequence of numbers corresponding to the number of distinct regions that are created by the lines. Show that sn
l1
sn
n2 n 2
1
n.3. 0 clude from this investigation? d. a. Explain why sn c. f2(x0). Find all values of x0 for which the sequence f1(x0). Evaluate s1. Repeat parts b and c if f(x) 3x(1 x).1 SEQUENCES
447
. A sequence of functions with general term fn(x) is defined recursively. Repeat part b for a variety of values for x1. Describe the behaviour of the sequence as n increases. l2.
and D4 are shown below. What happens as n gets large? 17. If the length of the side in the original triangle is 1.b. D2. n 2 satisfies this recursive definition. d. D1. an. Verify that an 3n 1.
448 C H A P T E R 1 2
. … satisfies the recursion an 3an–1. n 2. Find a formula for the general term bn. Dn is produced by replacing the middle third of all lines in Dn 1 by two line segments of equal length. The second term D2 is formed by replacing the middle third of each side by two other line segments of the same length as shown. Find a formula for An. The Koch snowflake is produced by recursively operating on a sequence of diagrams. a2. c. f. The first term in the sequence D1 is an equilateral triangle with side length 1. Explain why the sequence a1. e. D3. develop a recursive definition for the length of the side bn of the small equilateral triangles in the nth diagram. the fraction of the area of the original triangle remaining in the nth diagram. Determine what happens to the area and perimeter of Dn as n gets large. ….
That is. if tn a (n 1)d. 2 PA RT I A L S U M S A N D S I G M A N OTAT I O N
449
. That is. t4 t5 … t25 c. We use a special notation called sigma notation to express Sn compactly. …. We write
n
Sn
i 1
ti
Note that is the upper case Greek letter sigma (the equivalent of our letter s). we reviewed the formulae for the sum of the first n terms of both arithmetic and geometric sequences. then the sum is Sn t1 t2 … tn
n [2a a
(n
1)d]
Creating a new sequence in which each term is the sum of the terms of a given sequence is so common that we use a special language and notation to describe it. …. t2. Solution a. we use lower case letters such as i. and k for the index. n 1. Its purpose is to tell us what terms to include in the sum. … is an arithmetic sequence. The i is called the index and the values below and above give the range of the index in the summation. S2 t1 t2. The sum of the first 25 odd-numbered terms.
EXAMPLE 1
For a sequence with general term tn. By custom. 2 — Partial Sums and Sigma Notation
In the introduction to this chapter. we determine the nth term Sn of the new sequence to be the sum of the first n terms of the given sequence. Sn is the sum of the terms ti for i 1. The sum is t4 c. chosen to remind us that we are constructing a sum. tn.Section 12.
t2 t5 t3
… … t5
t50
i 1 25
ti tj
j 4 25
t25 … t49
t2k
k 1
1
The index in sigma notation can be any letter. For instance. …. The sum of the first 50 terms. There is nothing magical about sigma notation—it is simply a convenient way to express the sum of terms of a sequence. 2. …. S1 t1. For any sequence of numbers t1. b. The sum is t1 b. write the following sums in a. j. n. 2. The sum is t1
50
notation.
1 2 . The notation tells us to add the terms ti as the index i ranges from 1 to n in steps of 1. Sn t1 t2 … tn
The terms Sn of the new sequence are called the partial sums of the terms of the original sequence.
The partial sum sn is
Sn
i 1 n
ai (5
i 1 n
2i)
n
5
i 1 i 1
2i
1 2 .8 2
3
4
5
6
7]
The rules of arithmetic apply to summation notation. n
1. express sn numbers from 1 to n. determine the sum
S40
i 1 40
(4i 4i
i 1 40
3)
40
3
i 1 40
4
i 1
i
i 1
3 2 3 40 120 … 3 40} [3 3 3 … 3]
4[1 3280 3160
. Solution
40
4n
3. Solution The ith term of the sequence is ai
n
ai in terms of
i 1 i 1
i. v2.7
b. The expression on the right reflects the fact that when we add. then
n n n n n
aui
i 1
a
i 1
ui
(ui
i 1
vi)
i 1
ui
i 1
vi
The expression on the left uses the fact that a is a common factor for each term in the sum. … are two sequences and a is a constant. … are the terms of an arithmetic sequence with first term 7 and
n n
common difference 2. un. If u1. we can switch the order of the terms without changing the total. vn. 4 40 2 41
EXAMPLE 6
If a1. 3
i 1
i
3[1 3 84
2
7. …. … and v1. …. u2. a2. an. ….
EXAMPLE 5
For the arithmetic sequence defined by f(n) of the first 40 terms. 2 PA RT I A L S U M S A N D S I G M A N OTAT I O N
451
. the sum of the natural
7
(i
1)2
5
2i.
….
EXAMPLE 8
By using the result of Example 7 with the sequence defined by tn
n
n3. …. The second includes all terms from the first to the nth.
Solution The sum of the differences is
n n n n
dj
j 1 j 1
(tj
1
tj)
j 1
tj
1
j 1
tj
The two sums on the right side are almost the same.
ai
n(a
d)
d
i 1
i.
n
i
i 1
n(n 2
1)
EXAMPLE 7
For any sequence t1. we have dj n3 (j 3n2 1)3 3n
j 1 n
dj
j 1
(n
1)3
13
n3
3n2
3n. determine
an expression for
i 1
i2. the difference between consecutive terms of the original sequence. and there are numerous other interesting sequence results that can be determined by using similar techniques. tn.n
5n
2
i 1
i
n n
For any arithmetic sequence with general term ai we can see that
i 1
a
(i
1)d
(a
d)
id. tn disappear so we have
n
dj
j 1
tn
1
t1
This result is useful. t2. …. The advantage of this derivation is that
you can work out the sum of the terms of an arithmetic sequence by remembering only the sum of the natural numbers. Hence. taking the difference.
j3
n
3j2 dj (3j2
3j
1 Substituting. define a new sequence with general term dn tn 1 tn. the terms t2. The first includes all terms from the second to the (n 1)st.
n
Solution Using the result from Example 7.
3j
n
1)
n
j 1 n
3
j 1
j2
3
j 1
j
j 1
1
452 C H A P T E R 1 2
. .
n
Evaluate
j 1
dj in terms of the original sequence.
Exercise 12. we can substitute to get an expression for
j2. a. n 1 2. several students simplified the expression by cancelling an xi from the numerator and denominator on the left side.2
Part A
Knowledge/ Understanding
1. 2 PA RT I A L S U M S A N D S I G M A N OTAT I O N
453
. the sum of the squares of the positive integers from 50 to 100 d. n 1 c. write the sum out explicitly. as shown
5
xi2
i 1 5
5
xi
xi
i 1
i 1
1 2 .n
Since
n j 1
j
n(n 2
1)
n
and
j 1
1
n. the sum of the first fifteen terms of the sequence with general term tn 3 2n 1. Expand the following sums expressed in sigma notation. Express the following sums in sigma notation. the sum of the first 30 odd-numbered terms of the sequence with general term tn. if you have doubts about the notation. On a statistics test.
j 3
(j
1)2
c.
i 1
1 i
Communication
3. 1 2 3 … 25 b. In any situation.
i 1
i
b.
j 1 n
3
j 1
j2
n3
2n3 2n3
3n2
6n2
3n
6n 2
3 n(n 2 1)
3n2 3n 2n
n
n
Then
j 1
j2
n(n
3n2 n 2 1)(2n 1) 6
We can evaluate many other partial sums using the same method. Do not be fooled by summation notation.
10 7 n
a. The exercises will give you lots of practice using sigma notation.
3
(n 1).
6. determine
1
c.
j 1
g2j
1
7. the sum of the first n terms for all n
8
n
1.. 11. Use the fact that n is the number of binary sequences of length n with k n n exactly k 1s to evaluate . x5 were a sequence of numbers.where x1. …. Consider three sequences u1. 9.) Part B 4. w2. Use the result to find the sum of the first n terms of the sequences 1. …. u2. b. v1. kk.. Find the sum of the first n terms of the sequence 9. Write an expression for A using sigma notation. . …. …. Explain why the students all lost marks for this simplification. k 0 k
454 C H A P T E R 1 2
. a. 99. Show that
n 1
bn
0. v2.. …. vn. and w1. Construct three such sequences of numbers with n 3 to see if the following is true.
n n
uivi
i 1 i 1
vi
n
n
uiwi
i 1 i 1
wi
Can we cancel the ui from the expression on the left? 5. …. the sum of the first ten terms b.. Consider the geometric sequence with general term gn
60
arn 1. … where k represents a digit from 2 to 8.. … . …. 999.. un. a100 is a sequence of 100 numbers with average value A. wn..
100
an
A for
c. kkk. (Hint: Pick any five numbers for the xi. For the geometric sequence with general term gn a. Suppose that a1. and k. 111. Consider the new sequence with terms defined by bn 1 n 100. a2.
gi Evaluate
i 1 120
. gi
i 61
Thinking/Inquiry/ Problem Solving
8.
16. Use the method of Example 8 to find a simple
1 i(i 1)
expression for Part C
. Consider the arithmetic sequence with general term an 1 3(n and the geometric sequence with general term gn 2n.
10
b. That is.
12. and so on.
13. . Prove that
n 1 j 1
jtj
j 1
sj. i . Consider the sequence tn n2. t1 a2. …. Use the method of Example 8 to find a simple
n
expression for the sum of the first n natural numbers
i 1
i.. n
1
a. Evaluate
i 1
gi and
i 1
ai. Find a simple expression for
i 1
ai2. t2 a4. Evaluate
Thinking/Inquiry/ Problem Solving
( 1)iti and
11. n 1.
10 10
1). Suppose that t1. t3 a8.
i 1
metic and a geometric sequence.
n
jgj. Suppose that tn
a
(n
1)d. Use the result of question 15 to evaluate j of a geometric sequence. n
n i 1
1.. n 1 and the sequence of differences dn tn 1 tn. n
n
1 are an arith( 1)isi. where gj is the general term
17. j
1.n n
1 and the sequence of differences
tn. n 1.
15.Knowledge/ Understanding
10. tn are the first n terms of a sequence and we construct a
n n n
new sequence with sj
j 1
tj. the square of the sum of the first n natural
j 1
2
1 2 .
n
14. Consider the sequence tn dn tn
1
1 . 2 PA RT I A L S U M S A N D S I G M A N OTAT I O N
455
. Find
i 1
ti. n
1 and sn
n i 1
arn 1. Use the method of Example 8 to show that the sum of the cubes of the first n natural numbers is numbers. Define a new sequence tn by selecting the gnth term from the arithmetic sequence. where ai
2i
3. n.
Once we have finished Step 2. t3 2(3) 1 7.3 — Mathematical Induction
In this section. We use mathematical induction to verify that tn 2n 1 is the correct formula. Now we use mathematical induction to prove that we have found the correct formula in the above recursion. …is true.
456 C H A P T E R 1 2
. We start by evaluating the first few terms of the sequence. We have t1 t2 2(1) 1 3. For instance. Since we have shown that S1 is true. Step 3: Combine the results of the first two steps to conclude that every statement in the sequence is true. we introduce a new method of proof with the peculiar name mathematical induction. Suppose that we want to show that every term in a sequence of statements S1. However. t5 2(15) 1. we have no idea if this is the correct formula for larger values of n. prove that the next term is also true. and so on. with t1 1. Why do these steps actually prove that all the statements are true? Step 1 verifies that the first statement is true. we use it recursively in Step 3. You will recall that we compared induction and deduction in the first chapter. n 2. t4 2(7) 1 15. S3 is also true. You can see the motivation for mathematical induction in the following activity. mathematical induction is a deductive method of proof.Section 12. given that S2 is true. Step 1: Show that the first statement in the sequence S1 is true. …. Step 2: Given that any one term in the sequence is true. There are three steps in the proof using mathematical induction. then Step 2 tells us that S2 is also true. Suppose a sequence is defined by the recursion tn 2tn 1 1. and using the result from Step 2 again. the formula tn 2n 1 gives the correct answers for n 5. S2. Find an expression for the general term and prove that the expression is correct. 1 31. Curiously. Now. we can easily check that the less obvious formula tn
12 18n 23n2 12 6n3 n4
also gives the correct answers for the first five terms. Sn.
Looking at the pattern.
n k 1. an.EXAMPLE 1
If tn
2tn
1
1. a2. we use Step 2 recursively to conclude that the formula is correct for n 2.
1 2 . 3 M AT H E M AT I C A L I N D U C T I O N
457
. One analogy for proof by mathematical induction is the task of devising a method to reach any rung on an infinite ladder. so the formula is correct for
Step 2: Now suppose it is given that the formula is correct for some value of n. we use induction to prove that the formula is correct. That is.
Proof Step 1: Substituting n n 1. A common use of mathematical induction is to prove the correctness of a formula for the partial sums of the terms of a sequence. we are given that tk 2k 1. That is. Step 3: Since the formula is correct for n 1 (Step 1). Step 3 puts Steps 1 and 2 together to give us a way to reach any rung on the ladder. prove that tn
2n
1 for all n
1. we have tk
1
2tk 2(2k 2k 1
1 1) 1 1
which is correct. … with partial sum sn a1 … an. n 1 and a postulated formula for sn.
1. That is. The proof is complete. if we are given a sequence a1. if we have already reached rung 4. then n 3. n
2 with t1
1. we get 21
1
1. say n k. and so on for all n.
Rung k+1 Rung k Rung k–1
Rung 6 Rung 5 Rung 4 Rung 3 Rung 2 Rung 1
Step 1 gets us on the first rung. Step 2 gives us a way to go from any rung to the next. To complete Step 2. one rung at a time. …. we can use Step 2 to get to rung 5. This completes Step 2. Using the recursion. we need to show that the formula is correct for the next value.
since
458 C H A P T E R 1 2
. in Step 2. Note that if Step 1 fails.EXAMPLE 2
Prove that 1
1!
2
2!
…
n
n!
(n
1)!
1 for all n
1. Step 3 is the concluding step and is the same in every application of mathematical induction. that is. Step 1 is usually easier. To many students. then we can connect consecutive statements. A key component of Step 2 is to find a connection between consecutive terms in the sequence of statements. say n sk 1 ak (k! 1) k k! k!(1 k) 1 (k 1)! 1
as required. We always write Step 3 to demonstrate to the reader the logic of the proof. Then. In any proof using mathematical induction. we apply Step 2 recursively to conclude that the formula is correct for all values of n 1.
Step 3: Given that the formula is correct for n 1. It is not true that every term in the sequence of statements is true. so the formula is correct k 1. then s1 for n 1. Step 1 puts us on the first rung. If the problem is to verify a formula for a partial sum. Then
Step 2: Suppose the formula is correct for some value of n. Combining the two steps and using recursion (Step 3) gives us a way to climb to every rung. 3. then S1 is false. See Exercise 10 for an example of this. 2. a proof by mathematical induction feels wrong because. 4. and we can stop. so we do it first. it seems that we assume the truth of what we are trying to prove.
Solution Note that the left side of the expression is the partial sum sn of the terms of the sequence with general term an n n! Our goal is to prove that sn (n 1)! 1 for all n 1. Some Observations about Proof by Induction 1. What we assume in Step 2 is that we have reached a particular rung. You might wonder if it is possible to carry out Step 2 if Step 1 fails. The formula is correct for k if it is correct for k
1. The ladder analogy is helpful. we prove that we can climb to the next rung. What we are trying to prove is that we can climb to any rung. Steps 1 and 2 depend on the particular result we are trying to prove. we can do Steps 1 and 2 in either order. to prove for all n that t1 t2 … tn f(n) where f(n) is a postulated function. Step 1: If n 1. sk a1 1 and (1 1)! 1 1.
8. we cannot use mathematical induction. This means that the statements can be ordered. We only use mathematical induction to establish the truth of a sequence of statements. We have an an
n3
1
n 3 n n 3n 3
(n (n 3 n3
1)3 3 1)3
(n n 1
1)
n3 n3 3n2
3n2 3n 3
1
n
1
n(n
1)
1 2 . it is more difficult to make the connection. a1 1 3 1 0. Sometimes students are tempted to try to use mathematical induction to prove theorems such as Prove that x2 2x 3 0 for all real values of x 1. Mathematical induction cannot be used here because for a given value of x. For Step 2. The reason for the name mathematical induction is that we often guess the answer based on an observed pattern. For a given sequence of statements.
n3 3 n
EXAMPLE 3
Use mathematical induction to prove that an Solution
3
is an integer for all n
1. 5. and use this method of proof to verify that our guess is correct. With this given information and a bit of algebra. then we can conclude that an an 1 (an an 1) is an integer. There is a first statement. we can take the kth term as given and then show that the (k 1)st term is true. 6. if we are not given a formula for a partial sum. when n 1. For example. it is easier to start from the (k 1)st term. which is an integer. we now prove that an is also an integer. there are often simpler and more illustrative methods.
For Step 1. we are given that an 1 (n 1) 3 (n 1) is an integer. Mathematical induction is only one method of proof. we assume that the (n 1)st statement is 3 true. 3 M AT H E M AT I C A L I N D U C T I O N
459
. then so is the next term. we show that if any one term in the sequence of statements is true. For partial sums. We cannot express this theorem as a sequence of statements.t1
t2
…
tn
(t1
t2
…
tn 1)
tn
For other problems (see Example 3). The first statement in the sequence is true. 7. there is no next term. If it is more convenient. We can take the given term as the (k 1)st term and show that the kth term is true. For Step 2. Note that if we can show that an an 1 is an integer. second statement. That is. Mathematical induction can be used only when we know the answer. and so on.
We start by verifying the first term in the sequence directly (Step 1). since the first is. Consider the sequence t1. ….
n3 3 n
EXAMPLE 3 (REVISITED)
Prove that an
is an integer for all n
1. explain why t2 2!.
460 C H A P T E R 1 2
. we know the first statement is true. Explain how the results from b and d can be used to conclude that tn n! for all n 1. Here we break the proof down into its simplest pieces.
1
is
Applying Step 3. There is a much easier way to prove the result in Example 3. In summary. We want to prove that tn n! for all n 1 using mathematical induction. we prove that any term in the sequence is true if we are given that the previous statement is true. we have shown that if an an integer. 3 we conclude that n 3 n is an integer for any value of n 1.which is always an integer for n an integer. tn defined recursively by t1 1 and tn n tn 1. Write clearly what is given for Step 2. What is the first step in the proof? b. Then in Step 3. We conclude that each statement is true. mathematical induction is a method of proof that can be used to show that a sequence of statements is true. so the given expression is always an integer. f. the third is true since the second is. Using the results from b and d. The proof is complete. and so on. Repeatedly applying Step 2. 1!? c. a. One of these must be divisible by 3.
Solution We can factor the numerator of an to get an
n(n2 1) 3 (n 1)n(n 3 1)
After factoring. e. Complete Step 2. then so is an. Is it true that t1 d. Hence. t2.
Exercise 12. For Step 2. we can see that the numerator is the product of three consecutive integers.3
Part B
Communication
1. we repeatedly use Step 2 to conclude that the second statement is true.
a. Explain why
a. we are given that Sn
1 1
c. Use mathematical induction to prove that an n 1. we are given that the formula is correct for the sum of the first
{n 1}[2a ({n 2 1} 1)d]
. Consider the sequence defined by en n2 n for n
for all
1. If tn n 3n 2n . 12 3 4 22 … … … (2)(3) (2n (3n n2 … 1) 2) n2
n(3n 1) 2 n(n 1)(2n 1) 6
d. Sn t1 t2 … tn. 1. Use mathematical induction to prove that the sum of the first n terms. For Step 2. prove that 1 for all n 1. (Hint: show that tn 1 tn is an integer for all n. (1)(2)
(n)(n
1)
n
n(n n 1
1 1 e. show that the formula is correct if n b. That is. Use mathematical induction to prove that every term in the sequence is even. Using mathematical induction. Combine Steps 1 and 2 to conclude that the formula is correct for all values of n 1. For Step 2. n 1 terms. n 1. use mathematical induction to prove that tn is an 6 integer for all n 1. Consider the arithmetic sequence with nth term tn a (n 1)d. 8.
a.2. The sequence an satisfies the recursion an
an
1
(n
1)2. 3 M AT H E M AT I C A L I N D U C T I O N
461
. where a and d are given constants. establish a connection between Sn Sn Sn 1 tn. is given by the formula Sn
n[2a (n 2 1)d]
for all n
1. Prove that n(n 5) is even for all n 1. Using mathematical induction. b. prove that the following statements are true for all n 1. (1)(2) … (n)(n1 1) (2)(3) f. 1 1! 2 2! … n n!
1)(n 3
2)
(n
1)!
1
3 2 5. Construct a simpler proof. n
n(n 1)(2n 6
2 with
1)
a1 0.
2
22
…
2n
1
2n
1
4. 7.
1 2 . d. For Step 1.
Knowledge/ Understanding
3.
Show that the formula is correct for the sum of the first n terms using this given information. 1 b. and Sn. 1 c.)
Knowledge/ Understanding
6.
Prove that the number of terms in the expansion is 2n for all n
1. a11. …. …. Suppose we have two sequences of numbers. In a puzzle called the Tower of Hanoi. 10. Consider expanding the product of n factors (a1 b1)(a2 into a sum of terms. na 1 b2) … (an bn) 1 0 . Show that Step 2 of the induction proof works but that Step 1 fails.
15. 1 0 . Suppose we try to prove that tn n! for all n 1. (a1 b1)(a2 b2) a1a2 a1b2 b1a2 b1b2. The disks are all of different radii and are initially placed on one peg in decreasing size from bottom to top. Prove using mathematical induction that (1 a)n 1 na for all n 1. …. 13. Suppose A is the matrix
12. Prove that An a 1
11. Prove that the sum of the cubes of three consecutive positive integers is always divisible by 9. b2. bn.
462 C H A P T E R 1 2
. The initial position for the game with four disks is shown in the diagram. Suppose a is a positive number. Consider the sequence with general term defined by the recursion vn 2vn 1 n for all n 1 with v1 2. n tn
1
for
a. Use mathn n n
ematical induction to prove that
i 1
(ai
bi)
i 1
ai
i 1
bi for all n
1. Prove that tn 2(n!) for all n 1. Mathematical induction can be used to provide a formal proof to statements that are obvious but difficult to prove. A sequence with general term tn is defined recursively by tn n 1 with t1 2. Prove that vn 5 2n for all n 1.
Communication
1
2
n
14. that can fit over one of three pegs.
a. Prove that every term in the expansion contains one letter from each of the n factors. and b1. a2. each with a hole in the centre. …. there are a number of disks.
Application
16. For example. What do you conclude about the statement tn n!? b.
b. an.9. Here is an example.
1 for n n2
n
d. Drawing conclusions from observed patterns can be dangerous. Using a spreadsheet or other program. Discuss the behaviour of s(n) as n gets large. How many terms are in the sequence 2k 1. 3 M AT H E M AT I C A L I N D U C T I O N
463
. 21. …. Can you conclude that tn is prime for all n c. 18. where n is the number of disks.The object of the game is to move all the disks to another peg. Disks are moved one at a time from one peg to another with the only restriction being that a larger disk can never be placed on a smaller one.) 2 1 and the partial sum t(n)
i 1 2
g(i).
b. 2 c. Suppose n straight lines are drawn in the plane so that no two are parallel and no three are concurrent (i. we need the formality of mathematical induction. Let Tn be the number of distinct regions that are formed. t2. show that t1. Verify that t40 is not prime. Prove that s(2k)
k for k 2
1. 2k 2k 1?
1
1.
n2 n 2 2
c. t39 are prime numbers. Construct two proofs to verify that 2n3 n 1. no three intersect in a common point). Therefore.
1 for all n n
1. 2n 1.
7. T2 Tn
1
1?
4. …. Show that T1 b. T3 n. Verify for any positive integer k that f (2k 1) f (2k 1 1) … f(2k 1) 1 (you do not need mathematical induction to show this). 20. (Remarkably. b. Let Mn be the minimum number of moves (a move corresponds to moving one disk from one peg to another).. Consider the sequence defined by tn 41 n n2 for n 1. a. Use mathematical induction to verify that Mn Part C
t chnology
e
17. a. t(n) approaches 6 as n
1 2 .
a. Show that Tn 2. Use mathematical induction to show that Tn 19.e. Consider the sequence f(n)
1 for n n
for all n
1. a.
3n2
n is divisible by 6 for all
n
1 and the partial sum s(n)
i 1
f(i). Show that M1 1 and M2 3. 2k
1
2. b. Consider the sequence g(n) Prove that t(n) gets large.
Now suppose we are given that in the product of n 1 binomials.Section 12. The algebraic expression a b is called a binomial. for example. where we are given that each term ti has exactly one symbol from each factor. we first need to show that this observation is true for the product of any number of binomials. The Binomial Theorem was known by the Islamic mathematician al-Karaji in the 10th century and was rediscovered by the British scientist Sir Isaac Newton in the 17th century. You will remember or can quickly deduce that (a (a (a b)1 b)2 b)3 a b a2 2ab b2 a3 3a2b 3ab2
b3
The binomial theorem gives the expansion in terms of powers of a and b for the expression (a b)n for any positive integer n. each term contains exactly one symbol from each factor. We write (a1 b1)(a2 b2) … (an 1 bn 1) t1 t2 t3 … tm. a1b2a3 has a from the first and third binomials and b from the second. The idea behind the theorem is very simple. we look at the Binomial Theorem. Here we include subscripts in the binomials so that we can see what happens in the expansion. It is remarkable that we can turn this algebra problem into a simple counting problem. each
Proof We use mathematical induction. For Step 1. also true for n 2 and n 3. the statement is true for n 1 and as shown above. For example. expanding (a b)n looks like a difficult algebra problem. At first glance. Consider. The first line of the expansion shows that the same result is also true for the product of two binomials. A bit of work gives (a1 b1)(a2 b2)(a3 b3) (a1a2 a1b2 b1a2 b1b2)(a3 b3) a1a2a3 a1a2b3 a1b2a3 a1b2b3 b1a2a3 b1a2b3 b1b2a3 b1b2b3 The key observation is that each of the terms on the right contains exactly one letter from each of the three binomials. To prove the Binomial Theorem. one of the most famous results in mathematics. Then
464 C H A P T E R 1 2
. expanding the product of three binomials (a1 b1)(a2 b2)(a3 b3).
b2)…(an
bn).
THEOREM
In the expansion of the n binomial factors (a1 b1)(a2 term contains exactly one symbol from each factor. The trick is to look carefully at how the multiplication of binomial factors works.4 — The Binomial Theorem
In this section.
Taking an a or b from each factor. the coefficient of a2b in the expansion is 1 3 3 3 . Thus the expansion simplify to an kbk. k
The first term in the expansion corresponds to selecting no bs and. Each of these selections gives a term a2b and. Thus. and b3 are 0 2 3 3 3 a 0 THE BINOMIAL THEOREM 3 2 ab 1 3 2 ab 2 3 3 b . a from each of the n factors. We use the same process to expand (a 3 b)n. There are n ways of selecting k bs from the n factors. Combining these terms we get k becomes n n n n 1 1 n n k k n n (a b)n a a b … a b … b 0 1 k n The general term is n n k k a b. We have (a b)3 (a b)(a b)(a b). the coefficients of a3. therefore. combining like terms. Having selecttwo.
12.
The expansion of (a n n a (a b)n 0
b)n is n n 1 1 a b 1
…
n n k k a b k
…
n n b n
Proof In the expansion of the Binomial Theorem (a b)n (a b)(a b) … (a b).(a1
b1)(a2
b2) … (an
bn)
[(a1 (an [t1 t1an
b1)(a2 bn) t2 … t1bn
b2)…(an tm](an … tman
1
bn–1)]
bn) tmbn
Each term in this expansion has exactly one symbol from each factor. Hence. and b3. The terms in the expansion are formed by picking either a or b from each of the n factors. This completes Step 2. there is exactly k n terms in the expansion that one way to select the (n k)as. . we can select one b in 1 ed the b. we have (a b)3 ab2.4 THE BINOMIAL THEOREM
465
. ab2. Now we look at expanding (a b)n. therefore. The question is how many? We get a2b by selecting b from one of the factors and a from the other 3 ways. we conclude that the theorem is true for all n 1. respectively. 3 . A term with k bs and (n k)as can be simplified to an kbk. Similarly. we get terms of the forms a3. The last term corresponds to choosing all bs from every factor and. For each of these. no as. Combining the two steps. We start with the special case n 3 to see how the proof works. and . a2b. there are k n n k k a b . there is only one way to choose as and that is from every factor b is not taken from. Since there are three factors.
EXAMPLE 1
Find the coefficients of x2. it is usual to leave the answer as given here. Here are some examples of the use of the Binomial Theorem.
…
Solution This identity is easy to prove using the Binomial Theorem. The coefficient of x4 is 4 4 Expressions such as this can always be simplified to a numerical value. find the value of n. respectively. we set k 4. x8.
Solution n n k k The general term in the expansion is (2x)k 2x. we get n
EXAMPLE 4
n Prove that the sum of the binomial coefficients is 0 2n. and xk are
x)20. For the coefficient of k 7 3 7 2 ( 3)4 23 . by trial and error. they are often called the binomial coefficients. 34 . 3 3 Then n(n 1)!(n 2) 20 3 n(n 1)(n 2) 120 6. and . Solution 7 The general term in the expansion is (2)7 k( 3x)k. k k n 3 n For the coefficient of x3. and xk in the expansion of (1 Solution From the binomial theorem we have 20 20 20 2 (1 x)20 x x 0 1 2 The coefficients of x2. so 2 160 or 20. x4.
EXAMPLE 3
If the coefficient of x3 in the expansion of (1
2x)n is 160. 2 8 k
EXAMPLE 2
Determine the coefficient of x4 in the expansion of (2 – 3x)7. Unless a numerical value is specifically requested. however. k 3.
…
20 k x k
…
20 20 x 20
20 20 20 .Because the symbols for n choose k appear so prominently in the expansion. . n 1 n 2 n n Solving. x8. We have n n n r n n (1 x)n x … x … x 0 1 r n
466 C H A P T E R 1 2
.
as required. we see the 0 n n = n. 0 1 n Exercise 28 at the end of this section gives another way to prove this result. reflecting the fact that n n 1 for all values of n 0. The sum of the (n 1)st row is 2n.n n n . we get 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Here we have listed the first seven rows of Pascal's triangle. corresponds to the binomial coefficients with n 4. we get an array of numbers called Pascal's triangle. This means that the fifth 0 row. It is named after its discoverer. Several patterns are apparent. Note that in general. sequence of natural numbers. If we write down the coefficients in the expansion of n n n r n n x … x … x (1 x)n 0 1 r n in a series of rows. Substitute x 1 to get (1 1)n 2n The last example gives one of many identities involving the binomial coefficients. If we replace the binomial coefficients by their values. Here are the first five rows. 0 0 1 0 2 0 3 0 4 0 We set the first row to 4 1 3 1 4 2 2 1 3 2 4 3 1 1 2 2 3 3 4 4
0 1 to preserve the symmetry.4 THE BINOMIAL THEOREM
467
. for example. one for each value of n. since 1 which restates the result of Example 4. In the interior of the triangle. the sum of the coefficients in the expansion of (a bx)n is found by setting x 1 to give (a b)n.
12. the great French mathematician and philosopher Blaise Pascal (1623–1662). we see that every number is the sum of the two closest numbers in the row above. Two sides of the triangle have only 1s.. . One number in from the edge.
then n k 1 n n + . Now we count these subsets in a second way. Every subset of size k does not contain A (Case 1) or n does contain A (Case 2). k k k 1 Since the two expressions are equal.For example. There are subsets of size k k that can be formed from the n letters other than A.
EXAMPLE 5 Pascal's Identity Theorem If n 1. A is in the subset. the coefficient of xk in the expansion of (1 x)n n 1 n n . In Exercise 25. We show below that this observation is generally 2 1 2 true. k k k 1
is
In the following exercises and problems. you are asked to consider another. the coefficient of xk is 1 1 . 1 k n. n ways. one of which is A. Equating the two counts gives the identity. There are Proof 2 Consider (1 (1 x) n 0
Proof 1 Suppose we have n
n
x)n
1
(1
…
x)(1 x)n n xk 1 k 1
n k x k
…
n n x k
1
From the binomial theorem. We provide two proofs. For case 2. k k 1 1 distinct letters. the coefficients of every power of x are n 1 n n equal. we explore more consequences of the Binomial Theorem. On the right side. It follows that .
4 3 3 .
468 C H A P T E R 1 2
. Hence the number The remaining k 1 letters can be selected in k 1 of subsets of size k that can be constructed from n 1 distinct letters is n n . k k 1
1 k subsets of size k that can be constructed from these letters. The two cases are disjoint.
Determine the coefficient of z41. prove that
is
n k 1 2x
. Prove
470 C H A P T E R 1 2
. In the expansion of (1 – x2)(1 17. By making use of the fact that (1 x)3(1 x)n (1 n n n n n 3 3 3 . a
15
contains
b.
18.
Thinking/Inquiry/ Problem Solving
20. d. Determine n. (2 3x)(1 3x)6 b. the third term is 189x2.
x)8. a. Use the fact
that S(x) is the sum of terms in a geometric sequence.
Thinking/Inquiry/ Problem Solving
22.9. the coefficient of x2 is 15. Which term in the expansion of 1 a a. In the expansion of (1 x)8 (1 (3 2x)(1
x x)12. x)n 1 nx for all n 21x 1.
15. Determine the coefficient of x4 in the expansion of a. a10 x)n. 10. Determine the coefficient of x3 in f(x) 11. the coefficient of x7 is 24. Determine the general term in the expansion of (z2 b. determine the coefficient of x5. r r 1 r 2 r 3 r 21.
14. The first three terms of a binomial expansion are 1 Determine the function that gives this expansion. Suppose r and s are roots of the quadratic equation x2 that r n sn is an integer for all values of n 1. (1 3a2
24
x
x2)(1
2x)7
13. In the expansion of (a
Communication
x)2n. 189 x2. Determine the value 2z5)10. Show that the coefficient of x k in the expansion of S(x) 1 (1 x) (1 x)2 … (1 x)n
1
x)n 3. Does z30 occur in this expansion? 16. In which term does z36 occur? e. Suppose x
0. c. Determine the first four terms in the expansion of (1 ascending powers of x. Determine the middle term.
1
0. In the expansion of (1 of n.
x2)(1
x2)8 in
2x)7. Explain why (1
19. Determine a.
12.
Use this problem to explain why n n without any calculation. 29.
x)n
(x
1)n to show that
n n–r
n for r
24. we used the binomial expansion of (1 x)n to prove that n n n … 2n. In Example 4. find the value of x
. 26.
Application
30. To see another way to prove this identity. where k k 0
m. How many binary sequences of length n with exactly k 1s can you construct? c.. How many binary sequences (a sequence with each term 0 or 1) of length n can you construct? b. Find the general term of this sequence. Suppose you have a set of n distinct objects that you want to divide into two disjoint subsets of size r and n r. If n 1. Consider Vandermonde's identity m n m n m n + k 0 k 1 k 1
1 n is 672. k
n. 0 1 n consider the following counting problems.
12. n–r r n 1 n n algebraically by 25. Use the expansion of (1 10 10 10 1 1 2 3 x)10 to prove that 10 10 … 9 10
0
27. 32. Use the fact that (1 0 r n. Show that the sum of all the entries in Pascal's triangle down to and including the nth row is 2n 1. a. 1 k n. 10. If the coefficient of x3 in the expansion of 2x n. prove that k k k 1 expanding and simplifying the binomial coefficients. 6. One diagonal (a line parallel to the edge) of Pascal's triangle begins 1. Use the expansion of (1 x)12 (1 x)12 to verify that 12 12 12 … 211 1 2 4 12 28.4 THE BINOMIAL THEOREM
471
. Combine the results of a and b to verify the identity.
m n . 3.23. Part C 31.. ….
The Binomial Theorem is true for a negative integer exponent. k k x)
1
36. Show that r1n r2n is rational for all n 1. 0.
38. Let S2(x) 1 2x 3x2 4x3 … (n 1)xn …. Prove the identity … 0 1 2 r n r 1 . Determine S1(x) xS1(x).
b. c are rational numbers. Use the fact that (1
x)m
n
(1
x)m(1
x)n to establish the identity. b. Repeat this division for (1 x) 1. recall that n n(n 1)(n 2)…(n k 1) . The expression n ( n)! has no meaning. n N. 0 k r? n n 1 n 2 n r b. (1 x) 2. However. Suppose that you have n 1 0s and r 1s. 37. How many binary sequences can you construct that end in exactly k 1s. we can see what the expansion of (1 1–x
x)
1
looks
like by dividing 1 by 1 x. r 34. 35. since the definition of ( n k)!k! k factorial requires that we use positive integers.
472 C H A P T E R 1 2
. (1 x) 2. The following questions will help you to see what happens if the exponent is negative. Do the division for five terms. Suppose that r1 and r2 are the roots of the quadratic equation ax2 bx c where a.
a. show that S1(x) (1 – x) 1. Let S1(x) 1 x x2 x3 … xn From this. and note that the series obtained appears to have an infinite number of terms.a. 33. Use the result from the previous Question to show that S2(x) (1 x) 2. Since (1
1 . Determine S2(x) xS2(x). …. Verify the identity by counting the number of subsets of size k that can be selected from a set of n red objects and m blue objects. Use this to prove that k! k n n k 1 ( 1)k .
40. Determine the first four terms and the general term for each of the following. (1
x) 2x)
2 4
c. Use n 2 in this expression for five terms and compare the expansion with the one you obtained in Question 38. (1 e. From the previous questions. (1 x) x)
1 4
b. (1 d. we can guess that the Binomial Theorem for negative exponent must give n n n (1 x) n 1 ( x) ( x)2 … ( x)k … 1 2 k n k 1 ( 1)k ( x)k k k 0 where n 1 and the series has an infinite number of terms. For Step 2. (1
x) 3x)
3 5
41. a. (1 f.39. by k 1 k x . assuming that Sn(x) (1 x)
n
n
12. Question 39 establishes the first step required for an inductive proof that the Binomial Theorem is true for a negative integer exponent. use the induction k k 0 method to prove that the theorem holds for all n 1.4 THE BINOMIAL THEOREM
473
.
the value of the coefficient in any given term in a binomial expression 8. determine. using the Binomial Theorem. evaluate sequence sums expressed using
6. interpret arithmetic and geometric sequences 4. using mathematical induction or other means. whether or not a given expression is true for all values of the variable 7. interpret recursive definitions of sequences and identify a sequence by writing out a number of terms 2. simplify expressions given in notation notation
5. determine.Key Concepts Review
You should be able to 1. use the Binomial Theorem in establishing simple combinatorial identities
474 C H A P T E R 1 2
. express sequences recursively 3.
. Investigate and Apply 1.. 3. Calculate the first six terms of the sequence given by t1 1. (2n 1)2 n 1
n
2." This is the sequence from the famous Tower of Hanoi puzzle. n 2. c. using mathematical induction.. Calculate the first 10 terms of the sequence t1 tn 2n 2 4
tn 2n
1 2 1 .
1 t .wrap-up
CHAPTER 12: RECURSIVE SEQUENCES
investigate and apply
Recursively defined sequences are used in many areas of applied mathematics where direct approaches to solving problems are not available. Calculate the squares of each of the terms from part a. tn 1 tn 1 t 2 . a. 7. as "each successive term is found by doubling and adding one. c. 3. What are the terms in the sequence from part a doing? This question demonstrates an example of Newton's Method.. Which sequence approaches pi faster?
INDEPENDENT STUDY
Investigate extensions to the Tower of Hanoi puzzle. n 2.
2 2. People are usually less likely to notice that the nth term can be found directly as 2n 1. ..
b. 4. Sometimes they provide a natural first step toward a direct solution. . . tn 1
b. 3. a. Prove.. 15... 3. Calculate the first 20 terms of the sequence t1 n 2.. that the recursive formula t1 tn 2tn 1 1.. . It is a recursive method for finding successively more accurate approximations to solutions of an equation (in this case the equation is x2 2). many people see the pattern in 1. For example. 3... is equivalent to tn 2n 1. 31.. 3. How many decimal places of pi have been computed? What method was used? What is the reason for calculating pi to so many decimal places? (For students who have studied calculus: What is Newton's Method? Why does it work?) What are recursive function calls in computer programming? How are they similar to recursively defined numerical sequences? ●
RICH LEARNING LINK WRAP-UP
475
. 2
n 1
1.. 2. .
An arithmetic sequence with general term an. The new sequences with general terms An and Gn are defined as the arithmetic average and geometric average of the first n terms a1. That is. According to the model.2% (that is.
2. b. will the population grow or shrink over time? c. the sum of the terms t1 t2 … t99 t3
1 t2
3( 1)n 1 . Determine if the corresponding sequences are arithmetic. an. Find expressions for An and Gn.
2. n 1. At the start of year 1. The emigration rate is 0. A sequence of numbers xn is defined by x1 1 and xn 2. Establish a recursive definition for un. Evaluate the first six terms of the sequence. find
b. prove that xn
6. …. a. b. 5.012 times the population size at the start of the year) and the death rate is 0. an. a2. the population is 30 000 000. An
a1 a2 n … an
. the number of births is 0. What is the value for the number of immigrants needed to maintain a constant population?
476 C H A P T E R 1 2
. Using mathematical induction. … is a sequence of positive numbers.
Gn
(a1
a2
…
an) n . … is a geometric sequence. the sum of the reciprocal of the terms t1 1
4. To model the population size of a country. The birth rate in any year is 1. geometric. …. For the geometric sequence with general term tn a. the sum of the odd numbered terms t1 c. n
1
1
a. Can a sequence be both arithmetic and geometric? Explain. n 2 t5 … … t99
1 t99
n
1. an. or neither. Repeat part a if a1. Suppose a1. b. a2. let un be the size at the end of the nth year. n 1.Review Exercise
1. n 1 has the sum of the first
10
two terms 5 and the sum of the third and fourth terms 17.
a. … is an arithmetic sequence. 3. …. Find
i 1
ai. Suppose that a1.3n
1
2
3xn 1. an.5% and the country admits 100 000 immigrants in any year. ….8%.
Explain why
n
tn
(n 2 1)n
t1. a new square is placed on top of the pile with one side the middle one third of the supporting square.7. A sequence xn is defined by x1 matical induction to prove
1 and xn
3xn
1. A point P in the plane starts at the origin (0. D2. … is constructed recursively by starting with the unit square. Using mathematical induction. b.
c. Combine the results of a and b to prove that the sum of the first n natural numbers is
i 2
i
. Establish recursions for the height and enclosed area of Dn. A second square is constructed on the top 1 of the first square 3 as shown. a.
10. At any step n. 0). di
i 2
b. Use mathe-
a. A sequence of diagrams D1.
Diagram 1
Diagram 2
At each step.
2i. 1) to the previous point. n
2. no term in the sequence is divisible by 3 b.
1
11.
1 (2n 1) (2n 1) 2n
n 1
. Show that di
n
(i 1)i. i 1 and let the differti ti 1 for all i 2. the term x2k is divisible by 4 for all k 1
REVIEW EXERCISE
477
. the point moves by adding the vector (2.. Consider the sequence with general term ti ence between two consecutive terms be di a. prove that for all positive integers n that
1 1 3 3 1 5
. How many moves are required before the point escapes the circle x2 y2 400? 8. What happens to the height and area as n gets large? 9.
then finding the standard average.. a2. an... Suddenly every student would look better against the average. we know that f (x) term in the sum and group like terms.THE CAUCHY SCHWARZ INEQUALITY
You have learned about setting up and solving many different kinds of equations in the mathematics that you have studied in school. b2. which applies to two sequences of numbers a1. an and b1. a2... The inequality is
n
aibi
i 1
2
n
n
ai2
i 1 i 1
bi2
This inequality guarantees. Mathematicians are also interested in studying inequalities. actually discovered by the Russian mathematician Bunyakovskii in 1859. that the correlation between two sets of numbers must fall between –1 and 1. This process could have interesting consequences if your mathematics teacher decided to use the geometric mean to report the class average on a test. bn. with no restrictions on positivity. and. .. Another famous inequality is the Cauchy Schwarz inequality. you may have seen the famous geometric–arithmetic inequality for a set of positive numbers a1... exponentiating with base e. Consider the function
n
f (x)
i 1
(ai
bix)2 0 for all values of x. if you calculate an average by first taking the natural logarithm of each number. finally.. we get
n n
f (x)
i 1
ai2
2
i 1
aibi x
i 1
bi2 x2
480 C H A P T E R 1 2
. Surprisingly. . . For example. you will get an answer smaller than the standard average. we can prove this inequality with the simple properties of quadratic functions. If we expand each
n
Since every term in the sum is a square. for example. which states that
n n
ln(ai) exp {
i 1
ai }
i 1
n
n
In words.
Since f (x)
i 1 n
(ai
n
bix)2
n
and every term in the sum is non-negative. b3). This surprising connection between a purely algebraic expression and the geometric notions of angle can be extended to higher dimensions (for example. Suppose we have two vectors a (a1. a3) and b (b1. f(x) is a quadratic function in x. Hence. That is
n
2
i 1
aibi
2
n
n
4
i 1
ai2
i 1
bi2
0
Simplifying and rearranging terms.
n
Equality holds only if the equation f(x)
0 has exactly one real root. rearranging the terms and taking the square root 1
a•b ab
a2b2
cos
1
In this case. For example.
E X T E N D I N G A N D I N V E S T I G AT I N G
481 481
. the corresponding quadratic equation f(x) 0 has at most one real root and the discriminant is less than or equal to 0. we have the Cauchy Schwarz inequality. the inequality is just a statement that the cosine of the angle between two vectors is always between –1 and 1. Then we can write the Cauchy Schwarz inequality as (a • b)2 or. a2. Equality is achieved when a is a multiple of b so that the angle between the vectors is 0. The cosine of the angle between two vectors is specified by exactly the same formula and is equivalent to the general version of the Cauchy Schwarz inequality. we must have ai the same multiple of bi for each term if equality holds. and the dot product and vector lengths are defined by making similar extensions. n 3). which is always greater than or equal to 0. Vectors are defined by adding additional coordinates.
The inequality and its properties have a geometric interpretation that you have already seen for n 3. b2.That is. if ai 2bi for each i. 1 i n then
i 1
aibi
2
ai2
i 1 i 1
bi2 .
a. a2. and
i 1
ai
60. …. {1. The sequence x1. 3 … (2n 1. … is defined by the recursion x1 n 2. x3 and x4. How many such sequences are there if a. Do not x
6. n n 1 n 8. the sequence starts and ends with a different digit? 4. Prove Pascal's identity r r r
n2 for all
1 . 5. A sequence of length 4 is used as a password. the sequence must have exactly three 1s? c. 1 4.Cumulative Review
CHAPTERS 10–12
1. 2. Find n(A B). 1. x2. …. …. b. prove that 1 n 1. how many are not perfect B?
2. Using mathematical induction. A binary sequence of length 6 is formed using any number of 0s or 1s. No letter or digit may be repeated. there are no restrictions? b.
60 i=1
ai
20. 3xn
1
7.n 1
9. an. 9} and any of the 26 letters selected from {a. xn r 1) n. Guess and prove a formula for xn. b. …. z}. ….
. Find the coefficient of x10 in the binomial expansion of 2x simplify your answer. In the set of numbers U squares?
3. In an arithmetic sequence a1. 1000}. The elements of the sequence can be any of the digits selected from {0. Find n(U). Let U be the set of all such passwords and A: the subset of passwords that start with a digit B: the subset of passwords that end with a digit a. What is the product rule used in counting arguments?
10 20
1 20 .
482 C U M M U L AT I V E R E V I E W C H A P T E R S 1 0 – 1 2
2. Evaluate x2. b. What passwords are in the subset A c.
Find
i 1
ai.
Explain why this argument is not correct. How many different sequences are possible if the largest term must be less than or equal to r. b. 3. To count the number of possible teams. Select one of the six grade 11 students. A mathematics club has five grade 12 students and six grade 11 students.10. 11. A sequence of length 5 is formed with terms selected from {1. …. The team must have at least one grade 11 student. Then pick the remaining two students 10 45 ways. 2. A sequence of length 7 or 8 is formed from the nine letters of the word Descartes. 1 r n?
C U M U L AT I V E R E V I E W C H A P T E R S 1 0 – 1 2
483
. A team of three students is to be picked for a contest. n} where no two terms have the same value. a club member uses the following argument. How many of these sequences have the two letters S consecutively? 12. so there are 6 45 270 from the other ten students in 2 possible teams a. Calculate the correct answer.
choose Construct. Join AD to complete the rectangle. Under Transform. and construct point E. hit multiply (*). Under the Transform menu. Then select point B. and then click on "Slope DB" in the sketch. Point on Object. and then click on "Slope CA" in the sketch. select line segment AB and point A. Slope. Can you think of another way? 2. Choose Measure. Hide Line. and then choose Transform. Note: There are a number of ways of creating a parallelogram. 2. and try moving the vertices to convince yourself that the shape remains a parallelogram when manipulated. and select point C. Join BC and AD to complete the parallelogram. Select point B and line segment AB. Measure the lengths and slopes as in Part 1 and make your hypotheses. Measure the side lengths (or distances between points) and make a hypothesis. Translate.
Part 2 — Rectangle
1. Select point B and then point C. Select line segments DB and CA and choose Measure.
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. Create another point. and then choose Perpendicular Line from the Construct menu. choose Mark Vector (B → C). Construct segments AC and BD and the point at intersection. With the line selected. choose Mark Vector (B → C). Select the line and choose Display. and then choose Transform. Translate. 3. Can you think of another way? 3. Construct a line segment and label it AB. Note: There are a number of ways of creating a rectangle. Construct line segment AB. Calculate. and then move the new point along the line to an appropriate position. E. select line segment AB and point A. and try moving the vertices to convince yourself that the shape remains a parallelogram when manipulated. the point at the intersection of the diagonals. You will have to label the new point and change the label to D by double clicking on the default name (with the label tool ). Construct segments DB and AC. 4. What is the purpose of this step? What are we hoping to find? 4. C.Part 1 — Parallelogram
1. hold down Shift ▲ .
CHAPTER 2 — REVIEW OF PREREQUISITE SKILLS.72 cm Distance A to CB = 6. Measure the lengths and slopes as in Part 1 and make your hypotheses. Hide Circle. Label the new point C and join BC to complete the rhombus. Point On Object. choose Mark Vector (A → B). Circle by Center + Point. select line segment DA and point D. What is the purpose of this step? Select the circle and choose Display. Is a rhombus the only quadrilateral that exhibits this property? By creating a random quadrilateral and moving the points. Translate. Can you think of any? 3. Under Transform. 5. Review of Prerequisite Skills. E. and then point B. try to find another quadrilateral with diagonals that share some of the properties shown here. Question 7. choose Construct.Part 3 — Rhombus
1. and choose Construct. Label this new point D and join it to A.75 cm h 2 = 2.53 cm h 3 = 1. Select point A. and then choose Transform. Why does the figure remain a rhombus when the vertices are moved around? Note: There are many other ways of creating a rhombus.44 cm
h1 P
h2
h3 B C
h 1 + h2 + h3 = 6.72 cm
(Height of Triangle)
490 T E C H N O L O G Y A P P E N D I X
. Select point A and then point B. QUESTION 7
Use Geometer's Sketchpad to investigate and construct equilateral heights as an extension to Chapter 2. With the circle still selected. 4.
A
h 1 = 2. Construct line segment AB. Construct segments AC and BD and the point at intersection. 2.
5. Use Measure. choose the Preferences option and follow these instructions: i) Under the Measurements category.
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. Construct the point of intersection of the new line and line segment BC. A and B. Create an equilateral triangle by first creating two points. h2. Select BC and point P. 2. • In the Display menu. Select point B and then Transform. Perpendicular Line. Select the three points and Construct.1 — CONGRUENCY IN TRIANGLES.
SET UP
• Choose the Graph menu. 3. Calculate to calculate the sum h1 h2 h3.1. Why does it work? Can you prove it?
SECTION 2. • Select the Snap To Grid option. Rotate. Manipulate point P and the vertices of the triangle to verify that the property continues to hold. and h3. Select point A and choose Transform. 7. select Text Format. Select point A and side BC and Measure. and join it to P. Create a point P somewhere inside the triangle. Open Geometer's Sketchpad and follow the instructions below. Label the lengths from P as h1. Select the line and use Display. Repeat this process on sides AB and AC. I N V E S T I G AT I O N 1
The following investigations will validate three different postulates we use when proving triangles congruent. Distance. and enter 60º. 6. Segment to complete the triangle. ii) Under Autoshow Labels for. then Construct. iii) Click OK. Mark Center 'A'. 4. Hide Line to simplify your sketch. select Points.
Measure the length of AB. Go to the Display menu and click Relabel Point. How? First click on the Arrow tool . Measure ∠CAB. While holding down Shift click on point C. let's create a second diagram identical to diagram CAB. Label the segment to have endpoints A and B (if they are not already). How? Click on the Line Segment tool . Here. left-click on one endpoint of the line segment to highlight it. You should see Length (Segment AB) = (your length). 2. Release the shift key and go to the Measure menu and click Angle. Using the Line Segment tool. choose the letter you want to represent your point. click away from your diagram anywhere on the sketchpad (to deselect the line). 3. 5. Choose an area away from your first diagram and do the following: 6. left-click on your line segment.
492 T E C H N O L O G Y A P P E N D I X
. Repeat the steps to label the other endpoint. How? First choose the Arrow tool. Put the on any point on the grid. draw a second line segment to another point on the grid. Draw a line segment between any two points on your grid. Click and drag your mouse to another point on the grid to make a line segment.2 cm
Now.INVESTIGATION 1 (SAS)
First you will construct two triangles. Go to the Measure menu and click Length. Now. draw a line segment from one point to another on your grid. and finally B.
Sample Sketch
A
Angle(CAB) = 53°
B
C
Length(Segment AC) = 3. From point A. Label this new endpoint C. Measure the length of AC. away from your first diagram. then A. ` 4. Now. 1.2 cm Length(Segment AB) = 3. How? First.
and finally E. How? First click on the Arrow tool. To measure DF. 13. Go to the Measure menu and click Length. Now. then D. How? To do this.2 cm Length(Segment AB) = 3. Label the endpoint F. simply click on an endpoint and drag). left-click on one endpoint of the line segment to highlight it. and at the same time make ∠FDE the same measurement as ∠CAB. 8. Complete the triangle in both diagrams by drawing in line segment CB in triangle CAB. Release the shift key and go to the Measure menu and select Angle. first select the Arrow tool. make the line segment DF the same length as AC. You should see Length (Segment DE) = (your length). 12.
Sample Sketch A Angle(CAB) = 53° B Sample Sketch F
C Length(Segment AC) = 3. From point D. and line segment FE in triangle FDE. click on point F. Now measure ∠FDE and line segment DF. Label the segment having endpoints D and E. Measure the length of DE. click on it and go to the Measure menu and select Length. Deselect your points by clicking anywhere on the sketchpad. Now. First click on the Arrow tool. By dragging point F around. While holding down the shift key.7.2 cm Length(Segment DE) = 3. Go to the Display menu and click Relabel Point. 11. left-click on the line. Drag your endpoint along until its length is exactly the same as AB in your other diagram. 10.2 cm
E Angle(FDE) = 53° D Length(Segment DF) = 3.2 cm
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. Adjust the length of DE by clicking on the Arrow tool and grabbing one endpoint of your line segment (to do this. 9. Repeat the steps to label the other endpoint. draw a second line segment.
Show that Heron's formula works using your two triangles as test cases. You can use the built-in Calculate function in the Measure menu to assist you. b) What did you do early in your construction to guarantee your two triangles would be congruent? c) List all the properties that exist between two triangles that are congruent. (Click on each while holding down the shift key). DID YOU KNOW? You can also calculate the area of a triangle using Heron's formula. and then • Choose the Graph menu. iii) Go back to the Measure menu and choose Perimeter. c are the side lengths of your triangle and .Answer the following questions: 1. I N V E S T I G AT I O N 2
INVESTIGATION 2 (SSS)
Open Geometer's Sketchpad. iv) Do the same for your other triangle. i) Go to the Construct menu and choose Polygon Interior. How does ∠ABC relate to ∠DEF? 4.
494 T E C H N O L O G Y A P P E N D I X
. How are the measurements of side CB and side FE related? 2. ii) Now go to the Measure menu and choose Area. a) Are these triangles congruent? Why or why not? Properly name the congruent triangles. This formula is based on the side lengths of a triangle.
SECTION 2. The formula is given by A s
a
s(s
b 2
a)(s
c
b)(s
c)
where a.1 — CONGRUENCY IN TRIANGLES. b. How does ∠ACB relate to ∠DEF? 3. Calculate the area and perimeter of each triangle by following these instructions: Select all three points of your triangle.
Using the line segment tool. Label the segment having endpoints D and E. Go to the Measure menu and click Length. draw a second line segment to another point on the grid. How? Click on the Line Segment tool . Measure the length of AC. Now. Here. Click and drag your mouse to another point on the grid to make a line segment. 7. left-click on one endpoint of the line segment to highlight it. Label this new endpoint C. 5. Using the Line Segment tool. connect B to C with a line segment. 2. Measure side BC.
TECHNOLOGY APPENDIX
495
.2 cm
C
4. ii) Under Autoshow Labels for. Choose an area away from your first diagram and follow these instructions: 6. How? First choose the Arrow tool. left-click on your line segment. Measure the length of AB. Now. Label the segment to have endpoints A and B. You should see Length(Segment AB) = (your length). iii) Click OK. let's create a second diagram identical to diagram CAB. draw a line segment between any two points on your grid. select Points. How? First click on the Arrow tool . choose the letter you want to represent your point. 3. choose the Preferences option and follow these instructions: i) Under the Measurements category. Finally.• Select the Snap To Grid option. • In the Display menu. select Text Format. Repeat the steps to label the other endpoint.
Length(Segment AB) = 4. Now.0 cm A
Sample Sketch F B
Length(Segment AC) = 3. Put the on any point on the grid. Construct two triangles using the following instructions: 1. Go to the Display menu and click Relabel Point. draw a line segment from one point to another on your grid. From point A.
From point D. 13. Go to the Display menu and click Relabel Point. Drag your endpoint along until its length is exactly the same as AC in your other diagram. 14. Measure the angles ∠BAC and ∠EDF. Adjust the length of DE by clicking on the Arrow tool and grabbing one endpoint of your line segment (to do this. How? First click on the Arrow tool.2 cm
Length(Segment DE) = 4. Measure the length of DE. Calculate the area and perimeter of each triangle by following these instructions: i) Select all three points of your triangle. simply click on an endpoint and drag). Finally. iv) Go back to the Measure menu and choose Perimeter. choose the letter you want to represent your point. 10. Here. Drag your endpoint along until its length is exactly the same as AB in your other diagram. draw a second line segment.0 cm
E
12. Measure DF. You should see Length(Segment DE) = (your length). 8. 9. move point F so that EF equals BC. Measure side EF. Repeat the steps to label the other endpoint. Adjust the length of DF by clicking on the Arrow tool and grabbing one endpoint of your line segment (to do this. left-click on the line. Now. Now. v) Do the same for your other triangle. simply click on an endpoint and drag).
496 T E C H N O L O G Y A P P E N D I X
. 11. Go to the Measure menu and click Length.How? First click on the Arrow tool. left-click on one endpoint of the line segment to highlight it. What do you notice about these angles? Predict what relationship might exist between a) ∠ABC and ∠DEF b) ∠BCA and ∠EFD Check your prediction by measuring these angles. connect E to F with a line segment. iii) Now go to the Measure menu and choose Area. Label the endpoint F. Now. ii) Go to the Construct menu and choose Polygon Interior.
D
Sample Sketch
F
Length(Segment DF) = 3.
15. Here. choose the letter you want to represent your point. Click OK. and then • Choose the Graph menu. choose the Preferences option and follow these instructions: i) Under the Measurements category. 3. 16.1) on your grid. 2. iii) Click OK. SOMETHING TO THINK ABOUT. Are these triangles congruent? Why or why not? Properly name the congruent triangles. I N V E S T I G AT I O N 3
INVESTIGATION 3 (ASA)
Open Geometer's Sketchpad. ii) Under Autoshow Labels for. What did you do early in your construction to guarantee your two triangles would be congruent? 17.1 — CONGRUENCY IN TRIANGLES. • What would be the relationship between two circles whose centre is the circumcentre of two congruent triangles? • What would be the relationship between two triangles whose vertices are the midpoints of the sides of two congruent triangles?
SECTION 2.. select Points. • In the Display menu.. Now. Go to the Display menu. and click Relabel Point. List all the properties that exist between two triangles that are congruent. Measure the length of AB. • Select the Snap To Grid option.
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. Label the segment to have endpoints A and B. select Text Format.1) and B( 5. How? Click on the Arrow tool . left-click on one endpoint of the line segment to highlight it. Repeat the steps to label the other endpoint. Construct two triangles using the following instructions: 1. draw a line segment between points A( 10. Using the Line Segment tool .
You should see Length(Segment AB) = (your length). Label this new endpoint C. 9. 1) on the grid. 1) on the grid. draw a second line segment to point ( 9. Note: We must maintain angle BAC and angle ABD when we do this. Measure ∠BAC. choose the letter you want to represent your point. left-click on your line segment.
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.0 cm Angle(BAC) = 63°
A B
Angle(ABD) = 45° Angle(AGB) = 72°
G D C
Now. draw a line segment from E(2.How? Click on the Arrow tool. let's create a second diagram that contains the same fixed angles and contained side. 4) to F(7. From point A. From point B.
Length(Segment AB) = 5. 7. 4. 10. 4) on your grid. Repeat the steps to label the other endpoint. Label the segment having endpoints E and F. How? First click on the Arrow tool. Label this new endpoint D. Here. Now. Go to the Display menu and click Relabel Point. draw a second line segment to point ( 7. Using the Line Segment tool. 5. Measure ∠ABD. and label this point G. 6. Measure ∠AGB. Click on the intersection of AC and BD with the Arrow tool. Go to the Measure menu and click Length. Let's complete the triangle by extending AC and BD until they intersect. Now. 8. We have just created two fixed angles and a contained side. left-click on one endpoint of the line segment to highlight it.
make an angle FEH that is equal to ∠BAC. Are the two triangles you created congruent? Why or why not? Name the congruent triangles. Show that the perimeter and area of each triangle is the same by constructing the Polygon Interior in ∆ABG and ∆EFK and then measuring the area and perimeter from the Measure menu. How? Click on the Arrow tool. You must maintain the angles FEH and EFJ as they were. Extend FJ and EH until they intersect.0 cm Angle(BAC) = 63° A Angle(ABD) = 45° Angle(AGB) = 72° A B J B K H F Length(Segment EF) = 5. 12. 18. Measure to check. 16. 21. Using the Line Segment tool. Click on the intersection point of FJ and EH using the Arrow tool.11. Make an angle EFJ that is equal to ∠ABD and going in the general direction of point H. Name all the corresponding congruent sides and angles in these triangles. List all the properties that exist between the two congruent triangles. 13. left-click on the line. 20. You should see Length(Segment EF) = (your length). Measure ∠EKF.0 cm Angle(FEH) = 63° Angle(EFJ) = 45°
H
J
14. What did you do early in your construction to guarantee the triangles would be congruent? 19. Label this point K.
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. Measure ∠EFJ to check. Measure the length of EF.0 cm Angle(FEH) = 63° Angle(EFJ) = 45° Angle(EKF) = 72°
G D C
17.
E F
Length(Segment EF) = 5. Go to the Measure menu and click Length. 15. Now. How does it compare with ∠AGB?
E Length(Segment AB) = 5.
At the end of this investigation you will be asked to i) list the properties of congruent quadrilaterals. Draw another line segment from ( 12. select Points. ii) explain how we prove that two quadrilaterals are congruent. 2. B( 12.
How? Using the Line Segment tool . Label the vertex at ( 9. F(1. 3) point A. 3). 2) to ( 2. Then draw a line segment from ( 9. • In the Display menu. • Select the Snap To Grid option. draw a line segment from ( 10. 1. select Text Format. 2). C( 9. 1) point B. 0) point D. G(4. 1) to ( 9. I N V E S T I G AT I O N
In this activity you will investigate the properties that exist between congruent quadrilaterals. Draw a quadrilateral having the following coordinates as its vertices: A( 10. iii) Click OK. and then • Choose the Graph menu. Create an identical quadrilateral on another part of the sketchpad using the coordinates E(3. 2) point C. ii) Under Autoshow Labels for. Open Geometer's Sketchpad. 3). choose the Preferences option and follow these instructions: i) Under the Measurements category.
F G E
H
500 T E C H N O L O G Y A P P E N D I X
. 3) to ( 12.
B C D A
3. 0). 0). 2). 1). and finally a line segment from ( 2.S E C T I O N 2 . 0). 1 — C O N G R U E N T Q U A D R I L AT E R A L S . 2 ). Label the vertices of this quadrilateral EFGH. Label the vertex at ( 10. Label the vertex at ( 2. 0) to ( 10. 1). H(11. Label the vertex at ( 12. 1). 3). D( 2.
which is also the centre of the construction circle. EXTENSION We have just shown that the two quadrilaterals ABCD and EFGH are congruent. Example 1. Measure the angles that the median forms at the top vertex to show that they are equal. 2. Is quadrilateral ABCD
quadrilateral EFGH? Why or why not?
5. 1. and joining the points on the circumference to each other and then to the centre of the circle.1. 3. Prove that the quadrilaterals ABCD and EFGH are congruent. This is the median.
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.4. Calculate the following measurements: a) ∠ABC b) ∠BCD c) ∠DAB d) ∠ABC e) AB f) BC g) CD h) DA _____________ _____________ _____________ _____________ and and and and and and and and ∠EFG ∠FGH ∠HEF ∠EFG EF FG GH HE _____________ _____________ _____________ _____________
_____________ _____________ _____________ _____________
_____________ _____________ _____________ _____________
6. Construct a midpoint on the base side of the triangle and join the midpoint to the top vertex of the triangle.1 — EXAMPLE 1
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2.
SECTION 2. What relationship exists between the area and perimeter of these corresponding quadrilaterals? 7. Calculate the area and perimeter of each quadrilateral using Polygon Interior in the Construct menu. Write out the properties that exist between corresponding congruent quadrilaterals. plotting any two points on its circumference. Construct an isosceles triangle by first constructing any circle.
3. Measure the lengths of the two sides BC and DC just constructed to show that they are equal. 2.
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.1. Measure the two angles formed by the diagonal AC and vertex C to show that they are also equal. and plot any point C on that angle bisector. Join that point to the other two points on the construction circle to form the required quadrilateral.4. Drag point C to show that these properties remain true for many such quadrilaterals.
m ECH = 43° m FCH = 43°
C
A
E
H D
F
B
QUESTION 2.7: 1.7
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties required to be proven in Question 2. Construct two equal sides of the quadrilateral AB and AD by constructing two radii of the same circle. Drag the centre of the construction circle to show that this property remains true for many isosceles triangles. Construct an angle bisector of the angle between the two equal sides constructed in step 1. 4.1.
11
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties required to be proven in Question 2. and join it to the other endpoints to form the quadrilateral with opposite sides equal.m BCA = 23° m DCA = 23° m BC = 6. 2. Construct any two joined sides of a quadrilateral using Segment in the Construct menu.11. and plot the point of intersection of the diagonals.91 cm m DC = 6.91 cm
B C
A
D
E
QUESTION 2. Plot this point.
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.1.1. Drag any vertex of the quadrilateral to show that the properties remain true for many such quadrilaterals. 4. The point of intersection of these circles is the opposite vertex of the required quadrilateral. Construct the diagonals of the quadrilateral. 1. Construct two circles with radii the lengths of the segments in step 1 and centred on the unattached endpoints of these segments. Measure the length of each part of the diagonals to show that the diagonals bisect each other. 3.
84 cm
D C
E
B A
S E C T I O N 2 .45 cm m AC = 7. Create the remaining sides of this parallelogram FGHE using the same method as in step 2.
504 T E C H N O L O G Y A P P E N D I X
. This completes the construction of the first parallelogram ABCD. Parallelogram Area Property Theorem. Create base FG equal in length to BC using a construction circle with centre F and radius BC. one above the other. To create the other parallelogram.2.04 cm m DB = 7. by first constructing a line. Plot the point of intersection of this line and the upper parallel line and label it D.81 cm AE = 7. Construct any line segment BC on the lower line.m AB = 9. 1. Construct any two parallel lines. and a line parallel to the first line through the plotted point. Construct a line parallel to AB that passes through the other endpoint C.04 cm CE = 2. 2 — PA R A L L E L O G R A M A R E A PROPERTY THEOREM
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2. 2.84 cm ED = 7. a point above the line. and join the endpoint B of this segment to any point A on the upper parallel line to form side AB.45 cm m CD = 9. 3. plot another point F on the lower parallel line.81 cm EB = 2.
and join the endpoint A of this segment to any point D on the upper parallel line to form side AD.47 cm2
A D C
Area EFGH = 32. Example 2.
Area DABC = 32. 5. This completes the construction of the parallelogram ABCD.2 — EXAMPLE 2
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2.2. 1. 2. Select each area and measure it using Area in the Measure menu. Construct a line parallel to AD that passes through the other endpoint B.47 cm2
E H
B
C
F
G
SECTION 2. Check that this property remains true for many such parallelograms by dragging one of the vertices of the first parallelogram or any parallel line. Construct any two parallel lines. Construct any line segment AB on the lower line. one above the other. a point above the line. The area will appear shaded. by first constructing a line. and a line parallel to the first line through the plotted point. Plot the point of intersection of this line and the upper parallel line and label it C. Construct the polygon interior of each parallelogram by selecting all of its vertices and using Polygon Interior in the Construct menu.
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.4. Verify that the two areas are equal.
Area EAB = 24.38 cm 2 Area DABC = 48. investigate the relationship between triangle areas rather than paralleloram areas.
506 T E C H N O L O G Y A P P E N D I X
.2
Use Geometer's Sketchpad to investigate and construct midpoints in a triangle as an extension to Question 2. Select each area and measure the area using the Measure tool.19 cm2 (Area FDE) 2 = 48. Construct the polygon interior of the triangle using the same method.2. The area will appear shaded. Construct the polygon interior of the parallelogram by selecting all of its vertices and using Polygon Interior in the Construct menu. 4. and join it to A and B to form the sides of the required triangle.2. In this case. Verify that the area of the parallelogram is twice the area of the triangle by using Calculate in the Measure menu to find the product two times the triangle area.3.2. 5. Drag any vertex of the parallelogram or either parallel line to verify that the property holds true for many parallelograms and triangles.38 cm2
C E D C
A
B
B
QUESTION 2. Construct any point E on the upper line.
3. How many hypotheses can you make from this single diagram? How many could you prove?
QUESTION 2. Hold the shift key and select Area ABC. Angle. Note which measurements change and which ones remain the same.6
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Question 2.10. Join the midpoint to the opposite vertex to form a median. 12. and then choose Measure. Construct the polygon interior of one of the smaller triangles by selecting all 3 vertices (one point being the midpoint) and using the Construct Interior tool. Manipulate the original triangle by moving point A. 2.
508 T E C H N O L O G Y A P P E N D I X
. Calculate the ratios of corresponding sides using Measure. 17. 4. you can display all of the angle measures. Select one of the vertices of the triangle and move it to various positions. 15. Add more data to your table by double clicking on the data in the table.6. Area DEF and the Ratio (Area ABC)/(Area DEF). Drag any vertex of the large triangle to check that the property holds true for many triangles. If you don't like the headings to the left of the table. What does this mean? 13.2. B. 11. 1. Calculate (as in step 8). Construct any triangle and plot the midpoint of one side. Area. A table will appear with one set of data in it. Repeat for the other. you can double click on them to give them a more meaningful name. Verify that the areas are equal. 16. The constructed interior will be shaded. By selecting three points in order and choosing Measure. Measure the area of each smaller triangle by selecting the area and using Measure. smaller triangle. 14. Add as many sets of data as you wish to the table.2. Tabulate. or C.
4.17
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Question 2. Construct an isosceles triangle using a construction circle with two radii drawn in it.14
cm2
QUESTION 2. 2.2.C
Area CDB = 32. Join these points of intersection to each other.17.
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. Show that this property remains true for many isosceles triangles by dragging any vertex of the triangle. Construct two perpendiculars from each equal side of the triangle to the opposite vertices. 1.14 cm2
D
A
B
Area DAB = 32. Construct the points of intersection of these perpendiculars with the triangle sides. Join the two points on the circumference to complete the triangle. 3. Verify that this line is parallel to the base of the triangle by measuring the slope of each line segment.2.
the midpoint of the segment. Join the two points on the circumference to complete the triangle. Construct any segment.A
E D
Slope DE = 0. 1. Plot the midpoint of the base of this isosceles triangle and join it to the opposite vertex. Plot a point anywhere on the right bisector and join it to the endpoints of the line segment.3 — The Right Bisector Theorem. Show that this property holds true in general by dragging the point on the right bisector.3 — THE RIGHT BISECTOR THEOREM
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2. Measure the lengths of the lines joining the point on the right bisector to the endpoints of the segment to verify that they are equal in length.11
B
SECTION 2.11
C B
Slope BC = 0. To demonstrate the converse. This is the right bisector line. and then the line perpendicular to the segment. through the midpoint.
510 T E C H N O L O G Y A P P E N D I X
. 4. 2. construct an isosceles triangle using a construction circle with two radii drawn in it. 3. 5.
Proof A
D
m DA = 10. 7. drag any vertex of the isosceles triangle.6. To verify that this line is the perpendicular bisector of the base. To show that this property holds true in general.28 cm
m DB = 10.28 cm
A
C
B
Proof B
A
m AEC = 90°
C E
m AED = 90°
D
B
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. measure the angles it makes with the base line.
drag P until the angles are equal and observe the equal segments. 4. 3. Construct any two points X and Y on the arms of this angle. To demonstrate Part 2.4 cm
P
YP = 2.
512 T E C H N O L O G Y A P P E N D I X
. Observe that the segments are equal.3 — Angle Bisector Theorem. Construct any angle ABC by constructing two rays with a common starting point B. Then. 1.9°
PX = 2. Observe that the angles are equal. through the points plotted on the arms of the angle. Measure the lengths of segments XP and YP and angles XBP and YBP.3 — ANGLE BISECTOR THEOREM
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2. or conversely.
Drag point P until the segments are equal and observe the equal angles. construct perpendiculars to the arms of the angle. 2.SECTION 2. To demonstrate Part 1 of the theorem. Plot the point of intersection of these two perpendiculars and label it P.4 cm
B Y C
m PBY = 13. drag point P until the angles are equal.8°
5. drag point P until the segments are equal in length.
X
A
m XBP = 13.
4 — T H E PA R A L L E L L I N E S T H E O R E M
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2. 2.3. Construct the third perpendicular bisector. Construct any triangle and plot the midpoints of each side.4 — The Parallel Lines Theorem. Show that this property holds true for many triangles by dragging any vertex of the triangle. and verify that it passes through the point of intersection of the other bisectors.QUESTION 2.11
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Question 2.
B
E
G F
A
D
C
S E C T I O N 2 . These are the perpendicular bisectors. 4.3. 3. 5. Plot the point of intersection of the perpendicular bisectors.
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. Construct any two non-intersecting line segments and a third line that crosses these two. 1.11. Construct perpendiculars to two of the sides through each midpoint. 1.
3.42°
G B
C H
m GHD = 65. To construct a triangle with one exterior angle. 5.6. Construct an angle bisector of the exterior angle ACD. Also measure the angles ABC and BAC. Drag a point on either of the non-intersecting lines until the alternate angles and the corresponding angles are equal. Drag point A until the slope of the angle bisector is the same as side AB. Observe that the two measured angles are equal. Measure any pair of alternate angles and corresponding angles. 4.
Drag point B until alternate angles are equal. draw the other two sides of the triangle AC and AB. 3.
514 T E C H N O L O G Y A P P E N D I X
. Plot the points of intersection of all the lines. 1. Measure its slope and the slope of the nonadjacent side AB.2. plot a point C on it. and observe that lines have equal slopes.074
A
F
m AGH = 65. first construct a long line segment BD.4. 2. Slope AB = -0.42°
D E
Slope CD = -0. and from that point. Show that the line segments are parallel by measuring the slopes of these line segments.6
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Question 2.4.074
QUESTION 2.
5 — The Triangle Area Property. Measure the lengths of the two triangle bases and calculate the ratio of the longer base to the shorter base using the Measure/Calculate menu. 6. Show that this property remains true for many triangles by dragging the upper line up or down and/or dragging one of the endpoints of one of the triangle bases. a point above the line. 4. Repeat this operation to construct another triangle beside the first that will have the same height. and measure the area of each triangle by selecting each interior and using the Measure/Area menu.Drag point A until lines have equal slopes. m BAC = 78°
A
Angle Bisector Slope m = 2. Construct the polygon interior of each triangle by selecting the vertices of each. 1.6
Slope AB = 1. Calculate the ratio of the larger area to the smaller area using the Measure/Calculate menu. Construct two parallel construction lines by first constructing a line. and observe that the ratios of the areas equal the ratio of the bases. parallel to the first line. 5.
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. and another line through the point. Construct the base of the first triangle by making a line segment on the lower parallel line and by joining the endpoints of this line segment to a point on the upper parallel line. 3.7
B
C
D
m ABC = 59°
m ACB = 43°
SECTION 2. and then observe that the triangle has two equal angles.5 — THE TRIANGLE AREA PROPERTY
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2. 2.
m HI = 9. and join these two points with segment ST.5 — The Triangle Proportion Theorem.5 (Area JFG)
J K
Triangles with equal heights have areas that are proportional to their bases. or conversely. drag S until the side ratios are equal and observe that the line segment ratios are also equal.5 (m FG)
E
Area KHI = 33. plot points S and T on each of two adjacent sides.
Drag point S until the ratios of side lengths are equal and observe equal slopes. Repeat for segments PT and TR. Drag point S until the slope of ST equals the slope of QR. Measure the slopes of segments ST and QR.0 cm
Q
Slope QR = -0. OR conversely. Measure the lengths of segments PS and SQ and calculate the ratio of their lengths. 3.022
R
516 T E C H N O L O G Y A P P E N D I X
.3 cm
T
PT = 1.
A F
G
H
B
I
SECTION 2.6 cm m FG = 6.068 TR TR = 5.9 cm Slope ST = -0.3 cm PS S = 1. 2. drag S until the slopes are equal and observe equal proportions. Construct any triangle PQR. indicating that these lines are parallel.068 SQ SQ = 4. 1. PS = 5.4 cm2 (Area KHI) = 1.3 cm (m HI) = 1.5 — THE TRIANGLE PROPORTION PROPERTY THEOREM
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2.022
P
PT = 5.7 cm2 Area JFG = 22. Observe that the side ratios are also equal.
5. While holding the shift key down. Calculate. Label the vertices.357
E
Slope FG = -0. 5. choose Construct. and choose appropriate display colours. B and C.SECTION 2. draw three points." Manipulate the triangle. pentagon. Choose Construct. and hexagon and calculate the area ratios.782
C
Slope HE = -0. Area. and then Construct. determine the ratio of the area of the outer triangle to the area of the "Midpoint triangle. Using Measure. Does the ratio appear to be always true? Can you prove it? 4. Point at Midpoint. Example 4. Construct the midpoint quadrilateral. 2.5 — EXAMPLE 4
Use Geometer's Sketchpad to construct and explore midpoint polygons similar to those in Section 2. Manipulate the figures by moving the vertices. You have created a "Midpoint" triangle. Segment. and after selecting points A.
Midpoints of quadrilateral form parallelogram
D F
Slope EF = 0. Construct. Segment again.782
A H B
G
Slope GH = 0. Polygon Interior and Measure. Do the same for triangle DEF. You may be surprised by what you find…. Can you make any general hypotheses regarding the ratio of areas of polygons formed this way?
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. 3.357
1.
5. DY. to finish the construction of parallelogram ABCD. a line parallel to AD through point B. and B.11. and AC. Example 4.
C D
AK = 4.5.5.11
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Question 2. Drag any vertex of the parallelogram to show that this property holds true for many parallelograms. respectively. KL. 3.QUESTION 2.3 cm KL = 4. C. Plot the point of intersection of XB with AC and label it K. iv) Plot the point of intersection of the last two lines constructed and label it C. and join points BX. and also plot the point of intersection of DY and AC and label it L. 1. v) Join points A. in that order.3 cm
A X K
L Y
B
518 T E C H N O L O G Y A P P E N D I X
.3 cm LC = 4. finally. 2. ii) Plot a point D above the construction line and construct a line through points AD. and LC to confirm that BX and DY trisect AC. D. Measure the lengths of AK. or as follows: i) Construct a construction line and construct a line segment AB on this line. Construct midpoints X and Y of sides AD and BC. Construct a parallelogram by using the method given in Section 2. iii) Construct a line parallel to AB through point D and.
Plot the point of intersection of the rays to determine the vertex of the new triangle.6 — SIMILARITY IN TRIANGLES. 4. Measure these two angles. 2. Measure the sides of the triangles. are the triangles similar? b) If two triangles are constructed such that two pairs of corresponding sides are proportional and the contained angles are equal. Using the ray tool. Measure the length of this side and the angle it makes with the base.SECTION 2. Also measure the length of the three sides of the triangle and move these measurements close to the sides they measure. Construct any small triangle with a horizontal base. I N V E S T I G AT I O N
Use Geometer's Sketchpad to explore the following problem involving similar triangles. Solution b): SAS~ 1. are the triangles similar? Write your predicted answer to each of the problems posed above. 3. with vertices that are the endpoints of this horizontal line segment. Measure the three angles of this triangle and move the measurements close to the angles they measure. and adjust the size of the angles until they equal the corresponding angles in the first triangle. Measure the three angles of this triangle and move the measurements close to the angles they measure. 2. Construct a second side for this triangle. Construct a horizontal base for a second triangle that is longer than the base of the first triangle. Construct any small triangle with a horizontal base. and calculate the ratios of corresponding sides. Construct a horizontal line segment that is longer than the base of your first triangle. Adjust the size of this angle until it equals the corresponding angle in the first triangle.
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. construct any angles directed upwards. and measure the angle at that vertex. Measure the length of this new base. Observe that this angle equals the corresponding angle in the first triangle. Solution a): AA~ 1. based on your knowledge of similar triangles. Observe that these are equal because the triangles are similar. Problem a) If two triangles are constructed such that two pairs of corresponding angles are equal. are the triangles similar? c) If two triangles are constructed such that all three pairs of corresponding sides are proportional.
3. choosing Dilate. 4. making sure that the angle equality is maintained. Measure the lengths of all the sides of both triangles and calculate the ratios of each pair of corresponding sides. showing that the sides of the triangles are proportional. Solution c): SSS~ 1. Measure the remaining angles of the second triangle and observe that the triangles are similar. Repeat this process for the other two vertices and join them with line segments to create a triangle A'B'C. Plot any point D below the small triangle. This will plot image point A'. Conclusions Write a summary of your findings in this investigation as they relate to the problem statement given. Now select point A and create the image of A under a dilation by using the Transform menu. Select point D and click on Mark Centre in the Transform menu to select this point as the centre of dilation. Adjust the length of the second side of the second triangle until the two ratios of the side lengths are equal. Complete the construction of the second triangle by joining the ends of the two sides previously constructed with a line segment. The following steps will use the dilation feature of Geometer's Sketchpad to construct a triangle with corresponding sides that are proportional. and click on OK. 2. include printouts of the sketches you made in this investigation. and entering a dilation factor of any value (Suggestion: use a a small factor like 2). 4.
520 T E C H N O L O G Y A P P E N D I X
. Observations In your report. Calculate the ratio of the length of the base of the second triangle to the length of the base of the first triangle. Also calculate the ratio of the length of the second side of the second triangle to the length of the corresponding side in the first triangle.3. Measure all the angles in both triangles to verify that the triangles are similar. Observe that these ratios are all equal. Construct any small triangle ABC.
First. plot any point D below the small triangle. Measure the lengths of the bases and altitudes of each triangle.6 — EXAMPLE 1
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2. construct any small triangle ABC.7 cm
A
G B
m AB = 4. drag any vertex of the original triangle. 5.000 (m CG) (m A'B') = 2. Example 1. choosing Dilate. Verify that the triangles are similar by measuring corresponding angles. and entering a dilation factor of any value (Suggestion: use a small factor like 2). Select point D and click on Mark Centre in the Transform menu to select this point as the centre of dilation. 4. Next. Construct the altitude from the base of each triangle by selecting the base and the opposite vertex and constructing a perpendicular. and calculate the ratios of the longer length to the shorter length for the bases and the altitudes to verify that they are equal.
C'
m C'F = 5. 1. and click on OK. Now select point A and create the image of A under a dilation by using the Transform menu.SECTION 2. 3.9 cm m CG = 2. To show that this property holds true in general. 2.5 cm (m C'F) = 2.000 (m AB)
A' C
F
B'
m A'B' = 8. Plot the points of intersection of each perpendicular with the bases of their respective triangles. One way to construct similar triangles is to use the dilation feature in Geometer's Sketchpad. This will plot image point A'.6. Repeat this process for the other two vertices and join them with line segments to create a triangle A'B'C' similar to ABC.4 cm
D
522 T E C H N O L O G Y A P P E N D I X
.
0 cm 2
A'
Area ABC = 8. Click on OK. 1. Verify that the triangles are similar by measuring corresponding angles.2 cm
D
B
= 4. This will plot image point A'.
(Area A'B'C') = 4. The interior of each triangle will be shaded. and entering a dilation factor of any small value. 6. 3. Measure each triangle's area by selecting the interior and using the Measure/Area tool. Calculate the ratio of the larger area to the smaller area.SECTION 2. iii) Select point D and click on Mark Centre in the Transform menu to select this point as the centre of dilation. Construct two similar triangles using the dilation method as follows: i) Construct any small triangle ABC. ii) Plot any point D below the small triangle.8 cm (m A'B') (m AB)
2
B'
m AB = 7.840
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.840 (Area ABC)
C'
Area A'B'C' =40.6. Construct the polygon interiors of the triangles by selecting each vertex and using the Polygon Interior tool in the Construct menu. and calculate the square of the ratio of the longer length to the shorter length using the Measure/Calculate tool. Example 2. Repeat using a different dilation factor.3 cm 2
A
C
m A'B' = 15. 4. Now select point A and create the image of A under a dilation by using the Transform menu. Verify that this ratio equals the ratio of the areas. Measure the lengths of the bases of each triangle. choosing Dilate. Repeat this process for the other two vertices and join them with line segments to create a triangle A'B'C' similar to ABC.6 — EXAMPLE 2
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Section 2. 5. 2. Show that this property holds true in general by dragging any vertex of either triangle.
900 (m AB) (Perimeter C'A'B') = 1. choosing Dilate.6. Measure the lengths of two corresponding sides of the similar triangles and calculate the ratio of the longer side to the shorter side. Observe that the ratio of the perimeters equals the ratio of the side lengths.900 (Perimeter ACB)
C'
Perimeter C'A'B' = 20. 2. Measure each triangle's perimeter by selecting the triangle's interior and using the Measure/Perimeter tool. and join them with line segments to create a triangle A'B'C' similar to ABC.0 cm
m AB = 4. and entering a dilation factor of any value. Construct the polygon interiors of the triangles by selecting each vertex and using the Polygon Interior tool in the Construct menu. 4.9 cm
A' C B'
m A'B' = 9. Click on OK. The interior of each triangle will be shaded.QUESTION 2.2 cm
B
A
Perimeter ACB = 11. Construct a pair of similar triangles using the dilation method as follows: i) First construct any small triangle ABC. Select point A and create the image of A under a dilation by using the Transform menu. Verify that the triangles are similar by measuring corresponding angles.8 cm
D
524 T E C H N O L O G Y A P P E N D I X
.9
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Question 2. ii) Plot any point D below the small triangle.
(m A'B') = 1.6.9: 1. Calculate the ratio of the larger perimeter to the smaller perimeter using the Measure/Calculate tool. 3. Repeat this process for the other two vertices. This will plot image point A'. Show that this is true in general by dragging any vertex of either triangle. iii) Select point D and click on Mark Centre in the Transform menu to select this point as the centre of dilation.
Construct a right-angled triangle ABC by first drawing segment AB and constructing a line perpendicular to AB through A. DC.000 cm2 (m BC) (m BD) = 29. plotting the point of intersection of this perpendicular and segment BC.4 cm m BC = 10.6. Show that all these relationships are true in general by dragging any vertex of the right-angled triangle. 6.6 cm m AC = 8.786 cm2 m AD = 4. Calculate the value of AD2 and verify that it equals the product (BD)(DC).4 cm
B D C A
b)iii) (m BA) 2 = 29.11B
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Question 2. Calculate the value of AB2 and verify that it equals the product (BC)(BD). 2. and labelling the point D. Show the Pythagorean theorem by calculating the sum AC2 paring it with BC2 AB2 and com-
8.
2 b)i) (m AD) = 21. 1.9 cm m BA = 5. Plot a point C anywhere on this perpendicular and construct segments AC and BC to complete the right triangle.11b. AB.QUESTION 2. and BC. 4.8 cm m DC = 7. BD.6 cm m BD = 2.269 cm2
2 b)ii) (m AC) = 79. 3.269 cm2
(m BD) (m DC) = 21.6. Measure the lengths of all the following segments using the Measure/Length tool: AD. 5. AC.785 cm2
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. Construct the altitude AD by constructing a perpendicular to BC through A. Calculate the value of AC2 and verify that it equals the product (BC)(DC). 7.786 cm2
(m BC) (m DC) = 79.000 cm2 c) (m AC) 2 + (m BA) 2 = 108.785 cm2 (m BC) 2 = 108.
6.8 cm FD = 2.12
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Question 2. and verify that the product 2(FD) equals BF using the Measure/Calculate tool. 2.12.15
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in Question 2.6.QUESTION 2. Measure segments BF and FD.4 cm FD 2 = 4.8 cm
C
D
F
A
E
B
QUESTION 2. choosing Dilate. Construct any triangle ABC and midpoints D and E of sides AC and AB.9 cm FE 2 = 5.
BF = 4. ii) Plot any point F below the small triangle. iii) Select the point F and click on Mark Centre in the Transform menu to select this point as the centre of dilation.6. Construct two similar pentagons using the dilation method as follows: i) Construct any small pentagon ABCDE. and join points C and E to form the other median CE. and entering a dilation factor of any small
526 T E C H N O L O G Y A P P E N D I X
.8 cm CF = 5.8 cm FE = 2. Show that this property is true for many triangles by dragging any vertex of the triangle. 1. 2. 1.15. Repeat these calculations for the other median. Join points B and D to form one median BD.6. Now select point A and create the image of A under a dilation by using the Transform menu. Plot the point of intersection of the medians and label it point F. respectively. 3.
Construct the polygon interiors of each pentagon by selecting all the vertices and using Construct. Construct any scalene triangle and measure the length of all three sides. and join them with line segments to create a pentagon A'B'C'D'E' similar to ABCDE. Plot the point of intersection of the perpendicular line with the opposite side. Show that this property is true for many similar pentagons by dragging any vertex of either pentagon to change the shape of the pentagon. Observe that this result equals the ratio of the areas.2 cm Area p1 = 15.6 cm
D C
A
B'
m AB = 6. 5. This will plot image point A'.
E' D'
Area p2 = 44. and calculate the square of the ratio of the longer side to the shorter side. QUESTION 15
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in the Chapter 2 Review Exercise. The interior area will be shaded. Perpendicular. Polygon Interior. Repeat for the other two altitudes. Measure the length of the altitude using Measure.value.4 cm2
(Area p2) (Area p1)
B
= 2. Repeat using a different dilation factor.5 cm2
A' E
C'
m A'B' = 10. 4. Measure the areas of each pentagon by selecting the interior and using Measure. Adjust side lengths so that none are equal (if necessary). Repeat this process for the other four vertices. 6. Construct an altitude to one side by selecting one side and the opposite vertex and using Construct. Measure the length of two corresponding sides of the two pentagons. Question 15. 2. Distance. Calculate the ratio of the larger area to the smaller area. Area. 7.
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. Click on OK.890
CHAPTER 2 REVIEW EXERCISE. 3. 1.890
2
(m A'B') (m AB)
F
= 2.
first draw any line segment. A more accurate method would be to construct an equilateral triangle by the following method. and observe that the triangle is equilateral. This may require some careful adjustments of more than one side. 4. Plot the point of intersection of the two circles. Repeat step 2 above. centred at the other endpoint. Join this point to the endpoints of the first segment to complete the construction. This method does not demonstrate that the converse is also true.9 cm AE = 10. then draw a circle with radius equal to the length of the line segment.9 cm A F
E
B
528 T E C H N O L O G Y A P P E N D I X
.20 cm m CA = 7.38 cm
D
B
E
A
F
C
Method B
C
D CF = 10. and then to construct and measure the altitudes as described earlier.3. centred at one endpoint. Adjust the side lengths of the triangle by dragging one side until all three sides are equal.9 cm DB = 10.42 cm m BC = 8. Conversely.
Method A
AE = 6. adjust the length of the altitudes by dragging one of the vertices until they are all equal.37 cm CD = 6.90 cm BF = 7.72 cm m AB = 8. and then construct another circle with the same radius. Observe that the altitude lengths are now also equal. To construct an equilateral triangle.
Join points A and D to form one median AD. and join points B and E to form the other median BE.870 cm2
F
G
A
E
C D B
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. 5. Measure the areas of the six interior triangles using Measure. of sides AC and BC. Question 17. Colour.870 cm2 Area DGB = 8. and join F to G and G to C. This will form the third median and complete the construction of the division of the triangle into six smaller triangles.870 cm2 Area GAF = 8. 1.870 cm2 Area CGD = 8. Plot the point of intersection of the medians and label it point G. 3. Show that this property holds true for many triangles by dragging any vertex of the outside triangle. 4.CHAPTER 2 REVIEW EXERCISE. Construct any triangle ABC and midpoints E and D.870 cm2 Area GFB = 8. Construct the midpoint F of side AB. 2. 6. Construct the polygon interior of each of the six small triangles by selecting its vertices and using the Construct/Polygon Interior tool. QUESTION 17
Use Geometer's Sketchpad to construct the following diagram that demonstrates the properties to be proven in the Chapter 2 Review Exercise. respectively. Observe that all six areas are equal. Area. The colour of each interior can be changed to a different colour by selecting the interior and using the Display.
Area EGA = 8.870 cm2 Area EGC = 8.
select point A. Develop a formal proof for this phenomenon. choose Construct. 1. 8. hold Shift. select the circle. and then select the midpoint and Construct. Perpendicular Line. Point on Object. Segment. Animate. Use the Circle tool any circle. 4. State a hypothesis illustrating what you have just seen. 10. Select the circle again and choose Construct. and then choosing Construct. Select line segment AB. hold the shift key. (Alternatively. similar to various examples and exercises in Section 3. and then choose Display. 9.1. Grab either A or B and swing it around the circle and witness what happens. or by selecting the points using the Arrow tool and the shift key. Point on Object again. Use the Label tool to label the newly created points A and B. Point at Midpoint. to construct
B
2. 7. With the circle selected. Join these two points using the Segment tool .SECTION 3. choose Construct. 3.
A
530 T E C H N O L O G Y A P P E N D I X
. With the segment selected. Slowly and watch it happen). 5.1 — PROPERTIES OF CIRCLES
Use Geometer's Sketchpad to investigate and construct chords in a circle. 6.
press 2nd x -1 to select EDIT. press 2nd x -1 . and so on. then row 2.) 2. and then press the button and choose the matrix to define. and then press means press the "cursor right" button). Example Define the matrix [C] to be the augmented matrix representing the system of equations below. press will clear. Note: For a negative number. 4 — S E T T I N G U P A M AT R I X U S I N G THE TI-83+
Defining a matrix
MATRX
1. use the white key in the bottom row. Input the values for row 1. Working with Matrix [A]: i) Enter the number of rows and then press ENTER . When you are finished entering values. etc.
Solution We need to define a 3
MATRX
i) Press 2nd x -1 to select the EDIT option. Input each element in the matrix by typing a value and then ENTER .
532 T E C H N O L O G Y A P P E N D I X
. ii) Scroll down using the key to [C]. type 3 and press ENTER . For the second dimension. press 2nd x -1 to display the MATRX EDIT ENTER ( menu. type 4 and press ENTER . Under NAMES. To set up your matrix. Press ENTER .
MATRX
(If you wish to define matrix [B] or [C]. iii) To define the first dimension. choose your defined matrix by scrolling down and then pressing ENTER ENTER .. Now press ENTER . To display your matrix. x 3y 2z 2x 5y z 3x 6y 2z 9 3 8 4 matrix. You will have Matrix [A] followed by its dimensions at the top of your screen. ii) Enter the number of columns and then press ENTER .S E C T I O N 8 .
QUIT
4. 3.
MATRX
2nd
MODE
and the screen
5.
MATRX
vi) Now display matrix [C] by pressing 2nd x -1 . press
and
MODE
. will reduce a matrix directly to reduced row echelon form. Begin by typing 1 and pressing ENTER .
Example Given the matrix [C] below. You should have the matrix shown below on your screen.
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.iv) Input the values of the matrix. ref( rref( will reduce a matrix to echelon form. then scroll down to [C] and press ENTER ENTER .
ENTER
. The commands are found under the MATRX MATH menu. 4 — R E D U C I N G A M AT R I X D I R E C T LY TO REDUCED ECHELON FORM OR ROW REDUCED ECHELON FORM
The TI-83+ has built-in commands that allow the user to reduce an augmented matrix directly to row echelon form or reduced row echelon form. ii) writing it in reduced row echelon form and writing the solution directly. solve the system with the TI-83+ by i) writing it in row echelon form and solving by back substitution. and then type
2nd
3 and press
QUIT
v) When you have entered your final value. and so on.
S E C T I O N 8 .
) • *row( (Used to multiply a row by a non-zero constant. EXAMPLE 1 Find the intersection of the two planes 2x 2y 5z 10 0 and 2x y 4z 7 0. You should have an augmented matrix on your screen like the one that follows. Print matrix [A] on your screen. (See Setting Up a Matrix Using the TI-83+).) • *row+( (Used to multiply a row by a non-zero constant and then add that row to another row. You saw how to make an augmented matrix to represent the planes and how to reduce this matrix through Gauss-Jordan elimination.) • row+( (Used to add rows together.4 you were introduced to the intersection of two planes and how a matrix could be used to describe the relationship of their intersection.J O R D A N E L I M I N AT I O N
In Section 8. Turn to the Chapter 8 Review Exercise and repeat question 22. First. • replace any row by the sum (or difference) of that row and a multiple of another row. For each part. 4 — S O LV I N G A M AT R I X U S I N G G A U S S . We will now use the TI-83+ to reduce the matrix and find the relationship between the planes. write the row echelon form of the matrix and the reduced row echelon form of the matrix. • interchange any rows.
Solution Define the matrix [A] as a matrix containing the coefficients from each equation.) Let us reduce the matrix given in Example 1 in Section 8.2.
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. The TI-83+ calculator provides commands to perform each of these operations. These commands are found in the MATRX MATH menu.
S E C T I O N 8 . recall that when performing Gauss-Jordan elimination we may do the following: • multiply any row by a non-zero constant. They are • rowSwap( (Used to interchange rows. Give the solution to each and the proper geometrical interpretation.4 using the TI-83+.
8. Type -1. name of the matrix. 6. 2). 7. (Remember to use the white
MATRX
( )
key on the calculator). row to be added). on the calculator. You should now have *row+( 1. follow these steps: 1. We will then replace row 2 with these new values. This says we will multiply row 1 of matrix [A] by 1 and then add this to row 2. [A]. Type 1. Close the bracket. 1 of our matrix. 5. 3.
2. Press the comma key . Type 2. Procedure
MATRX
i) Press 2nd x -1 .1. Press the comma key on the calculator. To do this we must multiply row 1 by 1 and then add this row to row 2. The format for this command is *row+( constant multiplier. 2) on your screen. Press
2nd x -1
and select [A]. ii) On your screen you should have *row+(. on our screen we want to have *row+( 1. We now want to obtain a zero in position 2.
536 T E C H N O L O G Y A P P E N D I X
. [A]. Press
ENTER
. 1. We accomplish this with the following steps: After *row+ (. 9.
4. then press
ENTER
. Press ENTER . Press the comma key. 1. So. and then cursor right to the MATH menu and down to F:*row+(. row to be multiplied.
and type 3. • Close the bracket. row to be multiplied)
MATRX
• Press
2nd
x -1
and select MATH. The format for this 3 command is *row( constant multiplier. To do this we will use the *row( command. In other words. press
MATRX
. Matrix [A] is now
2.
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.
MATRX
. You should now have *row(1/3. selecting [A].
STO➨ 2nd x -1
ENTER
ENTER
. • Press
ENTER
. • Type 2.
MATRX
Now store this result in matrix [A] by pressing and then pressing ENTER ENTER . press the • Press the comma key. select [A].)
• Press
2nd
x -1
and select [A]. we multiply row 2 by 1 . You should have the following matrix on your screen:
STO➨
2nd
x -1
. matrix name. 2) on your screen. To do this.
• Press the comma key. and then press • Type 1.You should have the following matrix on your screen:
iii) This operation does not actually change matrix [A]. and press equal to the matrix above. (This is one third. then press
ENTER
.
ENTER
• Cursor down to E:*row( . [A].
key. We now want to divide row 2 by 3.
x 4
1 t and y 2
1
3t. in all cases.
S E C T I O N 8 .The matrix on your screen should look like this:
You can now write the corresponding equations from the reduced matrix and describe the relationship between the intersection of the planes.
539
TECHNOLOGY APPENDIX
. row m) • used to multiply row m in the defined matrix by a non-zero constant. row n)
STO➨
2nd
x -1
select [your defined matrix to store in]. storing the result in row n. In this case. Remember. then
ENTER
ENTER
. 3. z t. row m. the defined matrix is not actually changed until after you perform the commands:
MATRX
1. the intersection is a line and we have written it below in parametric form. storing the result in row n.J O R D A N COMMANDS USING THE TI-83+
SUMMARY OF GAUSS-JORDAN COMMANDS USING THE TI-83+ • used to interchange rows m and n in the defined matrix. 2. rowSwap( matrix name. 4. *row+( constant. t
R
Extension: Use a 3-D graphing program such as Winplot or Zap-a-Graph and graph these planes to see their line of intersection. storing that result in row m. row n) • used to multiply row m. 4 — S U M M A RY O F G A U S S . matrix name. matrix name. in the defined matrix. row n) • used to add row m to row n in the defined matrix. by a non-zero constant and then to add row m to row n. row m. row m. row+( matrix name. *row( constant.
and 8 from Exercise 8. 7. complete questions 3.5. 2.25 and B 6 10 15 . Use the TI-83+ calculator to give the geometrical interpretation of the system of equations in question 1 of Exercise 8.
540 T E C H N O L O G Y A P P E N D I X
. 1. a b The determinant of a 2 2 matrix [A] is defined to be c d det A a b c d ad cb
Example Calculate the determinant of the matrix H Solution 1 det H (2)(0.25) 4. 25
Solution 2 Find the determinant for H using the TI-83+ graphing calculator. we will use determinants to investigate the conditions in which a system of n linear equations in n variables has a unique solution.4 and complete question 6 using the TI-83+ calculator and the row reducing commands available. Also we will look at using determinants to investigate whether sets of vectors are linearly dependent or independent. Using the TI-83+ calculator. Return to Exercise 8.EXERCISE 1. 4. we will look at the determinants of matrices having sizes 2 2 or 3 3. 3.4. For our investigation.
S E C T I O N 8 . Set up matrix H and define it as [A]. 5 — M AT R I C E S A N D D E T E R M I N A N T S
In this activity.5 ( 1)(4) det B ( 6)( 25) 0 (10)(15) 2 4 1 0.
2. Create a matrix A that contains the components of a and b written horizontally. Calculate the determinant of this matrix. 3 8 Your matrix should look like . 12 32 3. What is the determinant? ACTIVITY 2 1. Are the vectors m (6, 5) and n 3, 5 dependent? Explain. 2
2. Create a matrix A that contains the components of m and n written horizontally. Calculate the determinant of this matrix. Your matrix should look like 3. What is the determinant? ACTIVITY 3 1. Are the vectors r (12, 5) and t (13, 1) dependent? Explain. 6 5 3 5 2 .
2. Create a matrix A that contains the components of r and t written horizontally. Calculate the determinant of this matrix. 12 5 Your matrix should look like . 13 1 3. What is the determinant?
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Conclusion
What can be said about the determinant of a matrix that contains dependent vectors? Construct a theorem that describes the determinant of a matrix containing dependent vector components.
SECTION 8.5 — CRAMER'S RULE FOR 2 ✕ 2 M AT R I C E S
Determinants can be used to determine whether a system of linear equations in two variables has a unique solution. We do this using Cramer's Rule.
Cramer's Rule Given the system of equations there is a unique solution if D If D x 0, then the unique solution is
D1 ,y D D2 , where D1 D
We can calculate the determinant using the TI-83+ by defining our 3 and using the det( function under the MATRX MATH menu. Example
Determine whether the given vectors in R3 are dependent or independent. u Solution We will represent the vectors in a 3 3 matrix and find the determinant using the TI-83+. If the determinant is equal to zero, then the vectors are dependent. 1. Set up a matrix to represent these vectors and define it as [A].
MATRX
(4, 2, 5); v
( 4, 2, 9); w
(4, 6, 19)
2. Press
2nd
x -1
and select the MATH menu.
ENTER
3. Select 1:det( and press
MATRX
.
ENTER
4. Press
2nd
x -1
, select [A], and then press
.
5. Close the bracket. 6. Press
ENTER
.
As you can see, the vectors are dependent because the determinant is equal to 0. We knew this to be the case, and we can write the vectors as a linear combination of one another: 2(4, 2, 5) ( 4, 2, 9) (4, 6, 19)
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As we did with systems of equations in R2, we can use Cramer's Rule for 3 matrices to find the unique solution to a system when it exists. Cramer's Rule
3
A system of three equations in three variables has a unique solution if and only if the 3 x 3 determinant of the coefficients is not zero. If this is the case, then the unique solution of the system given in ax dx gx by ey hy cz fz iz k1 k2 k3 is x
D1 ,y D D2 ,z D D3 , D
ber, disregarding the sign. Written as x. For example, 3 3, 4 4, and 0 0. Acceleration: the rate of change of velocity with respect to time. Acute Angle: a positive angle measuring less than 90º. Algebraic Equation: an equation of the form f (x) 0 where f is a polynomial algebraic function and only algebraic operations are required to solve it. Algorithm: derived from the name of a ninth-century Persian author, Abu Ja'far Mohammed ibn al Khowarizmi. A step-by-step description of a solution to a problem. Altitude: the line segment drawn from one vertex of a triangle perpendicular to the opposite side. The three altitudes of a triangle intersect at the orthocentre. Anagram: a rearrangement of all the letters of a word to form a new word. Analog: a device that uses physical quantities rather than digits for storing and processing information. Angle: given two intersecting lines or line segments, the amount of rotation about the point of intersection (the vertex) required to bring one into correspondence with the other. Angle of Inclination (of a line): the angle α, 0 α 2 , that a line makes with the positive x-axis. Also known as the angle of slope or gradient of a line. Arc: a portion of a curve. For a circle, an arc is a portion of the curved line (circumference) that encloses the circle. A line drawn through a circle may divide the circumference into two unequal arcs: a major and a minor arc. Arithmetic Progression (Sequence): an ordering of numbers or terms where the difference between consecutive terms is a constant. Arithmetic Series: the sum of the indicated terms of an arithmetic sequence. Assumption: a statement that is to be accepted as true for a particular argument or discussion. Assymetric: unbalanced, without symmetry. Asymptote: a straight line is an asymptote of a curve if the curve and the line approach indefinitely close
together but never meet. Augmented Matrix: a matrix made up of the coefficient matrix and one additional column containing the constant terms of the equations to be solved. Axiom: a statement assumed to be true without formal proof. Axioms are the basis from which other theorems and statements are deduced through proof. Axis: a line drawn for reference in a coordinate system. Also, a line drawn through the centre of a figure. Axis of Symmetry: a line that passes through a figure in such a way that the part of the figure on one side of the line is a mirror reflection of the part on the other side. Basis Vectors: a set of linearly independent vectors such that every vector in that vector space can be expressed as some linear combination of the basis vectors. In the Cartesian coordinate system, the basis ˆ ˆ ˆ vectors i , j , and k form a basis for the two- or threedimensional spaces in which vectors exist. Biconditional Statement: a statement in which the truth of either part of the statement depends upon the truth of the other (expressed as p↔q). p is true if and only if q is true or q is true if and only if p is true. Binomial: an algebraic expression with two terms. For example, 2x 3y is a binomial. Binomial Theorem: the expansion in terms of powers of a and b for the binomial (two-termed) expression (a b)n. It was first discovered by the Islamic mathematician al-Karaji in the tenth century, and later rediscovered by Newton in the seventeenth century. Cardioid: a plane curve traced by a point on a circle rolling on the outside of a circle of equal radius. Cartesian Coordinate System: a reference system in two-dimensional space, consisting of two axes at right angles, or three-dimensional space (three axes) in which any point in the plane is located by its displacements from these fixed lines (axes). The origin is the common point from which each displacement is measured. In two-dimensional space, a set of two numbers or coordinates is required to uniquely define a position; in three-dimensional space, three coordinates are required.
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545
Cartesian (Scalar) Equation of Line: an equation
of the form Ax By C 0 where the vector (A, B) is a normal to the line. There is no Cartesian Equation of a line in three-dimensional space. Cartesian (Scalar) Equation of a Plane: an equation of the form Ax By Cz D 0 where the vector (A, B, C) is normal to the plane. Central Angle (of a circle): an angle subtended by an arc of the circle that has the centre of the circle as its vertex and the radii of the circle as its sides. Centroid: the centre of mass of a figure. The centroid of a triangle is the point of intersection of the three medians. Chord: a line segment joining points on a curve. In a circle, the maximum length of a chord is the length of the diameter. Circle: the locus of a point that moves so that it is always a constant distance (the radius) from a fixed point (the centre). Also, the set of all points in the plane that are equidistant from a fixed point.
Circle, Equation in Standard Form:
(x a)2 (y b)2 r2, where the centre of the circle is (a, b) and the radius is r.
Circle, Equation in General Form:
x2 y2 2gx 2fy c 0. The standard form of the equation can easily be derived from the general form and vice versa. Circumcentre: the point at which the perpendicular bisectors of the sides of a triangle meet, or the centre of the circumscribed circle that passes through the three vertices of the triangle. Circumference: the boundary line enclosing a figure or the length of that line. Usually applied to the boundary line of a circle, where the length of the circumference or perimeter is 2 r. Clockwise Rotation: a rotation in the same direction as the movement of the hands of a clock. Coefficient Matrix: a matrix whose elements are the coefficients of the unknown terms in the equations to be solved by matrix methods. Collinear: lying in the same straight line. Two vectors are said to be collinear if and only if it is possible to find a non-zero scalar, a, such that x au. Common Difference: the difference between any two consecutive terms in an arithmetic sequence. For example, in the sequence 5, 9, 13, 17, …, the common difference is 4.
Common Ratio: the ratio of consecutive terms in a geometric sequence. For example, in the sequence 3, 12, 48, 192, …. the common ratio is 4. Combinatorics: the branch of mathematics that deals with systematic ways of counting the number of arrangements in which a set of objects can be arranged. Commutative: the property that, for certain binary mathematical operations, the order does not matter. Addition and multiplication are commutative operations. Complement of a Set: If U is the universal set and A is any subset, the complement of A consists of all elements of U not in A. The set A plus its complement equals the universal set U. Complementary Angle: two angles are called complementary if the two angles add up to a right angle. Complex Number: a number of the form z a bi where a and b are real numbers and i 1. Complex Plane: a graphical method of representing a complex number z a bi, where the real part a is plotted along the horizontal real axis and its imaginary part b along the vertical imaginary axis of a Cartesian coordinate system. The complex number is displayed as the ordered pair (a, b) in the coordinate plane or as a vector drawn from the origin to the point (a, b). Concurrency: the condition where lines meet together at a common point. In a triangle, each of the medians, altitudes, angle bisectors, and perpendicular bisectors of the sides are concurrent. Concyclic Points: points that lie on a circle. Conditional Statement: a statement in which the first part of the statement implies the second part. Expressed as p → q (if p is true then q is true). Congruency: the condition of being equal in size and shape. Two figures are said to be congruent if one of them can be made to coincide at every point with the other by either a translation or rotation in space. Conjecture: a generalization or educated guess made using inductive reasoning. Conjugate Axis: the imaginary axis of symmetry of a hyperbola that is perpendicular to the transverse axis. Conjugate Complex Numbers: two complex numbers of the form a bi and a bi. The product of the two numbers is a real number.
546 G L O S S A RY
Converse: a statement formed from another state-
Deductive Reasoning: a method of reasoning that
ment by interchanging the subject and the predicate. The result of the interchange may not necessarily be true. For example, if p → q, the converse, q → p, may be true or false. Convex Polygon: a polygon in which each of the interior angles is less than 180º. Coordinates: a set of numbers that uniquely define the position of a point with respect to a frame of reference. Two coordinates are required in two-dimensional space; three in three-dimensional space. Coordinate System: a frame of reference used for describing the position of points in space. See Cartesian Coordinate System. Consistent: a linear system is said to be consistent if it has at least one solution. If there are no solutions, the system is said to be inconsistent. Coplanar: points or lines lying in a plane are said to be coplanar. Three points uniquely define a plane. Corollary: a theorem that follows so obviously from the proof of some other theorem that virtually no further proof is required. Cosine Law: a formula relating the lengths of the three sides of a triangle and the cosine of any angle in the triangle. If a, b, and c are the lengths of the sides and A is the magnitude of the angle opposite a, then a2 b2 c2 2bc cos A. Two other symmetrical formulas exist involving expressions for the other two sides. Counterclockwise Rotation: a rotation in the opposite direction of the movement of the hands of a clock. Cross Product (Vector): a vector quantity that is perpendicular to each of two other vectors and is defined only in three-dimensional space. Cube: the three-dimensional Platonic solid that is also called a hexahedron. The cube is composed of six square faces that meet each other at right angles, and has eight vertices and 12 edges. Cyclic Polygon: a polygon with vertices upon which a circle can be circumscribed. Since every triangle has a circumcircle, every triangle is cyclic. Cyclic Quadrilateral: a quadrilateral whose four vertices lie on a circle. The four vertices are concyclic points. Cylinder: a three-dimensional solid of circular crosssection in which the centres of the circles all lie on a single line of symmetry.
allows us to prove a statement to be true by progressing from the general to the particular. Degree: the unit of angle measure defined such that an entire rotation is 360º. The degree likely derives from the Babylonian year, which was composed of 360 days (12 months of 30 days each). The degree is subdivided into 60 minutes per degree and 60 seconds per minute since the Babylonians used a base 60 number system. Diagonal: a line connecting two non-adjacent vertices of a polygon. Diameter: a line segment joining two points on the circumference of a circle or sphere and passing through the centre. Dilatation: a transformation that changes the size of an object. Direct Proportion: two quantities x and y are said to be in direct proportion if y kx where k is a constant. This relationship is commonly written as y x. Direction Angles (of a vector): the angles that a vector makes with the x-, y-, and z-axes, respectively, where the angles lie between 0º and 180º. Direction Cosines (of a vector): the cosines of the direction angles of a vector. Direction Numbers (of a line): the components of the direction vector of a line. If the direction vector is normalized into a unit vector, the resulting components represent the direction cosines of the line. Direction Vector (of a line): a vector that determines the direction of a particular line. Discriminant: in the quadratic formula, the value under the square root sign: b2 4ac. It is used to determine the nature of the roots of an equation. Disjoint Sets: two sets that have no elements in common. If set A and set B have no elements in common, then A B 0. Displacement: a translation from one position to another, without consideration of any intervening positions. The minimal distance between two points. Distance: the separation of two points measured in units of length, or the length of the path taken between two points, not necessarily the minimal distance (displacement). Dot (Scalar) Product: the multiplication of two vectors resulting in a scalar quantity. It is calculated by multiplying the magnitude of each of the two vectors by the cosine of the angle between the vectors with the tails of the two vectors joined together.
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Element of a Set: any member of the set. Equiangular Triangles: three-sided figures where
Geographic Profiling: a field developed by mathe-
the three contained angles are equal to each other. (See Similar Triangles). Equilibrant Force: a force equal in magnitude but acting in the opposite direction to the resultant force. It exactly counterbalances the resultant force, resulting in a state of equilibrium. Equilibrium, State of: a state of rest or uniform motion of an object that will continue unless the object is compelled to change position by the action of an outside force. Equivalent: equal in value. Explicit: precisely and clearly expressed or readily observable; leaving nothing to be implied. Exponent: the notation bp means the product of p factors of b where b is the base and p the exponent of the power bp. Factorial: for any positive integer n, the product of all the positive integers less than or equal to n. Factorial n is denoted by n!. n! n(n 1)(n 2) …3.2.1. 0! is defined as 1. Fibonacci Numbers: the sequence of numbers 1, 1, 2, 3, 5, 8, 13, 21, … in which each number is the sum of the two previous ones. x1 1, x2 1, xn 1 xn xn 1 n 2. Force: a physical influence that causes a change in the direction of a physical object. Formula: a mathematical equation relating two or more quantities. Fractal: a curve of surface that contains more but similar complexity the closer one looks. Frame of Reference: a fixed arrangement used for describing the position of points or objects. Frequency: the number of occurrences within a given time period (usually 1 second) or the ratio of the number of observations in a statistical category to the total number of observations. Gaussian Elimination: a matrix method used to solve a system of linear equations, in which all elements below the main diagonal are made 0 by row reduction, and the resulting lines are considered as equations. Gauss-Jordan Elimination: a matrix method used to solve a system of linear equations, in which all elements on the main diagonal are made 1, and other elements above or below the main diagonal are made 0 using row reduction.
548 G L O S S A RY
maticians to help police determine where perpetrators of crime are likely to live. Geometric Progression (Sequence): a succession of numbers in which each consecutive number is found by multiplying the previous number by a fixed multiplier (the common ratio). Geometric Series: the indicated sum of the terms of a geometric sequence. Geometry: the branch of mathematics that deals with the shape, size, and position of figures in space. Golden Ratio: the division of a line segment AB by AP an interior point P so that AB . It follows AP PB that AP PB
1 5 2
, which is a root of x2
x
1
0.
Gradient: the slope or steepness of a line or curve. Gravity: the force of attraction exerted by one object
on another.
Hexagon: a six-sided polygon. Hypotenuse: the side opposite the right angle in a
right-angled triangle. It is always the longest of the three sides. Hypothesis: a concept that is not yet verified but that, if true, would explain certain facts or phenomena. Identity: a mathematical statement of equality that is true for all values of the variables. For example, sin2 cos2 1 is an identity, true for all values of the variable. Implicit: implied but not directly expressed. Inconsistent: a linear system of equations that has no solution. Indirect Proof (Proof by Contradiction): an approach where all possible outcomes are listed and all but one are eliminated through an intelligent reasoning process. Inductive Reasoning: a method of reasoning that allows us to prove a statement to be true by progressing from specific examples of data or collected evidence to a general conclusion. Intercept: the directed distance along an axis from the point of origin to a point of intersection of the graph of a curve with that axis. Intersection (of sets): a subset corresponding to the elements common to two sets. The intersection of set A and set B is denoted by A B and contains only elements present in both sets. Irrational Number: a real number that cannot be expressed as the ratio of two integers.
Isosceles: having two sides of equal length Iteration: a method of evaluating a function where
Newton's First Law of Motion: an object will
an initial value is calculated, and each subsequent term is calculated based on the output from the previous term. Lever Arm: the distance along the shaft from the axis of rotation to the point at which the force is applied. Linear Combination (of vectors): an expression that consists only of scalar multiples of vectors (for example, au bv cw, a, b, c R). Linear Dependence (of vectors): a set of vectors u, v, w, x... is linearly dependent if a linear combination of them (for example, au bv cw dx ...) produces the zero vector [0] and not all of a, b, c and d… are zero. Linear Independence (of vectors): a set of vectors u, v, w, x... is linearly independent if the only linear combination au bv cw dx ... that produces the zero vector [0] is the one in which all of the scalars a, b, c, d… are zero. Linear System (of equations): a set of two or more linear equations. A system of linear equations may have a unique solution, an infinite number of solutions, or no solution. Locus: a set of points that satisfy a given condition or the path traced out by a point that moves according to a stated geometric condition. See Circle. Magnitude: the property of relative size or extent. The magnitude of a vector is the length of the vector from the tail to the head. Mathematical Induction: a system of reasoning applied to certain theorems about integers, leading from specific facts to general conclusions. Matrix: a rectangular (or square) array of numbers set out in rows and columns. The numbers are called elements. The number of elements is the product of the number of rows multiplied by the number of columns. Median: the middle term of a sequence of numbers arranged in ascending order. If the sequence has an even number of terms, the median is the average of the two middle terms. Median Line of a Triangle: the line in a triangle drawn from a vertex to the midpoint of the opposite side. The three medians of a triangle intersect at the centroid.
remain in a state of rest or equilibrium unless it is compelled to change that state by the action of an external force. Normal: perpendicular; any vector that is perpendicular to a line is called the normal to the line. Obtuse Angle: an angle that measures greater than 90º and less than 180º. Origin: the point of intersection of the coordinate axes drawn in a Cartesian coordinate system. Orthocentre: the intersection of the three altitudes drawn in a triangle. Orthogonal: meeting at right angles. Palindrome: a sequence of symbols that reads the same from either end (for example, the number 1331 or the word level). Parallel: being everywhere equidistant but not intersecting. Parallelogram: a quadrilateral with opposite sides that are parallel. Parallelepiped: a box-like solid, the opposite sides of which are parallel and congruent parallelograms. Parameter: a variable that permits the description of a relation among other variables (two or more) to be expressed in an indirect manner using that variable. Parametric Equation: an equation in which the coordinates are each expressed in terms of quantities called parameters (for example, x r cos , y r sin 0). , the parameter, may assume any positive value. Pascal's Triangle: a triangular array of numbers where each number in a particular row is equal to the sum of the two numbers in the row immediately above it. The triangle was studied by Pascal (1623–1662), although it had been described 500 years earlier by Chinese mathematician Yanghui and the Persian astronomer-poet Omar Khayyám. It is known as the Yanghui triangle in China. Pentagon: a five-sided polygon. Perfect Square: a number that can be expressed as the product of two equal factors. Perimeter: the length of the boundary enclosing a figure. Permutation: an ordered arrangement or sequence of a set of elements or objects. Perpendicular: a straight line at right angles to another line.
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Plane: a flat surface, possessing the property that the line segment joining any two points in the surface lies entirely within the surface. Polygon: a closed plane figure consisting of n points (vertices) where n 3 and corresponding line segments. A polygon of three sides is a triangle; of four sides, a quadrilateral; and so on. Polyhedron: a solid bounded by plane polygons. Position Vector: a vector drawn from the origin to the point marking the head of the vector. Prime Number: a natural number (counting number) having no factors except itself and 1. The first primes are 2, 3, 5, 7, 11, 13, … . It is not common to include 1 among the prime numbers. Prism: a polyhedron with two congruent and parallel faces. Probability: the ratio of the number of favourable outcomes to the total number of possible outcomes. Projection: a mapping of a geometric figure formed by dropping a perpendicular from each of the points onto a line or plane. Proof by Contradiction (Indirect proof): an approach where all possible outcomes are listed and all but one are eliminated. Pyramid: A polyhedron with one face (the base) as a polygon and all the other faces as triangles meeting at a common vertex (the apex). A right pyramid is a pyramid for which the line joining the centroid of the base and the apex is perpendicular to the base. Pythagorean Theorem: in any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Quadrant: any one of the four areas into which a plane is divided by two orthogonal coordinate axes. Radius: a line segment that joins a point on the circumference of a circle to the centre. Ratio: a number or quantity compared with another. It is usually written as a fraction or with the symbol [:]. Rational Number: a number that can be expressed as an integer or as a quotient of integers (a fraction). Real Number: any rational or irrational number. Rectangle: A parallelogram in which the angles are right angles. Recursion: a method of defining sequences in which the first term is defined and each subsequent term is determined by a process applied to preceding terms. See Fibonacci Numbers for an example.
Reduced Row-Echelon Form: a matrix derived by the method of Gauss-Jordan elimination that permits the solution of a system of linear equations. Reflection: a transformation of a point, line, or figure that results in a mirror image of the original. Resultant Force: the single force that has the same net effect of a group of several forces. Rhomboid: a parallelogram with adjacent sides not equal. Rhombus: a parallelogram having equal sides. The diagonals of a rhombus are at right angles to each other Scalar: a quantity having magnitude only. Quantities having magnitude and direction are called vectors. Scalar Dot Product: the multiplication of two vectors resulting in a scalar quantity. It is the multiplication of the magnitude of each of the two vectors by the cosine of the angle between them as they are joined at their tails. Scalene Triangle: a triangle with no sides equal. Secant: a line segment that cuts through a circle or other figure. In a circle, the portion of the secant inside the circle is called a chord. Sector: the part of a circle that is bounded by two radii and the included arc. Segment of a Circle: the part of a circle bounded by a chord and the arc subtending the chord. Semicircle: the part of a circle bounded by the diameter and an arc. Sequence: a set of numbers arranged in order according to some rule. Series: the sum of the terms of a sequence. Set: a collection of objects. The individual members of a set, called elements, share some property or rule that determines whether each element is in the set. The elements of a set are unique. Set Theory: the systematic study of the properties of sets. Sierpinski's Triangle: a simple fractal resulting from the recursive manipulation of an equilateral triangle. Sigma Notation: a convenient method to express the sum of the terms of a sequence. For example,
n
Sn
i 1
ti, where
is the upper-case Greek S indicat-
ing sum. The i is called the index and the values 1 and n give the range (inclusive) of the index in summation.
550 G L O S S A RY
Similarity: two plane figures are similar if the angles of one, taken in order, are respectively equal to the angles of the other, in the same order, and the corresponding sides are proportional. Similar Triangles: two triangles are similar if the angles of one, taken in order, are respectively equal to the angles of the other, in the same order, and the corresponding sides are proportional. Sine Law: the theorem that relates the lengths of sides of a triangle to the sines of the angles opposite those sides. In a triangle with sides of lengths a, b, and c and angles opposite those sides A, B, and C, a b c sin A sin B sin C . Skew Lines: non-intersecting, non-parallel lines in
space. Two lines are skew if and only if they do not lie in a common plane. Slope: the steepness of a line or curve. In the plane, the slope is equal to the tan , where is the angle of inclination. Sparse System: a linear system of equations involving a large number of equations, many having coefficients equal to zero. Speed: the rate of change of distance with respect to time but without reference to direction. The average speed is the distance travelled divided by the travel time. Velocity is the quantity used when direction is indicated. Sphere: the set of points in space at a given distance (the radius) from a fixed point (the centre). In Cartesian coordinates, the equation of a sphere is x2 y2 z2 r2. Square: a rectangle having all sides equal. Statistics: a branch of mathematics dealing with the systematic collection and arrangement of large numbers of observations together with ways of drawing useful conclusions from such data. Subset: a subset A of a set B is a set whose elements are all elements of B. A is called a proper subset if it does not contain all the elements of B. If it contains all the elements of B, it is called an improper subset of B. Supplementary Angles: two angles whose sum is two right angles or 180º. Symmetric Equation (of a line): the equation of a line determined by eliminating the parameter from the parametric equations of a line. Symmetry: an attribute of a shape; exact correspondence of form on opposite sides of a dividing line (axis of symmetry) or plane.
Tangent: a line segment drawn to a figure that touches that figure at one and only one point. Theorem: a statement that has been proved to be true, provided certain hypotheses (axioms) are true. Such a statement might not be deemed to be a theorem unless it is considered worthy of special attention by the mathematics community. Torque: the action of a force that causes an object to turn rather than to change position. Transformation: a change in the size, shape, or position of a figure. Examples of transformations are translations, reflections, rotations, and dilatations. Translation: a transformation that changes only the position of a figure. A transformation that maps each point (x, y) on the figure to a new point (x a, y b) where a and b are components of the translation vector (a, b). Transpose (of a matrix): the interchange of rows and columns in a matrix. Row 1 becomes Column 1, and so on. Transverse Axis: the real axis of symmetry of a hyperbola. See Conjugate Axis. Trapezium: a quadrilateral with neither pair of opposite sides parallel. See Trapezoid. Trapezoid: a four-sided planar figure in which two of the opposite sides are parallel. Trigonometry: the study of the properties of trigonometric functions and their applications to various mathematical problems. Trigonometric Functions: the sine (sin), cosine (cos), tangent (tan), and their inverses, cosecant (csc), secant (sec), and cotangent (cot). Also called circular functions. Union (of sets): the set of elements made up of the elements of a pair of sets. The union of set A and set B is denoted by A B and contains all elements present in both sets. Elements found in each set are found only once in the resulting set. Unit Vector: a vector with a magnitude of 1. Such vectors are denoted with a carat [ˆ] sign placed over ˆ ˆ ˆ the symbol. For example, i , j , and k are unit vectors in the direction of the x-, y-, and z-axes. Universal Set: the set of all possible elements of the type being counted. The universal set is designated as U. Variable: a quantity, represented by an algebraic symbol, that can take on any one of a set of values.
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Vector Cross Product: a vector quantity that is perpendicular to each of two other vectors and is defined only in three-dimensional space. the boundary of the universal set is a rectangle. to enable the study of common properties of many different mathematical objects. Vector Space: an abstract system. The direction of the vector is the direction of the arrow drawn from the tail to head in reference to the basis vectors of the coordinate system. A polynomial of the nth degree has n zeros or roots. Sets and subsets are represented as circles. including vectors.
552 G L O S S A RY
. A directed line segment consisting of two points: the tail (initial point) and the head (end point).Varignon Parallelogram: a parallelogram formed
by joining the midpoints of the four sides of a quadrilateral. Zero Matrix: a matrix in which all the elements are zero. Velocity (Relative): the velocity of an object that an observer measures when he perceives himself to be stationary (at rest). Its direction is undefined. Weight: the vertical force exerted by a mass (of a body) as a result of the force of gravitation. The distance between the tail and the head is the magnitude of the vector. first developed by Peano. Zero Vector: the zero vector [0] has zero magnitude. Zero of a Function: the value(s) of x for which the function f(x) 0. Velocity: the distance travelled per unit time where the direction as well as the magnitude (speed) is important. Venn Diagram: a picture used to display the universal set and the relationship between selected subsets. Work: the action of a force on an object causing a displacement of the object from one position to another. Vector: a quantity possessing magnitude and direction. | 677.169 | 1 |
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Unformatted text preview: 18.022 Lecture notes The course is divided into 6 parts: Part 1 (Lectures I VII): Euclidean Spaces and Vector Algebra Part 2 (VIII XIII): Differential Calculas for Scalar Fields and Functions of Several Real Variables. Part 3 (XIV XVII): Multiple Integrals Part 4 (XVIII XXII): Line Integrals and Surface Integrals in Scalar and Vector Fields Part 5 (XXIII XXVIII): Vector Integral Calculus in Two and Three Dimensions Part 6 (XXIX XXXV): Linear Algebra in Multivariable Calculus Lecture I The Three-dimensional Space We refer to the Euclidean two- and three-dimensional spaces as E 2 and E 3 , respectively. Euclidean spaces have a measure of distance between points; for every two points P and Q, we denote it by d(P, Q). This measure satisfies the following two laws: i) For any two given points P and Q, d(P, Q) = 0 if and only if P = Q. ii) For any three given points P, Q and R, d (P, R) d (P,Q) + d (Q, R)....
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This note was uploaded on 09/24/2011 for the course MATH 1802 taught by Professor Duorg during the Two '04 term at Macquarie. | 677.169 | 1 |
Mukilteo PrecalculusMax H.Namrata P.
...It allows one to perform numerical calculations much faster compared to programming languages like C, C++ and Java. We can visualize the results using graph plotting. I have been using Matlab software to solve mathematical problems such as Linear equations, non-linear equations, First order differential equations, second order differential equations.
Spencer H | 677.169 | 1 |
NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002 Visit us at
Preface
We feel happy and honoured while presenting this book "Advanced Mathematics" for engineering students studying in B. Tech. IV Semester (EE and EC Branch) of Rajasthan Technical University and all Indian Universities. In this book we have presented the subject matter in very simple and precise manner. The treatment of the subject is systematic and the exposition easily understandable. All standard examples have been included and their model solutions have also been given. This book falls into five units: In first and second unit we have discussed the Numerical Analysis. The unit I deals with Finite Difference—Forward, Backward and Central difference, Newton's formula for Forward and Backward differences, Interpolation, Stirling's formula and Lagrange's interpolation formula. Solution of nonlinear equations in one variable by Newton-Raphson method, Simultaneous algebraic equation by Gauss and Regula-Falsi method, Solution of simultaneous equations by Gauss elimination and Gauss Seidel methods, Fitting of curves (straight line and parabola of second degree) by method of least squares are also discussed. In unit II, we have discussed Numerical differentiation, Numerical Integration, Trapezoidal rule, Simpson's one-third and three-eighth rules. Numerical solution of ordinary differential equations of first order, Picard's method, Euler's and modified Euler's methods. Miline's method and Runga-Kutta fourth order method, Simple linear difference equations with constant coefficients are also discussed in the unit. Unit III deals with the special functions, Bessel's functions of first and second kind, Simple recurrence relations, Orthogonal property of Bessel's transformation and generating functions. Legendre's function of first kind, Simple recurrence relations, Orthogonal property and generating function are also discussed. In unit IV, the basic principles of probability theory is given in order to prepare the background for its application to various fields. Baye's theorem with simple applications, Expected value, Theoretical probability distributions—Binomial, Poisson and Normal distributions are discussed. Unit V deals with Lines of regression, concept of simple Co-relation and Rank correlation. Ztransforms, its inverse, simple properties and application to difference equations are also discussed. We are grateful to New Age International (P) Limited, Publishers and the editorial department for their commitment and encouragement in bringing out this book within a short span of period. AUTHORS
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Acknowledgement
We are thankful to Prof. L. K. Maheshwari, Vice-Chancellor, Prof. R. K. Mittal, Deputy Director (Administration), Prof. G. Raghurama, Deputy Director (Academic) of Birla Institute of Technology & Science (BITS), Pilani for their encouragement and all over support in completing the book. The authors are also highly thankful to Dr. P. S. Bhatnagar, Director, BK Birla Institute of Engineering & Technology (BKBIET), Pilani for his motivation, time to time support and keen interest in the project. Dr. S. R. Singh Pundir, Reader, D. N. College Meerut, deserves for special thanks.Thanks are also due to Mr. Anil and Mr. Rahul of BKBIET for providing necessary help during the project. We also place our thanks on record to all those who have directly or indirectly helped us in completion of the project. At the last but not in the least we are very much indebted to our family members without whom it was not possible for us to complete this project in time. Thanks are also due to M/s NEW AGE INTERNATIONAL (P) LTD. PUBLISHERS and their editorial department. AUTHORS
UNIT I
ANALYSIS-I NUMERICAL ANALYSIS-I
In this unit, we shall discuss finite differences, forward, backward and central differences. Newton's forward and backward differences interpolation formula, Stirling's formula, Lagrange's interpolation formula. The unit is divided into five chapters: The chapter first deals with the forward, backward, central differences and relation between them, fundamental theorem of the difference calculus, factorial notation and examples. Chapter second deals with interpolation formula of Newton's forward, Newton's backward, Stirling's for equally width of arguments, and Lagrange's formula for unequally width of arguments. Chapter third deals with solution of linear simultaneous equation by Gauss elimination and Gauss-Seidel method. Chapter fourth deals with solution of algebraic and transcendental equation by Regula-Falsi and Newton-Raphson method. Chapter fifth deals with fitting of curves for straight line and parabola of second degree by method of least squares.
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CHAPTER
1
Calculus of Finite Differences
INTRODUCTION Numerical analysis has great importance in the field of Engineering, Science and Technology etc. In numerical analysis, we get the result in numerical form by computing methods of given data. The base of numerical analysis is calculus of finite difference which deals with the changes in the dependent variable due to changes in the independent variable.
1.1. FINITE DIFFERENCES Suppose the function y = f(x) has the values y0, y1, y2, ...... yn for the equally spaced values x = x0, x0 + h, x0 + 2h, ..... x0 + nh. If y = f (x) be any function then the value of the independent variable 'x' is called argument and corresponding value of dependent variable y is called entry. To determine the value of y and
dy for some intermediate values of x, is based on the principle of finite difference. Which requires dx three types of differences.
1.2. FORWARD DIFFERENCES The differences y1 – y0, y2 – y1, ..... yn – yn – 1 are called the first forward differences of the function y = f (x) and we denote these difference by ∆ y0, ∆ y1 ..... ∆ yn , respectively, where ∆ is called the descending or forward difference operator. In general, the first forward differences is defined by ∆yx = yx + 1 – yx The differences of the first forward differences are called the second forward differences and denoted by ∆2y0, ∆2y, etc.
3
Interpolation
INTRODUCTION Suppose y = f (x) be a function of x and y0, y1, y2, ....., yn are the values of the function f (x) at x0, x1, x2, ...., xn respectively, then the method to obtaining the value of f (x) at point x = xi which lie between x0 and xn is called interpolation. Thus, interpolation is the technique of computing the value of the function outside the given interval. If x = xi does not lie between x0 and xn then computing the value of f (x) at this point is called the extra polation. The study of interpolation depends on the calculus of finite difference. In this chapter, we shall discuss Newton-Gregory forward and backward interpolation, Lagrange's, Stirling's interpolation formula and method of finding the missing one and more term.
2.1. TO FIND ONE MISSING TERM Method 1. Suppose one value of f (x) be missing from the set of (n + 1) values (i.e., n values are given) of x, the values of x being equidistant. Let the unknown value be X. Now construct the difference table. We can assume y = f (x) to be a polynomial of degree (n – 1) in x, since n values of y are given. Now equating to zero the nth difference, we get the value of X. Method 2. Suppose one value of f (x) be missing from the set of (n + 1) values (i.e., n values are given) of x, the values of x being equidistant. Then we can assume y = f (x) to be a polynomial of degree (n – 1) in x ∴ or or or ∆n f (x) = 0 (E – I)n f (x) = 0 (En – nC1 E n–1 + nC2 E n–2 – ..... + (–1)n I) f (x) = 0 En f (x) – nC1 E n–1 f (x) + nC2 E n–2 f (x) – ..... + (– 1)n f (x) = 0
21
= 6.5 – 0.81 – 0.225 – 0.025625 f (1.6) = 5.439375. Example 8. The population of a town in the decennial census were as under estimate the population for the year 1895 and 1925.
Year x Population f(x) (In thousands) 1891 46 1901 66 1911 81 1921 93 1931 101
Solution of Linear Simultaneous Equations
INTRODUCTION Simultaneous linear equations occur in the field of science and engineering like as analysis of a network under sinusoidal steady-state condition, determination of the output of a chemical plant and finding the cost of reaction, the analysis of electronic circuits having a number of invariant element etc. We solved the system of simultaneous linear equation by matrix method or by Cramer's rule. But these methods are fail for large system. In this chapter we discuss some direct and iterative method of solutions.
The system of equations given above is said to be homogeneous if all the bi (i = 1, 2 ..... m) are zero. Otherwise, it is called as non-homogeneous system. The solution of such equations can be obtained by 1. Determinant method 2. Matrix inversion method 3. Direct methods (i) Gauss elimination method 4. Indirect methods (i) Tacobi iterative method (ii) Gauss-Seidel iterative method (iii) Relaxation method But in the present chapter we shall discuss only Gauss elimination and Gauss-Seidel method. (ii) Gauss-Jordan method (iii) Triangularization method
3.2. GAUSS ELIMINATION METHOD In this method, the unknowns from the system of equations are eliminated successively such that system of equations is reduced to an upper triangular system from which the unknowns are determined by back substitution. We proceed stepwise as follows. Consider the system of equations
Again eliminating x2 from (vi) with the help of (v). Divide (v) by 6 and then this equation is subtracted after multiplies by 5 from (vi), we get x1 – x2 + x3 = 6 6x2 – x3 = – 9 x3 = 3 Substitute the value of x3 into (viii), we get
−9+3 =–1 6 and again substitute the values of x2 and x3 into (vii), we get
Eliminating x from (ii) and (iii) equation by subtracting 6 and 20 times of first equation respectively, we get 1 1 x+ y+ z=1 ...(iv) 2 3 y+z=–6 ...(v)
16 z = – 20 3 Now eliminating y from (vi) by subtracting 5 times of (v), we get
4.3. ROOT OF THE EQUATION The value of x which satisfying the equation f (x) = 0 is called the root of the equation. The roots of the linear, quadratic, cubic, or biquadratic equations are obtained by available methods, but for transcendental equation or higher degree equation can not solved by these methods easily. So those types of equation can be solved by numerical methods such as bisection, secant, NewtonRaphson, Regula-Falsi method etc. In this chapter, we shall discuss only Regula-Falsi and Newton-Raphson method.
Thus, the first approximation of the root is given by x1 = x0 + h = x0 –
f ( x0 ) f ′( x 0 )
Similarly taking x1 as initial approximation, to be the better approximation of the root x2 is obtained as x2 = x1 –
f ( x1 ) f ′( x1 )
f ′(x0) ≠ 0
Proceeding in the same way we get better approximation of the root is given by xn+1 = xn –
f ( xn ) , f ′( x n )
n = 0, 1, 2, 3 .....
This is known as Newton-Raphson formula.
4.5. REGULA-FALSI METHOD This is the oldest method for finding the real root of the equation f (x) = 0. In this method we take two points x0 and x1 such that f (x0) and f (x1) are of opposite signs i.e., f (x0) f (x1) < 0. The root must lie in between x0 and x1 since the graph y = f (x) crosses the x-axis between these two points. Now equation of the chord joining the two points A[x0, f (x0)] and B[x1, f (x1)] is
Y A(x0, f(x0)) x3 x2 O x4 X D(x3, f(x3)) C(x2, f(x2)) B(x1, f(x1))
54
ADVANCED MATHEMATICS
y − f ( x 0 ) f ( x1 ) − f ( x 0 ) = x − x0 x1 − x 0
...(4.2)
In this method the curve between the points A [x0, f (x0)] and B[x1, f (x1)] is replaced by the chord AB by joining the points A and B and taking the point of intersection of the chord with the x-axis as an approximation to the root which is given by putting y = 0 in (4.2). Thus, we have x2 = x0 –
( x1 − x 0 ) f ( x0 ) f ( x1 ) − f ( x 0 )
If now f (x0) and f (x2) are of opposite signs, then the root lies between x0 and x2. Then replace the part of curve between the points A(x0, f (x0)) and C(x2, f (x2)) by the chord joining these points and this chord intersect the x-axis then we get second approximation to the root which is given by x3 = x0 –
( x2 − x 0 ) f ( x0 ) f ( x2 ) − f ( x0 )
Curve Fitting
INTRODUCTION Curve fitting have a great importance in the field of statistics (not only theoretical but also practical) Engineering and Science.
5.1. SCATTER DIAGRAM Let (xi, yi): i = 1, 2 ... n be n sets of numerical values of two variables x and y. If we plot these n sets on the graph then we get a diagram. This resulting diagram is called the scatter diagram.
Y
O
X
5.2. CURVE FITTING Let (x1, y1) (x2, y2) ..... (xn, yn) are the n numerical values of two variables x (independent) and y (dependent). By the scatter diagram, we get an approximate relation between these two variables called empirical law. Curve fitting means that relationship between two variable in the form of equation of the curve from the given data. In other words, the method of obtaining an equation of best fit is called curve fitting. Methods of Curve Fitting Following are some methods for fitting a curve: 1. Graphical method 3. Method of group averages 2. Method of least squares 4. Method of moments.
63
But in this chapter, we shall discuss only method of least squares for straight line and parabola.
64
ADVANCED MATHEMATICS
5.3. METHOD OF LEAST SQUARES Let us consider a set of m observations (xi, yi) : i = 1, 2, 3..... m of two variables x and y. We wish to fit a polynomial of degree n. Assume that such curve is y = a + bx + cx2 + ..... qxn of these values. ...(5.1) In order to determine a, b, c ..... q such that it represents the curve of best fit. Now we apply the principle of least squares so put x = x1, x2 ..... xm in (5.1), we get
2 n y′ = a + bx1 + cx1 .... qx1 1 2 n y′ = a + bx2 + cx2 + ..... + qx2 2
These equations are known as normal equations. On solving these normal equations, we get the values of constants a, b, c ..... q.
Note 1: If n = 1, then the curve to be fitted is a straight line y = a + bx whose normal equations are Σy = ma + bΣx Σxy = aΣx + bΣx2 which can be solved for a and b. Note 2: If n = 2, then the curve to be fitted is a parabola of second degree y = a + bx + cx2 whose normal equations are Σy = ma + bΣx + cΣx2 Σxy = aΣx + bΣx2 + cΣx3 Σx2y = aΣx2 + bΣx3 + cΣx4 which can be solved for a, b and c.
U | | V | | | W
...(5.3)
CURVE FITTING
65
5.4. WORKING RULE TO FIT A STRAIGHT LINE TO GIVEN DATA BY METHOD OF LEAST SQUARES (1) Let the equation of straight line be y = a + bx. (2) From the given data we calculate Σx, Σy, Σx2, and Σxy. (3) Putting above values in normal equations ma + bΣx = Σy aΣx + bΣx2 = Σxy. (4) Solving equations (i) and (ii) for a and b. (5) Putting the values of a and b in y = a + bx we get equation of straight line of best fit.
...(i) ...(ii)
5.5. WORKING RULE TO FIT A PARABOLA TO THE GIVEN DATA BY METHOD OF LEAST SQUARES (1) Let the equation of the parabola be y = a + bx + cx2. (2) From the given data we calculate Σx, Σy, Σx2, Σx3, Σx4, Σxy and Σx2y. (3) Putting above values in normal equations ma + bΣx + cΣx2 = Σy aΣx + bΣx2 + cΣx3 = Σxy aΣx2 + bΣx3 + cΣx4 = Σx2y. (4) Solving equations (i), (ii) and (iii) for a,b and c. (5) Putting the values of a and b in y = a + bx + cx2 we get equation of parabola of best fit.
UNIT II
ANALYSIS-II NUMERICAL ANALYSIS-II
In this unit, we shall discuss numerical differentiation, numerical integration and numerical solution of ordinary differential equation of first order. The chapter first deals with differentiation of function is solved by first approximation with the help of interpolation formula and differentiating this formula as many times as required. The chapter second deals with integration of function by Trapezoidal rule, Simpson's "1/3" rule and Simpson's "3/8" rule. Chapter third deals with solution of ordinary differential equation of first order by Euler's method, Euler's modified method, Picard's method, Milne's method, and Runge-Kutta method.
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CHAPTER
1
Numerical Differentiation
INTRODUCTION Numerical differentiation is the process of obtaining the value of the derivative of a function from a set of numerical values of that function 1. If the argument are equally spaced. (a) We will use Newton-Gregory forward formula. If we desire to find the derivative of the function at a point near to beginning. (b) If we desire to find the derivative of the function at a point near to end then we will use Newton-Gregory backard formula. (c) If the derivative at a point is near the middle of the table we apply stirling difference formula. 2. In case the argument are unequally spaced then we should use Newton's divided differenc formula.
Note 1. If we want to determine the value of the derivatives of the function near the beginning of arguments, we use Newton's forward formula. 2. If derivative required near the end of arguments, we use Newton's backward formula. 3. If derivative required at the middle of the given arguments, we apply and central difference formula. 4. And we use Newton's divided difference formula when argument are not equally spaced.
Numerical Integration
INTRODUCTION Numerical integration is the process of obtaining the value of a definite integral from a set of numerical values of the integrand. The process to finding the value of the definite integral I =
function of a single variable, is called as numerical quadrature. If we apply this for function of two variables it is called mechanical cubature. The problem of numerical integration is solved by first approximating the function f (x) by a interpolating polynomial and then integrating it between the desired limit. Thus
Solution. Divide the interval (0, 6) into six parts each of width h = 1 and compute the value of y=
1 at each point of sub-interval. These values are given below 1 + x2
1 1 + x2
1 0.5 0.2 0.1 0.0588 0.0385 0.027
6. Use Simpson's one-third rule to find the approximate area of the cross-section of a river 80 meter wide, the depth y at a distance x from one bank being given by the following table: x y 0 0 10 4 20 7 30 9 40 12 50 15 60 14 70 8 80 3
7. Evaluate
z
4
0
e x dx , by Simpson's rule using the data.
x ex
0 1
1 2.72
2 7.39
3 20.39
4 54.60
and compare it with the actual value. 8. Calculate an approximate value of the integral third rule (iii) Simpson's three-eighth rule.
z
π/ 2
sin x dx by (i) Trapezoidal rule (ii) Simpson's one-
0
NUMERICAL INTEGRATION
101
9. Use Simpson's one-third rule to prove that loge 7 is approximately 1.9587 using
Ordinary Differential Equations of First Order
INTRODUCTION In this chapter, we will discuss the important methods of solving ordinary differential equation of first order having numerical coefficients and given boundary or initial conditions
F i. e., dy = f ( x, y) given y( x ) = y I numerically. H dx K
0 0
These method also useful to solve those types of problem related to first order differential equation which cannot be integrated analytically.
dy = x2 + y2 – c2 dx Some important methods are:
3.1. EULER'S METHOD This is simplest and oldest method was devised by Euler. It illustrates, the basic idea of those numerical methods which seek to determine the change ∆y in y corresponding to a small increase in the arguments x. Consider the differential equation
dy = f (x, y) dx with initial condition y = y0 when x = x0, i.e., y(x0) = y0
y′ =
...(3.1)
102
ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER
103
We wish to solve (3.1), for the values of y at x = xi where xi = x0 + ih, i = 1, 2, 3, ...... Now integrate (3.1), we have y1 = y0 +
Equation (3.7) in which the unknown function y appears under the integral sign, is called an integral equation. In this method, the first approximation y(1) is obtained by replacing y by y0 in f(x, y) in R.H.S. of (3.7) and integrating w.r.t. x, we get i.e., y(1) = y0 +
z
x
x0
f ( x, y ) dx
...(3.7)
The second approximation y(2) is obtained by replacing y by y(1) in f(x, y) in R.H.S. of (3.7) and integrating w.r.t. x, we get y(2) = y0 +
z
x
x0
f ( x, y0 ) dx
...(3.8)
Proceeding in the same way we obtain y(3), y(4), ....., yn–1 and yn where y(n) = y0 +
x x0
with y(0) = y0 we repeat the steps till the two value of y becomes same to the desired degree of accuracy.
4.2. ORDER OF DIFFERENCE EQUATION The order of the difference equation is the difference between the largest and smallest arguments occurring in the difference equation divided by the unit of argument. Thus, the order of the difference equation =
Largest argument − Smallest argument Unit of argument
The order of the difference equation yn + 2 − 7 yn = 5 is
( n + 2) − n = 2. 1
119
120
ADVANCED MATHEMATICS
4.3. DEGREE OF DIFFERENCE EQUATION The highest degree of yn's in the difference equation is called the degree of the difference equation. Example 1. The order and degree of yn+2 + 4yn+1 + 4yn = 2n are
n+2−n = 2 and 1 respectively. 1 n+3−n = 3 and 2 1
4.4. SOLUTION OF DIFFERENCE EQUATION Any function which satisfies the given difference equation is called the solution of the difference equation. A solution in which the number of arbitrary constants is equal to the order of the difference equation is called general solution of the difference equation. A solution which is obtained from the general solution by assigning particular values is called particular solution.
Equation (4.4) is known as auxiliary equation. This equation has k roots which we take m1, m2,...,mk. Here some cases arise. Case I. When roots are all real and distinct If m1, m2 ..... mk are real and distinct roots of auxiliary equation (4.4), then the solution is yn = c1(m1)n + c2(m2)n ...... + ck(mk)n. Case II. When some of the roots are equal If two roots be equal i.e., m1 = m2, then the solution is yn = (c1 + c2n) (m1)n + c3(m3)n + ...... + ck(mk)n. If three roots be equal i.e., m1 = m2 = m3, then the solution is yn = (c1 + c2n + c3n2) (m1)n + c4(m4)n +......+ ck(mk)n If k roots be equal i.e., m1 = m2 = m3 ...... = mk , then the solution is yn = [c1 + c2n + c3n2 +.....+ ck nk–1] (m1)n. Case III. When the roots are complex number We know that if complex roots occur then they must be conjugate complex number i.e., if (α + iβ) is the root then (α – iβ) is also a root where α and β are real. Then the solution is yn = c1(α + iβ)n + c2(α – iβ)n which can be written as yn = γ n (A1 cos nθ + A2 sin nθ) where γ= θ=
α 2 + β2
tan–1 (β/α).
and A1 = c1 + c2, A2 = i(c1 – c2)
Case IV. When some of the complex roots are equal Let the root (α ± i β) becomes twice then the solution.
where a0, a1, a2, ..... an are all constants is known as non-homogeneous linear difference equation with constant coefficient. The general solution of (4.5) consists of two parts, complementary function and particular integral Complementary function is the general solution of the homogeneous equation i.e., left hand side 1 R(n). of (4.5) and particular integral = φ( E ) Rules for Obtaining the Particular Integral The particular integral (P.I.) = Case I. When R(n) = an ∴ P.I. =
= [φ (1 + ∆)]–1 nk. First, we expand [φ(1 + ∆)]–1 in ascending power of ∆ by the Binomial theorem as far as the term in ∆k, then express nk in the factorial notation and distribute it each term of the expansion. Case IV. When R(n) = an F(n), where F(n) being a polynomial in n. ∴ P.I. =
UNIT III
SPECIAL FUNCTION
Exponential, Logarithmic, trigonometric, hyperbolic, etc., are the elementary functions. While Bessel's functions, Legendre's polynomial, Laguerre polynomial, Hermite polynomial. Chebyshev polynomial, Beta function, Gamma function, Error function, etc., are the special functions. In this unit we shall discuss only two types of special functions, Bessel's and Legendre's. Chapter first deals with the Bessel's function of first and second kind, orthogonal property of Bessel's function recurrence relations and generating functions. Chapter second deals with the Legendre's function of first kind, orthogonal property of Legendre's function recurrence relations, and generating function.
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CHAPTER
1
Bessel's Functions
1.1. BESSEL'S EQUATION
d2y dy +x + (x2 – n2) y = 0 is called the Bessel's differential 2 dx dx equation. The solution of Bessel's equation is called Bessel's function. It is also known as cylindrical and spherical function.
The differential equation of the form x2 Bessel's functions are appear in (1) the elasticity (6) electrical fields (2) the fluid motion (7) potential theory (3) planetary motion (5) dynamical astronomy (4) the oscillatory motion of a hanging chain (8) Euler's theory of circular membrane.
Since the equation (1.4) is an identity, therefore the coefficients of various powers of x must be zero. ∴ Equating to zero the coefficients of the lowest power of x (i.e. xm) in (1.4), we get a0 [m2 – n2] = 0 But ∴ have a1[(m + 1)2– n2] = 0 But (m + 1)2 – n2 ≠ 0 for m = ± n given by (1.5) ∴ a1 = 0 ar+2 [(m + r + 2)2 – n2] + ar = 0 or or ar+2 = – ar+2 = – Putting r = 1 in (1.6), we get a3 = – Again equating to zero the coefficient of the general term (i.e. xk+r+2) in (1.4), we get m2 a0 ≠ 0 – n2 = 0 or m=±n xm+1) ...(1.5) in (1.4), we
Now equating to zero the coefficients of the next lowest degree term of x (i.e.
which is an identity. Equating to zero the coefficient of the lowest power of x(i.e. xm–2), we have a0m2 = 0 But a0 ≠ 0 therefore m2 = 0, a1(m + 1)2 =0 ∴ m=0 xm–1), ...(1.10) we have Now equating to zero the coefficient of the next lowest power of x give (i.e. But (m + 1) ≠ 0 for m = 0 given by (1.10) Now equating to zero the coefficient of the general term (i.e. xm+r), we have ar+2 (m + r + 2)2 + ar = 0
16. Prove that [Pn(x)] < 1 when – 1 < x < 1. 17. Let Pn (x) be the Legendre polynomial of degree n, show that for any function f (x) for which the nth derivative is continuous.
ANSWER
13. 0.
UNIT IV
STA STATISTICS AND PROBABILITY
Probability theory is used in many situation which involve an element of uncertainty. It is used to make important decision on business, to determine premiums on insurance policies, to determine demand in inventory control. Apart from it is used in Engineering. Science, Social Science, Genetics etc. Probability theory is also used in medical sciences. In this unit, we shall discuss some basic idea of probability, addition and multiplication theorems of probability, Baye's theorem with simple applications. Expected value, Theoretical probability distributions—Binomial, Poisson and normal distributions and applications of probability in many area of human activities. Lines of regression, correlation and rank-correlation are also discussed. Chapter 1 deals with basic idea of probability, theorems of probability and Baye's theorem with their applications. Chapter 2 deals with random variable, probability distribution: Binomial, Poisson and Normal distributions. In chapter 3, the lines of regression, concept of simple correlation and rank-correlation will be discussed. In the 16 Century an italian mathematician Jerome Cardon (1501–1576) wrote the first book on the subject of probability theory ''Book on Games of Chance'. Besides Cardon, Pascal (1623–1662). Fermat (1601–1665), J. Bernoulli (1654–1705). De Moiure (1667–1754), Chebychev (1821–1894), A.A. Markov (1856–1922) and A.N.Kolmogorov (1903 ......) gives outstanding contributions in probability theory. Probability theory is designed to estimate the degree of uncertainty regarding the happening of a given phenomenon. The word probable itself indicates such a situation. Its dictionary meaning is ''likely though not certain to occur''. Hence, when a coin is tossed a tail is likely to occur but may not occur. When a die is thrown, it may or may not show the number 6.
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CHAPTER
1
Theory of Probability
INTRODUCTION Before we discuss formal definition of probability, first we shall define certain terminologies and notations, which are used in defining probability.
1.1. TERMINOLOGY AND NOTATIONS 1. Random experiment: Any experiment whose outcomes cannot be predicted or determined in advance is a random experiment. OR A random experiment is an experiment whose outcomes or result is not unique and therefore cannot be predicted with certainty. Tossing of a coin, head or tail may occur, throwing a die, 1, 2, 3, 4, 5 or 6 may appear and drawing a card from a well shuffled pack of cards are examples of random experiments. 2. Sample space and sample point: The set of all possible outcomes of a random experiment is called the sample space and each element of a sample space is called a sample point. If a die is thrown, it will land with anyone of its 6 faces pointing upward, resulting in anyone of the numbers 1, 2, 3, 4, 5, 6 appearing on the top face. Hence, the number of each face is a possible result. We write S = {1, 2, 3, 4, 5, 6} The set S is a sample space of throwing a die and 1, 2, 3, 4, 5, 6 are called sample point. When two dice is thrown, then the sample space consists of the following 36 points:
LM (1, 1), 1), MM((2,, 1), 3 S = ( 4, 1), MM(5, 1), NM(6, 1),
(1, 2), ( 2, 2), (3, 2), ( 4, 2), ( 5, 2), (6, 2),
(1, 3), (2, 3), (3, 3), ( 4, 3), ( 5, 3), (6, 3),
179
(1, 4), (2, 4), (3, 4), ( 4, 4), (5, 4), (6, 4),
(1, 5), ( 2, 5), (3, 5), ( 4, 5), ( 5, 5), (6, 5),
(1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)
OP PP PP QP
180
ADVANCED MATHEMATICS
Note. The set of all possible outcomes of a single performance of a random experiment is exhaustive events or sample space.
3. Event: Any subset A of a sample space S is called an event. Consider the experiment of tossing of a coin, we have S = {H, T} Here are 4 subsets of S, φ, {H}, {T}, {H, T}. Each subset of S is an event. 4. Simple event: An event is said to be simple event, if it has only one sample point. In tossing of a coin, the events {H} and {T} are simple events. 5. Compound event: An event is said to be compound event, if it has more than one sample point. When two coins are tossed once then sample space (S) = {HH, HT, TH, T T}. Each event of S is compound event. 6. Complement of an event: The complement of an event A with respect to the sample space S is the set of all elements of S which are not in A.
S
A =S–A
7. Favourable events:The number of cases favourable to an event in a trial is the number of outcomes which entail the happening of the event. For example, in throwing of two dice, the number of cases favourable to getting the sum 6 is (1, 5), (2, 4), (3, 3), (3, 3), (4, 2), (5, 1), i.e. 6. 8. Equally likely events: Events are said to be equally likely, if there is no reason to expect any one in preference to any other. For example, In throwing a die, all the six faces are equally likely to come. 9. Mutually exclusive events: Two events A and B are said to be mutually exclusive events iff A ∩ B = φ. If a die is thrown, we will get either an 'even number' or an 'odd number'. The event A ''getting an even number'' and event B "getting an odd number''. We say that A and B are mutually exclusive events because both the events cannot occur at the same time. We have A = {2, 4, 6} Then, A∩B=φ and B = {1, 3, 5}
In tossing of a coin, we will get either a head or tail. The event A ''getting a head'' and event B ''getting a tail''. We say that A and B are mutually exclusive events because both the events cannot occur at the same time. We have Then, A = {H} and B = {T} A∩B=φ
Note. 1. Simple events of a sample space are always mutually exclusive. 2. The events, which ensure the required happening is called favourable events.
10. Odd in favour of an event and odd against an event: Let there are m outcomes favourable to a certain event and n outcomes are not favourable to the event in a sample space, then
THEORY OF PROBABILITY
181
odd in favour of the event = and odd against the event =
m n n . m
11. Permutation: A permutation is an arrangement of objects in a definite order. The number of permutations of n objects used r at a time, denoted by nPr is
nP r
=
n! (n − r ) !
12. Combination: A combination is a selection of objects without regard to order. The number of combinations of n distinct objects selected r at a time, denoted by nCr is
nC r
=
n! r ! (n − r ) !
1.2. DEFINITIONS The chance of happening of an event when expressed quantitatively is called probability. However, we give three definitions of probability: (1) Classical or Mathematical definition of probability. (2) Statistical or Empirical definition of probability. (3) Axiomatic definition of probability. 1. Classical or Mathematical definition of probability: If an event A can happen m ways out of possible n mutually exclusive and equally likely outcomes of a random experiment then probability of A is P(A) =
Number of favourable cases m = n Total numbers of possible cases
The probability of not happening of A is P( A ) = 1 – Thus,
Note. 1. 2. 3. 4.
n−m m = n n
P(A) + P( A ) = 1, 0 ≤ P(A) ≤ 1 and 0 ≤ P( A ) ≤ 1.
If P(A) = 1, then A is called certain event. If P(A) = 0, then A is called impossible event. This definition is fail when outcomes are not equally likely and number of outcomes is infinite. The probability of an event is a number between 0 and 1. If occurrence is certain, its probability is 1. If the event cannot occur, its probability is 0.
182
ADVANCED MATHEMATICS
2. Statistical or Empirical definition of probability: If trial be repeated for a large number of times, say n, under the same conditions, and a certain event A happen on m times then the probability of the event A is P(A) = lim where, the limit is unique and finite. 3. Axiomatic definition of probability: Let S be a finite sample space and A be any event in S, then probability of A is defined by following three conditions. (i) For every event A in S, 0 ≤ P(A) ≤ 1 (ii) P(S) = 1. (iii) If A and B are mutually exclusive events in S, then P(A ∪ B) = P(A) + P(B).
n→ ∞
Theorem 2: The probability of the complementary event A of A is given by P( A ) = 1 – P(A). Proof: Since, A and A are mutually disjoint events, so that A∪ A =S ⇒ ⇒ ⇒ ⇒ P(A ∪ A ) = P(S) P(A) + P( A ) = P(S) =1 P( A ) = 1 – P(A) P( S ) = P(φ)
Note. 1. If φ is the impossible event and S is the sample space then S = φ
SOLVED EXAMPLES
Example 1: What is the chance that a leap year, selected at random, will contain 53 Sundays? Solution: We know that a leap year consists of 366 days and contains 52 complete weeks and 2 days over. Combinations of these two days are as follows: (1) Monday, Tuesday (3) Wednesday, Thursday (5) Friday, Saturday (7) Sunday, Monday Of these seven P(S) = 7 are likely cases only and last two are favourable i.e. n(A) = 2. The required probability = (2) Tuesday, Wednesday (4) Thursday, Friday (6) Saturday, Sunday
n( A) 2 = . n( S ) 7
Example 2: From a pack of 52 cards, one is drawn at random. Find the probability of getting a king. Solution: From a pack of 52 cards 1 card can be drawn in 52 ways, i.e. n(S) = 52. Number of ways in which a card can be king = 4 i.e. n(A) = 4 The required probability =
n( A) 4 1 = = . n( S ) 52 13
186
ADVANCED MATHEMATICS
Example 3: What is the probability of throwing a number greater than 3 with an ordinary die whose faces are numbered from 1 to 6. Solution: There are 6 possible ways in which the die can fall, and all of these there (4, 5, 6) are favourable to the event required i.e. ⇒ n(S) = 6 P(> 3) = and n(A) = 3
n( A) 3 1 = = . n( S ) 6 2
Example 4: A coin is tossed. If it shows head, we draw a ball from a bag consisting of 3 blue and 4 white balls; if it shows tail, we throw a die. Describe the sample space of the experiment. Solution: Let us denote the blue balls by B1, B2, B3 and white balls by W1, W2, W3, W4. The sample space of the experiment is S = {HB1, HB2, HB3, HW1, HW2, HW3, HW4, T1, T2, T3, T4, T5, T6} Example 5: From a pack of 52 cards there are drawn at random. Find the chance that they are a king, a queen and a knave. Solution: From a pack of 52 cards a draw of 3 can be made in 52C3 ways. Thus, n(S) = 52C3 In a pack of 52 cards are 4 kings, 4 queens and 4 knaves. A king can be drawn in 4C1 ways, a queen in 4C1 ways and a knave in 4C1 ways. Each of these ways be withdrawn in 4C1 × 4C1 × 4C1 ways. Thus, The required probability n(A) = 4C1 × 4C1 × 4C1 P(A) =
Example 7: If from a lottery of 30 tickets, marked 1, 2, 3, ...... four tickets be drawn, what is the chance that marked 1 and 2 are among them. Solution: We have, n(S) = 30C4 (four tickets can be selected) When 2 tickets are to be included always, remaining two can be selected in n(A) = 28C2 The required chance P(A) =
28C 2
ways, i.e.
n( A) = n( S )
28 30
C2 2 . = C2 145
Example 8: A manufacturer supplies cheap quarter horse power motors in a hot of 25. A buyer before taking a lot, tests a random sample of 5 motors and accepts the lot if they are all good, otherwise he rejects the lot. Find the probability that (i) he will accept a lot containing 5 defective motors. (ii) he will reject a lot containing only one defective motor. Solution: The buyer can choose a random sample of 5 motors out of 25 in 25C5 ways. (i) The buyer will accept a lot, if in his sample all the motors are good. That means his sample consists of 5 motors from 20 nondefective motors. So the number of ways of selecting the sample for acceptance of the lot, will be 20C5 The required probability
20 ! C 5 ! 15 ! = 25 5 = 25 ! C5 5 ! 20 !
20
=
20 × 19 × 18 × 17 × 16 2584 = = 0.292 25 × 24 × 23 × 22 × 21 8855
(ii) When he is rejecting a lot containing one defective motor, so his sample will contain the only defective motor and 4 others which are chosen from the 24 non-defective motors. So number of ways of selecting this sample is = 24C4 × 1C1 The required probability
24 ! 5 1 C4 4 ! 20 ! = 25 = = = = 0.2 25 ! 25 5 C5 5 ! 20 !
24
EXERCISE 1.1
1. What is the chance of throwing a 3 with an ordinary die? 2. What is the chance of that a non-leap year should have fifty three Sundays? 3. An integer is chosen at random from the first two hundred digits. What is the probability that the integer chosen is divisible by 6 or 8.
188
ADVANCED MATHEMATICS
4. In a class of 66 students 13 are boys and the rest are girls. Find the probability that a student selected will be a girl. 5. A bag contains 7 white and 12 black balls. Find the probability of drawing a white ball. 6. A card is drawn from an ordinary pack and a gambler bets that it is a spade or an ace. What are the odds against his winning this bet? 7. From a set of 17 cards, number 1, 2, 3, ...... 16, 17, one is drawn at random. Show that the chance that its number is divisible by 3 or 5 or 7 is
9 . 17
8. A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. A disc drawn at random from the bag. Calculate the probability that it will be (i) red (ii) yellow (iii) blue (iv) not blue. 9. In the odds in favour of an event are 4 to 5, find the probability that it will occur. 10. In a single throw of two dice, find the probability of getting a total of 9 or 11. 11. Find the probability of drawing either an ace or a spade or both from a pack of cards. 12. Probability that a boy will pass an examination is that at least one of them passes examination?
2 2 and that for a girl it is . What is the probability 5 5
ANSWERS
1.
1 6 7 19 1 6
2.
1 7
3.
1 4 4 2 1 2 , , , 9 9 3 3 19 . 25
4.
53 66 4 9
5.
6. 9 to 4
8.
9.
10.
11.
4 13
12.
1.5. INDEPENDENT EVENTS Two events are said to be independent events, if the occurrence of one does not effect the occurrence of the other. OR Two events A and B are said to be independent events iff P(A ∩ B) = P(A) . P(B). When a die is thrown, let A be the event ''number appearing is a multiple of 3'' and B be the event ''number appearing is even''. We have A = {3, 6}, B = {2, 4, 6} and P(A) =
2 1 = 6 3
THEORY OF PROBABILITY
189
P(B) =
3 1 = 6 2
1 6
A ∩ B = {6} P(A ∩ B) = Using,
P(A ∩ B) = P(A) . P(B)
1 1 1 = × 6 3 2
Hence, A and B are independent events.
1.6. CONDITIONAL PROBABILITY Let A and B be events such that P(B) ≠ 0. The conditional probability of A given B, denoted by P(A/B), is defined by P(A/B) =
P( A ∩ B) P( B)
where, P(A/B) is the probability of occurrence of A given that B has already happened. Theorem: If the two events A and B are independent events, then P(A/B) = P(A) and P(A ∩ B) = P(A) . P(B) we have, and P(A/B) = P(B/A) = P(B/A) = P(B) Proof: We know that if A and B are given to be independent events, then
1.7. MULTIPLICATIVE THEORY OF PROBABILITY OR THEOREM OF COMPOUND PROBABILITY If there are two events, the respective probability of which are known then the probability that both will happen simultaneously is the product of the probability of one and the conditional probability of the other, given that the first has occurred, i.e. P(A ∩ B) = P(A) P(B/A) or P(B ∩ A) = P(B) . P(A/B)
190
ADVANCED MATHEMATICS
Proof: Let out of n outcomes, m1 be the number of cases favourable to the happening of A.
m1 . n Let m2 be the number of cases favourable to the happening of B.
∴ Probability of happening of the event A = P(A) =
∴ Conditional probability of B, given that A has happened = P(B/A) =
m2 m1
Thus, the number of cases favourable to the happening of both A and B are m2 out of n. ∴ P(A ∩ B) =
m m m2 = 1 × 2 = P(A) . P(B/A). n m1 n
Note. 1. If A and B are independent events, then P(B/A) = P(B) ∴ P(A ∩ B) = P(A) . P(B) 2. P(A ∩ B) is also written as P(A and B) or P(AB). 3. P(A ∪ B) is also written as P(A or B). 4. Term 'independent' is defined in terms of ''probability of events'' whereas mutually exclusive is defined in terms of ''events''. Moreover, mutually exclusive events never have an outcome common, but independent events do have an outcome common, provided each event is non-empty. Clearly 'independent' and 'mutually exclusive' do not have the same meaning. 5. If two events A and B are independent, then (i) A and B are independent event. (ii) A and B are independent event. (iii) A and B are independent event.
Example 2: A die is rolled twice and the sum of the numbers appearing on them is observed to be 7. What is the conditional probability that the number 2 has appeared at least once? Solution: Let A = getting the number 2 at least once B = getting 7 as the sum of the numbers on two dice.
Example 3: A family has two children. What is the conditional probability that both are boys given that at least one of them is a boy? Solution: Let the sample space S is given by S = {(b, b), (b, g), (g, b), (g, g)} Let A denote the event that both children are boys, B the event that at least one of them is a boy and all outcomes are equally likely, then the desired probability is
1 1 P( A ∩ B) P(A/B) = = 4 = . 3 3 P( B) 4 Example 4: Ajeet can either take a course in computers or in maths. If Ajeet takes the computer 1 course, then he will receive an A grade with probability ; if he takes the maths course then he will 2
receive an A grade with probability
1 . Ajeet decides to base his decision on the flip of a fair coin. What 3 is the probability that Ajeet will get an A in maths?
Solution: Let C be the event that Ajeet takes maths and A denote the event that he receives on A in whatever course he takes, then the desired probability is P(A ∩ C ) = P(C ) P(A/C ) =
1 1 1 × = 2 3 6
Example 5: Ashu and Ankit appear in an interview for two vacancies in the same post. The probability of Ashu's selection is
1 1 and that of Ankit's selection is . What is the probability that 7 5
(i) both of them will be selected? (iii) only one of them is selected?
(ii) none of them will be selected?
Solution: Let A denote the Ashu is selected and B denote the Ankit is selected then A and B are independent. 1. P(both of them will be selected) = P(A ∩ B) = P(A) P(B) =
Example 6: The probability that a teacher will give an unannounced test during any class meeting is
1 . If a student is absent twice, what is the probability that he will miss at least one test? 5
Solution: Let T1 be the event of I test held on his first day of absence and T2 be the event of II test held on second day of his absence. Since, T1 and T2 are independent, then the required probability P(probability that he will miss at least one test) P(T1 ∪ T2) = P(T1) + P(T2) – P(T1 ∩ T2) = P(T1) + P(T2) – P(T1) P(T2) =
1 1 1 1 2 1 9 = . + − × = − 5 5 5 5 5 25 25
Example 7: Three groups of children contain 3 girl and 1 boy; 2 girls and 2 boys; 1 girl and 3 boys respectively. One child is selected at random from each group. Find the chance that there selected comprise 1 girl and 2 boys. Solution: Let A, B and C be the three groups. Given that A Boys Girls 1 3 B 2 2 C 3 1
Let B1, B2, B3 be the events of selecting a boy from A, B and C group respectively and let G1, G2, G3 be the events of selecting a girl from A, B and C group respectively. Then B1, B2, B3, G1, G2, G3 are independent events such that
Example 8: A and B are two independent witness in a case. The probability that A will speak the truth is x and the probability that B will speak the truth is y. A and B agree in a certain statement. Show that the probability that this statement is true:
(ii) P(A ∩ B), if A and B are independent events. 2. In the two dice experiment, if A is the event of getting the sum of numbers on dice as 11 and B is the event of getting a number other than 5 on the first die, find P(A ∩ B). Are A and B independent events? 3. If A and B are independent events, then show that (i) A , B (iii) A , B are also independent events. 4. A problem of statistics is given to three students A, B and C whose chances of solving it are 1/2, 3/4 and 1/4 respectively. What is the probability that the problem will be solved? 5. A bag contains 3 red and 5 black balls and a second bag contains 6 red and 4 black balls. A ball is drawn from each bag. Find the probability that (i) both are red, (iii) one is red and one is black. 6. A bag contains 19 tickets, numbered from 1 to 19. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both the tickets will show an even number. 7. A bag contains 5 white, 7 red and 8 black balls. If four balls are drawn one by one without replacement, find the probability of getting all white balls. 8. An article manufactured by a company consists of two parts A and B. In the process of manufacture of part A, 9 out of 100 are likely to be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part B. Find the probability (i) that the assembled part will not be defective (ii) that the assembled part will be defective. 9. A bag contains 4 white and 6 black balls. If two balls are drawn in succession, what is the probability that one is white and other is black? 10. A bag contain 10 white and 15 black balls. Two balls are drawn in succession. What is the probability that first is white and second is black? (ii) both are black, (ii) A, B and
Proof: By the definition of conditional probability P( Bk ∩ A) P(Bk/A) = P( A) By the theorem of total probability P(A) =
...(1)
∑
i =1
n
P( Bi ∩ A) =
∑
i =1
n
P( Bi ) P( A / Bi )
...(2)
Also from the theory of multiplication P(Bk ∩ A) = P(Bk) P (A/Bk) Putting the values of P(A) and P(Bk ∩ A) from (2) and (3) in equation (1), we get P(Bk/A) = ...(3)
P( Bk ) P( A / Bk )
∑
i =1
n
.
P( Bi ) P( A / Bi )
SOLVED EXAMPLES
Example 1: In 1888 there will be three candidates for the position of Director—Dr. Singhal, Dr. Mehra and Dr. Chatterji—whose chances of getting the appointment are in the proportion 4 : 2 : 3 respectively. The probability that Dr. Singhal, if selected, will abolish co-education in the college is 0.3. The probability of Dr. Mehra and Dr. Chatterji doing the same are respectively 0.5 and 0.8. What is the probability that co-education will be abolished in the college? Solution: Let B1, B2 and B3 be the probability of being appointed of Dr. Singhal, Dr. Mehra and Dr. Chatterji respectively. Let the probability of co-education being abolished be A. Then, by the theorem of total probability, we have P(A) = P(B1) P(A/B1) + P(B2) P(A/B2) + P(B3) P(A/B3)
2 4 3 × (0.3) + × (0.5) + × (0.8) 9 9 9 2 1 4 23 + + = . = 15 9 15 45 Example 2: An urn I contains 3 white and 4 red balls and an urn II contains 5 white and 6 red balls. One ball is drawn at random from one of the urn and is found to be white, find the probability that it was drawn from urn I.
=
Solution: Let
B1: the ball is drawn from urn I B2: the ball is drawn from urn II A: the ball is white
THEORY OF PROBABILITY
197
To find P(B1/A). By Baye's theorem P(B1/A) =
P( B1 ) P( A / B1 ) P( B1 ) P( A / B1 ) + P( B2 ) P( A / B2 )
1 2 3 7
...(1)
Since, two urn are equally likely to be selected, P(B1) = P(B2) =
P(A/B1) = Probability of a white ball is drawn from urn I = P(A/B2) = Probability of a white ball is drawn from urn II = By (1), we have
5 11
1 3 × 33 2 7 = . P(B1/A) = 1 3 1 5 68 × + × 2 7 2 11 3 and the probability that another 5 2 person can hit the same target is . But the first person can fire 8 shoots in the time the second person 5 fires 10 shoots. They fire together, what is the probability that the second person shoots the target.
Example 3: The probability that a person can hit a target is Solution: Let A denote the event of shooting the target and B1 denote the event that the first person shoots the target and B2 denote the event that the second person shoots the target. Therefore, P(B2/A) is the probability that the second person shoots the targets. Now, we have P(A/B1) =
3 2 , P(A/B2) = 5 5
It is given that the ratio of the shoots of the first person to those of the second person in the same time is
8 4 i.e. . 10 5
Example 4: The contents of urns I, II and III are as follows: 1 White, 2 black and 3 red balls, 2 White, 1 black and 1 red balls, and 4 white, 5 black and 3 red balls. One urn is chosen at random and two balls drawn. They happen to be white and red. What is the probability that they come from urns I, II or III? Solution: Let B1: Urn I is chosen; B2: Urn II is chosen; B3: Urn III is chosen and A: the two balls are white and red. To find P(B1/A), P(B2/A) and P(B3/A). Now,
1 . Since, three urns are equally likely to be selected. 3 P(A/B1) = Probability of a white and a red ball are drawn from urn I
2 P( A ∩ B1 ) 1 P(B1/A) = = 7 = 4 P( A) 2 7 Example 6: In a bolt factory, machines M1, M2 and M3 manufacture respectively 25%, 35% and 40% of the total of their output 5, 4 and 2 percent are defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by machine M1, M2 and M3?
Solution: Let A, B, C denote the events that the bolt was manufactured by machines M1, M2 and M3 respectively and let D denote the event of its being defective. Then P(A) = 0.25, P(B) = 0.35, P(C) = 0.40 The probability that a defective bolt is drawn from those manufactured by M1 is
5 = 0.05 100 4 2 = 0.04, P(D/C) = = 0.02 Similarly, P(D/B) = 100 100 By Baye's theorem, we have
Example 7: In a certain college 25% of boys and 10% of girls are studying mathematics. The girls constitute 60% of the student body (a) what is the probability that mathematics is being studied? (b) If a student is selected at random and is found to be studying mathematics, find the probability that student is a girl? (c) a boy? Solution: Given that probability of a boy P(B) = and Probability of a girl P(G)
60 3 = 100 5 Probability that maths is studied given that the student is a boy 25 1 = P(M/B) = = 100 4 10 1 Similarly, P(M/G) = = 100 10 (a) Probability that math is studied 40 2 = 100 5
=
P(M) = P(G) P(M/G) + P(B) P(M/B) =
3 1 2 1 4 . . + . = 5 10 5 4 25
(by theorem on total probability)
(b) By Baye's theorem, we have Probability that a math student is a girl
EXERCISE 1.3
1. A bag X contains 2 white and 3 red balls and a bag Y contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag Y. 2. Three urns contain 6 red, 4 black; 4 red 6 black; 5 red, 5 black balls respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it drawn from the first urn. 3. A businessman goes to hotels H1, H2, H3, 20%, 50%, 30% of the time, respectively. It is known that 5%, 4%, 8% of the rooms in H1, H2, H3 hotels have faulty plumbing (i) Determine the probability that the businessman goes to hotel with faulty plumbing (ii) what is the probability that business's room having faulty plumbing is assigned to hotel H3? 4. There are three boxes containing respectively 1 white, 2 red, 3 black balls; 2 white, 3 red, 1 black balls; 3 white, 1 red, 2 black balls. A box is chosen at random and from it two balls are drawn at random. The two balls are one red and one white. What is the probability that they come from the (i) first box, (ii) second box (iii) third box. 5. An urn contains 10 white, 9 black, 8 red and 3 blue balls. Balls are drawn one by one at random from the urn until 2 blue balls are obtained at the 11th draw. Find the probability of drawing 2 blue balls upto 11th draw. 6. Suppose the supply of transistors is produced by three systems S1, S2 and S3. Further, suppose that S1 produces 20%, S2 produces 30% and S3 produces 50% of the supply and that the defective (D) rates for three systems S1, S2, S3 are respectively 0.01, 0.02 and 0.05. If a transistor is randomly selected from the supply and is found to be defective, find (i) P(S1/D) (ii) P(S2/D) (iii) P(S3/D). 7. The chance that a doctor will diagnose a disease correctly is 70%. The chance that a patient dies by his treatment after correct diagonals is 35% and the chance of death by wrong diagnosis is 80%. If a patient dies after taking his treatment, what is the chance that the disease was diagnosed correctly. 8. Companies B1, B2, B3 produce 30%, 45% and 25% of the cars respectively. It is known that 2%, 3% and 2% of the cars produced from B1, B2 and B3 are defective. (i) What is the probability that a car purchased is defective? (ii) If a car purchased is found to be defective, what is the probability that this car is produced by company B3 ?
ANSWERS
1.
25 52 2 11
2. (ii)
2 5 6 11
3. (i) 0.054 (iii)
(ii) 5. 7.
4 9 19 406 49 97
4. (i)
3 11
6. (i) 0.061 8. (i) 0.0245
(ii) 0.182 (ii)
(iii) 0.757
10 . 49
202
ADVANCED MATHEMATICS
1.10. BINOMIAL THEOREM Let the probability of the happening of an event in one trial is p and q = 1 – p be the probability that the event fails in one trial. Let us find out the probability of exactly r successes in n trails. The chance that an event happens at least r times in n trials is given P = pn + nC1 pn–1q + nC2 pn–2q2 + ... + nCr pr qn –r Hence, the probability that the event will happen exactly r times in n trails is the (n + 1)th term in the expansion of (q + p)n.
1.11. MULTINOMIAL THEOREM If a dice has f faces marked with 1, 2, 3, ...... f, the probability of throwing a total p with n dice is given by =
Note. If n dice are different with faces f1, f2, f3, ......, fn, then the required chance is =
1.12. RANDOM VARIABLE A variable, which takes a definite set of values with a definite probability associated with each value is called a random variable. Hence, if a variable x takes the values x 1, x 2 , x 3, ......, x n with respective probability p1, p2, p3, ......, pn so that
∑
i =1
n
pi = 1 . Then a discrete probability distribution is defined. The function
P(x), which has respective probabilities p1, p2, p3, ......, pn for x1, x2, x3, ......, xn is known as the probability function of x. The variable x in such a case is called the random variable.
SOLVED EXAMPLES
Example 1: In a single throws, with a pair of dice, what is the chance of throwing doublets or not. Solution: When throw a pair of dice, we get only 6 doublets namely (1, 1); (2, 2); (3, 3); (4, 4); (5, 5); (6, 6). The chance of throwing a doublet in one throw And the chance of not throwing a doublet =
6 1 = 62 6
5 1 = 6 6
=1–
204
ADVANCED MATHEMATICS
Example 2: A dice is thrown five times. Find the probability of 3 coming up (i) exactly 3 times (ii) at least 3 times. Solution: The probability of 3 coming up = Let p = 1/6 and q = 5/6. The probability of 3 successes in 5 trails is
5C 3
Coeff. of x 14 in (1 − 4 x 6 + 6x 12 ......) × (1 + 4 x + 10 x 2 + ...... + 16x 8 + ...... + 680 x 14 + ......) = 64 5 680 − 660 + 60 80 = 4 = 4 81 6 6 Example 4: A person throws two dice, one the common cube, and the other regular tetrahedron, the number in the lowest face being taken in the case of a tetrahedron. What is the chance that the sum of the numbers thrown is not less than 5.
= Solution: We know that the common cube has 6 faces and the regular tetrahedron has 4 faces. Hence, 6 × 4 is the total number of ways in which the cube and the tetrahedron can fall. The favourable ways of getting a sum not less than 5 = the sum of the coeff. of x5, x6, ......, x10 in (x + x2 + ...... + x6) (x + x2 + x3 + x4) = the sum of the coeff. of x5, x6, ......, x10 in (x2 + 2x3 + 3x4 + 4x5 + 4x6 + 4x7 + 3x8 + 2x9 + x10) =4+4+4+3+2+1 ∴ The required chance =
3 18 = 6×4 4
THEORY OF PROBABILITY
205
Example 5: Four tickets marked 00, 01, 10, 11 respectively are placed in a bag. A ticket is drawn at random five times, being replaced each time. Find the probability that the sum of the numbers on tickets thus drawn is 23. Solution: Given that four tickets marked 00, 01, 10, 11 respectively are placed in a bag. The four tickets can be drawn five times in 45 ways. The favourable number of ways for obtaining a sum = Coeff. of x23 in (x0 + x1 + x10 + x11)5 = Coeff. of x23 in (1 + x)5 (1 + x10)5 = Coeff. of = 100 Hence, the required probability = x23 in (1 + 5x + 10x2 + 10x3 + 5x4 + x5) (on factorization) (1 + 5x10 + 10x20 + ......)
100 25 = . 5 256 4 1 (b + a) m − (b − a) m . 2 (a + b) m
Example 6: If m things are distributed among a men and b women, show that the chance that the number of things received by men is odd, is
Solution: Given that m things are distributed among a men and b women. The probability that men may get a thing is
a . a+b b . a+b
The probability that a women may get a thing is
If out of m things 'a' men get only one thing and other things go to the women, then the probability for men is =
mC 1
Similarly, the probability that 'a' men may get three things, five things, ...... are respectively
mC 3
FG a IJ FG b IJ H a + bK H a + bK
m −3
m −1
FG a IJ FG b IJ H a + b K H a + bK
3
, mC5
FG a IJ FG b IJ H a + b K H a + bK
5
m −5
, ......
Let A be the event the number of things got by men is odd and Ai denotes the event 'men get ith thing' i = 1, 3, 5, 7, ...... Thus, the probability that the number of things received by men is odd is, P(A) = P(A1) + P(A3) + P(A5) + ...... =
mC 1
FG a IJ FG b IJ H a + b K H a + bK
m −1
+
mC 3
FG a IJ FG b IJ H a + b K H a + bK
3
m −3
+
mC 5
FG a IJ FG b IJ H a + b K H a + bK
5
m −5
+ ......
206
ADVANCED MATHEMATICS
=
1 ( a + b) m 1 ( a + b) m
m
C1 ab m −1 + m C3 a 3 b m −3 + m C5 a 5 b m − 5 + ......
m
=
LM 1 (b + a) N2
−
1 (b − a ) m 2
OP Q
=
1 ( b + a) m − ( b − a) m . . 2 ( a + b) m
Example 7: From a bag containing 5 one rupee coins and 3 coins of 20 paisa each, a person is allowed to draw 2 coins indiscriminately. Find the value of his expectation. Solution: Given that a bag containing 5 one rupee coins and 3 coins of 20 paisa each. Probability of drawing 2 rupees =
5 8
Example 8: A person draws cards one by one from a pack until he draws all the aces. How many cards he may be expected to draw? Solution: Let a person has to make n draws for all aces. It means that in n – 1 draws, three aces and in nth, one ace. The probability of such an occurrence
4
The least square number of draws he has to make is 4 and the maximum number 52. Hence, n range 4 to 52. The expected number of draws =
52
∑
n=1
n.4
( n − 1) ( n − 2) ( n − 3) 49 × 50 × 51 × 52
=
4 49 × 50 × 51 × 52
LM n NM∑
52 n =1
4
−6
∑
n =1
52
n 3 − 11
∑
n =1
52
n2 − 6
∑ nP . P
52 n =1
O Q
Example 9: A makes a bet with B of 5S to 2S that in a single throw with two dice he will throw seven before B throws four. Each has a pair of dice and they throw simultaneously until one of them wins, equal throws being disregarded. Find B's expectation. Solution: The chance of 7 throwing with two dice = =
Example 11: What is the expectation of the number of failures preceding the first success in an infinite series of independent trails with constant probability of successes. Solution: The probabilities of success in first, second, third, ...... trails respectively are p, qp, q2p, q3p, ...... Let X be the no. of failures Then, the expected number of failures preceding the first success E(X ) = 0 . p + 1 . qp + 2q2p + ...... = qp [1 + 2q + 3q2 + ...... ∞], q < 1 =
qp (1 − q ) 2 q 1− p = . p p
[3 p = 1 – q]
=
EXERCISE 1.4
1. If on average 1 vessel in every 10 is wrecked find the probability that out of 5 vessels expected to arrive, 4 at least will arrive safely. 2. A teacher claims that he could often tell while his students were still in their first year whether they will obtain I, II, III divisions or fail in their final examinations. To demonstrate his claim, he forecasts the fates of 8 students. Find the probability of his being correct in 5 cases. 3. Five coins whose faces are marked, 2, 3 are thrown; what is the chance of obtaining a total of 12? 4. Determine the probability of throwing more than 8 with 3, perfectly symmetrical dice. 5. If three symmetrical dice are thrown, calculate the probability that the sum of numbers is 12.
THEORY OF PROBABILITY
209
6. A and B in turns toss an ordinary die for a prize of Rs. 44. The first to toss a 'six' wins. If A has first throw, what is his expectation? 7. A bag contains 2 white and 3 black balls. Four persons A, B, C and D in the order named each draw one ball and do not replace it. The person to draw a white ball receives Rs. 200. Determine their expectations. 8. A box containing 2n tickets among which nCi bear the number i, (i = 0, 1, 2, ...... n). A group of m tickets is drawn, what is the expectation of the sum of numbers? 9. A person draws 2 balls from a bag containing 3 white and 5 black balls. If he is to receive Rs. 10 for every white ball which he draws and Rs. 1 for every black ball. What is his expectation? 10. Two players of equal skill A and B play a set of games; they leave off playing when A wants 3 points and B wants 2. If the stake is Rs. 16, what share ought each to take?
ANSWERS
1. 5.
45927 50000
2.
189 8192
3.
5 6
4. 8.
7 20 , 27 27 mn 2
25 216 9. 8.75
6. 24 10. 5, 11
7. 80, 60, 40, 20
CHAPTER
2
Theoretical Distributions
In this chapter, we shall discuss some probability distribution such as Binomial, Poisson, Normal distribution with their applications. In theoretical distribution, normal distribution is most important distribution. Binomial and Poisson distributions are discrete probability distribution and Normal distribution is continuous probability distribution. Before we discuss formal theoretical probability distribution, first we shall define certain terminologies and notations, which are used in defining theoretical distributions.
2.1. TERMINOLOGY AND NOTATIONS 1. Random variable: A variable, which takes a definite set of values with a definite probability associated with each value is called a random variable. 2. Continuous random variable: A random variable X is said to be a continuous random variable, if it takes all possible values between its limits. For example, height of a tree, weight of a school children etc. 3. Discrete random variable: A random variable X is said to be a discrete random variable, if it takes only finite values between its limits, for example, the number of student appearing in a festival consisting of 400 students is a discrete random variable which can assume values other than 0, 1, 2, ......, 400. 4. Continuous probability distribution: Let X be a random variable which assume values in the interval [– ∞, ∞], then the probability in (a, b) is defined by P(a ≤ x ≤ b) =
2.2. BINOMIAL DISTRIBUTION Binomial distribution is the extension of the theorem of probability. Binomial distribution was discovered by James-Bernoulli in 1700. Let X be the random variable denote the number of successes in these n trials. Let p and q be the probabilities of success and failure one any one trial.Then in the n independent trails the probability that there will be r successes and n – r failures is given by P(X = r) = nCr pr qn – r , r = 0, 1, 2, ......, n The probability distribution of the random variable X is therefore given by
X P(X)
nC 0
0 p0 qn
nC 1
1 p1 qn–1
nC 2
2 p2 qn–2
.......r
nC r
.........n
nC n
pr qn–r
pn qn-n
Hence, the probability distribution is called the binomial distribution because for r = 0, 1, 2, ......, n, are the probabilities of the successive terms of the binomial expansion of (q + p)n.
Note. 1. The constant n, p, q are called parameters of the distribution. 2. Also denoted b(r ; n, p) = nCr pr qn – r, r = 0, 1, 2, ......, n 3. For N set of n trial the successes 0, 1, 2, .....r, ....., n are given by N(q + p)n, which is called binomial distribution.
1 1 × = 3. 2 2 4 Example 3: The mean and variance of a binomial distribution are 4 and 3 respectively. Find the probability of getting exactly six successes in this distribution.
Solution: The mean of binomial distribution and the variance of binomial distribution Using (1) and (2), we have npq = 3 and by (1), ⇒
= np = 4 (given) = npq = 3 (given)
3 4
...(1) ...(2)
4.q=3 ⇒ q=
3 1 = 4 4 1 np = 4 ⇒ n × = 4 ⇒ n = 16 4
p=1–q=1–
216
ADVANCED MATHEMATICS
The probability of 6 success = 16C6 4 tails.
FG 1 IJ FG 3IJ H 4K H 4K
6
10
=
8008 × 310 . 416
Example 4: In 256 sets of 12 tosses of a coin, in how many cases one can expect 8 heads and
1 1 and q be the probability of tail i.e., q = . 2 2
Solution: Let p be the probability of head i.e., p =
n = 12, N = 256. The binomial distribution is N(q + p)n = 256
FG 1 + 1 IJ H 2 2K
12
The probability of 8 heads and 4 tails in 12 trial is
495 4096 The expected number of such cases in 256 sets
12C 8
FG 1 IJ FG 1 IJ H 2K H 2K
8
4
=
= 256 ×
495 = 30.9 ≈ 31 (nearly). 4096
Example 5: During war, 1 ship out of 9 was sunk on an average in making a voyage. What was the probability that exactly 3 out of a convoy of 6 ships would arrive safely ? Solution: Let p be the probability of a ship arriving safely i.e., p=1–
1 8 = 9 9 n = 6, N = 1
then q = 1 – p =
1 9
The binomial distribution is N (q + p)n =
FG 1 + 8 IJ H 9 9K
6
The probability that exactly 3 ship arrive safely = 6C3
FG 8 IJ FG 1 IJ H 9K H 9K
3
3
= 20 ×
10240 512 = . 6 96 9
Example 6: The probability that a pen manufactured by a company will be defective is 12 such pens are manufactured, find the probability that (i) exactly two will be defective. (ii) at least one will be defective. (iii) none will be defective.
1 . If 10
THEORETICAL DISTRIBUTIONS
217
Solution: Given that the probability of a defective pen =
1 = 0.1 10
Then the probability of a non-defective pen is = 1 – 0.1 = 0.9 (i) The probability that exactly two pen will be defective = 12C2 (0.1)2 (0.9)10 = 0.2301 (ii) The probability that at least two will be defective = 1 – {Probability that either none or one is non-defective} = 1 – {12C0 (0.9)12 + 12C1 (0.1) (0.9)11} = 1 – 0.6588 = 0.3412 (iii) The probability that none will be defective = 12C12 (0.9)12 = 0.2833. Example 7: Six dice are thrown 729 times. How many times do you expect at least three dice to show a five or six ? Solution: Let p be the chance of getting 5 or 6 with one die then p=
1 , 3
q=1– N = 729
1 2 = 3 3
n = 6,
Since, dice are in sets of 6 and there are 729 sets. The binomial distribution is N(q + p)n = 729 The required number is = 729
FG 2 + 1IJ H 3 3K
3 3
6
R C F 1I F 2 I S H 3K H 3 K T
6 3
+ 6 C4
F 1I F 2 I H 3K H 3 K
4
2
+ 6 C5
F 1I F 2 I + H 3K H 3 K
5
6
C6
F 1I U H 3K V W
6
{160 + 60 + 12 + 1} = 233. 36 Example 8: A perfect cubical die is thrown a large number of times in set of 8. The occurrence of 5 or 6 is called a success. In what proportion of sets do you expect 3 successes. Solution: Let p = the chance of occurrence 5 or 6 with one die = q=1–p=1–
1 2 = , 3 3
1792 100 = 27.3%. × 6591 N Example 9: In 800 families with 4 children each how many families would be expected to have (i) 2 boys and 2 girls (ii) at least one boy (iii) no girl (iv) at least two girls ? Assuming that equal probabilities for boys and girls.
Solutioin: Since, the probabilities for boys and girls are equal Then let p = probability of having a boy = and Here q = probability of having a girl = n = 4, N = 800
1 2
1 2
The binomial distribution is N (q + p)n = 800
FG 1 + 1 IJ H 2 2K
4
.
(i) The expected number of families having 2 boys and 2 girls = 800 4C2
800 [6 + 4 + 1] = 550. 24 Example 10: The probability that a men aged 60 years will live up to 70 years is 0.65. What is the probability that out of ten men now 60 years, at least 7 would live up to 70 years ?
=
Solution: Given that the probability of a men aged 60 years will live up to 70 years is = 0.65 Then p = 0.65, q = 1 – p = 1 – 0.65 = 0.35 n = 10 Let X be the number of men who live up to 70 years. The probability that out of 10 men, r men will up to 70 years is P (X = r) = nCr pr qn – r = 10Cr (0.65)r (0.35)10 – r The probability of at least 7 men would live up to 70 years. P(X ≥ 7) = P(X = 7) + P(X = 8) + P (X = 9) + P (X = 10) = 10C7 (0.65)7 (0.35)3 + 10C8 (0.65)8 (0.35)2 + 10C9 (0.65)9 (0.35)1 + 10C10 (0.65)10 = 0.2523 + 0.1756 + 0.0725 + 0.0135 = 0.5140. Example 11: Assuming that on the average one telephone number out of 15 called between 2 p.m. and 3 p.m. on week days is busy. What is the probability that if 6 randomly selected telephone numbers are called (i) not more than three, (ii) at least three of them will be busy ? Solution: Let p be the probability of a telephone number out of 15 called between 2 p.m. and 3 p.m. on week days is busy, i.e., p =
EXERCISE 2.1
1. The mean and variance of binomial distribution are 4 and
4 respectively. Find P(X ≥ 1). 3
2. The items produced by a firm are supposed to contain 5% defective items. What is the probability that a sample of 8 items will contain less than 2 defective items ?
THEORETICAL DISTRIBUTIONS
221
3. Find the binomial distribution for which the mean is 4 and variance is 3. 4. Bring out the fallcy in the statement: The mean of a binomial distribution is 3 and variance is 4. 5. If 10% bolts produced by a machine are defective, determine the probability that out of 10 bolts chosen at random (i) 1 (ii) none (iii) at least one (iv) at most two bolts, will be defective. 6. An irregular six-faced die is thrown, and the expectation that in 100 throws it will give five even numbers is twice the expectation that it will give four even numbers. How many times in 10,000 sets of 10 throws would you expect it to give no even number ? 7. The incidence of occupational disease in an industry is such that the workmen have a 25% chance of suffering from it. What is the probability that out of six workmen 4 or more will contact the disease ? 8. A box contains 'a' red and 'b' black balls, 'n' balls are drawn. Find the expected number of red balls drawn. 9. Assuming that half the population are consumers of rice so that the chance of an individual being a consumer is
1 and assuming that 100 investigators, each take ten individuals to see whether they are 2
consumers, how many investigators do you except to report that three people or less are consumers ? 10. Find the most probable number of heads in 99 tossing of a biased coin, given that the probability of a head in a single tossing is
3 . 5
11. In litters of 4 mice the number of litters which contained 0, 1, 2, 3, 4 females were noted. The figures are given in the table below: Number of female mice Number of litters 0 8 1 32 2 34 3 24 4 5 Total 103
In the chance of obtaining a female in a single trial is assumed constant, estimate this constant of unknown probability. 12. The probability of a man hitting a target in
1 . He fires 7 times. What is the probability of his hitting the 4
target at least twice. How many times must he fires so that the probability of his hitting at least one is greater than
2 ? 3
13. Eight coins are tossed at a time, 256 times. Number of heads observed at each throw are recorded and the results are as given below. Find the expected frequency and fit a binomial distribution. What are the theoretical values of the mean and standard deviation ? Calculate also the mean and standard deviation of the observed frequencies. 14. The probability that a bomb droped from a plane strike the target is
1 . If six such bombs are droped find 5
the probability that at least two will strike the target, and exactly two will strike the target.
222
ADVANCED MATHEMATICS
15. In a precision bombing attack there is a 50% chance that anyone bomb will strike the target. Two direct hits are required to destroy the target completely. How many bombs must be dropped to give a 99% chance or better of completely destroying the target ? 16. If a coin is tossed N times, where N is very large even number. Show that the probability of getting exactly
2.4. POISSON'S DISTRIBUTION Poisson distribution was discovered by a french mathematician Simeon Denis Poisson in 1837. Poisson distribution is also a discrete probability distribution of a discrete random variable, which has no upper bound. Poisson distribution is a limiting form of the binomial distribution (q + p)n under the following conditions: (i) n → ∞, i.e., the number of trials is indefinitely large. (ii) p → 0, i.e., the constant probability of success for each trial is indefinitely small. (iii) np is a finite quantity, say m.
m m , q = 1 – , where m is a positive real number. n n Poisson distribution deals with situations explained below: (i) Number of suicides or deaths by heart attack in 1 minute.
Thus, p = (ii) Number of accidents that take place on a busy road in time t. (iii) Number of printing mistakes at each unit of the book. (iv) Number of cars passing a certain street in time t. (v) Emission of radioactive particles. (vi) Number of faulty blades in a packet of 1000. (vii) Number of person born blind per year in a certain village. (viii) Number of telephone calls received at a particular switch board in 1 minute.
THEORETICAL DISTRIBUTIONS
223
The probability distribution of a random variable X is said to be a Poisson distribution, if the random variable assumes only non-negative values and its probability distribution is given by P(X = r) =
mr e − m , r = 0, 1, 2, ....... r!
where, m is called the parameter of the distribution and m > 0 The probability of r successes in a binomial distribution P(X = r) = nCr pr qn – r =
Example 2: Six coins are tossed 6400 times. Using the Poisson distribution, what is approximate probability of getting six heads x times. Solution: Let p be the probability of getting all the six head in a throw of six coins. i.e., p=
1 1 = . 6 64 2
Here, n = 6400
1 = 100 64
The mean of Poisson distribution = np = 6400 ×
The probability of getting all six heads x times according to Poisson distribution is
e − 100 (100) x . x!
Example 3: Show that for a Poisson's distribution γ1 γ2 σ m = 1, when σ and m are the standard deviation and mean respectively. Solution: In Poisson distribution γ1 = and The standard deviation the mean Now, we have
Hence, the mean and variance of X are each equal to 1. Example 5: In a Poisson distribution P(x) for x = 0 is 0.1. Find the mean given that loge 10 = 2.3026. Solution: The Poisson distribution is P(x) =
m x e −m , x = 0, 1, 2, ....... , m > 0 x!
...(1)
Given that for x = 0 P(x) = 0.1 Then by (1), we have 0.1 = ⇒
m0e − m 0!
⇒ e–m = 0.1
⇒ em = 10
m = loge 10 = 2.3026.
Example 6: If 2% electric bulbs manufactured by a company be defective, find the probability of (i) 0 (ii) 1 (iii) 2 (iv) 3 defectives in a lot of 100 bulbs. Solution: Let p be the probability of electric bulbs manufactured by a company be defective. i.e., Given that, we have, p=
2 . 100 n = 100 2 = 2 = m (mean) 100
np = 100 ×
Let X be the number of electric bulbs manufactured by a company be defective. Then, P(X = r) =
Example 7: A Poisson distribution has a double mode at x = 4 and x = 5. Find the probability that x will have either of these values P(x = 4) = P(x = 5). Solution: The Poisson distribution is P(X = r) = Given that ⇒ We know that, Then
Example 10: If the probability that an individual suffers a bad reaction from injection of a given serum is 0.001. Determine the probability that out of 2000 individual (i) exactly 3, (ii) more than 2 individuals suffer from bad reaction, (iii) none.
THEORETICAL DISTRIBUTIONS
231
Solution: Let p be the probability that an individual suffers a bad reaction from injection of a given serum i.e., Given that we have p = 0.001 Since, p is very small so we used Poisson's distribution. n = 2000 np = 0.001 × 2000 = 2
Example 11: In a certain factory turning Lexi blades, there is a small chance, 0.002 for any blade to be defective. The blades are in packets of 10. Use Poisson distribution to calculate the approximate number of packet containing no defective, one defective and two defective blades respectively in a consignment of 10,000 packets. Solution: Let p be the probability that a blade is defective i.e., p = 0.002 Since, p is very small so Poisson's distribution is used. Here, n = 10 then m = np = 10 × 0.002 = 0.02. Let X denote the number of defective blades in a packet of 10. Then P(X = r) =
EXERCISE 2.2
1. If X be a Poisson variate such that 3P(X = 3) = 4P(X = 4), find P (X = 7 ). 2. Criticise the following statement: The mean of a Poisson distribution is 7, while the standard deviation is 6. 3. The probability that a man aged 50 years will die within a year is 0.01125. What is the probability that of 12 such men at least 11 will reach their fifty first birthday ?
THEORETICAL DISTRIBUTIONS
233
4. A car-hire-firm has two cars, which if hires, out day by day. The number of demands for a car on each day is distributed as Poisson distribution with mean 1.5. Calculate the proportion of days on which neither car is used and the proportion of days on which some demand is refused {e–1.5 = 0.2231}. 5. If m is the parameter of a Poisson variate, show that the probabilities that the value of the variate taken at random is even or odd are e– m cosh m and e–m sinh m. 6. Letters were received in an office on each of 100 days. Assuming the following data to form a random sample from a Poisson's distribution. Fit the distribution and calculate the expected frequencies taking e– 4 = 0.183. Number of letters Frequencies 0 1 1 4 2 15 3 22 4 21 5 20 6 8 7 6 8 2 9 0 10 1
7. The frequency of accidents per shift in a factory is given in the following table. Calculate the mean number of accidents per shift. Find the corresponding Poisson distribution and compare with actual observations. Accidents per shift Frequencies 0 192 1 100 2 24 3 3 4 1 Total 320
8. A manufacturer of coffer pins knows that 5 per cent of his product is defective. If he sells coffer pins in boxes of 100 and guarantees that not more than 4 pins will be defective, what is the approximate probability that a box will fail to meet the guaranteed quality ? (e–5 = 0.0067). 9. A telephone switch board handles 600 calls on the average during a rush hour. The board can make a maximum of 20 connections per minute. Use Poisson distribution to estimate the probability that the board will be over during any given minute [e–10 = 0.00004539]. 10. Suppose that a book of 600 pages contains 40 printing mistakes. Assume that these errors are randomly distributed throughout the book and x, the number of errors per page has a Poisson distribution. What is the probability that 10 pages selected at random will be free of errors ? 11. Find the probability that at most defective fuses will be found in a box of 200 fuses, if experience show that 2 per cent of such fuses are defective. 12. An insurance company found that only 0.01% of the population is involved in a certain type of accident each year. If its 1000 policy holders were randomly selected from the population. What is the probability that not more than two of its clients are involved in such an accident next year ? {e–0.1 = 0.9048}. 13. Red blood cell deficiency may be determined by examining a specimen of the blood under a microscope. Suppose a certain small fixed volume contains on the average 20 red cells for normal persons. Using Poisson distribution, obtain the probability that a specimen from a normal person will contain less than 15 red cells. 14. Show how the Poisson distribution
mr e − m , (r = 0, 1, 2, 3, ......) r!
can be regarded as the limiting case of the binomial distribution. Hence or otherwise obtain the mean and the variance of the Poisson distribution, assuming the variance of the binomial distribution.
which is the recurrence relation for the moments of normal distribution.
2.11. AREA UNDER THE NORMAL CURVE The probability density function of the normal distribution is given by f (x) =
1 e 2π σ
−1 x − m σ 2
F H
I K
2
,
– ∞ < x < ∞.
f(z)
If x is a normal random variable with mean m and standard deviation σ, then the random variable z =
FG x − m IJ H σ K
has the normal distribution with mean 0 and the standard deviation 1. The random variable z is called the standard normal random variable. The probability density function of standard normal random variable z is given by f (z) =
O Z
1 − 2 z2 , –∞<z<∞ e 2π
1
Note 1. The graph of f (z) is famous 'bell shaped' curve. 2. If f(z) is the probability density function for the normal distribution, then P(z1 ≤ z ≤ z2) =
Where m is the mean and σ is standard deviation. (4) The curve is bell-shaped and symmetry about the line x = m i.e., y-axis. (5) As x increases then f(x) decreases rapidly, the maximum probability at point x = µ is given by f (x) or [P(x)]max =
1 2π σ
z
X2
X1
− 1 H e 2 2π σ
1 x−m σ
F
I K
2
dx
(6) Since, f (x) is probability density function, so it never be negative, i.e., no portion of the curve lies below the x-axis. (7) x-axis is an asymptote of the curve. (8) The point of inflexion of the normal curve are x = ± σ when the origin is taken at the mean. (9) Mean deviation about mean =
SOLVED EXAMPLES
Example 1: Prove that for normal distribution the mean deviation from the mean equal to standard deviation nearly. Solution: Let m and σ be the mean and standard deviation of the normal distribution respectively. By definition of mean deviation =
4 of 5
Example 3: Assume the mean height of soldiers to be 68.22 inches with a variance of 10.8 (inches)2. How many soldiers in a regiment of 1000 would you expect to be over 6 feet tall? Given that the area under the standard normal curve between Z = 0 and Z = 0.35 is 0.1368; and between Z = 0 and Z = 1.15 is 0.3746. Solution: Given that mean (m) = 68.22, variance (σ2) = 10.8 and standard deviation We have ⇒ (σ) = Z=
10.8 X −m σ
Z=
X − 68.22 10.8
X − 68.22 3.286
⇒
Z=
When X = 72, Now
Z=
72 − 68.22 = 1.15 3.286
P (X ≥ 72) = P (Z ≥ 1.15) = 0.5 – 0.3749 = 0.1251
The probability of getting a soldier above six feet is 0.1251. Hence, the number of soldiers who are over 6 feet tall. = 1000 × 0.1251 = 125.1 = 125 (nearly). Example 4: Assuming the resistance of the resistors to be normally distributed with mean 100 ohms and standard deviation 2 ohms, what percentage of resistors will have resistance between 98 ohms to 102 ohms. Solution: Given that mean (m) = 100 ohms standard deviation (σ) = 2 ohms We have Z=
X −m σ
Hence, 68.26% of resistors will have resistance between 98 ohms to 102 ohms. Example 5: In a Normal distribution 31% items are under 45 and 8% are over 64. Find the mean and standard deviation of the distribution. Solution: Let m and σ be the mean and standard deviation respectively.
45 − m σ 64 − m If x = 64, Z= σ 45 − m The area between 0 and = 0.50 – 0.31 = 0.19 σ From the table, for the area 0.19,
Example 6: The distribution of weekly wages for 500 workers in a factory is approximately normal with the mean and students derivation of Rs. 75 and Rs. 15. Find the number of workers who receive weekly wages: (i) more than Rs. 90 (ii) less than Rs. 45.
242
ADVANCED MATHEMATICS
Solution: The normal distribution is f (x) = Given that we have (i) when X = 90, N = 500,
Show that the total probability is 1. Solution: The total probability =
z
∞
−∞
f ( x ) dx =
1 2π σ
z
∞
−∞
e
−
1 x−m σ 2
FG H
IJ K
2
dx
Let
x−m 2σ
=t
= = = Now let ⇒
σ 2 2π σ 1 π 2 π
z z
z
⇒ dx =
∞
2 σ dt
−∞
e − t dt
2
2
∞
−∞ ∞
e − t dt
2
0
e − t dt
y
{As f (– t) = f (t)}
t2 = y or t = 2t dt = dy = =
2 π 1 π 1 π
=
z z z
∞
0
e− y
dy 2 y
∞
0 ∞
y − 1/ 2 e − y dy
1 −1
0
y2
e − y dy
LMBy Gamma integral OP MN e x dx = ΓnPQ
=
1 1 × π =1 Γ(1/2) = π π
z
∞
−x
n −1
0
THEORETICAL DISTRIBUTIONS
247
EXERCISE 2.3
1. Let X be a normal variate with mean 30 and standard deviation 5, find the probabilities that (i) 26 ≤ X ≤ 40, (ii) X ≥ 45 and (iii) | X – 30 | > 5 2. Students of a class were given anaptitute test. Their marks were found to be normally distributed with mean 60 and standard deviation 5. What percentage of students scored more than 60 marks? 3. 1000 light bulbs with a mean life of 120 days are installed in a new factory, their length of life is normally distributed with standard deviation 20 days. How many bulbs will expire in less than 90 days? 4. On a final examination in mathematics, the mean was 72, and the standard deviation was 15. Determine the standard scores of students receiving grades. (i) 60 (ii) 72 (iii) 93. 5. In a male population of 1000, the mean height is 68.16 inches and standard deviation is 3.2 inches. How many men may be more than 6 feet (72 inches)? 6. For a certain normal distribution the first moment about 10 is 40 and the fourth moment about 50 is 48. What is the arithmetic mean and variance of the normal distribution. 7. In a normal distribution whose mean is 2 and standard deviation 3, find a value of the variate such that the probability of the interval from the mean to the value is 0.4115. Find another value such that the probability for the interval from x = 3.5 to that value is 0.2307. 8. A random variable x is normally distributed with m = 12 and standard deviation 2. Find P(9.6 < x < 13.8) given that for
x x = 0.9, A = 0.3159 and for = 1.2, A = 0.3849. σ σ
9. There are six hundred students in P.G. class and the probability for any student to need a particular book on statistics from the University Library on one day is 0.05. How many copies of the book should be kept in the University Library, so that the probability may be greater than 0.90 that none of the student needing a copy from the Library has to come back disappointed. 10. If shells are classified as A, B, C according as the length breadth index as under 75, between 75 and 80, or over 80, find approximately (assuming that distribution is normal) the mean and standard deviation of a series in which A are 58%, B are 38%, and C are 4% being given that f (t) =
1 2π
11. Prove that for the normal distribution, the quartile deviation, the mean deviation and the standard deviation are approximately in the ratio 10 : 12 : 15.
Correlation and Regression
Modern age is an age of planning. Economic planning means direction and control of economic resources both current and potential to meet the social and economic objectives of the state. In absence of statistics planning cannot be imagined. All economic plans of a country are based on statistical data of economic activity of that country. Statistics is the branch of scientific method which deals with data obtained by counting or measuring the properties of population of natural phenomena. Statistics is a branch of science which deals with the collection of data. Classification and tabulation of numerical facts as the basis for explanation, description and comparison of phenomena. Modern statistics takes into consideration however, biological, astronomical and physical as well as social phenomena. Statistics methods are used in every department of human activity. Psychology, education, public health, agriculture, business, economics and administration. Statistics methods are also useful to stock-exchange brokers, bankers, speculators and public utility. Concerns like water works, electric supply companies, railways etc. Statistics analysis of railway working is very important in their expansion programmes. The records on the basis of demand in several parts of the year, railways authorities make adequate provision for services by increasing the length of trains providing special trains such as Jammu, Banglore, Goa, Mumbai etc. in summer. Therefore, in every business (big or small) used of statistical data in any form.
3.1. FREQUENCY DISTRIBUTION Frequency distribution is a way to represent the large amount of data according to quantitative characteristics frequency distribution helps the management of the bank for announcing new saving policies for the benefit to its customers.
3.2. BIVARIATE FREQUENCY DISTRIBUTION The study of relationship between two or more variables is one of the most important problems in life. In a bivariate distribution, we find the relationship between the two variables. For example, if we find the heights and weights of 10th class of student, we shall get a bivariate distribution. Here, one variable is related to height and other variable is related to weight.
248
= N , then it define a frequency distribution which is called a bivariate frequency distribution.
3.3. CORRELATION The relationship between two variables such that a change in one is accompanied by change in the other in such a way that an increase in one is accompanied by an increase or decrease in the other, is called a correlation. For example, the number of policemen and the number of crimes, the number of trains and the number of passengers. If there is no relationship indicated between two variables then they are said to uncorrelated or statistically independent. For example, marks in statistics of a child and weights of her mother. It is not necessary that one unit of change in one variable is followed by one unit of change in other variable. It is also not essential that the ratio should be the same in all cases.
3.4. POSITIVE CORRELATION If an increase or decrease in the values of one variable corresponds to an increase or decrease in other variable respectively, then the correlation is said to be positive correlation. For example, age of husbands and wives are known to have a positive correlation. If an increase or decrease in the values of one variable is always followed by a corresponding and proportional increase or decrease in other variable, then the correlation is said to be perfect positive correlation.
3.5. NEGATIVE CORRELATION If an increase or decrease in the values of one variable corresponds to a decrease or increase in other variable respectively, then the correlation is said to be negative correlation. For example, the number of fourth class workers in a college and the number of duty room in a college. If an increase or decrease in the values of one variable is always followed by a corresponding and proportional decrease or increase in other variable, then the correlation is said to be perfect positive correlation.
3.6. LINEAR OR NON-LINEAR CORRELATION When the variations in the values of two variables are in a constant ratio, then correlation is said to be linear correlation. This type of relationship represented by a linear equation of the form Y = aX + b.
250
ADVANCED MATHEMATICS
Otherwise, if the ratio of variation in the values of two variables are not constant, then correlation is said to be non-linear correlation. This type of relationship represented by y = a + bx + cx2 etc.
Note: If the change in one variable has a very little effect in the other variable then there is no correlation.
3.7. COEFFICIENT OF CORRELATION The coefficient of correlation measures the degree of relationship between two or more variable. The coefficient of correlation varies between –1 to +1. If the coefficient of correlation reaches unity then it is perfect. The +1 indicates perfect positive correlation and –1 indicates perfect negative correlation. Whenever the value of coefficient of correlation is zero then there is no correlation.
3.9. KARL PEARSON'S COEFFICIENTS OF CORRELATION Karl Pearson (1867–1936), a great statistician and biologist, developed a formula for the calculation of coefficient of correlation. The correlation coefficient between two variables X and Y denoted by γ, defined as γ (X, Y ) =
Hence, two variables X and Y are uncorrelated. Now let two random variables X and Y with probability density function
R1 , | f (x) = S 4 |0 , T
Now, E(X ) =
−2<x<2 otherwise
1 4
and Y = X 2
z
2
−2
x f ( x ) dx =
and
E(XY ) = E(X 3 ) =
1 x4 = 4 4
LM OP N Q
z
z
2
−2
x dx = 1 4
1 x2 4 2
2
−2
x 3 f ( x ) dx =
z
LM OP N Q
2
=
−2
1 [4 − 4] = 0 8
2
−2
x 2 dx
2
=
−2
1 [16 − 16] = 0 16
∴
Cov (XY ) = E(XY ) – E(X ) . E(Y ) = 0
i.e., the correlation coefficient is zero but the variables are dependent, by the relation Y = X2. Theorem 3: The correlation coefficient is independent of change of origin and scale. Proof: The correlation coefficient, between two variables X and Y is γ (X, Y ) =
3.10. PROBABLE ERROR The coefficient of correlation also has a probable error. It is that amount when added to and subtracted from the coefficient of correlation would give such limits within which we can reasonably expect the values of coefficient of correlation to vary. The formula for probable error is P.E. = (0.6745) ×
1− γ2 . n
The following rules are observed for probable error: (i) If γ < 0.30, then it is insignificant. (ii) If γ < P.E, then it is not at all significant. (iii) If 0.4 < γ < 0.6, then correlation cofficient is normal. (iv) If γ > 0.9, then it is highly significant. (v) If γ > 6 P.E, then it is significant i.e., correlation is certain.
CORRELATION AND REGRESSION
255
3.11. CORRELATION COEFFICIENT FOR A BIVARIATE FREQUENCY DISTRIBUTION For a bivariate frequency distribution
Σf UV −
γ=
LMRΣf U T NMS
2
(Σf U ) 2 − Σf
U RΣf V VS WT
Σf U Σf V Σf
2
( Σf V ) 2 − Σf
UOP VQP W
1/ 2
where, f is the frequency of a particular rectangle in the correlation table, and U=
3.12. RANK CORRELATION OR SPEARMAN'S COEFFICIENT OF RANK CORRELATION Charles Edward Spearman (1906) a great psychologist and statistician, developed a formula for the calculation of coefficient of correlation, which is called the ''Rank correlation coefficient or Spearman's correlation coefficient. Let us suppose that sometimes the condition is such type which cannot be measured quantitatively, but can be ranked among themselves. For example, it is possible for a class-teacher to arrange his students in ascending or descending order of intelligence, even though intelligence cannot be measured quantitatively. There are so many attributes which are incapable of quantitative measurement such as honesty, character, morality etc. Suppose (xi, yi), i = 1, 2, 3, ......, n be the ranks of n individuals of a group corresponding to two characteristics A and B. Assuming that no two individuals are equal in either classification, each individual takes the values 1, 2, 3, .... n. Now,
X= σ2 X 1 n +1 =Y (1 + 2 + 3 + ...... + n ) = n 2
Here, we see that in X-series, 75 occurs 2 times. 64 occurs 3 times and in Y-series, 68 occurs 2 times. If any two or more individuals have same value with respect to characteristics X and Y, then Spearman's rank formula is fails. For these common ranking the correct formula for γ, to Σ d 2 we add
m( m 2 − 1) for each value repeated, where m is the number of times a value occurs. As in given above 12 example,
3.14. SCATTER DIAGRAM OR DOT DIAGRAM In this method the values of two variables (series) are plotted on a graph, taking one on X-axis and the other along Y-axis. There is one dot for one pair of series, there being as many dots on the graph as there pairs. This collection of dots is called a scatter diagram or dot diagram.
3.15. REGRESSION The study of regression has a great importance in physical sciences, where the data are generally in functional relationship. The word 'Regression' was used by sir Francis Galton (1822–1911) in a paper entitled 'Regression towards Mediocrity in Hereditary stature' in his studies connected with the relation between the heights of fathers and heights of sons. Knowledge of regression analysis comes in handy in
270
ADVANCED MATHEMATICS
finding the probable value of one variable for a given value of the other, when the two variables are known to be correlated. Thus, if we know that two series is relating to supply and price are correlated we can find out what would be the effect on price, if the supply of commodity is increased or decreased to a particular level.
3.16. LINE OF REGRESSION Regression is the estimation for prediction of unknown values of one variable from known values of another variable. The line which describes the average relationship between two variables is known as line of regression. The equations describing the regression lines are called regression equations. A regression equation is the algebric expression of regression line. A regression line is one which shows that the best mean values of one variable corresponding to mean values of the other variable. With two series x and y, if the two regression lines range themselves along two straight lines, the correlation between x and y is linear. If the two straight lines of regression coincide, correlation is perfect. If the two lines cut each other at right angles, correlation is zero. Let Y = aX + b be the equation of the line of the best fit of x. Changing the origin to ( X , Y ), where X is the mean of x-series and Y is the mean of y-series. Let x, y be the deviations from the respective means X and Y . ∴ x = X – X, y = ax + b where, x=X– X and y = Y – Y y= Y −Y
Then, equation Y = aX + b is changed in the form
Let (xc, yc) be any point. Then the difference between this point and the above line is yc – axc – b Hence U, the sum of the squares of such distance is U = Σ ( y – ax – b)2, for all values of c. Now for U is minimum, we choose a and b such that
If m1 and m2 are the slops of the lines and θ is the angle between them, we have
m − m1 tan θ = 2 = 1 + m1m2
=
σy 1− γ2 σx γ 1+ σ2 y σ2 x
F GH
I JK F 1 − γ I σ σ = G H γ JK . σ + σ
2 x y 2 x
2 y
.
when γ = 0, tan θ = ∞, θ =
π i.e., the two lines of regression are perpendicular to each other or variables 2
are uncorrelated. When γ = ± 1, tan θ = 0, θ = 0 or π i.e., the lines coincide and therefore is a perfect correlation between the two variables x and y. The estimated value of y is the same for all values of x or vice-versa. Example 2: For two variables x and y with the same mean, the two regression equations are b 1−a . Find also the common mean. y = ax + b and x = αy + β. Show that = β 1− α Solution: Suppose the common mean is m. The given two regression equations are y = ax + b and and x = αy + β y – m = a(x – m) x – m = α(y – m) Comparing equation (3) and (4) with (1) and (2), we get b = m (1 – a) and β = m(1 – α) Dividing (5) by (6), we get ...(5) ...(6) ...(1) ...(2) ...(3) ...(4)
= (1 – γ)2 Σ y2 Since, U be the sum of squares then Σ y2 will not be negative. ⇒ 1 – γ2 ≥ 0 or γ2 ≤ 1 Hence, γ lies between – 1 and + 1. Example 4: Show that the coefficient of correlation is the geometric mean between the two regression coefficients. Solution: The regression coefficient of y on x = γ The regression coefficient of x on y = γ
Also find the equation of the lines of regression. 5. The lines of regression for a data are given as under: 2y – x – 50 = 0, 3y – 2x – 10 = 0. Show that the regression estimate of y for x = 150 is 100, whereas the regression estimate of x corresponding to y = 100 is 145. Explain the difference. 6. The variables x and y are connected by the equations ax + by + c = 0, show that the correlation between them is – 1, if the sign of a and b are like, and + 1 if they are different. 7. A study of prices of a certain commodity at Hapur and Kanpur yield the following data: Hapur Rs Average price per kilo Standard deviation γ between prices at Hapur and Kanpur 2.463 0.326 0.774 Kanpur Rs 2.797 0.207
Estimate from the above data the most likely price (i) At Kanpur corresponding to the price of Rs. 3.052 per kilo at Hapur. (ii) At Hapur corresponding to the price of Rs. 2.334 per kilo at Kanpur. 8. Heights of fathers and sons are given in the inches Height of Father (X) Height of Son (Y) 65 67 66 68 67 64 67 68 68 72 69 70 71 69 73 70
From the two lines of regression and calculate the expected average height of the son when the height of the father is 67.5 inches.
280
ADVANCED MATHEMATICS
9. The equations of two regression lines obtained in a correlation analysis of 60 observations are 5x = 6y + 24 and 1000 y = 768x – 3608. What is the correlation coefficient and what is its probable error ? Show that the ratio of the coefficient of variability of x to that of y is
UNIT V
VARIA ARIATIONS CALCULUS OF VARIATIONS AND TRANSFORMS
In this chapter, we deal with calculus of variations and problems of determining the maximum or minimum values of a definite integral involving in a certain function, strong variation, weak variation, simple variation problems and Euler's equation are also discussed. Calculus of variations has a great importance to solving the problems of dynamics of rigid bodies, as a unifying influence in mechanics, vibration problems and as a guide in the mathematical interpretation of many physical phenomena. First in 1744 Euler, discovered the basic differential equation for a minimizing curve. The calculus of variations also deals with minimum problems depending upon surfaces, analytical mechanics, optimization of orbits and the study of extrema of functionals. In differential calculus, we discuss the problems of maxima and minima of functions. The calculus of variations is concerned with maximizing or minimizing functionals.
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CHAPTER
1
Calculus of Variations
1.1. FUNCTIONAL A real valued function f whose domain is the set of real functions is called a functional. Let us consider the problem to determine a curve y = y(x) through two points P(x 1, y1) and Q(x 2, y2 ) for which y(x1) = y1, y(x2) = y2 such that
Y
Q(x2, y2)
is a minimum.
z
z
x2
x1
1+
F dy I H dx K
2
dx
...(1.1)
O
P(x1, y1) X
In general terms, to determine the curve y = y(x), where a given function f x, y,
F H
dy dx
I K
y(x1) = y1 and y(x2) = y2 such that for
or f (x, y, y′),
f ( x, y, y ′ ) dx is maximum or minimum
x2
x1
...(1.2)
Hence, an integral such as (1.1) and (1.2), which assumes a definite value for functions of the type y = y(x) is known as functional.
Note. A function y = y(x), which extremizes a functional is known as extremal or extremizing function.
1.2. EULER'S EQUATION A necessary condition for I=
to be an extremum (maximum or minimum) is that
∂f d ∂f − = 0, ∂y dx ∂y ′
z
x2
x1
f ( x, y, y ′) dx
with y(x1) = y1 and y (x2) = y2
FG IJ H K
where y′ =
dy dx
This is called Euler's equation.
283
284
Proof: Given that
I=
Let y = y(x) be the curve joining two points P(x1, y1) and Q(x2, y2) which makes the given function I an extremum. Let Y = y(x) + αg(x) ...(1.4) be a neighbouring curve joining these point P and Q, where α is a small parameter but independent of x and g(x) is an arbitrary differentiable function. At the end point P and Q, g (x1) = 0 = g (x2) When α = 0, neighbouring curve (1.4) becomes Y = y(x), which is extremal If we replace y by Y in equation (1.3) then, we get
z
ADVANCED MATHEMATICS
x2
x1
f ( x, y, y ′ ) dx
...(1.3)
Y
Y = y(x
) ) + ag(x
Q(x2, y2)
y(x)
P(x1, y1) y1 O
y=
y2
X
...(1.5)
z
x2
x1
f ( x, Y , Y ′ ) dx =
= I(α) which is a function of α I (α) =
x2 x1
Hence,
The functional I (function of α) is a maximum or minimum for α = 0, when I′(α) = 0 Differentiating equation (1.6) under the integral sign by Leibnitz's rule, we have ...(1.7)
Here f is a function of x, y and y′ Differentiating f with respect to x, we get
z
x2
x1
f ( x , y, y ′ ) dx
df ∂f ∂f dy ∂f dy ′ = + + dx ∂x ∂y dx ∂y ′ dx
or Now consider
d d ∂f ∂f ∂f y′ = y′ + y ′′ dx dx ∂y ′ ∂y ′ ∂y ′
FG H
df ∂f ∂f ∂f = + y′ + y ′′ dx ∂x ∂y ∂y ′
...(1.13) ...(1.14)
IJ K
FG IJ H K
Subtracting (1.14) from (1.13), we get
df d d ∂f ∂f ∂f ∂f − y′ y ′ − y′ = + dx ∂y ′ dx dx ∂y ′ ∂x ∂y
FG H
IJ K
FG IJ H K
286
ADVANCED MATHEMATICS
R FG IJ S H K T d F ∂f GH f − y ′ ∂∂yf ′ IJK − ∂x = 0 dx
Hence
d d ∂f ∂f ∂f ∂f f − y′ − = y′ − dx ∂y ′ ∂x ∂y dx ∂y ′
FG IJ U H KV W
(Using Euler's equation)
d ∂f ∂f f − y′ − =0 dx ∂y ′ ∂x
LM N
OP Q
...(1.15)
2. Since Let Then
∂f is also a function of x, y, y′. ∂y′
∂f = h (x, y, y′) ∂y′
...(1.16)
d ∂f dh ∂h ∂h dy dh dy ′ + + = = ∂x ∂y dx dy ′ dx dx ∂y ′ dx
FG IJ H K
= =
∂ ∂f ∂ ∂f ∂ ∂f y′ + y ′′ + ∂x ∂y ′ ∂y ∂y ′ ∂y ′ ∂y ′
FG IJ H K
FG IJ H K
FG IJ H K
∂2 f ∂2 f ∂2 f + y′ + y ′′ ∂x∂y ′ ∂y∂y ′ ∂y ′ 2
...(1.17)
Using (1.12) and (1.17), we get
∂f ∂2 f ∂2 f ∂2 f − + y′ + y ′′ = 0 ∂y ∂x∂y ′ ∂y∂y ′ ∂y ′ 2
or
R S T
U V W
y ′′
∂2 f ∂2 f ∂2 f ∂f =0 + y′ + − 2 ∂y ′ ∂y∂y ′ ∂x∂y ′ ∂y
...(1.18)
1.4. SOLUTION OF EULER'S EQUATIONS There are several situations in which we can easily be obtain the solution of Euler's equation in following cases: Case I: When f is independent of x then By equation (1.15), we have
∂f d f − y′ =0 ∂y ′ dx
∂f = 0. ∂x
FG H
IJ K
CALCULUS OF VARIATIONS
287
Integrating, we get
f – y′
∂f = Constant ∂y′
...(1.19)
Case II: When f is independent of y then
d ∂f dx ∂y ′
By equation (1.12), we have
FG IJ = 0 H K
∂f =0 ∂y
Integrating, we get
∂f = Constant ∂y′
∂f = 0. By equation (1.21), we have ∂y′
...(1.20)
Case III: When f is independent of y′ then
∂f =0 ∂y
...(1.21)
Case IV: When f is independent of x and y then
∂f ∂f = 0 and =0 ∂y ∂x
∂2 f = 0 and ∂x∂y′
By equation (18), we have y″ If
∂2 f =0 ∂y∂y′
∂2 f =0 ∂y′ 2
...(1.22)
∂2 f ≠ 0 then y″ = 0 ∂y′ 2
y = ax + b. ...(1.23)
which gives a solution of the form
Note: Any function which satisfies Euler's equation is called a Extremal. Extremal is obtained by solving the Euler's equation.
d (δy′) dx Here, if δy is small then the variation is called strong variation and if δy′ is small, then the variation is called weak variation.
where
δy =
1.6. ISOPERIMETRIC PROBLEMS The problem of maximum or minimum with constraints, it is required to determine the maximum or minimum of a function of several variable g (x1, x2, ......, xn) where the variables x1, x2, ......., xn are connected by some given condition or relation called a constraints. In calculus of variations, in some problems, it is necessary to determine extremum of an integral while one or more integrals involving the same variables and the same limits are to be kept constant. Such type problems are called isoperimetric problems and mostly solved by the method of Lagrange's multipliers. For example, to determine the shape of the closed curve of the given perimeter enclosing maximum area. Let the isoperimetric problem consists of determine a function f (x) which extremizes the functional I=
Subject to the condition that the another integral J=
x2 x1
To solve this type of problem, we use the method of Lagrange's multipliers and consider an integral H= where
z z
x2
x1
f ( x, y, y ′) dx
...(1.24)
g( x, y, y ′ ) dx = C
...(1.25)
h = f + λg and λ is the Lagrange's multiplier.
z
x2
x1
h( x , y, y ′ ) dx
...(1.26)
For H is extremum, if I is extremum because J is constant. Then, the modified Euler's equation is given by
Applying the given boundary condition in (iii), we get y (0) = 1 = 0, y(1) = 2 = – which is not possible
1 , again contradiction. 2 Hence, these are no extremal because it is not satisfy the boundary condition.
Example 3: Prove that the shortest distance between two points in a plane is a straight line. Solution: Let P(x1, y1) and Q(x2, y2) be the two given points and s is the length of the arc joining these points. Then with S=
z z
x2 x1
ds =
x2
x1
1+
F dy I H dx K
2
dx =
z
x2
x1
1 + y ′ 2 dx
...(i)
y(x1) = y1, y(x2) = y2. If S satisfies the Euler's equation then it will be minimum. The Euler's equation is In equation (i), Then
∂f d ∂f − ∂y dx ∂y ′
FG IJ = 0 H K
...(ii)
Y
f = 1 + y′2
∂f = 0, ∂y ∂f 1 = (1 + y′2)–1/2 (2y′) ∂y′ 2
Q(x2, y2) s P(x1, y1) O X
Putting these values of
d F 0– GH dx G
∂f ∂f and in (ii), we get ∂y ∂y′
y′ 1 + y′
y′
2
I JJK = 0
or
d dx
Integrating both sides, we get
1 + y′ 2
F GGH
y′ 1 + y′
2
I =0 JJK
= a (Constant)
Squaring both sides, we get y′2 = a2 (1 + y′2) or or or which is a straight line. y′2 = or y′2 (1 – a2) = a2 or y′2 = m2
Example 4: Find the path on which a particle in a absence of friction, will slide from one point to another in the shortest time under the action of gravity. Solution: Suppose the particle start sliding on the curve OA from O with zero velocity. Let OP = S and t be the time taken from O to P. Using the principle of work and energy, we have K.E. at P – K.E. at O = Work done in moving the particle from O to P. ⇒ or or or
1 mv2 – 0 = mgy 2 v2 = 2gy
Now the geodesic on the sphere r = a is the curve for which s is minimum. From (ii), we have f = a 1 + sin 2 θ . φ ′ 2 Which is a function of θ and φ′ while φ is absent. Then the Euler's equation reduces to
∂f = c1 (Constant) ∂φ ′ a sin 2 θ . φ ′ a (1 + sin 2 θ . φ ′ 2 )
Since, the curve passes through the origin O and A for which y = 0, so c1 = 0 πy2 + or
2 πyλ 1 + y′
2
=0
or y +
2λ 1 + y′2
=0
y 1 + y ′ 2 = – 2λ
squaring both sides, we get
dy 1+F I H dx K
or
2
4λ2 = 2 y
= dx
or
dy = dx
4λ2 − y 2 y2
y dy 4λ2 − y 2
Integrating both sides, we get – or At 0, from (iv), we get Then by (iv), we have Squaring both sides, we get which is a circle Hence, on revolving the circle about x-axis, the solid formed is a sphere. Example 8: Show that the functional
2. Find a complete solution of the Euler's equation for 3. Find the extremal of the functional I = 4. Find the extremals of the functional 5. Solve the Euler's equation for 6. Solve the Euler's equation for
Z-Transform
INTRODUCTION In this chapter, we shall discuss a new type of transform, Z-transform. The progress of communication engineering is based on discrete analysis. Z-transform plays an important role in solving difference equation which represent a discrete system. Thus, the study of Z-transform is necessary part for engineers and scientists. Z-transform has many properties like Laplace transform but Z-transform operates on a sequences {un} of discrete integer-(½) valued arguments (n = 0, ± 1, ± 2, .....) and Laplace transform operates on a continuous function that is the main difference between these transforms. | 677.169 | 1 |
Showing 1 to 30 of 36
1.1 Sequences
Learning objectives:
To define an infinite sequence of real numbers and to
discuss its convergence and divergence.
To define the boundedness of a sequence and to prove
that every convergent sequence is bounded.
To define Non-decreasing an
3.2
Limits and Continuity
Learning objectives:
To define limits of function of two and three variables
To define the continuity of a function of two and three
variables at a point
AND
To practice the related problems
Limits and Continuity
In this module
3.4
Homogeneous Functions and Eulers Theorem
Learning objectives:
To define a homogeneous function of two and three
variables.
To state and prove Eulers Theorem for functions of two
variables.
AND
To practice the related problems.
In this module we def
3.6.
Derivatives of composite functions and implicit
functions (Chain Rule)
Learning objectives:
To discuss the chain rule for functions two and three
independent variables.
To discuss two and three variable Implicit differentiation
AND
To practice the re
3.3
Partial Derivatives
Learning objectives:
To define partial derivatives.
To calculate partial derivatives.
To discuss the second and higher order partial derivatives.
To state mixed derivative theorem.
AND
To practice the related problems.
Partial
3.1
Preliminaries
Learning objectives:
To defined a real valued function of independent
variables.
To study the level curves of a function of two variables and
level surfaces of a function of three variables.
To defined the domain and range of function
3.7
JACOBIANS
Learning objectives:
To define the Jacobian of a coordinate transformation.
To study properties of Jacobians.
AND
To practice the related problems.
Jacobians arise naturally when there is a coordinate
transformation. This concept is named
3.9
Maxima and Minima
Learning objectives:
To define local extrema of functions of two variables
To state second derivative test for local extreme values.
AND
To practice the related problems
Maxima and Minima
Continuous functions of two variables atta
3.10
Lagranges Method of Multipliers
Learning objectives:
To find the extremum of a function subject to given set of
constraints by the method of Lagranges Multipliers.
AND
To practice the related problems
Lagranges Method of Multipliers
In many practica
3.8
Taylors Expansion
Learning objectives:
To state and prove Taylors Theorem for functions of two
variables
To compute the maximum absolute error for the linear and
quadratic approximations of f (x, y) about a point
AND
To practice the related problem
FINITE DIFFERENCES
Assume that we have a table of values , ,
=, 0, 1, 2 2 , . of
any function
,( the values of
being equally spaced, i.e.
) ,
=, 0, 1, 2 . . Suppose that we are required to
,
recover the values of
for some intermediate values of , or to
.
NEWTONS INTERPOLATION FORMULAE
Interpolation
The statement
= ( ),
means: Corresponding to every value of in the range
, there exists one or more values of . Assuming that
( ) is single-valued and continuous and that it is known
explicitly, then the values
NEWTON-RAPHSON METHOD
This method is generally used to improve the result
obtained by one of the previous methods. Let
be an
approximate root of ( ) = and let 0 )
be the
correct root so that ( 0 ) = Expanding (
by
Taylors series, we obtain
(
+
(
(
) =
Neg
THE METHOD OF FALSE POSITION
This is the oldest method for finding the real root of a
nonlinear equation ( ) = 0 and closely resembles the
bisection method. In this method, also known as regular
falsi or the method of chords, we choose two points and
such
Solution of Algebraic and
Equations: Bisection Method
Transcendental
Introduction
In scientific and engineering studies, a frequently occurring
problem is to find the roots of equations of the form.
( )=0
(1)
If ( ) is a quadratic, cubic or a biquadratic
CENTRAL DIFFERENCE INTERPOLATION FORMULAE
In the preceding module, we derived and discussed
Newtons forward and backward interpolation formulae,
which are applicable for interpolation near the beginning
and end respectively, of tabulated values. We shall,
4.8
NUMERICAL INTEGRATION
The general problem
stated as follows.
( ,
,( ), ) , , ( ,
not known explicitly, it
the definite integral
of numerical integration may be
Given a set of data points
( , where ( is
of a function
is required to compute the value of
3.5
Differentiability
Learning objectives:
To state and prove Increment theorem for functions of
two variables.
To define the differentiability of a function of two
variables at a point.
To prove Differentiability implies Continuity.
AND
To practice t
2.9
Solution of Euler-Cauchy equation
Learning objectives:
To learn the method of solving Euler Cauchy equation
To study the equations reducible to Euler Cauchy form
and their solutions
AND
To practice the related problems
Solution of Euler-Cauchy equa
1.4
Integral Test
Learning objectives:
To state and prove integral test for the convergence and
divergence of series of positive real numbers.
To study convergence and divergence of
And
To practice the related problems.
, the -series.
Integral Test
In
1.3 Infinite Series
Learning objectives:
To define the convergence and divergence of an infinite
series of real numbers.
To discuss the convergence and divergence of Geometric
Series of real numbers.
To state the
term test for divergence.
To prove the
1.5
Comparison Tests
Learning objectives:
To state and prove the comparison and limit comparison
tests for the convergence of the series of real numbers.
And
To practice the related problems.
Comparison Tests
We have learnt how to determine the converge
1.6
The Ratio and Root Tests
Learning objectives:
To learn the Ratio Test and the Root Test for the
convergence and divergence of series of real numbers.
And
To practice the related problems.
The Ratio and Root Tests
This module is devoted for the study
1.7
Alternating Series, Absolute and Conditional
Convergence
Learning objectives:
To state and prove the alternating series test for the
convergence of alternate series.
To discuss Absolute and Conditional convergence of series
of real numbers.
To disc
1.2
Limits of Sequences
Learning objectives:
To state the rules for the algebra of convergent sequences.
To study the sequence versions of
Sandwich Theorem
Continuous Function Theorem
LHopitals Rule
To prove some commonly occurring limits.
And
To p
1.8. Power Series
Learning objectives:
To define the Power series and to prove the convergence
theorem for Power series.
To find the Interval of convergence and radius of
convergence of a Power series.
To state the Term by Term differentiation and inte
1.9
Taylor and Maclaurin Series
Learning objectives:
To define the Taylor series and Maclaurin series generated
by a function at a point.
To define a Taylor polynomial of a given order generated
by a function at a point.
To state and prove the Taylors
2.3.
Exact First Order Differential Equations
Learning objectives:
To state the necessary and sufficient condition for the
differential equation , + , to be exact
To find the integrating factors and general solution of nonexact differential equations.
A | 677.169 | 1 |
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Linear Algebra: A First Course presents an introduction to the fascinating subject of linear algebra. As the title suggests, this text is designed as a first course in linear algebra for students who have a reasonable understanding of basic algebra. Major topics of linear algebra are presented in detail, with proofs of important theorems provided. Connections to additional topics covered in advanced courses are introduced, in an effort to assist those students who are interested in continuing... Topic: Linear algebra
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These pages are a collection of facts (identities, approximations, inequalities, relations, ...) about matrices and matters relating to them. It is collected in this form for the convenience of anyone who wants a quick desktop reference . Lecture Notes Collection FreeScience.info ID767 Obtained from Topics: Linear Algebra, "
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While I began researching for this book on linear algebra, I was a little startled. Though, it is an accepted phenomenon, that mathematicians are rarely the ones to react surprised, this serious search left me that way for a variety of reasons. First, several of the linear algebra books that my institute library stocked (and it is a really good library) were old and crumbly and dated as far back as 1913 with the most 'new' books only being the ones published in the 1960s. Topic: linear algebra
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Chapter 12 of the Precalculus text and Chapter 12 of the Honors Algebra 2 Trig text deal with matrices. Here we go over some of the applications of the wonderful world of the matrix. After some calculus, you may enjoy taking a course in Linear Algebra! Topics: matrix, linear algebra
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Because any enumeration of a set of simple roots is related to any other enumeration by some �nite number of permutations, the relationship between any two Cartan matrices for a root system is an isomorphism by the conjugate product of permutation matrices. Topics: Maths, Linear Algebra and Geometry, Linear Algebra, Mathematics Source:
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This section explains three Fourier series: sines, cosines, and exponentials e^(ikx) . Square waves (1 or 0 or _1) are great examples, with delta functions in the derivative. We look at a spike, a step function, and a ramp�and smoother functions too. Topics: Maths, Linear Algebra and Geometry, Linear Algebra, Mathematics Source:
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We shall associate a �Cartan matrix� to the system _ _ _ and derive some properties of this matrix. An abstract Cartan matrix will be any square matrix with this list of properties. It turns out that an abstract Cartan matrix essentially determines the root system. In this paper we will work toward making this statement precise and proving it. The problem of classi�cation of root systems is reduced then to the classi�cation of the Cartan matrices. Topics: Maths, Linear Algebra and Geometry, Linear Algebra, Mathematics Source:
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This summary begins with two lists that use most of the key words of linear algebra. The �rst list applies to invertible matrices. That property is described in 14 di_erent ways. The second list shows the contrast, when A is singular (not invertible). Topics: Maths, Linear Algebra and Geometry, Linear Algebra, Mathematics Source:
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Now we are ready to introduce the notion of a re�ection in a Eu_clidean space. A re�ection in V is a linear mapping s : V_ V which sends some nonzero vector _ _ V to its negative and �xes pointwise the hyperplane H_ orthogonal to _. To indicate this vector, we will write s = s_. The use of Greek letters for vectors is traditional in this context. Topics: Maths, Linear Algebra and Geometry, Linear Algebra, Mathematics Source:
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A basic introduction to the Calculus and Linear Algebra. The goal is to make students mathematically literate in preparation for studying a scientific/engineering discipline. The first week covers differential calculus: graphing functions, limits, derivatives, and applying differentiation to real-world problems, such as maximization and rates of change. The second week covers integral calculus: sums, integration, areas under curves and computing volumes. This is not meant to be a comprehensive... Topics: ars digita, calculus, linear algebra | 677.169 | 1 |
Mathematics
Qualification:
A Level
Duration:
2 years
Level:
3
Curriculum Manager:
Dr David Lynch
What is Mathematics?
It's the last 5 minutes of the match and Eric Dier is about to take a free kick that could take England to the World Cup finals. The crowd is crying out for him to score. What speed and direction does he need to kick the ball?
The ice cap at the North Pole is melting. When warmer climes reach further north, the ice covering the frozen wastes of Russia will recede to expose peat that will release a massive amount of CO2. Is this global warming?
To reduce packaging costs, what size should 400g cylindrical tins of baked beans be to minimise the amount of steel used? What speed does a sky diver reach before pulling her parachute?
If you invest £1000 every year into a savings account paying 3% interest per annum, how much money will be in the account after 5 years, and how long will it take for you to save £20,000?
Maths will allow you to answer these and many other questions!
What Mathematics will I do?
The new linear A-level Maths course is made up of two-thirds Pure Maths, one-sixth Mechanics and one-sixth Statistics. All the content is compulsory, i.e. there are no options.
In Pure Maths you extend your skills in algebra, trigonometry and geometry from GCSE and you will be introduced to new areas of Maths such as calculus, sequences and series, and exponentials. Calculus will help you work out the most efficient size of tin cans and how fast sky divers reach. Sequences and series will allow you to work out the sum of an infinite number of numbers and work out your savings account balance or your mortgage payments. And exponentials will enable you to make predictions for the world's population in 2050. Algebra is the basis of almost all A-level Maths, building on the skills you learn at GCSE (such as quadratics and simultaneous equations), and subsequently forms the biggest part of the course.
Statistics will give you the skills to analyse the proportion of CO2 in the atmosphere using the Normal Distribution to determine if there is a significant increase that could contribute to global warming. You will be able to determine whether the brain mass of mammals has a correlation with body size and how strong the correlation is. Using measures of spread you will be able to identify the differences in the birth and death rates of less economically developed countries to target money that will improve the economy and health of the population.
In Mechanics you will learn about Newton's Laws of Motion, including forces, moments, projectiles and friction. Newton's Laws are used to work out the acceleration of a car pulling a caravan up a hill. The moment of a force is the turning effect of that force, and is used when calculating where two kids, of different weights, need to sit to balance a see-saw. When a ball is kicked, projectile motion is used to calculate its path. And the Police use knowledge of friction and the length of skid marks to work out whether a car, involved in a crash, was speeding at the time of the accident.
What's different about the new linear A-level Mathematics?
In the new A-level Maths there is a bigger emphasis on problem solving, modelling and reasoning. Modelling in mathematics is about starting with a real life situation, making assumptions to simplify and deciding which features are important and which are not. This allows mathematics to be used to provide information about the real situation. You will also need to work with large data sets, using spreadsheets to help you interpret the data. And, because the course is linear, all the exams are at the end of the 2nd year.
I've heard about Further Maths. What is this?
If you are a good Mathematician and likely to get a grade 7, or preferably 8 or 9, at GCSE plus a range of other good grades, you should consider taking A-level Mathematics and Further Maths. See Further Mathematics for more information.
Progression
Mathematics is highly valued in a very wide range of careers and professions as it provides good evidence of the ability to think clearly and logically. It is accepted almost everywhere as a good entry qualification for practically every subject at degree level. It is often a requirement for anyone interested in Physics, Engineering, Statistics, Computing, Economics and other scientific and business related careers.
If you are considering studying Mathematics, Engineering or Physics at university, you should also consider studying Further Maths, which would add extra depth and breadth to your understanding in preparation for these degree courses. Studying Further Mathematics is seen as an indication of high academic aptitude which is often essential when making applications to certain courses and universities. Please see the A-level Further Maths subject leaflet to find out more.
Entry requirements
Students should normally have achieved an A*- C grade profile at GCSE. For GCSE English and Maths where a new grading system has been introduced, a Grade 4 is equivalent to a Grade C.
For A-level Maths you must have at least a Grade 6 (preferably a 7 or above) in GCSE Maths. In addition, you need to enjoy Maths – especially algebra!
If you are expecting a grade 7 or above, you could consider taking A-level Maths and Further Maths.
Course costs
Students will be expected to provide their own stationery and textbooks. All students must also have a scientific calculator with natural display that can calculate Binomial & Normal probabilities. The best option will be the Casio FX-991EX (around £20), but it isn't available in the UK yet. Some students also buy a graphics calculator, such as the Casio FX-9750GII (around £60). In addition, all students must purchase the A-level Maths Student Course Materials via the college's Online Store. This will be about £5 and includes interactive ICT resources and exercise books to write your notes in.
If the costs of equipment, materials and trips may cause you financial hardship, you may wish to read through details of our financial support scheme on our website. | 677.169 | 1 |
Abstract:
Programs included are: Section E, Additional Mathematics Prograns GLPSA1-two phase simplex method of linear programming. TMFCEV-time function evaluation (engineering applications). CALC-calculator for numbers of up to 100 digits. CXEXP-raise a complex number to a real or complex power. GSIMEQ-solves simultaneous linear equations. CXARTH-performs vector arithmetic operations. POLY-finds polynomial to approximate a table of x-y data. POLFIT-performs multiple regression/coffelation analysis. FREQ-finds number of data points (frequency) within limits; data can be in a file or from terminal. RANDOM-generates random numbers with user-specified sample and population sizes. ANVAR3-computes analysis of variance table for a 2-way classification of variables design in which a single observation is made for each combination of levels. ANVAR4-computes analysis of variance table for a 2-way classification of variables factorial design with replicated observations. GANOVA-computes analysis of variance table for a 2-way classification of variables; data is entered down columns. MULTX-performs a least squares curve fit to one of seven functions; operation is interactive at the terminal. POLSUB-exercises students in polynomial subtraction operations. GINTLP-solves linear programniing problems with variables of values 1 and 0. SIPRAC-exercises students in mathematical operations on signed numbers. KR20-item analysis and Kuder- Richardson formula 20. DE10R-solves a first order differential equation (Runge-Kutta). DE20R-solves second order differential equation (Runge-Kutta). SUNSET-computes Greenwich Mean Time (or other time) for sunrise and sunset given a particular week and a given latitude and longitude. WAVES-plots effects of changing wavelength, amplitude and phase on two waves and their sum. SPCTRA-plots optical absorption spectra of two species equilibrium mixtures. FACrRL-produces any desired factorial and preceding factorials. SQRZ-finds the square root of a complex number. BISQAR-drill for students on squaring binomials.
Section F, Teacher Assistance Programs AVERG1-averages and curves grades. GRADE-given the number of questions on a test, prints list of percent score vs. number of questions right or wrong. STAT-performs statistical analysis of student grades. SCORES-computes mean, standard deviation, and standard scores for student grades. | 677.169 | 1 |
Differential Equations are the language in which the laws of nature are expressed. Understanding properties of solutions of differential equations is fundamental to much of contemporary science and engineering. Ordinary differential equations (ODE's) deal with functions of one variable, which can often be thought of as time. Topics include: Solution of first-order ODE's by analytical, graphical and numerical methods; Linear ODE's, especially second order with constant coefficients; Undetermined coefficients and variation of parameters; Sinusoidal and exponential signals: oscillations, damping, resonance; Complex numbers and exponentials; Fourier series, periodic solutions; Delta functions, convolution, and Laplace transform methods; Matrix and first order linear systems: eigenvalues and eigenvectors; and Non-linear autonomous systems: critical point analysis and phase plane diagrams | 677.169 | 1 |
Build essential skills while having fun with Home Workbooks! Now updated with fun, colorful pages and engaging art, each book measures 7" x 9.25" and is filled with 64 pages of age-appropriate activities, puzzles, and games. These teacher-approved books are perfect for home, school, summer breaks, and road trips! Skills covered include one- and two-digit addition and subtraction, problem solving, and more! An incentive chart and 140 full-color stickers are also included to help parents or teachers track student progress. Home Workbooks are available for prekindergarten through grade 3 students, and feature titles in a wide variety of skill areas to suit any need.
Fortunately, there's Schaum's. This all-in-one-package includes more than 650 fully solved problems, examples, and practice exercises to sharpen your problem-solving skills. Plus, you will have access to 25665 fully solved problems
Concise explanations of all geometry concepts
Support for all major textbooks for geometry courses
Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum's to shorten your study time--and get your best test scores!
Don't be tripped up by trigonometry. Master this math with practice, practice, practice!
Practice Makes Perfect: Trigonometry is a comprehensive guide and workbook that covers all the basics of trigonometry that you need to understand this subject. Each chapter focuses on one major topic, with thorough explanations and many illustrative examples, so you can learn at your own pace and really absorb the information. You get to apply your knowledge and practice what you've learned through a variety of exercises, with an answer key for instant feedback. Offering a winning solution for getting a handle on math right away, Practice Makes Perfect: Trigonometry is your ultimate resource for building a solid understanding of trigonometry fundamentals.
Most math and science study guides are a reflection of the college professors who write them-dry, difficult, and pretentious.
The Humongous Book of Trigonometry Problems is the exception. Author Mike Kelley has taken what appears to be a typical trigonometry workbook, chock full of solved problems more than 750! and made notes in the margins adding missing steps and simplifying concepts and solutions, so what would be baffling to students is made perfectly clear. No longer will befuddled students wonder where a particular answer came from or have to rely on trial and error to solve problems. And by learning how to interpret and solve problems as they are presented in a standard trigonometry course, students become fully prepared to solve those difficult, obscure problems that were never discussed in class but always seem to find their way onto exams | 677.169 | 1 |
: Pearson New International Edition
ISBN :
9781292024714
Publisher :
Pearson Education Limited
Author(s) :
Sullivan, Michael
Publication Date :
29 Jul 2013
Edition :
9
Overview
Mike Sullivan's 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, Algebra and Trigonometry has evolved to meet today's course needs, building on these hallmarks by integrating projects and other interactive learning tools for use in the classroom or online.
New Internet-based Chapter Projects apply skills to real-world problems and are accompanied by assignable MathXL exercises to make it easier to incorporate these projects into the course. In addition, a variety of new exercise types, Showcase Examples, and video tutorials for MathXL exercises give instructors even more flexibility, while helping students build their conceptual understanding. | 677.169 | 1 |
Functions and Their Graphs
Polynomial and Rational Functions
Exponential and Logarithmic Functions
Trigonometric Functions
Analytic Trigonometry
Additional Topics in Trigonometry
Sequences, Series and Proof by Induction
Topics in Analytic Geometry
Limits and Introduction to Calculus
^^sections from my pre-calc class. Remember how to solve equations and the types of graphs, conic sections and such. Also, this question is in the language and culture section | 677.169 | 1 |
Item description for An Introduction to Riemann Surfaces, Algebraic Curves and Moduli Spaces (Theoretical and Mathematical Physics) by Martin Schlichenmaier...
This book gives an introduction to modern geometry. Starting from an elementary level the author develops deep geometrical concepts, playing an important role nowadays in contemporary theoretical physics. He presents various techniques and viewpoints, thereby showing the relations between the alternative approaches.
At the end of each chapter suggestions for further reading are given to allow the reader to study the touched topics in greater detail.
This second edition of the book contains two additional more advanced geometric techniques: (1) The modern language and modern view of Algebraic Geometry and (2) Mirror Symmetry.
The book grew out of lecture courses. The presentation style is therefore similar to a lecture. Graduate students of theoretical and mathematical physics will appreciate this book as textbook. Students of mathematics who are looking for a short introduction to the various aspects of modern geometry and their interplay will also find it useful. Researchers will esteem the book as reliable 87 Martin Schlichenmaier
Martin Schlichenmaier is full professor for mathematics at the University of Luxemburg. He has held several teaching and research positions in the mathematics department of the University of Mannheim | 677.169 | 1 |
Math 320
Our Math 320 page contains four topics that instructors at Solano believe our students need to already know in order to be successful in this course. If you find that you need more than just a quick review of these topics, you need to enroll in Math 310, which starts students at the beginning of the mathematics courses offered at Solano Community College, and has no prerequisites.
Here's what you need to do:
1) Watch each video (they are between five and ten minutes each).
2) Work on the sample problems linked underneath each video.
3) Check your solutions using the link underneath each video.
4) At the bottom of the page is sample final for our Arithmetic course, Math 310. If you take this exam, and find that you have a good knowledge of all the material, you have a good chance of being successful in Math 320. | 677.169 | 1 |
This valuable learning tool applies the principles and practices of basic math to everyday business problems and situations. Using a reality-based, practical approach, it guides future business managers through easy-to-follow explanations, worked examples, and exercises—making sure that one concept is mastered before progressing to the next. Chapter topics include working with whole numbers, fractions, and decimal numbers; banking records; payroll; business measurements; percent; insurance; simple interest; promissory notes; installment loans; consumer credit; compound interest and present value; discounts; markup; inventory and turnover; depreciation; taxes; investments; financial statement analysis; and statistics. For individuals preparing for a successful business career, and personal financial success.
REVIEWS for Practical Business Math | 677.169 | 1 |
This practical guide eases you into the subject of probability using familiar items such as coins, cards, and dice. As you progress, you will master concepts such as addition and multiplication rules, odds and expectation, probability distributions, and more. You'll
It's a no-brainer! You'll learn about:
Classical probability
Game theory
Actuarial science
Addition rules
Bayes' theorem
Odds and expectation
Binomial distribution
Simple enough for a beginner, but challenging enough for an advanced student, Probability Demystified, Second Edition, helps you master this essential subject.
Allan G. Bluman is professor emeritus at the Community College of Allegheny County, South Campus. He taught mathematics and statistics in high school, college, and graduate school for more than 35 years. Mr. Bluman is the recipient of "An Apple for the Teacher Award" for bringing excellence to the learning environment and the "Most Successful Revision of a Textbook" award from McGraw-Hill. He is the author of three mathematics textbooks and several highly successful books in the DeMYSTiFieD series, including Pre-Algrebra DeMYSTiFieD, Math Word Problems DeMYSTiFieD and Business Math DeMYSTiFieD.1780971 Professional 2012-03884121
Descrizione libro McGraw-Hill Professional Publishing. Paperback. Condizione libro: New. Paperback. 272 pages. Stack the odds in your favor for mastering probabilityDont leave your knowledge of probability to chance. Instead, turn to Probability Demystified, Second Edition, for learning fundamental concepts and theories step-by-step. This practical guide eases you into the subject of probability using familiar items such as coins, cards, and dice. As you progress, you will master concepts such as addition and multiplication rules, odds and expectation, probability distributions, and more. Youll Its a no-brainer! Youll learn about: Classical probabilityGame theoryActuarial scienceAddition rulesBayes theoremOdds and expectationBinomial distribution Simple enough for a beginner, but challenging enough for an advanced student, Probability Demystified, Second Edition, helps you master this essential subject. This item ships from multiple locations. Your book may arrive from Roseburg,OR, La Vergne,TN. Paperback. Codice libro della libreria 9780071780971 | 677.169 | 1 |
Overview
About The Product
PRINCIPLES OF ENGINEERING will help your students better understand the engineering concepts, mathematics, and scientific principles that form the foundation of the Project Lead the Way® (PLTW) Principles Of Engineering course. Important concepts and processes are explained throughout using full-color photographs and illustrations. Appropriate for high school students, the mathematics covered includes algebra and trigonometry. The strong pedagogical features to aid comprehension include: Case Studies, boxed articles such as Fun Facts and Points of Interest, Your Turn activities, suggestions for Off-Road Exploration, connections to STEM concepts, Career Profiles, Design Briefs, and example pages from Engineers' Notebooks. Each chapter concludes with questions designed to test your students' knowledge of information presented in the chapter, along with a hands-on challenge or exercise that compliments the content and lends itself to exploration in the classroom. Key vocabulary terms that align with those contained in the PLTW POE course are highlighted throughout the book and emphasized in margin definitions.
Features
This text is designed to serve as an information resource to help your POE students develop a greater understanding of the POE curriculum concepts and achieve the POE curriculum performance objectives.
Written by a POE Master Teacher and reviewed by professional engineers, POE teachers, and PLTW curriculum writers, it provides comprehensive coverage and support of the topics contained in the PLTW POE curriculum as no other textbook on the market does.
Allows learning to extend beyond the classroom by providing standard and detailed technical information on the topics that serve as the foundation for the premiere STEM-based, pre-engineering curriculum in the U.S.
Packed with step-by-step examples of applied mathematics and physics that are algebra- and trigonometry-based.
Applications of engineering principles are plentifully illustrated with full-color graphics.
Your Turn activities and Extra Mile exercises in every chapter give your students the opportunity to apply what they are learning and extend their education through hands-on projects.
Engineering career profiles throughout the chapters provide role models and inspiration for aspiring students.
About the Contributor
AUTHORS
Brett Handley
Brett Handley is a Technology Education teacher at Wheatland-Chili Middle School/High School, a Project Lead The Way (PLTW) school, where he teaches the PLTW middle school Gateway To Technology program and the following PLTW high school Pathway To Engineering classes: IED/DDP, CIM, and POE. He is a former PLTW Associate Director of Curriculum, has served as an IED and POE Master Teacher, and authored several IED, CIM and POE end-of-course exams. Mr. Handley also served as a reviewer of the ENGINEERING DESIGN: AN INTRODUCTION textbook. Mr. Handley has a Bachelor of Science Degree in K-12 Technology Education from the State University of New York College at Oswego, and a Master of Science Degree in Professional Studies with a concentration in Engineering Education from the Rochester Institute of Technology.
Craig Coon
Brockport Central High School
David M. Marshall
David M. Marshall is a Technology Education teacher at West Irondequoit Central School District; one of the first Project Lead The Way school districts in the US. Mr. Marshall teaches the PLTW middle school Gateway To Technology program and the following PLTW high school Pathway To Engineering classes: IED/DDP, DE, and POE. He has served as a POE Master Teacher since 1996, has authored several POE end-of-course exams, and has written curriculum materials for the POE, DE, and CIM courses. Mr. Marshall has a Bachelor of Science Degree in K-12 Technology Education and a Master of Science Degree in Education from the State University of New York College at Oswego. | 677.169 | 1 |
Document
Info
Targeted Math | Everyday Algebra Plus
This lesson plan is one of 12 that was created for teachers to use with their students who need to brush up on their math skills in order to pass the 2002 GED® Math Test. To see all 12 lesson plans, go to Targeted Math Instruction for the 2002 GED® Test.
Patterns are used by everybody everyday. Finding patterns is a math skill. Mathematics studies patterns in numbers and shapes to tell us more about patterns and what they mean. Where we find a pattern or constant relationship between two variables, we call it a function. For example, y is a function of 2x. When we chart the solutions on a coordinate plane we get a "picture" of the set of possible answers to this equation for the range of values for "x" or "y." This picture or graph allows us to draw conclusions and make predictions based upon the slope of the line. Graphing an equation gives us a picture. Seeing the picture gives us a deeper understanding of the numbers. Look for patterns to solve problems, and look for patterns among solutions to see the bigger picture | 677.169 | 1 |
Where is calculus used?
I've had high school AP calculus and took the AB and BC exams, which my college decided, this September, covered my Calculus I and II requirements. So I took a two-semester calculus-based statistics course, and I had to learn some stuff on the fly, but I got through it with good grades. Now, next year, I am thinking of taking Differential Equations. I signed up for it, actualy, but now I'm not so sure I want to take it.
Mainly, I only want to take courses that I directly see the use of. I'm a computer science major at the moment. Logic, discrete math, any computer science course, and linear algebra are some of these that I really see the use of and want to take (and am taking). I do not _have_ to take Differential Equations; my main reason for signing up for it was that I'd seen other people on these forums doing differential equations and I wanted to know what it's about.
But now I'm thinking, will I ever use differential equations outside of the course? So my question here is, what is the relevance of calculus with differential equations to other types of math? For example, topology. I'm trying to absorb as much computer-related math as I can, and I don't want to bother taking something that won't be useful later. Anyway, if I don't see it as important then there's an off chance that I might not do well in it. Are differential equations mainly a tool for physicists and physical engineers, or do they have broader scope?
differential equations are about deriving information about a phenomenon, from a knowledge of the way it changes.
for example, knowing that falling objects speed up at a constant rate due toi the pull of gravity enabled galileo to compute the exact speed and distance travelled by any falling object after any amoiunt of time.
knwoing more precisely the inverse square law for gravity of bodies that are further apart enabled newton to predict the orbits of planets.
knowing the rate of population growth can ebnable you to rpedict the population of the earth in the future, and knowing the interest rate and inflkation rate of moneyu can enable you to predict how much wealth you will have at old age, or what retirement savings you will need. it eanbles you to see the lies in the ads for lottery payoffs of huge sums, that conveniently omit such factors as the time value of money.
I have a friend in forestry who used integral calculus to estimate the dollar value and "btu's" of fallen trees over a large area.
I have another friend who uses differential geometry to design cardboard test models to measure the aerodynamics of automobiles of various shapes at a car company.
i think diff eq is really useful for physics, but all the engineers take it too, so it must be useful for them. I really like diff eq as well. i didn't think it was too hard. i thought it was an even more practical application of calculus. you also use lots and lots of various math stuff in it, so it was a good review of everything i've learned thus far. i'd say you ought take it. its pretty straight forward, and i think if you're a smart fella, it should likely be an easy A. well... not a terribly difficult A at least....
But now I'm thinking, will I ever use differential equations outside of the course?
Hello Bicycle. You know, so many people live their lives in a sort of haze about why things happen around them the way they do. They don't have a clue and it causes grief in their lives. I believe differential equations answer many questions about the world, why things happen the way they do, not just in physics, but biology, life, and society. Now I'm not saying differential equations is the answer to everything, but you know what, I use to wonder why about a lot of things. I don't anymore.
. I believe differential equations answer many questions about the world, why things happen the way they do, not just in physics, but biology, life, and society. Now I'm not saying differential equations is the answer to everything,...
But that's obvious, saltydog; that would have made every solution of diff.eqs equal to the constant function 42.
(Congrats with the new medal, BTWI once read an article about magnifying the Mandelbrot Set to world-record extents. They were trying to find the fine filigree which connected the child sets to the main set. However, the more they magnified it, the smaller the filigree would become . . . like that burried treasure story: the deeper they dug, the deeper it would sink.
I think the Universe is a lot like that. Infinite regression punctuated every so often with singularities which changes the rules and makes our concepts on one side, no longer relevant on the other side. Swimming in ice is my favorite analogy.
Why calc? One word: Money. And, if you integrate that one word, you get this large area known as the Stock Market.
I wonder, how many stock signals are derived from calc equations? Too many for one person to know (I bet). Think of it this way: if the stock market is a collection of continuous processes (where each stock is one processes of its own), then what tools do we have to measure the movement? Of course, the answer is, in part, our subject at hand.
As for Neural Nets: the summation formula for a back-prop neural net is rooted in e!
I remember in my CS degree thinking of the algorithm as an end in-and-of itself (rather like art). After all, CS is a fantastic thing all by itself. But, eventually, I got hungry. Using my degree to make money seemed like a cheapening of the pure algorithm, but eventually, theory has to turn into cash or we don't eat. Not to belie the art of our trade, but using the computer to apply math to make money is what it is all about -- and calc is part of that equation.
If that's not good enough, then I speak in the name of my professor, Dr. Scott Sigman, as he would tell us -- in essence -- that knowledge is worth having for its own sake, and sometimes it's just worth it to be educated! The ubiquitous plea for applicability may say more about us than about the "intangible" subject -- this last part is me, but I think it gets at Scott's point.
Unfortunately, there are always interesting classes to take, and sometimes you need to make difficult choices about what you have time for. If you have a more specific plan about what you want to do after you graduate, that will help.
There's also the option of learning it later on, independently of a university class. Learning doesn't have to end once you stop going to school. Independent study is especially important for grad students, since classroom learning can only go so far.
Staff: Mentor
One needs to master differential equations and integral calculus before one attempts numerical analysis. In my organization, we develop complex models of large and small structures with which we perform predictive analysis using nonlinear finite element analysis (FEA).
We start with the basic linear and partial differential, and integral equations for phenomenon like heat transfer, mass (fluid or gas) flow and stress-strain (constitutive models), both steady-state and time dependent.
Do we use calculus? Oh, yeah. We have a theory manual which is devoted to descriptions of the basic differential and integral equations which are the basis of our FEA models.
"What are some of the universities that require a course in DEs for completion of a pure computer science degree?"
To which dfan responded:
dfan said:
MIT is one.
Are you talking about a pure Computer Science degree or something that includes Electrical engineering or Engineering? Of course anything that includes engineering is likely going to have a course in DEs.
Regarding Waterloo, I don't see any requirements or any mention of a course in DEs in the link below.
I'm not saying a course in differential equations isn't useful, I'm probably going to take one myself at some point. It just seems to me to be a bit rude for someone to exclaim "what kind of institution doesn't require a course in DEs for comp sci majors?!" in response to the OP, when in fact that vast majority of universities -- perhaps even all, don't. | 677.169 | 1 |
Recent Advances in Numerical Methods features contributions from distinguished researchers, focused on significant aspects of current numerical methods and computational mathematics. The increasing necessity to present new computational methods that can solve complex scientific and engineering problems requires the preparation of this volume with actual new results and innovative methods that provide numerical solutions in effective computing times. Each chapter will present new and advanced methods and modern variations on known techniques that can solve difficult scientific problems efficiently. | 677.169 | 1 |
Power Point presentation, 12 slides, Explaining the meaning of the basic concepts used in set theory, based on IB Mathematical Studies Syllabus. For a preview of the power point copy the following link on your browser: | 677.169 | 1 |
Category Archives: high schoolThe obsession with Al Gebra and manipulations has used up loads of time which could have been spent on
1. Parameters.
The sudden appearance of the word "parameter" in High School : "Interpret expressions for functions in terms of the situation they model. 5. Interpret the parameters in a linear or exponential function in terms of a context."
The idea of a parameter is basic to the study of functions and relationships. At the start the equation y = mx + b has four letters in it. x and y are variables. What on earth are m and b? Numbers? Fixed numbers? Variable numbers, but not as variable as variables? No, they are parameters for the line. For a given line they are fixed, but for different lines one or both are different.
(When I was at school we, that is the kids, used to call them "variable constants")
2. Parametric representation of curves and relationships.
For example a circle. With parameter θ a point (x,y) on the unit circle is described by x = cos(θ), y = sin(θ)
and a parabola, parameter a, point on curve given by x = a, y = a2
and for a lot of curves the only neat way.
It also allows for ease in programming graphics of curves.
3. Polar coordinates. The ONLY mention of the word "polar" is with regard to representation of complex numbers. With no way of simple plotting them ?????
How about the function representation of a circle as r = 2 ??
There are others!
It was admitted at the time of development of the CCSSM that too much time was spent on K-8, and HS math was a rough job – so why can it not be modified ??????? | 677.169 | 1 |
Long/DeTemple/Millman's Mathematical Reasoning for Elementary Teachers presents the mathematical content needed for teaching within the context of the elementary classroom, giving future teachers the motivation they need while also showing them the bigger picture of when they will use and teach the concepts. The program endeavors to answer the frequently-asked question "Why are we learning this?" by going beyond skill explanations and showing the ways that these concepts are implemented in the future classroom and what types of questions children may ask. Now updated to include the Common Core State Standards for Mathematics, the text imparts mathematical reasoning skills, a deep conceptual understanding, and a positive attitude to those who aspire to be elementary or middle school teachers.
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This text is intended for a transition or introduction to proof and proving in undergraduate mathematics. Many of the elements needed for this transition are here, including predicate and propositional logic. The index is provided and extensive.
Accuracy rating: 5
I have contacted the author about one typographical error I found during my reading, but it is error-free for the majority of the textbook.
Relevance/Longevity rating: 5
I love the content of this textbook. Since this topic is relevant for many aspiring mathematicians, the text will live a long life.
Clarity rating: 4
I believe that the clarity of the text is wonderful, with the exception of one section. I thought that section 5.3, "Mathematical Writing," could have been worded a bit differently or presented to the reader with more discussion. It seemed like I was scolded during the section. I understand the intention of the section, and I praise the author for putting in a section like this since most mathematics textbooks do not, but it seemed to be "good or bad" instead of what the author stated, "good or bad writing is sometimes a matter of opinion." Perhaps the author could include some statements to have the reader read, like run-on sentences in english, and determine these rules for his/herself.
Consistency rating: 5
The text is consistent in framework and terminology. I found no discrepancies while reading.
Modularity rating: 4
There are certain unavoidable parts of the text that are self-referential. This is common with mathematics. I do enjoy that many sections of the text are about techniques of proving, which highlight the technique's importance in mathematics.
Organization/Structure/Flow rating: 5
I really enjoyed the topics in the text, including Chapters 4-10. The proving techniques are the cornerstones to mathematics. I will be incorporating these sections in my future courses due to the elegant way the author has handled these techniques. This would get a higher rating if I could.
Interface rating: 3
The interface was fine. It was a PDF version of a proof textbook. What I wanted from the text was to somehow incorporate the additional nuances that make eBooks slightly better than paper. For example, in the index, I would have liked hyperlinks back to the page where the term was defined. Perhaps I am being critical, but I think that eBooks or PDF versions should have these small refinements.
Grammatical Errors rating: 5
This text contains no grammatical errors from what I've observed.
Cultural Relevance rating: 4
There is one question for which I contacted the author on. On page 47, after the section on statements and truth of statements, there is an exercise to determine if a sentence is a statement, and if it is, determine if that statement is true. The statement is #15: "In the beginning, God created the heaven and the earth." In my opinion, this question may lead to discussions that stray from the original task. Other than this, I found the text to be not culturally insensitive or offensive.
Comments
Although there are certain aspects of the text I would modify, these aspects are minute compared to the amount of understanding and depth that the author introduces. The author has clearly taken his time in developing a textbook that can be accessible and transferrable to subsequent courses. I highly recommend any transition-to-proof or introduction-to-proof course be taught with this textbook.
I use this book for a "Discrete Mathematics for Educators" course. The students are all prospective middle and high school teachers, and the main goals are to prepare them for upper level mathematics courses involving proofs, and to give them a brief introduction to discrete mathematics. This book covers all of the needed proof techniques and gives interesting examples for them. I do use Chapter 3 (combinatorics) and add on some graph theory later on in the course. Thus, I would say it does a very nice job of both introducing students to proof and to intro number theory and combinatorics.
Accuracy rating: 5
After two semesters of teaching from it I have not found an error.
Relevance/Longevity rating: 5
Because of the content, this book passes the longevity test. We will not need to prepare students with introductions to other proof techniques (except perhaps proof by computer?), though additional introductory discrete topics would be great additions for me, though I am using the text for a specific course that goes beyond the scope of the book's intentions.
Clarity rating: 5
Very clear and well organized, and defines all new terminology. As a book used to transition students to upper level mathematics, this book does a very nice job of calling out mathematical language norms and writing norms.
Consistency rating: 5
Very consistent.
Modularity rating: 5
The author provides a nice suggested organization at the beginning, but I have deviated a bit and this book is fine for that. I skipped the chapter on combinatorics and have not used those examples in the proofs so far. After the first exam, we will do some combinatorics, and then go back and prove things about combinatorics and add in inductive proof techniques. The book's structure definitely allows for these sorts of easy changes.
Organization/Structure/Flow rating: 5
Very well organized. I especially like the advanced organizer that provides suggested exam timing w/r/t the chapters. I am deviating a bit this semester from the given order, but the book makes this easily doable, and it is still well organized even with the order mixed up a bit.
Interface rating: 5
No issues here. I would love for hyperlinks to be added, so that you could click on the table of contents to get to chapters (for example). It is very easy to just do a search for terms to get there quickly, but this would be a great addition.
Grammatical Errors rating: 5
No errors.
Cultural Relevance rating: 5
Not applicable? So - it is vacuously highly culturally relevant.
Comments
I really enjoy this book and love that it is free for my students. I've asked my students if they find the book useful and many have said they rely heavily on it. Also, since it is free I feel find going "off script" for a while - when I use an expensive text, I feel like I should make the most of the text for the students. But bc it is free I don't feel that pressure. That said, I don't find myself often deviating from the text's content because it meets my needs.
Reviewed by Roberto Munoz-Alicea, Instructor/Academic Support Coordinator, Colorado State University, on 1/8/2016.
This textbook covers an excellent choice of topics for an introductory course in mathematical proofs and reasoning. The book starts with the basics read more
This textbook covers an excellent choice of topics for an introductory course in mathematical proofs and reasoning. The book starts with the basics of set theory, logic and truth tables, and counting. Then, the book moves on to standard proof techniques: direct proof, proof by contrapositive and contradiction, proving existence and uniqueness, constructive proof, proof by induction, and others. These techniques will be useful in more advanced mathematics courses, as well as courses in statistics, computers science, and other areas. The book ends with additional topics in relations, functions, and cardinality of sets. There is a preface, an introduction, an index, and solutions to selected exercises.
Accuracy rating: 5
While spending a few hours reading the book, I did not find any inaccuracies. The definitions, theorems, and examples given, as well as the notation used, are good, standard, and well presented. For instance, I like how the book explains the differences among theorems, lemmas, corollaries, and propositions, since students sometimes are confused by such labels.
Relevance/Longevity rating: 5
The material covered in this textbook is very relevant and fundamental in mathematics, and this book covers all of the main topics. Relevance and longevity are not issue.
Clarity rating: 5
The book is quite clear in explaining the various topics covered, particularly when it comes to set theory and basic proof techniques. I was impressed by how easy to read and well organized this textbook is. Furthermore, the examples and figures are outstanding.
Consistency rating: 5
The book is consistent in its use of definitions, diagrams, and terminology. Any redundancy, especially in terms of definitions, can be useful to preserve modularity.
Modularity rating: 4
This book's modularity is good, for an introduction to proofs course. One could rearrange the order in which sections and topics in each chapter are covered, although it would be more challenging to rearrange chapters II, III, and IV without covering chapter I first. Chapter IV could be covered before chapters II and III. Also, mathematical induction could be covered before other proof techniques.
Organization/Structure/Flow rating: 4
Even though certain sections and topics could be rearranged, given the textbook's modularity, I think that the order in which the topics are covered is very logical. The fundamentals of set theory, logic, and counting techniques are covered in chapter I. These concepts are needed in order to cover proof techniques in chapters II and III. While mathematical induction could be covered before other proof techniques, it still works well to have it covered at the end of Chapter III. Relations, functions, and cardinality follow in chapter IV.
Interface rating: 4
Interface is not an issue for this book. The diagrams, charts, boxes, tables, headings, and the use of boldface and italic font to indicate definitions and other key concepts, are very helpful to better organize the material. One of my favorite diagrams is the one used to explain how mathematical induction works. One way to improve diagrams and figures would be to label all of them, to make them easier to refer to.
Grammatical Errors rating: 4
There are no obvious grammatical errors, as far as I could see. I would only suggest to avoid the use of apostrophes in expressions such as "it's" or "we've". Instead, write "it is" or "we have", as these expressions are better suited for professional writing.
Cultural Relevance rating: 5
Cultural relevance is not an issue for this textbook.
Comments
This book is excellent for an introduction to mathematics proofs course. Not only does it cover all of the main topics for such a course, but it also discusses mathematical writing, which is key when it comes to making mathematical concepts clear. Many students might know how to prove theorems or solve equations, but might not use correct mathematical notation. The book is very useful to prepare students for courses such as advanced calculus, which is a proof-intensive course. The numerous examples and diagrams used are useful, not only to make the material easier to understand, but also to motivate students to learn more. I would recommend this textbook to any instructor who teaches introduction to mathematical proofs, and to any student who is being exposed to this subject for the first time or needs to review this material.
Table of Contents
Chapter 1: Fundamentals
Chapter 2: How to Prove Conditional Statements
Chapter 3: More on Proof
Chapter 4: Relations, Functions and Cardinality
About the Book
This is a book about how to prove theorems.
Until this point in your education, you may have regarded mathematics primarily as a computational discipline. You have learned to solve equations, compute derivatives and integrals, multiply matrices and find determinants; and you have seen how these things can answer practical questions about the real world. In this setting, your primary goal in using mathematics has been to compute answers.
But there is another approach to mathematics that is more theoretical than computational. In this approach, the primary goal is to understand mathematical structures, to prove mathematical statements, and even to invent or discover new mathematical theorems and theories. The mathematical techniques and procedures that you have learned and used up until now have their origins in this theoretical side of mathematics. For example, in computing the area under a curve, you use the fundamental theorem of calculus. It is because this theorem is true that your answer is correct. However, in your calculus class you were probably far more concerned with how that theorem could be applied than in understanding why it is true. But how do we know it is true? How can we convince ourselves or others of its validity? Questions of this nature belong to the theoretical realm of mathematics. This book is an introduction to that realm.This text has been used in classes at: Virginia Commonwealth University, Lebanon Valley College, University of California - San Diego, Colorado State University, Westminster College, South Dakota State University, PTEK College - Brunei, Christian Brothers High School, University of Texas Pan American, Schola Europaea, James Madison University, Heriot-Watt University, Prince of Songkla University, Queen Mary University of London, University of Nevada - Reno, University of Georgia - Athens, Saint Peter's University, California State University, Bogaziçi University, Pennsylvania State University, University of Notre Dame
About the Contributors
Author(s)
Richard Hammack, PhD is an Associate Professor in the Department of Mathematics and Applied Mathematics at Virginia Commonwealth University. He received his PhD in mathematics from the University of North Carolina at Chapel Hill. | 677.169 | 1 |
In this book we generate graphic images using the software Mathematica thus providing a gentle and enjoyable introduction to this rather technical software and its graphic capabilities. The programs we use for generating these graphics are easily adaptable to many variations. These graphic images are enhanced by introducing a variety of different... more...
The presentation of this book is on the comprehensible application of techniques for the approximation of the mathematical problems that are frequently observed in physical sciences, engineering technology and mathematical physics. The acceptance of the technique for the solution has been justified from mathematical point of view. The Software required... more...
Need to learn MATHEMATICA? Problem SOLVED! Take full advantage of all the powerful capabilities of Mathematica with help from this hands-on guide. Filled with examples and step-by-step explanations, Mathematica Demystified takes you from your very first calculation all the way to plotting complex fractals. Using an intuitive format, this book... more...
Useful Concepts and Results at the Heart of Linear Algebra A one- or two-semester course for a wide variety of students at the sophomore/junior undergraduate level A Modern Introduction to Linear Algebra provides a rigorous yet accessible matrix-oriented introduction to the essential concepts of linear algebra. Concrete, easy-to-understand... more...
The digital era has dramatically changed the ways that researchers search, produce, publish, and disseminate their scientific work. These processes are still rapidly evolving due to improvements in information science, new achievements in computer science technologies, and initiatives such as DML and open access journals, digitization projects, scientific... more...
Mathematica by Example, Revised Edition presents the commands and applications of Mathematica, a system for doing mathematics on a computer. This text serves as a guide to beginning users of Mathematica and users who do not intend to take advantage of the more specialized applications of Mathematica. The book combines symbolic manipulation, numerical... more...
Mastering Mathematica®: Programming Methods and Applications presents the mathematical results and turn them into precise algorithmic procedures that can be executed by a computer. This book provides insight into more complex situations that can be investigated by hand. Organized into four parts, this book begins with an overview of the use of a pocket... more... | 677.169 | 1 |
Arce, et al
Mathematics of Investment begins with an Objective and a Definition of Terms.,Objectives guide both the learner and theprofessor on what they will learn and what they are expected to do in each chapter. Definition of Terms provides easy access to information/definition. | 677.169 | 1 |
The Mechanics CD is ideal for students who are taking those courses as part of a linear or modular course in A-Level Mathematics. For those on modular courses, the Mechanics CD will provide support for a single or double module. The CD provides an excellent grounding for Mechanics by offering clear and concise teaching material, short tests where the user's performance is evaluated and sample exam-style questions, which are accompanied by worked solutions. Navigation through the CD is straightforward, making the programs simple and enjoyable to use.
Topics covered:
Basic Mechanics
Kinematics
Dynamics
Work and Energy
Momentum
Projectiles
Circular Motion
Planar Concurrent Forces
Equilibrium
3-Force Problems
Friction
Moments
Simple Harmonic Motion | 677.169 | 1 |
Math Start
Program Overview
MATH Start is an intensive 8-week program for incoming CUNY students who want to increase their math proficiency before starting credit classes. The program enrolls students who have not passed both math sections of the CUNY Assessment Test (Pre-Algebra and Algebra).
The pre-college math instruction focuses on complex topics in algebra and helps students maximize their understanding through in-depth study of core math concepts in an interactive, supportive learning environment. In addition, students attend a weekly college success seminar to help them develop their academic identity and learn about college structures and campus resources.
Students pay only $35 for Math Start and receive free MetroCards to attend.**
Math Start helps students:
Eliminate or reduce remedial math needs
Prepare for success in college math coursework
Allows students up to two opportunities to meet CUNY's proficiency standards
**MetroCards for students to participate in Bronx Community College and Hostos Community Colleges' Math Start programs are made possible due to the generous support of The Carroll and Milton Petrie Foundation. | 677.169 | 1 |
Geometrical Methods of Mathematical the methods of modern differential geometry have become of considerable importance in theoretical physics and have found application in relativity and cosmology, high-energy physics and field theory, thermodynamics, fluid dynamicsMore... valuable for those with backgrounds in physics and applied mathematics who desire an introduction to the subject. Having studied the book, the reader will be able to comprehend research papers that use this mathematics and follow more advanced pure-mathematical expositions.
Preface
Some basic mathematics
The space R[superscript n] and its topology
Mappings
Real analysis
Group theory
Linear algebra
The algebra of square matrices
Bibliography
Differentiable manifolds and tensors
Definition of a manifold
The sphere as a manifold
Other examples of manifolds
Global considerations
Curves
Functions on M
Vectors and vector fields
Basis vectors and basis vector fields
Fiber bundles
Examples of fiber bundles
A deeper look at fiber bundles
Vector fields and integral curves
Exponentiation of the operator d/d[lambda]
Lie brackets and noncoordinate bases
When is a basis a coordinate basis?
One-forms
Examples of one-forms
The Dirac delta function
The gradient and the pictorial representation of a one-form
Basis one-forms and components of one-forms
Index notation
Tensors and tensor fields
Examples of tensors
Components of tensors and the outer product
Contraction
Basis transformations
Tensor operations on components
Functions and scalars
The metric tensor on a vector space
The metric tensor field on a manifold
Special relativity
Bibliography
Lie derivatives and Lie groups
Introduction: how a vector field maps a manifold into itself
Lie dragging a function
Lie dragging a vector field
Lie derivatives
Lie derivative of a one-form
Submanifolds
Frobenius' theorem (vector field version)
Proof of Frobenius' theorem
An example: the generators of S[superscript 2]
Invariance
Killing vector fields
Killing vectors and conserved quantities in particle dynamics
Axial symmetry
Abstract Lie groups
Examples of Lie groups
Lie algebras and their groups
Realizations and representations
Spherical symmetry, spherical harmonics and representations of the rotation group
Bibliography
Differential forms
The algebra and integral calculus of forms
Definition of volume -- the geometrical role of differential forms
Notation and definitions for antisy mmetric tensors
Differential forms
Manipulating differential forms
Restriction of forms
Fields of forms
Handedness and orientability
Volumes and integration on oriented manifolds
N-vectors, duals, and the symbol [epsilon][subscript ij...k]
Tensor densities
Generalized Kronecker deltas
Determinants and [epsilon][subscript ij...k]
Metric volume elements
The differential calculus of forms and its applications
The exterior derivative
Notation for derivatives
Familiar examples of exterior differentiation
Integrability conditions for partial differential equations
Exact forms
Proof of the local exactness of closed forms
Lie derivatives of forms
Lie derivatives and exterior derivatives commute
Stokes' theorem
Gauss' theorem and the definition of divergence
A glance at cohomology theory
Differential forms and differential equations
Frobenius' theorem (differential forms version)
Proof of the equivalence of the two versions of Frobenius' theorem
Conservation laws
Vector spherical harmonics
Bibliography
Applications in physics
Thermodynamics
Simple systems
Maxwell and other mathematical identities
Composite thermodynamic systems: Caratheodory's theorem
Hamiltonian mechanics
Hamiltonian vector fields
Canonical transformations
Map between vectors and one-forms provided by [characters not reproducible] | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
Study faster, learn better-and get top grades with Schaum's Outlines
Millions of students trust Schaum's OutlinesUse Schaum's Outlines to:
Brush up before tests
Find answers fast
Study quickly and more effectively
Get the big picture without spending hours poring over lengthy textbooks
Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum's to shorten your study time-and get your best test scores!
This Schaum's Outline gives you:
A quick, easy-to-follow guide to mathematical topics required for important concept development in physics
More than 1500 fully solved problems from both the physics and mathematics point of view
Synopsis
Learn to use math concepts to unravel physics problems
"Schaum's Outline of Mathematics for Physics Students" helps you to apply mathematical concepts to your studies and shows you how these concepts operate in physics problems. The book includes both fully solved problems and supplementary practice problems.
Synopsis
Confusing Textbooks?
Missed Lectures?
Tough Test QuestionsPhillip A. Schmidt, Ph.D., is the program coordinator for secondary education at the Teachers College of Western Governors University. Robert Steiner, Ph.D., is the project director of seminars on science at the American Museum of Natural History.
Table of Contents
Part I: Algebra and Geometry
Chapter 1: Introduction to Algebra
Chapter 2: Functions
Chapter 3: Graphs of Functions
Chapter 4: Linear Equations
Chapter 5: Simultaneous Linear Equations
Chapter 6: Quadratic Functions and Equations
Chapter 7: Inequalities
Chapter 8: The Locus of an Equation
Chapter 9: The Straight Line
Chapter 10: Families of Straight Lines
Chapter 11: The Circle
Part II: Pre-Calculus and Elementary Calculus
Chapter 12: Rational and Polynomial Functions
Chapter 13: Trigonometric Functions
Chapter 14: Exponential and Logarithmic Functions
Chapter 15: Complex Numbers
Chapter 16: The Calculus of Single-Variable Functions: A Mathematical Approach
Chapter 17: The Calculus of Single-Variable Functions: A Physics Approach | 677.169 | 1 |
Algebra 2 and PreCalculus Word Wall
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
3.26 MB | 42 pages
PRODUCT DESCRIPTION
Included are 40 digital posters (list of words below) for a vocabulary word wall in Algebra 2/PreCalculus! I use this product in my classroom as part of a larger display. As we cover a topic, I post the word wall poster. Then, before each summative test, we use the word wall as a place for review. Also, I recommend including the images in class lectures and on student work; the more exposure students can have the more likely they will remember these words later. Each of the words includes some sort of imagery that defines the word to help students remember these topics. These images will help any level of Algebra 2 and PreCalculus students and some can be used on a basic level in other courses as well as a content review in a Calculus class.
*Not included are trig identities because I find it better to use my "Trig Tool Box" to help students remember how to verify and organize trig identities. | 677.169 | 1 |
A Friendly Mathematics Competition: 35 Years of Teamwork in Indiana
"A Friendly Mathematics Competition" tells the story of the Indiana College Mathematics Competition (ICMC) by presenting the problems, solutions, and results of the first 35 years of the ICMC. The ICMC was organized in reaction to the Putnam Exam - its problems were to be more representative of the undergraduate curriculum, and students could work on them in teams. Originally participation was originally restricted to the small, private colleges and universities of the state, but was later opened up to students from all of the schools in Indiana. The competition was quickly nicknamed the "Friendly" Competition because of its focus on solving mathematical problems, which brought faculty and students together, rather than on the competitive nature of winning. Organized by year, the problems and solutions in this volume present an excellent archive of information about what has been expected of an undergraduate mathematics major over the past 35 years. With more than 245 problems and solutions, the book is also a must buy for faculty and students interested in problem-solving. The index of problems lists problems in: Algebraic Structures; Analytic Geometry, Arclength, Binomial Coefficients, Derangements, Differentiation, Differential Equations, Diophantine Equations, Enumeration, Field and Ring Theory, Fibonacci Sequences, Finite Sums, Fundamental Theorem of Calculus Geometry, Group Theory, Inequalities, Infinite Series, Integration, Limit Evaluation, Logic, Matrix Algebra, Maxima and Minima Problems, Multivariable Calculus, Number Theory, Permutations, Probability, Polar Coordinates, Polynomials, Real Valued Functions Riemann Sums, Sequences, Systems of Equations, Statistics, Synthetic Geometry, Taylor Series, Trigonometry, and Volumes. | 677.169 | 1 |
Mathematics - AllConferenceAlert
The main idea behind these conferences on Mathematics is to bring along the scholars, engineers, researchers, scientists, and practitioners from across the globe in order to present and exchange ideas on ongoing researches on this particular field of Mathematics. Besides that these conference also provide a scope for the dignitaries to exchange their own ideas and methodologies as well as their application experiences directly form face to face, so as to establish further business/ research relations for future collaborations. | 677.169 | 1 |
Prior to the nineteenth century, algebra meant the study of the solution of polynomial equations. By the twentieth century it came to encompass the study of abstract, axiomatic systems such as groups, rings, and fields. This presentation provides an account of the history of the basic concepts, results, and theories of abstract algebra.The development of abstract algebra was propelled by the need for new tools to address certain classical problems that appeared unsolvable by classical means. A major theme of the approach in this book is to show how abstract algebra has arisen in attempts to solve some of these classical problems, providing a context from which the reader may gain a deeper appreciation of the mathematics involved.Mathematics instructors, algebraists, and historians of science will find the work a valuable reference. The book may also serve as a supplemental text for courses in abstract algebra or the history of mathematics. | 677.169 | 1 |
Live online classes follow the Saxon Math 3 scope and sequence but the actual class time is filled with very game-like activities in a mix of individual and group activities. Live class activity time is followed by online practice quizzes and a weekly assessment.
Required Text: Saxon 3, ed 1 is an exact match, other editions are fine too. On this one, go for a teacher's manual instead of the student text. There is too much the student text doesn't show and I want families to be able to track all the skills we are working on. Find it used HERE at Amazon 54 student text completedQuestions are drawn randomly from a large poole of questions organized by question type. The exact question may differ between our sets and the text, but the topic order will match edition 3. You can use the 2nd edition, but remember that some variation in topic order will be there if you do.
Description:
Lessons are recorded and often interactive in format.
The online homework assignments draw questions from a large pool 1/2 student text
Questions are drawn randomly from a large pool of questions organized by question type. The exact question may differ between our sets and the text, but the topic order will match edition 3.
Age/Grade: 9th is the traditional grade level, mature middle school students are welcome too
Live Class Schedule: Currently full
M through R, 2:30 pm CST or T through F, 11 am CST
First day: September 6 Thanksgiving Break: November 21 - 25 Christmas Break: December 19 - 30 (This may change to last week of December and first week of January) Easter Break: April 10 - 14 Last Day: April 28
Description:
Integrated Geometry - By the time students get to the 20th lesson in Advanced Math and have been in the series since Algebra 1, they will be able to count 1 geometry credit.
Lessons are recorded and often interactive in format. This class has a live m online classroom option that opens on August 1. It fills fast, usually within 24 hours. The course is also available in AYOP (at your own pace) year-round without a seat limit how this text and have been in the series since Algebra 1, they will be able to count 1 geometry credit. Advanced Mathematics - Second Edition
Highly Recommended Text: Saxon Advanced Mathematics Solutions Manual - Second Edition (This works out each problem step by step. It is more than just an answer key. Advanced Math is a very challenging course, and the Solutions Manual will be very helpful to parents as you work with your students.)
Questions are drawn randomly from a large poole of questions organized by question type. The exact question may differ between our sets and the text, but the topic order will match edition 2.
Introduction to Logic is a thorough course covering the basics of the topic. Students participating in this course should be prepared for fairly advanced thought concepts. There is a reasonable amount of reading in the text with a significant number of key terms and concepts covered. Online in-class lecture supports the text material. Students have the benefit of comprehension measurement tools (quizzes and tests). The rigorous nature of the course may suit high school students best. | 677.169 | 1 |
The basic concepts of relativity theory are conveyed through worked and unworked examples in this text, which requires only elementary algebra and emphasizes physical principles and concepts. 1985 edition. | 677.169 | 1 |
ENGR Documents
Showing 1 to 30 of 46
Chapter 2
Matrices
2.1 Operations with Matrices
To represent matrices in any one of the following three ways
1. A matrix can be denoted by an uppercase letter such as A,
B, C,
2. A matrix can be denoted by a representative element
enclosed in brackets, s
Chapter 3
Vector Spaces
3.1 Vectors in the Rn
A vector is characterized by two quantities length and
direction.
Vectors in the Plane
Example 1
Vectors in the Plane
Example 2
Adding Two Vectors in the Plane
If c is positive, then cv has the same direction
Chapter 4
Systems of Forces and Moments
4.1
Two-Dimensional Description of
the Moment
The magnitude of the moment of the force F about point O
is the product Fd, where d is the perpendicular distance
from O to the line of action of the force.
The direct
Chapter 5
Determinants
5.1 The Determinant of the Matrix
Example 1
The Determinant of a Matrix of Order 2
Example 2
Matrix
Finding the Minors and Cofactors of a
Example 3
The Determinant of a Matrix of Order 3
Example 4
The Determinant of a Matrix of Orde
Chapter 1
Linear Systems
1.1
Introduction to System of Linear Equations
Linear Equations in n Variables
The equation of a line in two-dimensional space has the form
a1 x + a2 y = b,
a1, a2, and b are constants.
This is linear equation in two variables x a
Chapter 4
Orthogonality
4.1 Length and Dot Product in R n
The definition of the length of a vector v in R2. If v = (v1, v2) is a
vector in the plane, then the length of v, denoted by
defined to be
v
, is
Example 1
The Length of a Vector in Rn
Example 1
Th
Chapter 10
Internal Forces and Moments
10.1 Axial Force, Shear Force, and
Bending Moment
The axial forces P, shear force V, and bending moment M
are an equivalent system representing the internal
forces and moment at a cross section of a beam. These
are t
Chapter 6
Structures in Equilibrium
6.1
Trusses
Structures that consist of straight bars pinned at the
ends and are supported and loaded only at the joints
where the members are connected are called trusses,
and a truss is a structure that consists of two
Chapter 7
Centroids and Center of Mass
7.1
Centroids of Areas
The moment about the y axis is :
The moment about the x axis is :
Figure 7.3
An arbitrary area A is in the x-y plane (Figure 7.3). Let
us divide the area into parts
positions of the parts by
Chapter 8
Moments of Inertia
8.1
Definitions
1. Moment of inertia about the x axis
I = y dA
2
x
where y is the y coordinate of the differential element of
area dA.
The moment of inertia can be expressed in terms of
the radius of gyration about the x axis
Chapter 6
Eigenvalues and Eigenvectors
6.1 Eigenvalues and Eigenvectors
Remark: The eigenvector cannot be zero, because A
is true for all real values of eigenvalue
=0
0
=
. An eigenvalue of
, however, is possible.
Example 1
0
Verifying Eigenvalues and Eig
Chapter 5
Objects in Equilibrium
5.1
The Equilibrium Equations
When an object acted upon by a system of forces
and moments is in equilibrium, then the following
conditions are satisfied:
1. The sum of the forces is zero.
F=0
2. The sum of the moments abo
Chapter 9
Friction
9.1
Theory of Dry Friction
Coefficients of Friction
The Static Coefficient:
The magnitude of the
maximum friction forces that can be exert between two
plane, dry surface in contact that are not in motion
relative to one another is
f =s
RICHLAND COLLEGE
School of Engineering Business and
Technology
COURSE SYLLABUS
ENGR 2301
Engineering Mechanics I - Statics
Richland College is determined to prepare the student with the knowledge and
skills you need to succeed in today's dynamic work envi
Chapter3
FORCES
3.1TypesofForces
Terminology
LineofAction: When a force is represented by a vector, the
straight line collinear with the vector is called the line of action of
the force.
SystemofForces:
A system of forces is a set of forces.
A system of f
Chapter2
Vectors
VECTOROPERATIONSANDDEFINITIONS
2.1 ScalarsandVectors
Scalarquantities have only magnitudes. These can be represented by real
numbers are called scalars.
Examples of scalar quantities are mass, volume, density, speed, energy,
temperature,
Chapter4
SystemsofForcesandMoments
4.1TwoDimensionalDescriptionoftheMoment
The magnitude of the moment of the force F about point O is the
product Fd, where d is the perpendicular distance from O to the
line of action of the force.
The direction of the mo
Chapter6
StructuresinEquilibrium
6.1
Trusses
Structures that consist of straight bars pinned at the ends and are
supported and loaded only at the joints where the members are
connectedarecalled trusses,and atrussisastructurethatconsistsof
twoforcememberon
Chapter2
Strain
2.1
Deformation
A force is applied to a body, it will tend to change the bodys shape and
size. These changes are referred to as deformation.
2.2
Strain
NormalStrain. The normal strain as the change in length of a line
per unit length.
avg | 677.169 | 1 |
Search Continuing Education
Search Continuing Education
UPG 031 - Math 3
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Course Description
This course is designed to prepare you for further university calculus courses. As a Math 31 equivalency, this course covers topics of calculus as well as essential precalculus topics, including: review of functions, review of algebra, limits and continuity, differentiation, applications of differentiation, integration, and applications of integration.
Course Details
Course Learning Outcomes
By completion of this course, successful students will be able to:
Develop a strategic overview of a mathematical problem and use problems solving skills to analyze the problem
Use calculus to model some real world problems mathematically and interpret solutions as well as communicate results to a general audience
Observe that there are several useful techniques to tackle a given mathematical problem: graphically, algebraically or numerically, then decide the most useful or most appropriate technique for solving the problem
Determine if using technology tools and/or software is or is not appropriate to solve or analyze a given problem
Recognize the vital role of calculus not only in Science and Engineering; but also in the wider world (Social, Health, Economics, Business...Etc.)
Prerequisites | 677.169 | 1 |
Introduction
This cluster is designed to introduce students with a strong interest in mathematics to several different advanced topics. Many of these topics would ordinarily only be seen at the advanced undergraduate level, but all lend themselves to an introductory course at the high school level. No prior experience in any of these topics is expected, but enthusiasm for and interest in mathematics is essential.
Core Course (4 Weeks)
Combinatorics
Enumerative combinatorics provides a way to count in complicated mathematical settings, and we will learn several important counting techniques in this course. In addition, we will learn some related ideas in elementary probability and number theory. The following is an example of a combinatorics problem that we will encounter in our four-week journey through this interesting area of mathematics:
If a teacher returns a test to her class of 10 students at random, what is the probability that no student gets his or her own test?
Supplementary Courses
Symmetry (1 Week)
One can analyse images and shapes in terms of their symmetrically repeating patterns. A symmetry of a geometric figure is a motion which does not change the figure's appearance. These motions form a group in the mathematical sense, because if one such motion is followed by another, the combined motion also does not change the figure's appearance. We will study point groups of motions which keep a figure's center fixed. For example, the symmetry group of a square has 8 motions: the rotations of 0, 90, 180, and 270 degrees about the center (the first of these does not actually move anything), and the reflections in the horizontal line, the vertical line, and the two diagonal lines through the center. We will also look at point groups for 3D figures, for example, of a soccer ball sewn from twelve black pentagons and twenty white hexagons, or a baseball cover sewn from two leather pieces along a curved line.
We will study frieze groups for repeating patterns along a strip, "wallpaper" groups for patterns repeating along two directions in a plane, and space groups for repeating patterns in space, which appear in crystals. Along the way, we will study regular and semi-regular polyhedra (3D shapes bounded by flat regular polygon faces), kaleidoscopic patterns generated by mirrors, and tilings which fill a plane by regular polygons, or by identically shaped tiles.
Introduction to Graph Theory (1.5 Weeks)
A graph is a collection of points, called vertices, connected by edges. Using graphs to describe information has applications from scheduling tournaments to cryptography. We'll look some of the at the basics of graph theory, including graph coloring problems, Euler's formula for graphs on surfaces, and graph planarity. Klein bottles will make a guest appearance when we look at graphs on non-orientable surfaces.
Knots, Links and the Topology of Space (1.5 Weeks)
From knotted strands of DNA, to tangled necklaces, to the basis of String Theory, knots and links arise in many different applications. We'll look at the mathematical theory of knots, and learn what knot invariants can tell us about them. We'll connect knot theory to the study of all 3-dimensional spaces. For example, we'll examine whether or not our universe is flat. We thought the earth was flat for a long time... what about the 3-dimensional universe we live in? If we send a rocket off into space programmed to go "straight", will it eventually come back to where it started, like what happens if you go "west" long enough starting at a point on the equator? We'll look at some of the possibilities by studying what is known about the geometry and topology of space. What possibilities are there for 3-dimensional spaces? Can you come to the "edge" of the universe? | 677.169 | 1 |
This book combines traditional mainstream calculus with the most flexible approach to new ideas and calculator/computer technology. It contains superb problem sets and a fresh conceptual emphasis flavored by new technological possibilities. Chapter topics cover functions, graphs, and models; prelude to calculus; the derivative; additional applications of the derivative; the integral; applications of the integral; calculus of transcendental functions; techniques of integration; differential equations; polar coordinates and parametric curves; infinite series; vectors, curves, and surfaces in space; partial differentiation; multiple integrals; and vector calculus. For individuals interested in the study of calculus.
Note: This ISBN is now printed in a two volume set Part A and Part B which is shrinkwrapped together.
REVIEWS for | 677.169 | 1 |
Summer Assignments
These are your summer assignments. SAS students are dedicated learners and learning does not stop for the summer. Please complete the assignments by the first day of the new school year.
If you have any further questions, please contact the appropriate teacher or counselor.
Summer Assignment By Subject
Incoming 9th Graders
Summer Reading Assignment
Choose any 2 books from the Summer Reading List (file below). For each book you read, fill out the appropriate Summer Reading Form (Book Report Fiction or Book Report Non-Fiction). Please bring the completed forms to class on your first day of school (September 13th).
Students may also want to consider reading books from this list for their own enjoyment.
In addition, in lieu of one of the books, students may also choose to participate in a museum or cultural visit with their parents. To do this, they must have proof of attendance and complete the Museum Visit Form.
Math
AP Calculus - Summer 2016
OPTION 1 Calculus AP Students are required to complete THE SUMMER PACKET to discuss and review at the start of the school year. You do not have to print every page, just work on the problems and bring work and questions.
All students are recommended to practice math one way or another. Mr. Lange has many resources for Algebra, Geometry, Algebra II, Precalculus (Trig/MA), and Statistics on his website and blog ( | 677.169 | 1 |
PDF Books Free Download
Mastering Essential Math Skills GEOMETRY
Try our newest product, NO-NONSENSE ALGEBRA.Comes with free online video tutorials/b title will soon be available in kindle. Geometry is lots of fun and applies to so many things in the real world. Now is an excellent time to learn all about geometry the easy way! This is a new title by America's math teacher and author, Richard W. Fisher. This book will provide students with all the essential geometry skills. Vocabulary, points, lines, planes, perimeter, area, volume, and the Pythagorean…" | 677.169 | 1 |
Mathematical Ideas Miller is available now for quick shipment to any U.S. location. This edition can easily be substituted for ISBN 0321977076 or ISBN 9780321977076 the 13th edition or even more recent edition. You will save lots of cash by
What does your math course have to do with the latest TV shows or Hollywood movies? Plenty-if you're using the right text. Mathematical Ideas, Twelfth Edition brings the best of Hollywood into the classroom through descriptions of video clips from popular cinema and television. Well-known author John Hornsby's innovative approach is enhanced with great care in this revision, and refined to serve the needs of you and your instructor. Streamlined and updated, it offers a modernized design, new bubble pointers for Example annotations, and much more. It retains the consistent features, friendly writing style, clear examples, and exercise sets for which this text is known.
One of the biggest issues college math instructors face is capturing and keeping student interest. Over the years, John Hornsby has refined a creative solution-- bringing the best of Hollywood into his mathematics classroom. Mathematical Ideas applies this same strategy of engaging students through video clips from popular cinema and television to the textbook. Alongside fresh data and tools, this Eleventh Edition uses up-to-the-minute images as well as old favorites of math being done in Hollywood. In addition, examples are clarified with additional annotations, chapter summaries are made more intuitive to aid review, and chapter tests now include specific section references, making it easier for students to refer back to topics that need more attention. With great care and effort, the...
Less
Home Shop About Us Policies Contact Us Mathematical Puzzles: A Connoisseur's Collection Textbooks, Education Collected over several years by Peter Winkler, of Bell Labs, dozens of elegant, intriguing challenges are presented in Mathematical Puzzles. The answers are easy to explain, but without this book, devilishly hard to find. Creative reasoning is the key to these puzzles. No involved computation or higher mathematics is necessary, but your ability to construct a mathematical proof | 677.169 | 1 |
It's targeted as a survival study guide for science and engineering students who got screwed (or screwed themselves) on their math education prior to college. I liked the look of it from the details on Amazon and bought it. I haven't made time to get into it yet though, so all I can say is that it still looks like the best option for self study.
Maybe I should be looking into taking a class somewhere though, given my lack of discipline as displayed above. We have an excellent community college system where I live. If I can find the time and maybe get my employer to pony up for it, I might take a look there.
OffMyMeds
Sunday, May 23, 2004
Sorry for the ridiculously wide link. I don't know why I thought it would wrap. Should a moderator decide to remove it, the book is called _Maths: A student's Survival Guide_ by Jenny Olive and is easy to find at Amazon.
OffMyMeds
Sunday, May 23, 2004
Out of curiosity, does the math you'd like to learn have any connection to programming you'd like to do (I'm not being critical, I'm curious)?
How far would you like to go?
Teaching oneself math requires discipline most people don't have. I guess the key is not cheating on reading things over, and not going any further until you've mastered what's already been presented in a book. That's why it's so terrible to fall behind in a math class: your lack of comprehension compounds itself as time goes on.
Also, be sure to do lots (more than you would in a class) of problems, including some challenging ones.
At football games in America, we chant: 3.14159, sec tan cos sin, gooooo math!
And no, it has nothing to do with my day job. I just feel I need it, because I've always had that stuck in the back of my mind: "I'm a dumb ass on math".
RP
Monday, May 24, 2004
Actually, RP, some universities offer remedial math courses, i.e., courses that most students would've normally taken in high school.
And you're not a dumb ass in anything. Just remember, "you're good enough, you're smart enough, and doggone it, people like you!"
Stuart Smalley
Monday, May 24, 2004
I'd second doing a university class. The one I have offers a "Intro To College Math" courses geared towards students who might not have high-school math they need.
Jimmy Jo-jo
Monday, May 24, 2004
Most community colleges will teach remedial math. Usually a lot cheaper than at a university, and you're probably more likely to find weekend or night classes.
Once you're up to speed on that you can look for a university that teaches discrete math.
T. Norman
Monday, May 24, 2004
Most universities have remedial math courses now as well. I remember reading somewhere about 30% of new university students need to take remedial math and/or english once they're in college, because of the bad high school educations they're getting.
Andrew Hurst
Monday, May 24, 2004
I'm currently going to a community college which offers math down to the pre-algebra level, and credits cost about half what the local state university costs. Quality of teachers varies, especially given that my school has 60% of its courses offered by part-time teachers (who then have to commute between several community colleges, or have "real jobs" somewhere else, to earn a living), but generally my experience has been that if you're ready to learn, you'll do well.
Sam Livingston-Gray
Monday, May 24, 2004
Just another $0.02 re: community colleges and why you get your money's worth:
I count the fact that the instructors are often part-time and have real jobs as one of the advantages. I would much rather learn from a person using their craft for practical applications in his daily life than someone ensconsed in the academia bubble. I've taken classes at two community colleges, one state university and two private universities (I'm good at starting school, not good at finishing it). The quality of intructors varied equally at each, but the best teachers were almost always the ones with a day job. Many of them weren't just competent, they were sharp and engaging.
Ok enough soapboxing. My point: if the name of the school isn't going to affect your job opportunities, save your money and take a few night classes at your local comunity college. If the instruction seems disappointing, drop the class and ask around for who the good instructors are.
OffMyMeds
Tuesday, May 25, 2004
Correction to subject-verb agreement error: "learn from a person using *his* craft for practical applications in his daily life." | 677.169 | 1 |
This best-selling title provides in one handy volume the essential mathematical tools and techniques used to solve problems in physics. It is a vital addition to the bookshelf of any serious student of physics or research professional in the field. The authors have put considerable effort into revamping this new edition.
* Updates the leading graduate-level text in mathematical physics * Provides comprehensive coverage of the mathematics necessary for advanced study in physics and engineering * Focuses on problem-solving skills and offers a vast array of exercises * Clearly illustrates and proves mathematical relations
New in the Sixth Edition: * Updated content throughout, based on users' feedback * More advanced sections, including differential forms and the elegant forms of Maxwell's equations * A new chapter on probability and statistics * More elementary sections have been deleted
"synopsis" may belong to another edition of this title.
Product Description:
Provides the essential mathematical tools and techniques used to solve problems in physics. This work includes differential forms and the elegant forms of Maxwell's equations, and a chapter on probability and statistics. It also illustrates and proves mathematical relations.
Review:
"As to a comparison with other books of the same ilk, well, in all honesty, there are none. No other text on methods of mathematical physics is as comprehensive and as complete...I encourage the students to keep their copies as they will need it and will find it an invaluable reference resource in later studies and research." - Tristan Hubsch, Howard University
Book Description Academic Press. Hardcover. Book Condition: New. 0120598720598760 | 677.169 | 1 |
Pricing and Purchase Info
*Explains the mathematics essential to flight, teaching basic principles and reasoning*Provides an understanding that allows pilots to utilize new technologies*Examines techniques of GPS (Global Positioning System), and other navigation forms, including calculations of distance and bearings*Covers chart construction, magnetic compasses...
From the Jacket
THE BOOK THAT SOLVES THE MYSTERIES OF NAVIGATIONNavigating is easier and safer when you truly understand how it works. This enjoyably readable, in-the-cockpit guide helps you build that base of understanding, without pain. Written by flight instructor/mathematician/computer expert/teacher James S. Wolper, Understanding Mathematics for ...
James S. Wolper is Professor of Mathematics at Idaho State University and a professional pilot and flight instructor. He is also on the computer science and environmental engineering faculties at ISU. He has a bachelor's degree in mathematics from Harvard University and a Ph.D. in mathematics from Brown University. His research has bee... | 677.169 | 1 |
Dedham, MA PrecalculusRobert Z.
...In general, the vocabulary standards focus on understanding words and phrases and their relationships. Of course, the specifics vary by grade level. For instance, eighth graders concentrate on the following skills:
? L.8.4. | 677.169 | 1 |
NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002 Visit us at
Dedicated to all my teachers
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Foreword The otherwise enjoyable subject becomes sometimes to the students a frightening nightmare and in the course of my long career in the field of education, I have found many students even becoming nervous at the mention of Mathematics. That is, in my experience, mostly because of the way it is taught by some teachers and the way some of the books are written. My esteemed colleague Smt. Veena, G.R. is not only a good teacher who would make the students to enjoy Mathematics, but also a good "Narrator" of Mathematics. Extremely resourceful and completely committed, Veena has brought into this book her direct experience in the class room as a teacher. In lucid style, she has written this book making it students-friendly, teacher-friendly and teachinglearning focused. While congratulating her for this good work, I wish that both the students and the teachers would welcome this book and make use of it.
ADHAKRISHNA PROF. K.E. RADHAKRISHNA Principal, Surana College, Bangalore-560004.
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Preface
This book is written as per the Syllabus of Basic Mathematics for II Year Pre-University Course of Karnataka. Each topic has been discussed exhaustively as per the requirements of the latest syllabus. The book has been written in a very simple style, to enable students to understand the subject effectively. More focus is given for a systematic approach to enhance the grasping of the subject so that all types of questions could be answered well in each chapter. I hope this book will satisfy all the requirements of the students for learning the subject successfully and getting through the examination with flying colours. My sincere thanks to Prof. K.E. Radhakrishna, Principal, Surana College, an eminent academician and educationist for his foreword. I also thank Ms. Sudha S., Department of Mathematics, Surana College for reading the manuscript and identifying the unforeseen computational errors. I am also thankful to Mr. R.K. Gupta, Chairman, Mr. Saumya Gupta, Managing Director, New Age International Pvt. Ltd., New Delhi and Mr. Vincent D'souza, Branch Manager, Mr. Babu V.R., Marketing Manager, Bangalore branch for accepting to publish the book. I sincerely welcome criticism, views and suggestions from readers.
EENA G.R. VEENA veena.gr@gmail.com
Circ 13. Circles
13.1 13.2 13.3 13.4 13.5 13.6 13.7 Definitions Equation of a Circle Point of Intersection of a Line and a Circle Equation of Tangent to the Circle x2 + y2 + 2gx + 2fy + c = 0 at the Point (x1, y1) on it Length of the Tangent from the Point (x1, y1) to the Circle x2 + y2 + 2gx + 2fy + c = 0. Condition for the Line y = mx + c to be a Tangent to the Circle x2 + y2 = a2 and point of contact Condition for the Line lx + my + n = 0 to be a Tangent to the Circle x2 + y2 + 2gx + 2fy + c = 0.
278
278 278 292 298 299 300 301
ara 14. Parabola
14.1 14.2 14.3 14.4 14.5 Introduction Parabola Equation of the Parabola in the Standard Form Different Forms of Parabola with Vertex (0, 0) Different Forms of Parabola with Vertex (h, k)
1
Mathematical Logic
1.1 INTRODUCTION:
Logic is the science dealing with principles of reasoning. We can find all the different ways of solving a problem by logical reasoning. The English Mathematician George Boole is the founder of mathematical logic. To express the principles of reasoning, a symbolic language has been developed. This symbolic language is called mathematical logic or symbolic logic. Mathematical logic finds application in switching circuits, digital computers and other digital devices.
1.2 PROPOSITIONS:
A proposition is a statement which in the given context is either true or false but not both. The propositions are denoted by small letters p, q, r... Examples: 1. Sum of two even integers is even integer. 2.
3 is a rational number.
TRUTH TRUTH VALUE: The truthness or falsity of a proposition is called its truth value. If a proposition is true it is denoted by 'T' and if it is false it is denoted by 'F'. Example: The truth value of 1. 5 + 6 = 11 is 'T'. 2. 'Asia is in India' is F. 3. 'Today is Sunday' is either 'T' or 'F' in the given context i.e., on a particular day it is only one of 'T' or 'F'.
1.3 LOGICAL CONNECTIVES AND COMPOUND PROPOSITIONS
Two or more simple propositions are connected by using the words 'and', 'or', 'if ... then', 'if and only if'. These words or phrases are called logical connectives. Any proposition containing one or more connectives is called a compound proposition. The simple propositions occurring in a compound proposition are called its components. TRUTH ABLE: TRUTH TABLE: The truth values of the compound proposition for all possible truth values of its components is expressed in the form of a table called truth table. For a compound proposition with only one proposition, Truth table consists of 2 possibilities (either T or F). For a compound proposition with two propositions truth table consists of 22 = 4 possibilities. For a compound proposition with 3 propositions truth table consists of 23 = 8 possibilities. CONJUNCTION: CONJUNCTION: If p and q are 2 simple propositions. Then the proposition 'p and q' is called the conjunction of p and q. It is denoted by p ∧ q. Example: If p : 7 is a prime number. q : 6 is an even number, then p∧q : 7 is a prime number and 6 is an even number. The truth value of the compound proposition p ∧ q depends on the truth values of p and q. Note that the conjunction of p and q is true only when both p and q are true, otherwise it is false.
Truth Table
p T T F F
q T F T F
p∧q T F F F
DISJUNCTION: DISJUNCTION: If p and q are 2 simple propositions, then proposition 'p or q' is called the disjunction of p and q. It is denoted by p ∨ q. Example: If p : 2 is rational number.
Mathematical Logic
3
q : 2 is odd number, then p ∨ q = 2 is rational or 2 is odd number. The truth value of p ∨ q depends on the truth values of p and q. Note that the disjunction of p and q is false only when both p and q are false. Otherwise it is true.
Truth Table
p T T F F
q T F T F
p∨q T T T F
CONDITIONAL (IMPLICATION) TION): CONDITIONAL (IMPLICATION): If p and q are two simple proposition, then the proposition if 'p ... then q' is known as conditional or implication. It is denoted by p → q or p ⇒ q. Example: If p : 6 is an even number. q : 6 is divisible by 2, then p → q : If 6 is an even number then 6 is divisible by 2. The truth value of p → q depends on the truth values of p and q. Note that p → q is false only when p is true and q is false.
Truth Table
p T T F F
q T F T F
p→q T F T T
BICONDITIONAL IMPLICATION EQUIV UIVALENCE): BICONDITIONAL (DOUBLE IMPLICATION OR EQUIVALENCE): If p and q are simple propositions, then the proposition 'p if and only if q' is called biconditional or double implication. It is denoted by p ↔ q. Example: If p : k is odd number. q : k2 is odd number, then p ↔ q : k is odd number if and only if (iff) k2 is odd number. Note that p ↔ q involves both the conditionals p → q and q → p.
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Basic Mathematics
∴
p ↔ q is p → q ∧ q → p
a
f a
f f a f
The biconditional p → q is true if p and q are both true or both false i.e., if p and q have same truth values. Otherwise it is false.
tab Truth table
p T T F F
q T F T F
p→q T F T T
q→ p T T F T
p ↔ q i. e., p → q ∧ q → p T F F T
a
NEGATION: NEGATION: If 'p' is a proposition then the proposition 'not p' is called negation of p. It is denoted by ~p. Example: If p : 6 is odd number then ~p : 6 is not an odd number. If p is true then ~p is false and if p is false then ~p is true.
tab Truth table
p T F
~p F T
WORKED EXAMPLES
I. Write the following propositions in symbols: 1. An integer is even if and only if it is divisible by 2. Solution: Let p : An integer is even.
q : It is divisible by 2. ∴ The given proposition is p ↔ q. 2. If 6 + 3 = 7, then 7 − 3 = 6 Solution: Let p : 6 + 3 = 7 q:7−3=6 Then the given proposition is p → q 3. I play chess or I study at home. Solution: Let p : I play chess q : I study at home.
Mathematical Logic
5
The given proposition: p ∨ q. 4. It is raining and the ground is wet. Solution: Let p : It is raining. q : The ground is wet. The given proposition : p ∧ q. 5. If it rains today then government declares a holiday and we are happy. Solution: Let p : It rains today q : Government declares a holiday r : we are happy. Given proposition: p → (q ∧ r). 6. If a number is not real then it is complex. Solution: Let p : A number is real. q : It is complex. Given proposition in symbols: ~p → q. 7. If Rama is intelligent or hardworking then logic is easy. Solution: Let p : Rama is intelligent q : Rama is hardworking r : Logic is easy. Given proposition: (p ∨ q) → r. 8. If 3 is not odd and 2 is not even then 7 is not odd or 8 is not even. Solution: Let p : 3 is odd q : 2 is even r : 7 is odd s : 8 is even. Given proposition: (~p ∧ ~q) → (~r ∨ ~s) II. Express each of the following compound propositions in sentences if p : The question paper is difficult. q : I get good marks. r : I can go abroad. 1. p ∧ ~q Solution: The question paper is difficult and I do not get good marks. 2. q → r If I get good marks, then I can go abroad. 3. (~p ∧ q) → r Solution: If the question paper is not difficult and I get good marks then I can go abroad. 4. r ↔ q Solution: I can go abroad iff I get good marks. 5. (p ∧ ~q) → ~r
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Basic Mathematics
Solution: If the question paper is difficult and I do not get good marks then I can not go abroad. III. If p, q, r are propositions with truth values T, F, T respectively then find the truth value of the following propositions: 1. p ∧ ~q
p T q F ~q T p∧ ~ q T
Explanation: Explanation: A compound proposition with 2 components will have 4 possibilities. Write 2 T and 2F under p, alternatively T, F under q to get all possible combinations. Now ~p is T when p is F and vice versa. ~q is T when q is F and vice versa. ~p → ~q is F only when ~p is T and ~q is F. Otherwise it is T. 3. p ↔ (p ∧ q)
p T T F F
4. p ∨ ~(~p ∨ q)
q T F T F
p∧q T F F F
p↔ p∧q T F T T
a
f
p T T F F
~p F F T T
q T F T F
~ p∨q T F T T
~ ~ p∨q F T F F
a
f
p∨ ~ ~ p ∨ q T T F F
a
f
Mathematical Logic
9
5. p ∧ (q ∨ r)
p T T T T F F F F
q T T F F T T F F
r T F T F T F T F
q∨r T T T F T T T F
p∧ q∨r T T T F F F F F
a f
Explanation: Explanation: A compound proposition having 3 components will have eight possibilities. Write 4T and 4F under p, 2T, 2F, 2T, 2F under q and alternately T and F under r to get all the possible combinations of truth values. Now q ∨ r is F only when both q and r are F. Otherwise it is T. p ∧ (q ∨ r) is T only when p is T and (q ∨ r) is T otherwise it is F. 6. (p → ~q) ↔ (q ∧ ~ r)
p T T T T F F F F
q T T F F T T F F
r T F T F T F T F
~q F F T T F F T T
p →~ q F F T T T T T T
~r F T F T F T F T
q∧ ~ r F T F F F T F F
a p →~ q f ↔ aq ∧ ~ r f
T F F F F T F F
1.4 TAUTOLOGY AND CONTRADICTION
A tautology is a compound proposition which is always true irrespective of the truth values of its components. A contradiction is a compound proposition which is always false irrespective of the truth values of its components. Note: To determine whether a given proposition is a tautology or a contradiction, construct its truth table. If its truth value for all possibility is True, then it is tautology. If its truth value for all possibility is False then it is contradiction. Otherwise it is neither tautology nor contradiction.
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Basic Mathematics
WORKED EXAMPLES
1. Prove that p ∧ ~p is a contradiction.
p T F
~p F T
p∧ ~ p F F
From the last column p ∧ ~p is a contradiction. 2. Prove that p ∨ ~p is a tautology.
p T F
~p F T
p∨ ~ p T T
From the last column p ∨ ~p is a tautology. 3. Find whether p → ~p is tautology or contradiction or neither
From the last column it is clear that (p ∧ ~q) ↔ q is neither tautology nor contradiction. 5. Prove that (p → q) ↔ (~q → ~p) is a tautology.
p T T F F
q T F T F
~q F T F T
~p F F T T
p→q T F T T
a xf
~ q →~ p T F T T
a yf
X↔Y T T T T
From the last column it is clear that (p → q) ↔ (~q → ~p) is a tautology.
Mathematical Logic 11
6. Prove that (p ∨ q) ∧ (~p ∧ ~q) is a contradiction.
p T T F F
q T F T F
~p F F T T
~q F T F T
p∨q T T T F
~ p∧ ~ q F F F T
a p ∨ q f ∧ a ~ p∧ ~ qf
F F F F
From the last column, the given proposition is contradiction. 7. Examine whether (p → q) ∧ (q → r) is a tautology or contradiction or neither
p T T T T F F F F
q T T F F T T F F
r T F T F T F T F
p→ q T T F F T T T T
q→r T F T T T F T T
a p → qf ∧ a q → r f
T F F F T F T T
From the last column it is clear that the given proposition is neither tautology nor contradiction. 8. Prove that
a p → qf ∧ a q → r f →
p→ q T T F F T T T T q→r T F T T T F T T
p → r is a tautology
p T T T T F F F F
q T T F F T T F F
r T F T F T F T F
a p → qf ∧ a q → r f
T F F F T F T T
p→ r T F T F T T T T
a p → qf ∧ a q → r f →
T T T T T T T T
p+r
From the last column clearly the given proposition is tautology.
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Basic Mathematics
1.5 LOGICAL EQUIVALENCE:
Two propositions X and Y are said to be logically equivalent if and only if they have same truth values and we write X ≡ Y. Note that X ≡ Y if and only if X ↔ Y is always true or X ↔ Y is tautology.
WORKED EXAMPLES:
I. Negate the following: 1. 6 is an odd number or 3 is an even number. Let p : 6 is an odd number. q : 3 is an even number. Given proposition in symbol is p∨q
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Basic Mathematics
We know
~ p ∨ q ≡ ~ p∧ ~ q
a
f
∴ Negation: 6 is not an odd number and 3 is not an even number. 2. He is rich and He is not happy Let p : He is rich. q : He is happy. Given: p∧~q We know
~ p∧ ~ q ≡ ~ p∨ ~ ~ q ≡~ p ∨ q
a
f
a f
∴ Negation: He is not rich or He is happy. 3. If the cow is big, then it is healthy. Let p : Cow is big. q : It is healthy. Given proposition in symbols: p → q We know
~ p → q ≡ p∧ ~ q
a
f
∴ Negation: Cow is big and it is not healthy. 4. If the triangles are not equiangular then the sides are not proportional. Let p : The triangles are not equiangular. q : The sides are not proportional. Given proposition: p → q Its negation: ~ ( p → q) But
~ p → q ≡ p∧ ~ q
a
f
Negation: The triangles are not equiangular and the sides are proportional. 5. 6 is even if and only if it is divisible by 2. Let p : 6 is even. q : 6 is divisible by 2. Given: p ↔ q. Its negation is ~(p ↔ q) But
~ p ↔ q ≡ p∧ ~ q ∨ ~ p ∧ q
a
f a
f a
f
∴ Negation: 6 is even and it is not divisible by 2 or 6 is not even and it is divisible by 2.
1.6 CONVERSE, INVERSE AND CONTRAPOSITIVE OF A CONDITIONAL
Let p → q be the given conditional then the conditional q → p is called converse of p → q. The conditional ~p → ~q is called inverse of p → q. The conditional ~q → ~p is called contrapositive of p → q. Note that contrapositive is converse of inverse.
4. If I work hard then I can score 90% and I can go for engineering. Solution: Let p : I work hard. q : I can score 90% r : I can go for engineering. Given p → (q ∧ r) Converse: q∧r→p i.e. If I can score 90% and I can go for engineering then I work hard. Inverse is ~p → ~(q ∧ r) i.e., ~p → ~ q ∨ ~r If I do not work hard then I can not score 90% or I cannot go for engineering. Contrapositive is ~(q ∧ r) → ~p i.e., (~q ∨ ~r) → ~p If I cannot score 90% or I cannot go for engineering then I do not work hard. 5. If e is not irrational and π is rational then 6 is not even or 2 is odd. Let X : e is not irrational and π is rational Y : 6 is not even or 2 is odd. Given X→Y Converse: Y→X If 6 is not even or 2 is odd then e is not irrational and π is rational. Inverse: ~X → ~Y If e is irrational or π is not rational Then 6 is even and 2 is not odd. Contrapositive: ~Y → ~X If 6 is even and 2 is not odd then e is irrational or π is not rational.
REMEMBER Otherwise
EXERCISE
I. Write the following compound propositions in symbols. Write follo ollowing propositions 1. If a triangle is equilateral then all the sides of the triangle are equal. 2. If I don't go to picnic then I will study at home. 3. Sun rises in the east and earth is not flat. 4.
2 is irrational or 5 is real.
5. A number is prime if and only if it is not composite. 6. If 2 + 2 ≠ 4 and 6 + 6 ≠ 12, then 4 + 7 = 6 or 5 + 3 = 9. 7. ABC is a right angled triangle if and only if one of the angle = 90° and square on the hypotenuse = sum of the squares on other 2 sides. 8. a + ib = x + iy iff a = x and b = y. II. If p, q and r are 3 propositions with truth values T, F and T respectively then find the truth are propositions truth values espectively find truth values of the following: follo ollowing: 1. 5.
VI. Negate the following compound propositions: Neg follo ollowing propositions: 1. 5 is odd and 6 is even. 2. Cow is not big or it is black. 3. If 2 lines are parallel then they do not intersect. 4. I will pass the examination iff the questions are easy. 5. p∧ ~ q 6. p →~ q 7. p ∧ q → r 8. p ↔~ q VII. Find the inverse converse and contrapositive of the following: inverse converse contrapositiv follo ollowing: 2 is even. 1. If x is even then x 2. If a2 + b2 = c2 and a2 = c2 then b2 = 0. 3. If a number is real then it is rational or it is irrational. 4. If Smitha gets a first class then she is either intelligent or hard working. 5. If 3 is not prime and 7 is not an odd number then 37 is not an even number or 73 is an odd number. 6. ~p → ~q 7. 8. 9. 10.
It is not odd or 6 is not even Cow is big and it is not black. 2 lines are parallel and they intersect I will pass the examination and the questions are not easy or I will not pass the examination and the questions are easy. ~p ∨ q p∧q ~p ∨ (q ∧ ~r) (p ∧ q) ∨ (~p ∧ ~q)
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VII. 1. Inverse: If x is not even then x2 is not even
Converse: If x2 is even then x is even.
Contrapositive: If x2 is not even then x is not even. 2. Inverse: If a2 + b2 ≠ c2 or a2 ≠ c2 Then b2 ≠ 0.
Converse: If b2 = 0 then a2 + b2 = c2 and a2 = c2. Contrapositive: If b2 ≠ 0 then a2 + b2 ≠ c2 or a2 ≠ c2.
3. Inverse: If a number is not real then it is not rational and it is not irrational.
Converse: If a number is rational or it is not irrational then it is real. Contrapositive: If a number is not rational and it is not irrational then it is not real.
4. Contrapositive: If Smitha is neither intelligent nor hardworking then she doesn't get a first class.
Converse: If Smitha is either intelligent or hardworking then she gets a first class. Inverse: If Smitha does not get first class then she is neither intelligent nor hardworking.
5. Contrapositive: If 37 is even and 73 is not odd then 3 is prime or 7 is odd.
Converse: If 37 is not even or 73 is odd then 3 is not prime and 7 is not odd. Inverse: If 3 is prime or 7 is odd then 37 is even and 73 is not odd.
6. Inverse: p → q Converse: ~q → ~p Contrapositive: q → p
2
Permutation and Combination
2.1 INTRODUCTION:
In our daily life we come across situations where we have to select or arrange certain things out of a given number of things. This selection or arrangement involves a principle known as fundamental principle which is illustrated by the following example. Suppose that in a auditorium there are 4 different entrance doors (say I1, I2, I3 and I4) and there are 5 different exit doors (say O1, O2, O3, O4 and O5). In how many ways can a person enter and leave the auditorium? If a person enters the auditorium through the door I1, he can go out by any one of the exit doors O1 O2 O3 O4 O5. So there are 5 ways of leaving the auditorium if the person enters it through door I1. Similarly corresponding to the entrance door I2 there are 5 ways of leaving the auditorium. Altogether 5 + 5 + 5 + 5 = 20 different ways. In general if there are m different entrance doors and n different exit doors, a person can enter and leave the auditorium in mn ways. This is fundamental principle.
2.2 FUNDAMENTAL PRINCIPLE:
Fig. 2.1
If one event can be done in m different ways and after it has been done in one of these ways, a second event (which is independent of the first) can be done in 'n' different ways then the two events together can occur in mn ways. The extension of this principle (also called the mnp ... principle) to the case of more than 2 events is obvious. Example: 1. A boy and a girl have to be selected from a group of 5 boys and 6 girls. In how may ways can the selection be made?
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Solution: Here First operation is selecting a boy from a group of 5 boys. This can be done in 5 ways. After this is done, the second operation is selecting a girl from 6 girls. This can be done in 6 ways. By fundamental principle, the total number of selections = 5 × 6 = 30 ways. 2. There are 4 candidates for the post of manager. 3 candidates for the post of officer and 5 for the post of clerk. In how many ways can these posts be filled? Solution: A manager may be selected in 4 ways. An officer may be selected in 3 ways and A clerk may be selected in 5 ways. By fundamental principle, the 3 posts together can be filled in 4 × 3 × 5 = 60 ways.
2.3 PERMUTATION AND COMBINATION:
Suppose 3 members (say a, b, c) went to a cinema where they get only 2 tickets. So it is required to select 2 members out of 3 members (a, b, c). The following selections are possible: ab, bc, ca. So there are 3 ways of selecting 2 members out of 3 members. In symbols this can be written as 3c2. It is called number of combination of 3 members taken 2 at a time. Now consider 3 symbols α, β and γ. If we wish to arrange these letters taken 2 at a time, we get the following arrangements αβ, βγ, γα, βα, γβ, αγ. These arrangements are called number of permutations of 3 symbols taken 2 at a time. In symbols this can be written as 3p2. Note that in arrangement order is important. Therefore αβ is different from βα, whereas in selection order is not important. Hence ab is same as ba and so it is regarded as only one selection.
2.5 PERMUTATION:
An arrangement of all or part of a set of objects in some order is called permutation. If the objects are arranged along a straight line it is called a linear permutation. If the objects are arranged around a circle then it is called circular permutation.
Permutation and Combination 23
2.5.1 Linear Permutation: Each of the different arrangements in a straight line that can be made by taking some or all of a number of things at a time is called linear permutation. The number of permutation of n distinct things taken r at a time is denoted by npr. 2.5.2 Value of npr: The number of permutation of n distinct things taken r at a time i.e. npr will be same as the number of ways in which r blank places can be filled up with n given objects. As the first place can be filled in by any one of the n objects, there are n ways of filling the first place. After having filled in the first place by any one of the n things, there are (n − 1) objects left. Hence 2nd place can be filled in (n − 1) ways. Similarly 3rd place can be filled in (n − 2) ways and so on. Hence rth place can be filled in (n − (r − 1)) ways i.e., (n − r + 1) ways.
2.5.3 Value of npn: The number of permutation of n distinct objects taken all at a time is npn. It is same as the number of ways in which n blank spaces can be filled up with n given objects.
Position of the object Number of ways
1st n
2 nd n −1
3 rd n−2
...... ......
n − 1th 2
n th 1
As the first place can be filled in by any one of the n objects there are n ways of filling the first place. After having filled in the first place, 2nd place can be filled in (n − 1) ways and so on. ∴ By fundamental principle,
n
pn = n n − 1 n − 2 ... 2 ⋅ 1
a fa f
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Basic Mathematics
n
∴ ∴ 2.5.4 Value of 0 : We have Put
n n
pn = n
n
Value of
pn = n .
pr =
n n−r n n−n
r = n.
pn =
n
But ∴ Cross-multiplying,
pn = n n= n 0
0=
n n
⇒
0 =1
WORKED EXAMPLES:
1. Evaluate: ( a)
5
p2
n
(b)
6
p3
(c)
7
p7
We have
pr =
n n−r
5
p2 =
5 = 5 × 4 = 20. 5−2
6 7
p3 = 6 × 5 × 4 = 120. p7 = 7 = 5040.
2. If npn = 720 find n. We have
n
pn = n = 720 given
a
f
Permutation and Combination 25
n = 6 × 5 × 4 × 3 × 2 ×1
n= 6
⇒ 3. If np2 = 72, find n. We have
n = 6.
2 3 4 5 6
720 360 120 30 6 1
= n (n − 1) = 72. n (n − 1) = 9 × 8 [By inspection: 72 = 9 × 8]. ⇒ n = 9. 4. In how many of the permutations of 7 things taken 4 at a time will (a) One thing always occur (b) One thing never occur? Solution: (a) Keeping aside the particular thing which will occur, the number of permutation of 6 things taken 3 at a time = 6p3 = 6 × 5 × 4 = 120. Now this particular thing can take up any one of the four places and so the total number of ways = 120 × 4 = 480 ways. (b) Leaving aside the particular thing which has never to occur, the number of permutation of 6 things taken 4 at a time = 6p4 = 6 × 5 × 4 × 3 = 360 ways. 5. How many 4 digit numbers can be formed with the digits 2, 4, 5, 7, 9. (Repetitions not being allowed). How many of these are even? Number of 4 digit numbers that can be formed with the digits 2, 4, 5, 7, 9 (without repetitions) = 5p4 = 5 × 4 × 3 × 2 = 120. Th H T U Since we require an even number, we must have 2 or 4 in the unit's place. After filling the unit's place by 2 or 4, the remaining 3 places (Ten's, Hundred's and Thousand's) can be filled by remaining digits 5, 7, 9, 2 or 4 in 4p3 ways 4 × 3 × 2 = 24 ways. ∴ Number of even numbers that can be formed 2 or 4 4 = 2× 4 p3 = 2 × 24 = 48
p3
np 2
6. How many numbers can be formed by using any number of digits 3, 1, 0, 5; no digit being repeated in any number: Solution: The number of single digit numbers = 3p1 (excluding zero) = 3. The permutation of 4 digits taking 2 at a time are 4p2 but 3p1 of these have zero in ten's place so reduce to single digit number. ∴ Number of 2 digit numbers = 4p2 − 3p1. Similarly number of 3 digit numbers = 4p3 − 3p2. Number of 4 digit numbers = 4p4 − 3p3.
= 3 + 9 + 18 + 18 = 48. 7. There are 4 Kannada books, 3 Hindi books and 5 English books. In how many ways can these be placed on a shelf if the books of the same language are to be together? Solution: Since the books of the same language are to be together. Let us consider the 4 Kannada books as 1 unit, 3 Hindi books as another unit and 5 English books as a different unit. Then we have to arrange
3 3 different units. This can be done in p3 = 3 = 6 ways. 4 But the 4 Kannada books remaining together can be arranged in p4 =
4 ways. Similarly, 3 Hindi
books can be arranged in 3 ways and 5 English books can be arranged among themselves in 5 ways. ∴ Number of permutations = 3 × 4 × 3 × 5
=3×2×4×3×2×1×3×2×5×4×3×2×1 = 6 × 24 × 6 × 120 = 103680 ways. 8. In how many different ways can 5 examination papers be arranged in a row so that the best and the worst papers may never come together. Solution: Without any restrictions the 5 exam papers can be arranged among themselves in 5p5 = 5 = 120 ways. Considering now, the best and the worst papers as a single paper, we have only 3 + 1 = 4 papers. Now these 4 papers can be arranged taking all at a time in 4 p3 = 4 × 3 × 2 = 24 ways. But in each of the 24 ways, the best and worst papers can be arranged among themselves in 2 = 2 ways. ∴ Total number of ways, where the best and the worst paper always come together = 24 × 2 = 48. Hence required number of arrangements where best and the worst paper never come together = 120 − 48 = 72 ways. 9. In how many ways can 5 boys and 3 girls be seated in a row so that (a) each girl in between 2 boys (b) no two girls sit together (c) all the girls are together. Solution: (a) First we arrange 5 boys in a row. This can be done in 5p5 = 5 = 120 ways. . Consider one such arrangement:
B1 ↓ B2 ↓ B3 ↓ B4 ↓ B5
Permutation and Combination 27
There are 4 places available for 3 girls, so that each girl is between 2 boys. The 4 places can be filled by 3 girls in 4p3 ways = 4 × 3 × 2 = 24 ways. For one arrangement of boys, there are 24 ways of arranging girls. ∴ For 120 arrangement of boys = 120 × 24 = 2880 ways. (b) Again we arrange 5 boys in a row. This can be done in 5 = 120 ways. Considering 1 such arrangement *B 1 *B 2 * B 3 *B 4 * B 5 * There are 6 places available for the girls so that no two girls are together. The 6 places can be filled by 3 girls in 6p3 ways = 6 × 5 × 4 = 120 ways. ∴Required number permutations = 120 × 120 = 14400 ways. (c) We require all the 3 girls to be together. So we consider 3 girls as one unit. Now 5 boys and 1 unit (of 3 girls) can be arranged in 5 + 1 = 6 = 720 ways. Again 3 girls, wherever they are together can be arranged in 3 = 6 ways. ∴ Number of permutations = 720 × 6 = 4320 ways. 10. How many different words can be formed with the letters of the word 'ORDINATE'. (a) Beginning with O (b) beginning with O and ending with E. (C) So that vowels occupy odd places. (a) Keeping 'O' in the first place. We can arrange the remaining 7 letters in 7 places in
7
p7 = 7 ways = 5040 ways.
O
7
p7
∴
Number of words that can be formed with the letters of word 'ORDINATE' that begins with 0 = 5040. (b) Keeping O in the first place and E in the last place, the remaining 6 letters can be arranged in 6 place in 6p6 = 6 = 720 ways.
O E
∴Total number of permutations = 24 × 24 = 576. 11. If the letters of the word MAKE be permuted and the words so formed be arranged as in dictionary what is the rank of the word? Solution: The alphabetical order of the letters is AEKM. If A is fixed in 1st place, the other 3 letters can be permuted in 3 = 3 × 2 = 6 ways. . The number of words begin with A = 6. Similarly the number of words which begin with E and K = 3 = 6 each. The words beginning with M are MAEK and MAKE. ∴ Rank of the word = 6 + 6 + 6 + 1 + 1 = 20. 12. Prove that the number of ways in which n books can be placed on a shelf when 2 particular books are never together is n − 2 ⋅ n − 1. Solution: Regarding the 2 particular books as one book, there are (n − 1) books now, which can be arranged in
n −1
a
f
pn −1 = n − 1 ways. Now these two books can be arranged in 2 ways. Therefore the
number of permutations in which 2 particular books are always together = 2 . n − 1 . The number of permutations of n books without any restriction = n . ∴ The number of permutations in which 2 particular books never occur together
= n − 2. n −1 = n ⋅ n − 1 − 2. n − 1
a = an − 2 f
= n −1 n − 2
f
n −1
Hence proved.
2.6 PERMUTATION OF THINGS OF WHICH SOME ARE ALIKE
Theorem: Theorem If x is the number of permutations of n things taken all at a time of which one set p are alike, another set q are alike and so on. Then prove that
x= n p ⋅ q ...
Permutation and Combination 29
Proof: Replacing p like objects by 'p' unlike objects, q like objects by 'q' unlike objects and so on, we arrive at a stage where all n objects are distinct. By permuting these unlike objects amongst themselves each of the x permutations would give rise to p ⋅ q ... permutations. Hence x permutations give rise to x ⋅ p ⋅ q ... . But the number of permutations of n distinct objects taken all at a time = n . ∴ ⇒
x ⋅ p ⋅ q ... = n x= n . p ⋅ q ...
WORKED EXAMPLES:
1. In how many ways can letters of the word 'INDIA' be arranged? Solution: There are 5 letters in the word 'INDIA' of which I occur 2 times. ∴ Number of words possible =
5 2
= 5 × 4 × 3 = 60. 2. In how many ways can the letters of the word 'PERMANENT' be arranged so that 2E's are always together. Solution: There are 9 letters in the word 'PERMANENT' of which E occurs 2 times, N occurs 2 times. If 2 E's are together, taking them as one letter we have to arrange 8 letters in which N occurs two times. ∴ Number of arrangements =
8 2
= 8 × 7 × 6 × 5 × 4 × 3 = 20,160 3. In how many ways can be letters of the word 'HOLLOW' be arranged so that the 2L's do not come together. Solution: There are 6 letters in the word HOLLOW in which L occurs 2 times, O occurs 2 times. ∴ Number of arrangement =
6 2⋅ 2 6× 5× 4×3×2 2 ×1× 2 ×1
=
= 180. If 2 L's are together, taking them as one letter, we have to arrange 5 letters in which O occurs 2 times. ∴ Number of arrangements in which 2 L's are together =
5 = 5× 4×3 2
∴
= 60. Number of arrangements in which 2L's do not come together = 180 − 60 = 120.
30
Basic Mathematics
4. Find the number of permutations of the word 'EXCELLENCE'. How many of these permutations (i) begin with E (ii) begin with E and end with C (iii) begin with E and end with E (iv) do not begin with E. Solution: The word 'EXCELLENCE' contains 10 letters of which E occurs 4 times C occurs 2 times, L occurs 2 times. Number of permutations =
10 4⋅ 2⋅ 2
=
10 × 9 × 8 × 7 × 6 × 5 = 37800. 2 ×1× 2 ×1
( i) Put E in the first place and arrange the rest. Now there are 9 letters of which E occurs 3 times, L occurs twice and C occurs twice. ∴ Number of permutations =
9 3⋅ 2 ⋅ 2 9×8×7×6×5× 4 2 ×1× 2 ×1
=
= 15,120. (ii) Put E in the first place and C in the last place, and arrange the rest. There are 8 letters of which E occurs 3 times, L occurs twice. ∴ Number of permutations =
8 3⋅ 2 8× 7×6×5×4 = 3360. 2
=
( iii) Put one E in the first place and another E in the last place and arrange the rest. There are 8 letters of which E occurs 2 times, L occurs twice and C occurs twice. ∴ Number of permutations =
8 2⋅ 2⋅ 2 8×7×6×5× 4×3 2 ×1× 2 ×1
=
= 5040. (iv) Number of permutations that do not begin with E = Total number of permutations. — Number of permutations that begin with E. = 37800 − 15120 = 22680. 5. How many numbers less than 3 millions can be formed using the digits of the number 2123343?
Permutation and Combination 31
Solution:
Million HTh TTh Th H T U
1 or 2
Since the number should be less than 3 millions, we can have either 1 or 2 in the million's place. Keeping 1 in the million's place and arranging the rest. There are 6 numbers 2, 2, 3, 3, 3, 4 of which 2 repeats twice, 3 occurs thrice. ∴ Number of numbers =
6 6×5× 4 = 2⋅ 3 2
= 60. Now keeping 2 in the million's place and arranging the rest. There are 6 numbers 1, 2, 3, 3, 3, 4 in which 3 occurs thrice. ∴ Number of numbers =
6 =6×5×4 3
= 120. Total number of numbers less than 3 millions. = 60 + 120 = 180.
2.7 CIRCULAR PERMUTATION:
If we have to arrange 4 letters A, B, C, D in a row. Then 2 of the arrangements would be ABCD, BCDA and we treat these two arrangements as different. But if we arrange along the circumference of the circle, the two arrangements and are one and the same.
So we conclude that circular permutations are different only when the relative order of the objects is changed; otherwise they are same. In circular permutation of n different things one thing is kept fixed and the balance (n − 1) things are arranged relative to it in n − 1 ways. If the clockwise and anticlockwise orders are distinguished, the required number of permutations = n −1 . ∴ Number of ways in which n persons can occupy the chairs in a round table = n − 1 . If the clockwise and anticlockwise orders are not distinguished then required number of permutations =
n −1 2
.
32
Basic Mathematics
∴ Number of ways in which n flowers or n beads are strung to form garland or necklace =
n −1 2
.
WORKED EXAMPLES:
1. In how many ways 5 people sit around a table? Solution: Fixing the position of one person, the remaining 4 persons can sit around a table in 4 ways = 4 × 3 × 2 × 1 = 24. ∴ 5 people can sit round a table in 24 ways. 2. In how many ways can 7 different jewels be strung into a necklace? Solution: Keeping one jewel fixed, remain 6 jewels can be arranged in 6 = 720 ways. Since clockwise and anticlockwise arrangements are same, Required number of permutations.
1 × 720 = 360 ways. 2
3. In how many ways can 7 people be arranged at a round table so that 2 particular persons always sit together. Solution: First, the two particular persons can be arranged in 2 = 2 ways. . Considering them as one fixed person, the remaining 5 persons can be arranged in 5 = 120 ways. ∴ The required number of permutation = 120 × 2 = 240. 4. A round table conference is to be held between delegates of 9 countries. In how many ways can they be seated if 2 particular delegates must not sit next to each other? Solution: The number of ways in which 2 particular delegates must not sit next to each other = Total number of permutations — Number of permutations in which 2 particular delegates sit next to each other. Now Total number of circular permutations of 9 delegates = 9 − 1 = 8 Considering 2 delegates as one fixed person, the remaining 7 delegates can be arranged in 7 ways. 2 delegates again can be arranged in 2 ways. ∴ Number of permutations in which 2 particular delegates sit next to each other = 7 × 2 ∴ Required number of permutation
= 8 − 7 × 2 = 8 7 − 7 × 2 = 7 8 − 2 = 6 7 = 30240
a f
5. In how many ways can 6 persons sit around a table so that all shall not have the same neighbours in any 2 arrangements? Solution: 6 persons can sit round a table in 5 = 120 ways. But each person will have the same neighbours in clockwise and anticlockwise arrangements.
Permutation and Combination 33
∴
Required number of ways =
5 120 = = 60. 2 2
6. In how many ways can 4 gentlemen and 4 ladies sit down together at a round table so that no two ladies may come together. Solution: Let the gentlemen first take up their seats. They can sit in 3 = 6 ways. When they have been seated, there remain 4 places for the ladies each between 2 gentlemen. Therefore the 4 ladies can sit in 4 places in 4 = 24 ways. ∴ Required number of ways = 3 × 4 = 6 × 24 = 144.
2.8 COMBINATION
Each of the different groups or selection which can be made by taking some or all of a number of things at a time (irrespective of the order) is called a combination. The number of ways of selection of n different things taken r at a time is called the number of combination of n different things taken r at a time. It is written as ncr. 2.8.1 Value of ncr: The number of combinations of n different things taken r at a time can be arranged in r ways. ∴
nc r
combinations will produce n cr × r permutations.
Now n cr × r = Number of permutations of n different things taken r at a time = npr. ∴ ⇒
n
is the total number of combination of n + 1 things taken r at a time Proof by analytic method: which is nothing but the combination that contain a particular thing (ncr) plus the combination that do not contain a particular thing (ncr − 1). i.e.,
n +1
9. In how many ways can 4 persons be selected from amongst 9 persons? How many times will a particular person be always selected? Solution: The number of ways in which 4 persons can be selected from amongst 9 persons
= 9c4 =
9 9×8× 7×6 = = 126. 9 − 4⋅ 4 4 × 3 × 2 ×1 8 = 56. 8 − 3⋅ 3
Let a particular person is selected always. Then we have to select 3 persons from the remaining 8 persons. This can be done in 8c3 ways =
10. A student has to answer 7 out of 10 questions in an examination. How many choices has he, if he must answer the first three questions. Solution: There are 10 questions of which a student must answer first 3 questions. Remaining 4 questions (3 he has to answer 7 questions) can be selected among 10 − 3 = 7 questions in 7c4 ways. ∴ Number of combinations = 7 c 4 =
12. Find the number of (a) Straight lines (b) triangles that can be drawn from 20 points of which 4 are collinear. Solution: Two points are needed for a straight line. If none of the 20 points are collinear then we would get 20c2 straight lines. But 4 points are given to be collinear. So we would not get 4c2 lines, instead we get only one straight line containing all the 4 points. ∴ Number of straight lines =
20
c2 − 4 c2 + 1
= =
20 4 − +1 20 − 2 ⋅ 2 4 − 2 ⋅ 2 20 × 19 4 × 3 − +1 2 2
= 190 − 6 + 1 = 185. (b) We need 3 non-collinear points for a straight line. If none of the 20 points are collinear then we would get 20c3 triangles. Since 4 points are given to be collinear, we would not get 4c3 triangles from these points. ∴ Number of triangles = 20c3 − 4c3
=
20 4 − 20 − 3 ⋅ 3 4 − 3 ⋅ 3
= 1140 − 4 = 1136. 13. A committee of 10 members is to be chosen from 9 teachers and 6 students. In how many ways this can be done if ( i) The committee contains exactly 4 students. (ii) There is to be a majority of teachers. ( iii) There are atleast 4 students. ( iv) There are at most 7 teachers. Solution: ( i) The committee contains exactly 4 students and 10 − 4 = 6 teachers. 4 students can be selected out of 6 students in 6c4 ways and 6 teachers out of 9 teachers can be selected in 9c6 ways.
= 15 × 84 = 1260. there is to be majority of teachers, the committee may consist of 6 teachers 4 students 7 teachers 3 students 8 teachers 2 students 9 teachers 1 student. Number of selection =
9 9 9 9
c6 × 6 c 4 = 1260 c7 × 6 c3 = 720 c8 × 6 c2 = 135 c9 × 6 c1 = 6
Total number of selections = 1260 + 720 + 135 + 6 = 2121. there is to be at least 4 students, the committee may consist of 4 students 6 teachers 5 students 5 teachers 6 students 4 teachers. Number of selections =
6 6 6
c4 × 9 c6 = 1260 c5 × 9 c5 = 756 c6 × 9 c 4 = 126
So Total number of selections = (iv) As ( a) ( b) (c ) ( d) 1260 + 756 + 126 = 2142. there is to be atmost 7 teachers, the committee may consist of 7 teachers and 3 students 6 teachers and 4 students 5 teachers and 5 students 4 teachers and 6 students.
14. A team of eleven is to be chosen out of 16 cricketers of whom 4 are bowlers and 2 others are wicket keepers. In how many ways can the team be chosen so that there are at least 3 bowlers and at least one wicket keeper. Solution: 4 Bowlers (a) (b) (c ) (d) Number of ways (a) (b) (c ) (d) ∴
4 4 4 4
Total number of ways = 960 + 420 + 840 + 252 = 2472. 15. Arun has 7 friends, 4 of them are boys and 3 are girls. His sister, Aalekya has 7 friends, 4 of them are girls and 3 of them are boys. In how many ways can they invite for a party of 3 girls and 3 boys. So that there are 3 of Arun's friends and 3 of Aalekya's friends.
∴ Total number of selections = 16 + 324 + 144 + 1 = 485. 16. How many diagonals are there in a octagon? Solution: Number of diagonals in octagon = 8 c2 − 8 = 20. [3 octagon has 8 sides]
REMEMBER:
• •
n
pr =
n n−r
•
n
n
p0 = 1,
n . p ⋅ q ...
pn = n
•
n
p
n −1 . 2
•
cr =
n n−r⋅ r
Permutation and Combination 43
• • • • •EXERCISE
I. Find the value of: 1. 10p3 2. 12p3 3. 15c8 4. 8c5 5. 14c10 II. Find n if 1. np2 = 90 2. nc3 = 20 III. Find r if 1. 6pr = 360 2. 13pr = 156 IV. Find n and r if 1. npr = 240 and ncr = 120 2. npr = 336 and ncr = 56 V. 1. If np4 = 12, np2 = 120, find n. 2. If np4 = 56, np2 = 120, find n. VI. 1. How many 3 digit numbers can be formed by using the digits 9, 7, 6, 5, 3, 2 (repetitions not allowed)? (a) How many of these are less than 400? (b) How many of these are multiples of 5? (c) How many of these are multiples of 2? 2. In how many ways can the letters of the word 'STRANGE' be arranged so that (a) The vowels never come together. (b) The vowels are never separated. 3. A shelf contains 6 Hindi books, 5 Kannada books and 8 English books. In how many ways can they be arranged so that (a) Hindi books are together? (b) Hindi books are together and Kannada books are together.
44
Basic Mathematics
4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.
(c) Books of the same languages are together. ( d) No two English books are together. How many arrangements of the letters of the word SUNDAY can be made if the vowels are to appear only in the odd places. How many numbers of four different digits can be formed using the digits 0, 1, 2, 3, 4, 5? How many of them are even? If the letters of the word GATE be permuted and the words so formed be arranged as in a dictionary what will be the rank of the word? Find the number of permutations of the letters of the word INSTITUTION, when all the letters are taken at a time. How many of them (i) Have 3T's together (ii) begin with 2N's. Find the number of permutations of the letters of the word ASSASSINATION. How many of them (i) have 3 A's together (ii) begin with 2 N's. Find the number of permutation of letters of the word TOMORROW. How many of them have (i) 3O's together. (ii) End with 2R's. Find the number of ways in which 6 different beads can be arranged to form a necklace. In how many ways can 5 persons sit around a table. In how many ways can 4 boys and 4 girls be seated round a table so that no two boys are adjacent. In how many ways can 7 persons sit around a table so that all shall not have the same neighbours in any two arrangements? A round table conference is to be held between delegates of 20 countries. In how many ways can they be seated if 2 participants may wish to sit together always. (i) If nc10 = nc6, find n. (ii) If 43cr − 6 = 43c3r + 1, then find r. From 8 lecturers and 4 students a committee of 6 is to be formed. In how many ways can this be done so that the committee contains (i) exactly 2 students (ii) atleast 2 students. How many (i) straight lines (ii) triangles are determined by 12 points, no three of which lie on the same straight line. How many (i) straight lines (ii) triangles are determined by joining 20 points in a plane of which 6 are collinear. In how many ways a student can choose 8 questions from a set of 12 questions if the questions 1 and 10 are compulsory. Find the total number of diagonals of a hexagon. Out of 3 books on maths, 4 books on Physics and 5 books on Chemistry, how many collections can be made, if each collection consists of ( i) exactly one book on each subject (ii) at least one book on each subject.
3
Probability
3.1 INTRODUCTION:
The term probability refers to the chance of happening or not happening of an event. The theory of probability provides a numerical measure of the elements of uncertainity. It enables us to take decision under conditions of uncertainity with a calculated risk. The theory of probability has its origin in the games of chance, related to gambling for instance throwing a dice or tossing a coin. Generally speaking, the probability of an event denotes the likelihood of its happening. The value of probability ranges between zero and one. If an event is certain to happen its probability would be 1 and if it is certain that the event wouldn't take place, then the probability of its happening is zero. Ordinarily in social sciences probability of the happening of an event is rarely 1 or 0. The reason is that in social sciences we deal with situation where there is always an element of uncertainity about the happening or not happening of an event.
3.2 TERMINOLOGY:
Before we give definition of probability, it is necessary that we familiarise ourselves with certain terms that are used in this context. (i) Random experiment: It is an experiment which if conducted repeatedly under homogeneous exper xperiment: conditions doesn't give the same result. The result may be any one of the various possible outcomes. For example: If a die is thrown it wouldn't always fall with number 3 up. It would fall in any one of six ways which are possible. ev (ii) Trial and event: The performance of a random experiment is called a trial and the outcome – an event. Event could be either simple or compound (or composite). An event is called simple if it corresponds to a single possible outcome. Thus in tossing a die, the chance of getting 3 is a simple event (Q 3 occurs in a die only once). However the chance of getting an odd number is compound (Q odd numbers are more than one — 1, 3 and 5). (iii) Exhaustive cases: All possible outcomes of an event are known as exhaustive cases. In the throw Exhaustiv of a single die the exhaustive cases are six, as the die has only 6 faces each marked with different
Probability
47
(iv)
(v)
(vi)
(vii)
numbers. Similarly the number of exhaustive cases in tossing 2 coins would be 4: HH, HT ,TH and TT (H-Head, T-tail). oura Favourable cases: The number of outcomes which result in the happening of a desired event are called favourable cases. Thus in a single throw of a die the number of favourable cases of getting an odd number are 3 (i.e. 1, 3 and 5). Mutually exclusiv xclusi Mutually exclusive cases: Two or more cases are said to be mutually exclusive if the happening of any one of them excludes the happening of all others in a single experiment. Thus in a throw of a single die, the events 5, 4 and 3 are mutually exclusive. Equally likel ely Equally likely cases: Two or more events are said to be equally likely if the chance of their happening is equal, i.e., there is no preference of any one event over the other. Thus in the throw of a die, the coming up of 1, 2, 3, 4, 5 or 6 is equally likely. Independent dependent ev Independent and dependent events: An event is said to be independent if its happening is not affected by the happening of the other events. So in the throw of a die repeatedly coming up of 5 on the first-throw is independent of coming up of 5 again in the second throw. However we are successively drawing cards from a pack without replacement, the event would be dependent.
3.3 DEFINITION OF PROBABILITY:
We can define probability in 3 ways. (i) Mathematical or classical definition. (ii) Statistical or Empirical definition. (iii) Subjective approach or set theoretic approach definition. i Mathematical or classical definition: If there are 'n' mutually exclusive, exhaustive and equally likely simple events in a trial, 'm' of them are favourable to the occurrence of an event A. Then probability or chance of occurrence of A equal to
P A =
af
Number of favourable cases Total number of all possible equally likely cases P A =
af
m . n
Note: 1. Since 0 ≤ m ≤ n we have 0 ≤ or
m ≤ 1. n
0 ≤ P A ≤ 1.
af
2. When m = n, P (A) = 1 and when m = 0, P (A) = 0, i.e. when m = n, the event A is certain and when m = 0, the occurrence of event A is an absolute impossibility. 3. The probability of non-occurrence of A is denoted by
48
Basic Mathematics
P A =
di
Number of unfavourable cases . Total number of all possible cases P A = = n m − n n m =1− P A n
di
n−m n
=1−
∴ ⇒
af P d A i = 1 − P a Af P a Af + P d A i = 1 .
4. The main disadvantage of mathematical method is that it fails when there are infinite number of possible outcomes and it cannot be applied to trials where the outcomes are not equally likely. ii Statistical or Empirical definition of probability: If a random experiment is repeated for an indefinitely large number of times under identical conditions, then the limiting value of the ratio of the number of times an event occur to the total number of trials is said to be the probability of occurrence of the event, provided the limit is a definite finite number. If T is the number of trials and event A occurs f times in 'T' trials, then the probability of occurrence of event A is given by probability = P A = lim
af
T →∞
F f I. HTK
We use this method when the elementary events are not equally likely and the exhaustive number of cases in a trial is infinite. The limitation of this method is that in practice an identical experimental condition doesn't exist while repeating a random experiment for a large number of times. Moreover the relative frequency i.e.
f may not attain a unique limiting value when T → ∞. T
iii Subjective probability or set theoretic approach: A set of points representing all possible elementary outcomes of a random experiment is called the sample space (S). The number of all possible sample points in the sample space S is represented by n (S). The definition of probability is based on the following assumptions. (i) Total number of elementary events in the sample space (S) is finite say N (ii) N elementary events of the experiment are equally likely.
P A =
af
Number of elementary events favourable to event A Total number of equally elementary events in S.
Probability
49
i.e.,
P A =
Number of a f n aaAff = Total number sample points in A. S. of sample points in n S
WORKED EXAMPLES:
1. If one card is drawn at random from a well shuffled pack of 52 cards. Then find the probability of each of the following. (a) Drawing an ace card, (b) Drawing a face card, (c) Drawing a diamond card, (d) Drawing either spade or hearts, (e) Not drawing an ace of hearts. Solution: (a) One card can be drawn out of 52 cards in 52c1 = 52 ways = n (S). One ace card can be drawn out of 4 ace cards in 4c1 = 4 ways = n (A). ∴Probability of drawing an ace card =
n A Number of favourable cases = Total number of all possible equally likely cases n S
af af
=
4 1 = . 52 13
(b) A face card can be drawn out of 12 face cards in 12c1 = 12 ways. ∴Number of favourable cases = 12 Total number of all possible equally likely cases = 52c1 = 52. ∴Probability of drawing a face card =
12 3 = . 52 13
13c 1
(c) A diamond card can be drawn out of 13 diamond cards in ∴Probability of drawing a diamond card
= 13 ways.
=
13 1 = . 52 4
(d) There are 13 spade and 13 hearts cards in a pack of cards. Either a spade or a heart can be drawn in 26c1 = 26 ways.
50
Basic Mathematics
∴Probability of drawing either a spade or a hearts card = (e) There is one ace of hearts. ∴ ∴ Probability of drawing an ace of hearts =
26 1 = . 52 2
1 . 52 1 51 . = 52 52
Probability of not drawing an ace of hearts = 1 −
2. Three balls are drawn at random from a bag containing 6 blue and 4 red balls. What is the probability that two balls are blue and one is red? Solution: The bag contains (6 + 4) = 10 balls. 3 balls can be drawn out of 10 balls in 10c3 = ∴Total number of cases = 120. Now 2 blue balls can be drawn out of 6 in 6c2 =
3. Three unbiased coins are tossed. What is the probability of obtaining (a) all heads (b) two heads (c) one head (d) atleast one head (e) atleast two heads (f) All tails. Solution: There are 23 = 8 mutually exclusive exhaustive and equally likely cases HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. (a) Probability of all heads =
If a student is selected at random from the entire group of 100 students, find the probability that his marks (i) is under 40. (ii) above 50 (iii) either between 31 to 40 or 41-50. (i) Total number of students = 100. Number of students obtaining marks less than 40 = 5 + 10 + 13 + 14 = 42. ∴ Required probability =
(iii) Number of students obtaining marks between 31 to 40 = 14 and number of students obtaining marks between 41 to 50 = 17. ∴ Total number of students whose score is either between 31 to 40 or 41 to 50 = 14 + 17 = 31. ∴ Required probability =
31 100
5. If a pair of dice is thrown, find the probability that the sum of digits is neither 7 nor 11. Solution: A pair of dice is thrown. ∴ n (S) = 6 × 6 = 36. Let A be an event of getting the sum 7 and B be an event of getting the sum 11. Then ∴
3.4 ADDITION RULE OF PROBABILITY:
Statement: Statement: If A and B are 2 events, then probability that at least one of them occurs is given by
P A∪ B = P A + P B − P A∩ B
Proof: Proof: Consider the Venn diagram. The shaded portion denotes A∪B, i.e., set of all outcomes where some of the outcomes are common to both A as well as B. So P (A∪B) is the probability of happening of atleast one of the events A and B.
U A A ∩ B B
a
f af af a
f
Fig. 3.1
P (A) + P (B) is the sum of all the probabilities in A and all the probabilities in B. So the probability in A ∩ B has been added twice in P (A) as well as in P (B). So we must subtract P (A ∩ B) once from P (A) + P (B) to obtain probabilities in A∪B.
∴
P A∪ B = P A + P B − P A∩ B
Since A and B are mutually exclusive, A and B are disjoint sets. ∴ ⇒ ∴ ⇒
a f P a A ∪ Bf = P a Af + P a Bf − 0 P a A ∪ Bf = P a Af + P a Bf
Probability
53
Note: 1. P (A∪B) means P (A or B) i.e., probability of happening of atleast one of the events A and B. P (A ∩ B) means P (A and B) i.e., probability of happening of both the events A and B. 2. P (A∪B∪C) is probability of happening of atleast one of the events A, B and C. It is given by addition rule as
P A∪ B∪C = P A + P B + P C − P A∩ B
a
f af af af a f − P a B ∩ C f − P aC ∩ Af + P a A ∩ B ∩ C f
This can be proved by writing the Venn diagram.
U A B
C
Fig. 3.2
WORKED EXAMPLES:
1. A ticket is drawn from a bag containing 25 tickets bearing number 1, 2, 3, ..., 24, 25. Find the probability of its bearing a number which is either even or a multiple of 3. Solution: The events 'even number' and 'a multiple of 3' are not mutually exclusive as there are some numbers which are even as well as multiples of 3. Ex: 6, 12, 24. ∴ P (an even number or a multiple of 3) = P (an even number) + P (a multiple of 3) − P (an even number and a multiple of 3)
[Since there are 12 even numbers from 1 to 25, 8 multiples of 3 and 3 numbers which are even as well as multiples of 3 from 1 to 25]. 2. What is the probability of getting either total of 7 or 11 when a pair of dice is tossed? Solution: Total outcomes when a pair of dice is tossed = 6 × 6 =36. The events 'a total of 7' and 'a total of 11' are mutually exclusive events. ∴ P (a total of 7 or 11) = P (a total of 7) + P (a total of 11).
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Basic Mathematics
= P {(6, 1), (5, 2), (4, 3) (3, 4) (2, 5) (1, 6)} + P {(6, 5) (5, 6)}
=
6 2 8 2 + = = . 36 36 36 9
3. The probability that a contractor will get a plumbing contract is 2/3 and the probability that he will not get an electric contract is 5/9. If the probability of getting atleast one contract is 4/5, what is the probability that he will get both the contracts. Solution: Let A be an event that a contractor gets plumbing contract. B be an event that a contractor gets electrical contract. Then Given:
P A =
af
2 3
5 P B = . 9 P B = 1− P B = 1− P A∪ B =
di
af
di
5 4 = . 9 9
a f 4 5 P a A ∩ Bf = ?
f
From addition rule,
P A∪ B = P A + P B − P A∩ B
a
f af af a 4 2 4 = + − P a A ∩ Bf 5 3 9
⇒
P A∩ B = =
a
f
2 4 4 + − 3 9 5
30 + 20 − 36 45
P A∩ B =
a
f
14 . 45 14 . 45
∴ Probability that a contractor will get both the contracts =
4. One card is drawn from a pack of 52 cards, what is the probability that the card drawn is neither red nor king. Solution: The event 'card drawn is red' and 'card drawn is king' is not mutually exclusive because there are two cards in the pack which are red as well as king. ∴ P (card drawn is red or king)
Probability
55
= P (card drawn is red) + P (card drawn is king) − P (card drawn is red and king) Since there are 26 red cards, 4 king cards and 2 cards which are red as well as king, P (Cards drawn is red or king),
5. A card is drawn at random from a well shuffled pack of 52 cards. What is the probability that it is a heart or a queen or black card. Solution: Let A be an event that 'the card drawn is heart'. B be an event that 'the card drawn is queen' and C be an event that 'the card drawn is black'. A, B and C are not mutually exclusive events. So
P A∪ B∪C = P A + P B + P C − P A∩ B
3.5 CONDITIONAL PROBABILITY:
Let us consider the following example. A fair dice is thrown and the number that appeared is even. What is the probability that the number 4 has appeared? Since it is given that 'the number appeared is even'. Possible outcomes are no longer {1, 2, 3, 4, 5, 6} but only {2, 4, 6}. Out of these 3 possible outcomes there is 1 outcome in favour of appearance of 4. ∴ Probability of appearance of 4, given that an even number has appeared =
1 . 3
Formally, if E is the event 'The number that appeared is 4' and F the event 'Number that appeared is even' then P (E/F) denotes the probability of E given that F has happened. ∴
1 P E F = . 3 The probability that an event B occurs subject to the condition that A has already occurred is called the conditional probability of occurrence of the event B. It is denoted by P (B/A).
a f
3.6 MULTIPLICATION RULE:
Statement: Statement: The probability of the simultaneous occurrence of 2 events is the product of the probability that one of the events will occur and the conditional probability that the other event will occur given that the first event has occurred. If A and B are two events then P (A ∩ B) is the probability of their simultaneous occurrence and it is given by
or
a f af a f P a A ∩ Bf = P a Bf ⋅ P a A Bf
P A∩ B = P A ⋅P B A
Proof: Proof: Let A be any event with sample points n (A), i.e., P (A) > 0. If n (S) is the total number of sample points in S and B = another event such that A and B are not disjoint sets. (i.e., A∩B ≠ φ). Let the sample points in A∩B be n (A∩B). Then from definition P A =
WORKED EXAMPLES:
1. A pair of dice is thrown and sum of the numbers on the two dice comes to be 7. What is the probability that the number 4 has come on one of the dice? Solution: Let the events A and B be such that Event B: Sum of numbers on the two dice is 7. Event A: The number 4 has come. Total outcomes when a pair of dice is thrown = 36 = n (S)
P B = P 6, 1 , 5, 2 , 4, 3 , 3, 4 , 2, 5 , 1, 6
a f ma f a f a f a f a f a fr 6 P a Bf = . 36
P A ∩ B = 4, 3 , 3, 4 =
To get P (A∩B) select the outcomes favourable to A from the outcomes that are favourable to B. ∴
a
f ma f a fr
2 . 36
From Multiplication theorem,
P A∩ B = P A ⋅P A B
a
f af a f
Probability
59
⇒
P A B =
a f P aPAa∩fBf A
2 2 1 P A B = 36 = = . 6 6 3 36
a f
∴ Probability that the number 4 has come on one dice given that sum of numbers on 2 dice is 7 =
1 . 3
2. Two cards are drawn from a pack of 52 cards with replacement (i.e., the second card is drawn after replacing the first card in the pack). Find the probability that (a) Both are ace, (b) First card is jack and second card is king, (c) One is king and other is queen. Solution: (a) In a pack of 52 cards, there are 4 ace cards. ∴ P (both cards are ace)
=
4 4 1 . × = 52 52 169
(b) P (First card is jack and second card is king)
=
4 4 1 . × = 52 52 169
[Q There are 4 jack and 4 king cards in a pack of 52 cards] (c) P [One card is king and other is queen] = P [First is king and 2nd is queen] ∪ P [First is queen and 2nd is king]
3. Two cards are drawn without replacement from a pack of 52 cards. What is the probability that (i) both are queen. (ii) both are diamond cards. (iii) one is king and the other is ace. Solution: We are taking 2 cards from 52 cards. This can be done in 52c2 ways. There are 4 queen cards from which we require 2 queen cards. This can be done in 4c2 ways.
4
∴
Required Probability =
52
c2 4×3 = c2 52 × 51
=
1 . 221
60
Basic Mathematics
OR
There are 4 queen cards in a pack of 52 cards. ∴ Probability of drawing first queen card =
4 52
Since the card drawn is not replaced, we are left with 51 cards and 3 queen cards. ∴ Probability of drawing 2nd queen card = ∴ Probability of both queen cards = (ii) Required probability =
13 52
3 51
4 3 1 . × = 52 51 221
c2 c2 13 × 12 1 = 2 ×1 = . 52 × 51 17 2 ×1
(Since there are 13 diamond cards).
(iii) There are 4 favourable choices to take out a king and 4 favourable choices to take out an ace. ∴ Number of favourable cases = 4c1 × 4c1 In total, there are 52 cards out of which any 2 cards can be taken. ∴ Required probability =
4
c1 × 4 c1 52 c2 = 4×4×2 8 . = 52 × 51 663
4. A lot contains 10 items of which 3 are defective. 3 items are chosen from the lot at random one after another without replacement. Find the probability that all the 3 are defective. Solution: Let A, B and C be the events of drawing defective items in the first, second and third drawing respectively. Hence Probability of all the three items being defective is given by
P A∩ B∩C = P A ⋅P B A ⋅P C A∩ B = = 3 2 1 × × 10 9 8 1 . 120 OR
a
f af a f a
f
3 defective items can be picked from 3 defective items in 3c3 ways.
Probability
61
3 items can be picked from 10 items in ∴
10c
3
ways.
3 10
Required probability =
c3 c3
1 6 = 10 × 9 × 8 10 × 9 × 8 3 × 2 ×1 = 1 . 120
5. Anil and Bharath appear in an interview for 2 vacancies. The probability of their selection being
1 1 and respectively. 7 5
Find the probability that (i) both will be selected (ii) only one is selected (iii) none will be selected (iv) atleast one of them will be selected. Solution: Let A: Anil be selected. B: Bharath be selected. Given
EXERCISE
1. A bag contains 100 tickets each bearing a distinct number from 1 to 100. A ticket is drawn from the bag. Find the probability that
Probability
63
(a) the ticket bears an odd number. (b) the ticket bears a number divisible by 3. (c) the ticket bears a number divisible by 3 or 5. 2. A pair of dice is thrown. Find the probability that the sum will be (a) equal to 4 (b) less than 4 (c) greater than 4. 3. A bag contains 4 white, 5 red and 6 green balls. 3 balls are drawn at random. What is the probability that (a) All are green (b) All are white. 4. A card is drawn from a pack of playing cards. What is the probability of drawing (a) black card (b) king card (c) diamond card. 5. A box has 3 silver and 2 gold coins. 2 coins are drawn at random. Find the probability that (a) both the coins are silver ones. (b) both are gold coins (c) one of them is a silver coin. 6. A problem in statistics is given to 3 students A, B and C. Their probabilities in solving it are
1 1 1 , and respectively. What is the probability that the problem would be solved? 8 6 4
7. The following table gives a distribution of wages of 1000 workers.
Wages Rs.
An individual is selected at random from the above group. What is the probability that his wages are (a) under Rs. 160 (b) above Rs. 200 (c) between Rs. 160 and 200. 8. One card is drawn from a pack of 52 cards. What is the probability that the card drawn is (a) either red or king (b) either king or queen 9. A ticket is drawn from a bag containing tickets bearing numbers 1 to 25. Find the probability that the number is either even or a multiple of 3. 10. A coin and a die are thrown. What is the probability of getting a head or an even number. 11. A box contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If one item is chosen at random, what is the probability that it is rusted or a bolt? 12. A die is rolled. If the outcome is an odd number, what is the probability that it is prime. 13. A pair of dice is rolled. If the sum on the 2 dice is 9. Find the probability that one of the dice showed 3. 14. If A and B are 2 events in a sample space S such that P A = Find (a) P (A∩B) (b) P A ∩ B . 15. 3 children are randomly selected from a class. What is the probability that (a) all 3 were born on Monday (b) 2 were born or Friday and the other on Tuesday (c) none were born on Wednesday.
af
d
i
1 5 3 , P B = , P A∪ B = . 2 8 4
d i
a
f
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Basic Mathematics
ANSWERS
1. (a) 2. (a) 3. (a) 4. (a) 5. (a) 6.
1 2 3 36 4 91 1 2 3 10
(b) (b) (b) (b) (b)
33 100 3 36 4 455 1 13 1 10
(c) (c)
47 100 30 36
(c) (c)
1 4 3 5
290 64 127 1000 28 52
(b) (b) 10. 13.
7. (a) 8. (a) 9. 12.
195 1000 2 13 3 4 1 2
(c)
678 1000
17 25 2 3
11.
5 8 1 8
(b)
14. (a)
1 4
4
Binomial Theorem
4.1 INTRODUCTION:
We know
a x + af a x + af a x + af
1 2 3
= x+a = x 2 + 2ax + a 2 = x 3 + a 3 + 3ax x + a
a
f
= x 3 + 3 x 2 a + 3xa 2 + a 3 .
The general expansion for (x + a)n where n is positive integer is given by Sir Isaac Newton and is known as Binomial Theorem.
4.2 STATEMENT OF BINOMIAL THEOREM:
If x and a are real numbers and n is any positive integer then
The number of terms in the expansion of (x + a)n is n + 1. (2) In the expansion of (x + a)n, from left to right the power of x decreases by 1 from one term to the next term and the power of a increases by 1. (3) In any term, the power index of x plus the power index of a = n. (4) The co-efficients C0, C1, C2 ... Cn are called binomial co-efficients. (5) The general term in the expansion is (6) Since
n n − cr x n −r ⋅ a r which is (r + 1)th term. Hence Tr +1 = n cr x n r ⋅ a r .
cr = n cn − r , the binomial co-efficients of first and last term are equal: Second and penultimate term are equal and so on. (7) If n is even then the number of terms in (x + a)n is (n + 1) which is odd. So there will be only one
middle term and it is TF n
I . If n is odd then the number of terms in (x + a) H2 K
+1
n
is (n + 1) which is
even. So there will be two middle terms. They are TF n +1 I and TF n +1
20. Using Binomial theorem prove that 6n − 5n always leaves the remainder 1 when divided by 25. We know, 6n = (1 + 5)n We have from Binomial theorem.
Binomial Theorem
81
a x + af a1 + 5f
n
= x n + nx n −1 ⋅ a +
n n −1 ⋅ x n − 2 ⋅ a 2 + ... + a n . 2!
a f
n
=1+ n⋅5 +
n n −1 2 n n −1 n − 2 3 ⋅5 + 5 + ... + 5 n . 2 3!
a f
a fa f
f
6 n = 1 + 5n +
6 n − 5n = 1 +
n n −1 2 ⋅ 5 + ... + 5n . 2
a f
n n −1 2 n n −1 n − 2 ⋅ 5 3 + ... + 5 n . 5 + 2 3!
a f
a fa
Now right hand side contains all terms containing 5n except 1st term 1. ∴ when RHS is divided by 25, it leaves the remainder 1. 21. Find the greatest term in the expansion of (x − y)20 when x = 12 and y = 4
− • The general term or (r + 1)th term is given by Tr +1 = n cr x n r ⋅ a r .
•Binomial Theorem
85 inEXERCISES
by theorem: I. Expand by using Binomial theorem: 1.
3.
5.
F x + 1I H xK FG 2 p − q IJ H q 2pK F y+ 1 I GH yJ K
4
2. 1 + xy
6
a
f
7
4.
5
F 2a − b I H 3K
6
II. Find the indicated term in the expansions: 1. 4th term in (2 + a)7
4
IV. Find the term independent of x in the following expansions: IV. term independent follo ollowing expansions: 1.
3.
5.
F 4a − 3 I GH 3 2a JK FG x − 2 IJ H2 xK F 2x + 1 I H xK
2
9
2.
Fx − 3 I H xK
4
10
10
2
4.
FG 2 x − 1 IJ H xK
15
10
2
6.
Fx − 1 I H xK
2
21
coeff V. Find the coefficient of: 1. x 23 in x 2 − x
d
i
20
4 4 2. x in x +
F H
1 x3
I K
b 3
15
1 1 4 3. 17 in x − 3 x x
5 5. x 2
I K 1 in F 2 x + I H xK
3 5
F H
15
6 3 4. a b in 2 a −
F H
I K
9
5
6.
1 1 in 2 − 4 x x
F H
I K
8
value VI. Find the value of: 1. 3. VII.
d2 + 3 i + d2 − 3 i d1 + 5 i + d1 − 5 i
5
5
2.
d
2 +1 −
i d
6
2 −1
i
6
5
F 1. Prove that in the expansion of G ay H
2
b + y
IJ K
25
. There is no term independent of y.
2. The second, 3rd and 4th term of expansion of (x + y)n are 108, 54 and 12 respectively. Find x, y and n. 3. Find the value of (1.01)5 correct to four decimal places. 4. Prove that the sum of odd binomial co-efficients of order n = 2n − 1. 5. Prove that the sum of binomial co-efficient of order n = 2n.
a rational fraction. If the degree of f (x) is less than degree of g (x) then it is called proper fraction. Otherwise it is called an improper fraction. By division, an improper fraction can always be reduced to the sum of a polynomial and a proper fraction. Examples: 1.
4x −1 is a proper fraction since degree of numerator (=1) is less than degree of denomi2 x 2 + 8x − 1 nator (=2).
Here we have expressed the sum of 2 proper fractions as a single proper fraction. The reverse process of expressing a single proper fraction as the sum of the two or more proper fractions is called as 'Resolving into partial fractions'. The following rules are used to resolve a proper fraction into partial fractions: 1. To each linear factor (ax + b) which occurs only once as a factor of the denominator, there corresponds a partial fraction of the form
A where A is constant. ax + b
Example:
a fa
a
2
x2 − 1 A B . = + 2x + 3 x + 1 2x + 3 x +1
f a f a
aax + bf
Ar
r
f
2. To each linear factor (ax + b) which occurs r times as a factor of the denominator, there corresponds r partial fractions of the form,
A1 A2 + ax + b ax + b
f aax + bf
+
A3
3
+ ... +
where A1 , A2 , Ar are constants.
Example:
x2 x 2x + 3
a
f
3
=
A B C + + 2x + 3 x 2x + 3
a
f a
f a2 x + 3f
2
+
D
4
.
Note, here corresponding to linear factor x which occur only once in denominator, we have taken only one constant A and corresponding to linear factor 2x + 3 which occur 3 times, we have taken 3 constants B, C and D. 3. To each non factorisable quadratic factor ax2 + bx + c which occur only once as a factor of denominator, there corresponds a partial fraction of the form constants.
Ax + B where A and B are ax + bx + c
2
Example:
d x + 1ia x − 1f
2
3x − 7
=
Ax + B C + 2 x +1 x −1
90
Basic Mathematics
Observe here, corresponding to x2 + 1, which is quadratic and non-factorisable we have taken Ax + B and corresponding to x − 1 which is linear we have taken one constant C. 4. To each repeated non-factorisable quadratic factor (ax2 + bx + c) which occurs r times as a factor of the denominator, there corresponds r partial fractions of the form.
If an improper fraction is given to resolve, first by dividing the numerator by denominator write the given fraction as the sum of the polynomial and the proper fraction. Then the proper fraction is resolved into partial fractions.
EXERCISE
Resolve into partial fractions: 1.
a x − 2fa x − 3f a fa a fa
1 x +1 x + 2 x + 3
8x − 1
2.
4x + 6 x2 − 1
3.
2x + 3 x − 3x + 2
2
4.
fa f
2
5.
a x − 1fd x dx
2
x 2 − 10 x + 13
2
− 5x + 6
i
6.
a
3x + 5 x+2 2 x−3 x2 + 1
2
fa f
−4
7.
9 x +1 x + 2 1 x −1 1+ x
f
4 x 2 + 3x − 1
8.
+ 2x + 1 x − 1
ia f
9.
a x − 2f d x
5x 2 + 1 x3 − 1
i
10.
a fa f
3x − 1
2
11.
3x + 2 x3 + x
12.
13.
a x + 2fd1 − x + x i
2
14.
a x − 2f d2 x + 3i
2
x2 + 2x + 4
15.
a x + 1fd x + 1i
2
x −1
16.
a
x2 − 2 x −4 x +3
fa f
2 x 2 + 3x + 2 17. x2 − x − 2
20.
x 3 + 7 x 2 + 17 x + 11 18. x 2 + 5x + 6
19.
2 x 2 − 3x − 4 x2 − x − 6
x 4 − 3 x 3 − 3x 2 + 10 x +1 2 x − 3
a fa f
ANSWERS
1.
3 5 + x −2 x +3
2.
5 1 − x −1 x +1 2 3 4 + − x −1 x − 2 x − 3
3.
7 5 − x − 2 x −1
1 1 1 4. 2 x + 1 − x + 2 + 2 x + 3
a f a
a f
2
5.
1 −14 6. 25 x + 2 + 5 x + 2
f a
f
+
14 25 x − 3
a f
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Basic Mathematics
9 9 9 7. x + 1 − x + 2 − x + 2
a
f
2
5 0 3 8. − 2 x + 1 + x + 1 2 + 2 x − 1
a f a f
a f
5 11 5 9. 16 x + 2 + 16 x − 2 + 4 x − 2
a
f
a
f a
f
2
1 1 1 10. 4 x − 1 + 4 1 + x + 2 1 + x
11.
a f a f a f
a f d i
2
2 −2 x + 3 + 2 x x +1
12.
2 3x + 1 + x −1 x2 + x + 1 x −1 + x + 1 x2 + 1
13.
x −1 + x + 2 1 − x + x2 1 2 + x+3 x−4 2 1 + x+2 x−3
12 13x + 4 − 14. 11 x − 2 11 2 x 2 + 3
17. 2 +
15.
16. 1 − 19. 2 −
20.
a f a f a x − 2f + 16 a17+ 1f − 4 a x111f x +
16 1 − 3 x − 2 3 x −1
18.
a x + 2f + x 1 2 + x 2 3 + +
17 . 16 x − 3
2
−
a f
6
Matrices & Determinants
6.1 INTRODUCTION:
The theory of matrices was developed in 1857 by the French mathematician Cayley. It was not well advanced till 20th century. But now a days matrices are powerful tool in modern mathematics having wide applications.
6.2 MATRIX:
A matrix is an arrangement of numbers in rows (horizontal lines) and columns (vertical lines). The arrangement is usually enclosed between square brackets [ ] or curved brackets ( ) or pairs of vertical lines || ||. The matrix is usually denoted by a capital letter. Order of a matrix = Number of rows × Number of columns. If a matrix has m rows and n columns, then Order = m × n (read as m by n)
Example: Matrix A =
LM1 N4
2 5
3 has the order 2 × 3. 6
OP Q
6.3 TYPES OF MATRICES:
matr trix: 1. Rectangular matrix: If the number of rows is not equal to number of columns in a matrix, then that matrix is called rectangular matrix.
Example:
LMa A= b MNc
1
1 1
a2 b2 c2
a3 b3 c3
a4 b4 c4
a5 b5 c5
OP PQ
3× 5
(a) Row matrix or Row vector: If a matrix has only one row, then it is called row matrix. Row matr trix Row vector:
Example:
X= 1
2
3 1×3
(b) Column matrix or column vector: If a matrix has only one column then it is called column matr trix vector: matrix.
106
Basic Mathematics
Example:
LM1OP Y= 2 MN3PQ
a1 A = b1 c1
3 ×1
Square matr trix: 2. Square matrix: If the number of rows is equal to number of columns in a matrix, then it is called square matrix.
Example:
LM MN
a2 b2 c2
a3 b3 c3
OP PQ
3× 3
In a square matrix, the entries from the left top corner to the right bottom corner are called principal diagonal elements. So in the above example a1 b2 c3 are Principal diagonal elements. (a) Diagonal matrix: If in a square matrix all the non-diagonal elements are zero, then it is called diagonal matrix.
Example:
A=
LM1 0OP , B = LM2 N0 2Q MN0 0
6 LM5 0OP , B = LM0 N0 5Q MN0
0 −1 0
0 0 3
OP PQ
(b) Scalar matrix: If in a diagonal matrix, all the diagonal elements are equal, then it is called scalar matrix.
Example:
A=
0 6 0
0 0 6
OP PQ OP PQ
(c) Identity matrix or unit matrix: If in a scalar matrix all the diagonal elements are equal to 1, then it is called unit matrix or identity matrix.
Example:
I2 =
LM1 0OP , I N0 1 Q
LMa MN 0 0 LMa MN b c
1
3
1 = 0 0
LM MN
0 1 0
0 0 1
(d) Upper triangular matrix: A square matrix is called upper triangular if all the non-diagonal triangular matr trix: elements below the principal diagonal are zeroes.
Null matrix: If each element of a matrix is zero then it is called null matrix or zero matrix.
Example:
A=
LM0 N0
0 0
0 0 , B= 0 0
OP Q
LM N
0 0
OP Q
6.4 ALGEBRA OF MATRICES:
1. Equality of matrices: Two matrices of same order are said to be equal iff the corresponding matr trices: elements are equal,
a1 i.e., b 1
LM N
a2 3 = b2 −1
OP LM Q N
4 iff a1 = 3, b1 = −1, a2 = 4, b2 = 0. 0
OP Q
2. Addition and subtraction of matrices: Two or more matrices of the same order can be added or Addition subtraction matr trices: subtracted. It is the matrix obtained by adding or subtracting the corresponding elements i.e., If
Matr multiplica trix ultiplication: Matrix multiplication: The product of 2 matrices exists only when the number of columns in 1st matrix is equal to the number of rows in the 2nd matrix.
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Basic Mathematics
If A is a matrix of order m × n and B is a matrix of order n × p. Then AB exists and is of order m × p. To get the elements of AB, the elements of 1st row of A multiplied by the corresponding elements of the first column of B and the products are added. The sum is the element in the first row, first columns of AB. Similarly other elements are obtained.
8. In a determinant if all the elements on one side of the principal diagonal are zeros, then prove that the value of the determinant is equal to the product of the elements in the principal diagonal.
a1 A= 0 Proof: Proof: Let 0
a2 b1 0
a3 b2 be the determinant in which all the elements to the left of princ1
A = a1b1c1
cipal diagonal are zeros. To prove:
Consider
a1 0 0
a2 b1 0
a3 b1 b2 = a1 0 c1
b2 c1
− a2
0 0
b2 c1
+ a3
0 0
b1 0
= a1 b1c1 − 0 − a2 0 + a3 0 = a1b1c1 . Hence proved.
b
g
af af
Note: The above properties which are proved for the rows also holds good for columns because |A| = |A′| from property 1. So • The determinant changes its sign when 2 of its columns are interchanged i.e.,
a1 b1 c1
a2 b2 c2
a3 a2 b3 = − b2 c3 c2
a1 b1 c1
a3 b3 c3 a1 b1 c1 a3 b3 = 0 . c3
a1 • The value of the determinant is zero when 2 of its columns are identical i.e., b1 c1
• If every element of any column of a determinant is multiplied by constant k. Then the whole
ka1 determinant is multiplied by k i.e., kb1 kc1
a2 b2 c2
a3 a1 b3 = k b1 c3 c1
a2 b2 c2
a3 b3 c3
• If each element of any column of a determinant is sum of two terms then the determinant can be expressed as the sum of 2 determinants.
i.e.,
a1 + x b1 + y c1 + z
a2 b2 c2
a3 a1 b3 = b1 c3 c1
a2 b2 c2
a3 x b3 + y c3 z
a2 b2 c2
a3 b3 c3
• The value of a determinant is not altered if to the elements of any column the same multiples of the corresponding element of any other column are added.
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Basic Mathematics
i.e.,
a1 b1 c1
a2 b2 c2
a3 a1 b3 = b1 c3 c1
a2 + ka3 b2 + kb3 c2 + kc3
a3 b3 c3
′ c2 = c2 + kc3
• If all the elements in a column of a determinant are zero then the determinant is zero.
6.8 MINOR, CO-FACTOR, ADJOINT AND INVERSE OF A SQUARE MATRIX:
Let [aij] be a square matrix of order n × n. Then minor of an element aij is the determinant obtained by deleting the row and the column containing it (i.e. ith row & jth column). If minors are multiplied by (−1)i + j i.e., with proper signs + or − we get co-factors. Adjoint of a matrix is the transpose of co-factor matrix. Illustration: 1. If A = minor minor minor minor
INVERSE OF A SQUARE MATRIX:
If A is a non-singular square matrix of order n × n then there exists a square matrix B of order n × n such that AB = BA = I where I is the identity matrix of order n × n. Here B is called inverse of A. It is denoted by A−1. ∴
A ⋅ A −1 = A − 1 ⋅ A = I Inverse of A can be found by using the formula A −1 = A −1 = adjoint of A . A adj A . A
6.9 CHARACTERISTIC EQUATION OF A SQUARE MATRIX:
If A is any square matrix of order n × n and I is identity matrix of the same order then |A − λI| = 0 where λ is a constant is called characteristic equation of a square matrix A. The roots of the equation |A − λI| = 0 (i.e., value of λ which satisfies this equation) are called characteristic roots or eigen values. Examples: 1. If A =
This is characteristic equation.
2 By Cayley Hamilton theorem 'A' satisfies this equation. So A − 4 A − I = 0 where I is identity matrix of same order as that of A. Now to verify Cayley Hamilton theorem, We have to verify A2 − 4A − I = 0.
Hence total salary = Rs. 8,24,500. 6. A salesman has the following record of sales during 3 months for 3 items A, B, C which have different rates of commission.
Month A
Jan Feb Mar 100 300 100
Sales of units B
100 200 200
Total Commission C
200 100 300 900 1000 1400
Find out the rates of commissions on items A, B and C. Solution: Let x, y and z denote the rates of commission in Rupees per unit for A, B and C items respectively. Then the data given can be expressed as a system of linear equation.
Hence price per unit of x = Rs. 1000, that for y = Rs. 3000 and for z = Rs. 5000. 8. A company is considering which of the 3 methods of production it should use in producing 3 products X, Y and Z. The amount of each product and produced by each method is as shown below.
Method
I II III
REMEMBER:
• Matrix is an arrangement of numbers in horizontal rows and vertical columns. • Two matrices of the same order are said to be equal if and only if the corresponding elements are equal. • It
Matrices & Determinants
173
• rows (or columns) of the determinant are added the value of the determinant remains • For a non-singular matrix A.
A −1 =
adj A . A
• If A is a square matrix and I is the identity matrix of the same order. • The characteristic equation: |A − λI| = 0. The values of λ obtained is called eigen values or characteristic roots. • Every square matrix satisfies its characteristic equation, |A − λI| = 0. This is Cayley Hamilton theorem. • The solution of system of equations
45. A man buys 8 dozens of mangoes; 10 dozens of apples and 4 dozen of bananas. Mangoes cost Rs. 18 per dozen, apple Rs. 9 per dozen and bananas Rs. 6 per dozen. Represent the quantities bought by a row matrix and prices by column matrix and hence find the total cost. 46. A company is considering which of the 3 methods of production it should use in producing 3 goods X, Y and Z. The amount of each good produced by each method is shown in the matrix.
Matrices & Determinants
179
I 4 II 5 III 5
LM MM N
X
Y 8 7 3
Z 2 1 9
OP PP Q
The vector [10 4 6] represents the profit per unit for the goods X, Y and Z in that order. Find which method maximises profit. 47. Matrix A and B give the daily sales and sale price of chocolates for a shopkeeper.
Bar one Dairy milk Five star
Mon Tue A= Wed Thu
LM6 MM2 1 MN1
3 3 2 4
1 0 4 3
OP PP PQ
Bar one 5 B = Dairy milk 6 Five star 7
LM OP MM PP NQ
Find the total revenue for four days. 48. A salesman has the following record of sales during 3 months for 3 items, A, B and C which have different rates of commission.
Months A
January February March 90 130 60
Sales of units B
100 50 100
Total commission in Rs. C
20 40 30 800 900 850
Find out the rates of commission on A, B and C. 49. The prices of the 3 commodities X, Y and Z are x, y and z per unit respectively. A purchases 4 units of z and 3 units of x and 5 units of y. B purchases 3 units of y and sells 2 units of x and 1 unit of z. C purchases 1 unit of x and sell 4 units of y and 6 units of z. In the process A, B and C earn Rs. 6000, 5000 and 13,000 respectively. Using matrices, find the prices per unit of the 3 commodities. (Note that selling the unit is positive earnings and buying the unit is negative earning). 50. Suppose the matrices X and Y represent the number of items of different kinds produced by 2 manufacturing units in one day.
4 3 = 5 and Y = 4 . X 6 5
LM17 OP MM22PP It represent the number of items produced by one unit in 2 days and another in 3 days N27Q
together.
7
Ratio and Proportions, Variations
7.1 INTRODUCTION:
Two quantities of the same kind can be compared either by subtraction method or by division method. In subtraction method we find how much more (or less) is one quantity than the other, and in division method, we find how many times (or what fractional part) is one quantity of the other. The quotient here is nothing but the ratio of the two quantities. For example, if I have Rs. 100 and you have Rs. 600 then we can compare the money by subtraction method and say 'You have Rs. 500 more than what I have'. Or we can compare by division method and say you have 6 times the money what I have. If the ratio of 2 mutual quantities are equal then they are said to be proportional.
7.2
RATIO:
A ratio is a relation or comparison between two quantities of the same kind. The comparison is made by considering what multiple, part or parts the first quantity is of the second. The ratio of 2 quantities x and y is denoted by x : y or the second term y is called consequent. Note: 1. A ratio is a pure number. Hence it has no units. 2. When the terms of the ratio are multiplied or divided by the same quantity the ratio is not altered. For instance 2 : 3 = 4 : 6 = 40 : 60 = 80 : 120... 3. If a : b and c : d are two ratios, then the ratio ac : bd is called their compound ratio. Example : The compound ratio of 5 : 2 and 3 : 7 is 5 × 3 : 2 × 7 i.e., 15 : 14. 4. If a : b is the given ratio then the ratio
N . The first term x is called antecedent and O
Ratio and Proportions, Variations
183
(i) a2 : b2 is called its duplicate ratio. (ii)
= : > is called its subduplicate ratio.
Number of 1 Re. coins = 420 Number of 50 ps. coins = 252 × 2 = 504 Number of 25 ps. coins = 168 × 4 = 672. 9. In a mixture of 35 litres, the ratio of milk and water is 4 : 1. If 7 litres of water is added to the mixture, then find the ratio of milk and water in the new mixture. Solution: Given milk : water = 4 : 1 Sum of terms = 4 + 1 = 5 ∴ Milk in 35 litres mixture =
14. Three utensils contains equal mixture of milk and water in the ratio 6 : 1, 5 : 2, and 3 : 1 respectively. If all the solutions are mixed together, find the ratio of milk and water in the final mixture. Solution: In 1st utensil milk : water = 6 : 1 ∴ and Quantity of milk in 1st utensil = Quantity of water in 1st utensil =
6 7 1 7
Similarly 2nd utensil contains
5 2 milk and water. 7 7 3 4 1 . 4
Similarly quantity of milk in 3rd utensil = and water ∴ If all solutions are mixed, =
If 2 ratios are equal then the 4 quantities comprising them form a proportion i.e. if the ratio a : b is equal to c : d, then 4 quantities a, b, c, d are in proportion. Example: Example 1, 2, 4, 8 are in proportion since 1 : 2 = 4 : 8. Extremes Note : 1. a : b = c : d is also denoted by a : b : : c : d. 2. In a proportion a : b = c : d, the first and the last terms i.e., a and d are called exa : b = c : d tremes and the second and 3rd terms i.e. b and c are called means. means 3. In every proportion, the product of the means is equal to the product of the exFig. 7.1 tremes. i.e., ad = bc Conversely, if 4 quantities a, b, c, d are such that ad = bc, then they are said to be in proportion. a : b = c : d ⇔ ad = bc is called rule of 3. This rule is used to solve a proportion when one of the terms is unknown. For example: 4 : 5 = x : 15 ⇒
5 x = 4 × 15
Quantities are said to be in direct proportion when an increase (or decrease) in one kind is accompanied by an increase (or decrease) in the other. 2 3 4 ... Example: Number of chocolates 1 Cost of chocolates 3 6 9 12 ... Here we notice as the number of chocolates increases cost of chocolates also increases. So number of chocolates is directly proportional to cost of chocolates.
INVERSE PROPORTION OR INVERSE VARIATION:
Quantities are said to be in inverse proportion when an increase (or decrease) in one kind is accompanied by decrease (or increase) in the other. 1 2 3 4... Example: Number of workers Number of days 12 6 4 3... Here we notice as the number of workers increase, number of days required to finish the work decreases and vice versa. So it is inverse proportion. Note: If a : b = c : d represent a direct proportion, then a : b = d : c or (b : a = c : d) represent an inverse proportion.
4 x − 2b : a + 3b = 9 : 11.
8. What number must be subtracted from each of 9, 11, 15 and 19, so that the difference will be proportional. Solution: Let the number subtracted be x, So that 9 − x, 11 − x, 15 − x and 19 − x will be in proportion. ∴
As the length of the cloth increases. Cost also increases. ∴ Length and Cost are directly proportional. To denote direct proportion, we use 2 arrows, with same direction. ∴
10 : 22 = 225 : x
10 x = 22 × 225
x=
22 × 225 10
x = 495. ∴ 22 mts of cloth costs Rs. 495. 15. If 60 men can complete a job in 12 days, how many days will 36 men take to complete the same job?
Solution:
Men 60 36
days 12 x
As the number of men increases, the days required to complete the job decreases ∴ Men and days are inversely proportional. To denote this we use 2 arrows with opposite direction.
60 : 36 = x : 12 36 x = 60 × 12
x=
60 × 12 = 20. 36
∴ 36 men can complete a job in 20 days. 16. If 10 men can earn Rs. 105 in 7 days, in how many days will 15 men earn Rs. 225? Solution:
Men 10 15
Money 105 225
days 7 x
Here number of days is unknown. Leaving money, or keeping money constant, let us first consider men and days. As the number of men increases, days required to complete the job decreases. ∴ Men and days are inversely proportional. Now, leaving men, let us consider money and days. As the days increase, money earned also increases. ∴ Money and days are directly proportional.
Here money is unknown. Taking carpenter and money alone, as the carpenters increase, the money earned by them also increases. So it is direct proportion. Taking days and money, as the day increases money also increases. So it is direct proportion. Now taking hours and money, as the number of hours increases, money also increases. ∴ It is direct proportion. So corresponding compound proportion is
5:8 6 : 12 = 3600 : x 9:6
5 × 6 × 9 × x = 8 × 12 × 6 × 3600
x=
8 × 12 × 6 × 3600 5×6×9
x = 7680. ∴ 8 carpenters earn Rs. 7680 in 12 days working 6 hours a day. 18. A contractor undertook to make 15 kms of roadways in 40 weeks. In 10 weeks 3 kms were completed by 180 men working 8 hours a day. Then the men agreed to work 1 hour a day overtime and some boys were engaged to assist them. The work was finished in the stipulated time (that is 40 weeks). How many boys were employed if the work of 3 boys is equal to that of 2 men. Solution:
Men 180 x weeks hrs. kms. 10 8 3
a40 − 10f = 30
a8 + 1f = 9
a15 − 3f = 12
200
Basic Mathematics
men and Kms are directly proportional. men and hrs. are inversely proportional. men and week are inversely proportional. ∴ Required compound proportion is
19. If 15 men build a wall 40 ft long, 2 and 1/2 ft. thick and 21 ft. height in 18 days working 10 and 1/2 hrs. each day. In how many days working 15 hrs. a day will 45 men build a wall 200 ft. long, 5 ft thick and 20 ft. height. Solution:
Men 15 45
Length 40 200
Breadth (thickness) 1 2 2 5
Height 21 20
hours 10 1 2 15
days 18 x
As the men increase, days decrease, so inverse proportion. As length increase, days also increase, as breadth increase, days also increase. As height increase, days also increase. So, It is direct proportion. As the number of hrs. increase, days required to construct decrease. So it is inverse proportion. ∴ Corresponding Compound proportion is:
x = 40
∴ 40 days are required. 20. If 12 pumps working 6 hours a day can draw 2000 gallons of water in 20 days, find in how many days will 20 pumps working 9 hours a day draw 3000 gallons of water? Solution:
Pumps 12 20
Hours 6 9
Quantity 2000 3000
Days 20 x
As the number of pumps increase, the days required decreases. So it is inverse proportion. As the number of hours increases the days required to draw water decreases. So it is inverse proportion. As the number of gallons increase, number of days required also increases, so it is direct proportion. ∴ Corresponding compound proportion is
12 : 20 6: 9 3000 : 2000
U | V | W
= x : 20
x × 20 × 9 × 2000 = 20 × 12 × 6 × 3000
x=
20 × 12 × 6 × 3000 20 × 9 × 2000
x = 12
∴ 12 days are required.
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Basic Mathematics
7.5
PROBLEMS ON TIME AND WORK:
1 . n
Note: 1. If A can do a piece of work in n days. Then work done by A in 1 day = 2. If B's 1 day's work =
1 . Then B can finish the work in x days. x
3. If A is twice as good a workman as B, then Ratio of work done by A and B = 2 : 1 Ratio of time taken by A and B = 1 : 2 4. If A can do a piece of work in x days, and B can do it in y days, then A and B working together
xy will do the same work in x + y days.
5. If A and B together can do a piece of work in z days and A alone can do it in x days then B alone can do it in
zx . x−z
SOLVED EXAMPLES:
1. Ram can reap a field in 6 days which Raju alone can reap in 8 days. In how many days both together can reap this field? Solution: Ram's 1 day's work = Raju's 1 day's work =
1 6 1 8 1 1 + 6 8
Ram and Raju's 1 day's work =
=
∴
4+3 7 = . 24 24 24 3 = 3 days. 7 7
Both together can reap the field in
OR
We know, if A can do a piece of work in x days and B can do it in y days then A and B together can do it in
xy days. x+y
Ratio and Proportions, Variations
203
∴
Ram and Raju together can reap the field in
6×8 days. 6+8
=
48 24 3 = = 3 days. 14 7 7
2. X and Y together can dig a trench in 10 days which X alone can dig in 30 days. In how many days Y alone can dig it? Solution: X and Y's 1 day's work =
1 10 1 30
X's 1 day's work = ∴ Y's 1 day's work =
1 1 − 10 30
3 −1 2 = 30 30
∴ Y alone can dig a trench in
30 = 15 days. 2
OR
We know, if A and B together can do a piece of work in z days and A alone can do it in x days then B alone can do it in ∴
zx days. x−z
Y alone can dig a trench in
30 × 10 30 − 10
=
300 = 15 days. 20
3. A and B can do a piece of work in 12 days; B and C in 15 days; C and A in 20 days. In how many days will they finish it together and separately? (A + B)'s 1 day's work = (B + C)'s 1 day's work = (C + A)'s 1 day's work =
=
∴A alone can finish the work in 30 days. Similarly, B's 1 day's work =
1 1 1 − = 10 15 30
1 1 1 − = 10 20 20
∴ B alone can finish the work in 20 days. Similarly, C's one day's work = ∴
1 1 1 − = 10 12 60
C alone can finish the work in 60 days.
OR
A, B and C can do together in
2xyz days xy + yz + zx
= 2 12 × 15 × 20 12 × 15 + 15 × 20 + 20 × 12
a
f
=
2 3600 2 3600 = = 10 days. 180 + 300 + 240 720
a
f
a f a f
A, B and C can together finish the work in 10 days. 4. A can do a piece of work in 25 days which B alone can finish in 20 days. Both work for 5 days and then A leaves off. How many days will B take to finish the remaining work.
Ratio and Proportions, Variations
205
Solution: A's work in 1 day =
1 25 1 20
B's work in 1 day =
∴
(A + B)'s 1 day's work =
1 1 + 25 20
(A + B)'s 5 day' s work = 5
LM 1 + 1 OP N 25 20 Q f
=
9 20 9 Note this step 20
Remaining work = 1 −
a
=
11 20
Now
1 work is done by B in 1 day. 20
∴ ∴
11 ×1 11 20 work will be done by B in = 11 days. 1 20 20
B takes 11 days to finish the remaining work. 5. X is thrice as good a work man as Y and is therefore able to finish the piece of work in 60 days less than Y. Find the time in which they can do it, working together. Solution: Ratio of work done by x and y in same time = 3 : 1 ∴ Ratio of time taken = 1 : 3 If Y takes y days to finish a work. Then X takes y − 60 days to finish. Now
y − 60 : y = 1 : 3 3 y − 60 = y
3 y − y = 180
a
f
2 y = 180 ⇒ y = 90
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Basic Mathematics
∴ Time taken by Y to finish the work = 90 days and time taken by X to finish the work = 90 − 60 = 30 days. ∴ X's 1 day's work = Y's 1 day's work = (X + Y)'s 1 day's work = ∴ Both X and Y can finish the work in 6. A can build a wall in 30 days which B alone can build in 40 days. If they build it together and get a payment of Rs. 1400 what is A's share and B's share? Solution: A's 1 day's work = B's 1 day's work = ∴ Ratio of their work =
1 30 1 90
1 1 2 + = 90 30 45
45 1 = 22 days. 2 2
1 30 1 40 1 1 : 30 40
=4:3 Total = 4 + 3 =7 A's share = B's share =
4 × 1400 = Rs. 800 7 3 × 1400 = Rs. 600 7
7. A can do a piece of work in 10 days, while B alone can do it in 15 days. They work together for 5 days and the rest of the work is done by C in 2 days. If they get Rs. 1200 for the whole work how should they divide the money? Solution: A's 1 day's work = B's 1 day's work = (A + B)'s 1 day's work =
1 10 1 15
1 1 1 + = 10 15 6
Ratio and Proportions, Variations
207
(A + B)'s 5 day's work = 5 Remaining work = 1 − Given: ∴ Now Consider
F 1I = 5 H 6K 6
(Note this step)
5 1 = 6 6
C completes rest of the work in 2 days. C's 2 day's work =
1 6
A's 5 day's work : B's 5 day's work : C's 2 day's work.
5
F 1 I : 5F 1 I : 1 H 10 K H 15K 6
1 1 1 : : 2 3 6
=3:2:1 Total = 3 + 2 + 1 = 6. A's share = B's share = C's share =
3 × 1200 = Rs. 600 6 2 × 1200 = Rs. 400 6 1 × 1200 = Rs. 200. 6
8. A certain number of men complete a piece of work in 60 days. If there were 8 men more the work could be finished in 10 days less. How many men were originally there? Solution: Let number of men = x 8 men more means x + 8 Given:
∴ 40 men were originally there. 9. 16 men or 28 boys can fence a farm in 40 days. In how many days will 24 men and 14 boys complete the same work? Solution: Given, 16 men's work ≡ 28 boy's work. i.e., 16 : 28 i.e., 8 : 14 ∴ 8 men ≡ 14 boys. Now 24 men and 14 boys ≡ 24 men + 8 men ≡ 32 men. Given:
Men 16 32
days 40 x
As the number of men increases, the days required to fence a farm decreases. ∴ It is inverse proportion. ∴
16 : 32 = x : 40
32 × x = 16 × 40
x=
16 × 40 = 20. 32
∴ 20 days are required to complete the work. 10. 2 men and 4 boys can do a work in 33 days. 3 men and 5 boys can do the same work in 24 days. How long shall 5 men and 2 boys take to finish it? Given: 2 men and 4 boys can do the work in 33 days. ⇒ 2 × 33 men and 4 × 33 boys can do it in 1 day. ∴ 66 men and 132 boys can do it in 1 day ...(1) Similarly 3 men and 5 boys can do the work in 24 days. ⇒ i.e., From (1) and (2)
66M + 132B ≡ 72M + 120B 132B − 120B ≡ 72M − 66M
3 × 24 men and 5 × 24 boys can do it in 1 day 72 men and 120 boys can do it in 1 day ...(2)
Ratio and Proportions, Variations
209
12B ≡ 6M M ≡ 2B.
∴ Now given: i.e., Now,
One man's work is equivalent to 2 boy's work. 2 men and 4 boys can finish the work in 33 days. 2M + 2M can finish the work in 33 days. 4 men can finish the work in 33 days. 5 men and 2 boys ≡ 5M + 1M = 6 Men.
Men 4 6
days 33 x
As men increases, days required to finish the work decreases so it is inverse proportion. ∴
4 : 6 = x : 33 6 x = 33 × 4 x= 33 × 4 == 22. 6
∴ 5 men and 2 boys can finish the work in 22 days.
7.6
Note :
PROBLEM ON TIME AND DISTANCE:
Distance Time
(1) Speed =
(2) x km hr = x ×
F H
5 mts sec 18 5 km hr. 18
(3) x mts sec = x ×
F H
I K
I K
WORKED EXAMPLES:
1. The distance between 2 stations A and B is 450 kms. A train starts at 4 p.m. from A and moves towards B at an average speed of 60 km/hr. Another train starts from B at 3 : 20 p.m. and moves towards A at an average speed of 80 km/hr. How far from A will the two trains meet and at what time?
210
Basic Mathematics
Solution: Let the trains meet at a distance x kms. from A. Let trains from A to B and B to A be X and Y respectively. Given: Speed of X = 60 km/hr Speed of Y = 80 km/hr. Time required to cover x kms. by X =
x 60 450 − x 80
Time required to cover (450 − x) kms. by Y = Difference between time =
3. A bullock cart has to cover a distance of 80 km in 10 hrs. If it covers half of the journey in time. What should be its speed to cover the remaining distance in the time left? Total distance = 80 kms. Solution: Total time = 10 hrs. Distance left =
Speed = 10 km/hr. 4. A man travels 360 km in 4 hrs. partly by air and partly by train. If he had travelled all the way by
4 of the time he was in train and would have arrived at his destination 5 2 hrs. early. Find the distance he travelled by air and train. Solution: Total time = 4 hrs.
air, he would have saved Given:
4 of total time in train = 2 hrs. 5 2×5 5 = hrs. 4 2
Total time in train =
Given: If 360 km is covered by air then time taken is 4 − 2 = 2 hrs. ∴ When
3 is spent in air, 2
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Basic Mathematics
Distance covered =
360 3 × = 270 kms. 2 2
Distance covered in train = 360 − 270 = 90 km. 5. An aeroplane started 30 minutes later than the scheduled time from a place 1500 km away from its destination. To reach the destination at the scheduled time, the pilot had to increase the speed by 250 km/hr. What was the speed of the aeroplane per hour during the journey? Solution: Let the time taken by aeroplane in later case = x hrs. We know
Note: 1. The word Alligation literally means linking. The rule takes its name from the lines or links used in working out questions on mixture. 2. Alligation method is applied for percentage value, ratio, rate, prices, speed etc. and not for absolute values. 3. Alligation is the rule that enables us to find the proportion in which two or more ingredients at the given price must be mixed to produce a mixture at a given price. Cost price of unit quantity of the mixture is called the mean price. 4. Rule of alligation: If 2 quantities are mixed in a ratio, then
WORKED EXAMPLES:
1. The price of first quality of rice is Rs. 16 per kg and that of second quantity rice is Rs. 10. In what ratio these two should be mixed so that the mixture can be sold for Rs. 12 per kg. Solution: Cost price of 1 kg cheaper Rice 10 (c) Mean Price 12 (m) (d − m) = 16 − 12 =4 m−c = 12 − 10 =2 C.P. of 1 kg dearer rice 16 (d)
10. Two numbers are in the ratio 5 : 8. If 9 is added to each then they are in the ratio 8 : 11. Find the numbers. 11. The ratio between the ages of Khan and Ranjith is 6 : 5 and the sum of their ages is 44 years. Find the ratio of their ages after 8 years. 12. One year ago the ratio of between Sarala and Saraswathi's salary was 3 : 4. The ratio of their individual salaries between last year's and this year's salaries are 4 : 5 and 2 : 3 respectively. At present the total of their salary is Rs. 4160. Find the salary of Sarala now. 13. The ratio between Sumit's and Prakash's age at present is 2 : 3. Sumit is 6 years younger than Prakash. Find the ratio of Sumit's age to Prakash's age after 6 years. 14. 8 labourers can build a wall in 6 days. In how many days, 12 labourers can do the same work? 15. 6 carpenters working 7 hrs. a day can complete 24 tables in 20 days. How many days will 12 carpenters working 6 hrs. a day take to complete 36 tables. 16. A, B, C start a business with investments of Rs. 25,000, Rs. 16,000 and Rs. 12,000 respectively. If the profit for the year amounts to Rs. 6850. Find the share of each partner. 17. Ram can reap a field in 9 days which Deepak alone can reap in 12 days. In how many days both together can reap this field.
1 18. A can do F I H 3K
rd
F 2I of the work in 5 days and B can do H K 5
th
of the work in 10 days. In how many
19.
20. 21. 22. 23. 24. 25. 26. 27. 28.
days both A and B together can do the work? Sunil completes a work in 4 days, whereas Dinesh completes the work in 6 days. Ramesh works 1 and 1/2 times as fast as Sunil. How many days it will take for the 3 together to complete the work. A can complete a job in 9 days, B in 10 days and C in 15 days. B and C start the work and are forced to leave after 2 days. Find the time taken to complete the remaining work. If 3 men or 4 women can construct a wall in 43 days. Then find the number of days that 7 men and 5 women take to construct it. 8 men can dig a pit in 20 days. If a man works half as much again as a boy then in how many days 4 men and 9 boys can dig a similar pit. A train leaves Meerut at 6 a.m. and reaches Delhi at 10 a.m. Another train leaves Delhi at 8 am. and reaches Meerut at 11 : 30 a.m. At what time do the 2 trains cross each another. A boy goes to school with a speed of 3 km/hr and returns to the village with a speed of 2 km/hr. If he takes 5 hrs. in all, find the distance between village and the school. In what proportion must Ragi at Rs. 3.10 per kg be mixed with Ragi at Rs. 3.60 per kg so that the mixture be worth Rs. 3.25 a kg. A man possessing Rs. 1000 lent a part of it at 6% S.I. and the other at 8% SI the yearly income is Rs. 75. Find the sum lent at 8% SI. Kantilal mixes 80 kgs of sugar worth Rs. 6.75 per kg with 120 kg worth Rs. 8 per kg. At what rate he sell the mixture to gain 20%. A jar contains a mixture of 2 liquids A and B in the ratio 7 : 5. When 9 litres of mixture is drawn off and the jar is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the jar initially?
220
Basic Mathematics
29. A sum of Rs. 41 was divided among 50 boys and girls each boy gets 90 paise and a girl 65 paise. How many boys are there?
8
Averages
8.1 INTRODUCTION:
Condensation of data is necessary in statistical analysis because a large number of big figures are not only confusing to mind but also difficult to analyse. In order to reduce the complexity of data and to make them comparable it is essential that the various phenomena which are being compared are reduced to one figure each. It is obvious that a figure which is used to represent a whole series should neither have the lowest value in the series nor the highest value but a value somewhere between these two limits, possibly in the centre where most of the items of the series cluster. Such figures are called measures of central tendencies or averages. Averages are usually of the following types: (a) Mathematical average (i) Arithmetic average or mean (ii) Geometric mean and (iii) Harmonic mean. (b) Average of position: (iv) Median and (v) Mode. Of the above mentioned five important averages, we are going to discuss in this book, about arithmetic average or Mean in detail.
8.2 ARITHMETIC AVERAGE OR MEAN:
Mean of the certain number of quantities which are all of equal weightage or importance is the figure obtained by dividing the total values of various items by their number. If the marks obtained by a student in five subjects are 56, 67, 82, 43 and 52, then to find the mean of these marks we shall add these marks and divide the total so obtained by the number of items which is 5. ∴
Mean =
56 + 67 + 82 + 43 + 52 5
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Basic Mathematics
Mean =
300 = 60. 5
In general, if x1, x2, x3, ... xn are the values in a data containing n items then their mean, denoted by
X is X =
x1 + x 2 + x 3 + ... + x n n
i.e.,
X=
∑ xi
i =1
n
n
If all the quantities are not of equal importance i.e. equal weight, we compute the weighted average as follows: If w1, w2, w3, ... wn are the weights associated to the values x1, x2 ... xn respectively the weighted average Xw is given by
Xw =
x1w1 + x 2 w2 + ... + x n wn w1 + w2 + ... + wn
Xw =
∑x w
i i =1 n
n
i
∑w
i =1
i
For example if Rama scores 35 in 2 subjects, 42 in 4 subjects and 72 in remaining 4 subjects. Then average marks scored by Rama
=
35 × 2 + 42 × 4 + 72 × 4 2+4+4 = 70 + 168 + 288 10 = 526 = 52.6. 10
Note: To compute arithmetic mean in a continuous series, the midpoints of the various class intervals are written down to replace the class intervals. Once it is done, there is no difference between a continuous series and discrete series.
8.3 COMBINED AVERAGE:
It is nothing but average of averages. Combined average is computed when the data set consists of different groups and average for each group is known.
Averages
223
If X1 , X2 ... Xn , are the average of groups 1, 2, ..., n and N1, N2, ... Nn are number of quantities in groups 1, 2, 3, ... n. Then combined average X1, 2, 3 ... n is given by
X1, 2, 3 ... n =
X1 N1 + X2 N2 + Xn N n N1 + N 2 + ... + N n
For example if the average marks of group of 5 students is 56 and average marks of another group of 6 students is 72 then combined average marks of 11 students.
= 10.7222. Average price of a book = Rs. 10.72. 5. The average marks of 15 students of class is 45. A student who has secured 17 marks leaves the class. Find the average marks of the remaining 14 students.
Averages
225
Solution: Average marks of 15 students = 45 ∴ Total marks = 45 × 15 = 675 If the one student with 17 marks leaves the class then total marks of remaining 14 Students = 675 − 17 = 658 Average marks of 14 students =
658 = 47. 14
6. The average age of 10 students is 6 years. The sum of the ages of 9 of them is 52 years. Find the age of the 10th student. Solution: Given: Average age of 10 students = 6 years Total age of 10 students = 10 × 6 = 60 years Given, the sum of ages of 9 of them = 52 years. ∴ Age of the 10th student = 60 − 52 = 8 years. 7. The average age of 10 students is 14 years. Among them, the average age of 5 students is 12 years. Find the average age of remaining students: Solution: Given: Combined Average X = 14.
N1 + N2 = 10 N1 = 5, N2 = 5, X1 = 12, X2 = ?
We have
X=
X1 N1 + X2 N 2 N1 + N 2 12 × 5 + X2 × 5 5+5 60 + 5 X2 10
14 = 14 =
140 = 60 + 5 X2 5 X2 = 140 − 60 = 80 X2 = 80 = 16. 5
∴ Average age of remaining 5 students = 16 years. 8. A shopkeeper purchased a certain number of dress materials at an average price of Rs. 190 each. The average price of 10 dress materials was Rs. 175 and that of remaining dress materials was Rs. 200. Find the total number of dress materials purchased. Solution: Combined mean X = 190 (Given)
10. A batsman realises that by scoring a century in the 11th innings of his test matches he has bettered his average of the previous 10 innings by 5 runs. What is his average after the 11th inning: Solution. Let the average runs in 10 innings be x Then total runs in 10 innings = 10x Average runs after 11th innings = x + 5 (Given) Total runs in 11th innings = (x + 5) 11 Also given a batsman scores a century in the 11th innings. ∴ Runs in 11 innings — Runs in 10 innings. = Runs in 11th innings = 100.
a x + 5f11 − 10 x = 100
11x + 55 − 10 x = 100
x = 100 − 55 x = 45. ∴ Average runs after 11th innings = x + 5 = 45 + 5 = 50. 11. Ms. Vani bought 17 books in a discount sale. The average price of books being Rs. 53. The average price of the eleven Kannada books is Rs. 71. If the prices of the remaining 6 English books form an increasing arithmetic progression with last term Rs. 25. Find the price of cheapest English book. Solution.
Given:
X = 53 X1 = 71, N1 = 11 X2 = ?, N2 = 6
We have
X=
N1 X1 + N 2 X2 N1 + N2
11 71 + 6 X2 11 + 6
53 =
a f d i
53 =
⇒
781 + 6 X2 17
781 + 6 X2 = 53 × 17 6 X2 = 901 − 781 6 X2 = 120 X2 = 20.
228
Basic Mathematics
∴ Total cost of 6 English books = 6 × 20 = Rs. 120. Given the cost of 6 English books form an increasing A.P. ∴ Costs are a, a + d, a + 2d, a + 3d, a + 4d and a + 5d. Given last term = 25 a + 5d = 25 5d = 25 − a Also sum of first n terms in an A.P. is
Sn =
∴
n 2a + n − 1 d 2 6 2a + 6 − 1 d 2
a f
a f
S6 =
S6 = 3 2 a + 5d
Substituting we get
120 = 3 2 a + 25 − a
120 = a + 25 3 40 = a + 25
40 − 25 = a ⇒ a = 15. ∴ Cost of the cheapest English book = Rs. 15. 12. At a place, the average temperatures from Monday to Thursday was 35°C and from Tuesday to Friday was 38°C. Find the day temperatures on Monday and Friday if the ratio of temperatures on Monday and Friday is 5:7. Solution: Average temperatures from Monday to Thursday = 35° (Given)
∴ Monthly expenses per student = Rs. 600. 14. Out of 3 numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, Find the 3 numbers. Solution. Let the 3 number be x, y and z. Given: Also given average = 56. ∴
16. Ten years ago, the average age of a family of 4 members was 24 years. Two children having been born, the average age of the family is same today. What is the present age of the youngest child if they differ in age by 2 years. Solution: Let the age of youngest child be x yrs. ∴ Age of next child = x + 2 years. 10 years ago Average age of 4 members = 24 Total age of 4 members = 24 × 4 = 96. After 10 years, Total age of 4 members = 96 + 4 × 10 [Each member's age increases by 10 years] = 96 + 40 = 136.
Averages
231
Now average of 6 members
=
136 + x + x + 2 = 24 given 6
136 + 2 x + 2 = 24 × 6 138 + 2 x = 144
a
f
2 x = 144 − 138
2x = 6 x=3
∴ Present age of the youngest child = 3 yrs. 17. The average ages of A and B is 42 yrs., that of B and C is 28 yrs. and that of C and A is 40 yrs. Find the ages of A, B and C. Solution: Let the ages of A, B and C be a, b and c. Given average of A and B = 42 ⇒
a+b = 42 2
...(1)
a + b = 84 Given: Average age of B and C = 28 years.
b+c = 28 2
b + c = 56
Also given average age of C and A is i.e., ⇒ Solving (1) and (2) we get
EXERCISES
1. A cricketer makes 72, 59, 101, 18 and 10 runs respectively in 5 matches played by him. Find his average score. 2. The average weight of a class of 24 students is 35 kgs. If the weight of the teacher is included, the average rises by 400 gms. Find the weight of the teacher. 3. The average marks of 15 boys of a class is 65 and 11 girls of the same class is 78. Find the average marks of the students. 4. 3 tests in English, 2 in Hindi, 4 in Kannada and 5 in Sociology are conducted. The average marks scored by Raju in English is 60, that in Hindi is 56 and in Kannada is 45. If the average marks of all the subjects taken together is 48. Then find the average marks scored by him in Sociology. 5. 10 shirts and 5 pants were bought for Rs. 6000. If the average price of a shirt is Rs. 450, then find the average price of a pant. 6. The average of 25 results is 18, that of first 12 is 14 and of the last 12 is 17. Then find the 13th result. 7. The average age of A, B, C, D five years ago was 45 years. By including X, the present age of all the five is 49 years. Find the present age of X. 8. A batsman makes a score of 87 runs in the 17th innings and thus increased his average by 3. Find his average after 17th inning. 9. Miss Radha bought 51 dress materials in a discount sale. The average price of a dress material being Rs. 318. The average price of 33 polyster dress materials is Rs. 426. If the prices of the remaining cotton dress materials form an increasing arithmetic progression with last term 150. Find the price of the cheapest cotton dress material. 10. The average temperature for Monday, Tuesday and Wednesday was 40°C. The average temperature for Tuesday, Wednesday and Thursday was 41°C. If the temperature on thursday was 42°C. Then find the temperature on Monday.
Averages
233
11. The average age of a husband and a wife was 23 years when they were married 5 years ago. The average age of the husband, the wife and a child who was born during the interval is 20 years now. How old is the child now? 12. The average age of 5 members of a committee is the same as it as 3 years ago, because an old member has been replaced by a new member. Find the difference between the ages of old and new member. 13. There were 35 students in a hostel. If the number of students is increased by 7, the expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Re 1. Find the original expenditure of the mess. 14. The average weight of A, B and C is 45 kg. If the average weight of A and B is 40 kgs and that of B and C is 43 kg. Then find the weight of B. 15. The average salary of 20 workers in an office is Rs. 1900 per month. If the manager's salary is added, the average salary becomes Rs. 2000 per month. What is the manager's annual salary?
9
Bill Discounting
9.1 INTRODUCTION:
Suppose a merchant A purchases goods worth say Rs. 50,000 from another merchant B at a credit for certain period say 6 months. Then B draws up a draft i.e. prepares a special bill called 'Hundi' or bill of exchange. On the receipt of the goods, A gives an agreement dually signed on the bill stating that he has accepted the bill and the money can be withdrawn from his bank account after 6 months of the date of bill. On this bill, there is an order from A to his bank asking to pay Rs. 50,000 to B after 6 months. If B needs the money of this bill earlier than 6 months, then B can sell the bill to a banker or a broker who pays him the money against the bill but somewhat less than the face value. Bill discounting is essentially lending by the banker against the bills and the bankers charges a certain interest for doing this service.
9.2 TERMINOLOGY:
Bill: The document which required an individual to pay a fixed amount after a fixed later date is called a bill. Discount: When a bill is cashed in advance of its date of maturity an amount is deducted by the money lender or bank from the amount of the bill due. This amount deducted is called Discount. True discount: The present value of a sum of money is that principal which, if placed on stipulated rate for a specified period will amount to that sum of money at the end of the specified period. The interest on the present value of the bill is called true discount. Banker' er's Banker's discount: The interest on the face value or amount of the bill is called Banker's discount. Banker' gain: er's Banker's gain: The difference between banker's discount and true discount is called Banker's gain.
Anil, BTM Layout, Bangalore Six months later pay me or my order the sum of Rs. 50,000 (Rupees fifty thousand) for value received. Anil Akram In the above bill of exchange. Drawer is Mr Akram so payee is Akram. Drawee or Acceptor is Mr. Anil. Drawing date: 10.2.2005 Bill amount or Face value of Bill = Rs. 50,000 = Nominal value of the bill. Legally due date: drawing date + Bill period + grace period of 3 days (It is customary to give 3 days grace period). So In the above bill, Legally due date is 13.8.2005
Date of drawing : 10 - 2 - 2005 Bill period : 0 - 6 - 0
Q
Grace period : 3 - 0 - 0 Legally due date : 13 - 8 - 2005
9.3 FORMULAE:
If F is the face value of the bill, BD is the banker's discount, t is the time in years, R = Rate of interest
7. A bill was drawn on March 8th at 7 months period and was discounted on May 18th at 5%. If the banker's gain is Rs. 3 find the true discount, the banker's discount, and the sum of the bill. Solution: Date on which the bill was drawn = March 8th i.e., 8−3 Period (7 months): 7 Grace period: 3−0
Bill Discounting 241
Legally due date: 11 − 10 i.e., Oct. 11th. Date on which the bill was discounted: May 18th Time for which the bill has yet to run:
May + June + July + Aug + Sep + Oct 13 30 31 31 30 11 = 146 days.
∴
t=
146 2 = yrs. 365 5 2 at 5% 5
Now Banker's gain = Simple interest on true discount i.e. Rs. 3 is the S.I. on T.D. for
Then ⇒
3= P=
P×5× 100
2 5
100 × 3 = Rs. 150 2 2 yrs. at 5% 5
Banker's discount = T.D. + S.I. on TD
= Rs. 150 + SI on 150 for 150 ×
150 +
2 ×5 5 100
= 150 + 3 = Rs. 153. OR
B. D.= TD + B. G. BD = 150 + 3 = Rs. 153.
Now
Sum =
BD × TD 153 × 150 = = Rs. 7650. BD − TD 153 − 150
8. A bill was drawn on the 10th July 1960 at 3 months after signed and was accepted on presentation on 1st Aug. 1960. It was discounted on 23rd Aug '69 at 5% p.a. simple interest to realise Rs. 2475. Find the face values of thee bill and banker's discount.
Drawing date : 10 - 7 -1960 Bill period : 0 - 3 - 0
Solution:
Grace period : 3 - 0 - 0 Legally due date : 13 -10 -1960
242
Basic Mathematics
Given: Bill was present on 1st Aug. 1969. Number of days for which banker's discount is paid =
9. The present worth of a bill due sometime hence is Rs. 1100 and the true discount on the bill is Rs. 110. Find the banker's discount and the extra gain the banker would make in the transaction. Solution: Here time and rate of interest are not given. Given: PW = Rs. 1100 TD = Rs. 110 We have Squaring
TD =
EXERCISE
1. A bill for Rs. 3500 due for 3 months was drawn on 27th March 2000 and was discounted at the rate of 7% on 18th April 2000. Find the banker's discount and discounted value of the bill. 2. The banker's discount and true discount on a sum of money due four months are respectively Rs. 510 and Rs. 500. Find the seem and the rate of interest. 3. The difference between BD and TD on a bill due after 6 months at 4% interest per annum is Rs. 20. Find the true discount bill discount and face value of the bill. 4. The banker's gain on a certain bill due 6 months hence is Rs. 10, the rate of interest being 10% p.a. Find the face value of the bill. 5. A banker pays Rs. 2340 on a bill of Rs. 2500, 146 days before the legally due date. What is the rate of discount charged by banker? 6. A bill for Rs. 1460 drawn at 3 months was discounted at 4% p.a. on 9th November for 1454.40. On what date the bill was drawn? 7. A bill was drawn on April 14th at 8 months after date and was discounted on July 24th at 5% p.a. If the banker's gain is Rs. 2, what is the face value of the bill. 8. Find the banker's discount and cash value of a bill for Rs. 3400/- drawn on April 25th 1996 at 7 months and discounted on September 16th, 1996 at 5%. 9. The banker's gain of a certain sum due 2 years hence at 5% per annum is Rs. 8. Find the present worth. 10. The present worth of a sum due sometimes hence is Rs. 576 and banker's gain is Re. 1. Find the true discount. 11. The banker's gain on a sum due 3 years hence at 5% is Rs. 90. Find the banker's discount. 12. The banker's discount on a bill due 1 year 8 months hence is Rs. 50 and true discount on the same sum at the same rate percent is Rs. 45. Find the rate of interest.
Bill Discounting 245
13. A bill for Rs. 3500 due for 3 months was drawn on 27th March 2000 and was discounted on 18th April 2000 at 7% rate of interest. Find the banker's discount and discounted value of the bill. 14. A bill for Rs. 2920 drawn at 6 months was discounted on 10.4.97 for Rs. 2916. If the discount rate is 5% per annum, on what date was the bill drawn? 15. If the difference of simple interest and true discount of a sum due one year 6 months, hence at 8% p.a. is Rs. 81.45. Find the sum.
10
Stocks and Shares
10.1 STOCK:
In order to meet the expenses of a certain plan or a big project. Loan is raised from the public at a certain fixed rate of interest. Bonds or promissory notes of a fixed value are issued for sale to the public. If a man purchases a bond of Rs. 1000 at which 5% interest has been fixed. Then the holder of such bond is said to have 'a Rs. 1000 stock at 5%'. Here Rs. 1000 is called the face value of the stock. Usually a period is fixed for the repayment of the loan i.e., the stock matures at a fixed date only. Now if the person holding a stock is in need of the money before the date of maturity of stock, he can sell the bond to some other person, where by the claim of interest is transferred to that person.
10.2 SHARES:
To start a big concern or a business a large amount of money is needed. This is usually beyond the capacity of one or two individuals. Therefore a group of individuals get together and form the company. The company issues a prospectus and invites the public to subscribe. The required capital is divided into equal small parts called shares, each of a particular fixed value. The person who possesses one or more share is called a share holder. Sometimes the company asks its share holders to pay some amount immediately and balance after some period. The total money raised immediately is called the paid up capital.
10.3 DISTINCTION BETWEEN STOCK AND SHARES:
The main distinction between stocks and shares is (i) Shares can be issued directly, but stocks cannot be issued directly. Firstly amount is collected on shares from the public. Stock certificates are issued only after collecting full amount of shares. Stocks are never issued unless the shares are issued and subscribed in full by the public. (ii) Shares need not be fully paid, but stocks must be fully paid. For example the value of a share is Rs. 100. This is known as face value of the share. A share of Rs. 100 each may be called up at Rs. 50 each by the company and the balance of Rs. 50 may be paid at later stage by the public as and when demanded by the company. Stock must be fully paid means face value of Rs. 100 on each share must be paid in full by the public in order to get the stock certificate from the company.
Stocks and Shares
247
10.4 TERMINOLOGY:
Debentur bentures: Debentures: Debentures are long term loans taken by the company from the public. Every person who lends such an amount is given a certificate of loan called debentures. Shares: Face Value of Shares: It is the price at which shares are first issued by a company. It is the price printed on the share certificate. Mark Market Price: It is the price at which the share can be brought or sold on the stock/share market. Par Value of Shares: When the shares are issued to the public at the face value, it is called par value Shares: of share. Example: When Rs. 10 share is issued at its face value, it is called par value. Abov Above Par: If the market value of the share is more than the face value, it is said to be above par or at premium. Example: When Rs. 10 shares are issued at Rs. 12 the shares are said to be issued at Rs. 2 premium. Below Below Par: If the market value of the share is less than the face value, it is said to be below par or it is said to be at discount. Example: When Rs. 100 shares are issued at Rs. 90. Then the shares are said to be issued at 10% discount. Dividend: Dividend: It is the portion of the profit of the company which is distributed to the share holders. The dividend is always calculated on the face value of the share. Dividends may be cash dividends or share dividends. Bonus shares are known as stock dividends. Ex-Dividend Cum-dividend Ex-Dividend and Cum-dividend Prices: Interest on bond is payable on pre-determined dates. If the bond is bought or sold on a date closer to the interest due date, the prices may be quoted Ex-interest or Cum-interest. If the price is ex-interest the selling price of it is not inclusive of interest. If it is quoted cum interest, the buyers will receive the interest amount. In the case of shares, share dividend are paid instead of interest. Cum dividend price quotations are usually higher than the Ex-dividend quotations. Yield: Actual dividend received by the actual amount invested in a stock or shares called yield. i.e., i.e.,
Yield = =
Dividend Amount invested Nominal interest Amount invested
Brokers and Brokerage: Buying and selling of stocks or shares is done through the person called Brokers oker Brokera oker 'brokers' at stock exchange. They charge certain amount called brokerage. Note that when stock/share is purchased, brokerage is added to the cost price and when stock or share is sold brokerage is subtracted from the selling price. hares: Kinds of Shares: A company may issue two kinds of shares. They are (i) Preference shares (ii) Equity shares (i) Preference shares: A preference share holder enjoys a preferential claim with regard to the Preference shares: efer payments of dividend and repayment of capital. The rate of dividend is fixed, but it is paid before profit is distributed to other members.
248
Basic Mathematics
shares: (ii) Equity shares: An equity share holder has no special rights. The rate of dividend is not fixed. It varies from year to year. An equity share holder is paid dividend only after the claims of preference share holders are satisfied. Quotation: Consider the statement "Government paper mills 11% shares at 110". This is a quotation. This means a share of the mill having face value Rs. 100 is available for sale at Rs. 110. This share fetches him a dividend of Rs. 11 every year.
WORKED EXAMPLES:
1. Find the cost of 80 shares at 5% if the market value of the share is 93 and its par-value is Rs. 100. If a person invests Rs. 37200 in such shares then find his annual income. Solution. Cost of 1 share = Rs. 93 Cost of 80 such shares = 80 × 93 = Rs. 7440. Now, the person has invested Rs. 37,200. By investing Rs. 93, the person gets 1 share. ∴ By investing Rs. 37200 the person gets
= 0.044811. So yield from 5 and ½% stock at 102 is greater. Hence it is a better investment. 4. Find the cash required to purchase Rs. 20,000 stock at 105 (brokerage ½%). Also find the annual dividend received if the company declares dividend of 8 and ½%. Cash required to purchase Rs. 100 stock. Solution.
= 105 +
F H
1 211 = 2 2 20,000 × 100 211 2
I K
∴ Cash required to purchase Rs. 20,000 stock =
=
20,000 × 211 = Rs. 21,100. 2 × 100
Annual dividend for Rs. 100 stock = 8 and ½ Rs.
250
Basic Mathematics
Annual dividend for Rs. 20,000 stock
=
20,000 × 8 100
1 2
= 200 ×
17 = Rs. 1,700. 2
5. A person has invested a certain sum of money in 13% stock at 96. He sold the investment when the market value went up to 101.5. He gained Rs. 1470 in this process. If he has paid the brokerage at 2% for all the transaction, what was the amount of cash investment and what was the stock value of the investment in the first instance. Solution. Let the amount invested = Rs. x In the first instance, Cost of 1 share = 96 + 2 (Brokerage) i.e., Cost of 1 share = Rs. 98. Now For Rs. 100 share — Rs. 98 is the amount received. For Rs. x share —
7 1 debentures at 131 . Find the change in his income a brokerage of % being charged on each 8 8 transaction. Vivek has Rs. 16,500 stock in 3%. Income from Rs. 100 stock = Rs. 3
∴ Income from Rs. 16,500 stock =
= Rs 505. ∴ Change in income = Rs. 505 − Rs. 495 = Rs. 10. ∴ Income is increased by Rs. 10. 7. Tulasi has invested Rs. 1,00,000 partly in 12% stock at 120 and partly in 15% stock at 75. If the total income from both is Rs. 15,000. Find the amount invested in 2 types of stocks. Solution. Let the amount invested in 12% stock at 120 be x. Then Amount invested in 15% stock at 75 = 1,00,000 − x
= Rs. 1825. ∴ Mr. Gauriprasad bought stock worth Rs. 1825. 9. Pusphak buys Rs. 2,000 shares paying 9% dividend. If he wants to have an interest of 12% on his money, then find the market value of each share.
Stocks and Shares
253
Solution. Dividend on Rs. 2000 share in 9% of 2000.
=
9 × 2000 = 180. 100
He wants to have 12% on his money. Rs. 12 is an income on Rs. 100 share. Rs. 180 is income on
180 × 100 = 1500. 12
∴ Market value of each share must be Rs. 1500 in order to have interest of 12% on his money. 10. A man invests some money partly in 3% stock at 96 and partly in 4% stock at 120. What is the ratio of money he must invest in order to get equal dividends from both. Solution. For an income of Re 1 in 3% stock at 96, investment = Rs. For an income of Re 1 in 4% stock at 120 investment = Rs. = Rs. 30. Ratio of investments = 32:30 = 16:15.
96 = Rs.32. 3
120 4
∴
REMEMBER:
• Yield = • • • •
Nominal interest Amount invested
When stock is purchased, brokerage is added to cost price. When stock is sold, brokerage is subtracted from selling price. Interest or dividend is paid on the face value of the stock or share not the market value. 4 and ½% stock at 96 means a stock whose face value is Rs. 100 is available at Rs. 96. Interest earned in 4 and ½. • Shares need not be fully paid but stock must be fully paid.
3. Find the cash realised by selling Rs. 2400, 5 and ½% stock at 5 premium, brokerage being 1/4%. 4. Which of the following is better investment? 6% at 94 or 8% at 110. 5. Ramu possesses 150 shares of Rs. 25 each, the dividend declared by the company is 12%. What is the dividend earned by him. If he sells the shares at Rs. 40 and reinvest the proceeds in 7% shares of par value Rs. 100 at Rs. 80, Find the change in his dividend income. 6. A man invested Rs. 6750 partly in 6% stock at 140 and partly in 5% stock at 125. Find his investment in cash if the income derived from both the investments is Rs. 280. 7. Mr. Vivek invested Rs. 2200 partly in 10% stock at 120 and partly in 12% stock at 96. Find his investment in each if the income derived from both the investments is Rs. 200. 8. A man invests some money partly in 6% stock at 96 and partly in 5% stock at 120. In what ratio, he must invest the money so as to get equal dividends from both. 9. Rs. 2780 is invested partly in 4% stock at 75 and 5% stock at 80. If the income from both investments are equal, find the investment in 5% stock. 10. Mr. Harish has invested a certain amount of money in 13% stock at 101. He sold it when market value went down to 96.5. He lost Rs. 3564 on this process. If he has paid the brokerage at 1 and ½% for all transactions, what was the amount of cash investment? What was the stock value of the investment in first instance.
11
Learning Curve
11.1 INTRODUCTION:
When an individual performs the same task repeatedly the second and subsequent time will provide experience to the individual and there by there will be an increase in the degree of efficiency in performance. This is due to the learning process. The learning effect improves productivity of individuals especially that of workers engaged in factories and industries.
11.2 LEARNING CURVE:
It is the curvilinear relationship between the decrease in average labour hours per unit with increase in the cumulative output. The assumption of learning curve theory is that every time total output of a product doubles, the cumulative average time taken to produce one unit decreases by a constant percentage. For example: 80% learning effect means that when the cumulative output is doubled, the cumulative average hours per unit will be 80% of the previous level.
11.3 THE LEARNING CURVE RATIO:
It is the ratio of average labour cost of first 2N units and average labour cost of first N units. It is also known as experience ratio or improvement ratio or efficiency ratio. ∴
Experience ratio =
Average labour cost of first 2 N units Average labour cost of N units
11.4 GRAPHICAL REPRESENTATION OF LEARNING CURVE:
We know 80% learning effect means that when the cumulative output is doubled the cumulative average hours per unit will be 80% of the previous level. Let us consider the effect of 80% learning in the form of a table taking cumulative time per unit as 100.
Taking the total output on x-axis and cumulative average time per unit on y-axis, we get the learning curve as shown in the figure.
x y
1 100
2 60
4 48
8 38.4
16 30.72
32 24.57
y Scale: x - axis : 1 Unit = 1 cm y - axis : 20 units = 1 cm
100 80 60 40 20 0
x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Fig. 11.1
The curve clearly indicates that there will be fast learning effect in the initial stages and after sometime there will be a steady state phase in which there is not any significant learning effect.
Learning Curve
257
11.5 LEARNING CURVE EQUATION:
If b = logarithmic of learning ratio to base 2, y = cumulative average time per unit, x = cumulative total number of units produced a = time for first unit, then the learning curve equation is given by y = axb. By considering log on both sides we get
b = −0.7368. 2. A worker takes 10 hrs. to produce the first unit of product. What is the cumulative average time per unit taken by him for the production of first two units? (Assuming the learning effect to be 80%). Also find the total time for producing the first two units. Solution: Given: a = 10 hrs. x=2 80% learning effect. We have learning curve equation y = axb
where
= 1.249. Time taken for producing the first 2 units = 8 × 2 = 16 hrs. 3. The time required to produce the first unit of a product is 1000 hrs. If the manufacturers experiences 80% learning effect, calculate the average time per unit and the time taken to produce altogether 8 units. Also find the total labour charges for the production of 8 units at the rate of Rs. 12.50 per hour. a = 1000 Solution: Given: x=8 ∴
Total time to produce 8 units = 4096 hrs. Cumulative average time per unit = 512 Total time to produce 8 units = 512 × 8 = 4096. Given Labour charges for 1 hr. = Rs. 12.50 ∴ Labour charges for 4096 hrs. = 4096 × 12.50 = Rs. 51,200. 4. An engineering company has won the contract for supplying aircraft engines of a new type. The prototype constructed to win the contract took 500 hrs. It is expected that there will be 90% learning effect. Estimate the labour cost of producing 8 engines of new order, if the labour cost is Rs. 40 per hourEXERCISE
1. 2. 3. 4. 5. Define the term learning curve ratio. What is learning curve? What do you mean by 80% learning effect. Find the index of learning for 80% learning effect. A worker requires 40 hrs. to produce first unit of a product. How much time is required for him to produce a total of 4 units. 6. A worker requires 20 hrs. to produce the first unit of a product. If the learning effect is 80% calculate how much time is required to produce 4 units. Also find the labour charges for the production of 4 units at the rate of Rs. 20 per hr. 7. A company requires 2134.2 hrs. to produce the first 40 machines. If the learning effect is 80%, find the number of hrs. required to produce next 120 machines. 8. The first unit of a product took 80 hrs. to manufacture. If the workers show 80% learning effect, find the total time taken to produce 8 units.
ANSWERS
1.
Average labour cost of first 2N units . Average labour cost of first N units
262
Basic Mathematics
2. Curvilinear relationship between the decrease in average labour hrs. per unit with increase in curve ratio. 3. When the cumulative output is doubled, cumulative average labour hrs./unit will be 80% of the previous level. 4. −0.3219 5. 108.4 hrs. 6. 48.8 hrs. 7. 3329.35 hrs. 8. 32.768 hrs.
12
Linear Programming
12.1 INTRODUCTION:
Linear programming is a powerful technique which can indicate a definite conclusion as to the best utilisation of available resources under given circumstances. Any industrial process may consists of a number of activities relating to capital to be employed, products to be made and sold, materials to be used: machines to be run; inventories to the stored and consumed or a combination of the above. Since utilisation of one affects the utilisation of another and due to the limitation of the total available resources, these activities are interdependent or interlocking. In such a situation, a large number of ways exists in which the available resources can be allocated to the competing demands. Linear programming enables us to arrange for that combination of resources which optimise the cost, production, profit etc. This technique was evolved by George B. Dantzig as a tool for planning the diversification activities of U.S. Airforce in 1947.
12.2 LINEAR PROGRAMMING:
Definition: Definition: Linear programming is a mathematical technique for determining the optimal allocation of resources and obtaining a particular objective when there are alternative uses of the resources: money, manpower, material, machine and other facilities. The objective in resource allocation may be cost minimisation or profit maximisation. The technique of linear programming is applicable to problems in which the total effectiveness can be expressed as a linear function of individual allocations and the limitations on resources give rise to linear equalities or inequalities of the individual allocations. Objective function: The objective function is a quantified statement in linear programming model of Objectiv what the best results or the best advantage which is aimed for as the objective of the resource allocation decision. Objective function will either be to maximise a value or to minimise a value. The objective function is expressed as a linear function of decision variables. It is usually stated as maximise or minimise a1x1 + a2x2 + ... +anxn where x1 x2 ... xn are the decision variables in the problem and a1, a2, ..., an are constant values for each variable. variab aria Decision variables: The decision variables in any activity is a variable which is competing with other activities for limited resources. These variables are inter-related linearly in terms of utilisation of resources.
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Constraints: Constraints: The resources like production capacity, manpower, time, space, technology, etc. are scarce and there are limitations on what can be achieved. These restrictions are a set of conditions which an optimal solution must satisfy. They are known as constraints. These are expressed as linear inequalities or equalities in terms of decision variables. Non-neg tivity Non-negativity conditions: All decision variables must assume non-negative values. If any of the variable is unrestricted in sign, a trick can be employed which will enforce the non-negativity without changing the original information of the problem.
12.3 SOLUTION TO LINEAR PROGRAMMING PROBLEM:
A set of values of decision variables x1 x2 ... xn which satisfies the constraints of linear programming problem is called solution. easible Feasible solution: Any solution to a linear programming problem which satisfies the non-negativity restriction of the problem is called a feasible solution to the Linear programming problem. Optimal solution: Any feasible solution which optimises (maximises or minimises) the objective function of a linear programming problem is called an optimal solution. Solution of linear programming by graphical method: If the objective function is a function of 2 variables only then the LPP can be solved by graphical method. One variable is taken along x-axis and another along y-axis. Since negative values are not allowed, the graph contains only first quadrant. That is, 0 and positive values of x and y are considered.
This is a straight line parallel to x-axis. All points below this line are represented by the inequality y < 2 and all points above this line are represented by y > 2 and the corresponding graphs are
y>2 y=2 2 y<2 1 2
Hence consider the points (0, 3) and (2, 0) and join them to get the graph of linear equation 3x + 2y = 6.
3 2 1
(0, 3)
3x +2 y= 6
(2, 0) 0 1 2 3
Fig. 12.4
Any value (x, y) which will fall in the shaded area in figure (a) is represented by the inequality 3x + 2y ≤ 6 and any value which will fall in the shaded area in figure (b) is represented by the inequality 3x + 2y ≥ 6.
Remark: (1) When there are several inequalities, which are true at the same time, the feasible region of combinations of values x and y must be a region where all the inequalities are satisfied. (2) Optimum solution to linear programming problem lies at one of the vertices only. Hence find the co-ordinates of each vertex and substitute in the objective function. The value for which the objective function is highest/least is the optimum solution for maximisation/minimisation.
WORKED EXAMPLES:
1. Write the following data in the form of LPP: Machine Chair Table Time M1 5 2 50 min M2 2 7 130 min The price of each chair is Rs. 120 and each table is Rs. 130. Solution: Let the number of chairs be x and number of tables be y. Time availability for M1 ≤ 50 and Time availability for M2 ≤ 130. ∴
5 x + 2 y ≤ 50 and
2. A company manufacturers 2 types of bulbs A and B by using 2 machines M1 and M2. One bulb of type A require 2 hrs. at machine M1 and 1 hr. at machine M2 and one bulb of type B requires
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267
one hr. at M1 and 2 hrs. at M2. The profit from each bulb of type A is Rs. 2 and that of type B is Rs. 3. The number of hrs. available per week on machines M1 and M2 are 20 hrs. and 30 hrs. respectively. Formulate the above problem as LPP. The aim of the company is to maximise the profit. Solution: Let the number of bulbs of type A = x and that of type B = y At machine M1, one bulb of type A requires 2 hrs. and one bulb of type B requires 1 hr. Also number of hrs. available per week on machine M1 is 20. ∴
2 x + 1y ≤ 20
Similarly at machine M2, one bulb of type A requires 1 hr., B requires 2 hrs. Also number of hours available per week = 30 hrs. ∴
Zmax = 1800 at x = 0 and y = 300. 5. A radio factory produces 2 different types of transistor radios (a) cheaper ordinary model and (b) an expensive special model. For greater efficiency the assembly and finishing operation and performed in 2 workshop the ordinary model requires 3 hrs. of workshop A and 4 hrs. of workshop B. While special model requires 6 hrs. of workshop A and 4 hrs. of workshop B. Due to limited resources and skilled labour only 180 hrs. of workshop A and 200 hrs. in workshop B. The factory makes a profit of Rs. 300 on each ordinary model and Rs. 400 on each special model. Formulate the above as a LPP and solve by graphical method, assuming the company expects maximum profit. Solution: Let the number of ordinary models = x and the number of special model = y.
Ordinary Workshop A: Workshop B: 3 4 Special 6 4 Total/hrs. 180 200
∴ Z is minimum when x = 15 and y = 20 ∴ Zmin = 2500. 8. A certain company wishes to plan its advertising strategy to reach certain minimum percentage of high and low income groups. Two alternatives are considered namely cinema commercials and magazines. Magazines advertising has an exposure for the high income group of 2% per page but only 1% per page exposure for the low income group. Cinema commercials exposes 3% of the low income group per show and only 1% of high income group per show. Magazines advertising
REMEMBEREXERCISE
1. Write the following data in the form of LPP: Machine Product A Product B Time M1 2 3 60 min. M2 4 5 100 min. The price of product A is Rs. 1000 each and product B is Rs. 1500 each. 2. Old hens can be brought at Rs. 200 each and the young ones at Rs. 500 each. The old hen lays 3 eggs and young one lays 5 eggs per week: If there are only Rs. 8000 available to spend on
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purchase, the hens should be bought in order to have a maximum profit per week, assuming that the house cannot accommodate more than 20 hens at a time. 3. Maximise: 30x + 20y subject to
10 x + 6 y ≤ 1000, 5 x + 4 y ≤ 600
x, y ≥ 0
4. Maximise 5x + 8y subject to
2 x + y ≤ 100 ; x + 2 y ≤ 200
x, y ≥ 0.
5. A company produces 2 products x and y each of the product require two operations one on machine A and the other on machine B. The machine hours required by these two products and the total hours available are given as follows:
Machine hrs. required X
A B 2 4
Product Y
5 3
Total machine hrs. available
19 17
Each unit of the product x and y makes a profit of Rs. 3 and Rs. 4. Find the optimal solution of the product to obtain the maximum profit. 6. Minimise Z = 4x + 4 subject to 3 x + 4 y ≥ 20, x + 5 y ≥ 15, x , y ≥ 0. . 7. Minimise Z = 10x + 6y subject to 2 x + y ≥ 60, 4 x + y ≥ 80, x1 y ≥ 0. . 8. Minimise 2 = 3x + 5y subject to 5 x + 2 y ≥ 0, x + y ≥ 4, x + 3 y ≥ 6, x1 y ≥ 0. . 9. An animal feed company must produce 200 kgs of mixture consisting of ingredients x1 and x2 daily. x1 costs Rs. 3 per kg and x2 Rs. 8 per kg. Not more than 80 kgs. of x, can be used and atleast 60 kgs. of x2 must be used. Find how much of each ingredients should be used if the company wants to minimise the cost. 10. A chemist provides his customers at least cost, the minimum daily requirement of 2 vitamins A and B by using a mixture of 2 products M and N. The amount of each vitamin in one gram of each product, the cost per gram of each product and minimum daily requirements are given below:
Vitamin A
M N Minimum requirement 6 2 12
Vitamin B
2 2 8
Cost per gram
20 ps. 16 ps.
Formulate the problem of finding the least expensive combination which provide the minimum requirement of the vitamins.
13
Circles
13.1 DEFINITIONS:
Circle is a locus of point which moves such that its distance from a fixed point is constant. The fixed point is called centre and the distance is called radius of the circle. centre The boundary line of a circle is called circumference circumference cumference. Note: (1) The distance between 2 points (x1, y1) and (x2, y2) is given by distance formula
bx
2
− x1
g + by
2
2
− y1
g
2
.
(2) Slope of the line joining (x1, y1) and (x2, y2) is
y2 − y1 x 2 − x1
(3) 2 lines are perpendicular if and only if the product of their slopes = −1 (4) Angle in a semi circle = 90°.
13.2 EQUATION OF A CIRCLE:
Equation circ whose centre at orig igin 1. Equation of the circle whose centre is at the origin and radius = r units: Let O (0, 0) be the centre of the circle and r be the radius of the circle. Let P (x, y) be any point on the circle.
y
P (x, y) r x′ O (0, 0) x
OP = Radius = Distance between O (0, 0) and P (x, y).
∴
r=
a x − 0f + a y − 0f
2
2
r = x 2 + y2
⇒
x 2 + y2 = r 2 .
y′
Fig. 13.1
Circles 279
This is the equation of the circle whose centre is at the origin and radius = r units. 2. Equation of the circle whose centre is at (h, k) and radius = r units: Equation circ whose centre at radius y Let C (h, k) be the centre of the circle. 'r' be the radius of the circle. P (x, y) be any point on the circle. r CP = Radius = Distance between C (h, k) and P (x, y)
P
(x,
y)
r=
Squaring,
a x − hf + a y − k f
2
2
2
C (h, k)
i.e.,
a f + a y − kf a x − hf + a y − k f = r .
r2 = x − h
2
2
2
0
x
2
Fig. 13.2
This is the equation of the circle whose centre is at the origin and radius = r units. Equation circ whic hich described 3. Equation of the circle which is described on line joining (x1, y1) and (x2, y2) as the diameter: Let A (x1, y1) and B (x2, y2) be the ends of the diameter of a circle. Let P (x, y) be any point on the circle. P (x, y) Join AP and BP. We know, 90° Angle in a semicircle = 90° ∴ So Slope of AP × Slope of BP = −1 Slope of line joining A (x1, y1) and P (x, y) =
4. Find the equation of the circle with centre (4, 3) and which passes through (0, 0). Solution: Centre = (4, 3) (Given) Circle passes through (0, 0). Distance between centre and any point on the circle = Radius. ∴ Distance between (4, 3) and (0, 0) = Radius
13.3 POINT OF INTERSECTION OF A LINE AND A CIRCLE:
Chord: Chord: A line which divides the circle into 2 parts is called chord. So chord is a line joining any 2 distinct points on the circle. angent: Tangent: A line which touches the circle at only one point and which is perpendicular to the radius drawn from the point of contact is called tangent. Note:
chord
13.6 CONDITION FOR THE LINE y = mx + cTO BE A TANGENT TO THE CIRCLE x2 + y2 = a2 AND POINT OF CONTACT
The line y = mx + c is a tangent to the circle x2 + y2 = a2 if and only if length of the perpendicular from the centre to the line is equal to radius. Now centre of the circle = (0, 0) Radius = a ∴ Length of perpendicular from (0, 0) to the line y = mx + c = Radius (i.e. mx − y + c = 0) ⇒
m 0 − 0 +c m2 + 1 c m2 + 1
af af
=a
=a
Circles 301
c m2 + 1
= ±a
y = mx + c
⇒ Squaring,
c = ±a m2 + 1
c2 = a 2 m2 + 1 .
This is the condition for the line y = mx + c to be a tangent to the circle x2 + y2 = a2.
(0, 0)
d
i
Fig. 13.18
∴
Equation of tangent = y = mx ± a m 2 + 1 Also we know that equation of tangent at (x1, y1) on
...(1) is
x2
+
y2
=
a2
xx1 + yy1 = a 2
⇒ Comparing (1) and (2) we get
yy1 = − xx1 + a 2 y1 − x1 a2 = = 1 m ±a m2 + 1
...(2)
Taking 1st and 3rd ratio,
y1 = ± a m2 + 1
Taking 2nd and 3rd ratio.
− x1 a =± m m2 + 1 am m2 + 1 − am m2 + 1
x1 = m
∴ The points of contact are
F GH
am m2 + 1
,
−a m2
I or F J GH +1K
,
a m2
I J +1K
out of the 2 points of contact, one will satisfy the equation of tangent.
• Any line is a tangent to the given circle if the length of the perpendicular14
Parabola
14.1 INTRODUCTION:
When a solid cone is cut by a plane, the curves which lies on the surface of the cone and the plane, are the curves – circle, parabola, ellipse and hyperbola. These curves are called as conic sections. Definitions: Let l be a fixed line and S be fixed point. A point P moves in a plane containing S and l such that its distance from S bears a constant ratio to its distance from the line l, i.e.,
P
M
SP = a constant. PM
The locus of the point P is called a conic. A conic is the locus of the point which moves such that the ratio of its distance from a fixed point in the plane to its distance from a fixed line in a plane is constant.
l
S
Fig. 14.1
The fixed point is called focus and the fixed line is called directrix and the constant ratio called eccentricity of the conic. If the eccentricity of the conic is less than 1 then the conic is called ellipse. If the eccentricity is greater than 1 then it is called hyperbola. If the eccentricity is equal to one then the conic is called parabola. Circle is regarded as the conic of eccentricity zero. It is a particular case of an ellipse.
SP is PM
14.2 PARABOLA:
Parabola is the locus of point which moves in a plane such that its distance from a fixed point is equal to its distance from a fixed line. The fixed point is called focus. The fixed line is called directrix.
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14.3 EQUATION OF THE PARABOLA IN THE STANDARD FORM:
Let S be the focus, l be the directrix. Draw SZ ⊥ l. Let 'O' be the midpoint of SZ. Let SZ = 2a. So that OS = OZ = a. Choose O as origin the 2 mutually perpendicular lines OX and OY as co-ordinate axes. Let P (x, y) be any point on the parabola. Draw PM ⊥r to l and PN ⊥r X-axis. Join PS.
l y P M
THE PARABOLA y2 = 4ax.
For the parabola y2 = 4ax we have the following: 1. Shape: • Since (0, 0) satisfy the equation y2 = 4ax, the curve passes through the origin. • The equation y2 = 4ax remains unchanged if we replace y by −y. So the curve is symmetric about x-axis. • For negative values of x, we get imaginary values for y. So the curve entirely lies on the right side of y-axis if a is +ve or a > 0. (and if 'a' is −ve, the curve entirely lies on the left side of y-axis). Hence the shape of the parabola is
L l x
O
S
L′
Fig. 14.2
2. 3. 4. 5. 6.
The point O(0, 0) is called vertex of the parabola. The line OX, the +ve x-axis is called the axis of the parabola. S is the focus and its co-ordinate = (a, 0). The line l is the directrix and its equation is x = −a. The line LSL′ which is perpendicular to axis and passing through the focus is called Latus rectum. L and L′ are ends of latus rectum. L and L′ has x-co-ordinate 'a'. To get y-co-ordinate, since L and L′ lie on the parabola y2 = 4ax it satisfies the equation. ∴
Parabola which open downward is any quadrant depending on value of h and k. Vertex : (h, k) Axis : x = h Focus = (h, −a + k) Equation of directrix: y = a + k Ends of Latus rectum: (±2a + h, −a + k) Equation of latus rectum: y = −a + k Length of latus rectum: 4a. Note: In any parabola, the focus is inside the curve (i.e. on the axis of the parabola) and directrix is away from the curve (i.e. directrix never meet the parabola). 2. The distance between the vertex and the focus is equal to the distance between the vertex and the directrix is equal to a. 3. The distance of a point on the parabola from the focus is called the focal distance of the point. 4. A chord drawn through the focus is called focal chord and the focal chord perpendicular to the axis is Latus rectum. perpendicular to the directrix. Distance between directrix and vertex = a.
15
Limits and Continuity
15.1 INTRODUCTION:
The discovery of calculus was done independently and almost during the same time by Sir Isaac Newton and Gottfried Wilhelm Leibnitz. Calculus is developed on the basis of a more fundamental concept called the Limit.
15.2 CONSTANTS AND VARIABLES:
A quantity which remains the same throughout any mathematical discussion is called a constant. A quantity which takes different values in any mathematical discussion is called a variable. For instance, if we increase the production by using more raw materials, the cost of the machine doesn't change, but the costs of raw materials, Labour sales change. So cost of the machine which remains the same value throughout is constant and cost of raw materials and labour which assumes any numerical value out of given set of values are variables. Constants are represented by a, b, c, d, e and variables are represented by u, v, w, x, y, z.
15.3 FUNCTION:
Let A and B be 2 non empty sets. A rule f which associates each element x of A to an unique element y of B is called a function. It is denoted by f : A → B. If x ∈ A is related to y ∈ B, then we write y = f (x). Algebraic function:
F 3x + 8 I or rational function G H 4 x − 1 JK
2 3
A function y = f(x) is said to be an algebraic function if f(x) is a polynomial function (eg: x2 + 3x − 1) or irrational function (e.g.: (1 − x)2/3).
From the above set of values, we can observe that the value of y is slightly less than 2 or slightly more than 2 when the value of x is slightly less than 1 or slightly more than 1. But when x = 1, y has no value. In otherwords, when x is very nearly equal to 1, y is very nearly equal to 2. In symbols this is expressed as lim i.e.
x →1
x2 − 1 =2 x −1
lim y = 2
x →1
Read it as the limit of y as x tends to 1 is 2. The statement simply imply when x ≈ 1 y ≈ 2 even though it doesn't exist when x = 1.
Limits and Continuity 345
Definition of Limit: A function f(x) is said to tend to a limit l as x tends to a if the numerical difference between f(x) and l can be made as small as we please by taking the numerical difference between x and a as very small. In other words f(x) is very nearly equal to l when x is very nearly equal to a. It is denoted by lim f x = l or
x→a
af
x →a
lt f x = l.
af
Left Hand and Right Hand Limits: When x → a, through the values which are smaller than 'a' then we write x → a − 0 this
x →a− 0
lim f x or lim f a − h is called Left Hand Limit [LHL] of a function f (x).
h→ 0
af
a
f
For example: if x → 3 by taking the values 2.9, 2.99, 2.999, 2.9999... then x → 3 − 0 or Simply x → 3− When x → a, through the values which are greater than 'a' then we write x → a + 0 this lim f x or lim f a + h is called Right Hand Limit [RHL] of a function f(x).
h→0
(i) If f (a) ≠ an indeterminate form then f (a) itself is the limit. (ii) If f (a) = an indeterminate form then suitable method / formula is applied to reduce f (x) so that it will not take an indeterminate form when x is replaced by a.
16
Differential Calculus
16.1 INTRODUCTION:
Differential calculus was discovered by Sir Isaac Newton of England and Wilhelm Leibnitz of Germany. It deals with the study of rate of change of one quantity with another.
f x + δx − f x exists and finite then the function y = f (x) is said to be differentiable at x δx f x + δx − f x is called derivative or differential co-efficient of y with respect to x. It is δx dy or y′ or y1 or f′(x). dx
∴
Let y = c where c is a constant. Let δx be an increment given to x. δy be the corresponding increment in y. But Since c is a constant function any change in x will not cause change in y. In other words δy = 0. ∴ ∴
δx → 0
16.4 RULES OF DIFFERENTIATION:
1. Derivative of Product of Constant and a Function: Let y = Ku where K is a constant and u is a function of x. Let δx be an increment given to x. δu be the increment in u and δy be the corresponding increment in y. ∴
y + δy = K u + δu
a
f
y + δy = Ku + Kδu
δy = Ku + Kδu − y
δy = Ku + Kδu − Ku
δy = Kδu
Divide by δx and take lim
δx→0
δx → 0
lim
δy δu = lim K ⋅ δx δx → 0 δx
⇒
dy du =K⋅ dx dx d du Ku = K ⋅ dx dx
i.e.,
a f
Hence derivative or differential co-efficient of constant multiple of a function is constant into derivative of the function. Examples: (a)
d d 4 log x = 4 ⋅ log x dx dx = 4⋅ 1 4 = . x x
a
f
a f
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(b)
d d x 8e x = 8 ⋅ e dx dx
d i
d i
= 8⋅ex
2. Derivative of Sum of 2 Functions: Let y = u + v where u and v are functions of x. Let δx be an increment given to x. δu, δv be the increments in u and v. δy be the corresponding increment in y. ∴
∴ Derivative or differential co-efficient of sum of 2 functions is derivative of first function plus the derivative of the second function. Examples: (a) If y = x4 + ex, then
dy d 4 = x + ex dx dx
d
i
=
d 4 d x x + e dx dx
d i
d i
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373
dy = 4 x 4 −1 + e x dx dy = 4x3 + e x dx
(b)
d log x + x 2 dx = = =
d
i
d i
d d 2 log x + x dx dx 1 + 2 x 2 −1 x 1 + 2x x
a f
3. Derivative of Difference of 2 Functions: Let y = u − v where u and v are functions of x. Let δx be an increment given to x. δu, δv be the increments in u and v. δy be the corresponding increment in y.
∴ Derivative of difference of 2 functions is the derivative of first function minus the derivative of the second function. Examples: (a)
d 3 x − x2 dx
d
i
= d 3 d 2 x − x dx dx
d i
d i
= 3 x 3−1 − 2 x 2 −1
3x 2 − 2 x
(b) If y = logx − ex, then
dy d = log x − e x dx dx = =
d
i
d i
d d x log x − e dx dx 1 − ex . x
a f
4. Derivative of Product of 2 Functions: Let y = uv where u and v are functions of x. Let δx be an increment given to x, δu, δv be the increments in u and v. Let δy be the corresponding increment in y.
y + δy = u + δu v + δv
a fa f δy = au + δufav + δv f − y
δy = uδv + vδu + δuδv
δy = uv + uδv + vδu + δuδv − uv
Divide by δx and take lim
δx→0
δx → 0
lim
δy uδv + vδu + δuδv = lim δ x δx → 0 δx = lim u
δx → 0
δv δu δuδv +v + δx δx δx
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375
As lim , δu and δv are small. Hence the product δu × δv is very very small. So the term
δx→0
δuδv can δx
be neglected. ∴ ⇒
δx → 0
lim
δy δv δu = lim u + v δx δx → 0 δx δx dy dv du =u +v dx dx dx
i.e.,
d dv du uv = u + v dx dx dx
a f
So Differential Co-efficient or derivative of product of 2 function is first function into derivative of 2nd function plus second function into derivation of the first function. i.e., = (I function) ⋅
Hence Derivative or differential co-efficient of quotient of 2 functions is Denominator into differential co-efficient of Numerator, minus Numerator into differential co-efficient of Denominator, whole divided by square of the denominator.
i.e.,
d Nr. = dx Dr.
d d aDr.f dx aNr.f − aNr.f dx aDr.f F I H K aDr.f
2
Nr.: Numerator Dr.: Denominator This rule is known as quotient rule. Examples:
16.5 DIFFERENTIATION OF COMPOSITE FUNCTIONS:
Chain rule: If y = g(u) and u = f (x) are 2 differentiable functions, then the composite function y = g[f (x)] can be differentiated by using chain rule, which can be stated as
16.6 DIFFERENTIATION OF IMPLICIT FUNCTIONS:
Consider the function, y = x2 +ex − logx. Here y is expressed as a function of x. i.e., y = f (x). Such functions are called explicit functions. Now consider the function xy + log y + ex = 0. Here y is not expressed explicitly as a function of x. Such functions are called implicit functions. They are of the form f (x, y) = 0.
dy dy , differentiate the given function shift all the terms containing to Left Hand Side and dx dx dy dy common and shift the co-efficient of to Right the remaining terms to Right Hand Side. Take dx dx Hand Side.
To find
16.7 DIFFERENTIATION OF PARAMETRIC FUNCTIONS:
If both x and y are expressed as a function of another variable say t then the function y = f (x) is said to be in parametric form. The variable t is called a parameter. To find
17
Application of Derivatives
Derivatives have many applications. To mention a few, we use differentiation in finding 1. Equation of tangent and normal. 2. Length of subtangent and subnormal 3. Angle between the curves 4. Rate measure i.e., variation with respect to time 5. Maximum and minimum values of the function. Derivatives are also used in finding marginal cost and marginal revenue.
17.1 DERIVATIVE AS A RATE MEASURE:
Velocity: Rate of change of displacement is called velocity. If v is the velocity, 's' is the displacement and s = f (t) Then
Velocity =
ds = f′ t dt
af
Accelera Acceleration: The rate of change of velocity is called acceleration. If 'a' is the acceleration v is the velocity. Then
a= dv d ds d 2s = 2 = dt dt dt dt
F I H K
Note: 1. Rate means differentiation with respect to time, t. Rate of increase of area =
So car stops after travelling a distance of 3 mts. 3. With usual notation, if s = at + b. Where a and b are constants then prove that velocity is constant and acceleration is zero. Solution: Given s = at + b diff. w.r.t. t.
d 2s = 0 3 a is constant. dt 2
∴ acceleration = 0 4. A square plate expands uniformly, the side is increasing at the rate of 6 mm/sec. What is the rate of increase in area when side is 13 mm. Solution: Given Rate of increase of side = 6 mm/sec. i.e.,
dS = 6 mm sec. dt dA when S = 13 mm. dt
A = S2
To find:
We have Area of square = side × side
diff. w.r.t. t.
dA dS = 2S ⋅ dt dt dA = 2 13 6 dt
a fa f
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dA = 156 mm 2 sec. dt
∴ Rate of increase of area = 156 mm2/sec. 5. A stone is dropped into a pond, waves in the form of circles are generated and the radius of the outermost wave increase at the rate of 2 mm/sec. How fast is the area increasing (a) when the radius is 5 mm (b) after 3 sec. Solution: If r is the radius at time t; Then given: To find (a) ( b)
dA dr = π 2r ⋅ dt dt dA = π 2 × 6 ⋅2 dt after 3 seconds dA = 24π dt after 3 sec
∴ After 3 seconds, Area is increasing at the rate of 36 π mm2/sec. 6. Water flows into a cylindrical tank of radius 4 mts at 80,000 cc/hr. How fast water level is raising? Solution: If r is the radius, h is the height. The volume V is given by
7. Sand is poured at the rate of 30 cc/sec and it forms a conical pile in which the diameter of the circular base is always equal to one third the height. At what rate height of the pile is increasing when the height is 30 cm. Solution: 2 × Radius = Diameter
8. A spherical snow ball is forming so that its volume is increasing at the rate of 8 cm/sec. Find the rate at which the radius is increasing when the snow ball is 2 cm in diameter. Also find the rate of increase in surface area. Solution. Given: To find: and
9. A ladder 13 ft long rests with its ends on a horizontal floor and against a smooth vertical wall. If the upper end is coming downwards at the rate of 1 ft/min. Find the rate at which the lower end moves, when the upper end is 5 ft from the ground. Solution: Let PQ be the ladder. At time t, Let OP = y, OQ = x From figure OP2 + OQ2 = PQ2 P y2 + x2 = 132 Given To find
10. An aeroplane at an altitude of 400 kms flying horizontally at 500 km/hr passes directly over an observer. Find the rate at which it is approaching the observer when it is 500 kms away from him. Solution. Let at time 't' 'A' be the position of the A x B aeroplane, B be the point on the path directly above the observer O. Let AB = x. OA = y. From fig. AB 2 + OB 2 = OA 2
y 400 kms
y is said to be stationary at a point if it neither increases nor decreases. At such point
If a continuous function increases upto a certain value and then decreases from that value, then that value is called a maximum value of a function. Similarly if continuous function decreases to a value and then increases, then that value is called minimum value of the function. Note that a continuous function may attain maxima/minima at several points or it may neither have maxima nor minima. In the figure, the points P, Q, R, S, are points of maxima. The points A, B, C, are points of minima. Between any 2 maxima there must be a minima and vice versa.
Application of Derivatives
439
Y S
X
Fig. 17.6
DERIVATIVE TEST FOR MAXIMA AND MINIMA:
Let y = f (x) be the given function. To find the maximum and / or minimum value of the function. We find
So the numbers are x = 10 and y = 20 − 10 = 10. Reqd. Numbers are: 10 and 10. 3. Find two numbers whose sum is 16 and sum of their cubes is minimum. Solution: Let the 2 numbers be x and y. Given: Their Sum = 16.
x + y = 16
⇒
y = 16 − x
...(1)
To find: x and y such that sum of their cubes is minimum. i.e.,
x 3 + y 3 Should be minimum.
442
Basic Mathematics
Let
P = x 3 + y3 P = x 3 + 16 − x
a
f
3
[from (1)]
For P to be extremum,
dP =0 dx
Now
P = x 3 + 16 − x
diff. w.r.t. x.
a
f
3
dP = 3x 2 + 3 16 − x dx dP = 3x 2 + 3 16 − x dx
2
a
f
2
⋅
d 16 − x dx
a
f
a f a−1f 0 = 3 x − 3 a16 − x f 0 = 3 x − d16 + x − 32 x i
2 2 2 2 2
0 = x 2 − 256 − x 2 + 32 x
0 = −256 + 32 x
32 x = 256
x=
256 16 = = 8. 32 2
So the numbers are x = 8 and y = 16 − 8 = 8 i.e. 8 and 8. 4. Prove that logx do not have maxima or minima. Proof: Proof: Let y = log x. For maxima or minima,
dy =0 dx y = log x dy 1 = dx x 0= 1 x
1 =0 x
Application of Derivatives
443
⇒ x has no finite value. ∴ logx do not have maxima or minima. 5. The product of 2 positive numbers is 36. Find the minimum value of their sum. Solution: Let the 2 number be x and y. Given: Their product = 36.
xy = 36
⇒ Their sum = x + y.
y=
36 x
S=x+ dS =0 dx S=x+
36 x
For S to be extremum
36 x
diff. w.r.t. x.
dS −1 = 1 + 36 2 dx x 0 = 1− 0=
⇒
F I H K
36 x2
x 2 − 36 x2
x 2 − 36 = 0 x 2 − 36 ⇒ x = ± 6 dS 36 =1− 2 dx x
diff. w.r.t. x.
Now
d2S = −36 −2 x −3 dx 2 d 2 S 72 = dx 2 x 3
d
i
72 d2S = 3 > 0 → positive. 2 dx at x = 6 6
444
Basic Mathematics
∴ x = 6 is a point of minima.
y=
∴ Minimum value of their sum = 6 +
36 36 = = 6. x 6
36 6
= 6 + 6 = 12.
6. Prove that a maximum rectangle that can be drawn with a constant perimeter is a square. Proof: Proof: Let x be the length and y be the breadth of a rectangle. Given: Perimeter = constant. Sum of all sides = constant.
x + y + x + y = 2k say 2 x + 2 y = 2k 2 x + y = 2k
x+y=k
a f
D
x
C
a
f
y
y
A
x
B
y=k−x
Area of rectangle = Length × Breadth
A = xy A=x k−x
Fig. 17.7
a
f
A = kx − x 2
For area to be extremum,
dA =0 dx
A = kx − x 2
diff. w.r.t. x.
dA = k 1 − 2x dx
0 = k − 2x
af
⇒
2x = k
x=
k 2
Consider
dA d2A = k − 2 x diff. again w.r.t. x, = −2 < 0 for all dx dx 2
values of x.
Application of Derivatives
445
∴ A attains maxima for all values of x. Now,
y=k−x=k−
k 2k − k k = = 2 2 2 k 2
Hence Length Breadth
x=y= 2x = 2 ⋅ 2y = 2 ⋅
k =k 2 k =k 2
∴ Rectangle has Length = Breadth. Hence it is a square. 7. Prove that the maximum rectangle that can be inscribed in a circle is a square.
D
x
C
y
y
A
x
B
Fig. 17.8
Let ABCD be the rectangle with length x and breadth y inscribed in a circle of radius say a. ∴ ∴ From right angled triangle, ABD,
OB = OD = a BD = 2 a
∴ Rectangle has maximum area when its length = 2 a and Breadth = 2 a , i.e., when its length = Breadth = 2 a , i.e., when it is a square. 8. What is the largest size rectangle that can be inscribed in a semicircle of radius r unit so that 2 vertices lie on the diameter. Solution: Let AB be the diameter of a semicircle. S R 2x Let 2 vertices P and Q of the rectangle PQRS lie on the diameter. Let
Maximum area = Length × Breadth =r 2× = r2.
9. A box is constructed from a square metal sheet of side 60 cm by cutting out identical squares from the four corners and turning up the sides. Find the length of the side of the square to be cut out so that the box is of maximum volume.
r 2
Application of Derivatives
449
Solution: Let the side of the cut out square be x. Now length of the box = 60 − 2x Breadth of the box = 60 − 2x Height of the box = x. Volume = Length × Breadth × height.
d2y > 0, then x = a is a point of minima. Minimum value of the function is y = f (a). dx 2 at x = a
if
d2 y is less than zero, x = b is a point of maxima. Maximum value of the function is dx 2 at x = b y = f (b).
d2y if is equal to zero, then x = c is called point of inflection. At x = c the function neither dx 2 at x = c attains maxima nor minima.
EXERCISE
1. The distance s is metres moved by a particle in 't' seconds is given by s = 45t + 11 t2 − t3. Find the time when the particle comes to rest? 2. The displacement s of a particle at time 't' seconds is given by 2t3 − 3t2 − 36 t + 90. Find the (a) velocity after 4 seconds (b) displacement and acceleration when the velocity vanishes (c) acceleration after 4 seconds. 3. With usual notation if. If s2 = at2 + 2bt + c, then prove that (a) the acceleration is inversely proportional to s3. (b) the acceleration is
a − v2 where v is the velocity. s
4. The equation of motion of a particle is given by s = 9t2 − t3. Find the displacement when velocity is zero and velocity when the acceleration is zero. 5. If s = at3 + bt, find a and b given that when t = 3 velocity is zero and acceleration is 14 units. 6. When breaks are applied to the moving car, the car travels a distance S feet in t seconds given by s = 20t − 40 t2. When does the car stop? 7. The side of a square sheet metal is increasing at 3 mm/min. At what rate is the area increasing when the side is 10 mm long. 8. A circular patch of oil spreads on water, the area is growing at the rate of 2 sq cm/hr. How fast are the radius and the circumference increasing when the diameter is 24 cm. 9. A drop of ink spreads over a blotting paper so that the circumference of the blot which is circular increases at the rate of 3 cm/min. Find the rate of increase of the radius and area when its circumference is 4π cm. 10. A stone is dropped into a pond, waves in the form of circles are generated and the radius of the outermost ripple increases at the rate of 2 mm/min. How fast is the area increasing when the radius is 5 mm, after 5 min? 11. A cylindrical tank is 10 mts in diameter, water is flowing in it at the rate of 24 m3/min. at what rate height of the water is rising?
Application of Derivatives
453
12. Water is flowing into a right circular cylindrical tank of radius 50 cm at the rate 500 π cc/min. Find how fast is the level of water rising? 13. A ladder 20 ft long rests with its ends on a smooth horizontal floor and against a smooth vertical wall. If the lower end is moved at the rate of 5 ft/min. Find the rate at which the upper end moves when the lower end is 12 ft. from the wall. 14. The radius of the sphere is decreasing at the rate of 3 cm/sec. Find the rate at which surface area is decreasing when radius is 12 cm. 15. The height of circular cone is 30 cm and it is constant. The radius of the base is increasing at the rate of 0.25 cm/sec. Find the rate of increase of volume of the cone when the radius is 10 cm. 16. A man 6 ft tall is moving directly away from a lamp at a height of 10 ft above the floor. If he is moving at the rate of 6 ft/sec find the rate at which the length of his shadow is increasing? 17. Find the maximum and minimum value of the function 4x3 − 15x2 + 12 x + 7. 18. Find the maximum and minimum value of the function x3 − 3x2 − 9x + 17. 19. Prove that the function xe−x attains maxima at x = 1 and its maximum value is 20. Prove that xx is minimum when x =
Prove that x3 − 3x2 − 9x + 9 has a point of inflection at x = 1 Prove that y = a−(x − b)2/5 has no point of inflection. Find two number whose sum is 24 and their product is maximum. The sum of two numbers is 20. If the product of square of the first and cube of the 2nd is maximum. Find the numbers. Prove that the area of the right angled triangle of given hypotenuse is maximum when the triangle is isoceles. Prove that among all rectangles of a given area the square has minimum perimeter. Prove that among all rectangles of given perimeter, the square is the one with shortest diagonal. What is the largest rectangle that can be inscribed in a semicircle of radius 1 so that two vertices lie on the diameter. A rectangular field is to be fenced off along the bank of a river, no fencing is required along the river. If the material for fence cost Rs. 8 per running foot for 2 ends and Rs. 12 per running foot for the side parallel to the river, find the dimension of the field of largest possible area that can be enclosed with Rs. 3600 worth of fences.
18.5 INTEGRATION BY PARTS:
From product rule in differentiation we have
d dv du uv = u ⋅ +v⋅ dx dx dx
Integrating w.r.t. x. we get
a f
z
d dv du uv = u dx + v dx dx dx dx uv = u dv + v du. v du dx dx
⇒ ⇒ By taking u as first function,
z z z z z z FH IK
a f
udv = uv −
dv as second function du as differential of 1st function v as the integral of the 2nd function we get
Integration
475
za
I function II function dx = I function
fa
f
a
f aII functionf dx − LMN
z
zz
II function ⋅
d I function dx dx
a
fOPQ
This is known as integration by parts. The success of the method of integration by parts depends on choice of 1st function and 2nd function. Logarithmic, algebraic and exponential functions should be taken in the same order of priority. [LIATE: log, Inverse trig., Algebraic, Trigonometric and Exponential] For example, to integrate xex, x is first function and ex is 2nd function where as to integrate xlogx, logx is first function and x is 2nd function.
definite integral of f (x) from a to b. Here a is called lower limit and b is called upper limit low limit. To evaluate definite integral, integrate the given function as usual. In the final answer substitute the upper limit value for x − lower limit value for x.
19.3 APPLICATION OF DEFINITE INTEGRALS TO FIND AREA:
One of the application of definite integrals is to find area. (i) Area enclosed by the curve y = f (x), the xaxis and the lines x = a and x = b is given by
A = y dx =
a
• Area enclosed by the curve y = f (x), X-axis and the lines x = a and x = b is given by
A = y dx =
a
z z af
b b a
f x dx .
EXERCISE
I. 1. Evaluate:
3.
5.
z zd z FGH
b a 3 1 4 1
a x + 1f dx
3 x − 2 x + 1 dx
2
2.
i
4.
1 x+ dx x
IJ K
6.
zd za zd
2 1 1 0 2 0
x 2 − 1 dx
i f
2 x − 1 dx
8
x 3 + 2 x 2 − x + 1 dx
i
502
Basic Mathematics
7.
9.
11.
13.
15.
17.
19.
zd zd za z za zd z
1 −1 2 −1 1 0 1 0 2 1 1 0 1 −1
x + 4 x − 2 dx
3
i
8.
x 3 + 3 x 2 + x dx
i
10.
x x + 1 x − 1 dx
fa f
12.
x+2
x2 + 4x + 3
dx
14.
log x x
f
2
dx
16.
1+ e
−x
i dx
18.
x 2 e − x dx
20.
z FH zd z a fa z z za f z
2
x2 +
1
1 − x dx x2
I K
2
6 x −5 + 2 x 4 + 4 dx
i
1 1
x + 1 x − 2 dx
f
0 e
0 e
1 dx x 1 + log x
log x dx
1 1
x2 1 − x
5 2
0 1
0
x dx x +1
21.
23.
e2
25.
za z za
0 −1 5 2
dx 1− x x + 2
fa
f
22.
dx 2 x + 4x + 3
24.
z z
1 0 1 0
x dx x +1
2
x3 dx e2 x
log x dx
f
2
e
II. 1. Find the area bounded by the curve y = x3, the x-axis and the lines x = 1 and x = 2. 2. Find the area bounded by the curve y = x2 − 4x, x-axis and the lines x = 1 and x = 3. 3. Find the area enclosed by the curve y =
x x-axis and ordinates x = 0 and x = 1. x +1
Definite Integrals
503
4. 5. 6. 7. 8. 9. 10. 11. 12.
Find the area bounded by the x-axis and the curve y = x2 − 7x + 10. Find the area bounded by the curve y = 4x − x2 − 3 with x-axis. Find the area bounded by the curve y = 4x − x2 and x-axis. Find the area between the curves y2 = 4x and y = 2x. Find the area between the curves y2 = 2x − 2 and the line y = x − 5. Find the area between the parabolas y2 = 4x and x2 = 4y. Find the area between the parabola y2 = 4ax and the line y = 2x. find the area between x2 = y and y2 = 8x. Find the area between the curves y = 11x − 2y − x2 and y = x.
ANSWERS
1.
a ab − af LMN b + 2 + 2 OPQ 28 3 4 3
7. −4
8− 3
2.
4 3 4 3
3. 20
4.
1 9 365 32
5.
20 3 1 4 1 e
6.
8.
9.
57 4
10.
3
11. −
12. −
13.
14. 2
d
2 −1
i
15.
alog 2f
3
16. −1
17. 2 −
18.
16 693 −19 3 + 8e 2 8
19.
e2 − 5 e
20. 1 − log 2
21.
2 log 2 3
22. log 2
23.
1 5 log 2 4
24. II. 1. 4. 7. 10.
2 25. 2e − e
15 sq units 4 9 sq units 2 1 2 a sq units 3 1 2 a sq units 3
2. 5.
22 sq units 3 4 sq units 3
3. 1 − log e 2 sq units 6. 9. 12.
b
g
64 sq units 3 16 sq units 3 4 sq units 3
8. 18 sq units 11.
8 sq units 3
20
Application of Calculus in Business
20.1 TERMINOLOGY:
Cost Function: The outflows usually raw materials, rent, utilities, pay of salaries and so forth form the total cost. It is sum total of all costs. Economists and accountants often define total cost as sum of 2 components. Total variable cost and total fixed cost. Total variable cost varies with the level of output. (Eg.: raw materials) whereas total fixed cost remains the same (for example rent). Hence Total cost = Variable cost + Fixed cost era Average Cost: Average Cost is the Cost per unit of the output. It is obtained by dividing total cost by the total quantity produced. If TC is the total cost of producing x units or q units, then average cost AC is given by, AC =
TC TC or . x q
Marg Marginal Cost: Marginal cost is the additional cost incurred as a result of producing and selling one more unit of the product. If TC is the total cost of producing x units or q units then the derivative
d d TC or TC repredx dq
sents the instantaneous rate of change of total cost, given a change in number of units (x or q) produced. ∴ Marginal Cost, MC =
d d TC or TC . dx dq
Revenue Revenue function: The money which flows into an organisation from either giving service or selling products is called as revenue. The most fundamental way of computing total revenue from selling a product is Total revenue = Price per unit × Quantity sold. If TR is the total revenue p is the price per unit and q or x is the quantity sold then TR = pq or TR = px.
Application of Calculus in Business
505
Marg re enue: Marginal revenue: It is the additional revenue derived from selling one more unit of a product. It is obtained by differentiating total revenue with respect to quantity demanded. If MR is the marginal revenue. TR is the total revenue and x or q is the No. of units produced. Then
MR =
d TR dx
a f
or
MR =
d TR . dq
Prof ofit Profit function: Profit for an organisation is the difference between total revenue and total cost. If TR is the total revenue and TC is the total cost then Profit = TR − TC. For many production situations, the marginal revenue exceeds the marginal cost, at lower level of output. As the level of output increases, the amount by which marginal revenue exceeds marginal cost becomes smaller. Eventually a level of output is reached at which marginal revenue = Marginal cost. Beyond this point marginal revenue is less than marginal cost and the total profit begins to decreases with added output. So from theoretical stand point, the profit maximization level of output can be identified by the criterion. Marginal revenue = Marginal cost. i.e., MR = MC. i.e.,
d d TR = TC dx dx
or
a f
d d TR = TC dq dq
Note: We know if
a f
d f x =g x dx
Then
By applying this to,
z af
af af
g x dx = f x + c where c is the constant of integration.
af
d TR = MR and dx d TC = MC dx
We get
So if we know marginal cost/marginal revenue (MC/MR) we can calculate, total cost or Total revenue (TC/TR) by using
za
MR dx = TR and
f
za
MC dx = TC.
f
506
Basic Mathematics
TC = MC dx or MR dq TR = MR dx or MR dq
z z
z z
WORKED EXAMPLES:
1. If the total cost function C(x) of a firm is given by C(x) = x3 − 3x + 7. Then find the average cost and marginal cost when x = 6 units. Solution: Given c(x) = x3 − 3x + 7. We have Average Cost = Average Cost =
5. The marginal revenue (in thousands of rupees) function for a particular, commodity is 4 + e−0.03x, where x is denotes the number of units sold. Determine the total revenue from the sale of 100 units given that e−3 = 0.05 (Approx). Solution: We know Total revenue =
Total revenue from the sale of
100
za
Marginal revenue dx
f
100 units =
100
=
za zd
0 0
MR dx
f
4 + e −0.03 x dx
i
e −0.03 x = 4x + −0.03
100
0
= 4 100 +
LM a f N
e −0.03×100 e0 − 4 0 + − 0.03 − 0.03
OP Q
LM a f N
OP Q
Application of Calculus in Business
509
= 400 −
LM N
e −3 1 − − 0.03 0.03
OP L Q MN
OP Q
400 −
∴
0.05 1 + = 400 − 1.66 + 33.33 = 431.667 0.03 0.03
Total revenue = 431.667 thousands of rupees. = Rs. 4,31,667 6. Find the total cost of producing 1000 electric bulbs if the marginal cost (in Rs. per unit) is given by C ′ x =
9. A T.V. manufacturer produces x sets per week at a total cost of Rs x2 + 1560x + 50,000. He is a monopolist and the demand function for this product is x = per set. What is the monopoly price in order to maximise the profit? Given: Total cost = x2 + 1560 x + 50,000 diff. w.r.t. x.
EXERCISE
1. If the marginal cost of a product is 3x2 − 4x where x is the output, find the total cost of producing 10 units. 2. If the marginal revenue function is given by
2 , find the total revenue function. 2x + 1
3. A company has revenue function given by R = 100q − q2. Find the marginal revenue function (where q is the output). 4. Find the marginal revenue for the demand function 3x − x2. 5. Find the average cost function for the marginal cost function x2 + 2x. 6. The demand function of a firm is given by P = 50 − 2x. Where P is the price per unit for x units. Determine the marginal revenue. 7. The cost function C = 5 +
48 + 3 x 2 where x is the number of articles produced. Find the minix
mum value of C. 8. If the marginal cost is given by x2 + 7x + 6 and fixed cost is Rs 2500, determine the total cost of producing 6 units. 9. If C is the total cost for producing x units of a product and the average cost function is given by 0.003 x2 − 0.04 x + 6 +
3000 , find the marginal cost function. x
514
Basic Mathematics
10. The total cost function of a firm is given by C(x) = 2x3 − x2 + 5x. Find the average and marginal cost function. 11. If the marginal revenue is given by 30 −
x3 , then find the total revenue function. 30
1 3 x . Where C is the total cost for 3 x units. Calculate the output at which the average cost is minimum. 13. The unit demand function is x = 25 − 2p where x is the number of units and p is the price. Let the average cost per unit be Rs 20. Find the price per unit that maximises the profit function. 14. The demand function of a company is given by P(q) = 800 − 0.4q and the total cost function is given by C(q) = 80 q + 8000 where q is the level of output,P is the price per unit. Find the level of output and the price charged which maximises the profit. 15. Find the total revenue, marginal revenue, marginal revenue and average revenue when the demand function is given by Q = 30 − 4P + P2 where P is price and Q is the quantity demanded. Also calculate the marginal revenue when P = 3.
12. The cost function of a firm is given by C x = 300 x − 10 x 2 + 16. A manufacturer has a total cost function C = 100 + 32 x . He would like to know at what level of output his marginal cost will be Rs. 2.00. 17. If the total cost function is C = 8x3 − 12x2 + 20x where x is the level of output, find x at which the average cost is minimised find the total cost for this output. 18. The marginal cost function of a firm is 150 − 10x + 0.2x2, where x is the output, find the total cost function, if the fixed cost is Rs 750, what is the average cost? 19. If the total revenue (R) and total cost (C) functions are given by R = 30x − x2 and C = 20 + 6x, find x if marginal revenue = marginal cost. Also find average cost and average revenue. 20. If the marginal revenue is given by 15 + 30 units.
10. Total cost of a commodity is given by C = 1/3 x3 − x2 + 3x + 10. Find the marginal cost. II. Answer any ten of the following: 11. If p is 2 is prime, q is 3 is even, r is
12. Find the number of words that can be formed using all the letters of the word "BOOKS". How many of them begin with 'B'. 13. There are 10 points out of which 3 are collinear. Find the number of straight lines that can be formed. 14. Two dice are thrown simultaneously. What is the probability of getting the sum 7? 15. Expand 2 x −
26. Prove that if each element of a row (or column) constant multiples of corresponding elements of other rows (or columns) are added, then the value of the determinant is unaltered. IV Answer any three of the following: . 4 × 3 = 12 27. If two men and four women can do a work in 33 days and 3 men and 5 women can do the same work in 24 days, how long shall 5 men and 2 women take to do the same work? 28. Calculate the total wages earned per week by 400 workers in a factory from the following data Daily wages 12.50 – 17.50 17.50 – 22.50 22.50 – 27.50 27.50 – 32.50 32.50 – 37.50 37.50 – 42.50 42.50 – 47.50 47.50 – 52.50 Number of Workers 51 38 42 59 80 60 50 20
29. A bill for Rs. 2725.75 was drawn on 03.06.1997 and made payable 3 months after due date. It was discounted on 15.06.1997 at 16% p.a. What is the discounted value of the bill and how much has the banker gained in this transaction. 30. A business men buys and sells chairs and tables. He has Rs. 3000/- to invest. A chair costs him
Examination Corner
519
Rs. 50/- and a table costs him Rs. 90/-. He has space which can accommodate at most 48 pieces. If he sells each chair for Rs. 200 and each table for Rs. 400/-. Find the number of chairs and tables he has to buy to obtain maximum profit. V . 31. Find the equation of the circle passing through the origin and having centre at (−3, 4). OR Derive the equation of a parabola in the form y2 = 4ax. VI. Answer any three of the following: 32. Prove that Lt 4
VII. Answer any two of the following: 10 × 2 = 20 36. (a) Two number are in the ratio 4:7. If 12 is added to each of them then the new ratio is 8:11. Find the numbers. (b) In how many ways 3 boys and 5 girls be arranged in a row so that (i) no two boys are together. (ii) all the girls are together. (c) Solve by matrix method. 2x + y + z = 7. x − y + 2z = 5 3x − 2y + 2z = 5. 37. (a) Find the equation of the circle with (−3, 5) and (6, 1) as the extremities of a diameter. (b)/- per hour. (c) The radius of a circular blot of ink is increasing at the rate of 3 cm per minute. Find the rate of increases of its area when its radius is 2 cms. What is the rate of increases of its circumference? 38. (a) Find the range in which the function f (x) = x2 − 6x + 5 is (1) increasing, (2) decreasing. (b) Find the area bounded by the curve y = x2 − 7x + 10 with x-axis.
z
1 d2y =− . 2 at 3 dx 2
e x x 3 dx.
2
520
Basic Mathematics
1 (c) Prove without expanding 1 1
a b c
a2 b2 = a − b b − c c − a . c2
a
fa fa f
***
MODEL QUESTION PAPER 2:
I. Answer all questions: 1 × 10 = 10 1. If the truth value of p and q are T and F respectively. Then what will be the truth value of
a p ∧ qf → a ~ p∨ ~ qf .
LM2 Nx OP Q
2. Find the number of permutation of the letters of the word BALLOON taken all together. 3. Find the mean proportional to 2:8. 4. Find x if 5. 6. 7. 8.
7 is singular. 14
A bill was drawn on 14.3.1997 for 3 months. When does the bill fall legally due? Find the yield by investing Rs. 1140 on 15% stock quoted at Rs. 95. Write any one step to formulate an LP problem. Find the radius of the circle 4x2 + 4y2 + 8x + 9y − 7 = 0.
x →1
9. Evaluate: lim
d
3
x −1
id
4
x −1
i f
2 × 10 = 20
10. If the marginal revenue function is given by 3x2 − 8, then find the total revenue function. II. Answer any ten of the following: 11. Construct the truth table for p∧ ~ q → ~ p . 12. How many four digits numbers greater than 2000 can be formed with the digits 0, 1, 2, 3, 4, 5. 13. Find the number of permutations of the letters of the word 'BANANA' taking all letters. How many of them begin with N. 14. Two cards are drawn at random from a pack of well shuffled 52 cards. What is the probability of getting king and queen cards. 15. Expand using binomial theorem: 3 +
2 x 2 + 3x + 2 . x2 − x − 2
26. If the elements of any row (or column) is multiplied by non zero constant K. Then prove that the value of the determinant is multiplied by K. IV Answer any three of the following: . 27. 16 men or 28 boys can fence a farm in 40 days. In how many days will 24 men and 14 boys complete the same work. 28. 3 tests in English, 2 in Hindi, 4 in Kannada and 5 in Sociology are conducted. The average marks scored by Rashmi in English is 60, that in Hindi is 56 and that in Kannada is 45. If the average marks of all the subjects and all the tests taken together is 48. Then find the averages marks scored by her in Sociology. 29. The difference between the bill discount and true discount on a certain sum of money due in 4 months in Rs. 10. Find the amount of the bill (or face value) if the rate of interest is 3% p.a. 30. Maximise Z = 7x + 4y subject to x + 3y ≤ 3, 6 x + 3 y ≤ 8 ; x ≥ 0, y ≥ 0. . V . 31. Derive the equation of circle which is described on the line joining (x1 y1) and (x2 y2) as the ends of diameter. OR Find the equation of parabola whose ends of latus rectum are (3, 2) and (3, −4). VI. Answer any three of the following: 32. Evaluate: (i) lim 4 ×3 = 12 (ii) lim 1 + 3 x
n →∞
37. (a) A circle with centre (2, −1) passes through (−1, 3). Find the equation of the circle. (b) The first sample batch is of 50 units of product. A took 80 hours to make. The company now wishes to estimate the total time taken to make 100 additional units. Solve the problem using 80% learning effect. (c) If the base of the triangle is 3 times the height and the height is decreasing at the rate of 4 cm/ sec. then find the rate of decrease of the area when height is 1 cm. 38. (a) Find 2 numbers whose sum is 20 and whose product is maximum. (b) Find the area bounded by the curve y2 = 4x and the line y = x.
a2 (c) Prove without expanding − ab ac ab −b 2 bc ac bc = 4 a 2 b 2 c 2 . −c2
***
MODEL QUESTION PAPER 3:
I. Answer all questions: 1 × 10 = 10 1. Write the contrapositive of 'If the triangle is not right angled then its two sides are not perpendicular'. 2. If nc2 = 105, then find n. 3. Find x and y if 2
r and s. 12. In how many ways can 3 girls and 4 boys sit round the table so that no two girls sit next to each other. 13. Find the number of triangles that can be formed from 12 points of which 4 are given to be collinear. 14. 2 dice are thrown simultaneously what is the probability of getting the sum 8.
26. A salesman has the following records of sales during 3 months for 3 items A, B, C which has different rates of commission.
Months A
Jan Feb Mar 90 130 60
Sales of units B
100 50 100
Find out the rates of commission on items A, B and C. IV Answer any three of the following: . 4 × 3 = 12 27. In a mixture of 35 litres, the ratio of milk and water is 6:1. If 2 litres of milk and 3 litres of water are added to the mixture, what is the new ratio of milk and water? 28. The average age of A and B is 18 yrs., that of B and C is 17 yrs. and that of C and A is 20 yrs. Find the ages of A, B and C. 29. The banker's gain on a certain bill due 6 months hence is Rs. 20. The rate of interest being 20% per annum. Find the face value of the bill. 30. A company produces 2 products x and y. Each of the product require two operations one on machine A and other on machine B. The machine hours required by these two products and the total hours available are given as follows:
Machine hrs. required
Product —–––––––———–––––––––––––––– x y
2 4 5 3
Total machine hrs. available
A B
19 17
Each unit of the product x and y makes a profit of Rs. 3 and Rs. 4. Find the optimal solution of the product to be produced to obtain the maximum profit. V . (4) 31. Find the equation of circle passing through (4, 0) and (0, 4) and its centre lies on 3x + 4y = 7. OR
(b) What do you mean by 80% learning effect? Find the index of learning for 80% learning effect. (4) (c) A ladder 5 mts. long rests with its ends on a horizontal floor and against a smooth vertical wall. If the upper end is coming downwards at the rate of 40 cm/sec. Find the rate at which the lower end moves when the upper end is 3 mts. from the ground. (4) 2/sec. Find the rate at which the perimeter 38. (a) The area of square is increasing at the rate of 2 cm is increasing when the side of the square is 16 cm. (2) (b) Find the area between the parabolas.
x 2 = 4 y and y 2 = 4 x.
(c) Prove without expanding
(4)
c 1+ a b a 1 + b c = 1 + a + b + c. a b 1+ c
***
(4)
526
Basic Mathematics
CHAPTERWISE ARRANGED QUESTION BANK 1. MATHEMATICAL LOGIC:
I. One mark questions (VSA): 1. Define a proposition. 2. Give the truth value of '4 + 3 ≥ 8'. 3. If p : 6 is an even number, 'q : 4 is odd', then write the true value of p ∨ q. 4. Write in symbols: The necessary and sufficient condition for a triangle to be equilateral is all its sides should be equal. 5. Write symbolically: '6 is even and 2 is not irrational'. 6. Negate: 'Cow is not big or it is black.' 7. If the truth values of p and q are T and F respectively then what will be the truth value of
a p ∧ qf → a p ∨ qf .
8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
If the truth values of p, q, r are T, T, F respectively, find the truth value of p → (~r → q). If the truth value of p → ~r is F, then find truth value of r. Write the truth table for p ∧ ~p. Write the truth table for p ∨ ~p. Write the converse of 'If a + 2 = 3, then a = 1.' Write the converse of 'If cows can fly then birds cannot fly'. Write the inverse of 'If he is rich then he is happy.' Write the contrapositive of 'If e is not irrational then 2 is rational'. Construct the truth table for p → ~p. Negate: 'If 2 is even then 3 is odd'. Negate: p → ~q. Negate: 'If 6 is even and 7 is odd, then 2 is prime.' Prove that p ≡ ~(~p).
II. Two marks questions (SA): 1. Construct the truth table for ~p ∧ q. 2. Construct the truth table for p → (~q ∧ p). 3. Negate: 'If a triangle is equilateral then its sides are equal and angles are equal.' 4. Negate: If x + y ≠ 6 and x ≠ 5, then y ≠ 0 or y ≠ 7. 5. Write the inverse and converse of: If a triangle is equilateral then its sides are equal and angles are equal. 6. Write the converse and contrapositive of 'If a number is real then it is either rational or irrational.' 7. Prove that ~[p → q] ≡ p ∧ ~q.
f a f 3. Prove that a p ∧ qf ∨ a p ∨ qf ↔ q is neither tautology nor contradiction.
4. Prove that the negation of disjunction of 2 proposition is logically equivalent to the conjunction of their negation. 5. Prove that
8. Define converse and contrapositive of a conditional and prove that a conditional and its contrapositive are logically equivalent. 9. Prove that the converse and inverse of a conditional are logically equivalent. 10. Prove that contrapositive of a conditional is converse of inverse of a conditional.
2. PERMUTATION, COMBINATION AND PROBABILITY:
I. One mark questions (VSA): 1. If nc4 = nc6, then find nc7. 2. Find the number of permutations of the letters of the word BOOKS, taking all letters. 3. How many of the arrangements of the word 'ABACUS' begin with A. 4. If nc2 = 105, find n. 5. If np2 = 30, find n. 6. If np5 = 24 nc4, then find n. 7. How many of the arrangements of the word 'CRICKET' begin and end with C. 8. State addition rule of probability for mutually exclusive events.
528
Basic Mathematics
9. A ticket is drawn from a bag containing tickets bearing numbers 1 through 25. Find the probability of its bearing a number which is a multiple of 3. 10. One card is drawn from a pack of 52 cards. What is the probability of getting a king card? II. Two marks questions (SA): 1. In how many ways 4 boys and 3 girls may be arranged for a photograph so that all the 3 girls are shown together? 2. In how many ways can 5 members forming a committee out of 10 be selected so that 2 particular members must be included? 3. Six students have taken an examination. In how many ways can the first 3 positions be declared? 4. In how many ways can the letters of the word HEXAGON be arranged so that H appears in the beginning. 5. There are 5 maths books, 4 accounts books and 3 economics books. In how many ways can the books be arranged so that the books of the same subject are together. 6. How many words can be formed by using all the letters of the word 'ALLAHABAD'? 7. 20 persons were invited to a party. In how many ways can they and host be seated at a circular table? 8. There are 12 points in a plane. 4 points are collinear. Find how many straight lines can be drawn? 9. A box contains 7 red, 6 white and 4 blue balls. How many selections of 3 balls can be made so that none is red. 10. How many diagonals are there in an octagon? 11. Three coins are tossed. What is the probability of getting. 1. Exactly 2 heads. 2. at least 2 heads. 12. A pair of dice is thrown. What is the probability that the sum of the numbers obtained is more than 10. 13. A card is drawn from a pack of 52 cards. What is the probability that it is either king card or a red card. 14. A pair of dice is rolled. If the sum on the 2 dice is 9, find the probability that one of the dice is showed 3. 15. Two dice are thrown, find the probability of a doublet. 16. Prove that
0 =1
III. Four marks questions (ET): 1. A student has to answer 8 out of 10 questions in an examination. How many choices has he? How many choices has he if he must answer the first 3 questions. 2. Prove that
n
pr =
n . n−r
Examination Corner
529
n 3. Prove that cr =
n . n−r⋅ r
4. Prove that ncr = ncn − r and hence find n if nc8 = nc4. 5. Prove that ncr + ncr − 1 = n + 1cr and verify this for n = 5 and r = 3. 6. A committee of 10 members is to be chosen from 9 teachers and 6 students. In how many ways this can be done if (1) the committee contains exactly 4 students. (2) there are atmost 7 teachers. 7. How many 4 digits numbers greater than 3000 can be formed with the digits 1, 2, 3, 5, 7 (repetitions of digits is not allowed). How many of them are even? 8. How many (i) Straight lines (ii) Triangles are determined by 12 points. No three of which lie on the same straight line. 9. There are 5 questions in part A and 4 in part B of a question paper. In how many ways can student answer 6 questions if he has to choose at least 2 from each part. 10. A man has 7 relatives. 4 of them are ladies and 3 are gentleman. His wife also has 7 relatives 3 of them are ladies and 4 are gentlemen. In how many ways can they invite 3 ladies and 3 gentleman to a dinner partly so that there are 3 of man's relatives and 3 of wives' relatives. 11. State and prove addition rule of probability. 12. What is the probability of getting neither 8 nor 11 when a pair of dice is tossed? 13. State and prove multiplication rule of probability. 14. Out of the numbers 1 to 120, one number is selected at random. What is the probability that it is divisible by 8 or 10. 15. A card is drawn from a pack of 52 cards. What is the probability that it is neither a red card nor a jack card.
Write the row matrix which represents the belongings of Rakshith. 15. A company sold 22 chairs, 15 tables in January and 16, 28 respectively in June. Represent the data in matrix form. 16. A business organisation has 2 distribution outlets. At the beginning of the month, the stock on hand of products A, B and C are given in matrix M and the quantities of A, B and C sold during the month are given in matrix N. What are the stocks on hands (in matrix form) at the end of the month?
represent? 13. Ajith buys 5 kgs of tomato, 2 kgs. of Beans and 8 kgs of carrot. If the cost of each is Rs. 5, Rs. 7 and Rs. 9 respectively, then find the total cost by matrix method. 14. A company sold 40 metal chairs, 20 wooden chairs and 10 plastic chairs. The selling price of a metal chair is Rs. 150 that of wooden chair is Rs. 500 and plastic chair is Rs. 300. Find the total revenue using matrix method. 15. Find the adjoint of
7. Find the missing term: x:4 = 27:12. 8. If a:b = 2:3, b:c = 4:5, then what is c:a? 9. If Ramu can do a piece of work in 8 days and Rithu can do the same work in 3 days, then in how many days, both together can do the same work? 10. Divide Rs. 180 in the ratio 1:2:3. II. Two marks questions (SA): 1. Two number are in the ratio 4:7. If 12 is added to each of them then the new ratio is 8:11. Find the numbers. 2. The prices of a scooter and a T.V. are in the ratio 3:2. If the scooter costs Rs. 6000 more than the T.V., then find the cost of TV. 3. A certain amount was divided between Kavitha and Reena in the ratio 4:3. If Reena's share was Rs. 2400, then find the amount divided. 4. What number should be added to each one of 6, 14, 18, 38 to make it equally proportionate. 5. Three friends divide Rs. 624 among themselves in the ratio friend. 6. Gold is 19 times as heavy as water and copper 9 times as heavy as water. Find the ratio in which these two metals be mixed so that the mixture is 15 times as heavy as water. 7. The ages of Vivek and Sumit are in the ratio 2:3. After 12 years their ages will be in the ratio 11:15. Find the age of Sumit. 8. If 60 men can complete a job in 12 days, how many days will 36 men take to complete the same job? 9. Find 3 numbers in the ratio 2:3:5, the sum of whose squares is 608. 10. Cycling
1 1 1 : : . Find the share of the third 2 3 4
3 of his usual speed, a student is 10 minute late to his class. Find his usual time to cover 4 the distance.
III. Four marks questions (ET): 1. If 2 men and 4 women can do a work in 33 days and 3 men and 5 women can do the same work in 24 days. How long shall 5 men and 2 women take to do the same work. 2. A can do a piece of work in 6 days and B alone can do it in 8 days. A and B under took to do it for Rs. 320, with the help of C, they finished it in 3 days. How much is paid to C? 3. 12 men or 18 women can reap a field in 14 days. In how many days 8 men and 16 women can reap it? 4. 4 men and 6 women finish a job in 8 days while 3 men and 7 women finish it in 10 days. In how many days 10 women working together will finish it? 5. 5 carpenters can earn Rs. 3600 in 6 days working 9 hours a day. How much will 8 carpenters earn in 12 days working 6 hours a day?
Examination Corner
537
6. Two trains start at the same time from Aligarh and Delhi and proceed towards each other at 16 km/hr and 21 km/hr respectively, when they meet, it is found that one train has travelled 60 km more than the other. Find the distance between 2 stations. 7. Kanthilal mixes 80 kgs of sugar worth Rs. 6.75 per kg with 120 kg worth of Rs. 8 per kg. At what rate shall he sell the mixture to gain 20%? 8. A can contains a mixture of 2 liquids A and B in the ratio 7:5, when 9 litres of mixture is drawn off and the can is filled with B, the ratio becomes 7:9. How many litres of the liquid A was contained by the can initially? 9. Two men undertake to do a piece of work for Rs. 400. One alone can do it in 6 days and the other in 8 days. With the help of a boy they finished it in 3 days. What is boy's share? 10. Adarsh buys 2 horses for Rs. 1350 and sells one at 6% loss and other at 7.5% gain and on the whole, he neither gains nor loses. What does each horse cost?
6. MATHEMATICS OF FINANCE: (Averages, Bill Discounting, Stocks & Shares, Learning Curve, LPP)
I. One mark questions (VSA): 1. A bill drawn for 4 months was legally due on 10-10-2003. Find the date of drawing the bill. 2. Define yield. 3. What is feasible region in LPP? 4. The average age of 10 students is 6 years. The sum of the ages of 9 of them is 52 years. Find the age of 10th student. 5. If 10 kgs of a commodity is purchased at Rs. 24 per kg and 5 kgs at Rs. 18 per kg, what is the average price of the commodity per kg? 6. Find the cost of Rs. 9100, 8 3 % stock at 92. 4 7. The present worth of a certain sum due sometime hence is Rs. 1600 and the true discount is Rs. 160. What is the banker's gain? 8. What is learning curve? 9. Find the index of learning for 80% learning effect. 10. What is the cost price of Rs. 100 share at 4 discount if the brokerage is 1/4%? II. Two mark questions (SA): 1. Find the interest earned on Rs. 2448.35 cash invested in 15% stock at 81.5, given that brokerage is 0.125%. 2. Find the market value of 5 and 1/4% stock, in which an income of Rs. 756 is derived by investing Rs. 14976, brokerage being 1/4%. 3. The average age of 10 students is 14 years. Among them, the average age of 4 students is 12 years. Find the average age of the remaining students. 4. Find the income derived from 44 shares of Rs. 25 each at 5 premium (brokerage 1/4 per share) the rate of dividend being 5%. Also find the rate of interest on the investment.
538
Basic Mathematics
5. A worker takes 20 hours to produce the first unit of product. What is the cumulative average time per unit taken by him for the production of first 2 units? III. Four marks questions (ET): 1. A bill for Rs. 2725 was drawn on 3-6-1997 and made payable 3 months after due date. It was discounted on 15-6-1997 at 16% p.a. What is the discounted value of the bill and how much has the banker gained in this transaction. 2. A business man buys and sells chairs and tables. He has Rs. 3000 to invest. A chair costs him Rs. 50 and a table costs him Rs. 90. He has space which can accommodate at most 48 pieces. If he sells each chair for Rs. 200 and each table for Rs. 400, find the number of chairs and tables he has to buy to obtain maximum profit. 3. per hour. 4. A bill of Rs. 5000 drawn on 10-04-96 at 3 months was discounted on 1-5-96 at 7% per annum. Find the discounted value of the bill and banker's gain. 5. A banker pays Rs. 2340 on a bill of Rs. 2500, 146 days before the legally due date. What is the rate of discount charged by the banker? 6. A bill for Rs. 2920 drawn at 6 months was discounted on 10-4-97 for Rs. 2916. If the discount rate is 5% per annum on what date was the bill drawn? 7. A batsman realises that by scoring a century in the 11th innings of his test matches he has bettered his average of the previous 10 innings by 5 runs. What is his average after the 11 innings? 8. Ms. Vidya bought 17 books in a discount sale. The average price of the books being Rs. 53. The average price of the eleven maths books is Rs. 71. If the prices of the remaining 6 accounts books form an increasing arithmetic progression with last term Rs. 25, find the price of the cheapest book. 9. Maximise Z = 10x + 30y subject to x + 2y ≤ 20. x + 5y ≤ 35 and x, y ≥ 0. Indicate the feasible region on the graph sheet. 10. Minimise Z = 20x + 10y subject to x + 2y ≤ 40, 3x + y ≥ 30, 4x + 3y ≥ 60, x ≥ 0, y ≥ 0. 11. Mr. Harsha has invested a certain amount of money in 13% stock at 101. He hold the investment when the market value went down to 96.5. He lost Rs. 3564 on this process. If he has paid the brokerage of 1.5% for all the transactions, what was the amount of cash investment and what was the stock value of the investment in the first instance. 12. A person has invested Rs. 4300 partly in 4.5% stock at 72 and partly in 5% stock at 95. If the total interest is Rs. 250, find his investment in each type of stock. 13. The time required to produce the first unit of a product is 1000 hrs. If the manufacturer experiences 80% learning effect, calculate the average time per unit and the time taken to produce altogether 8 units. Also find the total labour charges for the production of 8 units at the rate of Rs. 10 per hour.
11. Define a parabola: what is the eccentricity of the parabola. 12. Find the length of latus rectum for the conic 4 y 2 + 3y + 6 x = 1 . III. Four marks questions (ET): 1. Derive the standard equation of the circle. Find its centre and radius.
8. With usual notation, S = at + b, where a and b are constants. Prove that velocity is constant and acceleration is zero. 9. Find 2 numbers whose sum is 10 and whose product is maximum. 10. If the total cost function c (x) of a firm is given by c (x) = x3 − 3x + 7, then find the average cost and marginal cost where x = 6 units. III. Four marks questions (ET): 1. If x 1 + y + y 1 + x = 0 and x ≠ y, then prove that
dy 1 =− dx 1+ x
a f
2
.
2. If x = at2, y = 2at, then prove that
1 d2y . =− 2 2at 3 dx
3. The radius of a circular blot of ink is increasing at the rate of 3 cm per minute. Find the rate of increase of its area when its radius is 2 cm. What is the rate of increase of its circumference? 4. Prove that the largest rectangle that can be inscribed in a circle of given radius is a square. 5. Find the derivative of x3n from first principle. 6. State and prove quotient rule in differentiation.
544
Basic Mathematics
7. If x m y n = x + y
a
f
m +n
, then prove that
5
dy y = . dx x
8. Find the derivative of
2
x from first principles.
9. If y = x x + e x , then find
dy . dx
10. Find the maximum and minimum value of the function 2 x 3 − 15 x 2 + 36 x + 10. 11. Prove that the area of the right angled triangle of given hypotenuse is maximum when the triangle is isosceles. 12. If total revenue function is R q = 5 +
af
96 + 6q 2 where q is the number of units manufactured. q
Find the maximum value of total revenue. 13. If total revenue function and total cost function are given by TR = 40x and TC = 2 + 4x respectively, then find at what level of output profit is maximised? 14. A TV manufacturers produces x sets per week at a total cost of Rs. x2 + 1560x + 50,000. He is
12000 − P where P is price per unit, what 179 is the monopoly price in order to maximise the profit?
a monopolist and the demand function for this is x =
8. Find the area bounded by the curve y = x2 − x with x-axis. 9. The marginal cost function of a firm is 150 − 10x + 0.2x2 where x is the output. Find the total cost function, if the fixed cost is Rs. 750. What is the average cost? 10. The marginal revenue function of a firm is 60x − x2 where x is the output. Find the total revenue function. III. Four marks questions (ET): 1. Evaluate:
GIST AND FORMULAE 1. MATHEMATICAL LOGIC: Otherwise • ~ (p ∧ q ) ≡ ~ p ∨ ~ q • ~ (p ∨ q) ≡ ~p ∧ ~q • ~ (p → q ) ≡ p ∧ ~ q • ~ (p ↔ q) ≡ (p ∧ ~ q) ∨ (~p ∨ q) • Converse of the conditional p → q is q → p • Inverse of the conditional p → q is ~p → ~q • Contrapositive of conditional p → q is ~q → ~p.
2. PERMUTATION AND COMBINATION:
• •
n
pr =
n n−r
•
n
n
p0 = 1,
n . p ⋅ q ...
pn = n
• n p −1 . 2
Examination Corner
547
• • • • • •
n
cr =
n n−r⋅ r • The general terms or (r + 1)th term is given by Tr +1 = n cr x n − r ⋅ a r . • isIf an improper fraction is given to resolve, first by dividing the numerator by denominator write the given fraction as the sum of the polynomial and the proper fraction, then the proper fraction is resolved into partial fractions.
6. MATRICES & DETERMINANTS:
• Matrix is an arrangement of numbers in horizontal rows and vertical columns. • Two matrices of the same order are said to be equal if and only if the corresponding elements are equal.
Examination Corner
549
• If • row (or column) of the determinant are added the value of the determinant remain For a non-singular matrix A.
A −1 =
adj A . A
If A is a square matrix and I is the identity matrix of the same order. The characteristic equation: |A − λI| = 0. The values of λ obtained is called eigen values or characteristic roots. Every square matrix satisfies its characteristic equation, |A − λI| = 0. This is Cayley Hamilton theorem. The solution of system of equations
a1 x + b1 y + c1 z = d1
• When stock is purchased, brokerage is added to cost price. • When stock is sold, brokerage is subtracted from selling price. • Interest or dividend is paid on the face value of the stock or share not the market value.
552
Basic Mathematics
• 4 ½% stock at 96 means a stock whose face value is Rs. 100 is available at Rs. 96. Interest earned in 4 ½. • Shares need not be fully paid but stock must be fully paid12. LINEAR PROGRAMMING• Any line is a tangent to the given circle if the length of the ⊥r ⊥r to the directrix. Distance between directrix and vertex = a.
15. LIMITS AND CONTINUITY:
• lim
x →a
x n − an = na n −1 x−a
• lim
ex − 1 =1 x→0 x ax − 1 = log e a x
• lim
x →0
• lim 1 +
n →∞
F H
1 n
I K
1 n
n
=e
• lim 1 + n
n→ 0
a f
=e
x →a
• A function y = f(x) is said to be continuous at x = a if lim− f x = f a = lim+ f x
x→a | 677.169 | 1 |
homework-01 - Modern Algebra and Geometry I Math 4000/6000...
Modern Algebra and Geometry I Math 4000/6000 (92-321) Homework 1 – Due Wednesday, January 11 1. Read the syllabus and the list of student learning outcomes posted on the course web page. ~ cdrup/teaching/math4000sp2012/index.html We will begin class on Wednesday, January 11 with a quiz assessing your knowledge of basic course policies. All of the quiz answers will be contained in the syllabus. The quiz will be open notes, so you are allowed (in fact, encouraged!) to print out the syllabus for use during the quiz. 2. Reflective writing assignment. After reading the syllabus and the list of expected student learn-ing outcomes, think about the following questions: • Why are you taking Math 4000/6000? • What are your goals for this course (academic, personal, career, etc.)? • What steps will you take to achieve those goals? • What was your last mathematics course? What did you do well in that course? • How do you plan to build on your success from your most recent mathematics course?
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This note was uploaded on 02/20/2012 for the course MATH 4000 taught by Professor Staff during the Spring '08 term at University of Georgia Athens. | 677.169 | 1 |
3000 Calculus
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The Trigonometry And Pre-Calculus Tutor is a 5 hour DVD course geared to fully prepare a student to enter university level Calculus. Most students that have trouble with Calculus discover quickly that the root cause of their difficulty is actually that they have never mastered essential material in Trigonometry. For instance, a Calculus textbook will assume that the student is comfortable converting between degrees and radians and can use the unit circle to mentally calculate the Sin of an angle without a calculator. These topics and many other essential concepts are presented in this DVD course with exceptional clarity so that the student will feel well prepared to move into Calculus and Physics | 677.169 | 1 |
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Unformatted text preview: The James R. C. Leitzel Lecture Increasing the number of mathematics majors: Lessons learned from working with the minority community by William Yslas Vlez Department of Mathematics University of Arizona Tucson, Arizona This Powerpoint presentation will appear on my website: Organization of this lecture How did I get started in working with minority students? How did my interaction with these students change over time and how did it change me? What can be learned from this experiment, an experiment that has had some success? How can the entire mathematics community use the results of this experiment to increase the number of mathematics majors? A very important variable in this issue of increasing the number of mathematics majors are the attitudes of: a) individual faculty b) mathematics departments c) our professional organizations Lessons Learned Good students need attention and advice. Provide timely information to students, help them to understand the system and future opportunities. The transition from high school to university is brutal. Students oftentimes choose engineering because they liked mathematics in high school. Appointm ents for Thursday, July 21 2005 Tim e Nam e Math Cour Ethnicity Class Major 9:00 M. Robles 124 H FR 9:20 9:40 10:00 I. Am ene 124 B FR MCB 10:20 10:40 11:00 D. Sinohui 124 H FR changed to chem 11:20 11:40 E. Jaram illo 124 H JR pre pharm 12:00 A. Ram sey 124 B FR Coe 12:20 S. Arvizu 124 H FR BMB 12:40 1:00 R. Enriquez 124 H FR NMS 1:20 H. Flores 124 H FR CVE 1:40 R. Garcia 124 H FR AEE 2:00 A. Valenzuela 124 H FR NMS 2:20 2:40 3:00 D. Kaigler 124 B FR PRCS 3:20 C. Huerta 124 H FR BIOCHEM 3:40 Y Huerta 223 H SO Engr. M ath Good students need attention and advice. Provide timely information. The transition from high school to university is rough. Students choose engineering because they like mathematics. The study of mathematics opens up many doors. Resources for calculus students If you are enrolling in first semester calculus, review your algebra and trigonometry before classes begin. Here are two websites to help you review: Algebra: Trigonometry: _studyguide.doc Here is the math departments webpage ( ) that contains the day-to-day calendars and suggested homework assignments. After you have reviewed your algebra and trigonometry, you should start looking at the homework sets. Try to complete the homework for the first chapter that will be covered in your course. If you are taking math 129 or math 223, here is the website ( dyguides.html ) that contains old final exam questions for these courses. If you are taking math 129, go over the exam questions for math 124/125....
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Mathematics
The Mathematics Department aims to help students simultaneously develop specific mathematical competencies and sharpen their general analytical reasoning skills. We believe that mathematics is worthy of study both because of its direct applications and due to its cultivation of mathematical modes of thinking, which provide powerful tools, useful across many fields.
Our standard course sequence is: Algebra I; Geometry; Algebra II/Trigonometry; Precalculus; Calculus. Our honors sequence is: Geometry Honors; Algebra II/Trigonometry Honors; Precalculus Honors; Calculus AB or BC. While we encourage students to challenge themselves, we appreciate that many factors contribute to a student's selected path. We appreciate the role of technology in teaching and learning, and we attempt to promote intentional uses of technology coupled with an understanding of the mathematics underlying the tool. Skills in mental arithmetic and approximation, for example, prevent a calculator from becoming an oracle.
As the learning of mathematics often happens at its own pace, we work closely with students both inside and outside of class hours. Our doors are open early and our classrooms are often full during the lunch time study period.
Algebra I is a comprehensive course that includes topics such as recognizing and using patterns, operations in the real number system, solving equations and inequalities, proportional reasoning and statistics, linear functions, absolute value functions, solving system of equations, laws of exponents and exponential equations, polynomials and factoring, quadratic functions, rational functions, square-root functions, trigonometry, probability and set theory, and transformations of functions. Algebraic thinking skills are developed through a variety of classroom activities, including group and individual investigations, hands-on manipulatives, real-world problem solving, projects, class discussions, note taking, and utilizing technology. The prerequisite for this course is Pre-Algebra.
This course builds students' understanding of the principles of Geometry. Students develop geometric reasoning skills such as, analyzing rigid motions, completing formal constructions, and writing proofs. Some of the topics in this course include: reasoning in geometry, using tools of geometry, discovering and proving triangle properties, discovering and proving polygon properties, discovering and proving circle properties, transformations and tessellations, area, the Pythagorean Theorem, volume, similarity, and trigonometry. The prerequisite for this course is Algebra I.
This course presents the topics of geometry and provides students with a solid preparation and firm foundation of concepts, techniques, and applications for more advanced mathematics. Students actively engage in a process of self-discovery, constructing meaning, and making connections to geometric concepts with greater depth of understanding and retention. Some of the topics in this course include: constructions, deductive and inductive reasoning, congruence, parallel lines, quadrilaterals, area, volume, circles, trigonometry, and non-Euclidean geometries. Accelerated mathematical thinking skills are developed through a variety of classroom activities, including group and individual investigations, class discussions, note taking, and proofs. The prerequisite for this course is Algebra I (with at least an A- average for the year) and teacher recommendation
This course follows the mathematics curriculum delineated by the National Council of Teachers of Mathematics. Increased emphasis is placed on reading comprehension, problem solving, and the use of functions as tools for modeling real-world situations. Some of the topics in this course include: linear and quadratic functions, polynomial equations, exponential and logarithmic functions, trigonometric functions, matrices, and systems of equations.
This course presents the topics of a college freshman course covering Algebra and Trigonometry and provides students with a solid preparation and firm foundation of concepts, techniques, and applications for more advanced college-level mathematics. Topics presented include: equations, inequalities, mathematical modeling, functions and their graphs, polynomial functions, rations functions with conics, logarithmic functions, basic trigonometry, analytic trigonometry, systems of equations, and matrices. Accelerated algebraic thinking skills are developed through a variety of classroom activities, including group and individual investigations, real-world problem solving, class discussions, note taking, and utilizing technology. The prerequisite for this course is Geometry Honors (with at least a B average for the year) or Geometry (with at least an A average for the year) and teacher recommendation.
This course provides helps students to solidify and develop their understanding of concepts and techniques from algebra, geometry and trigonometry, thereby preparing students for future math courses including calculus and statistics. Topics covered include: trigonometric functions, complex trigonometry, polynomial, rational, exponential and logarithmic functions, combinatorics, recursion, analytic geometry, probability and statistics, and an introduction to the ideas of calculus. The prerequisite for this course is Algebra II Trigonometry.
This course continues to provide students with a solid preparation and firm foundation of concepts, techniques, and applications for more advanced college-level mathematics. It additionally prepares them for AP Calculus BC. Topics include polynomial, rational, exponential and logarithmic functions, trigonometric functions, applications of trigonometry, vectors, matrices, sequences, series, probability, analytic geometry (including 3-D), conics, polar and parametric functions, and limits. Accelerated algebraic thinking skills are developed through a variety of classroom activities, including group and individual investigations, real-world problem solving, class discussions, note taking, and utilizing technology. The prerequisite for this course is Algebra II Trigonometry Honors (with at least a B average for the year) and teacher recommendation.
This course begins by extending the study of functions and introduces students to functions, limits and derivatives. In the second semester, this course is devoted to extending these topics to include more complex families of functions and to introducing the integral calculus. The goals of this course are to: (a) support and strengthen the students' understanding of the fundamental concepts underlying calculus (b) develop the students' problem-solving skills (c) provide students with a solid conceptual introduction to calculus. This course is primarily designed to prepare students for an introductory college-level calculus course. The prerequisite for this course is Precalculus.
This course follows the Advanced Placement Calculus AB syllabus. It presents the topics of a college freshman course in differential and integral calculus that include: limits and continuity, derivatives, applications of derivatives, the definite integral, differential equations and mathematical modeling, applications of definite integralsThis course follows the Advanced Placement Calculus BC syllabus. It presents the topics of a college freshman course in differential and integral calculus that include: limits and continuity, derivatives, applications of derivatives, the definite integral, differential equations and mathematical modeling, applications of definite integrals. Additionally, it presents parametric, polar and vector functions, and polynomial approximations and seriesMultivariable Calculus Honors is a second year calculus course, which involves the extension of calculus to two or more variables. This course will be run in a seminar style with participants expected to contribute to the discussion and presentation of the material. This course will also involve a number of self-directed projects. The prerequisite for the course is the completion of AP Calculus AB (with a 4 or higher on the AP exam) or AP Calculus BC (with a 3 or higher on the AP exam).
This course introduces students to the core concepts of programming while simultaneously teaching students how to program. It covers fundamental programming ideas like functions, control structure, variables and assignments, etc as well as basic abstract data structures like vectors, queues, stacks, etc. It will also touches on the topics of data abstraction, recursion and algorithms. The prerequisite for this course is Algebra I.
This course provides students with an understanding of the basics of statistics to help them both in other areas of study and in every day life. The three main topics of the course are analyzing data, producing data and chance. This course focuses on developing an understanding of statistics through examining its applications, and activities play a large role in this class on a daily basis. The prerequisite for this course is Algebra I.
This course, which follows the College Board AP Statistics Course requirements, introduces students to the major concepts and tools for collecting, analyzing, and drawing conclusions from data. Students learn to conduct statistical analyses, reflect on what these concepts and techniques mean, and identify how they were derived. Students are challenged to develop their critical thinking and general analytical skills. The prerequisite for this course is Precalculus or Algebra II Trigonometry (with at least an B+ average for the year) and teacher recommendation (which may take into account ACT scores as well as reading and writing proficiency). | 677.169 | 1 |
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INTRODUCTORY
LINEAR ALGEBRA
AN APPLIED FIRST COURSE
E I G H T H E D I T I O N
INTRODUCTORY
LINEAR ALGEBRA
AN APPLIED FIRST COURSE
Bernard Kolman
Drexel University
David R. Hill
Temple University
Upper Saddle River, New Jersey 07458
Library of Congress Cataloging-in-Publication Data
Kolman, Bernard, Hill, David R.
Introductory linear algebra: an applied first course-8th ed./ Bernard Kolman, David R. Hill
p. cm.
Rev. ed. of: Introductory linear algebra with applications. 7th ed. c2001.
Includes bibliographical references and index.
ISBN 0-13-143740-2
1. Algebras, Linear. I. Hill, David R. II. Kolman, Bernard. Introductory linear algebra
with applications. III. Title.
QA184.2.K65 2005
512'.5--dc22 2004044755
Executive Acquisitions Editor: George Lobell
Editor-in-Chief: Sally Yagan
Production Editor: Jeanne Audino
Assistant Managing Editor: Bayani Mendoza de Leon
Senior Managing Editor: Linda Mihatov Behrens
Executive Managing Editor: Kathleen Schiaparelli
Vice President/Director of Production and Manufacturing: David W. Riccardi
Assistant Manufacturing Manager/Buyer: Michael Bell
Manufacturing Manager: Trudy Pisciotti
Marketing Manager: Halee Dinsey
Marketing Assistant: Rachel Beckman
Art Director: Kenny Beck
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Art Editor: Thomas Benfatti
Creative Director: Carole Anson
Director of Creative Services: Paul Belfanti
Cover Image: Wassily Kandinsky, Farbstudien mit Angaben zur Maltechnik, 1913, St¨ adische Galerie im Lenbachhaus, Munich
Cover Image Specialist: Karen Sanatar
Art Studio: Laserwords Private Limited
Composition: Dennis Kletzing
c 2005, 2001, 1997, 1993, 1988, 1984, 1980, 1976 Pearson Education, Inc.
Pearson Prentice Hall
Pearson Education, Inc.
Upper Saddle River, NJ 07458
All rights reserved. No part of this book may
be reproduced, in any form or by any means,
without permission in writing from the publisher.
Pearson Prentice Hall
R
is a trademark of Pearson Education, Inc.
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
ISBN 0-13-143740-2
Pearson Education Ltd., London
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Pearson Education Malaysia, Pte. Ltd.
• •
To the memory of Lillie
and to Lisa and Stephen
B. K.
To Suzanne
D. R. H.
• •
CONTENTS
Preface xi
To the Student xix
1 Linear Equations and Matrices 1
1.1 Linear Systems 1
1.2 Matrices 10
1.3 Dot Product and Matrix Multiplication 21
1.4 Properties of Matrix Operations 39
1.5 Matrix Transformations 52
1.6 Solutions of Linear Systems of Equations 62
1.7 The Inverse of a Matrix 91
1.8 LU-Factorization (Optional) 107
2 Applications of Linear Equations
and Matrices (Optional) 119
2.1 An Introduction to Coding 119
2.2 Graph Theory 125
2.3 Computer Graphics 135
2.4 Electrical Circuits 144
2.5 Markov Chains 149
2.6 Linear Economic Models 159
2.7 Introduction to Wavelets 166
3 Determinants 182
3.1 Definition and Properties 182
3.2 Cofactor Expansion and Applications 196
3.3 Determinants from a Computational Point of View 210
4 Vectors in R
n
214
4.1 Vectors in the Plane 214
4.2 n-Vectors 229
4.3 Linear Transformations 247
vii
viii Contents
5 Applications of Vectors in R
2
and R
3
(Optional) 259
5.1 Cross Product in R
3
259
5.2 Lines and Planes 264
6 Real Vector Spaces 272
6.1 Vector Spaces 272
6.2 Subspaces 279
6.3 Linear Independence 291
6.4 Basis and Dimension 303
6.5 Homogeneous Systems 317
6.6 The Rank of a Matrix and Applications 328
6.7 Coordinates and Change of Basis 340
6.8 Orthonormal Bases in R
n
352
6.9 Orthogonal Complements 360
7 Applications of Real Vector
Spaces (Optional) 375
7.1 QR-Factorization 375
7.2 Least Squares 378
7.3 More on Coding 390
8 Eigenvalues, Eigenvectors, and
Diagonalization 408
8.1 Eigenvalues and Eigenvectors 408
8.2 Diagonalization 422
8.3 Diagonalization of Symmetric Matrices 433
9 Applications of Eigenvalues and
Eigenvectors (Optional) 447
9.1 The Fibonacci Sequence 447
9.2 Differential Equations (Calculus Required) 451
9.3 Dynamical Systems (Calculus Required) 461
9.4 Quadratic Forms 475
9.5 Conic Sections 484
9.6 Quadric Surfaces 491
10 Linear Transformations and Matrices 502
10.1 Definition and Examples 502
10.2 The Kernel and Range of a Linear Transformation 508
10.3 The Matrix of a Linear Transformation 521
10.4 Introduction to Fractals (Optional) 536
Cumulative Review of
Introductory Linear Algebra 555
Contents ix
11 Linear Programming (Optional) 558
11.1 The Linear Programming Problem; Geometric Solution 558
11.2 The Simplex Method 575
11.3 Duality 591
11.4 The Theory of Games 598
12 MATLAB for Linear Algebra 615
12.1 Input and Output in MATLAB 616
12.2 Matrix Operations in MATLAB 620
12.3 Matrix Powers and Some Special Matrices 623
12.4 Elementary Row Operations in MATLAB 625
12.5 Matrix Inverses in MATLAB 634
12.6 Vectors in MATLAB 635
12.7 Applications of Linear Combinations in MATLAB 637
12.8 Linear Transformations in MATLAB 640
12.9 MATLAB Command Summary 643
APPENDIX A Complex Numbers A1
A.1 Complex Numbers A1
A.2 Complex Numbers in Linear Algebra A9
APPENDIX B Further Directions A19
B.1 Inner Product Spaces (Calculus Required) A19
B.2 Composite and Invertible Linear Transformations A30
Glossary for Linear Algebra A39
Answers to Odd-Numbered Exercises
and Chapter Tests A45
Index I1
PREFACE
Material Covered
This book presents an introduction to linear algebra and to some of its signif-
icant applications. It is designed for a course at the freshman or sophomore
level. There is more than enough material for a semester or quarter course.
By omitting certain sections, it is possible in a one-semester or quarter course
to cover the essentials of linear algebra (including eigenvalues and eigenvec-
tors), to show how the computer is used, and to explore some applications of
linear algebra. It is no exaggeration to say that with the many applications
of linear algebra in other areas of mathematics, physics, biology, chemistry,
engineering, statistics, economics, finance, psychology, and sociology, linear
algebra is the undergraduate course that will have the most impact on students'
lives. The level and pace of the course can be readily changed by varying the
amount of time spent on the theoretical material and on the applications. Cal-
culus is not a prerequisite; examples and exercises using very basic calculus
are included and these are labeled "Calculus Required."
The emphasis is on the computational and geometrical aspects of the sub-
ject, keeping abstraction to a minimum. Thus we sometimes omit proofs of
difficult or less-rewarding theorems while amply illustrating them with exam-
ples. The proofs that are included are presented at a level appropriate for the
student. We have also devoted our attention to the essential areas of linear
algebra; the book does not attempt to cover the subject exhaustively.
What Is New in the Eighth Edition
We have been very pleased by the widespread acceptance of the first seven
editions of this book. The reform movement in linear algebra has resulted in a
number of techniques for improving the teaching of linear algebra. The Lin-
ear Algebra Curriculum Study Group and others have made a number of
important recommendations for doing this. In preparing the present edition,
we have considered these recommendations as well as suggestions from fac-
ulty and students. Although many changes have been made in this edition, our
objective has remained the same as in the earlier editions:
to develop a textbook that will help the instructor to teach and
the student to learn the basic ideas of linear algebra and to see
some of its applications.
To achieve this objective, the following features have been developed in this
edition:
xi
xii Preface
New sections have been added as follows:
• Section 1.5, Matrix Transformations, introduces at a very early stage
some geometric applications.
• Section 2.1, An Introduction to Coding, along with supporting material
on bit matrices throughout the first six chapters, provides an introduc-
tion to the basic ideas of coding theory.
• Section 7.3, More on Coding, develops some simple codes and their
basic properties related to linear algebra.
More geometric material has been added.
New exercises at all levels have been added. Some of these are more
open-ended, allowing for exploration and discovery, as well as writing.
More illustrations have been added.
MATLAB M-files have been upgraded to more modern versions.
Key terms have been added at the end of each section, reflecting the in-
creased emphasis in mathematics on communication skills.
True/false questions now ask the student to justify his or her answer, pro-
viding an additional opportunity for exploration and writing.
Another 25 true/false questions have been added to the cumulative review
at the end of the first ten chapters.
A glossary, new to this edition, has been added.
Exercises
The exercises in this book are grouped into three classes. The first class, Ex-
ercises, contains routine exercises. The second class, Theoretical Exercises,
includes exercises that fill in gaps in some of the proofs and amplify material
in the text. Some of these call for a verbal solution. In this technological age,
it is especially important to be able to write with care and precision; therefore,
exercises of this type should help to sharpen such skills. These exercises can
also be used to raise the level of the course and to challenge the more capa-
ble and interested student. The third class consists of exercises developed by
David R. Hill and are labeled by the prefix ML (for MATLAB). These exer-
cises are designed to be solved by an appropriate computer software package.
Answers to all odd-numbered numerical and ML exercises appear in the
back of the book. At the end of Chapter 10, there is a cumulative review of
the introductory linear algebra material presented thus far, consisting of 100
true/false questions (with answers in the back of the book). The Instructor's
Solutions Manual, containing answers to all even-numbered exercises and
solutions to all theoretical exercises, is available (to instructors only) at no
cost from the publisher.
Presentation
We have learned from experience that at the sophomore level, abstract ideas
must be introduced quite gradually and must be supported by firmfoundations.
Thus we begin the study of linear algebra with the treatment of matrices as
mere arrays of numbers that arise naturally in the solution of systems of linear
equations—a problemalready familiar to the student. Much attention has been
devoted from one edition to the next to refine and improve the pedagogical
aspects of the exposition. The abstract ideas are carefully balanced by the
considerable emphasis on the geometrical and computational foundations of
the subject.
Preface xiii
Material Covered
Chapter 1 deals with matrices and their properties. Section 1.5, Matrix Trans-
formations, newto this edition, provides an early introduction to this important
topic. This chapter is comprised of two parts: The first part deals with matri-
ces and linear systems and the second part with solutions of linear systems.
Chapter 2 (optional) discusses applications of linear equations and matrices to
the areas of coding theory, computer graphics, graph theory, electrical circuits,
Markov chains, linear economic models, and wavelets. Section 2.1, An Intro-
duction to Coding, new to this edition, develops foundations for introducing
some basic material in coding theory. To keep this material at a very elemen-
tary level, it is necessary to use lengthier technical discussions. Chapter 3
presents the basic properties of determinants rather quickly. Chapter 4 deals
with vectors in R
n
. In this chapter we also discuss vectors in the plane and
give an introduction to linear transformations. Chapter 5 (optional) provides
an opportunity to explore some of the many geometric ideas dealing with vec-
tors in R
2
and R
3
; we limit our attention to the areas of cross product in R
3
and lines and planes.
In Chapter 6 we come to a more abstract notion, that of a vector space.
The abstraction in this chapter is more easily handled after the material cov-
ered on vectors in R
n
. Chapter 7 (optional) presents three applications of real
vector spaces: QR-factorization, least squares, and Section 7.3, More on Cod-
ing, new to this edition, introducing some simple codes. Chapter 8, on eigen-
values and eigenvectors, the pinnacle of the course, is now presented in three
sections to improve pedagogy. The diagonalization of symmetric matrices is
carefully developed.
Chapter 9 (optional) deals with a number of diverse applications of eigen-
values and eigenvectors. These include the Fibonacci sequence, differential
equations, dynamical systems, quadratic forms, conic sections, and quadric
surfaces. Chapter 10 covers linear transformations and matrices. Section 10.4
(optional), Introduction to Fractals, deals with an application of a certain non-
linear transformation. Chapter 11 (optional) discusses linear programming, an
important application of linear algebra. Section 11.4 presents the basic ideas
of the theory of games. Chapter 12, provides a brief introduction to MATLAB
(which stands for MATRIX LABORATORY), a very useful software package
for linear algebra computation, described below.
Appendix A covers complex numbers and introduces, in a brief but thor-
ough manner, complex numbers and their use in linear algebra. Appendix B
presents two more advanced topics in linear algebra: inner product spaces and
composite and invertible linear transformations.
Applications
Most of the applications are entirely independent; they can be covered either
after completing the entire introductory linear algebra material in the course
or they can be taken up as soon as the material required for a particular appli-
cation has been developed. Brief Previews of most applications are given at
appropriate places in the book to indicate how to provide an immediate appli-
cation of the material just studied. The chart at the end of this Preface, giving
the prerequisites for each of the applications, and the Brief Previews will be
helpful in deciding which applications to cover and when to cover them.
Some of the sections in Chapters 2, 5, 7, 9, and 11 can also be used as in-
dependent student projects. Classroom experience with the latter approach has
met with favorable student reaction. Thus the instructor can be quite selective
both in the choice of material and in the method of study of these applications.
xiv Preface
End of Chapter Material
Every chapter contains a summary of Key Ideas for Review, a set of supple-
mentary exercises (answers to all odd-numbered numerical exercises appear
in the back of the book), and a chapter test (all answers appear in the back of
the book).
MATLAB Software
Although the ML exercises can be solved using a number of software pack-
ages, in our judgment MATLAB is the most suitable package for this pur-
pose. MATLAB is a versatile and powerful software package whose cor-
nerstone is its linear algebra capability. MATLAB incorporates profession-
ally developed quality computer routines for linear algebra computation. The
code employed by MATLAB is written in the C language and is upgraded as
new versions of MATLAB are released. MATLAB is available from The Math
Works, Inc., 24 Prime Park Way, Natick, MA 01760, (508) 653-1415; e-mail:
info@mathworks.com and is not distributed with this book or the instruc-
tional routines developed for solving the ML exercises. The Student Edition
of MATLAB also includes a version of Maple, thereby providing a symbolic
computational capability.
Chapter 12 of this edition consists of a brief introduction to MATLAB's
capabilities for solving linear algebra problems. Although programs can
be written within MATLAB to implement many mathematical algorithms, it
should be noted that the reader of this book is not asked to write programs.
The user is merely asked to use MATLAB (or any other comparable soft-
ware package) to solve specific numerical problems. Approximately 24 in-
structional M-files have been developed to be used with the ML exercises
in this book and are available from the following Prentice Hall Web site:
These M-files are designed to transform
many of MATLAB's capabilities into courseware. This is done by providing
pedagogy that allows the student to interact with MATLAB, thereby letting the
student think through all the steps in the solution of a problem and relegating
MATLAB to act as a powerful calculator to relieve the drudgery of a tedious
computation. Indeed, this is the ideal role for MATLAB (or any other similar
package) in a beginning linear algebra course, for in this course, more than in
many others, the tedium of lengthy computations makes it almost impossible
to solve a modest-size problem. Thus, by introducing pedagogy and reining in
the power of MATLAB, these M-files provide a working partnership between
the student and the computer. Moreover, the introduction to a powerful tool
such as MATLAB early in the student's college career opens the way for other
software support in higher-level courses, especially in science and engineer-
ing.
Supplements
Student Solutions Manual (0-13-143741-0). Prepared by Dennis Kletzing,
Stetson University, and Nina Edelman and Kathy O'Hara, Temple University,
contains solutions to all odd-numbered exercises, both numerical and theoret-
ical. It can be purchased from the publisher.
Instructor's Solutions Manual (0-13-143742-9). Contains answers to all
even-numbered exercises and solutions to all theoretical exercises—is avail-
able (to instructors only) at no cost from the publisher.
Optional combination packages. Provide a computer workbook free of
charge when packaged with this book.
Preface xv
Linear Algebra Labs with MATLAB, by David R. Hill and David E.
Zitarelli, 3rd edition, ISBN 0-13-124092-7 (supplement and text).
Visualizing Linear Algebra with Maple, by Sandra Z. Keith, ISBN 0-13-
124095-1 (supplement and text).
ATLAST Computer Exercises for Linear Algebra, by Steven Leon, Eugene
Herman, and Richard Faulkenberry, 2nd edition, ISBN 0-13-124094-3
(supplement and text).
Understanding Linear Algebra with MATLAB, by Erwin and Margaret
Kleinfeld, ISBN 0-13-124093-5 (supplement and text).
Prerequisites for Applications
Prerequisites for Applications
Section 2.1 Material on bits in Chapter 1
Section 2.2 Section 1.4
Section 2.3 Section 1.5
Section 2.4 Section 1.6
Section 2.5 Section 1.6
Section 2.6 Section 1.7
Section 2.7 Section 1.7
Section 5.1 Section 4.1 and Chapter 3
Section 5.2 Sections 4.1 and 5.1
Section 7.1 Section 6.8
Section 7.2 Sections 1.6, 1.7, 4.2, 6.9
Section 7.3 Section 2.1
Section 9.1 Section 8.2
Section 9.2 Section 8.2
Section 9.3 Section 9.2
Section 9.4 Section 8.3
Section 9.5 Section 9.4
Section 9.6 Section 9.5
Section 10.4 Section 8.2
Sections 11.1–11.3 Section 1.6
Section 11.4 Sections 11.1–11.3
To Users of Previous Editions:
During the 29-year life of the previous seven editions of this book, the book
was primarily used to teach a sophomore-level linear algebra course. This
course covered the essentials of linear algebra and used any available extra
time to study selected applications of the subject. In this new edition we
have not changed the structural foundation for teaching the essential lin-
ear algebra material. Thus, this material can be taught in exactly the same
manner as before. The placement of the applications in a more cohesive
and pedagogically unified manner together with the newly added applica-
tions and other material should make it easier to teach a richer and more
varied course.
xvi Preface
Acknowledgments
We are pleased to express our thanks to the following people who thoroughly
reviewed the entire manuscript in the first edition: William Arendt, University
of Missouri and David Shedler, Virginia Commonwealth University. In the
second edition: Gerald E. Bergum, South Dakota State University; James O.
Brooks, Villanova University; Frank R. DeMeyer, Colorado State University;
Joseph Malkevitch, York College of the City University of New York; Harry
W. McLaughlin, Rensselaer Polytechnic Institute; and Lynn Arthur Steen, St.
Olaf's College. In the third edition: Jerry Goldman, DePaul University; David
R. Hill, Temple University; Allan Krall, The Pennsylvania State University at
University Park; Stanley Lukawecki, Clemson University; David Royster, The
University of North Carolina; Sandra Welch, Stephen F. Austin State Univer-
sity; and Paul Zweir, Calvin College.
In the fourth edition: William G. Vick, Broome Community College; Car-
rol G. Wells, Western Kentucky University; Andre L. Yandl, Seattle Univer-
sity; and Lance L. Littlejohn, Utah State University. In the fifth edition: Paul
Beem, Indiana University-South Bend; John Broughton, Indiana University
of Pennsylvania; Michael Gerahty, University of Iowa; Philippe Loustaunau,
George Mason University; Wayne McDaniels, University of Missouri; and
Larry Runyan, Shoreline Community College. In the sixth edition: Daniel
D. Anderson, University of Iowa; J¨ urgen Gerlach, Radford University; W. L.
Golik, University of Missouri at St. Louis; Charles Heuer, Concordia Col-
lege; Matt Insall, University of Missouri at Rolla; Irwin Pressman, Carleton
University; and James Snodgrass, Xavier University. In the seventh edition:
Ali A. Dad-del, University of California-Davis; Herman E. Gollwitzer, Drexel
University; John Goulet, Worcester Polytechnic Institute; J. D. Key, Clem-
son University; John Mitchell, Rensselaer Polytechnic Institute; and Karen
Schroeder, Bentley College.
In the eighth edition: Juergen Gerlach, Radford University; Lanita Pres-
son, University of Alabama, Huntsville; Tomaz Pisanski, Colgate University;
Mike Daven, Mount Saint Mary College; David Goldberg, Purdue University;
Aimee J. Ellington, Virginia Commonwealth University.
We thank Vera Pless, University of Illinois at Chicago, for critically read-
ing the material on coding theory.
We also wish to thank the following for their help with selected portions of
the manuscript: Thomas I. Bartlow, Robert E. Beck, and Michael L. Levitan,
all of Villanova University; Robert C. Busby, Robin Clark, the late Charles
S. Duris, Herman E. Gollwitzer, Milton Schwartz, and the late John H. Staib,
all of Drexel University; Avi Vardi; Seymour Lipschutz, Temple University;
Oded Kariv, Technion, Israel Institute of Technology; William F. Trench, Trin-
ity University; and Alex Stanoyevitch, the University of Hawaii; and instruc-
tors and students from many institutions in the United States and other coun-
tries, who shared with us their experiences with the book and offered helpful
suggestions.
The numerous suggestions, comments, and criticisms of these people
greatly improved the manuscript. To all of them goes a sincere expression
of gratitude.
We thank Dennis Kletzing, Stetson University, who typeset the entire
manuscript, the Student Solutions Manual, and the Instructor's Manual. He
found a number of errors in the manuscript and cheerfully performed miracles
under a very tight schedule. It was a pleasure working with him.
We thank Dennis Kletzing, Stetson University, and Nina Edelman and
Preface xvii
Kathy O'Hara, Temple University, for preparing the Student Solutions Man-
ual.
We should also like to thank Nina Edelman, Temple University, along
with Lilian Brady, for critically reading the page proofs. Thanks also to Blaise
deSesa for his help in editing and checking the solutions to the exercises.
Finally, a sincere expression of thanks to Jeanne Audino, Production Ed-
itor, who patiently and expertly guided this book from launch to publication;
to George Lobell, Executive Editor; and to the entire staff of Prentice Hall for
their enthusiasm, interest, and unfailing cooperation during the conception,
design, production, and marketing phases of this edition.
Bernard Kolman
bkolman@mcs.drexel.edu
David R. Hill
hill@math.temple.edu
TO THE STUDENT
It is very likely that this course is unlike any other mathematics course that
you have studied thus far in at least two important ways. First, it may be your
initial introduction to abstraction. Second, it is a mathematics course that may
well have the greatest impact on your vocation.
Unlike other mathematics courses, this course will not give you a toolkit
of isolated computational techniques for solving certain types of problems.
Instead, we will develop a core of material called linear algebra by introduc-
ing certain definitions and creating procedures for determining properties and
proving theorems. Proving a theorem is a skill that takes time to master, so at
first we will only expect you to read and understand the proof of a theorem.
As you progress in the course, you will be able to tackle some simple proofs.
We introduce you to abstraction slowly, keep it to a minimum, and amply il-
lustrate each abstract idea with concrete numerical examples and applications.
Although you will be doing a lot of computations, the goal in most problems
is not merely to get the "right" answer, but to understand and explain how to
get the answer and then interpret the result.
Linear algebra is used in the everyday world to solve problems in other
areas of mathematics, physics, biology, chemistry, engineering, statistics, eco-
nomics, finance, psychology, and sociology. Applications that use linear alge-
bra include the transmission of information, the development of special effects
in film and video, recording of sound, Web search engines on the Internet, and
economic analyses. Thus, you can see how profoundly linear algebra affects
you. A selected number of applications are included in this book, and if there
is enough time, some of these may be covered in this course. Additionally,
many of the applications can be used as self-study projects.
There are three different types of exercises in this book. First, there are
computational exercises. These exercises and the numbers in them have been
carefully chosen so that almost all of them can readily be done by hand. When
you use linear algebra in real applications, you will find that the problems are
much bigger in size and the numbers that occur in them are not always "nice."
This is not a problem because you will almost certainly use powerful software
to solve them. A taste of this type of software is provided by the third type of
exercises. These are exercises designed to be solved by using a computer and
MATLAB, a powerful matrix-based application that is widely used in industry.
The second type of exercises are theoretical. Some of these may ask you to
prove a result or discuss an idea. In today's world, it is not enough to be
able to compute an answer; you often have to prepare a report discussing your
solution, justifying the steps in your solution, and interpreting your results.
xix
xx To the Student
These types of exercises will give you experience in writing mathematics.
Mathematics uses words, not just symbols.
How to Succeed in Linear Algebra
• Read the book slowly with pencil and paper at hand. You might have to
read a particular section more than once. Take the time to verify the steps
marked "verify" in the text.
• Make sure to do your homework on a timely basis. If you wait until the
problems are explained in class, you will miss learning how to solve a
problem by yourself. Even if you can't complete a problem, try it any-
way, so that when you see it done in class you will understand it more
easily. You might find it helpful to work with other students on the mate-
rial covered in class and on some homework problems.
• Make sure that you ask for help as soon as something is not clear to you.
Each abstract idea in this course is based on previously developed ideas—
much like laying a foundation and then building a house. If any of the
ideas are fuzzy to you or missing, your knowledge of the course will not
be sturdy enough for you to grasp succeeding ideas.
• Make use of the pedagogical tools provided in this book. At the end of
each section we have a list of key terms; at the end of each chapter we
have a list of key ideas for review, supplementary exercises, and a chapter
test. At the end of the first ten chapters (completing the core linear algebra
material in the course) we have a comprehensive review consisting of 100
true/false questions that ask you to justify your answer. Finally, there is
a glossary for linear algebra at the end of the book. Answers to the odd-
numbered exercises appear at the end of the book. The Student Solutions
Manual provides detailed solutions to all odd-numbered exercises, both
numerical and theoretical. It can be purchased from the publisher (ISBN
0-13-143742-9).
We assure you that your efforts to learn linear algebra well will be amply
rewarded in other courses and in your professional career.
We wish you much success in your study of linear algebra.
INTRODUCTORY
LINEAR ALGEBRA
AN APPLIED FIRST COURSE
1
C H A P T E R
LINEAR EQUATIONS
AND MATRICES
1.1
LINEAR SYSTEMS
A good many problems in the natural and social sciences as well as in en-
gineering and the physical sciences deal with equations relating two sets of
variables. An equation of the type
ax = b,
expressing the variable b in terms of the variable x and the constant a, is
called a linear equation. The word linear is used here because the graph of
the equation above is a straight line. Similarly, the equation
a
1
x
1
+a
2
x
2
+· · · +a
n
x
n
= b, (1)
expressing b in terms of the variables x
1
, x
2
, . . . , x
n
and the known constants
a
1
, a
2
, . . . , a
n
, is called a linear equation. In many applications we are given
b and the constants a
1
, a
2
, . . . , a
n
and must find numbers x
1
, x
2
, . . . , x
n
, called
unknowns, satisfying (1).
A solution to a linear equation (1) is a sequence of n numbers s
1
, s
2
, . . . ,
s
n
, which has the property that (1) is satisfied when x
1
= s
1
, x
2
= s
2
, . . . ,
x
n
= s
n
are substituted in (1).
Thus x
1
= 2, x
2
= 3, and x
3
= −4 is a solution to the linear equation
6x
1
−3x
2
+4x
3
= −13,
because
6(2) −3(3) +4(−4) = −13.
This is not the only solution to the given linear equation, since x
1
= 3, x
2
= 1,
and x
3
= −7 is another solution.
More generally, a system of m linear equations in n unknowns x
1
, x
2
,
. . . , x
n
, or simply a linear system, is a set of m linear equations each in n
unknowns. A linear system can be conveniently denoted by
a
11
x
1
+ a
12
x
2
+ · · · + a
1n
x
n
= b
1
a
21
x
1
+ a
22
x
2
+ · · · + a
2n
x
n
= b
2
.
.
.
.
.
.
.
.
.
.
.
.
a
m1
x
1
+ a
m2
x
2
+ · · · + a
mn
x
n
= b
m
.
(2)
1
2 Chapter 1 Linear Equations and Matrices
The two subscripts i and j are used as follows. The first subscript i indi-
cates that we are dealing with the i th equation, while the second subscript j is
associated with the j th variable x
j
. Thus the i th equation is
a
i 1
x
1
+a
i 2
x
2
+· · · +a
i n
x
n
= b
i
.
In (2) the a
i j
are known constants. Given values of b
1
, b
2
, . . . , b
m
, we want to
find values of x
1
, x
2
, . . . , x
n
that will satisfy each equation in (2).
A solution to a linear system (2) is a sequence of n numbers s
1
, s
2
, . . . , s
n
,
which has the property that each equation in (2) is satisfied when x
1
= s
1
,
x
2
= s
2
, . . . , x
n
= s
n
are substituted in (2).
To find solutions to a linear system, we shall use a technique called the
method of elimination. That is, we eliminate some of the unknowns by
adding a multiple of one equation to another equation. Most readers have
had some experience with this technique in high school algebra courses. Most
likely, the reader has confined his or her earlier work with this method to lin-
ear systems in which m = n, that is, linear systems having as many equations
as unknowns. In this course we shall broaden our outlook by dealing with
systems in which we have m = n, m < n, and m > n. Indeed, there are
numerous applications in which m = n. If we deal with two, three, or four
unknowns, we shall often write them as x, y, z, and w. In this section we use
the method of elimination as it was studied in high school. In Section 1.5 we
shall look at this method in a much more systematic manner.
EXAMPLE 1 The director of a trust fund has $100,000 to invest. The rules of the trust state
that both a certificate of deposit (CD) and a long-term bond must be used.
The director's goal is to have the trust yield $7800 on its investments for the
year. The CD chosen returns 5% per annum and the bond 9%. The director
determines the amount x to invest in the CD and the amount y to invest in the
bond as follows:
Since the total investment is $100,000, we must have x + y = 100,000.
Since the desired return is $7800, we obtain the equation 0.05x + 0.09y =
7800. Thus, we have the linear system
x + y = 100,000
0.05x + 0.09y = 7800.
(3)
To eliminate x, we add (−0.05) times the first equation to the second, obtain-
ing
x + y = 100,000
0.04y = 2800,
where the second equation has no x term. We have eliminated the unknown
x. Then solving for y in the second equation, we have
y = 70,000,
and substituting y into the first equation of (3), we obtain
x = 30,000.
To check that x = 30,000, y = 70,000 is a solution to (3), we verify that
these values of x and y satisfy each of the equations in the given linear system.
Thus, the director of the trust should invest $30,000 in the CD and $70,000 in
the long-term bond.
Sec. 1.1 Linear Systems 3
EXAMPLE 2 Consider the linear system
x − 3y = −7
2x − 6y = 7.
(4)
Again, we decide to eliminate x. We add (−2) times the first equation to the
second one, obtaining
x − 3y = −7
0x + 0y = 21
whose second equation makes no sense. This means that the linear system (4)
has no solution. We might have come to the same conclusion from observing
that in (4) the left side of the second equation is twice the left side of the first
equation, but the right side of the second equation is not twice the right side
of the first equation.
EXAMPLE 3 Consider the linear system
x + 2y + 3z = 6
2x − 3y + 2z = 14
3x + y − z = −2.
(5)
To eliminate x, we add (−2) times the first equation to the second one and
(−3) times the first equation to the third one, obtaining
x + 2y + 3z = 6
− 7y − 4z = 2
− 5y − 10z = −20.
(6)
We next eliminate y from the second equation in (6) as follows. Multiply the
third equation of (6) by
_
−
1
5
_
, obtaining
x + 2y + 3z = 6
− 7y − 4z = 2
y + 2z = 4.
Next we interchange the second and third equations to give
x + 2y + 3z = 6
y + 2z = 4
− 7y − 4z = 2.
(7)
We now add 7 times the second equation to the third one, to obtain
x + 2y + 3z = 6
y + 2z = 4
10z = 30.
Multiplying the third equation by
1
10
, we have
x + 2y + 3z = 6
y + 2z = 4
z = 3.
(8)
4 Chapter 1 Linear Equations and Matrices
Substituting z = 3 into the second equation of (8), we find y = −2. Substi-
tuting these values of z and y into the first equation of (8), we have x = 1.
To check that x = 1, y = −2, z = 3 is a solution to (5), we verify that
these values of x, y, and z satisfy each of the equations in (5). Thus, x = 1,
y = −2, z = 3 is a solution to the linear system (5). The importance of the
procedure lies in the fact that the linear systems (5) and (8) have exactly the
same solutions. System (8) has the advantage that it can be solved quite easily,
giving the foregoing values for x, y, and z.
EXAMPLE 4 Consider the linear system
x + 2y − 3z = −4
2x + y − 3z = 4.
(9)
Eliminating x, we add (−2) times the first equation to the second one, to
obtain
x + 2y − 3z = −4
− 3y + 3z = 12.
(10)
Solving the second equation in (10) for y, we obtain
y = z −4,
where z can be any real number. Then, from the first equation of (10),
x = −4 −2y +3z
= −4 −2(z −4) +3z
= z +4.
Thus a solution to the linear system (9) is
x = r +4
y = r −4
z = r,
where r is any real number. This means that the linear system (9) has infinitely
many solutions. Every time we assign a value to r, we obtain another solution
to (9). Thus, if r = 1, then
x = 5, y = −3, and z = 1
is a solution, while if r = −2, then
x = 2, y = −6, and z = −2
is another solution.
EXAMPLE 5 Consider the linear system
x + 2y = 10
2x − 2y = −4
3x + 5y = 26.
(11)
Sec. 1.1 Linear Systems 5
Eliminating x, we add (−2) times the first equation to the second and (−3)
times the first equation to the third one, obtaining
x + 2y = 10
− 6y = −24
−y = −4.
Multiplying the second equation by
_
−
1
6
_
and the third one by (−1), we have
x + 2y = 10
y = 4
y = 4,
(12)
which has the same solutions as (11). Substituting y = 4 in the first equation
of (12), we obtain x = 2. Hence x = 2, y = 4 is a solution to (11).
EXAMPLE 6 Consider the linear system
x + 2y = 10
2x − 2y = −4
3x + 5y = 20.
(13)
To eliminate x, we add (−2) times the first equation to the second one and
(−3) times the first equation to the third one, to obtain
x + 2y = 10
− 6y = −24
−y = −10.
Multiplying the second equation by
_
−
1
6
_
and the third one by (−1), we have
the system
x + 2y = 10
y = 4
y = 10,
(14)
which has no solution. Since (14) and (13) have the same solutions, we con-
clude that (13) has no solutions.
These examples suggest that a linear system may have one solution (a
unique solution), no solution, or infinitely many solutions.
We have seen that the method of elimination consists of repeatedly per-
forming the following operations:
1. Interchange two equations.
2. Multiply an equation by a nonzero constant.
3. Add a multiple of one equation to another.
It is not difficult to show (Exercises T.1 through T.3) that the method of
elimination yields another linear system having exactly the same solutions as
the given system. The new linear system can then be solved quite readily.
6 Chapter 1 Linear Equations and Matrices
As you have probably already observed, the method of elimination has
been described, so far, in general terms. Thus we have not indicated any rules
for selecting the unknowns to be eliminated. Before providing a systematic
description of the method of elimination, we introduce, in the next section, the
notion of a matrix, which will greatly simplify our notation and will enable us
to develop tools to solve many important problems.
Consider now a linear system of two equations in the unknowns x and y:
a
1
x + a
2
y = c
1
b
1
x + b
2
y = c
2
.
(15)
The graph of each of these equations is a straight line, which we denote by
l
1
and l
2
, respectively. If x = s
1
, y = s
2
is a solution to the linear system
(15), then the point (s
1
, s
2
) lies on both lines l
1
and l
2
. Conversely, if the point
(s
1
, s
2
) lies on both lines l
1
and l
2
, then x = s
1
, y = s
2
is a solution to the
linear system (15). (See Figure 1.1.) Thus we are led geometrically to the
same three possibilities mentioned previously.
1. The system has a unique solution; that is, the lines l
1
and l
2
intersect at
exactly one point.
2. The system has no solution; that is, the lines l
1
and l
2
do not intersect.
3. The system has infinitely many solutions; that is, the lines l
1
and l
2
coin-
cide.
Figure 1.1
y
x
(b) No solution
l
1
l
2
y
x
(a) A unique solution
l
1
l
2
y
x
(c) Infinitely many solutions
l
1
l
2
Next, consider a linear system of three equations in the unknowns x, y,
and z:
a
1
x +b
1
y +c
1
z = d
1
a
2
x +b
2
y +c
2
z = d
2
a
3
x +b
3
y +c
3
z = d
3
.
(16)
The graph of each of these equations is a plane, denoted by P
1
, P
2
, and P
3
,
respectively. As in the case of a linear system of two equations in two un-
knowns, the linear system in (16) can have a unique solution, no solution, or
infinitely many solutions. These situations are illustrated in Figure 1.2. For a
more concrete illustration of some of the possible cases, the walls (planes) of
a room intersect in a unique point, a corner of the room, so the linear system
has a unique solution. Next, think of the planes as pages of a book. Three
pages of a book (when held open) intersect in a straight line, the spine. Thus,
the linear system has infinitely many solutions. On the other hand, when the
book is closed, three pages of a book appear to be parallel and do not intersect,
so the linear system has no solution.
Sec. 1.1 Linear Systems 7
Figure 1.2
(a) A unique solution
P
1
P
2
(c) Infinitely many solutions (b) No solution
P
3
P
2
P
1
P
3
P
1
P
3
P
2
EXAMPLE 7 (Production Planning) Amanufacturer makes three different types of chem-
ical products: A, B, and C. Each product must go through two processing
machines: X and Y. The products require the following times in machines X
and Y:
1. One ton of A requires 2 hours in machine X and 2 hours in machine Y.
2. One ton of B requires 3 hours in machine X and 2 hours in machine Y.
3. One ton of C requires 4 hours in machine X and 3 hours in machine Y.
Machine X is available 80 hours per week and machine Y is available 60 hours
per week. Since management does not want to keep the expensive machines
X and Y idle, it would like to know how many tons of each product to make
so that the machines are fully utilized. It is assumed that the manufacturer can
sell as much of the products as is made.
To solve this problem, we let x
1
, x
2
, and x
3
denote the number of tons
of products A, B, and C, respectively, to be made. The number of hours that
machine X will be used is
2x
1
+3x
2
+4x
3
,
which must equal 80. Thus we have
2x
1
+3x
2
+4x
3
= 80.
Similarly, the number of hours that machine Y will be used is 60, so we have
2x
1
+2x
2
+3x
3
= 60.
Mathematically, our problem is to find nonnegative values of x
1
, x
2
, and x
3
so
that
2x
1
+ 3x
2
+ 4x
3
= 80
2x
1
+ 2x
2
+ 3x
3
= 60.
This linear system has infinitely many solutions. Following the method
of Example 4, we see that all solutions are given by
x
1
=
20 − x
3
2
x
2
= 20 − x
3
x
3
= any real number such that 0 ≤ x
3
≤ 20,
8 Chapter 1 Linear Equations and Matrices
since we must have x
1
≥ 0, x
2
≥ 0, and x
3
≥ 0. When x
3
= 10, we have
x
1
= 5, x
2
= 10, x
3
= 10
while
x
1
=
13
2
, x
2
= 13, x
3
= 7
when x
3
= 7. The reader should observe that one solution is just as good as the
other. There is no best solution unless additional information or restrictions
are given.
Key Terms
Linear equation Solution to a linear system No solution
Unknowns Method of elimination Infinitely many solutions
Solution to a linear equation Unique solution Manipulations on a linear system
Linear system
1.1 Exercises
In Exercises 1 through 14, solve the given linear system by
the method of elimination.
1. x + 2y = 8
3x − 4y = 4
2. 2x − 3y + 4z = −12
x − 2y + z = −5
3x + y + 2z = 1
3. 3x + 2y + z = 2
4x + 2y + 2z = 8
x − y + z = 4
4. x + y = 5
3x + 3y = 10
5. 2x + 4y + 6z = −12
2x − 3y − 4z = 15
3x + 4y + 5z = −8
6. x + y − 2z = 5
2x + 3y + 4z = 2
7. x + 4y − z = 12
3x + 8y − 2z = 4
8. 3x + 4y − z = 8
6x + 8y − 2z = 3
9. x + y + 3z = 12
2x + 2y + 6z = 6
10. x + y = 1
2x − y = 5
3x + 4y = 2
11. 2x + 3y = 13
x − 2y = 3
5x + 2y = 27
12. x − 5y = 6
3x + 2y = 1
5x + 2y = 1
13. x + 3y = −4
2x + 5y = −8
x + 3y = −5
14. 2x + 3y − z = 6
2x − y + 2z = −8
3x − y + z = −7
15. Given the linear system
2x − y = 5
4x − 2y = t ,
(a) determine a value of t so that the system has a
solution.
(b) determine a value of t so that the system has no
solution.
(c) how many different values of t can be selected in
part (b)?
16. Given the linear system
2x + 3y − z = 0
x − 4y + 5z = 0,
(a) verify that x
1
= 1, y
1
= −1, z
1
= −1 is a solution.
(b) verify that x
2
= −2, y
2
= 2, z
2
= 2 is a solution.
(c) is x = x
1
+ x
2
= −1, y = y
1
+ y
2
= 1, and
z = z
1
+ z
2
= 1 a solution to the linear system?
(d) is 3x, 3y, 3z, where x, y, and z are as in part (c), a
solution to the linear system?
17. Without using the method of elimination, solve the
linear system
2x + y − 2z = −5
3y + z = 7
z = 4.
18. Without using the method of elimination, solve the
linear system
4x = 8
−2x + 3y = −1
3x + 5y − 2z = 11.
19. Is there a value of r so that x = 1, y = 2, z = r is a
solution to the following linear system? If there is, find
it.
2x + 3y − z = 11
x − y + 2z = −7
4x + y − 2z = 12
Sec. 1.1 Linear Systems 9
20. Is there a value of r so that x = r, y = 2, z = 1 is a
solution to the following linear system? If there is, find
it.
3x − 2z = 4
x − 4y + z = −5
−2x + 3y + 2z = 9
21. Describe the number of points that simultaneously lie in
each of the three planes shown in each part of Figure 1.2.
22. Describe the number of points that simultaneously lie in
each of the three planes shown in each part of Figure 1.3.
P
3
P
2
P
1
(a)
P
1
P
3
P
2
(b)
(c)
P
3
P
1
P
2
Figure 1.3
23. An oil refinery produces low-sulfur and high-sulfur fuel.
Each ton of low-sulfur fuel requires 5 minutes in the
blending plant and 4 minutes in the refining plant; each
ton of high-sulfur fuel requires 4 minutes in the blending
plant and 2 minutes in the refining plant. If the blending
plant is available for 3 hours and the refining plant is
available for 2 hours, how many tons of each type of fuel
should be manufactured so that the plants are fully
utilized?
24. A plastics manufacturer makes two types of plastic:
regular and special. Each ton of regular plastic requires
2 hours in plant A and 5 hours in plant B; each ton of
special plastic requires 2 hours in plant A and 3 hours in
plant B. If plant A is available 8 hours per day and plant
B is available 15 hours per day, how many tons of each
type of plastic can be made daily so that the plants are
fully utilized?
25. A dietician is preparing a meal consisting of foods A, B,
and C. Each ounce of food A contains 2 units of protein,
3 units of fat, and 4 units of carbohydrate. Each ounce of
food B contains 3 units of protein, 2 units of fat, and 1
unit of carbohydrate. Each ounce of food C contains 3
units of protein, 3 units of fat, and 2 units of
carbohydrate. If the meal must provide exactly 25 units
of protein, 24 units of fat, and 21 units of carbohydrate,
how many ounces of each type of food should be used?
26. A manufacturer makes 2-minute, 6-minute, and
9-minute film developers. Each ton of 2-minute
developer requires 6 minutes in plant A and 24 minutes
in plant B. Each ton of 6-minute developer requires 12
minutes in plant A and 12 minutes in plant B. Each ton
of 9-minute developer requires 12 minutes in plant A
and 12 minutes in plant B. If plant A is available 10
hours per day and plant B is available 16 hours per day,
how many tons of each type of developer can be
produced so that the plants are fully utilized?
27. Suppose that the three points (1, −5), (−1, 1), and (2, 7)
lie on the parabola p(x) = ax
2
+bx +c.
(a) Determine a linear system of three equations in three
unknowns that must be solved to find a, b, and c.
(b) Solve the linear system obtained in part (a) for a, b,
and c.
28. An inheritance of $24,000 is to be divided among three
trusts, with the second trust receiving twice as much as
the first trust. The three trusts pay interest at the rates of
9%, 10%, and 6% annually, respectively, and return a
total in interest of $2210 at the end of the first year. How
much was invested in each trust?
Theoretical Exercises
T.1. Show that the linear system obtained by interchanging
two equations in (2) has exactly the same solutions as
(2).
T.2. Show that the linear system obtained by replacing an
equation in (2) by a nonzero constant multiple of the
equation has exactly the same solutions as (2).
T.3. Show that the linear system obtained by replacing an
equation in (2) by itself plus a multiple of another
equation in (2) has exactly the same solutions as (2).
T.4. Does the linear system
ax + by = 0
cx + dy = 0
always have a solution for any values of a, b, c, and d?
10 Chapter 1 Linear Equations and Matrices
1.2
MATRICES
If we examine the method of elimination described in Section 1.1, we make
the following observation. Only the numbers in front of the unknowns x
1
, x
2
,
. . . , x
n
are being changed as we perform the steps in the method of elimina-
tion. Thus we might think of looking for a way of writing a linear system
without having to carry along the unknowns. In this section we define an ob-
ject, a matrix, that enables us to do this—that is, to write linear systems in
a compact form that makes it easier to automate the elimination method on a
computer in order to obtain a fast and efficient procedure for finding solutions.
The use of a matrix is not, however, merely that of a convenient notation. We
now develop operations on matrices (plural of matrix) and will work with ma-
trices according to the rules they obey; this will enable us to solve systems of
linear equations and solve other computational problems in a fast and efficient
manner. Of course, as any good definition should do, the notion of a matrix
provides not only a new way of looking at old problems, but also gives rise to
a great many new questions, some of which we study in this book.
DEFINITION An m × n matrix A is a rectangular array of mn real (or complex) numbers
arranged in m horizontal rows and n vertical columns:
A =
_
_
_
_
_
_
_
_
_
a
11
a
12
· · · · · · a
1 j
· · · a
1n
a
21
a
22
· · · · · · a
2 j
· · · a
2n
.
.
.
.
.
.
· · · · · ·
.
.
.
· · ·
.
.
.
a
i 1
a
i 2
· · · · · ·
✻
✛
j th column
i th row a
i j
· · · a
i n
.
.
.
.
.
.
.
.
.
.
.
.
a
m1
a
m2
· · · · · · a
mj
· · · a
mn
_
¸
¸
¸
¸
¸
¸
¸
_
. (1)
The ith row of A is
_
a
i 1
a
i 2
· · · a
i n
_
(1 ≤ i ≤ m);
the jth column of A is
_
_
_
_
a
1 j
a
2 j
.
.
.
a
mj
_
¸
¸
_
(1 ≤ j ≤ n).
We shall say that A is m by n (written as m × n). If m = n, we say that A is
a square matrix of order n and that the numbers a
11
, a
22
, . . . , a
nn
form the
main diagonal of A. We refer to the number a
i j
, which is in the i th row and
j th column of A, as the i, jth element of A, or the (i, j) entry of A, and we
often write (1) as
A =
_
a
i j
_
.
For the sake of simplicity, we restrict our attention in this book, except
for Appendix A, to matrices all of whose entries are real numbers. However,
matrices with complex entries are studied and are important in applications.
Sec. 1.2 Matrices 11
EXAMPLE 1 Let
A =
_
1 2 3
−1 0 1
_
, B =
_
1 4
2 −3
_
, C =
_
_
1
−1
2
_
_
,
D =
_
_
1 1 0
2 0 1
3 −1 2
_
_
, E =
_
3
_
, F =
_
−1 0 2
_
.
Then A is a 2 × 3 matrix with a
12
= 2, a
13
= 3, a
22
= 0, and a
23
= 1; B is
a 2 × 2 matrix with b
11
= 1, b
12
= 4, b
21
= 2, and b
22
= −3; C is a 3 × 1
matrix with c
11
= 1, c
21
= −1, and c
31
= 2; D is a 3 ×3 matrix; E is a 1 ×1
matrix; and F is a 1 × 3 matrix. In D, the elements d
11
= 1, d
22
= 0, and
d
33
= 2 form the main diagonal.
For convenience, we focus much of our attention in the illustrative ex-
amples and exercises in Chapters 1–7 on matrices and expressions containing
only real numbers. Complex numbers will make a brief appearance in Chap-
ters 8 and 9. An introduction to complex numbers, their properties, and exam-
ples and exercises showing how complex numbers are used in linear algebra
may be found in Appendix A.
A 1 ×n or an n ×1 matrix is also called an n-vector and will be denoted
by lowercase boldface letters. When n is understood, we refer to n-vectors
merely as vectors. In Chapter 4 we discuss vectors at length.
EXAMPLE 2 u =
_
1 2 −1 0
_
is a 4-vector and v =
_
_
1
−1
3
_
_
is a 3-vector.
The n-vector all of whose entries are zero is denoted by 0.
Observe that if A is an n ×n matrix, then the rows of A are 1×n matrices
and the columns of A are n × 1 matrices. The set of all n-vectors with real
entries is denoted by R
n
. Similarly, the set of all n-vectors with complex
entries is denoted by C
n
. As we have already pointed out, in the first seven
chapters of this book we will work almost entirely with vectors in R
n
.
EXAMPLE 3 (Tabular Display of Data) The following matrix gives the airline distances
between the indicated cities (in statute miles).
_
_
_
London Madrid New York Tokyo
London 0 785 3469 5959
Madrid 785 0 3593 6706
New York 3469 3593 0 6757
Tokyo 5959 6706 6757 0
_
¸
_
EXAMPLE 4 (Production) Suppose that a manufacturer has four plants each of which
makes three products. If we let a
i j
denote the number of units of product i
made by plant j in one week, then the 4 ×3 matrix
_
_
_
Product 1 Product 2 Product 3
Plant 1 560 340 280
Plant 2 360 450 270
Plant 3 380 420 210
Plant 4 0 80 380
_
¸
_
12 Chapter 1 Linear Equations and Matrices
gives the manufacturer's production for the week. For example, plant 2 makes
270 units of product 3 in one week.
EXAMPLE 5 The wind chill table that follows shows how a combination of air temperature
and wind speed makes a body feel colder than the actual temperature. For
example, when the temperature is 10
◦
F and the wind is 15 miles per hour, this
causes a body heat loss equal to that when the temperature is −18
◦
F with no
wind.
◦
F
15 10 5 0 −5 −10
mphThis table can be represented as the matrix
A =
_
_
__
¸
_
.
EXAMPLE 6 With the linear system considered in Example 5 in Section 1.1,
x + 2y = 10
2x − 2y = −4
3x + 5y = 26,
we can associate the following matrices:
A =
_
_
1 2
2 −2
3 5
_
_
, x =
_
x
y
_
, b =
_
_
10
−4
26
_
_
.
In Section 1.3, we shall call A the coefficient matrix of the linear system.
DEFINITION A square matrix A =
_
a
i j
_
for which every term off the main diagonal is zero,
that is, a
i j
= 0 for i = j , is called a diagonal matrix.
EXAMPLE 7
G =
_
4 0
0 −2
_
and H =
_
_
−3 0 0
0 −2 0
0 0 4
_
_
are diagonal matrices.
Sec. 1.2 Matrices 13
DEFINITION A diagonal matrix A =
_
a
i j
_
, for which all terms on the main diagonal are
equal, that is, a
i j
= c for i = j and a
i j
= 0 for i = j , is called a scalar
matrix.
EXAMPLE 8 The following are scalar matrices:
I
3
=
_
_
1 0 0
0 1 0
0 0 1
_
_
, J =
_
−2 0
0 −2
_
.
The search engines available for information searches and retrieval on the
Internet use matrices to keep track of the locations of information, the type of
information at a location, keywords that appear in the information, and even
the way Web sites link to one another. A large measure of the effectiveness
of the search engine Google
c
is the manner in which matrices are used to
determine which sites are referenced by other sites. That is, instead of directly
keeping track of the information content of an actual Web page or of an indi-
vidual search topic, Google's matrix structure focuses on finding Web pages
that match the search topic and then presents a list of such pages in the order
of their "importance."
Suppose that there are n accessible Web pages during a certain month.
A simple way to view a matrix that is part of Google's scheme is to imagine
an n ×n matrix A, called the "connectivity matrix," that initially contains all
zeros. To build the connections proceed as follows. When it is detected that
Web site j links to Web site i , set entry a
i j
equal to one. Since n is quite large,
about 3 billion as of December 2002, most entries of the connectivity matrix
A are zero. (Such a matrix is called sparse.) If row i of A contains many ones,
then there are many sites linking to site i . Sites that are linked to by many
other sites are considered more "important" (or to have a higher rank) by the
software driving the Google search engine. Such sites would appear near the
top of a list returned by a Google search on topics related to the information
on site i . Since Google updates its connectivity matrix about every month, n
increases over time and new links and sites are adjoined to the connectivity
matrix.
The fundamental technique used by Google
c
to rank sites uses linear
algebra concepts that are somewhat beyond the scope of this course. Further
information can be found in the following sources.
1. Berry, Michael W., and Murray Browne. Understanding Search Engines—
Mathematical Modeling and Text Retrieval. Philadelphia: Siam, 1999.
2.
3. Moler, Cleve. "The World's Largest Matrix Computation: Google's Page
Rank Is an Eigenvector of a Matrix of Order 2.7 Billion," MATLAB News
and Notes, October 2002, pp. 12–13.
Whenever a new object is introduced in mathematics, we must define
when two such objects are equal. For example, in the set of all rational num-
bers, the numbers
2
3
and
4
6
are called equal although they are not represented
in the same manner. What we have in mind is the definition that
a
b
equals
c
d
when ad = bc. Accordingly, we now have the following definition.
DEFINITION Two m×n matrices A =
_
a
i j
_
and B =
_
b
i j
_
are said to be equal if a
i j
= b
i j
,
1 ≤ i ≤ m, 1 ≤ j ≤ n, that is, if corresponding elements are equal.
14 Chapter 1 Linear Equations and Matrices
EXAMPLE 9 The matrices
A =
_
_
1 2 −1
2 −3 4
0 −4 5
_
_
and B =
_
_
1 2 w
2 x 4
y −4 z
_
_
are equal if w = −1, x = −3, y = 0, and z = 5.
We shall now define a number of operations that will produce new matri-
ces out of given matrices. These operations are useful in the applications of
matrices.
MATRIX ADDITION
DEFINITION If A =
_
a
i j
_
and B =
_
b
i j
_
are m × n matrices, then the sum of A and B is
the m ×n matrix C =
_
c
i j
_
, defined by
c
i j
= a
i j
+b
i j
(1 ≤ i ≤ m, 1 ≤ j ≤ n).
That is, C is obtained by adding corresponding elements of A and B.
EXAMPLE 10 Let
A =
_
1 −2 4
2 −1 3
_
and B =
_
0 2 −4
1 3 1
_
.
Then
A + B =
_
1 +0 −2 +2 4 +(−4)
2 +1 −1 +3 3 +1
_
=
_
1 0 0
3 2 4
_
.
It should be noted that the sum of the matrices A and B is defined only
when A and B have the same number of rows and the same number of columns,
that is, only when A and B are of the same size.
We shall now establish the convention that when A + B is formed, both A
and B are of the same size.
Thus far, addition of matrices has only been defined for two matrices.
Our work with matrices will call for adding more than two matrices. Theorem
1.1 in the next section shows that addition of matrices satisfies the associative
property: A + (B + C) = (A + B) + C. Additional properties of matrix
addition are considered in Section 1.4 and are similar to those satisfied by the
real numbers.
EXAMPLE 11 (Production) A manufacturer of a certain product makes three models, A, B,
and C. Each model is partially made in factory F
1
in Taiwan and then finished
in factory F
2
in the United States. The total cost of each product consists of
the manufacturing cost and the shipping cost. Then the costs at each factory
(in dollars) can be described by the 3 ×2 matrices F
1
and F
2
:
F
1
=
_
_
Manufacturing
cost
Shipping
cost
32 40
50 80
70 20
_
_
Model A
Model B
Model C
Sec. 1.2 Matrices 15
F
2
=
_
_
Manufacturing
cost
Shipping
cost
40 60
50 50
130 20
_
_
Model A
Model B
Model C
The matrix F
1
+F
2
gives the total manufacturing and shipping costs for each
product. Thus the total manufacturing and shipping costs of a model Cproduct
are $200 and $40, respectively.
SCALAR MULTIPLICATION
DEFINITION If A =
_
a
i j
_
is an m×n matrix and r is a real number, then the scalar multiple
of A by r, r A, is the m ×n matrix B =
_
b
i j
_
, where
b
i j
= ra
i j
(1 ≤ i ≤ m, 1 ≤ j ≤ n).
That is, B is obtained by multiplying each element of A by r.
If A and B are m × n matrices, we write A + (−1)B as A − B and call
this the difference of A and B.
EXAMPLE 12 Let
A =
_
2 3 −5
4 2 1
_
and B =
_
2 −1 3
3 5 −2
_
.
Then
A − B =
_
2 −2 3 +1 −5 −3
4 −3 2 −5 1 +2
_
=
_
0 4 −8
1 −3 3
_
.
EXAMPLE 13 Let p =
_
18.95 14.75 8.60
_
be a 3-vector that represents the current prices
of three items at a store. Suppose that the store announces a sale so that the
price of each item is reduced by 20%.
(a) Determine a 3-vector that gives the price changes for the three items.
(b) Determine a 3-vector that gives the new prices of the items.
Solution (a) Since each item is reduced by 20%, the 3-vector
0.20p =
_
(0.20)18.95 (0.20)14.75 (0.20)8.60
_
=
_
3.79 2.95 1.72
_
gives the price reductions for the three items.
(b) The new prices of the items are given by the expression
p −0.20p =
_
18.95 14.75 8.60
_
−
_
3.79 2.95 1.72
_
=
_
15.16 11.80 6.88
_
.
Observe that this expression can also be written as
p −0.20p = 0.80p.
16 Chapter 1 Linear Equations and Matrices
If A
1
, A
2
, . . . , A
k
are m × n matrices and c
1
, c
2
, . . . , c
k
are real numbers,
then an expression of the form
c
1
A
1
+c
2
A
2
+· · · +c
k
A
k
(2)
is called a linear combination of A
1
, A
2
, . . . , A
k
, and c
1
, c
2
, . . . , c
k
are
called coefficients.
EXAMPLE 14 (a) If
A
1
=
_
_
0 −3 5
2 3 4
1 −2 −3
_
_
and A
2
=
_
_
5 2 3
6 2 3
−1 −2 3
_
_
,
then C = 3A
1
−
1
2
A
2
is a linear combination of A
1
and A
2
. Using scalar
multiplication and matrix addition, we can compute C:
C = 3
_
_
0 −3 5
2 3 4
1 −2 −3
_
_
−
1
2
_
_
5 2 3
6 2 3
−1 −2 3
_
_
=
_
_
_
_
−
5
2
−10
27
2
3 8
21
2
7
2
−5 −
21
2
_
¸
¸
_
.
(b) 2
_
3 −2
_
− 3
_
5 0
_
+ 4
_
−2 5
_
is a linear combination of
_
3 −2
_
,
_
5 0
_
, and
_
−2 5
_
. It can be computed (verify) as
_
−17 16
_
.
(c) −0.5
_
_
1
−4
−6
_
_
+ 0.4
_
_
0.1
−4
0.2
_
_
is a linear combination of
_
_
1
−4
−6
_
_
and
_
_
0.1
−4
0.2
_
_
.
It can be computed (verify) as
_
_
−0.46
0.4
3.08
_
_
.
THE TRANSPOSE OF A MATRIX
DEFINITION If A =
_
a
i j
_
is an m ×n matrix, then the n ×m matrix A
T
=
_
a
T
i j
_
, where
a
T
i j
= a
j i
(1 ≤ i ≤ n, 1 ≤ j ≤ m)
is called the transpose of A. Thus, the entries in each row of A
T
are the
entries in the corresponding column of A.
EXAMPLE 15 Let
A =
_
4 −2 3
0 5 −2
_
, B =
_
_
6 2 −4
3 −1 2
0 4 3
_
_
, C =
_
_
5 4
−3 2
2 −3
_
_
,
D =
_
3 −5 1
_
, E =
_
_
2
−1
3
_
_
.
Sec. 1.2 Matrices 17
Then
A
T
=
_
_
4 0
−2 5
3 −2
_
_
, B
T
=
_
_
6 3 0
2 −1 4
−4 2 3
_
_
,
C
T
=
_
5 −3 2
4 2 −3
_
, D
T
=
_
_
3
−5
1
_
_
, and E
T
=
_
2 −1 3
_
.
BIT MATRICES (OPTIONAL)
The majority of our work in linear algebra will use matrices and vectors whose
entries are real or complex numbers. Hence computations, like linear combi-
nations, are determined using matrix properties and standard arithmetic base
10. However, the continued expansion of computer technology has brought to
the forefront the use of binary (base 2) representation of information. In most
computer applications like video games, FAX communications, ATM money
transfers, satellite communications, DVD videos, or the generation of music
CDs, the underlying mathematics is invisible and completely transparent to
the viewer or user. Binary coded data is so prevalent and plays such a central
role that we will briefly discuss certain features of it in appropriate sections of
this book. We begin with an overview of binary addition and multiplication
and then introduce a special class of binary matrices that play a prominent role
in information and communication theory.
Binary representation of information uses only two symbols 0 and 1. In-
formation is coded in terms of 0 and 1 in a string of bits.
∗
For example, the
decimal number 5 is represented as the binary string 101, which is interpreted
in terms of base 2 as follows:
5 = 1(2
2
) +0(2
1
) +1(2
0
).
The coefficients of the powers of 2 determine the string of bits, 101, which
provide the binary representation of 5.
Just as there is arithmetic base 10 when dealing with the real and complex
numbers, there is arithmetic using base 2; that is, binary arithmetic. Table 1.1
shows the structure of binary addition and Table 1.2 the structure of binary
multiplication.
Table 1.1
+ 0 1
0 0 1
1 1 0
Table 1.2
× 0 1
0 0 0
1 0 1
The properties of binary arithmetic for combining representations of real
numbers given in binary form is often studied in beginning computer science
courses or finite mathematics courses. We will not digress to review such
topics at this time. However, our focus will be on a particular type of ma-
trix and vector that contain entries that are single binary digits. This class of
matrices and vectors are important in the study of information theory and the
mathematical field of error-correcting codes (also called coding theory).
∗
A bit is a binary digit; that is, either a 0 or 1.
18 Chapter 1 Linear Equations and Matrices
DEFINITION An m × n bit matrix
†
is a matrix all of whose entries are (single) bits. That
is, each entry is either 0 or 1.
A bit n-vector (or vector) is a 1 ×n or n ×1 matrix all of whose entries
are bits.
EXAMPLE 16 A =
_
_
1 0 0
1 1 1
0 1 0
_
_
is a 3 ×3 bit matrix.
EXAMPLE 17 v =
_
_
_
_
_
1
1
0
0
1
_
¸
¸
¸
_
is a bit 5-vector and u =
_
0 0 0 0
_
is a bit 4-vector.
The definitions of matrix addition and scalar multiplication apply to bit
matrices provided we use binary (or base 2) arithmetic for all computations
and use the only possible scalars 0 and 1.
EXAMPLE 18 Let A =
_
_
1 0
1 1
0 1
_
_
and B =
_
_
1 1
0 1
1 0
_
_
. Using the definition of matrix addition
and Table 1.1, we have
A + B =
_
_
1 +1 0 +1
1 +0 1 +1
0 +1 1 +0
_
_
=
_
_
0 1
1 0
1 1
_
_
.
Linear combinations of bit matrices or bit n-vectors are quite easy to com-
pute using the fact that the only scalars are 0 and 1 together with Tables 1.1
and 1.2.
EXAMPLE 19 Let c
1
= 1, c
2
= 0, c
3
= 1, u
1
=
_
1
0
_
, u
2
=
_
0
1
_
, and u
3
=
_
1
1
_
. Then
c
1
u
1
+c
2
u
2
+c
3
u
3
= 1
_
1
0
_
+0
_
0
1
_
+1
_
1
1
_
=
_
1
0
_
+
_
0
0
_
+
_
1
1
_
=
_
(1 +0) +1
(0 +0) +1
_
=
_
1 +1
0 +1
_
=
_
0
1
_
.
From Table 1.1 we have 0 + 0 = 0 and 1 + 1 = 0. Thus the additive
inverse of 0 is 0 (as usual) and the additive inverse of 1 is 1. Hence to compute
the difference of bit matrices A and B we proceed as follows:
A − B = A +(inverse of 1) B = A +1B = A + B.
We see that the difference of bit matrices contributes nothing new to the alge-
braic relationships among bit matrices.
†
A bit matrix is also called a Boolean matrix.
Sec. 1.2 Matrices 19
Key Terms
Matrix n-vector (or vector) Scalar multiple of a matrix
Rows Diagonal matrix Difference of matrices
Columns Scalar matrix Linear combination of matrices
Size of a matrix 0, the zero vector Transpose of a matrix
Square matrix R
n
, the set of all n-vectors Bit
Main diagonal of a matrix Google
c
Bit (or Boolean) matrix
Element (or entry) of a matrix Equal matrices Upper triangular matrix
i j th element Matrix addition Lower triangular matrix
(i, j ) entry Scalar multiplication
1.2 Exercises
1. Let
A =
_
2 −3 5
6 −5 4
_
, B =
_
_
4
−3
5
_
_
,
and
C =
_
_
7 3 2
−4 3 5
6 1 −1
_
_
.
(a) What is a
12
, a
22
, a
23
?
(b) What is b
11
, b
31
?
(c) What is c
13
, c
31
, c
33
?
2. If
_
a +b c +d
c −d a −b
_
=
_
4 6
10 2
_
,
find a, b, c, and d.
3. If
_
a +2b 2a −b
2c +d c −2d
_
=
_
4 −2
4 −3
_
,
find a, b, c, and d.
In Exercises 4 through 7, let
A =
_
1 2 3
2 1 4
_
, B =
_
_
1 0
2 1
3 2
_
_
,
C =
_
_
3 −1 3
4 1 5
2 1 3
_
_
, D =
_
3 −2
2 4
_
,
E =
_
_
2 −4 5
0 1 4
3 2 1
_
_
, F =
_
−4 5
2 3
_
,
and O =
_
_
0 0 0
0 0 0
0 0 0
_
_
.
4. If possible, compute the indicated linear combination:
(a) C + E and E +C (b) A + B
(c) D − F (d) −3C +5O
(e) 2C −3E (f) 2B + F
5. If possible, compute the indicated linear combination:
(a) 3D +2F
(b) 3(2A) and 6A
(c) 3A +2A and 5A
(d) 2(D + F) and 2D +2F
(e) (2 +3)D and 2D +3D
(f) 3(B + D)
6. If possible, compute:
(a) A
T
and (A
T
)
T
(b) (C + E)
T
and C
T
+ E
T
(c) (2D +3F)
T
(d) D − D
T
(e) 2A
T
+ B
(f) (3D −2F)
T
7. If possible, compute:
(a) (2A)
T
(b) (A − B)
T
(c) (3B
T
−2A)
T
(d) (3A
T
−5B
T
)
T
(e) (−A)
T
and −(A
T
)
(f) (C + E + F
T
)
T
8. Is the matrix
_
3 0
0 2
_
a linear combination of the
matrices
_
1 0
0 1
_
and
_
1 0
0 0
_
? Justify your answer.
9. Is the matrix
_
4 1
0 −3
_
a linear combination of the
matrices
_
1 0
0 1
_
and
_
1 0
0 0
_
? Justify your answer.
10. Let
A =
_
_
1 2 3
6 −2 3
5 2 4
_
_
and I
3
=
_
_
1 0 0
0 1 0
0 0 1
_
_
.
If λ is a real number, compute λI
3
− A.
20 Chapter 1 Linear Equations and Matrices
Exercises 11 through 15 involve bit matrices.
11. Let A =
_
_
1 0 1
1 1 0
0 1 1
_
_
, B =
_
_
0 1 1
1 0 1
1 1 0
_
_
, and
C =
_
_
1 1 0
0 1 1
1 0 1
_
_
. Compute each of the following.
(a) A + B (b) B +C (c) A + B +C
(d) A +C
T
(e) B −C
12. Let A =
_
1 0
1 0
_
, B =
_
1 0
0 1
_
, C =
_
1 1
0 0
_
, and
D =
_
0 0
1 0
_
. Compute each of the following.
(a) A + B (b) C + D (c) A + B +(C + D)
T
(d) C − B (e) A − B +C − D
13. Let A =
_
1 0
0 0
_
.
(a) Find B so that A + B =
_
0 0
0 0
_
.
(b) Find C so that A +C =
_
1 1
1 1
_
.
14. Let u =
_
1 1 0 0
_
. Find the bit 4-vector v so that
u +v =
_
1 1 0 0
_
.
15. Let u =
_
0 1 0 1
_
. Find the bit 4-vector v so that
u +v =
_
1 1 1 1
_
.
Theoretical Exercises
T.1. Show that the sum and difference of two diagonal
matrices is a diagonal matrix.
T.2. Show that the sum and difference of two scalar
matrices is a scalar matrix.
T.3. Let
A =
_
_
a b c
c d e
e e f
_
_
.
(a) Compute A − A
T
.
(b) Compute A + A
T
.
(c) Compute (A + A
T
)
T
.
T.4. Let O be the n ×n matrix all of whose entries are
zero. Show that if k is a real number and A is an n ×n
matrix such that k A = O, then k = 0 or A = O.
T.5. A matrix A =
_
a
i j
_
is called upper triangular if
a
i j
= 0 for i > j . It is called lower triangular if
a
i j
= 0 for i < j .
_
_
_
_
_
_
_
_
_
_
a
11
a
12
· · · · · · · · · a
1n
0 a
22
· · · · · · · · · a
2n
0 0 a
33
· · · · · · a
3n0 0 0 · · · 0 a
nn
_
¸
¸
¸
¸
¸
¸
¸
¸
_
Upper triangular matrix
(The elements below the main diagonal are zero.)
_
_
_
_
_
_
_
_
_
_
a
11
0 0 · · · · · · 0
a
21
a
22
0 · · · · · · 0
a
31
a
32
a
33
0 · · · 0 0
a
n1
a
n2
a
n3
· · · · · · a
nn
_
¸
¸
¸
¸
¸
¸
¸
¸
_
Lower triangular matrix
(The elements above the main diagonal are zero.)
(a) Show that the sum and difference of two upper
triangular matrices is upper triangular.
(b) Show that the sum and difference of two lower
triangular matrices is lower triangular.
(c) Show that if a matrix is both upper and lower
triangular, then it is a diagonal matrix.
T.6. (a) Show that if A is an upper triangular matrix, then
A
T
is lower triangular.
(b) Show that if A is a lower triangular matrix, then
A
T
is upper triangular.
T.7. If A is an n ×n matrix, what are the entries on the
main diagonal of A − A
T
? Justify your answer.
T.8. If x is an n-vector, show that x +0 = x.
Exercises T.9 through T.18 involve bit matrices.
T.9. Make a list of all possible bit 2-vectors. How many are
there?
T.10. Make a list of all possible bit 3-vectors. How many are
there?
T.11. Make a list of all possible bit 4-vectors. How many are
there?
Sec. 1.3 Dot Product and Matrix Multiplication 21
T.12. How many bit 5-vectors are there? How many bit
n-vectors are there?
T.13. Make a list of all possible 2 ×2 bit matrices. How
many are there?
T.14. How many 3 ×3 bit matrices are there?
T.15. How many n ×n bit matrices are there?
T.16 OFF.
T.17 ON.
T.18. A standard light switch has two positions (or states);
either on or off. Let bit matrix
A =
_
_
1 0
0 1
1 1
_
_
represent a bank of light switches where 0 represents
OFF and 1 represents ON.
(a) Find a matrix B so that A + B will represent the
bank of switches with the state of each switch
"reversed."
(b) Let
C =
_
_
1 1
0 0
1 0
_
_
.
Will the matrix B from part (a) also "reverse" that
state of the bank of switches represented by C?
Verify your answer.
(c) If A is any m ×n bit matrix representing a bank of
switches, determine an m ×n bit matrix B so that
A + B "reverses" all the states of the switches in
A. Give reasons why B will "reverse" the states
in A.
MATLAB Exercises
In order to use MATLAB in this section, you should first read
Sections 12.1 and 12.2, which give basic information about
MATLAB and about matrix operations in MATLAB. You are
urged to do any examples or illustrations of MATLAB
commands that appear in Sections 12.1 and 12.2 before
trying these exercises.
ML.1. In MATLAB, enter the following matrices.
A =
_
_
5 1 2
−3 0 1
2 4 1
_
_
,
B =
_
_
4 ∗ 2 2/3
1/201 5 −8.2
0.00001 (9 +4)/3
_
_
.
Using MATLAB commands, display the following.
(a) a
23
, b
32
, b
12
(b) row
1
(A), col
3
(A), row
2
(B)
(c) Type MATLAB command format long and
display matrix B. Compare the elements of B
from part (a) with the current display. Note that
format short displays four decimal places
rounded. Reset the format to format short.
ML.2. In MATLAB, type the command H = hilb(5); (Note
that the last character is a semicolon, which
suppresses the display of the contents of matrix H.
See Section 12.1.) For more information on the hilb
command, type help hilb. Using MATLAB
commands, do the following:
(a) Determine the size of H.
(b) Display the contents of H.
(c) Display the contents of H as rational numbers.
(d) Extract as a matrix the first three columns.
(e) Extract as a matrix the last two rows.
Exercises ML.3 through ML.5 use bit matrices and the
supplemental instructional commands described in Section
12.9.
ML.3. Use bingen to solve Exercises T.10 and T.11.
ML.4. Use bingen to solve Exercise T.13. (Hint: An n ×n
matrix contains the same number of entries as an
n
2
-vector.)
ML.5. Solve Exercise 11 using binadd.
1.3
DOT PRODUCT AND MATRIX MULTIPLICATION
In this section we introduce the operation of matrix multiplication. Unlike
matrix addition, matrix multiplication has some properties that distinguish it
from multiplication of real numbers.
22 Chapter 1 Linear Equations and Matrices
DEFINITION The dot product or inner product of the n-vectors a and b is the sum of the
products of corresponding entries. Thus, if
a =
_
_
_
_
a
1
a
2
.
.
.
a
n
_
¸
¸
_
and b =
_
_
_
_
b
1
b
2
.
.
.
b
n
_
¸
¸
_
,
then
a · b = a
1
b
1
+a
2
b
2
+· · · +a
n
b
n
=
n | 677.169 | 1 |
Name:
1
MATH 317 - Practice midterm
Elon Lindenstrauss Spring 2005
Please read carefully the following instructions: Time: 80 min. Write your name on the top of EVERY page. Answer ALL parts of ALL questions. This is a closed book exam. You may NOT use any | 677.169 | 1 |
William J. Mueller
Department of Mathematics
University of Arizona
Tucson, AZ 85721 mueller@math.arizona.edu
The authors are participants in the Connected Curriculum Project, a World Wide Web project with home page at:
and materials available on servers at Cal Poly San Luis Obispo, Duke University, and Montana State University. The project is supported by National Science Foundation grant DUE-9555407.
The Connected Curriculum Project
CCP is the result of a happy meeting of a new medium the World Wide Web with some older ideas. Those actively involved in the calculus reform movement have generally shared the following three goals: To break down the barriers between disciplines to study mathematics in the context of real problems. To engage our students in active learning. To help our students and ourselves develop a sense of ownership of the concepts and techniques discussed in our classes. Almost all of lower division mathematics was developed to study problems from the real world the same problems that most of our students hope eventually to be able to solve. By studying mathematics in context, students learn more about doing mathematics, learn more about the world around them, and are better prepared to use mathematics effectively outside the mathematics classroom. Mathematics is an essential language for studying other subjects, and it should be routinely used across the curriculum. Our first goal is Mathematics Across the Curriculum.
The architecture of the World Wide Web is ideally suited for accomplishing all three of these goals. Web browsers are designed to encourage very active and interactive engagement with material. We use a browser as a conductor, orchestrating a mix of text, graphics, movies, and sound with helper applications, such as Mathematica, Maple, Mathcad, and TI Graph Link. In a typical session, students go back and forth between their browser window, their CAS window, and "hands-on" laboratory equipment, exploring real phenomena, mathematical models, and the mathematics underlying the models at the same time.
The hypertext structure of web material emphasizes links and connections rather than compartmentalization. Students starting with material in one subject will find themselves following links to other subjects. Hypertext also emphasizes choice. By choosing applications and examples that interest them, and by selecting background material they need, students construct their own individualized courses, tailored to their own interests and background.
The malleability of web-based material encourages multiple authorship and, most importantly, multiple ownership. Web-based courses are living courses, evolving as our world evolves. Our goal is to create and maintain a large volume of highly interconnected material that will support learning across the curriculum with bridges rather than barriers.
Design, Implementation, and Learning Issues
Over the past few years we have benefitted from a study of the ways materials can and should be presented on a computer screen. Issues such as use of color, placement, and font face and size, and arrangement of text, interactive mathematics, and graphics are important factors in the pedagogical success of interactive modules. The CCP materials are designed to translate sound design principles to a web-based environment in which the author has only limited control over the final rendering of the display.
Our materials guide exploration by a variety of tools, e.g., computer algebra systems such as Maple, Mathematica, or Mathcad, Java-based interactions, and CBL data gathering. Our live presentation shows how to tie all these activities together through the web and how to use related web-based materials as part of the exploration.
How does use of these materials in a course affect the development of the course and the student approach to learning? As we outline in the next section, the materials may be used in a variety of ways. What effects do these different approaches and combinations of them have on learning? Our live examples illustrate the lessons we have learned.
Uses of CCP Materials
Our materials are being and will be used in various ways: As supplements to existing courses that use currently available textbooks. To support new courses that draw material from several different fields. By students on their own to supplement and complement material in their courses, and to help them see connections between courses and fields that are often overlooked in a departmentalized university.
These materials can also be used in different settings: In regularly scheduled laboratory components for traditionally structured classes. In self-scheduled laboratory components for traditionally structured classes, with work at various sites on campus or at home or in dorms wherever students have web access, plus a helper application and (if needed) lab equipment. In courses where some lecturing is replaced by group work, with students at workstations that include computers and lab equipment, and with an instructor circulating to work with students in their small groups. By students who choose to use CCP as an additional resource working alone or in groups at home, in dorms, or in open labs.
Contributing Authors and Participating Sites
In addition to the authors of this paper, participants in and contributors to CCP include:
In the future we expect to accept contributions to a Connected Curriculum Library that will comprise peer-reviewed and professionally-edited materials residing on a single site, with several mirror sites. For cover mathematical topics from agriculture, architecture and enviromental web-based versions of all the Project CALC labs.
The focus of the Montana State Site is on interactive texts, "case studies." Through links to many applications, the texts seek to break down barriers between disciplines and to encourage student and teacher "ownership" of individual courses.
Summary
The Connected Curriculum Project is producing innovative interactive materials in areas of mathematics from precalculus to linear algebra and differential equations, as well as materials that illustrate the use of mathematics in other disciplines across the curriculum. Some of the units are designed to be used as integral parts of courses others are free-standing reference modules for review and enrichment. | 677.169 | 1 |
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Publication Date :
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Edition :
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Course Name: Quantitative Literacy
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What is the essence of the similarity between forests in a graph and linearly independent sets of columns in a matrix? Why does the greedy algorithm produce a spanning tree of minimum weight in a connected graph? Is it possible to test in polynomial time whether a matrix is totally unimodular?These questions form the basis of Matroid theory. The study of matroids is a branch of discrete mathematics with basic links to graphs, lattices, codes, transversals, and projective geometries. Matroids are of fundamental importance in combinatorial optimization and their applications extend into electrical engineering and statics. This book falls into two parts: the first provides a comprehensive introduction to the basics of matroid theory, while the second treats more advanced topics. The book contains over five hundred exercises and includes, for the first time in one place, short proofs of all but one of the majortheorems in the subject. The final chapter lists sixty unsolved problems and describes progress towards their solutions.
REVIEWS for Matroid | 677.169 | 1 |
Product Description:
The last five years have seen an immense growth in the use of symbolic computing and mathematical software packages such as Maple. The first three chapters of this book provide a user-friendly introduction to computer-assisted algebra with Maple. The rest of the book then develops these techniques and demonstrates the use of this technology for deriving approximate solutions to differential equations (linear and nonlinear) and integrals. In each case, the mathematical concepts are comprehensively introduced, with an emphasis on understanding how solutions behave and why various approximations can be used. Where appropriate, the text integrates the use of Maple to extend the utility of traditional approximation techniques. Advanced Mathematical Methods with Maple is the ideal companion text for advanced undergraduate and graduate students of mathematics and the physical sciences. It incorporates over 1000 exercises with different levels of difficulty, for which solutions are provided on the Internet.
REVIEWS for Advanced Mathematical Methods with Maple | 677.169 | 1 |
If formulas have always been your nemeses, this book is for you! Thoroughly practical and authoritative, this book brings together, in three parts, thousands of formulas, rules, and figures to simplify, review, or to refresh the users memory of what he/she studied in school.
From the success of Maths Formulas app, Physics Formulas has been developed and released to help users quickly refer to any Physics formulas for their study and work. This app displays most popular formulas in seven categories: Mechanics, Electricity, Thermal physics, Periodic motion, Optics, Atomic physics, Constants This is the essential app for everybody especially students, engineers and scientists.
Students and research workers in mathematics, physics, engineering and other sciences will find this compilation of more than 2000 mathematical formulas and tables invaluable. They will see quickly why half a million copies were sold of the first edition! All the information included is practical – rarely used results are excluded. Topics range from elementary to advanced-from algebra, trigonometry and calculus to vector analysis, Bessel functions, Legendre polynomials and elliptic integrals. Great care has been taken to present all results concisely and clearly. Excellent to keep as a handy reference!
An introduction to a variety of perturbation techniques for ordinary differential equations, this work outlines applications through specific examples. Mathematicians, engineers, and applied scientists will find its exposition entirely accessible | 677.169 | 1 |
Matrix Calculator This calculator performs all matrix, vector operations. You can add, subtract, find length, find dot and cross product, check if vectors are dependant. For every operation, calculator will generate a detailed explanation
Scientific Calculator SP This is a scientific calculator, which you can use for addition, subtraction, multiplication & other | 677.169 | 1 |
Mathematics for Elementary Educators 1 Flashcards
Mathematics for Elementary Educators 1 Advice
Mathematics for Elementary Educators 1 Advice
Showing 1 to 3 of 6
I really enjoyed this course. I feel as a student of education and with plans of being a elementary school teacher it is very important to have a solid understanding of elementary mathematics.
Course highlights:
This class starts really basic and moves forward. I found the math lattice to be very interesting and a great tool. This was something that I feel is relatively new and was not part of our schools curriculum when I was a child.
Hours per week:
6-8 hours
Advice for students:
My advice would be to do all of your homework as well as the extra problems available. Open communication with your instructor is also key to being successful in this course.
Course Term:Fall 2016
Professor:Course mentor group
Course Tags:Math-heavyGreat Intro to the Subject
Aug 24, 2016
| Would recommend.
Pretty easy, overall.
Course Overview:
Great class with lots of review. It helps to fill in any background knowledge that might have been missed in high school
Course highlights:
Lot''s of review in this class. It helps anyone who may have struggles with some concepts in high school. It does a great job of giving students lots of practice to make sure they understand the concepts and lots of chances to review what has been learned.
Hours per week:
0-2 hours
Advice for students:
Work hard and pay attention to details.
Course Term:Fall 2014
Professor:Don't Know
Course Required?Yes
Course Tags:Background Knowledge ExpectedGreat Intro to the SubjectMany Small Assignments
Aug 24, 2016
| Would recommend.
Pretty easy, overall.
Course Overview:
The course was fun and informative. It makes sure that the student knows what they need to in order to be able to teach the math material to others.
Course highlights:
This course was a review of the material that I learned in high school. It was a great review and helped to build on the concepts that I already knew | 677.169 | 1 |
Graduation Requirment 3.0 credits. Students watch online lectures with illustrated examples, complete assigned exercises, take quizzes to test concept comprehension and if they've progressed satisfactorily, take an exam at the end of each unit. Students must demonstrate proficiency by completing unit exams with an 80% score before continuing to the next unit. Students may watch the online lectures repeatedly if they are struggling with a specific math concept and instructors have online office hours to take questions from students using chat software. (Note: an 'A' designation after the course name means first semester of curriculum and 'B' indicates the second semester.)
Course List
MATH 021 JH
-
Pre-Algebra 1A
Description
JH Requirement--Grade 7
This course provides the student with the opportunity to master concepts in whole numbers and fractionsJH Requirement--Grade 8 Prerequisite: Pre-Algebra 1
This course provides the student with the opportunity to master concepts of mathematics. The class starts off with a long unit to introduce you to geometry and continues with the skills of measurements. Student prepare for future studies as they master the basics of arithmetic. Students will be challenged to master these skills as they work through an individualized math program from arithmetic to algebra.
Unit 6 Geometry
Unit 7 Measurement
Materials Needed:
Math Notebook (Graph paper)
Scanner
Texts
MATH 041 PRE-ALG
-
Adv Pre-Algebra - A
Description
JH Advanced Requirement--Grade 7
Will take Algebra 1 as an 8th Grader
This course provides the student with the opportunity to master concepts in whole numbers, fractions and decimals. The class also covers these basic math skills: percentages, basic geometry, measurements, graphing, statistics and an introduction to algebra | 677.169 | 1 |
PaGE: Mathematics (X.MATH)
X.MATH-400 Developing Mathematical Ideas: Building a System of Tens
Credits: 2
Participants will explore the base-ten structure of the number system, consider how that structure is exploited in multi-digit computational procedures, and examine how basic concepts of whole numbers reappear when working with decimals. They will study the various ways children naturally tend to think about separating and combining numbers and what children must understand in order to work with numbers in these ways.
This course provides opportunities for participants to examine the actions and situations modeled by the four basic operations. The course will begin with a view of young children's counting strategies as they encounter word problems, moves to an examination of the four basic operations on whole numbers, and revisits the operations in the context of rational numbers.
Applies to requirement(s): Meets No Distribution Requirement M. Flynn Restrictions: This course is limited to Mount Holyoke MTHTE.MAT students only
X.MATH-402 Developing Mathematical Ideas: Examining Features of Shape
Credits: 2
Participants examine aspects of two-dimensional and three-dimensional shapes, develop geometric vocabulary, and explore both definitions and properties of geometric objects. The seminar includes a study of angle, similarity, congruence, and the relationships between three-dimensional objects and their two-dimensional representations. Participants examine how students develop these concepts through analyzing print and video cases as well as reading and discussing research articles.
Participants will work with the collection, representation, description, and interpretation of data. They will learn what various graphs and statistical measures show about features of the data, study how to summarize data when comparing groups, and consider whether the data provides insight into the questions that led to data collection.
Applies to requirement(s): Meets No Distribution Requirement V. Bastable, M. Riddle Restrictions: This course is limited to Mount Holyoke MTHTE.MAT students only
Participants will examine different aspects of size, develop facility in composing and decomposing shapes, and apply these skills to make sense of formulas for area and volume. They will also explore conceptual issues of length, area, and volume, as well as their complex interrelationships.
Participantstables, or rules. Participants examine how students develop these concepts through analyzing print and video cases as well as reading and discussing research articles.Connecting Arithmetic to Algebra (CAA) is a year-long M. Flynn, The department Restrictions: This course is limited to Mount Holyoke MTHTE.MAT students only Instructor permission required. Coreq: X.MTHED-460. Notes: This is a year-long online course. At the conclusion of the 2-semester sequence, final letter grades will be awarded for both segments of the sequence.
X.MATH-462 Fostering Algebraic Reasoning
Fall and Spring. Credits: 3 | 677.169 | 1 |
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Unformatted text preview: Department of Mathematics Student Waterloo Catholic District School Board MFM1PI Ministry Document : The Ontario Curriculum Grades 9 and 10: Mathematics 2005 Prerequisite(s): Grade 8 Math Credit(s): 1.0 Course Description: This course enables students to develop an understanding of mathematical concepts related to introductory algebra, proportional reasoning, and measurement and geometry through investigation, the effective use of technology, and hands-on activities. Students will investigate real-life examples to develop various representations of linear relations, and will determine the connections between the representations. They will also explore certain relationships that emerge from the measurement of three-dimensional figures and two- dimensional shapes. Students will consolidate their mathematical skills as they solve problems and communicate their thinking. Overall Expectations for Student Learning Through this course, students will be expected to demonstrate knowledge, skills and values related to the...
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This note was uploaded on 04/19/2011 for the course MTH 110 taught by Professor Helenius during the Fall '08 term at Grand Valley State University. | 677.169 | 1 |
Mathematics learning in Stage 5
The arrangement of content in Stage 5 acknowledges the wide range of achievement of students in Mathematics by the time they reach the end of Year 8. Three substages of Stage 5 (Stages 5.1, 5.2 and 5.3) have been identified and made explicit in the syllabus:
Stage 5.1 is designed to assist in meeting the needs of students who are continuing to work towards the achievement of Stage 4 outcomes when they enter Year 9
Stage 5.2 builds on the content of Stage 5.1 and is designed to assist in meeting the needs of students who have achieved Stage 4 outcomes, generally by the end of Year 8
Stage 5.3 builds on the content of Stage 5.2 and is designed to assist in meeting the needs of students who have achieved Stage 4 outcomes before the end of Year 8.
Students studying some or all of the content of Stage 5.2 also study all of the content of Stage 5.1. Similarly, students studying some or all of the content of Stage 5.3 also study all of the content of Stage 5.1 and Stage 5.2. Content written in different substages within Stage 5 may be studied continuously. For example, students may study the content of the Linear Relationships substrand in Stage 5.1, followed immediately by the content of the Linear Relationships substrand in Stage 5.2, or by the content of the Linear Relationships substrand in Stage 5.2 and the Linear Relationships substrand in Stage 5.3. The following diagram illustrates the relationship between the substages in Stage 5.
A large variety of 'endpoints' is possible in Stage 5. For example, some students may achieve all of the Stage 5.2 outcomes and a selection of the Stage 5.3 outcomes by the end of Year 10.
When planning learning experiences for students in Years 9 and 10, teachers need to consider the courses of study that their students plan to follow beyond Stage 5. The table below outlines Stage 5 content recommendations in relation to current Stage 6 Mathematics Board Developed Courses. Other students may study mathematics in Stage 6 through Board Endorsed Courses or Life Skills. The diagram contained in the section titled The place of the Mathematics K–10Syllabus in the K–12 curriculum shows the relationship between Stage 5 Mathematics and the various courses offered in Stage 6.
Intended Stage 6 Board Developed Course
Recommended Stage 5 content (minimum)
Preliminary Mathematics General/ HSC Mathematics General 2
◊
(substrands identified by ◊)
All substrands of Stage 5.1 and the following Stage 5.2 substrands:
Financial Mathematics
Non-Linear Relationships
Right-Angled Triangles (Trigonometry)
Single Variable Data Analysis
Mathematics ('2 Unit')
§
(substrands identified by §)
All substrands of Stage 5.1 and Stage 5.2 and the following Stage 5.3 substrands:
Algebraic Techniques
Surds and Indices
Equations
Linear Relationships
and at least some of the content of the following Stage 5.3 substrands:
Non-Linear Relationships
Trigonometry and Pythagoras' Theorem
Properties of Geometrical Figures
Mathematics Extension 1
#
(optional substrands identified by #)
All substrands of Stage 5.1, Stage 5.2 and Stage 5.3, including the optional Stage 5.3 substrands: | 677.169 | 1 |
Project #68775 - Linear Algebra application homework
18 Mathematics Tutors Online
The link shows 14 possible assignments, only one needs to be completed. is preferred, because it seems the easiest conceptually, and I understand it the best. As far as the writing portion of the grade goes, I can write it myself if given a general understanding of the work done. | 677.169 | 1 |
Not a C Minus is a comprehensive study aid for senior high school Mathematics. It covers topics such as calculus, probability, finance and trigonometry, and uses a conversational, informal teaching style. Every topic is explained in detail, with sample questions and worked solutions.
- How to tutor students the right way.
- How to find your clients.
- How to make tutoring lessons effective and fun for students.
- How to praise your students the right way.
- How to start and build your tutoring business.
- How to raise your tutoring fee.
- Make an impact in a student's life.
- Help students achieve better grades and re-ignite their passions for education and life. | 677.169 | 1 |
Description
From rainbows, river meanders, and shadows to spider webs, honeycombs, and the markings on animal coats, the visible world is full of patterns that can be described mathematically. Examining such readily observable phenomena, this book introduces readers to the beauty of nature as revealed by mathematics and the beauty of mathematics as revealed in nature.
Generously illustrated, written in an informal style, and replete with examples from everyday life, Mathematics in Nature is an excellent and undaunting introduction to the ideas and methods of mathematical modeling. It illustrates how mathematics can be used to formulate and solve puzzles observed in nature and to interpret the solutions. In the process, it teaches such topics as the art of estimation and the effects of scale, particularly what happens as things get bigger. Readers will develop an understanding of the symbiosis that exists between basic scientific principles and their mathematical expressions as well as a deeper appreciation for such natural phenomena as cloud formations, halos and glories, tree heights and leaf patterns, butterfly and moth wings, and even puddles and mud cracks.
Developed out of a university course, this book makes an ideal supplemental text for courses in applied mathematics and mathematical modeling. It will also appeal to mathematics educators and enthusiasts at all levels, and is designed so that it can be dipped into at leisure.
About the author
John A. Adam is Professor of Mathematics at Old Dominion University, coeditor of "A Survey of Models for Tumor-Immune System Dynamics", and a regular contributor to leading journals in applied mathematics.
SimilarBecause of the increasing demands and complexity of undergraduate physics courses (atomic, quantum, solid state, nuclear, etc.), it is often impossible to devote separate courses to the classic wave phenomena of optics, acoustics, and electromagnetic radiation. This brief comprehensive text helps alleviate the problem with a unique overview of classical wave theory in one volume. By examining a sequence of concrete and specific examples (emphasizing the physics of wave motion), the authors unify the study of waves, developing abstract and general features common to all wave motion. The fundamental ideas of wave motion are set forth in the first chapter, using the stretched string as a particular model. In Chapter Two, the two-dimensional membrane is used to introduce Bessel functions and the characteristic features of waveguides. In Chapters Three and Four, elementary elasticity theory is developed and applied to find the various classes of waves that can be supported by a rigid rod. The impedance concept is also introduced at this point. Chapter Five discusses acoustic waves in fluids. The remainder of the book offers concise coverage of hydrodynamic waves at a liquid surface, general waves in isotropic elastic solids, electromagnetic waves, the phenomenon of wave diffraction, and other important topics. A special feature of this book is the inclusion of additional material designed to encourage the serious student to investigate topics often not covered in lectures. Throughout, the mathematics is kept relatively simple (mostly differential equations) and is accessible to advanced undergraduates with a year of calculus. In addition, carefully selected problems at the end of each section extend the coverage of the text by asking the student to supply mathematical details for calculations outlined in the section, or to develop the theory for related cases. Impressively broad in scope, Physics of Waves offers a novel approach to the study of classical wave theory — a wide-ranging but thorough survey of an important discipline that pervades much of contemporary physics. The simplicity, breadth, and brevity of the book make it ideal as a classroom text or as a vehicle for self-study.
Fluid mechanics, the study of how fluids behave and interact under various forces and in various applied situations—whether in the liquid or gaseous state or both—is introduced and comprehensively covered in this widely adopted text. Revised and updated by Dr. David Dowling, Fluid Mechanics, 5e is suitable for both a first or second course in fluid mechanics at the graduate or advanced undergraduate level.Along with more than 100 new figures, the text has been reorganized and consolidated to provide a better flow and more cohesion of topics.Changes made to the book's pedagogy in the first several chapters accommodate the needs of students who have completed minimal prior study of fluid mechanics.More than 200 new or revised end-of-chapter problems illustrate fluid mechanical principles and draw on phenomena that can be observed in everyday life
The multidisciplinary field of fluid mechanics is one of the most actively developing fields of physics, mathematics and engineering. In this book, the fundamental ideas of fluid mechanics are presented from a physics perspective. Using examples taken from everyday life, from hydraulic jumps in a kitchen sink to Kelvin–Helmholtz instabilities in clouds, the book provides readers with a better understanding of the world around them. It teaches the art of fluid-mechanical estimates and shows how the ideas and methods developed to study the mechanics of fluids are used to analyze other systems with many degrees of freedom in statistical physics and field theory. Aimed at undergraduate and graduate students, the book assumes no prior knowledge of the subject and only a basic understanding of vector calculus and analysis. It contains 32 exercises of varying difficulties, from simple estimates to elaborate calculations, with detailed solutions to help readers understand fluid mechanics.
Fluid Mechanics, Second Edition deals with fluid mechanics, that is, the theory of the motion of liquids and gases. Topics covered range from ideal fluids and viscous fluids to turbulence, boundary layers, thermal conduction, and diffusion. Surface phenomena, sound, and shock waves are also discussed, along with gas flow, combustion, superfluids, and relativistic fluid dynamics. This book is comprised of 16 chapters and begins with an overview of the fundamental equations of fluid dynamics, including Euler's equation and Bernoulli's equation. The reader is then introduced to the equations of motion of a viscous fluid; energy dissipation in an incompressible fluid; damping of gravity waves; and the mechanism whereby turbulence occurs. The following chapters explore the laminar boundary layer; thermal conduction in fluids; dynamics of diffusion of a mixture of fluids; and the phenomena that occur near the surface separating two continuous media. The energy and momentum of sound waves; the direction of variation of quantities in a shock wave; one- and two-dimensional gas flow; and the intersection of surfaces of discontinuity are also also considered. This monograph will be of interest to theoretical physicists.
Many of the world's most common processes and interactions are governed by the laws of thermodynamics and mechanics. While the transfer, release, or absorption of heat often accompany chemical reactions or seem inherent to mechanical systems, they are also familiar to anyone who has ever spent time outdoors on a warm day or touched a hot plate. Likewise, any physical body—large or small, solid or fluid—is subject to a wide range of forces that trigger motion. This detailed compendium explores the foundations and laws of both thermodynamics and mechanics as well as the lives of those individuals who helped advance these fundamental areas of physics.
"The material is superbly chosen and brilliantly written … the language is clear, direct and rigorous … a superb addition to the library of any physicist." — Physics Today.
This lucidly written text emphasizes optics and acoustics, with considerable emphasis placed on establishing a close connection between mathematical expressions and the associated physical ideas. Beginning with an inductive derivation of the equation for transverse waves on a string, the student is led through successively more complex subjects as acoustic plane waves, boundary-value problems, polarization, optical anisotropy, interference, diffraction, dispersion, and three-dimensional waves. Designed for a one-semester intermediate level undergraduate course (although there is easily enough material here for two semesters), Wave Phenomena is accessible to students with calculus up to the level of partial differentiations. Mathematical techniques, beyond the most elementary ones, are evolved, when needed, in the book itself. Moreover, the treatment is so thorough that students could work through the book themselves with minimal help from an instructor. There are a large number of interesting, well-chosen problems at the end of each chapter (with solutions for about half of them), and several appendixes providing additional mathematical apparatus: the representation of sinusoidal functions by complex numbers, frequently used trigonometric identities, one-dimensional hydrodynamic equations in Eulerian form, and more. Although geared to undergraduate physics majors, the comprehensiveness, rigor, and clarity of this text make it an ideal reference for almost any physicist in need of a review of optics or acoustics.
Like its predecessor, 200 Puzzling Physics Problems, this book is aimed at strengthening students' grasp of the laws of physics by applying them to situations that are practical, and to problems that yield more easily to intuitive insight than to brute-force methods and complex mathematics. The problems are chosen almost exclusively from classical, non-quantum physics, but are no easier for that. They are intriguingly posed in accessible non-technical language, and require readers to select an appropriate analysis framework and decide which branches of physics are involved. The general level of sophistication needed is that of the exceptional school student, the good undergraduate, or the competent graduate student; some physics professors may find some of the more difficult questions challenging. By contrast, the mathematical demands are relatively minimal, and seldom go beyond elementary calculus. This further book of physics problems is not only instructive and challenging, but also enjoyable entertaining | 677.169 | 1 |
About this product
Description
Description
Written for graduate and advanced undergraduate students in engineering and science, this classic book focuses primarily on set theory, algebra, and analysis. Useful as a course textbook, for self-study, or as a reference, the work is intended to familiarize engineering and science students with a great deal of pertinent and applicable mathematics in a rapid and efficient manner without sacrificing rigor. The book is divided into three parts: set theory, algebra, and analysis. It offers a generous number of exercises integrated into the text and features applications of algebra and analysis that have a broad appeal. | 677.169 | 1 |
Engineering Mathematics - Numerical Analysis
About the Course
Numerical Analysis, also known as- Numerical Methods and Statistics; is a subject which is taught in engineering disciplines, management studies and in B.Sc Math Honours. Do you know you can solve many real life problems easily if you know Numerical Methods? It's a very interesting subject and is quite easy too if you can understand it!
In this tutorial the tutor will explain almost all Numerical Methods topics you ever need to pass your exam! He'll tell you how to solve them easily step-by-step and will you give you many shortcut tips so that sometimes you can directly get the answer without doing the whole math! | 677.169 | 1 |
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
0.22 MB | 4 pages
PRODUCT DESCRIPTION
I designed these lessons to teach my students about geometric figures (the 7th in a series of 9). This lesson can be purchased as a complete bundled unit at a discounted price under the listing Features of Functions Complete Bundled Unit.
This lesson focuses on: quick review on writing rations and proportions given a context, given a graph and a table of values for two linear functions finding x and y values and adding the functions together and graphing the result, and rewriting equations into slope-intercept form | 677.169 | 1 |
Preview
Features
Offers a flexible presentation that allows instructors to match the text to their syllabus
Describes exciting applications of number theory to various problems in cryptography
Discusses recent developments
Includes "check your understanding" problems that test students on fundamental ideas, with solutions at the end of each chapter
Provides numerous examples as well as standard and computational exercises, with selected answers or hints in an appendix
Reviews the prerequisites in an appendix
Solutions manual available upon qualifying course adoption
Summary
Elementary Number Theory takes an accessible approach to teaching students about the role of number theory in pure mathematics and its important applications to cryptography and other areas.
The first chapter of the book explains how to do proofs and includes a brief discussion of lemmas, propositions, theorems, and corollaries. The core of the text covers linear Diophantine equations; unique factorization; congruences; Fermat's, Euler's, and Wilson's theorems; order and primitive roots; and quadratic reciprocity. The authors also discuss numerous cryptographic topics, such as RSA and discrete logarithms, along with recent developments.
The book offers many pedagogical features. The "check your understanding" problems scattered throughout the chapters assess whether students have learned essential information. At the end of every chapter, exercises reinforce an understanding of the material. Other exercises introduce new and interesting ideas while computer exercises reflect the kinds of explorations that number theorists often carry out in their research.
Author(s) Bio
James S. Kraft teaches mathematics at the Gilman School. He has previously taught at the University of Rochester, St. Mary's College of California, and Ithaca College. He has also worked in communications security. Dr. Kraft has published several research papers in algebraic number theory. He received his Ph.D. from the University of Maryland.
Lawrence C. Washington is a professor of mathematics and Distinguished Scholar-Teacher at the University of Maryland. Dr. Washington has published extensively in number theory, including books on cryptography (with Wade Trappe), cyclotomic fields, and elliptic curves. He received his Ph.D. from Princeton University.
Reviews
"This is a nice introduction to elementary number theory, designed for use in a basic undergraduate course. It can be used also for advanced high school students taking an accessible approach for an independent study. The book underlines the role of number theory in pure mathematics and its applications to cryptography and other areas." —Zentralblatt MATH 1322 | 677.169 | 1 |
Math Skills Maintenance Workbook: Course 3 - Taschenbuch
Internationaler Buchtitel. In englischer Sprache. Verlag: GLENCOE SECONDARY, 80 Seiten, L=277mm, B=214mm, H=5mm, Gew.=150gr, [GR: 27240 - TB/Didaktik/Methodik/Schulpädagogik/Fachdidaktik], [SW: - Textbooks], Kartoniert/Broschiert, Klappentext:..McGraw-Hill Education:
In order for their skills to remain fresh, students need opportunities to practice the math skills that they have learned in previous courses. Math Skills Maintenance contains pages of practices for various basic math skills. Each page is geared to one or more previously-learned skills. McGraw-Hill Education, Books, Math Skills Maintenance Workbook, Course 3 Books Math Skills Maintenance Workbook: Course 3 McGraw-Hill Schule & Lernen / Lehrermaterialien / Didaktik & Methodik 978-0-07-860733-2, McGraw-Hill/Glencoe Fremdsprachige Bücher / Englische Bücher 978-0-07-860733-2, Glencoe Secondary Long
McGraw-Hill
Titel:
Math Skills Maintenance Workbook: Course 3
ISBN-Nummer:
0078607337 | 677.169 | 1 |
Highschool Math Education with JavaScript
I have been talking with some of my coworkers, and a highschool math teacher friend of mine about developing a math curriculum for highschoolers based on JavaScript. The premise is that JavaScript programming is cool because your can build web applications with it, and implementing algorithms with JavaScript functions is more meaningful then memorizing textbook algorithms. | 677.169 | 1 |
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
2.33 MB | 48 pages
PRODUCT DESCRIPTION
Prepare your students for higher levels of thinking and the world of Algebra application. The major question from students has always been "when will I ever use this in real life". This set of activities is a step-by-step lesson in analyzing information and formulating a conclusion.
I created these lessons to reinforce the definitions of dependent and independent variable, the ability to generate and graph data, calculate an equation that best fits the data, project a future value, and formulate a conclusion.
Before assigning any of the activities, the teacher should make sure that the student can perform operations with positive and negative numbers and have discussed plotting points on a graph. Also if you have access to a computer or computer lab I would suggest that you download the "How to incorporate a spreadsheet software" guide and introduce your students to its application.
An objective for each activity is provided as well as a solution. Use of a graphing calculator or spreadsheet software such as Excel may be an optional activity to generating the graph.
Common core standards met using this product include CCSS Math. 6.EE.B.5, 7.EE.B.3, 7.EE.B.4, 7.EE.B.7, 7.EE.B.8.
Some additional activities you may wish to check out: (Click on the product you wish to view) | 677.169 | 1 |
9780817643102 Combinatorial Problems
"102 Combinatorial Problems" consists of carefully selected problems that have been used in the training and testing of the USA International Mathematical Olympiad (IMO) team. Key features: * Provides in-depth enrichment in the important areas of combinatorics by reorganizing and enhancing problem-solving tactics and strategies * Topics include: combinatorial arguments and identities, generating functions, graph theory, recursive relations, sums and products, probability, number theory, polynomials, theory of equations, complex numbers in geometry, algorithmic proofs, combinatorial and advanced geometry, functional equations and classical inequalities The book is systematically organized, gradually building combinatorial skills and techniques and broadening the student's view of mathematics. Aside from its practical use in training teachers and students engaged in mathematical competitions, it is a source of enrichment that is bound to stimulate interest in a variety of mathematical areas that are tangential to combinatorics. | 677.169 | 1 |
About this product
Description
Description
We present here the first of ten units of study used by the author to train beginning graduate students and advanced undergraduates for further study in the general topic of combinatorial algorithms. This unit of study focuses on mathematical concepts used to linearly order sets of objects. Our general approach is to focus on the geometric theory of algorithms by which we mean the use of graphical or pictorial ways to understand what is going on. This approach is both fun and powerful, preparing you to invent your own algorithms for a wide range of problems. | 677.169 | 1 |
Kirk Lowery mentioned Sage to me and with mathematics being fundamental to IR, it seemed like a good resource to mention. Either for research, using one of the course books or satisfying yourself that algorithms operate as advertised.
You don't have to take someone's word on algorithms. Use a small enough test case that you will recognize the effects of the algorithm. Or test it against another algorithm said to give similar results.
I saw a sad presentation years ago when a result was described as significant because the manual for the statistics package used said it was significant. Don't let that be you, either in front of a client or in a presentation to peers.
Sage is built out of nearly 100 open-source packages and features a unified interface. Sage can be used to study elementary and advanced, pure and applied mathematics. This includes a huge range of mathematics, including basic algebra, calculus, elementary to very advanced number theory, cryptography, numerical computation, commutative algebra, group theory, combinatorics, graph theory, exact linear algebra and much more. It combines various software packages and seamlessly integrates their functionality into a common experience. It is well-suited for education and research.
The user interface is a notebook in a web browser or the command line. Using the notebook, Sage connects either locally to your own Sage installation or to a Sage server on the network. Inside the Sage notebook you can create embedded graphics, beautifully typeset mathematical expressions, add and delete input, and share your work across the network.
The following showcase presents some of Sage's capabilities, screenshots and gives you an overall impression of what Sage is. The examples show the lines of code in Sage on the left side, accompanied by an explanation on the right. They only show the very basic concepts of how Sage works. Please refer to the documentation material for more detailed explanations or visit the library to see Sage in action.
In all fairness to Mathematica, the hobbyist version is only $295 for Mathematica 8. With versions for Windows (XP/Vista/7) Max OS X (Intel) and Linux. There is a reason why people want to be like…some other software. Mathematica has data mining capabilities and a host of other features. I am contemplating a copy of Mathematica as a Christmas present for myself.
Do note that all of the Fortune 50 companies use Mathematica. The hobbyist version allows you to add an important skill set that is relevant to a select clientele. Not to mention various government agencies, etc.
Should a job come along that requires it, I can simply upgrade to a professional license. Why? Well, I expect people to pay my invoices when I submit them. Why shouldn't I pay for software I use on the jobs that result in those invoices?
Don't cut corners on software. Same goes for the quality of jobs. It will show. If you don't know, don't lie, say you don't know but will find out. Clients will find simple honesty quite refreshing. (I can't promise that result for you but it has been the result for me over a variety of professions.) | 677.169 | 1 |
Home Shop About Us Policies Contact Us Schaum's Outline of Intermediate Algebra, Second Edition (Schaum's Outlines) Textbooks, EducationSEYMOUR LIPSCHUTZ - Schaum's Outline of Theory and Problems of Linear Algebra (S SEYMOUR LIPSCHUTZ - Schaum's Outline of Theory and Problems of Linear Algebra (Schaum's OutlinesThe super-condensed guide to college algebra combines the authority of Schaum's with quick-study approach to help busy students achieve better grades. This new edition is enhanced with a free-access online diagnostic test: an extensive set of review questions that pinpoint weaknesses and ensure full mastery of the subject.About the Book Schaum's Easy Outlines of College Algebraprovides busy students with a powerful tool to review the subject rapidly. By paring textbook subject matter down to the essentials, this handy guide makes every minute of study time count.Packed with quick study tips and at-a-glance tables and diagrams, this book is perfect for test preparation, pre-exam review, and last-minute cram situations. It combines the academic authority that the Schaum's name is known for,...
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The ideal review for the hundreds of thousands of college students who enroll in college algebra courses--now featuring free online videos Schaum's Outline of College Algebra is a complete, concise guide to college algebra, for students in 4-year colleges, 2-year colleges, or high school taking courses variously titled College Algebra, Precalculus, Algebra I, Algebra II, Linear Algebra, Arithmetic and Topics in Algebra, or Functions and Graphs. Features Outline format supplies a concise guide to the standard college course in college algebra 1,940 fully worked problems of varying difficulty Hundreds of additional practice problems Supplements the major leading textbooks in college algebra
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The ideal review for your elementary mathematics 2,500 fully worked problems of varying difficulty Clear, concise explanations of arithmetic, algebra, and geometry Outline format supplies a concise guide to the standard college courses in elementary mathematics Appropriate for the following courses: Basic Mathematics, Elementary Mathematics, Introduction to Mathematics | 677.169 | 1 |
In a detailed and comprehensive introduction to the theory of plane algebraic curves, the authors examine this classical area of mathematics that both figured prominently in ancient Greek studies and remains a source of inspiration and a topic of research to this day.
A beautiful and relatively elementary account of a part of mathematics where three main fields - algebra, analysis and geometry - meet. The book provides a broad view of these subjects at the level of calculus, without being a calculus book. Its roots are in arithmetic and geometry, the two opposite poles of mathematics, and the source of historic conceptual conflict. | 677.169 | 1 |
Topology of Surfaces - nouveau livre
[EAN: 9780387941028], Neubuch, [PU: Springer Okt 1997], DIFFERENZIALTOPOLOGIE ( TOPOLOGIE ); -, This item is printed on demand - Print on Demand Titel. - This book aims to provide undergraduates with an understanding of geometric topology. Topics covered include a sampling from point-set, geometric, and algebraic topology. The presentation is pragmatic, avoiding the famous pedagogical method 'whereby one begins with the general and proceeds to the particular only after the student is too confused to understand it.' Exercises are an integral part of the text. Students taking the course should have some knowledge of linear algebra. An appendix provides a brief survey of the necessary background of group theory. 296 pp. Englisch
Kinsey, L. Christine
Titre:
Topology of Surfaces
ISBN:
9780387941028
This book aims to provide undergraduates with an understanding of geometric topology. Topics covered include a sampling from point-set, geometric, and algebraic topology. The presentation is pragmatic, avoiding the famous pedagogical method "whereby one begins with the general and proceeds to the particular only after the student is too confused to understand it." Exercises are an integral part of the text. Students taking the course should have some knowledge of linear algebra. An appendix provides a brief survey of the necessary background of group theory. | 677.169 | 1 |
Math 2221: Linear Algebra and its Applications
Professor Mike Zabrocki
Description:
In this course, we learn the basic concepts of linear algebra which has
various applications. The computation aspects of linear algebra are emphasised
in Math 2221 and the applications are mostly reserved for Math 2222.
Throughout this
course, you should be able to master performing matrix operations and apply
them to solve linear systems. In addition, we cover the following topics:
Linear transformations, matrix operations, determinants, vector spaces, eigenvalues and
eigenvectors. Applications will be introduced as time permits.
The first quiz will be on Friday, January 11. The Midterm will be on
Friday, February 22. Check the web page reguarly for announcements of
subsequent quizzes and the final.
Lecture Schedule
Topics
Remarks
Friday, January 4
row reduction, 2 and 3-d geometry, section 1.1, 1.2
Monday, January 7
row reduction section 1.2
Wednesday, January 9
vectors, matrix equations, section 1.3, 1.4
Friday, January 11
section 1.5 solution sets to equations
Quiz 1
Monday, January 14
section 1.7 linear independence
return Quiz 1
Wednesday, January 16
linear independence
Friday, January 18
section 1.7 linear independence
Quiz 2
Monday, January 21
section 1.8 linear transformations
Wednesday, January 23
section 1.9 matrix linear trans, 1-1 and onto
return Quiz 2
Friday, January 25
section 2.1, 1-1 and onto
Monday, January 28
section 2.2 transpose, matrix multiplication
Wednesday, January 30
section 2.3, 2.4-invertible matrices
Friday, February 1
class cancelled due to snow
Monday, February 4
quiz 3
Wednesday, February 6
section 2.4 block matricies
return quiz 3
Friday, February 8
section 2.5 LU decomposition
February 11,13,15,18
Reading week/family day
no class
Wed, February 20
section 3.1 determinants
Fri, February 22
yes, there is a ...
Midterm
Mon, February 25
section 3.2,3.3 determinants, Cramer's rule
Wed, February 27
section 3.2,3.3 determinants
return midterm
Fri, February 29
section 3.2,3.3 determinants
Mon, March 3
section 4.1, vector spaces
Wed, March 5
section 4.1, vector spaces
Fri, March 7
quiz #4
Mon, March 10
section 4.2, column and null space
return quiz 4
Wed, March 12
section 4.3- linear independence, span
Fri, March 14
section 4.3, 4.4- basis
Mon, March 17
section 4.4- more bases
Wed, March 19
quiz #5
Fri, March 21
Good Friday
no class
Mon, March 24
section 4.5, 4.6 dimension and row space
return quiz #5
Wed, March 26
section 4.7 change of basis
Fri, March 28
section 5.1, 5.2 eigenvalues, eigenvectors
Mon, March 31
section 5.2, 5.3 eigenstuff and diagonalization
Wed, April 2
quiz #6
Thurs, April 3
section 5.3, 5.4 eigenstuff
Fri, April 4
Review
Announcements
(4/21/08) I am posting the unofficial final grades for the class here.
(4/16/08) I meant to get solutions for the practice
posted at least a little earlier. Sorry about
that. No one has even asked me for answers... Just so you are aware, many of the
answers that I provide on this set of solutions are not unique. You can always choose
different bases and this often changes the rest of the answers.
(4/3/08) I made up a practice for the final. The practice involves a lot of
computation so might take longer than 3 hours to do from beginning to end, but
it has a lot of the elements that we covered this term (but not all, just look
back at some of your quizzes). You can download
a copy of the practice here. I will eventually
post solutions
for this practice after I have given you enough of a chance to work on it. If you have
been working on the practice before I have a chance to post solutions
and would like me to check your answer then feel free
to send me an email and I can let you know if your answer is right or not.
(3/19/08) I received an email yesterday that online course evalations for this course
can be done at
Please take the time to give your feedback. You should have received
an email announcement about this.
(3/19/08) Quiz today see you there. It covers sections 4.1 through 4.4.
(3/7/08) I just got the final exam schedule and it seems to be set.
Our exam is April 18th at 9am and will be held in VH 3003.
(3/5/08) As far as I can tell we have classes today. I will be covering
more about vector spaces. There will be a quiz on Friday on Chapter 3 and
the beginning of Chapter 4.
(2/29/08) We have a makeup day, Thursday April 3 because of Family Day and
Friday April 4 because of Good Friday. We do not have a day scheduled for
the snow day that we had.
(2/20/08) Tentatively the exam for this class has been scheduled for Friday, April
18 from 9am-12pm.
(2/20/08) The midterm covers everything including determinants. This means
all sections that we covered in the book (see the table above for the complete
list).
(2/1/08) 10:40am Yikes! Snow. I just checked the
York website. It looks like
we have a snow day. The quiz will be rescheduled for Monday.
Current Status
Effective 11:30 am Friday 1 February 2008, the University has adopted Weather Emergency
Procedures at the Keele and Glendon Campuses and at both the Miles S. Nadal Management
Centre and Osgoode Hall Law School Downtown Centre. All day classes have been
cancelled and examinations postponed. University offices and Libraries are closed.
For information about possible cancellation or postponement of a scheduled special
event at the University, please telephone the Faculty, department or office responsible.
The suspension of normal operations at York University will remain in effect until such
time as the University lifts the Weather Emergency status.
(1/23/08) I noticed that I scheduled the midterm during reading week. Don't forget
that not only will we not have class during Feb 11-15 but also no class on Feb 18
due to the new and improved Ontario holiday schedule. So the midterm will now be
on Februrary 22.
(1/23/08) Finished the grading of the quiz. About half of you did well, the other half not.
If you need help on this stuff please come and talk with me about it BEFORE the next quiz.
The next quiz will be a week from Friday.
(1/14/08) Quiz again this Friday. 2 reasons. (1) You need to be experts on row
reduction as fast as possible. The quizzes were just OK. Most of the time an
error in row reducing was accompanied by an error in translating between matrices
and equations and we have to get rid of both types of errors. (2) Linear independence
is a major topic (even if it is just one section in the book). Note the (SG) study guide
note on page 70 of the book.
(1/9/08) Quiz this Friday. Do as practice the exercises from section 1.2 and 1.3.
(1/7/08) There were fewer people on the first day than I expected, but we have
started already. Read the first two sections on Chapter 1 of Lay's book on Linear
Algebra if you weren't there. We will have the first quiz on Friday about row reduction
and solving equations. | 677.169 | 1 |
Algebra 1 Help
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