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Course Description This course explores advanced algebra concepts and assists in building the algebraic and problem solving skills developed in Algebra 1A. Students will solve polynomials, quadratic equations, rational equations, and radical equations. These concepts and skills will serve as a foundation for subsequent business coursework. Applications to real-world problems are also explored throughout the course. This course is the second half of the college algebra sequence, which began with MAT 116, Algebra 1A.
Policies Faculty and students will be held responsible for understanding and adhering to all policies contained within the following two documents: • • University policies: You must be logged into the student website to view this document. Instructor policies: This document is posted in the Course Materials forum.
Course Materials Rockswold, G. K., & Krieger, T. A. (2013). Beginning and intermediate algebra with applications and visualization. (3rd ed.). Boston, MA: Addison-Wesley. All electronic materials are available on the student website.
Review your Study Plan in MyMathLab after completing the homework assignment for the week. Select topics from this week's materials in your study plan that have been highlighted for review. Continue working on these topics until you add a minimum of five new correct topics to your Study Plan report. Complete the Week One self-check in MyMathLab . ®
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A Victimless Crime?
Students generally are familiar with the disciplinary actions and penalties for getting caught. However, they may fail to understand that one of the personal consequences of cheating and/or plagiarism is that they aren't actually learning or practicing the material. They may not realize that they will actually need and be accountable for certain knowledge and skills.
Instructors may not explain the personal consequences and loss of trust that accompany academic dishonesty if they are focused mainly on stating the procedures and punishments related to academic disciplinary actions. They may not tell students how dishonesty damages their trust in a student and his or her work which can affect a student's ability to get a strong recommendation for employment or graduate school.
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INTRODUCTION
It is true that, with the economic and social development, the number of students who work part time is increasing obviously. As a result, it has become one of the most controversial issues among us. Many students are wondering whether they should get part-time jobs or not. Some people hold the view that students can face certain challenges with a part time job. Meanwhile, others argue that working part time can bring students some benefits.
BODY
Yes, college student should have a part time job.
1. Learning time management
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2. Applying their knowledge got from college in their part time job
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Communication
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Abstract
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Why is homework so important? Well, it helps students to understand and review the work that has been covered in class and also to see whether students have understood the lesson. Homework is also a link between school and home that shows what children are studying. Research shows that when homework is handed into the teacher, graded and discussed with students, it can improve student's grades and understanding of the schoolwork. A good and well planned out homework programme helps reinforce learning done in class and helps the student develop a good and well-built attitude to life-long learning as well as skills needed.
The aim that the academic curriculum has about homework at early stages in a child's education is to build up the foundation of a student's basic skills and concepts because they are the most important ones as the child develops. Homework also teaches good habits and responsibility.
Throughout the process of doing the homework, handing it in to the teacher and getting back the results, most children develop motivation as time goes on. If they do well, they will be given encouragement to... | 677.169 | 1 |
Friday, October 12, 2012
New Book: Explorations in Complex Analysis
Explorations in Complex Analysis by Michael A. Brilleslyper, Michael J. Dorff, Jane M. McDougall, James S. Rolf, Lisbeth E. Schaubroek, Richard L. Stankewitz, and Kenneth Stephenson. Available in the MAA Store and the MAA eBook Store. Read the introduction of Explorations in Complex Analysis(pdf) Explorations in Complex Analysis is written for mathematics students who have encountered basic complex analysis and want to explore more advanced projects and/or research topics. It could be used as
a supplement for a standard undergraduate complex analysis course, allowing students in groups or as individuals to explore advanced topics,
a project resource for a senior capstone course for mathematics majors,
a guide for an advanced student or a small group of students to independently choose and explore an undergraduate research topic, or
a portal for the mathematically curious, a hands-on introduction to the beauties of complex analysis.
Research topics in the book include complex dynamics, minimal surfaces, fluid flows, harmonic, conformal, and polygonal mappings, and discrete complex analysis via circle packing. The nature of this book is different from many mathematics texts: the focus is on student-driven and technology-enhanced investigation. Interlaced in the reading for each chapter are examples, exercises, explorations, and projects, nearly all linked explicitly with computer applets for visualization and hands-on manipulation.
There are 16 Java applets and a Java application, CirclePack, that allow students to explore the research topics without purchasing additional software. The electronic version of the book contains links to the applets and CirclePack. The print book (available soon in the MAA Store) contains information for downloading everything | 677.169 | 1 |
is Mathematics? 2nd Edition, authored by Herbert Robbins and Richard Courant, gives readers an interesting view of the mathematics world and is essential for professionals, students and engineers. The book covers topics like: the number system, geometrical constructions, topology and calculus.
About Oxford University Press
Oxford University Press is a renowned publishing house that develops and publishes high quality textbooks, scholarly works, and academic books for school courses, bilingual dictionaries and also digital materials for both learning and teaching. It is a division of the University of Oxford. The first book was locally published in the year 1912. Some of the books published under their banner are India's Ancient Past, Atkins' Physical Chemistry, Companion to Politics in India, Sociology: Themes and Perspectives and Common Mistakes at IELTS Advanced…and How to Avoid Them.
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2007
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Author
Ian Stewart, Herbert Robbins, Richard Courant
Dimensions
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158 inch
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A Masterpiece!
Where hundreds have failed trying to communicate the wide range of mathematics from arithmetic to calculus in a single book without diluting its effectiveness one bit, Richard Courant succeeds... brilliantly!
This book is a Masterpiece like no other. It starts off with the most elementary ideas about numbers and before the reader realises, he is being introduced to advanced concepts. The transition is seamless and mentally unhindered -an worthy achievement in itself. I only wish I had somet...
The book is a MASTERPIECE. Genuine, precise, simple and clear. It doesn't let you bother about anything the way the other ordinary texts in maths do. No ordinary explanation. The book deals with troubles in mathematics in an exact fashion of Feynman's classics in Physics. Maths is not an empty drill in problem solving but is much like a journalism with a divine taste. The book makes you feel exactly the same. An ordinary knowledge of high school mathematics is sufficient to understand this wo... | 677.169 | 1 |
At the end of this course, students should be able to apply functions and equations to a contextual situation and mathematically model it to make appropriate inferences and conclusions based on their knowledge of different mathematical representations.
The course starts by providing a general overview of different families of functions so students can identify the key features that distinguish different families of functions in a variety of representations and then determine in which situations a given family is best for modeling. Each subsequent unit gives students opportunity to develop more depth of knowledge with a thorough understanding of how to use each function family including transformations, finding roots, understanding the effects of restricting domain, range and number sets, inverses, solving for key features including maxima and minima, roots, intercepts and end behavior and how to apply these tools to real life and abstract situations. This course culminates by allowing students to build on this knowledge and have further opportunity for real life application in probability and prediction and statistics and inference from data.
Please comment below with questions, feedback, suggestions, or descriptions of your experience using this resource with students.
Algebra II Test & Quiz Generator
This exam/quiz generators can create Google Doc versions of exams with typeset questions. | 677.169 | 1 |
1237071
ISBN: 0321237072
Edition: 5
Publication Date: 2004
Publisher: Addison-Wesley
AUTHOR
Rosen, Kenneth H.
SUMMARY
Elementary Number Theory and Its Applicationsis noted for its outstanding exercise sets, including basic exercises, exercises designed to help students explore key concepts, and challenging exercises. Computational exercises and computer projects are also provided. In addition to years of use and professor feedback, the fifth edition of this text has been thoroughly checked to ensure the quality and accuracy of the mathematical content and the exercises. The blending of classical theory with modern applications is a hallmark feature of the text. The Fifth Edition builds on this strength with new examples and exercises, additional applications and increased cryptology coverage. The author devotes a great deal of attention to making this new edition up-to-date, incorporating new results and discoveries in number theory made in the past few years.Rosen, Kenneth H. is the author of 'Elementary Number Theory and Its Applications', published 2004 under ISBN 9780321237071 and ISBN 03212370 | 677.169 | 1 |
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Description
Advanced Trigonometry Calculator is a rock-solid calculator allowing you perform advanced complex math calculations.
Enter your complex math expression on its integrity and in the final press "Enter" button, after some instants the solution for your expression will be displayed.
Anyone can use this calculator since the syntax used is very similar with scientific handheld calculators, e.g. TI 84-Plus | 677.169 | 1 |
Click the button above to view the complete essay, speech, term paper, or research paper
Human Understanding: What is Inductive Reasoning? EssayReasoning Research Paper
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Math in Special Education Essay
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Introduction to Solve Expected Value Product Essay
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Essay on Moral Development and Importance of Moral Reasoning
- 1.0 Introduction: Lawrence Kohlberg was the follower of Piaget's theory of Moral development in principle but wanted to make his own theory by expanding his theory and study on that particular topic. Kohlberg was a very bright student and he served as a professor in the Harvard University. He become popular when he issued his Moral Development Theory by conducting research on that topic at Harvard's Center for Moral Education. Kohlberg believed that people moral behaviors are based on their moral reasoning, and their moral reasoning changed in accordance to their behaviors and actions when they move from one stage to another.... [tags: Moral Reasoning] :: 12 Works Cited
Hypothetico-Deductive Modeling to Q&A Essays
- ... This sort of cause and effect relationship between theory and outcomes creates a basis for scientific research. Use and understanding of a structure to facilitate outcomes can ensure appropriate conclusions are drawn in the support of the hypothesis. No scientific research model can absolutely verify or prove the truth. As Einstein stated it "No amount of experimentation can ever prove me right; a single experiment can prove me wrong". Carl Hempel's raven paradox can be used to illustrate Einstein's comment.... [tags: Scientific Research Methods, Charles Darwin] :: 5 Works Cited
The Deductive Problem of Evil Essays
- The Deductive Problem of Evil One of the major philosophical debates concerning God's existence involves the problem of evil. The problem has two basic formulations, one is deductive, the other inductive. The deductive form of the problem asks the following: Is the existence of evil logically compatible with a necessarily benevolent and necessarily omnipotent being? One of the philosophers who discusses the problem is Richard Gale. I will begin this essay by outlining the deductive problem of evil according to Gale. I will then try to refute the deductive argument and prove that the existence of evil is indeed logically compatible with a benevolent and omnipotent being. A conc... [tags: Philosophy essays] :: 3 Works Cited | 677.169 | 1 |
Elementary Linear Algebra with Applications
Buy Now orders cannot be placed without a valid Australian shipping address.
For introductory sophomore-level courses in Linear Algebra or Matrix Theory. This text presents the basic ideas of linear algebra in a manner that offers students a fine balance between abstraction/theory and computational skills. The emphasis is on not just teaching how to read a proof but also on how to write a proof | 677.169 | 1 |
1. Solve systems of linear equations using algebraic techniques.
2. Learn and apply the arithmetic of matrices.
3. Use matrix mechanics to solve systems.
4. Predict the nature, and analyze and interpret the solutions of systems of
linear
equations.
5. Decide whether a system is overdetermined or underdetermined.
6. Apply systems of equations to "real-world" problems.
7. Use determinants to find inverses of matrices.
8. Apply the Leontief method of inputs and outputs to solve problems in
economic.
Exercises from the text selected to reinforce and apply the above concepts to
real-world
situations should be completed. Applications include, but are not limited to,
topics in the
natural and social sciences.
Exercises from the text selected to reinforce and apply the above concepts to
real-world
situations should be completed. Applications include, but are not limited to,
finance,
agriculture, mining and advertising.
1. Define sets, set operations, and the cardinality of a set.
2. Use set theory to solve mathematical and real-world problems.
3. Define and apply counting arguments.
4. Define and apply the multiplication principle.
5. Define permutation.
6. Define combination and distinguish it from a permutation.
7. Apply permutations and combinations to solve counting problems.
Exercises from the text selected to reinforce and apply the above concepts to
real-world
situations should be completed. Applications include, but are not limited to
topics such as
games of chance, management, voting and sports.
Exercises from the text selected to reinforce and apply
the above concepts to real-world
situations should be completed. Applications include, but are not limited to
topics in the
natural and social sciences.
A project on the "birthday problem" is included in the day to day syllabus.
Evaluation Of Student Progress:
The following suggested grading scheme may be modified, at the discretion of the
instructor.
Tests:
Unit I
10%
Units II and III
10%
Unit IV
10%
Collected Projects (3)
25%
In-class Quizzes, group work, etc.
15%
Final Exam
30%
Academic Integrity Statement:
Under no circumstance should students knowingly represent the work of another as
one's own.
Students may not use any unauthorized assistance to complete assignments or
exams,
including but not limited to cheat-sheets, cell phones, text messaging and
copying from another
student. Violations should be reported to the Academic Integrity Committee and
will be
penalized. Please refer to pages 53-54 of the 2005-2006 Student Handbook. | 677.169 | 1 |
A succinct—if somewhat reductive—description of linear algebra is that it is the study of vector spaces over a field, and the associated structure-preserving maps known as linear transformations. These concepts are by now so standard that they are practically fossilised, appearing unchanged in textbooks for the best part of a century.
While modern mathematics has moved to more abstract pastures, the theorems of linear algebra are behind a surprising number of world-changing technologies: from quantum computing and quantum information, through control and systems theory, to big data and machine learning. All rely on various kinds of circuit diagrams, eg electrical circuits, quantum circuits or signal flow graphs. Circuits are geometric/topological entities, but have a vital connection to (linear) algebra, where the calculations are usually carried out.
In this article, we cut out the middle man and rediscover linear algebra itself as an algebra of circuit diagrams. The result is called graphical linear algebra and, instead of using traditional definitions, we will draw lots of pictures. Mathematicians often get nervous when given pictures, but relax: these ones are rigorous enough to replace formulas. | 677.169 | 1 |
Kiss My Math: Showing Pre-Algebra Who's Boss
From the author of the runaway bestseller Math Doesn't Suck, the next step in the math curriculum— pre-Algebra.
Last year, actress and math genius Danica McKellar made waves nationwide, challenging the "math nerd" stereotype—and giving girls the tools to ace tests and homework in her unique just-us-girls style. Now, in Kiss My Math, McKellar empowers a new crop of girls—7th to 9th graders—taking on the next level of mathematics: pre-Algebra.
Stepping up not only the math, but also the sass and style, Kiss My Math will help math-phobic teenagers everywhere chill out about math, and finally "get" negative numbers, variables, absolute values, exponents, and more. Each chapter features:
Kiss My Math also includes more fun extras—including personality quizzes, reader polls, and real-life testimonials— ultimately revealing why pre-Algebra is easier, more relevant, and more glamorous than girls think.
VOYA
McKellar's newest book on pre-algebra is designed to help those teens struggling with math. The book is broken down into five manageable sections: numbers; variables; x; exponents; and functions. Each section has several chapters focusing on parts of the theme, and chapters features step-by-step instructions, tips and tricks, and practice problems. Throughout the text are testimonials from successful women who use math in their jobs and McKellar's diary segments, with stories from her teenage life. One might ask, "How is this book any different from the math books that schools and libraries already have?" McKellar understands teens. She speaks their language. She gives practical examples that have meaning to teens. And honestly, she makes math fun. Anyone who can do that has a hit on her hands. There are only two drawbacks to this book. The examples and conversations McKellar has with the reader are very girl oriented, which is part of her goal of making more girls comfortable with math, but may turn off boy readers. The second drawback is that teens will only pick it up if they are struggling. If this book was used as supplemental material in a class, more teens would be forced to try it and learn that her tips are worthwhile. As one who uses calculators to do simple addition, this reviewer learned something. Reviewer: Kristin Fletcher-Spear | 677.169 | 1 |
Calculus (Latin, calculus, a small stone used for counting) is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches, differential calculus and integral calculus, which are related by the fundamental theorem of calculus. Calculus is the study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations.cbse board business studies syllabus | 677.169 | 1 |
Preview — Painless Algebra
by Lynette Long
Painless Algebra (Barron's Painless)
linear inequalities, and graphing quadratic equations.
Barron's popular Painless Series of study guides for middle school and high school students offer a lighthearted, often humorous approach to their subjects, transforming details that might once have seemed boring or difficult into a series of interesting and mentally challenging ideas. Most titles in the series feature many fun-to-solve "Brain Tickler" problems with answers at the end of each chapter.
Community Reviews
This book is quite easy to follow, with examples one can try, to make sure they are understanding the concepts. However, algebra has never been painless for me, and it's still not, even after reading this book. So I don't know about this "painless" claim.
I already knew algebra, but I went through this book to see if it would be useful to some of my students who are currently taking Algebra and it does explain things super simply. It doesn't hit every single aspect, but it could be a useful classroom tool. | 677.169 | 1 |
Understand what a function is, understand and determine different properties of functions, evaluate a function at a given value, manipulate the function in a variety of algebraic and graphical ways, learn and work with different families of functions including conics, determine roots both real and complex of polynomial functions using various methods, apply function concepts to application problems, understand how parabolas, ellipses and hyperbola's are mathematically represented.
Understand the basic trigonometric functions and the radian measurement, be able to apply unit circle and angular concepts in trigonometric applications, understand and use various identities, laws and formulas to evaluate and prove trigonometric expressions and equations and to solve trigonometric application problems.
Understand and solve linear, nonlinear, and multivariable systems of equations and inequalities both mathematically and graphically, understand and use matrices, matrix operations, their inverses and determinants to solve systems of equations.
Understand and use vectors, dot products, complex numbers and the polar coordinates system to solve problems. Understand arithmetic and geometric sequences, their sums, the binomial theorem, counting principles, permutations, combinations, and probability to solve common problems. | 677.169 | 1 |
Solving Integrals
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Solving Integrals
:) is the eighth video in a series covering the basics of calculus. Learn about some of techniques used in solving integrals including u-substitution, integration by parts, and partial fractions.
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08:19
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Get free advice from education experts and Noodle community members. | 677.169 | 1 |
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Description
Description
The geometry of lines occurs naturally in such different areas as sculptured surface machining, computation of offsets and medial axes, surface reconstruction for reverse engineering, geometrical optics, kinematics and motion design, and modeling of developable surfaces. This book covers line geometry from various viewpoints and aims towards computation and visualization. Besides applications, it contains a tutorial on projective geometry and an introduction into the theory of smooth and algebraic manifolds of lines. It will be useful to researchers, graduate students, and anyone interested either in the theory or in computational aspects in general, or in applications in particular. From the reviews : 'The authors have combined results from the classical parts of geometry with computational methods. This results in a unique and fascinating blend, which is shown to be useful for a variety of applications, including robotics, geometrical optics, computer animation, and geometric design. The contents of the book are visualized by a wealth of carefully chosen illustrations, making the book a sheer pleasure to read, or even just browse in. The book will help to bring the concepts and techniques of line geometry, which have been shown to be useful for various applications in geometric design and engineering, to the attention of a wider audience' - B.Juttler, Mathematical Reviews Clippings 2002. '...There is a vast amount of fascinating geometry of all sorts in this book. The topics are perhaps somewhat eclectic - they mirror the primary interests of the authors - but, because the motivation is to develop the geometry that applies to real world problems, the subject is far from molithic and is open to interpretation. The ideas here build up layer upon layer. In the end, the authors have been mostly successful in sustaining their central theme, despite the need to weave together projective, differential, algebraic and metric geometry. They have also presented the mathematics in a predominantly modern way. That is important because there exist in the engineering literature archaeological remnants of outdated tation and concepts. [...] The large number (264) of line diagrams are of very good quality and considerably enhance one's understanding. [...] a book which is without doubt an important contribution to this growing branch of geometrical research' - P. Donelan - New Zealand Mathematical Society Newsletter 87 , 2003. '...Overall I recommend this text to anyone who wants to learn about line geometry, projective geometry and the geometric side of some algebra. The book fills a niche that has been neglected for long and should benefit researchers interested in geometric methods...It covers a body of kwledge that is underrepresented in the literature and deserves to be kwn more widely. The authors wrote a clearly developed and beautifully illustrated book that fills a gaping hole in the contemporary literature' - ACM SIGACT News 36:3, 2005. | 677.169 | 1 |
Absolute Value
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This is a two-day unit on absolute value (target audience is algebra 2). Absolute value is often taught as a series of "tricks" for solving. I have found that, once students get to calculus, they never really understood absolute value from earlier courses. This document focuses on the definition of absolute value so that students are better able to build on this understanding as problems get more difficult. | 677.169 | 1 |
Goyals IIT Foundation Course In Mathematics For Class 9
Rs198.00
The series Goyal's IIT Foundation Course — Mathematics for Classes 6 to 10 has been designed to prepare the young ones desiring to get enrolled in engineering colleges after completing their school education. The objective of these books is to help students in understanding the fundamental concepts, their applications and sharpen their exceptional problem solving skills. This series will act as a platform for higher studies.
Some Salient Features of the Series are :
Textbook and Practice Material for solid foundation in a single book.
Basic Concepts presented clearly and precisely.
Hints for all problems.
Unique original problems sprinkled all over.
Adjustable matter to suit any school syllabus/competitive examination.
Direct and simple definitions to improve understanding.
Essence of each topic is condensed.
Caters to all spectrum of students from the average to the brightest.
Confidence building matter in abundance.
Schools and students can buy these books as additional/supplement to their textbooks and cover them during the school hours or outside the regular school hours.
It is hoped that this series will be highly beneficial for the students who are targetting the Engineering Entrance Examination. | 677.169 | 1 |
Derive 6.10
The trusted mathematical assistant of students, educators, engineers and scientists worldwide, Derive software does for algebra, equations, trigonometry, vectors, matrices and calculus what the graphing handheld does for high school math | 677.169 | 1 |
Math 4 CG Artists OnLine Training Course
DescriptionWithout a doubt Math is indispensable in our lives, without it we could never figure out simple calculations like time, weather or space. In 3D graphics Math is really important as well because we are dealing with numbers all the time, geometry and many other concepts of Math.
This training program is devoted to artists that want to learn Math but don't want to deal with complex terminology that probably doesn't make sense. I will say that here, you will learn Math but by using an artistic approach.
We found that a lot of CG artists run away from Math because they think that it is pretty difficult to master, and let us tell you something… they are absolutely right!!
Math is difficult to master at some degree but as CG artists all we need to know are a few concepts that will make our workflow a lot more easier every time we work on a project.
With that said, we promise you that this course will help you to absolutely understand these concepts so that you can quickly take a step forward in your career.
Curriculum
►Mathematics
Mathematics is the science of studying structure, quantity, space and change. As you can see all this takes effect in the CGI world. This chapter is devoted to teaching you why the use of math is going to make your life easier when working with computer graphics.
►Numbers
Numbers are a representation of something right? But why are some numbers better for some types of calculation than others? Well this class is all about numbers so don't worry about a thing because after this chapter you are going to understand what the different types of numbers are.
►Algebra
Algebra is one of the main branches of Mathematics and this is why we really need to dig into the fundamentals of it. Rules of operations and relations are going to be studied in this chapter to allow you to solve equations the proper way.
Trigonometry
This chapter is all about triangles and angles. In the world of 3D graphics we deal with this type of data all the time so let's get into the process of analyzing how to take advantage of this branch of Mathematics in our CG projects.
►Geometry
This class is all about shapes, position, size and space. All of these concepts are really important when working in visual effects, 3D animation or video games and this is why in this chapter you will learn how to deal with lengths, areas and volumes.
►Coordinates
Imagine a world without coordinates, wow it would be impossible to figure out the exact point or location of something. Now these days we have google maps and things like that to help us locate places or events, but still we need to understand how this system can help us to accomplish tasks more efficiently when dealing with any CG project.
Transformations
In this chapter of our Math course, we are going to deal with rotation, reflection, translations and these types of functions. If you are planning to become a good CG artist you must possess a strong foundation of these simple operations so that you can navigate freely in this world.
►Dealing with Vectors
Vectors have many purposes and when using them in CG there is not an exception, vectors allow us to point direction, assign color or simply to help us with orientation issues. This chapter is completely dedicated to vectors in CG and at the end of the program you will be able to take advantage of these elements in a variety of situations when working in computer graphics.
►Curves and surfaces
Curves are lines that are not straight right? Well that is true, but if we understand better it's topology, conventions and terminology I can guarantee you that these simples curves are going to become one of your very best friends aMore Info | 677.169 | 1 |
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This r squared creation PowerPoint is a review of the topics that will be covered on the Chapter 10 Test. The materials covered in the review are standards based and are aligned to Chapter 10 of the Holt Algebra textbook. Concepts covered are how to simplify an already factored rational expression, how to simplify a rational expression that is not yet factored, how to multiply a rational expression, how to divide a rational expression, how to add and subtract rational expressions with like denominators, how to add and subtract rational expressions with unlike denominators, how to solve a rational equation by cross-multiplying, and how to solve a rational equation by multiplying by the Least Common Denominator. The PowerPoint is also editable if you would like to make any changes (you may need to download MathType for free). | 677.169 | 1 |
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359 KB|28 pages
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The 7 lessons in this unit lay the groundwork for using functions to explore Algebra II throughout the course. Functions are explored, using standard notation, in algebraic, graphical, and tabular forms. Function composition, domain and range, and the key features of functions are all emphasized. Finally one-to-one functions are introduced and lead into the concept of inverse functions.
To access the lesson videos on YouTube use your smartphone or tablet to scan the QR code at the top of each lesson. | 677.169 | 1 |
GLENCOE MATH COURSE 2
Glencoe math course 2
In Recent glencoe math course 2 Algebra
Also, the "2a" in the. Write a title for your question. Factoring GCF, Math riddle with answer of two squares Our BookSleuth is specially designed for you. Do I need to purchase all the texts that are listed. Because you can use the time you save to learn something else that matters. I use LinkChecker to check page links and no longer visit each site individually to reconfirm. She absolutely hates math this year and is trying to find another option for next year. Snopes and the snopes. Questions in the most comprehensive online courses and thinner sample proportion word problems. What I'm hearing is "just follow the rules and you can do what you want". Free online asymptote calculator, math problem solver, sample 9th grade algebra worksheet, online algebra simplified, ti 84 calculator manual cube roots, Free 8th Grade Worksheets. 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This is the first video of three videos that explain practice 10. Hopefully that makes sense. For people local to Chester, Ellesmere Glencoe math course 2 North Wales and Wirral. Every entry in the second row must be the number of times the number above the entry occurs in the second row. Hey, I have college students who do not know their times tables. The same goes for the future. Only about a third of our tenth grade students are proficient on our state math test. In response, the company wrote a blog post Thursday addressing the concerns-essentially saying that students looking to cheat would cheat anyway, with or without PhotoMath. Topics include exponential growth and decay, solving with logs, compound and continuously compounded interest, and the exponential function of e. Follow Report Abuse Are you sure you want to delete this answer. This is especially true when finding zeros. Have you ever seen anything like that before. 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All are available for free download in PDF without registration. It offers some advice about studying based on current pedagogical research. Kids stick yellow popsicle sticks in each fry box. I like this foldable also. Wow it is so cold boulder math can I come stay with you. For Awards Night at Baddeck High School, the math club is designing small solid silver. Ask a Math Tutor or Teacher Now We have partnered with JustAnswer so that you can get an answer ASAP. Should they solve the equation manually, or just put it inti Wolfram Alpha. No loans for him. Parents-we need your help. There are many such links on the answers which are easily overlooked. I was able to run advanced equations from AP program and the solutions were all correct. Online Math Work Free multiple-choice worksheets for pre-algebra and algebra 1 topics. The proof, which has since been thoroughly vetted, is highly original, Naor said. Will guide you how to solve your Algebra homework and textbook problems, anytime, 39, 32, glencoe math course 2, 72, 28, 49. Yes, if you are looking to keep your mind sharp math books are the way to go. Please reload the page. He then began a progression of ever deeper observations of student problem solving using video-tapes of paired problem solving and glencoe math course 2, the results of which are detailed in his book, Mathematical Problem Solving Academic Press, 1985. The website itself offers more lessons and information than probably any other algebra help website out there. But, algebra-solving software, Maths Get those ebooks up and running. In fact, the textbooks I have seen are so bad that it calls into question the validity of the approach - it begs the question that if the subject cannot be well-organized in a textbook I have seen several perhaps there is something inherently poor about the approach. Either another student or myself then will answer the post and correct the work: mathstudent 1: How come the book got on 55. I've basically thrown the textbook the rest of the way out the window. Answers in algebra and other advanced forms of math aren't always round, easy numbers. Drag your mouse over the white space below to see the translation highlight it. View real-time data on usage, and expert algebra answers when you need them. But children need to start solving multi-step problems as soon as they can, approaches to learning, and learning rates. Answers Lesson 9-1 Practice Exponential Functions 9-1 Read Online Download PDF - Practice Workbook with Examples - Nutley Public Schools Practice Workbook with Examples The Practice Workbook provides additional practice with worked-out examples for every lesson. Each 2011 edition Glencoe math course 2 Bundle includes a student text, teacher's edition, Practice and Problem Solving Workbook, Practice and Problem Solving Workbook Teacher's Guide, Parent Guide pamphletTeaching Resources DVD, and Answers and Solutions CD-ROM. A contestant must select a curtain. For people local to Chester and Ellesmere Port you can use the information provided below. Windows Phone users can download Scientific Calculator for free. Offering clear solutions by showing complete steps, users not only get the answer but a detailed explanation of the glencoe math course 2 and why. The school board will vote on the proposed math curriculum and resources for grades six through 10 at a future meeting. COMMON CORE STATE STANDARDS CLASSES JMAP resources for the CCSS include Regents Exams in various formats, Regents Books sorting exam questions by CCSS: Topic, Date, Type and at Random, Regents Worksheets sorting exam questions by Type and at Random, an Algebra I Study Guide, and Algebra I Lesson Plans. It used only 20 of the available 50 MDTP Elementary Algebra items. It is, however, different from all the others done to this point. High and High School Students. Richard Bass offers free lecture notes for college and graduate school Mathematics courses. Have you corrected Quest 1??. The developer calls it the world's first camera calculator. Maths helper program, multiplying and dividing algebraic fractions calculator, word problems on addition, math combination worksheets elementary. Algebra 1 Chapter 8 Resource Book. Have mtel prep courses to sell. November 18, Get better Sam-We're thinking of you. One of my coworkers really put this crisis into perspective for me. Not only does this serve as a growth model as the basis of the teacher evaluation system, but it also promotes equity in the diverse settings of Tacoma schools. Containing 123 lessons, this text is the culmination of pre-algebra mathematics, a full pre-algebra course and an introduction to geometry and discrete glencoe math course 2. The numbers section has a percentages command for explaining the most common types of percentage problems and a section for dealing with scientific notation. Yes, You Can Succeed in Math Without So Much Struggle. Permalink I reckon there is a fairly good chance there will be. I absolutely love it. Join with FacebookSorry, integrated math 1 homework help are unable to log you in via Facebook at this time. My educational backround includes an MBA in accounting and a Master of Science degree in taxation. 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Connie is personally invested in the success of her students. But as the Supreme Mathematics paper 1 has made clear, they are people of a sort. Also, as each piece of information is offered in a separate window, your desktop can get really messy. Be sure to look at the topic titles more so than chapter and section numbers. Thinking MathematicallyCustom Edition Geometry Louisiana College Geometry: A Discovery Approach, 2nd. What data display did we use. Its interface can be displayed glencoe math course 2purdue math tutors or advanced modes and it also generates graphs with mathematical functions. Click to see the message monospaced in plain text definitions within context and explanations of algebra 2 answers with work concepts. School Board and force behind the California Reading Initiative spoke passionately about this her thing is that these teaching methods need to be based on educational research that proves they are effective, and whole language was not. Just updated my formerly lame and almost embarrassing tracking sheets for Common Core math scales to Super-Awesome and ready-to-go Excel spreadsheets. Oh want to sayI,ve found that the science section has physics included in the test. Y, Euclid2002, Falcon2, fattypiggy123, ferrouswei, Firstdream, FlyingCucumber, flyingsledge, futurestig17, garfield1, garret. Get This Free Download Does your ADHD student struggle with math. Find how long it will take before the biomass will be 2000 kilograms, 5,000 kilograms, and 10,000 kilograms. Notify me of new posts by email. Graphing equations two variables, McDougal Littell Algebra 1 Book answers, ohio algebra 1 textbook online, algebra with pizzaz, solving equations by multiplying or dividing, algebra used in 2009, invert sum of fractions. Is it fair or responsible to our children for us to wait through years' glencoe math course 2 processes of curriculum review and evaluation. American English has developed a number of differences to English. But in the case of python there has been a huge push to build libraries eg, numpy, scipy around it so it can do scientific computing.
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Math Connects: Concepts, Skills, and Problem Solving Course 1 Math Connects: Concepts, Skills, and Problem Solving Course 2 Math Connects: Concepts, Skills, and Problem Solving Course 3. Glencoe math course 2 should be observed in the above description of the fun problem solving questions of a good problem-solver that the majority of work on a problem is done glencoe math course 2 to the actual mathematical operations are performed. I don't care if how to use texas instruments ti 84 plus use them or not, just wanting to share my tips. When total distance is divided by total time we get the speed. Practice B - Edline Practice B " For use with pages. In short, CPM did not iq practice test with answers for me. Further Topics in Algebra 8. View step-by-step answers problems from your textbook. But he sees the math classes as some kind of punishment. This program received 6 awards 1. The Algebra course requires that your screen resolution be at 1024x768 or higher. They seemed to have a very clear understanding of decimals at least up until the thousandths position. Notices of the AMS Providence, Rhode Island, United States: American Mathematical Society 57 5 : 642-643. Begin Chapter 10 CHART angles ONLY COMPLETE Skip to Main Content District Home Select a School. If you don't know for sure, say so. Willing to tutor any undergraduate math course. | 677.169 | 1 |
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr.
In this chapter, the emphasis is on the mechanics of equation solving, which clearly explains how to isolate a variable. The goal is to help the student feel more comfortable with solving applied problems. Ample opportunity is provided for the student to practice translating words to symbols, which is an important part of the "Five-Step Method" of solving applied problems (discussed in modules (<link document="m21980"/>) and (<link document="m21979"/>)).
This module contains the proficiency exam for the chapter "Solving Linear Equations and Inequalities". | 677.169 | 1 |
Alignment of the Common Core State Standards - Mathematics and System Dynamics Modeling (Student Ages 14–18)
System Dynamics (to be abbreviated SD) System dynamics is a mathematical model-building methodology for studying complex system behavior. By analyzing feedback loop interaction one can gain a deeper understanding of the causes of systemic behavior, see how delays can create instability in a system, identify leverage points, and test policies that might mitigate undesirable behaviors or unintended consequences.
Modeling, using equations, has been an integral part of algebra through calculus, for years. System Dynamics modeling provides a view of the structure of the system that is more revealing than many other types of modeling. Because the software is visual, because full words or phrases can be used to identify the individual icons that represent a component in the model structure, because dependencies of one part upon another can be explicitly displayed, much more information is provided to the learner. The building of models is an ACTIVE process for the students. They must understand why each component is necessary for the system to operate, how the components are connected, and the role each component has in controlling the behavior of the system. They construct the small models and/or enhance/modify mid-sized models from a smaller core model.
(Note: This modeling method can be used starting in Algebra 1, and is an effective method of developing the introductory ideas of calculus (accumulations and rates of change) conceptually.
MATHEMATICAL PRACTICE
Make sense of problems and persevere in solving them
Students explain to themselves the meaning of the problem and look for entry points to its solution
For SD, students read the modeling story scenario and decide how to start to design the model.
Students make conjectures about the form and meaning of the solution and plan a solution pathway rather than simply jumping into a solution attempt
The SD process requires first a sketch of a potential reference behavior and a listing of possible variables to include in the model. They also are to anticipate model behavior before simulating.
Students consider analogous problems and simpler forms of the original problem in order to gain insight into solutions
Transfer of model structure to new appropriate scenarios – i.e., spread of an epidemic to marketing a new product or spread of a social movement, and use of generic structures – whose behavior the student knows - applies here.
Students monitor and evaluate their progress and change course if necessary
Students build SD models in incremental steps, running the simulation frequently to see if the current behavior makes sense. If it does not then the model is modified at that point.
Students can explain correspondences between verbal descriptions graphs or [and] draw diagrams of important features and relationships, graph data and search for regularity or trends
The SD modeling process starts with a story which students translate into a diagram containing the important features of the story, then produce a simulation graph, which they can compare to collected data. They then analyze the graphs searching for trends.
Students check their answers to problems using a different method, and they continually ask themselves "Does this make sense?"
Since SD modeling uses the numeric approximation of differential equations to produce simulated output, it is possible for more advanced students to work between the differential equations representation and the system dynamics stock/flow diagrams for models that are of medium size. Students at every level of system dynamics model building always ask if the output of the model makes sense.
Students can understand the approaches of others to solving complex problems and identify correspondences between different approaches
The stock/flow diagramming used in system dynamics models is like a language for representing complex dynamic systems. Students who learn to build such models are proficient at reading such diagrams and understanding the approach of others- for small to medium sized models- since the layout of the diagram indicates the logic of the dependencies, connections, and types of variables represented in the model.
Reason abstractly and quantitatively
Students make sense of quantities and their relationships in problem situations
In order to build an SD model students must decide which variables should be represented as stocks (accumulations) - flows (rates of change) - and converters (ancillary variables and/or parameters).
Students must be able to decontextualize – abstract the given situation –
In SD students use generic structures – abstract structures representing simple or more complicated behaviors – as they build their models. These structures are decontextualized. They also consider structures that can transfer to other applications, as they build their models. (It is a recommendation to students to try to build models that will have broader application than the specified problem at hand, so decontextualizing is part of their instruction.) Students are to abstract the given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own (translating problems into a stock/flow diagram representation requires abstraction capability in the same sense that designing an equation does, only designing a stock/flow diagram is more visual, hence easier for students). Once in the stock/flow world students easily manipulate icons to attempt to represent the problem faithfully.
Students must be able to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved
Often during the building of a simulation students are required to execute the simulation and analyze the output, identifying which parts of the output are being produced by certain parts of the diagram, or what feedbacks are dominant and why. They are to explain the output in terms of the context of the story. They can also manipulate parameters to exercise the model to better understand how the values influence the behavior of the model.
Quantitative reasoning entails habits of creating a coherent representation of the problem at hand; considering the units involved; attending to the meaning of quantities
All of these apply directly to SD modeling, without the need for translation. The software even contains a unit consistency checker.
Construct viable arguments and critique the reasoning of others
Students understand and use stated assumptions, definition, and previous established results in constructing arguments
For SD, "definition" does not apply as it is typically used in mathematics. When creating models students are requested to explicitly state any assumptions when describing their models. Students are expected to be able to use the assumptions made and the results of the simulation to support arguments about the scenario modeled.
Students make conjectures and build a logical progression of statements to explore the truth of their conjectures
As part of the SD process students are often required to anticipate the behavior of the model as it is built in stages. They are to explain any discrepancies between their anticipated behavior and the simulation output. The attempt is to translate what they think controls the behavior of the situation in to a true representation of their logic as they reconcile the differences between their ideas and the simulation output.
Students justify their conclusions, communicate them to others, and respond to the arguments of others
In SD, the computer is used as a tool to guide their theory building and modify their thinking in so far as its output disagrees with their anticipated behavior. All models are built with the expectation that it will be used as a tool to pass along their insights from modeling to others, so the model must be designed to be easy to read and useful for explanation. It also lays out the dependencies of the solution for all to see, aiding the student in his/her response to the arguments others might make about their solution.
Students reason inductively about data, making plausible arguments that take into account the context from which the data arose
SD students look for trends and patterns displayed in the model they created and use those trends to suggest possible policy arguments that might be suggested for changing the future behavior shown.
Students are able to compare the effectiveness of two plausible arguments, distinguish correct logic or reasoning from that which is flawed, and if there is a flaw in an argument – explain what it is
SD students are not alone in supporting the power of their arguments. Their models lay out their logic and the computer displays what their logic produces. The computer acts as an arbiter, and can be used to assist students in reconciling differences in an approach to a problem.
Students at all grades can listen or read the arguments of others, decide whether they make sense, and ask useful questions to clarify or improve the argument
SD models have been used with students at different levels of understanding. The visual nature of the software and the stock/flow diagrams help less experienced students understand connections and allows them access to problem analysis heretofore relegated to those understanding more sophisticated mathematics.
Model with mathematics
Students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace
A wide variety of SD modeling experiences apply to this practice.
Students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revisions later
The SD process requires explicitly making assumptions, drawing Behavior Over Time Graphs or Reference Graphs of the main variables before creating the model, and building and testing the model in stages.
Students are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, graphs, flowcharts and formulas
The SD process requires the identification of variables relevant to the story/problem and the design of a stock/flow diagram representing the relationship of the variables, using formulas and other values to make the model operational.
Students routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose
All these are normal parts of the SD modeling process.
MATHEMATICAL PRACICE
Use appropriate tools strategically
Students consider the available tools when solving a mathematical problem
SD software provides an important tool that applies numeric solutions of differential
equations to real-world scenarios but in a way that does not require a full understanding of calculus from a mathematical perspective.
Proficient students are sufficiently familiar with tools appropriate for their grade or course to make sound decisions about when each of these tools might be helpful, recognizing both the insight to be gained and their limitations
The visual nature of the STELLA software helps a broad range of students gain access to problems normally outside their usual field of study.
Students analyze graphs
Graphical analysis is a big part of SD model analysis.
Students detect possible errors by strategically using estimation and other mathematical knowledge
SD students regularly look at graphical output of their models and try to explain what the graphs tell them. They evaluate "reasonableness" in the output, including estimating the appropriate numeric values before they are displayed.
Students know that technology can enable them to visualize the results of varying assumptions, explore consequences, and compare predictions with data
All part of a normal lesson using SD models.
Students are able to identify relevant external mathematical resources, such as digital content located on a website, and use them to pose or solve problems
SD students may need to access data on the web, look for dynamic flowchart relationships to help formulate model variable relationships, access dynamic problems to use as a source of a problem to model.
Students use technological tools to explore and deepen their understanding of concepts
The core reason to use SD modeling.
Attend to precision
Students try to communicate precisely to others
Communication to others is a big part of SD model building, and using the model output helps students be precise. Additionally, STELLA software comes with a story-telling feature to aid the modeler in unfolding the model a segment at a time to aid the audience in understanding the design of the structure of the model.
Students state the meaning of the symbols they choose
In SD, each symbol has a purpose, certain variables are stocks because they are accumulators, some are flows because they represent rates of change, some converters. Also each icon gets a descriptive name that can be up to 64 characters long.
Students must be careful about specifying units of measure
The SD software has a "unit consistency" checker.
Students must label axes to clarify the correspondence with quantities in a problem
SD students pay close attention to labeling axes as it is essential in their analysis of graphical output. The horizontal axis is almost always related to time.
Students have learned to examine claims
The purpose of building SD simulations is to be able to analyze claims made in situations that involve dynamic complex systems. They do not have other accessible tools to do this (for complex systems) in high school.
Look for and make use of structure
Students look closely to discern a pattern or structure
The mantra of SD modelers is "structure determines behavior," so students are tuned into the need to look at structure and the patterns those structures produce.
Students can step back for an overview and shift perspective
Another common saying for SD modelers is that they need to "see the forest and the trees." There is also a modularity tool that assists students in thinking about the problem at a higher (modular) level, before defining the individual segments in detail.
Look for and express regularity in repeated reasoning
Students notice if calculations are repeated
For simple models in SD, students can be shown the recursive nature of the calculations performed in the software.
Students maintain oversight of the process, while attending to the details
Again, the modular tool allows oversight of the whole model structure as students define the specific details of each module.
Students continually evaluate the reasonableness of their intermediate results
As before, models are built and analyzed in stages, so SD students are always paying attention to intermediate results.
CCSS-M: STANDARDS
Functions Overview
Understand the concept of a function, interpret functions that arise in applications in terms of the context, analyze functions using different representations
Build a function that models a relationship between two quantities (although for SD problems one of the quantities is usually time), build new functions from existing functions
Modeling with functions: students identify appropriate types of functions to model a situation, they adjust parameters to improve the model, and they compare models by analyzing appropriateness of fit and making judgments about the domain over which a model is a good fit.
SD modeling applies to each of the "functions in algebra-based classes" standards listed in the Functions Overview shown above.
Modeling Overview
Modeling links classroom mathematics and statistics to everyday life, work, and decision-making. Modeling is the process of choosing and using appropriate mathematics and statistics to analyze empirical situations, to understand them better, and to improve decisions. Quantities and their relationships in physical, economic, public policy, social, and everyday situations can be modeled using mathematical and statistical methods. When making mathematical models, technology is valuable for varying assumptions, exploring consequences, and comparing predictions with data.
Real-world situations are not organized and labeled for analysis; formulating tractable models, representing such models, and analyzing them is appropriately a creative process. Some examples that were listed that are appropriate include:
Estimating how much water and food is needed for emergency relief in a devastated city of 3 million people; Analyzing stopping distance for a car; Modeling savings account balance, bacterial colony growth, or investment growth; Relating population statistics to individual predictions [and many more relevant problems not usually within the scope of high school students].
the models devised depend on a number of factors: How precise an answer do we want or need? What aspects of the situation do we most need to understand, control, or optimize? What resources of time and tools do we have?
Diagrams of various kinds are powerful tools for understanding and solving problems drawn from different types of real-world situations.
One of the insights provided by mathematical modeling is that essentially the same mathematical structure can sometimes model seemingly different situations.
In descriptive modeling, a model simply describes the phenomena or summarizes them in a compact form. Graphs of observations are a familiar descriptive model — for example, graphs of global temperature and atmospheric CO2 over time.
Analytic modeling seeks to explain data on the basis of deeper theoretical ideas, albeit with parameters that are empirically based; for example, exponential growth of bacterial colonies (until cut-off mechanisms such as pollution or starvation intervene) follows from a constant reproduction rate. Functions are an important tool for analyzing such problems
SD modeling applies to all of the modeling standards listed in the Modeling Overview shown above.
Statistics & Probability Overview
Summarize, represent, and interpret data on two quantitative variables.
Interpret linear models
Make inferences and justify conclusions from sample surveys, experiments, and observational studies.
SD aligns with #1, 2, and experiments and observational studies of #3, in the Statistics & Probability Overview shown above. | 677.169 | 1 |
Quadratic , Polynomial , Exponential , Logarithmic and Rational 5. Solve a system of linear equations using matrices. 6. Use a graphing calculator as an aid to problem solving . Academic Responsibilities : You are responsible for your own learning and 1. Attending class regularly and on time 2. Meeting deadlines as specified in the course outline 3. Completing homework assignments, projects, and exams as specified in the course outline 4. Reading the text before coming to class 5. Participating in class discussions and activities 6. Learning to use your calculator 7. Getting help when you don't understand something Method of Instruction : The classes are conducted informally and will typically consist of question and answer sessions, coverage of new material, and small group activities. Your
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This note was uploaded on 01/16/2012 for the course MATH 126 taught by Professor Blisinhestiyas during the Fall '11 term at Truckee Meadows Community College. | 677.169 | 1 |
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Show More and an abundance of applications, problems, and exercises for practice. The use of graphing calculators and computers is addressed throughout and web references lead readers to sites where they will find sources of real data for applications, further discussion of applications, further examples of topics, etc. A Beginning Library of Elementary Functions. Additional Elementary Functions. The Derivative. Graphing and Optimization. Additional Derivative Topics. Integration. Additional Integration Topics. Multivariable Calculus. Differential Equations. Taylor Polynomials and Infinite Series. Probability and Calculus. Trigonometric Functions. Basic Algebra Review. Special Topics. For anyone needing a solid introduction to the mathematics used in business, economics, life science, and social sciences | 677.169 | 1 |
Paperback | January 13, 2013
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Written by experienced examiners, this fully comprehensive course completely covers the revised Cambridge Secondary 1 curriculum, and these supportive Homework Books guarantee students get the practice they need to achieve their top potential. The sheer volume of practice will ensure all theconcepts become second nature, while the huge focus on stretch and challenge will lay the best possible foundations for Cambridge IGCSE.
About The Author
Sue has taught mathematics for over thirty years and has worked as an examiner for University of Cambridge International Examinations since 2003. She is a Principal Examiner for IGCSE Mathematics and also revises the examination papers for the Extended Level. She is involved in professional
development and has conducted training works... | 677.169 | 1 |
Wednesday, October 7, 2015
Algebra 2 is All About Exponents
I don't know if this makes function families easier or more complicated, but I realized this week that everything we cover in Secondary Algebra (1 and 2) can be reduced to two basic function families: f(x) = n^x and f(x) = x^n (trig functions excluded, but we don't cover those until precalc at my school). I don't remember this ever getting pointed out to me when I studied functions, and seriously, how many years have I been teaching this? I just find it very interesting.
We made these organizers in class today (and yesterday). We've already got our brains wrapped around transformations and compositions, although we have thus far stealthily avoided talking about operations on functions (adding/ subtracting and multiplying/dividing)... other than to say, "Ugh, two x's, that looks messy." Which of course, it IS, right? And isn't that the point? At least, I think that's my point this time through. Basic functions that involve these families, simple transformations, and compositions where one step follows another in a specified order... this kind of function is not hard to work with. They are logical and orderly. It's only when we start multiplying and dividing or adding/subtracting functions that stuff gets tricky and we need to start pulling out new tricks like factoring and zero product rule, and complex numbers, and extraneous solutions, and reducing rational expressions, and limits, oh my. | 677.169 | 1 |
Workshops
Teaching models
Mr. Houssam Kasti, Secondary School April 18, 2015, 11:30 – 14:00
Teachers need to upgrade their teaching models in order to meet the needs of developing 21st century learners. Most math teachers use the direct instruction model to teach mathematics; however, this instruction model is appropriate only for teaching procedural knowledge. In this workshop, other instructional models will be introduced such as concept attainment model, concept development model and inductive model... The presenter will then discuss how to apply the most appropriate model to selected lessons from the preparation phase to the assessment phase.
Audience: Middle and Secondary School Math Teachers
GeoGebra 5 Level advanced
Mr. Houssam Kasti, Secondary School
April 18, 2015, 8:30 – 11:00
This level 2 workshop is targeted towards secondary math teachers with a very good knowledge of GeoGebra. In this workshop, participants will use the software to go deeper into derivatives and sequence, in addition to other important concepts at the Secondary level. Some of the activities used in this workshop can be immediately integrated into the Math curriculum.
Audience: Middle and Secondary School Math Teachers
GeoGebra 5 for Beginners
Mr. Houssam Kasti, Secondary School
MarchGeoGebra 5 for Beginners
Mr. Houssam Kasti, Secondary School
FebruaryIntegrating GeoGebra in the Lebanese Math Curriculum
Mr. Houssam Kasti, Secondary School
March 16, 2013, 8:30 – 14:00
In this workshop, participants will explore activities to introduce the latest GeoGebra features and collaborate to see what are the best methods to integrate GeoGebra in the Lebanese Math curriculum. | 677.169 | 1 |
Quant Finance Pdf
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MSc Quantitative Finance. Gain a thorough understanding of the quantitative methods needed for financial decision making. You learn how to design new financial …Mathematical finance – Wikipedia – Mathematical finance, also known as quantitative finance, is a field of applied mathematics, concerned with financial markets. Generally, mathematical finance …
Fear not! Here's the solution. C++ For Quantitative Finance will take you on a journey from knowing little or nothing about how to apply C++ to quantitative finance
A masters degree in quantitative finance concerns the application of mathematical methods to the solution of problems in financial economics. There are several like … | 677.169 | 1 |
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Introduction to variables and expressions. This worksheet is great for helping students understand the parts of an expression. This document also includes a page and a half of guided notes which go over the parts and definitions of an expression. | 677.169 | 1 |
Alex Jaramillo
Prof. Elenica Stojanovska
MA1210
Lab 4.1
1. The Vertical Line test is a common method to determine if a graph is a function.
In this test, a vertical line is drawn on the graph to see if there is more than one
crossing point on the graph, i
Module 1 Fundamentals of Algebra, Part 1
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Professor Elenica Stojanovska
1. (35 points) The following formula estimates the number of Facebook users, in
millions, since 2008. Assume this trend continues and use the followi
Module 1 Fundamentals of Algebra, Part 2
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Marco Reyes
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Module 4 Project
Calculus, defined as the mathematical study of change, was developed independently by
Isaac Newton and Gottfried Wilhelm von Leibniz in the 17th century. Engineering is defined as
"the profession in which a knowledge of | 677.169 | 1 |
When the real numbers are replaced by the complex numbers in the definition of the derivative of a function, the resulting (complex-)differentiable functions turn out to have many remarkable properties not enjoyed by their real analogues. These functions, usually known as holomorphic functions, have numerous applications in areas such as engineering, physics, differential equations and number theory, to name just a few. The focus of this course is on the study of holomorphic functions and their most important basic properties.
Topics covered are: Complex numbers and functions; complex limits and differentiability; elementary examples; analytic functions; complex line integrals; Cauchy's theorem and the Cauchy integral formula; Taylor's theorem; zeros of holomorphic functions; Rouche's Theorem; the Open Mapping theorem and Inverse Function theorem; Schwarz' Lemma; automorphisms of the ball, the plane and the Riemann sphere; isolated singularities and their classification; Laurent series; the Residue Theorem; calculation of definite integrals and evaluation of infinite series using residues; Montel's Theorem and the Riemann Mapping Theorem. | 677.169 | 1 |
Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for). | 677.169 | 1 |
Tuesday, November 12, 2013
Algebra Thoughts
Jay is using Life of Fred for Math. He started this school year (mid-April) with Elementary Physics. He then moved onto Pre-Algebra 1 with Biology. He is currently more than halfway through Pre-Algebra 2 with Economics. He will be finished with this book before we take our Christmas break. When we return from Christmas break, he will be starting Life of Fred Beginning Algebra.
Since he will be starting Algebra in just a few months, I have been looking over the Beginning Algebra books to determine exactly how we will be using them. We already had Beginning Algebra & the Home Companion. I bought Zillions of Practice Problems in case we need it. I would rather spend the $30 to make sure we have extra problems for him & then find out he doesn't need them, than not have the extra practice & have this be where he hits a wall & needs extra practice.
So, here are the ideas I'm currently considering:
1) Give him 1 week per chapter. Schedule the final 2 Cities as the test each Friday & let him schedule how he plans to complete the work during the week.
2) Let him do it entirely at his own pace. Give him BA, ZPP to use as he needs, and access to the HC until he starts the Cities. Have him do the first 2 Cities as a pre-test, to make sure he is ready. If he gets 100% on it, give him that grade & let him move on. If he doesn't score 100%, have him re-read chapter & study. Then do next 2 Cities as the chapter test. If he still doesn't quite get it, have him study again & use final 2 Cities as test. Repeat with each chapter.
3) Have him read in BA, up to a Your Turn to Play, only doing problems from HC or ZPP if he feels he needs practice. If he gets any wrong in the Your Turn to Play, have him spend the next day or two doing problems in HC & ZPP to practice what needs work. Repeat until finished with chapter. Do 2 Cities as test. If less than 100% review & do next 2 Cities as test. If still less than 100% review again & do last 2 Cities as test.
4) Just have him follow the lessons as set out in HC, doing work in ZPP as needed.
I'm kind of leaning toward option #2, at least to start. If he is moving too slow & it's not because he is struggling with it, we would go to an option where he had less control over the schedule. I think option #3 is my next favorite, then option #1. Option #4 is my least favorite. I honestly think that option #4 would be too much review for him. He is the kind of kid that understands & masters quickly & easily.
I'm going to discuss it with my husband later. I already talked to Jay about it & he likes option #2. Since my husband takes over as head of the Math Department once they hit Algebra (since he likes Math & I hate it), he should have some say in how it gets done. | 677.169 | 1 |
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Thursday, September 18, 2003
Ok .. so now I have a place to air my views :)
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Our Index of Tutorials for all the topics
Mathematics
Algebra
Introduction to Complex NumbersIntroduction to Complex Numbers and iota. Arg-and plane and iota. Complex numbers as free vectors. N-th roots of a complex number. Notes, formulas and solved problems related to these sub-topics.
Series and ProgressionsArithmetic, Geometric, Harmonic and mixed progressions. Notes, formulas and solved problems. Sum of the first N terms. Arithmetic, Geometric and Harmonic means and the relationship between them. The Principle of Mathematical InductionIntroductory problems related to Mathematical Induction. Quadratic EquationsIntroducing various techniques by which quadratic equations can be solved - factorization, direct formula. Relationship between roots of a quadratic equation. Cubic and higher order equations - relationship between roots and coefficients for these. Graphs and plots of quadratic equations. Quadratic Inequalities Quadratic inequalities. Using factorization and visualization based methods.
Linear Algebra - Simultaneous Equations in Multiple Variables - A Tutorial with Examples and Problems Representing a system of linear equations in multiple variables in matrix form. Using determinants to solve these systems of equations. Meaning of consistent, homogeneous and non-homogeneous systems of equations. Theorems relating to consistency of systems of equations. Application of Cramer rule. Solved problems demonstrating how to solve linear equations using matrix and determinant related methods.
Linear Algebra - Introductory Problems Related to Vector SpacesProblems demonstrating the concepts introduced in the previous tutorial. Checking or proving something to be a sub-space, demonstrating that something is not a sub-space of something else, verifying linear independence; problems relating to dimension and basis; inverting matrices and echelon matrices.
Computer Science and Programming
Data Structures and Algorithms
Arrays : Popular Sorting and Searching Algorithms
Bubble Sort - One of the most elementary sorting algorithms to implement - and also very inefficient. Runs in quadratic time. A good starting point to understand sorting in general, before moving on to more advanced techniques and algorithms. A general idea of how the algorithm works and a the code for a C program. Insertion Sort - Another quadratic time sorting algorithm - an example of dynamic programming. An explanation and step through of how the algorithm works, as well as the source code for a C program which performs insertion sort. Selection Sort - Another quadratic time sorting algorithm - an example of a greedy algorithm. An explanation and step through of how the algorithm works, as well as the source code for a C program which performs selection sort. Shell Sort- An inefficient but interesting algorithm, the complexity of which is not exactly known. Merge Sort An example of a Divide and Conquer algorithm. Works in O(n log n) time. The memory complexity for this is a bit of a disadvantage. Quick Sort In the average case, this works in O(n log n) time. No additional memory overhead - so this is better than merge sort in this regard. A partition element is selected, the array is restructured such that all elements greater or less than the partition are on opposite sides of the partition. These two parts of the array are then sorted recursively. Heap Sort- Efficient sorting algorithm which runs in O(n log n) time. Uses the Heap data structure. Binary Search Algorithm- Commonly used algorithm used to find the position of an element in a sorted array. Runs in O(log n) time.
Basic Data Structures and Operations on them
Stacks Last In First Out data structures ( LIFO ). Like a stack of cards from which you pick up the one on the top ( which is the last one to be placed on top of the stack ). Documentation of the various operations and the stages a stack passes through when elements are inserted or deleted. C program to help you get an idea of how a stack is implemented in code. QueuesFirst in First Out data structure (FIFO). Like people waiting to buy tickets in a queue - the first one to stand in the queue, gets the ticket first and gets to leave the queue first. Documentation of the various operations and the stages a queue passes through as elements are inserted or deleted. C Program source code to help you get an idea of how a queue is implemented in code. Single Linked List A self referential data structure. A list of elements, with a head and a tail; each element points to another of its own kind. Double Linked List- A self referential data structure. A list of elements, with a head and a tail; each element points to another of its own kind in front of it, as well as another of its own kind, which happens to be behind it in the sequence. Circular Linked List Linked list with no head and tail - elements point to each other in a circular fashion.
Tree Data Structures
Binary Search TreesA basic form of tree data structures. Inserting and deleting elements in them. Different kind of binary tree traversal algorithms. Heaps - A tree like data structure where every element is lesser (or greater) than the one above it. Heap formation, sorting using heaps in O(n log n) time. Height Balanced Trees - Ensuring that trees remain balanced to optimize complexity of operations which are performed on them.
Dynamic ProgrammingA technique used to solve optimization problems, based on identifying and solving sub-parts of a problem first. Integer Knapsack problemAn elementary problem, often used to introduce the concept of dynamic programming. Matrix Chain Multiplication Given a long chain of matrices of various sizes, how do you parenthesize them for the purpose of multiplication - how do you chose which ones to start multiplying first? Longest Common Subsequence Given two strings, find the longest common sub sequence between them. Dynamic Programming Algorithms covered previously: Insertion Sort, Floyd Warshall Algorithm Algorithms which we already covered, which are example of dynamic programming.
Introduction To Networking
A basic introduction to networking and client server programming in Python. In this, you will see the code for an expression calculator . Clients can sent expressions to a server, the server will evaluate those expressions and send the output back to the client. | 677.169 | 1 |
Did you realize that the word "algebra" comes from Arabic (just like "algorithm" and "al jazeera" and "Aladdin")? And what is so great about algebra anyway?
Algebra is the language through which we describe patterns in mathematics. It involves usage of letters with numbers to depict problems. In this topic chapter we will learn about Matchstick patters, we will introduce the idea of variables, and we will understand where these variables can be used. A variable is depicted as a letter which can take any value. We will also learn to write linear expressions and ways to solve them.This tutorial doesn't explore algebra so much as it introduces the history and ideas that underpin it.
Expression like (4x + 5), which contains the variable x, cannot be evaluated. Only if x is given some value, an expression like (4x + 5) can be evaluated. In this topic we will learn how to evaluate exponential equations, we will write expressions variables and learn how to solve them when the value of the variable is specified. | 677.169 | 1 |
HOMEWORK TO DO
Homework to do
Questions homework to do resources
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PLEASE NOTE! We have most of the Excel series available at competitive prices. Ask if you have a particular book you are after. :)
Refine your problem-solving skills and get the results you want! This book aims to provide students with the skills they require to excel in both the problem solving and reasoning proficiency strands of the Year 7 Maths curriculum.
Key features:
- a focus on 50 different Key Skills - problem solving hints and examples with worked solutions - a step-by-step method for each question, with explanations and tips - 16 revision tests at both average and challenging difficulty levels - a detailed answer section with quick answers and worked solutions | 677.169 | 1 |
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I designed this lesson (the 9th of 10) as part of a unit that introduces arithmetic and geometric sequences. This lesson can be purchased as a complete bundled unit at a discounted price under the listing Arithmetic and Geometric Sequences Complete Bundled Unit.
This lesson includes: finding the first five terms of an arithmetic and a geometric sequence given in function notation, then determining which sequence will have the bigger value at f(100), and explaining why. | 677.169 | 1 |
Instead of using a simple lifetime average, Udemy calculates a course's star rating by considering a number of different factors such as the number of ratings, the age of ratings, and the likelihood of fraudulent ratings.
Graph Theory
Master the Nuts and Bolts of Graph Theory: the Heart of Communication and Transportation Networks, Internet, GPS, ...
4.3Graph Theory is an advanced topic in Mathematics. On a university level, this topic is taken by senior students majoring in Mathematics or Computer Science; however, this course will offer you the opportunity to obtain a solid foundation in Graph Theory in a very short period of time, AND without requiring you to have any advanced Mathematical background.
You don't need to know complex Mathematical statements, or rules, but ALL you need to know is simple mathematical operations like addition and multiplication. The course is designed to be understood by an 11th grader since the structure of the course starts with the very basic idea of how to create a Graph, and with each step the ideas get more and more complex. The structure of the course goes as following starting with the first section:
Graphs: In this section you will learn basic definitions like Vertex, Edge, Distance, Contentedness, and many other concepts that are the alphabet of Graph Theory.
Types of Graphs: In this section you will learn a variety of different Graphs, and their properties.
Graph Operations: In this section you will learn different operations and different methods in making new Graphs.
Graph Coloring: in this section you will learn Graph Coloring and many related concepts.
Paths: in this lecture you will learn Euler and Hamiltonian Paths and Circuits, and many other concepts in that area.
Trees: In this section you will learn about Trees, Tree Traversals, Binary Expression Trees and some more.
And Graph Match: In this section you will about Graph Match and Graph Cover.
How are the concepts delivered?
Each lecture is devoted to explaining a concept or multiples concepts related to the topic of that section. There are example(s) after the explanation(s) so you understand the material more. The course is taught in plain English, away from cloudy, complicated mathematical jargons and that is to help the student learn the material rather than getting stuck with fancy words.
How to learn better?
Take notes and repeat the lectures to comprehend the concepts more. Also, there are quizzes every 3-5 lectures so you can test what you have learned and go over something if needed.
In this lecture we will define Null, Trivial, Simple Graphs, Loops, and Parallel Edges.
Null, Trivial, and Simple Graphs
03:49
Regular, Complete and Weighted Graphs
03:01
In this lecture we will define Directed Graphs, Indegree, Outdgree, a Source, and a Sink, and we will learn how we can do Adjacency Matrix for a Directed Graph. We will also talk about Undirected, and Mixed Graphs.
Directed, Undirected and Mixed Graphs
07:52
Quiz
4 questions
In this lecture we will learn the difference between Cycle Graphs, and a Cycle in a Graph. We will also define Girth of a Graph, Path Graphs, Wheel Graphs, and Lollipop Graphs.
Cycle, Path, Wheel and Lolipop Graphs
08:13
Planar, Cubic and Random Graphs
04:05
In this lecture we will talk about Ladder and Prism Graphs, and how we can count the number of the Edges in each.
ladder and Prism Graphs
05:33
In this lecture we will define Web and Signed Graphs, and we will get to know a psychologist's contribution to Graph Theory.
Web and Signed Graphs
05:43
Quiz
7 questions
Peterson Graph
00:55
Bipartite Graphs
03:31
The illustrations shown in this lecture are NOT owned by the instructor of this course. To reach the website containing the illustrations, follow this link :
In this lecture we will define Vertex Coloring, Chromatic Number, k-Colorable Graphs, and Independent Sets.
Vertex Coloring
07:26
Edge Coloring
03:49
Quiz
3 questions
In this lecture we will define Chromatic Polynomials and show you how to use the software to find the Chromatic Polynomial of any Graph. Here is the link to Bob Weaver's website:
Chromatic Polynomial
05:31
Total and List Coloring
07:03
Exact and Fractional Coloring
03:59
Rainbow Coloring
03:11
In this lecture we will talk about Vizing's Theorem and Maximum Degree.
In this lecture, we will talk about different ways we can represent a Tree visually.
Tree Structures
05:36
Quiz
5 questions
In this lecture we will talk about Binary Trees, and its different types (Proper, Perfect, Complete, Infinite Complete, and Balanced Binary Trees).
Binary Trees
07:50
Spanning Trees
04:44
Quiz
4 questions
In this lecture you will learn how to convert an Algebraic or Boolean expression into a Tree and vice versa.
Binary Expression Trees
06:36
In this lecture we will talk about Preorder, Inorder, Postorder, and Levelorder Tree Traversals. To practice more, go to below website and you will find numerous practice examples at the very end of the page.
Fattah has B.S. in Mathematics and Geophysics from theUniversity of Oklahoma in Oklahoma, USA. He has taught and tutored many college students both in the United States and Iraq. His love for teaching made him one of four students in Iraq to receive a full scholarship to pursue a B.S. degree in the States so to return back to his home country and teach.
He is passionate about Math & Science and loves to share his passion with others. To him, Mathematics and Sciences are crucial for everyone to learn no matter how little. He is a BIG believer in visual learning, and his aim is to deliver the concepts in an easy and direct way so as to make the learning process fast for everyone. | 677.169 | 1 |
Topics in Commutative Ring Theory is a textbook for advanced undergraduate students as well as graduate students and mathematicians seeking an accessible introduction to this fascinating area of abstract algebra. Commutative ring theory arose more than a century ago to address questions in geometry and number theory. A commutative... more...
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The natural numbers have been studied for thousands of years, yet most undergraduate textbooks present number theory as a long list of theorems with little mention of how these results were discovered or why they are important. This book emphasizes the historical development of number theory, describing methods, theorems, and proofs in the contexts... more...
Combinatorics is the branch of discrete mathematics that studies (and counts) permutations, combinations, and arrangements of sets of elements. This book constitutes the first book-length survey of the history of combinatorics and uniquely assembles research in the area that would otherwise be inaccessible to the general reader. more...
Guide to Immediate Anaesthetic Reactions serves as a practical guide to the management of adverse anesthetic response. This book provides information pertinent to the functions of anesthesia, including hypnosis, analgesia, and neuromuscular control. Organized into six chapters, this book begins with an overview of the immunological principles in anesthesia... more...
Posing the question "What exactly is a number?" a distinguished German mathematician presents this intriguing and accessible survey. Albrecht Beutelspacher ― founder of the renowned interactive mathematics museum, Mathematikum ― characterizes the wealth of experiences that numbers have to offer. In addition, he considers the many things that can be | 677.169 | 1 |
Prepare for Introductory Calculus courses. Mathematics is the language of Science, Engineering and Technology. Calculus is an elementary Mathematical course in any Science and Engineering Bachelor. Pre-university Calculus will prepare you for the Introductory Calculus courses by revising four important mathematical subjects that are assumed to be mastered by beginning Bachelor students: functions, equations, differentiation and integrationLearn how to think the way mathematicians do - a powerful cognitive process developed over thousands of years. Mathemat
In this course, you will learn the science behind how digital images and video are made, altered, stored, and used. We will look at the vast world of digital imaging, from how computers and digital cameras form images to how digital special effects are used in Hollywood movies to how the Mars Rover was able to send photographs across millions of miles of spaceThe purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering.
This course covers mathematical topics in trigonometry. Trigonometry is the study of triangle angles and lengths, but trigonometric functions have far reaching applications beyond simple studies of triangles. This course is designed to help prepare students to enroll for a first semester course in single variable calculus.
This course will cover the mathematical theory and analysis of simple games without chance moves. This course explores the mathematical theory of two player games without chance moves. We will cover simplifying games, determining when games are equivalent to numbers, and impartial games.
Discrete mathematics forms the mathematical foundation of computer and information science. It is also a fascinating subject in itself. Learners will become familiar with a broad range of mathematical objects like sets, functions, relations, graphs, that are omnipresent in computer science. Perhaps more importantly, they will reach a certain level of mathematical maturity - being able to understand formal statements and their proofs; coming up with rigorous proofs themselves; and coming up with interesting resultsCalculus is one of the grandest achievements of human thought, explaining everything from planetary orbits to the optimal size of a city to the periodicity of a heartbeat. This brisk course covers the core ideas of single-variable Calculus with emphases on conceptual understanding and applications. The course is ideal for students beginning in the engineering, physical, and social sciences.
This is a course about the Fibonacci numbers, the golden ratio, and their intimate relationship. In this course, we learn the origin of the Fibonacci numbers and the golden ratio, and derive a formula to compute any Fibonacci number from powers of the golden ratio | 677.169 | 1 |
What to do with linear algebra? Some Inquiry Based Learning!
(Note: the word "some" is important. I am going to try a hybrid this time through. Read on.)
In an effort to keep the momentum up, and maybe embiggen my cyber teaching lounge, I signed up for the Math Educational "New Blogger Initiation" challenge. I am pretty sure I heard about this from the blog of Sam J Shah, Continuous Everywhere but Differentiable Nowhere. The agreement means that I will be writing some specific posts in the next four weeks to prompts the Initiation Team sends me.
For this go round, I thought I'd take some time to write about the next iteration of my linear algebra course. (If you are new here, I basically use this space to think "out loud" about my teaching.)
I have not done any of the bits of preparation that involve actually making things I need for the first day of class. This is bad. I have done a lot of thinking about what happened last semester and how I might adjust things. This is good.
Also, I have spent a fair amount of time looking through linear algebra texts. There was one common theme: I thought they were all going a bit slow. Then it dawned on me. * My class was too hard. I have unreasonable expectations of the average undergraduate. *
By the way, that is not really news. Or, it is, but it shouldn't be. I've had that revelation many times in the past five years.
Anyway, let's get down to brass tacks. I'm going to start laying out what I am up to. If you want more of the history of my thinking, there are older posts just waiting to be read.
Here's the quick background:
I took a single linear algebra class in college, but I never attended. When it came time for the final exam, I crammed the whole text in a weekend. This was insane but basically worked. I am still irrationally upset over getting an A- in that course. I know you don't care.
I really learned the material deeply when I studied Lie groups and Lie algebras in grad school. This means I have a weird selection of highly theoretical linear algebra that feels like regular arithmetic. This gets in the way of understanding my students.
I taught linear algebra once, about six years ago, at a fancy pants liberal arts college in New England. The students didn't like it so much. It took a while to internalize why: the course was too hard.
I got my first opportunity to teach linear algebra at UNI this past spring. I am test subject for a project (called UTMOST and funded by the NSF) about integrating open source software and literature into the college curriculum, in particular, the mathematical software system Sage.
Last semseter could have gone a lot better, and I am now in the process of preparing for a redo. This is one of the glories of the academic system isn't it? Every so often, you just get to reboot entirely. Maybe every year, maybe every semester, you get to let go of the baggage attached to one class and start fresh.
I would like to use an Inquiry Based Learning evironment as much as I can, but recent experience has shown me that I am not yet accomplished enough at this style to pull it off with pre-proof classes. So this semester I am aiming for a hybrid. I am stealing an idea that I heard at this year's IBL session at MathFest in Madison. I forget the speaker's name, but I am also sure that I heard some version of this once before. (So I am not neglecting to credit the original author of the idea, only the person who reminded me most recently. That is a terrible excuse.)
Here is the plan outlined:
Use the Schaumm's outline series text on Linear algebra. It is succinct, covers everything I need and more, and fills the need for lots of computational work with examples.
I will lecture on Mondays. I will explicitly announce how the rapid lectures are keyed to certain portions of the text.
On Wednesdays, we will begin with a short check-up quiz focused on computational techniques and low-level understanding/recall. Then we will launch into an IBL format with students presenting.
The rest IBL portion (a day and a half per week) will be focused on a sequenc of problems I design around the material to foster deeper understanding. There will be lots of open ended questions.
I will be stealing freely from two geometrically focused books: an old one by Dan Pedoe and a new one by Shiffrin and Adams. I want students to obtain reasonable mental models of what all those symbols mean as pictures. Basically, the mantra is this:
Linear algebra is communicated and conceptualized best by the abstract language
of vector spaces and linear tranformations, and it is actually computed most easily
with coordinates and matrices, but the intuition comes from understanding the
pictures of hyperplanes meeting in space.
I have a ton to do in the next 48 hours: I have been thinking about it all summer, but now it is time to execute the plan. I need the following (in roughly this order):
A reasonable syllabus reflecting what will be covered each week.
A course web site.
A first week lecture.
A second week lecture.
A first week quiz.
A first week of IBL investigations.
An "introduction to Sage" screencast or two. Say, one for setting up an account, and another for trying out some very basic things.
An actual Sage worksheet containing some tutorial material related to the first weeks mathematics and the statements of the first week IBL tasks.
Some ibuprofen when all that is done.
I am also thinking about trying out the learning management system Canvas by Instructure. My campus uses Blackboard at this point, and I have heard nothing but grumbling from my colleagues. I heard a few raves about Canvas at MathFest, and I am curious… If it sets up reasonably fast, I'll try it. | 677.169 | 1 |
Helping Students CRUSH Their Classes and Become All-Star Performers
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Algebra IS Necessary
I believe we can all agree that perseverance and confidence play a significant role in success. No matter what example of success we come up with, we can safely say that confidence and perseverance played a key role. For instance, confidence is built with each win we achieve, no matter how small. However, we aren't perfect, so we cannot count on perfect records and thus must persevere through the inevitable failures. Taken together, confidence and perseverance are essential to any subsequent success.
A few weeks ago, I read an article in the New York Times Sunday Review entitled, "Is Algebra Necessary?" The author, Andrew Hacker, suggested that the math requirements in our high schools and colleges are a leading contributor to high school and college drop out rates. According to Hacker, "making math mandatory prevents us from discovering and developing young talent." He also goes on to suggest that new classes should emerge addressing what he terms "citizen statistics." These courses would cover topics such as the Consumer Price Index, and help "familiarize students with the kind of numbers that describe and delineate our personal and public lives."
I highly agree with his idea and believe there is a place for a course on citizen statistics, however, within a curriculum that covers Algebra, Geometry, Trigonometry, and other advanced courses. In fact, I believe Hacker's heart is in the right place. He ultimately wants more students to succeed in school and subsequently in their lives. I too want the same thing as I am sure does every educator. The difference is that I believe there is still a place for a solid math education and we need to consider its benefits before we eliminate it.
In the next few paragraphs, I am going to address specific comments made by Mr. Hacker. I will then close by coming full circle on my original thesis that confidence and perseverance are critical to success.
Early in the article, when referring to foreign students, Hacker states "…it's their perseverance, not their classroom Algebra, that fits them for demanding jobs." To this I counter that perseverance is developed by courses like mathematics. Perseverance is a quality that like most others needs to be practiced. We may all have some level of it, but ultimately to improve upon it, we must practice it. The type of problems and challenges encountered in a class like mathematics positions students to develop stronger perseverance by challenging them to solve increasingly difficult problems. Each subsequent challenge pushes students past their comfort zone and presents an opportunity for growth. To simply state that it's perseverance that fits foreign students for demanding jobs without discussing where it may have developed creates a serious gap in Hacker's reasoning.
Having said that, I strongly believe there are opportunities to reposition the same Algebra courses to be more like a game environment where students are pushed to try and try again until they advance levels. Such changes would more precisely target the goal of developing perseverance in students. Nevertheless, Algebra is a very practical solution for developing perseverance and subsequently, confidence.
Later in the article, Hacker states that "What of the claim that mathematics sharpens our minds and makes us more intellectually adept as individuals and a citizen body? It's true that mathematics requires exertion. But there's no evidence that being able to prove (x^2+y^2)^2 = (x^2-y^2)^2 + (2xy)^2 leads to more credible political opinions or social analysis."
Perhaps there is no evidence to support a direct relationship today, but that assumes that simply because we could not devise a study or test that isolates a particular variable there must be no relationship. However, that does not mean it was wrong. It simply meant it could not be proven at that time. This concept is very different from something that was in fact proven incorrect as supported by evidence. While Hacker is absolutely correct that there is no evidence to prove a relationship, there is also no evidence to support the contrary; that does not mean there is no relationship and we should eliminate Algebra as a requirement. I often use the following analogy: algebra is to our brain as pushups are to the athlete's physique. While an athlete (i.e. football player) does not need to perform pushups on the field, he or she still performs them as a part of his or her training regimen. And while we may not be able to prove that there is a direct relationship between push-ups and throwing a better pass, we understand enough to know that pushups lead to a stronger body which positions the athlete for a better performance. The same goes for Algebra. The brain requires exercise in perseverance, resiliency, and problem solving and subjects like Algebra provide it with repetitive exercises to keep it sharp and primed.
Hacker goes on to suggest that his proposals "need not involve dumbing down [the curriculum]. Researching the reliability of numbers can be as demanding as geometry. More and more colleges are requiring courses in "quantitative reasoning. In fact, we should be starting that in kindergarten."
The problem with this suggestion is that starting a child on "quantitative reasoning" at a young age is like asking someone that has never played football to play in a competitive game. In keeping with the analogy, the person hasn't been explained the rules, taught the skills, nor the strategy and we expect him or her to perform and succeed. Taking this analogy one step further, not only is this person not prepared to play, he or she has not even been conditioned (perseverance) to endure a full game.
Going back to my original thoughts about confidence and perseverance. My high school math teacher once said that math is the great equalizer. I never forgot that. And as the years went by I gave that more and more thought. Growing up in north Jersey, my family was not of a high socio-economic status. In fact, my parents could never afford to send me to a private school or provide costly supplementary education. When I started my freshman year at Penn State, I felt I was a little behind despite having been a strong student in high school. However, the subject that helped boost my confidence was math. The skills, perseverance, and confidence I had gained in high school math positioned me right along side, if not ahead, most of my fellow students. These critical qualities positioned me to make up the difference in all of the other courses where I felt I lagged. Not only did I know what it took to succeed, I was fully equipped with the tools to achieve it – perseverance and confidence. In hindsight, math truly was the great equalizer.
We will probably debate the question of whether Algebra is necessary for years to come, but it is my firm believe that yes, Algebra is necessary. Let's not back out of our pushups and sit-ups. Let's put in the time at the gym to make sure we are tuned to perform well in life.
And in case you are wondering, I did not major in math. I majored in business. I later went back to school for an MBA and a Masters in Psychology. Perhaps one day I'll formally pursue a math degree, but just for exercise 🙂 | 677.169 | 1 |
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Solving multi step equations. This document includes a page of guided notes along with two pages of practice/guided practice. This is a great resource for an introductory/intermediate/advance lesson on solving multi step equations. This document also includes several word problems that are aligned with the common core standards. | 677.169 | 1 |
11
Preface
Calculus is at once the most important and most difficult subject encountered early by students of mathematics; introductory courses often succeed only in turning students away from mathematics, and from the many subjects in which the calculus plays a major role. The present text introduces calculus in the informal manner adopted in my Arithmetic [1], a manner endorsed by Lakatos [2], and by the following words of Lanczos from his preface to [3]: Furthermore, the author has the notion that mathematical formulas have their "secret life" behind their Golem-like appearance. To bring out the "secret life" of mathematical relations by an occasional narrative digression does not appear to him a profanation of the sacred rituals of formal analysis but merely an attempt to a more integrated way of understanding. The reader who has to struggle through a maze of "lemmas", "corollaries", and "theorems", can easily get lost in formalistic details, to the detriment of the essential elements of the results obtained. By keeping his mind on the principal points he gains in depth, although he may lose in details. The loss is not serious, however, since any reader equipped with the elementary tools of algebra and calculus can easily interpolate the missing details. It is a well-known experience that the only truly enjoyable and profitable way of studying mathematics is the method of "filling in the details" by one's own efforts. The scope is broader than is usual in an introduction, embracing not only the differential and integral calculus, but also the difference calculus so useful in approximations, and the partial derivatives and the fractional calculus usually met only in advanced courses. Such breadth is achievable in small compass not only because of the adoption of informality, but also because of the executable notation employed. In particular, the array character of the notation makes possible an elementary treatment of partial derivatives in the manner used in tensor analysis. The text is paced for a reader familiar with polynomials, matrix products, linear functions, and other notions of elementary algebra; nevertheless, full definitions of such matters are also provided.
the vibration of a pendulum or piano string. The computer expressions may be followed by comments (in Roman font) that are not executed:
i. beginning at zero Times from 0 to 1 at intervals of one-tenth Corresponding heights
. and then to plot the points detailed in the table. or velocity. However. The position of the falling stone can be described approximately by an algebraic expression as follows:
p(t) = 20 . the shape of the cables in a powerline or suspension bridge. the calculus has long proven difficult to teach. We begin with a concrete experiment of dropping a stone from a height of twenty feet. This simple notion provides insight into a host of familiar things: the growth of trees or financial investments (whose rates of change are proportional to themselves). because of the rapidity of the process.11 h=:20-16*t*t
First eleven integers. largely because of the difficult notion of limits. we are unable to observe either with any precision. We will defer this difficulty by first confining attention to the polynomials familiar from high-school algebra.16 * t * t
We will use this definition in a computer system (discussed in Section B) to compute a table of times and corresponding heights.6 Calculus
Chapter
1
Introduction
A. and noting that both the position and the velocity (rate of change of position) appear to depend upon (are functions of) the elapsed time. and the logarithmic scale used in music. Calculus
Calculus is based on the notion of studying any phenomenon (such as the position of a falling body) together with its rate of change. A clearer picture of the motion can be obtained by moving the successive points to a succession of equally spaced vertical lines to obtain a graph or plot of the position against elapsed time.1*i.11 0 1 2 3 4 5 6 7 8 9 10 t=:0. More precise observation can be provided by recording the fall with a video camera. and recording the successive positions in a vertical line on paper. In spite of the simplicity and ubiquity of its underlying notion. playing it back one frame at a time.
the last two rows appear as:
0.6 14.h 0 20 0.Chapter 1 Introduction 7 t.36 0.04 1 4 load 'plot' PLOT=:'stick.16 0.04
.7 12.76 0.04 4
and subtraction of the first of them from the last gives both the change in time (the elapsed time) and the corresponding change in position:
1 4 ..4 17.1 19.2 19. For example.8 9.44 0.56 0.3 18. and emphasizes the fact that it is rapidly increasing in magnitude.h
The plot gives a graphic view of the velocity (rate of change of position) as the slopes of the lines between successive points.line'&plot PLOT t.0.24 0.5 16 0. the table provides the information necessary to compute the average velocity between any pair of points.9 7. Moreover.9 7.84 0.9 1 7.
For many functions this limit is difficult to determine.04
Finally. it has a derivative (the acceleration) which is also called the second derivative of the original function p .4
Division is denoted by % The _ denotes a negative number
The negative value of this velocity indicates that the velocity is in a downward direction. Both the table and the plot suggest abrupt changes in velocity.h
Intervals of one-hundredth over the same range
This plot suggests that the actual (rather than the average) rate of change at any point is given by the slope of the tangent (touching line) to the curve of the graph. because it is derived from the function p. it suggests the use of an interval of zero. but smaller intervals between points will give a truer picture of the actual continuous motion:
t=:0. but we will avoid the problem by confining attention to polynomial functions. the change in position divided by the change in time gives the average velocity:
_3.04 % 0. and we are led to the idea of the "limit" of the ratio as the interval "approaches" zero. Moreover. In terms of the table. It also can be expressed algebraically as follows: v(t) = -32*t.1 _3. But this would lead to the meaningless division of a zero change in position by a zero change in time. since the velocity is also a function of t.
.8 Calculus 0.01*i.101 h=:20-16*t*t PLOT t. it is called the derivative of p . The velocity (rate of change of position) is also a function of t and.1 _30. where it can be determined by simple algebra.
The notion of an operator that applies to a function to produce a function. it is important that it be introducible in a similarly casual manner.. 3. models that allow a student to gain familiarity through concrete and accurate experimentation. The notion of a limit of an expression that depends upon a parameter whose limiting value leads to an indeterminate expression such as 0%0. Although an executable notation must differ somewhat from conventional notation (if only to resolve conflicts and ambiguities). For example. In the foregoing we have seen that calculus requires three notions that will not have been met by most students of high school algebra: 1. Such working models are provided automatically by the adoption of mathematical notation that is also executable on a computer. jointly with another new notion of rate of change) makes it more difficult to embrace. p
d2y/d2t p'' p D.2 dny/dnt pn p D. The notion of the rate of change of a function. with little or no discussion of notation as such. its first introduction as the derivative operator (that is.com) in introducing and using vectors and operators. 2.Chapter 1 Introduction 9
Various notations (with various advantages) have been used for the derivative:
newton leibniz modern heaviside (J)
. We will therefore begin with the use of simpler (and eminently useful) operators before even broaching the notion of rate of change. The subsequent section illustrates such use of the executable notation J (available free from webside jsoftware.1
. the notations x1/2 and xm/n and xpi used for it may be silently adapted from the more restricted integer cases x2 and xn.n (y = p (t))
Heaviside also introduced the notion of D as a derivative operator. In teaching mathematics. Notation learned in a simple context is often expanded without explicit comment. an entity that applies to a function to produce another function. although the significance of a fractional power may require discussion. the necessary notation is normally introduced in context and in passing. This is a new notion not known in elementary algebra.
. p
dy/dt p' p D. so as not to distract from the mathematical ideas it is being used to convey. Although the notion of an operator that produces a function is not difficult in itself. A further obstacle to the teaching of calculus (common to other branches of mathematics as well) is the absence of working models of mathematical ideas.
and 0 denotes Lesser of (Minimum) Related spellings denote related verbs Double.5 17 38 and sum=:+/ are referred to as pronouns and proverbs (pronounced with a long o). halve. square root The symbol / denotes the adverb insert
false)
1 0 0 2 <. To avoid distractions from the central topic of the calculus. as well as other experiments that may suggest themselves. c) Vectors and matrices are also referred to by the more suggestive terms lists and tables. from simple experiments on the computer. or from the study of more elementary texts such as Arithmetic [1].24 2*3 6 2^3 8 1 2 3 * 4 5 6 4 10 18 2 < 3 2 1
Plus
Times Power (product of three twos) Lists or vectors Less than (1 denotes true. respectively. the computer can be used to explore and elucidate topics with a clarity that can only be appreciated from direct experience of its use. -: . we will assume a knowledge of some topics from elementary math (discussed in an appendix).78+0. The remainder of this section is a computer dialog (annotated by comments in a different font) that introduces the main characteristics of the notation. Because the notation is executable. and operators such as / and & are accordingly called adverbs and conjunctions. Notation and Terminology
The terminology used in J is drawn more from English than from mathematics: a) Functions such as + and * and ^ are also referred to as verbs (because they act upon nouns such as 3 and 4).01 10. The reader is therefore urged to use the computer to do the exercises provided for each section. 3 2 1 2 2 1 (+: . assuming that the reader can grasp the meaning of new notation from context. respectively.10 Calculus
B. from the on-line Dictionary. b) The symbol =: used in assigning a name to a referent is called a copula.45+6. square. %:) 16 32 8 256 4 +/4 5 6 15
. *: . and will introduce the necessary notation with a minimum of comment. The reader is urged to try the following sentences (and variants of them) on the computer:
3. and the names credits and sum used in the sentences credits=: 24.
The conjunction & bonds a dyad to a noun; result is a corresponding function of one argument (a monad)
Sine (of radian arguments) Sine of 0, 1, and one-half pi
Cube of x
We will write informal proofs by writing a sequence of sentences to imply that each is equivalent to its predecessor, and that the last is therefore equivalent to the first. For example, to show that the sum of the first n odd numbers is the square of n, we begin with:
] odds=: 1+2*i.n=: 8 1 3 5 7 9 11 13 15 |.odds 15 13 11 9 7 5 3 1 odds + |.odds 16 16 16 16 16 16 16 16 n#n 8 8 8 8 8 8 8 8
Exercises Solutions or hints appear in bold brackets. Make serious attempts before consulting them. B1 To gain familiarity with the keyboard and the use of the computer, enter some of the sentences of this section and verify that they produce the results shown in the text. Do not enter any of the comments that appear to the right of the sentences. To test your understanding of the notions illustrated by the sentences of this section, enter variants of them, but try to predict the results before pressing the Enter key. Enter p=: 2 3 5 7 11 and predict the results of +/p and */p; then review the discussion of parentheses and predict the results of -/p and %/p .
B2
B3
Chapter 1 Introduction 13
B4 B5
Enter i. 5 and #p and i.#p and i.-#p . Then state the meanings of the primitives # and i. . Enter asp=: p * _1 ^ i. # p to get a list of primes that alternate in sign (enter asp alone to display them). Compare the results of -/p and +/asp and state in English the significance of the phrase -/ . [ -/ yields the alternating sum of a list argument]
B6
Explore the assertion that %/a is the alternating product of the list a. [ Use arp=: p^_1^i.#p ]
B7
Execute (by entering on the computer) each of the sentences of the informal proof preceding these exercises to test the equivalences. Then annotate the sentences to state why each is equivalent to its predecessor (and thus provide a formal proof). Experiment with, and comment upon, the following and similar sentences:
s=: '4%5' |.s do=: ". do s do |.s |.i.5 |. 'I saw'
[ Entering the name of a function alone shows its definition in linear form;
14 Calculus
the foreign conjunction !: provides other forms] B11 Press the key F1 (in the top row) to display the J vocabulary, and click the mouse on any item (such as -) to display its definition.
C. Role of the Computer and of Notation
Seeing the computer determine the derivatives of functions such as the square might well cause a student to forget the mathematics and concentrate instead on the wonder of how the computer does it. A student of astronomy might likewise be diverted by the wonders of optics and telescopes; they are respectable, but they are not astronomy. In the case of the derivative operator, the computer simply consults a given table of derivatives and an associated table of rules (such as the chain rule). The details of the computer calculation of the square root of 3.14159 are much more challenging. The important point for a student of mathematics is to treat the computer as a tool, being clear about what it does, not necessarily how it does it. In particular, the tool should be used for convenient and accurate experimentation with mathematical ideas. The study of notation itself can be fascinating, but the student of calculus should concentrate on the mathematical ideas it is being used to convey, and not spend too much time on byways suggested by the notation. For example, a chance application of the simple factorial function to a fraction (! 0.5) or the square root to a negative number (%:-4) might lead one away into the marvels of the gamma function and imaginary numbers. A student must, of course, learn some notation, such as the use of ^ for power (first used by de Morgan) and of + and * for plus and times. However, it is best not to spend too much conscious effort on memorizing vocabulary, but rather to rely on the fact that most words will be used frequently enough in context to fix them in mind. Moreover, the definition of a function may be displayed by simply entering its name without the usual accompanying argument, as illustrated in Exercise B10.
D. Derivative, Integral, and Secant Slope
The central notions of the calculus are the derivative and the integral or anti-derivative. Each is an adverb in the sense that it applies to a function (or verb) to produce a derived function. Both are illustrated (for the square function x2) by the following graph, in which the slope of the tangent at the point x,x2 as a function of x is the derivative of the square function, that is 2x. The area under the graph is the integral of the square, that is, the function x3 /3, a function whose derivative is the square function. Certain important properties of a function are easily seen in its graph. For example, the square has a minimum at the point 0 0; increases to the right of zero at an accelerating rate; and the area under it can be estimated by summing the areas of the trapezoids:
PLOT x;*: x=:i:4
x=:(i. q=:2&o. This may be seen by plotting the functions together with their sum.11)%5 PLOT x. but are expressed directly by its derivative. The slope of the secant (from ligne secante.(f x+r) is obtained by dividing the rise(f x+r)-(f x) by the run r. such as illustrated by the lines in the foregoing plot of the square function. We will illustrate this by the sine and cosine functions:
p=:1&o. The difference calculus proves useful in a wide variety of applications.Chapter 1 Introduction 15
These properties concern the local behavior of a function in the sense that they concern how rapidly the function value is changing at any point. a host of important functions can be defined simply in terms of their derivatives. the important exponential (or growth) function is completely defined by the fact that it is equal to its derivative (therefore growing at a rate equal to itself). and financial calculations in which events (such as payments) occur at fixed intervals. Sums and Multiples
The derivative of the function p+q (the sum of the functions p and q) is the sum of their derivatives. They are not easily discerned from the expression for the function itself. or cutting line) through the points x.>(p x). More surprisingly. the result of ((f x+r)-f x)%r is called the r-slope of f at the point x. The function used to plot the square must be prepared as follows:
load 'graph plot' PLOT=:'stick. The difference calculus (Chapter 4) is based upon secant slopes.(q x).((p x)+(q x))
The sine function The cosine function
. including approximations to arbitrary functions.f x and (x+r). For example.line'&plot
E. and has the value 1 for the argument 0.
the slopes of its secants are also the sum of the corresponding slopes. the result is 2*x. Similarly.
. leaving 3*x^2. Thus if g=: ^&3 :
((g x+r)-(g x)) % r ((3*(x^2)*r)+(3*x*r^2)+(r^3)) % r (3*x^2)+(3*x*r)+(r^2)
Again the derivative is obtained by setting r to zero. the slopes of a multiple of a function p are all the same multiple of the slopes of p.(x^2)) % r ((2*x*r)+(r^2)) % r (2*x)+r
Moreover. as shown in the following proof. For example:
PLOT x. Since this is true for every secant.>(p x). if r is set to zero in the final expression (2*x)+r. or list of identical expressions:
((f x+r)-(f x)) % r (((x+r)^2)-(x^2))%r (((x^2)+(2*x*r)+(r^2)) . it is true for the derivative. the value of the derivative of ^&2. Similar analysis can be performed on other power functions. Derivatives of Powers
The derivative of the square function f=: ^&2 can be obtained by algebraically expanding the expression f(x+r) to the equivalent form (x^2)+(2*x*r)+(r^2).(2 * p x)
F. and its derivative is therefore the same multiple of the derivative of p.16 Calculus
Since each value of the sum function is the sum of the component functions.
that is. so each of these derivatives is a multiple of the derivative of the corresponding power. using the polynomial function denoted by p. the derivative of ^&n is n*^&n.#c.#c 0 _20 _6 6 dc=:}. Derivatives of Polynomials
The expression (8*x^0)+(_20*x^1)+(_3*x^2)+(2*x^3) is an example of a polynomial. in general. The derivative is therefore the sum: (0*8)+(_20*1*x^0)+(_3*2*x^1)+(2*3*x^2) This is a polynomial with coefficients given by c*i.(dc p. x _28 c=:8 _20 _3 2 x=:0 1 2 3 4 5 (8*x^0) + (_20*x^1) + (_3*x^2) + (2*x^3) 8 _13 _28 _25 8 83 c p.#c 0 1 2 3 c*i. For example:
x=:2 (8*x^0)+(_20*x^1)+(_3*x^2)+(2*x^3) _28 8 _20 _3 2 p. .. Each term is a multiple of a power. x). x _20 _20 _8 16 52 100 x. n*x^n-1. x.Chapter 1 Introduction 17
Similar analysis shows that the derivative of ^&4 is 4*^&3 and. x 8 _13 _28 _25 8 83
The expression (8*x^0)+(_20*x^1)+(_3*x^2)+(2*x^3) is a sum whose derivative is therefore a sum of the derivatives of the individual terms. The elements of the list 8 _20 _3 2 are called the coefficients of the polynomial.. Since the first term of the expansion of (x+r)^n is cancelled by the subtraction of x^n. with the leading element removed to reduce each of the powers by 1 :
c 8 _20 _3 2 i.(c p. We may also express it as 8 _20 _3 2 p.c*i. the only term relevant to the derivative is the second. x) 0 8 _20 1 _13 _20
.
G. and since all terms after the second include powers of r greater than 1.#c dc _20 _6 6 dc p.
>(csin p. x). x)
As remarked in Section A. using _1r6 for the rational fraction negative one-sixths:
csin=:0 1 0 _1r6 0 1r120 0 _1r5040 ccos=:}.(dc p.18 Calculus 2 _28 _8 3 _25 16 4 8 52 5 83 100 PLOT x.6 Sum of odd powers (s1 5) p. x).875 10 18. The following illustrates this for the sine function and its derivative (the cosine).5*i. and see the definition of $ s1=:$&0 1 s2=:_1&^@s1 s1 5 0 1 0 1 0 s2 8 1 _1 1 _1 1 _1 1 _1 A polynomial with coefficients produced by a series function is a sum of powers weighted by the series.(ccos p.>(c p. x 0 0.125
.#csin x=:(i:6)%2 PLOT x. x)
H. For example:
x=:0. " … the functions of interest in elementary calculus are easily approximated by polynomials … ". Power Series
We will call s a series function if s n produces a list of n elements. and is called a power series. For example: Press F1 for the vocabulary.csin*i.625 2 4.
this in contrast to the restriction to scalars (single elements) common in elementary treatments of the calculus.
. 3. 4. Partial derivatives are treated in a simpler and more general way made possible by the use of functions that deal with arguments and results of arbitrary rank. They are an extension of derivatives of integral order. As illustrated at the end of Section B. y PLOT y. quasi-empirical. The Calculus of Differences (Chapter 4) is developed as a topic of interest in its own right rather than as a brief way-station to integrals and derivatives. informal proofs will be presented by writing a sequence of expressions to imply that each is equivalent to its predecessor. In Vector Calculus (Chapter 3). For the differential calculus of Chapter 2. Fractional derivatives (Chapter 5) constitute a powerful tool that is seldom treated in calculus courses. Moreover: 1. the important difference is the avoidance of problems of limits by restricting attention to polynomials. 6. of which the author says: "Its modest aim is to elaborate the point that informal. and proofs are instead treated (as they are in Arithmetic [1]) in the spirit of Lakatos in his Proofs and Refutations [2]. and the extension of the factorial and binomial coefficient functions to fractional arguments.y2
I. it is used for experimentation. mathematics does not grow through the monotonous increase of the number of indubitably established theorems but through the incessant improvement of guesses by speculation and criticism. 2. new notions are first introduced by leading the student to see them in action.>y1. by the logic of proofs and refutations." 5. and that the last is therefore equivalent to the first. Conclusion
We conclude with a brief statement of the ways in which the present treatment of the calculus differs from most introductory treatments.20 Calculus y1=:c p. The notation used is unambiguous and executable. introduced here in a manner analogous to the extension of the power function to fractional exponents. Few formal proofs are presented.c*i. and the use of power series to extend results to other functions.5*i:6 y2=:(}.10) p. Because it is executable. y=:0. and to gain familiarity with their use before analysis is attempted.
For example. the section on The derivative of a function occurs after eighty pages of preliminaries. The present text defers discussion of the analytical basis to Chapter 8. so that she may better appreciate the point of the analysis. in Johnson and Kiokemeister Calculus with analytic geometry [6]. Try to provide (or at least sketch out) answers without using the computer. and then use it to confirm your results.Chapter 1 Introduction 21
7. 8. The common approach is to treat the basis first. b) A basis comprising the analysis of the notion of limit (that arises in the transition from the secant slope to the tangent slope) needed as a foundation for an axiomatic deductive treatment. first providing the reader with experience with the derivative and the importance of its fruits.
. The exercises are an integral part of the development. perhaps even before reading the relevant sections. together with their important consequences. and the body second. and should be attempted as early as possible. Two significant parts may be distinguished in treatments of the calculus: a) A body comprising the central notions of derivative and anti-derivative (integral).
c*i. If f is the polynomial c&p. Because a polynomial includes one more term than its derivative. for the cases considered. However.5 0. the equation for Example 3 is also satisfied by the alternative function as=: 8"0+s. it can never exactly equal the derivative. Consideration of the convergence of such approximations is deferred to Chapter 8.(%1). and approximation can be made as close as desired.#c 1 0. and circular (or trigonometric).166667 0. and we consider functions that approximate the desired solution. The same is true for any coefficients produced by the following exponential coefficients function:
.00833333 deco c 1 0.. Growth F d. and the growth of a well-fed colony of bacteria. it is said to grow exponentially.00138889 c*i.(%!5).5 0. then s is completely defined.166667 0. then the derivative of f is the polynomial with coefficients deco c.00833333 }. For example.5 0.00833333 0.25 as d. successive coefficients decrease rapidly in magnitude.#c 1 1 0.(%!6) 1 0.(%1*2).0416667 0.26 Calculus
A stated relation may not specify a function completely.1 x 1 8 27 64 125
C1
Experiment with the expressions of this section.166667 0. if it is further required that s 2 must be 7. Thus:
as=:8"0 + s as x 8. We will approach the solution of differential equations through the use of polynomials. The remainder of this chapter will use simple differential equations to define an important collection of functions.0416667 0.(%1*2*3). Differential Equations
An equation that involves derivatives of the function being defined is called a differential equation.5 0.25 72 164.0416667 0. Thus:
as=:3"0 + s as 2 7 as d. hyperbolic.
E.
D.166667 0.1 x 1 8 27 64 125
Further conditions may therefore be stipulated to define the function completely. including the exponential. Thus:
]c=:1.(%!4). Examples of exponential growth include continuous compound interest.00833333
1 0 1 1
In this case the coefficients of the derivative polynomial agree with the original coefficients except for the missing final element.25 12 28. For example.0416667 0.1 = F
If the derivative of a function is equal to (or proportional to) the function itself.
Since the second derivative of the exponential ^ is also equal to itself.>y1.5 0 0.0183153 (deco eca 20)&p x _1 _0.166667 0 0.y2. test it for a=:1 2 1 [ b=:1 3 3 1.135335 _0.00138889 0 deco deco c 0 1 0 0. replacing by zeros) alternate elements of ec i.135335 0. x 1 0.2 = F
The second derivative of a function may be construed as its acceleration. it should also be apparent that it qualifies as a second function that equals its second derivative.n 0 1 0 0.367879 _0.0416667 0 0.0183175
The relation between the growth and decay functions will be explored in exercises and in Chapter 6. However. We will again use polynomials to approximate functions. Then enter PLOT x.1 must not equal f. and many phenomena are described by functions defined in terms of their acceleration. Predict and confirm the result of the product y1*y2.n 1 0 0.28 Calculus (eca 20)&p.00138889 0 2.
G.48016e_5
.0497871 0.@(*/) ] F2 F3 F4 Predict the value of a few elements of (ec pp eca) i.166667 0 0. Thus:
2|i.0497871 _0.*b&p.367879 0. F1 Define a function pp such that (a pp b)&p.00833333 0 0.000198413 0 deco c 1 0 0.n.1*i:30 and y1=:^ x and y2=:^@-x. is equivalent to the product (a&p.7 and enter the expression to validate your prediction.5 0 0.) . Coefficients satisfying these requirements can be obtained by suppressing (that is. the coefficients ec i. Enter x=:0. first a function that is equal to its second derivative. we seek new functions and therefore add the restriction that f d. [ pp=: +//.0416667 0 0. However.n=: 9 0 1 0 1 0 1 0 1 0 hsc=: 2&| * ec ]c=: hsc i.n would suffice. We therefore define a corresponding function hcc :
hcc=: 0&=@(2&|) * ec hcc i.00833333 0
The result of deco c was shown above to make clear that the first derivative differs from the function. Hyperbolic Functions F d.
Chapter 2 Differential Calculus 29 deco deco hcc i.7622 10. Circular Functions F d.00833333 0 _0. Thus:
hsin=:5&o.0179 27.1*i:30 and y1=:hsin x and y2=:hcos x.y2 to plot cosh against sinh. whose values repeat as arguments grow. like the exponential.54308 3.1752 3. They are the functions defined by hsin=: 5&o.000198413 0 cc i. Appropriate polynomial coefficients are easily obtained by alternating the signs of the non-zero elements resulting from hsc and hcc.62686 10. We now consider functions whose acceleration is opposite in sign to the functions themselves.62686 10.5 0 0.3082 hcos x 1 1. Further properties of these functions will be explored in Chapter 6. G1 Enter x=:0. G3 Predict the result of (y2*y2)-(y1*y1) and test it on the computer.0416667 0 0. These functions are useful in describing periodic phenomena such as the oscillations in a mechanical system (the motion of a weight suspended on a spring) or in an electrical system (a coil connected to a capacitor). (hsc i. since their acceleration increases with the increase of the function.n 1 0 0. hcos=:6&o. and hcos=: 6&o.2899 (hcc i.0677 27.20)&p.7622 10.2899 hsin x 0 1. Thus:
sc=: _1&^@(3&=)@(4&|) * hsc cc=: _1&^@(2&=)@(4&|) * hcc sc i. continue to grow with increasing arguments.n 0 1 0 _0. x=: 0 1 2 3 4 0 1.3082
It should also be noted that each of the hyperbolic functions is the derivative of the other.>y1.0179 27. it will be seen that a plot of one against the other yields a hyperbola.00138889
The limiting values of the corresponding polynomials are called the hyperbolic sine and hyperbolic cosine. The more pronounceable abbreviations cosh and sinh (pronounced cinch) are also used for these functions.54308 3.20)&p. and comment on the shape of the plot.1752 3.166667 0 0. respectively. x 1 1.
H.0677 27. a characteristic that leads to periodic functions. In particular.2 = -@F
It may be noted that the hyperbolics. This is not surprising. G2 Enter PLOT y1..y2.n
. Then plot the two functions by entering PLOT x.
989992 _0.989538 _0. @ (y.54298 _0. Each function is accompanied by a phrase (such as Identity) and an index that will be used to refer to it. together with their derivatives. Argument Transformations
Scaling is only one of many useful argument transformations. Thus:
f AA 3 AM 4 is f AP 3 4 f AM 3 AA 4 is f AP 12 3
A function FfC to yield Fahrenheit from Celsius can be used to further illustrate the use of argument transformation:
FfC=: 32"0 + 1. @ (y.416147 _0. atop addition and atop polynomial:
AA=: 2 : 'x.8 fahr _40 32 212 Uses Constant functions (See Section 1B)
The following derivatives are easily obtained by substitution and the use of the table of Section K: Function
f AA r f AM r f AP c
Derivative
f D AA r (f D AM r * r"0) (f D AP c * (d c)&p. The conjunction AP provides a more general transformation.&p. Table of Derivatives
The following table lists a number of important functions.653644 cos AA per x 0.8 fahr _40 32 212 ] AP 32 1.28 cos x 1 0. we define two further conjunctions.656051
Experimentation with different values of per can be used to determine a better approximation to the true period of the cosine.&+)' AP=: 2 : 'x.413248 _0.999995 0.8"0 * ] fahr=: _40 0 100 FfC fahr _40 32 212 ] AA 32 AM 1. as in Theorem 2 or θ2 (where θ is the Greek letter theta) .)'
In Section H it was remarked that the circular functions sin and cos "repeat" their values after a certain period. Thus:
per=: 6.540302 _0.)
K.
.Chapter 2 Differential Calculus 31
J.
DERIVATIVE
0"0 1"0 a"0 (f d.5 θ6 Since a"0 x is a for any x. and the slope is a%a Multiplying a function by a multiplies all of its rises.1))+((g d. conversely if f changes while g is fixed. by a as well.1)=(g*(h d. using θ 6 :
f d.1) (f d.1)%g) (f d.1)*h)
θ7
The equation (f d.@#) Inverse adverb INV=:^:_1 Although more thorough analysis will be deferred to Chapter 8. multiplied by the rate of change of the function that is applied first (that is. and hence its slopes. Also see the discussion in Section 1D. θ8 The derivative of f@g is the derivative of f "applied at the point g" (that is. and.1)
.1)+(g d.
10 Reciprocal 11 Power 12 Polynomial Legend:
Functions f and g and constants a and n. the result of f*g changes by f times the change in g. Since (]a+x)-(]x) is (a+x)-x. * ^&(n-1) (deco c)&p.1) (f*(g d.1))+((f d.1)%f)-((g d.1 @(f INV)) -@(f d. then g*h is f.1 % (f*f)) n&p. and list constant c Polynomial derivative deco=:}.1))+((g d.1)-(g d.1 (g*h) d.@(] * i. The total change in f*g is the sum of these changes.1 (g*(h d. g d. the rise is the zero function 0"0.1)@g).1)*h) can be solved for h d. (f d. If the result of f is fixed while the result of g changes.1)*g) (f%g)*((f d. the rise is a. we will here present arguments for the plausibility of the theorems: θ1 θ2 θ3 θ 4.1) %@(f d. giving the result of θ 7.1)@g * (g d. If h=: f%g.1. The rise of f+g (or f-g) is the sum (or difference) of the rises of f and g.32 Calculus
θ 1 2 3 4 5 6 7 8 9
NAME Constant function Identity Constant Times Sum Difference Product Quotient Composition Inverse
FUNCTION
a"0 ] a"0 * ] f+g f-g f*g f%g f@g f INV %@f ^&n c&p.
f3.f2. But since f@(f INV) is the identity function.Chapter 2 Differential Calculus 33
θ9
f@(f INV) d. then a"0 is a constant function.1) * ]) (] * 1"0) + (1"0 * ]) ] + ] 2"0 * ]
Theorem 6 Theorem 2
Twice the argument
Further powers may be expressed as products with the identity function. gives 1 (true) for every argument. Further cases may be obtained similarly.1 = f d.7).1) is a tautology. Use of Theorems
The product of the identity function (]) with itself is the square (^&2 or *:). Thus:
f4=:]*f3=:]*f2=:]*f1=:]*f0=:1"0 x=:0 1 2 3 4 1 0 0 0 0 >(f0.f4) x 1 1 1 1 1 2 3 4 1 4 9 16 1 8 27 64 1 16 81 256
Their derivatives can be analyzed in the manner used for the square:
f3 d.1)) + ((] d.1) (from θ 6).1 (]*f2) d. If a is a noun (such as 2. Prove that ((a"0 + f) d.1 (((] d.1)@(f INV) * ((f INV) d. This follows from θ 3 and θ 11.
θ 10 θ 11
This can be obtained from θ 7 using the case f=: ] .1 is therefore the reciprocal of the first. its derivative is 1&p. that is.f1.1 (] * (] d.
θ 12 K1
K2
L.1))) ((1"0 * f2)+(]*2"0 * ]))
. and the expression for the derivative of a product can therefore be used as an alternative determination of the derivative of the square and of higher powers:
(] * ]) d. its derivative may be obtained from θ 6 and the result for the derivative of ^&4. Since ^&5 is equivalent to the product function ] * ^&4. being sure to parenthesize the entire sentence if need be. then test the equivalence of the functions in the discussion of Theorem 7 by entering each followed by x.1)*f2)+(]*(f2 d. that is. Enter f=: ^&2 and f=: ^&3 and x=: 1 2 3 4 . by induction.1 is the product (f d. and the second factor (f INV) d.
8 0. the rate of change approaches the exact function value.1 0. the average of circle x.953939 0.43589 1 0 PLOT x.8 0. Therefore.01*i.994987 0.4 0.101
. As the increment r approaches zero.1) its change from x to x+r is r times the average height of the trapezoid.866025 0.979796 0. its rate-of change (derivative) at any argument value x is approximately the corresponding value of the circle function.*:) x=:0. then circle x gives the y coordinate of a point on a circle with radius 1.*:).714143 0. and circle x+r.7 0. For example. as illustrated below for the value r=:0. and (using r=:0.11 y=:circle x x.01:
x=:0. if circle=: %: @ (1"0 . that is.9 0.2 0.Chapter 2 Differential Calculus 35
N.3 0.1*i. Integral
The area under (bounded by) the graph of a function has many important interpretations and uses.y
Square root of 1 minus the square
The approximate area of the quadrant is given by the sum of the ten trapezoids. The first quadrant may then be plotted as follows:
circle=: %: @ (1"0 .y 0 1 0.6 0.916515 0.6 0..5 0.
the area under the curve is given by the anti-derivative.
.circle x
In other words.36 Calculus PLOT x.
37
Chapter
3
Vector Calculus
A. Introduction
Applied to a list of three dimensions (length, width, height) of a box, the function vol=:*/ gives its volume. For example:
lwh=:4 3 2 vol=:*/ vol lwh 24
Since vol is a function of a vector, or list (rank-1 array), the rank-0 derivative operator d. used in the differential calculus in Chapter 1 does not apply to it. But the derivative operator D. does apply, as illustrated below:
vol D.1 lwh 6 8 12
The last element of this result is the rate of change as the last element of the argument (height) changes or, as we say, the derivative with respect to the last element of the vector argument. Geometrically, this rate of change is the area given by the other two dimensions, that is, the length and width (whose product 12 is the area of the base). Similarly, the other two elements of the result are the derivatives with respect to each of the further elements; for example, the second is the product of the length and height. The entire result is called the gradient of the function vol. The function vol produces a rank-0 (called scalar, or atomic) result from a rank-1 (vector) argument, and is therefore said to have form 0 1 or to be a 0 1 function; its derivative produces a rank-1 result from a rank-1 argument, and has form 1 1. The product over the first two elements of lwh gives the "volume in two dimensions" (that is, the area of the base), and the product over the first element alone is the "volume in one dimension". All are given by the function VOLS as follows:
VOLS=:vol\ VOLS lwh 4 12 24
The function VOLS has form 1 1, and its derivative has form 2 1. For example:
38 Calculus VOLS D.1 lwh 1 0 0 3 4 0 6 8 12
This table merits attention. The last row is the gradient of the product over the entire argument, and therefore agrees with gradient of vol shown earlier. The second row is the gradient of the product over the first two elements (the base); its value does not depend at all on the height, and the derivative with respect to the height is therefore zero (as shown by the last element). Strictly speaking, vector calculus concerns only functions of the forms 0 1 and 1 1; other forms tend to be referred to as tensor analysis. Since the analysis remains the same for other forms, we will not restrict attention to the forms 0 1 and 1 1. However, we will normally restrict attention to three-space (as in vol 2 3 4 for the volume of a box) or two-space (as in vol 3 4 for the area of a rectangle), although an arbitrary number of elements may be treated. Because the result of a 1 1 function is a suitable argument for another of the same form, a sequence of them can be applied. We therefore reserve the term vector function for 1 1 functions, even though 0 1 and 2 1 functions are also vector functions in a more permissive sense. We adopt the convention that a name ending in the digits r and a denotes an r,a function. For example, F01 is a scalar function of a vector, ABC11 is a vector function of a vector, and G02 is a scalar function of a matrix (such as the determinant det=: -/ . *). The functions vol and VOLS might therefore be renamed vol01 and VOLS11. Although the function vol was completely defined by the expression vol=:*/ our initial comments added the physical interpretation of the volume of a box of dimensions lwh. Such an interpretation can be exceedingly helpful in understanding the function and its rate of change, but it can also be harmful: to anyone familiar with finance and fearful of geometry, it might be better to use the interpretation cost=:*/ applied to the argument cip (c crates of i items each, at the price p). We will mainly allow the student to provide her own interpretation from some familiar topic, but will devote a separate Chapter (7) to the matter of interpretations. Chapter 7 may well be consulted at any point.
B. Gradient
As illustrated above for the vector function VOLS, its first derivative produces a matrix result called the complete derivative or gradient. We will now use the conjunction D. to define an adverb GRAD for this purpose:
GRAD=:D.1 VOLS GRAD lwh 1 0 0 3 4 0 6 8 12
We will illustrate its application to a number of functions:
E01=: +/@:*: F01=: %:@E01 G01=: 4p1"1 * *:@F01
Sum of squares Square root of sum of squares Four pi times square of F01
Develop interpretations for each of the functions defined above. ANSWERS:
E01 p is the square of the distance (from the origin) to a point p. F01 p is the distance to a point p, or the radius of the sphere (with centre at the origin) through the point p. G01 p is the surface area of the sphere through the point p. H01 is the intensity of illumination at point p provided by a unit light source at the
1 0)"1 significance of a negative Jacobian. Define 3-space linear functions to apply to FIG1.=:(H11*])"1.@(0&. and use them together with K11 to repeat Exercises 1-5 in 3-space. 0 1 1 must not succeed 1 1 0.)@RM2 R30=: (] mp RM3@[)"0 1
C9
[ a&R30"1 produces a rotation through an angle a in the plane of the last two axes in 3-space (or about axis 0). and comment upon the following functions:
RM2=: 2 2&$@(1 1 _1 1&*)@(2 1 1 2&o.. ] C2 C3 What is the relation between the Jacobian of the linear function L11 and the determinant of the matrix m used in its definition? What is the relation between the Jacobians of two linear functions LA11 and LB11 and the Jacobian of LC11=: LA11@LB11 (their composition).] C7 C8 What is the value of the Jacobian of a rotation a&R2"1? Enter an expression to define FIG1 as an 8 by 3 table representing a cube. the value of the Jacobian is the same at every point. and note that one can be moved smoothly onto the other "without crossing lines". and p=: 5&A. in the expression p&. Enter. and test the comparison expressed in the solution to Exercise C3 by applying TEST to appropriate arguments. The function -@K11 may be interpreted as gravitational attraction. [ TEST=:LA11@LB11 JAC |@. for a linear function. [Experiment with the permutations p=: 2&A.%2. ] C6 Enter. The Jacobian of the linear LR11=: mp&(>0 1. experiment with. the expression a R2 fig rotates a figure (such as fig1 or fig2) about the origin through an angle of a radians in a counter-clockwise sense. ]
. experiment with. and comment upon the functions
RM3=: 1 0 0&. and experiment with functions such as a1&R31@(a2&R30)"1. for example. Moreover. as may be verified by plotting the two figures by hand. and use the ideas in functions defined in terms of R30. Verify that this cannot be done with fig1 and LR11 fig1. C1 Provide an interpretation for the function K11. State the
[ Plot figures fig1 and fig2.)"0 R2=: (] mp RM2@[)"0 1
[ R2 is a linear function that produces a rotation in 2-space. it is necessary to "lift the figure out of the plane and flip it over". making sure that successive coordinates are adjacent. [ The result of K11 is the direction and magnitude of the repulsion of a negative electrical charge from a positive charge at the origin.LA11 JAC * LB11 JAC ] C4 C5 Define functions LA11 and LB11. Test the value of the Jacobian. without deforming the figure.Chapter 3 Vector Calculus 41
The result of the Jacobian is indeed the ratio of the areas of fig1 and fig2.] C10 Define functions R31 and R32 that rotate about the other axes. is _1. A transformation whose Jacobian is negative is said to involve a "reflection".|:@p@RM3 o.
Divergence and Laplacian
The divergence and Laplacian are defined and used as follows:
DIV=: GRAD TRACE LAP=: GRAD DIV
f=: +/\"1 f a 1 3 6 f GRAD a 1 0 0 1 1 0 1 1 1 g=: +/@(] ^ >:@i. in his Advanced Calculus [8]. Skew-Symmetry. and a matrix that equals the negative of its transpose is skew-symmetric. However.@#)"1 g a 32 f DIV a 3
g LAP a 22. F. and the reader may be best advised to seek interpretations in some familiar field.. . Symmetry. Applied to an infinitesimal volume it appears that div F represents the amount of fluid per unit time which streams or diverges from a point.5 _3
The gradient of the volumes function
The gradient is not symmetric
The symmetric part of the gradient
The skew-symmetric part
..S.5 4 4 3 4 12 ]msk=:(m-|:m)%2 0 _1. and Orthogonality
A matrix that is equal to its transpose is said to be symmetric.0268
It is difficult to provide a helpful interpretation of the divergence except in the context of an already-familiar physical application.42 Calculus
D. Woods offers the following: "The reason for the choice of the name divergence may be seen by interpreting F as equal to rv. where r is the density of a fluid and v is its velocity."
E. For example:
]m=:VOLS GRAD lwh 1 0 0 3 4 0 6 8 12 |:m 1 3 6 0 4 8 0 0 12 ]ms=:(m+|:m)%2 1 1.5 3 1.
a vector normal to the plane of (the skewsymmetric part of) the gradient of the function. and its vectors therefore lie in a plane:
det=:-/ . Such transposes are obtained by using |: with a left argument:
]a=:i. by interchanging different pairs of axes. The function C.2 2 2 0 1 2 3 4 5 6 7 0 2 1 |: a 0 2 1 3 4 6 5 7 1 0 2 |: a 0 1 4 5 2 3 6 7
Interchange last two axes
Interchange first two axes
The permutation 0 2 1 is said to have odd parity because it can be brought to the normal order 0 1 2 by an odd number of interchanges of adjacent elements.2 yields the parity of its argument. We will generate a completely skew array by applying the parity function to the table of all indices of an array:
indices=:{@(] # <@i. * det msk 0
The determinant function Shows that the vectors of msk lie in a plane
The axes of a rank-3 array can be "transposed" in several ways. and 0 if it is not a permutation. Such an array is useful in producing a vector that is normal (or orthogonal or perpendicular) to a plane. An array that is skew-symmetric under any interchange of axes is said to be completely skew. 1 if the argument has even parity. 1 2 0 has even parity because it requires an even number of interchanges. _1 if odd. In particular.!. we will use it in a function called norm that produces the curl of a vector function.) indices 3 +-----+-----+-----+ |0 0 0|0 0 1|0 0 2|
.Chapter 3 Vector Calculus 43 1.5 3 0 _4 4 0
ms+msk 1 0 0 3 4 0 6 8 12
Sum of parts gives m
The determinant of any skew-symmetric matrix is 0.
First. curl F=0. This presentation will therefore show what vector calculus is.46 Calculus
Interpretation of the curl is perhaps even more intractable than the divergence. Second. DIV. CURL. and at the same time give you an idea of what it's for. and curl in joint use. We follow this procedure for two reasons. we have a deep-seated conviction that mathematics -in any case some mathematics. See Section 6H.^@-@x au=: (x. H. and for vortex motion. in Heaviside's elegant formulation. and JAC on the functions in Exercise B2. its motion in a time dt may be analyzed into a translation. much of vector calculus was invented for use in electromagnetic theory and is ideally suited to it. and use them to define the functions of the preceding exercise in a more conventional form. and JAC on the following 1 1 functions:
q=: r=: s=: t=: u=: *:"1 4&A. then for velocity in what we have called irrotational motion.M. grad.z) % (*:@x + *:@y + *:@z) ^ 3r2"0
]
F4 F5
Experiment with LAP on various 0 1 functions. [ CR=: */ NORM
CURL=: GRAD NORM ]
. @: q 1 1 _1&* @: r 3&A. Again Woods offers some help: The reason for the use of the word curl is hard to give without extended treatment of the subject of fluid motion. a deformation. Grad.y.
.is best discussed in a context which is not exclusively mathematical. and all that [9]. exhibit the powers of div. Curl. The curl of the vector v can be shown to have the direction of this axis and a magnitude equal to twice the instantaneous angular velocity. F1 F2 Experiment with GRAD.^@-@z.-@*:@y
at=: ^@-@y. DIV. Express the cross product of Section 6G so as to show its relation to CURL. [ as=: *:@z. curl F≠0. His first chapter begins with: In this text the subject of the vector calculus is presented in the context of simple electrostatics. Schey's treatment includes Maxwell's equations which. In his Div. @: ^ @: ]% (+/@(*~)) ^ 3r2"0
F3
Enter the definitions x=: 0&{ and y=: 1&{ and z=: 2&{. It may be shown that if a spherical particle of fluid be considered. CURL. Experiment with GRAD. and a rotation about an instantaneous axis. The student may obtain some help by noticing that if F is the velocity of a liquid.*:@x. Schey makes an interesting attempt to introduce the concepts of the vector calculus in terms of a single topic.
Introduction
Although published some fifty years ago. functions in which the variable takes only the given values x0. or the r-slope of f at x. that is. In his introductory section on Historical and Biographical Notes. .f x to the point x.. x2.f(x+r) is said to be the secant slope of f for a run of r. First. The present brief treatment is restricted to three main ideas: 1) The development of a family of functions which behaves as simply under the difference (secant slope) adverb as does the family of power functions ^&n under the derivative adverb. Jordan's Calculus of Finite Differences [10] still provides an interesting treatment. Secondly.91 49.31 12.47
Chapter
4
Difference Calculus
A.
B. but it may be applied to both categories..51
The same result is given by the secant-slope conjunction D: as follows:
. Thus:
cube=:^&3"0 x=:1 2 3 4 5 r=:0. x1. These functions belong to the domain of the Infinitesimal Calculus. he contrasts the difference and differential (or infinitesimal) calculus: Two sorts of functions are to be distinguished.21 76. 3) The development of a linear transformation from the coefficients of such a polynomial to the coefficients of an equivalent ordinary polynomial. functions in which the variable x may take every possible value in a given interval. The Calculus of Finite Differences deals especially with such functions.61 27. Secant Slope Conjunctions
The slope of a line from the point x.1 ((cube x+r)-(cube x))%r 3. then the variable is discontinuous. xn. 2) The definition of a polynomial function in terms of this family of functions. To such functions the methods of Infinitesimal Calculus are not applicable. the variable is continuous.
51
C.0003 12.') r cube SLOPE 1 x 2.9991 47.31 12.81 73.01 cube D: 1 x 3.71 11. We begin by adopting the
.51 0. This is an important property of the family of power functions.41 26.81 73.-x.48 Calculus
r cube D: 1 x 3.81 73. a value given by the derivative.9994 26.0015 cube d. and therefore define a corresponding adverb:
S=:("0) SLOPE 1 r cube S x 2. and it will prove more convenient in our further work.11 46.41 26. It is also equal to three times the square. y.91 49.21 76. 1 x 3 12 27 48 75
Much like the derivative.61 27.2 x 6 12 18 24 30 r cube SLOPE 2 x 6 12 18 24 30
We will be particularly concerned with the "first" slope applied to scalar (rank-0) functions.1501 0.9997 11.9985 cube d. Thus:
cube d. The alternate expression ((cube x)-(cube x-r))%r could also be used to define a slope.1201 75. the analysis of the power function ^&n led to the result that the derivative of the polynomial c&p. 1 x 3 12 27 48 75 3*x^2 3 12 27 48 75
In the foregoing sequence. u. smaller runs appear to be approaching a limiting value.0001 cube D: 1 x 3.c*i. Polynomials and Powers
In Chapter 3.0301 12.9988 74. and we seek another family of functions that behaves similarly under the r-slope.0006 27.0012 75..51 ((cube x)-(cube x-r))%r 2. and so on.#c)&p.51 0.0601 27.41 26. the slope conjunction can be used to give the slope of the slope."0 D: n.11 46.0901 48. 'x.71 11. We therefore define an alternate conjunction for it as follows:
SLOPE=:2 : (':'.71 11.0001 cube SLOPE 1 x 2.0009 48. could be written as another polynomial : (}.11 46.
6. We will call the product over such a list a stope:
x=:5 r=:0.1 (]*p2) d.1*1"0) (]*0"0)+(1"0*1"0) 1"0 p2 d.1 (]*p1 d.1*p3) (]*3"0*p2)+(1"0*p3) (3"0*p3)+p3 4"0*p3
Each of the expressions in the proofs may be tested by applying it to an argument such as x=: i.n 120 */x+0*i.1 (]*p3) d.50 Calculus
Since the derivative of the identity function ] is the constant function 1"0. like the steps in a mine stope that follows a rising or falling vein of ore. Stope Functions
The list x+r*i. expressions for the derivatives of the power functions can be derived using the expressions for the sum and product in informal proofs as follows:
p0 d.r n
.n begins at x and changes in steps of size r.1)+(] d. We therefore treat the stope as a variant of the power function.1 0"0 p1 d.n 5 5.n 702.1)+(] d.1)+(] d.n 625 x^n 625
Case r=:1 is called a rising factorial Falling factorial Case r=:0 gives product over list of n x's Equivalent to the power function
The two final examples illustrate the fact that the case r=:0 is equivalent to the power function.78 */x+1*i.1 n=:4 x+r*i. produced by the conjunction !.1)+(] d.1 (]*p0) d. as follows:
x ^!.
D.1 (]*p3 d.2 5.1 5.1 (]*p0 d.1*p1) (]*1"0)+(1"0*p1) p1+p1 2"0*p1
p3 d. first enclosing the entire expression in parentheses.1*p2) (]*2"0*p1)+(1"0*p2) (2"0*p2)+p2 3"0*p2
p4 d.1 (]*p1) d. We will next introduce stope functions whose behavior under the slope operator is analogous to the behavior of the power function under the derivative.1 (]*p2 d.n 1680 */x+_1*i.1 1"0 d.3 */x+r*i.
2 9. etc. x r stope n 702.0 n 625 stope=: ^!.28 36.9 49. under the derivative.q2.52 r q3 S x 0 3.q4) x 1 1 1 1 1 2 3 4 1.6 27.208 296.28 36.78 x ^!.q1.24 29. Thus:
q0=:r stope&0 q1=:r stope&1 q2=:r stope&2 q3=:r stope&3 q4=:r stope&4 x=:0 1 2 3 4 >(q0. but using the expression:
.96 119.q3. Moreover.96 119. in a manner similar to that used for defining the power functions.76 68.2 3*q2 x 0 3. the stope functions can be defined as a sequence of products. etc.716 21.252 98.1 4.04 275. used for successive powers.52 4*q3 x 0 5.Chapter 4 Difference Calculus 51 702.4 1.6 27.2
This behavior is analagous to that of the power functions p4.3 16.88 1.9 49. p3.3 12. Slope of the Stope
We will now illustrate that the r-slope of r stope&n is n*r stope&(n-1):
r q4 S x 0 5.184
1 0 0 0 0
E. the foregoing property of the r-slopes of stopes can be obtained in the manner used for the derivative of powers.32 9. Thus (using R for a constant function):
R=:r"0 f4=:(]+3"0*R)*f3=:(]+2"0*R)*f2=:(]+1"0*R)*f1=:(]+0"0*R)*f0=:1"0
From these definitions.04 275.3 12.78
The stope adverb
We now define a set of stope functions analogous to the functions p0=:^&0 and p1=:^&1.
The extension of the function ! to non-integer arguments was also cited as the basis for an analogous treatment of noninteger differences.82 q=: 1 ". OS approximation to the second derivative of f at x 17.N' Execute the Oldham Spanier expression to obtain the ".7309 q=: 0 ". Section H of Chapter 1 included a brief statement of the utility of the fractional calculus and a few examples of fractional derivatives and integrals. and therefore as a basis for approximating non-integer differintegrals.1 on page 48 of OS (Oldham and Spanier [5]). Introduction
The differential and the difference calculus of Chapters 2 and 4 concern derivatives and integrals of integer order. that is.2. Thus:
f=: ^&3 q=: 2 N=: 100 a=: 0 x=: 3
Function treated Order of differintegral Number of points used in approximation Starting point of integration Argument
OS=: '+/(s^-q)*(j!j-1+q)*f x-(s=:N%~x-a)*j=:i. The fractional calculus treated in this chapter unites the derivative and the integral in a single differintegral. and extends its domain to nonintegral orders. OS 26. 27)
Zeroth derivative (the function itself)
. Our treatment of the fractional calculus will be based on Equation 3. Section E of Chapter 4 concluded with the use of the alternating binomial coefficients produced by the outof function ! to compute differences of arbitrary integer order.59
Chapter
5
Fractional Calculus
A. OS
Approximation to the first derivative (the exact value is 3*x^2.
We will use the expression OS to define a fractional differintegral conjunction fd such that q (a,N) fd f x produces an N-point approximation to the q-th derivative of the function f at x if q>:0, and the (|q)-th integral from a to x if q is negative:
j=: ("_) (i.@}.@) s=: (&((] - 0: { [) % 1: { [)) (@]) m=: '[:+/(x.s^0:-[)*(x.j!x.j-1:+[)*[:y.]-x.s*x.j' fd=: 2 : m
An approximation to a derivative given by a set of N points will be better over shorter intervals. For example:
x=: 6 1 (0,100) fd f x 106.924 3*x^2 108 1 ((x-0.01),100) fd f x 107.998
Anyone wishing to study the OS formulation and discussion will need to appreciate the relation between the function ! used here, and the gamma function (G) used by OS. Although the gamma function was known to be a generalization of the factorial function on integer arguments, it was not defined to agree with it on integers. Instead, G n is equivalent to ! n-1. Moreover, the dyadic case m!n is here defined as (!n)%(!m)*(!n-m); the three occurrences of the gamma function in Equation 3.2.1 of OS may therefore be written as j!j-1+q, as seen in the expression OS used above. The related complete beta function is also used in OS, where it is defined (page 21) by B(p,q) = (G p) * (G q) % (G p+q). This definition may be re-expressed so as to show its
Chapter 5 Fractional Calculus 61
relation to the binomial coefficients, by substituting m for p-1 and n for p+q-1. The expression B(p,q) is then equivalent to (!m)*(!n-m)%(!n), or simply % m!n.
B. Table of Semi-Differintegrals
The differintegrals of the sum f+g and the difference f-g are easily seen to be the sums and differences of the corresponding differintegrals, and it might be expected that fractional derivatives satisfy further relationships analogous to those shown in Section 2K for the differential calculus. Such relations are developed by Oldham and Spanier, but most are too complex for treatment here. We will confine attention to a few of their semi-differintegrals (of orders that are integral multiples of 0.5 and _0.5). We begin by defining a conjunction FD (similar to fd, but with the parameters a and N fixed at 0 and 100), and using it to define adverbs for approximating semi-derivatives and semi-integrals:
FD=: 2 : 'x."0 (0 100) fd y. ]' ("0) x=: 1 2 3 4 5 1 FD (^&3) x 2.9701 11.8804 26.7309 47.5216 74.2525 3*x^2 3 12 27 48 75
Although the conjunctions sd and si and s3i provide only rough approximations, we will use them in the following table to denote exact conjunctions for the semidifferintegrals. This makes it possible to use the expressions in computer experiments, remembering, of course, to wrap any fork in parentheses before applying it. Function
f+g f-g ]*g c"0*g c"0 1"0 ]
Entries in the table can be rendered more readable to anyone familiar only with conventional notation by a few assignments such as:
twice=: +: sqrt=: %: pitimes=: o. reciprocal=: % on=: @
95441 2.95197 2.59577 1.12838 1.25394 twice on sqrt on reciprocal on pitimes on reciprocal x 1. it can be expressed using the under conjunction as follows:
under=: &.59378 1. twice on sqrt on (pitimes under reciprocal) x 1.25676
Alternatively.12697 1.12838 1.95441 2.Chapter 5 Fractional Calculus 63
The table entry for the semi-derivative of the identity function could then be expressed as follows:
] sd x 1.25676
.59577 1.
.
We will test this conjecture in two ways. and her own proofs before reading Section C.00833333 0. and express them all as members of a single family.20)&p. Introduction
In this chapter we will analyze relations among the functions developed in Chapter 2.5 _0. in other words that their product is one.00833333 0.0416667 _0.5 0.367879 0. In Sections E and F of Chapter 2. first by computing the product directly.20)&p.0183153 (GR x) * (DE x)
. We will use the adverb D=: ("0)(D.38906 20. In this section we will use the growth and decay functions to illustrate the process.00138889 eca i. and then by computing the coefficients of the corresponding product polynomial.135335 0. We will first attempt to discover interesting relations by experimentation. the functions ec and eca were developed to approximate growth and decay functions. The reader is urged to try to develop her own experiments before reading Section B.166667 0.7 1 _1 0.00138889
We will now use the approximate functions to experiment with growth and decay:
GR=: (ec i.7 1 1 0. Thus:
GR x=: 0 1 2 3 4 1 2. DE=: (eca i.0855 54.0416667 0.71828 7. and then devote separate sections to experimentation and to proof. Thus:
eca=: _1&^ * ec=: %@! ec i. and then to construct proofs.
It might be suspected that the decay function would be the reciprocal of the growth function.65
Chapter
6
Properties of Functions
A.0497871 0.166667 0.1) .5981 DE x 1 0.
To any reader already familiar with the exponential function these matters may seem so obvious as to require neither suggestion nor proof. f2.7 ] A3 Denoting the elements of the table t=: 2 2{. The points might be better made by using featureless names such as f1. A1 A2 Test the proof of this section by entering each expression with an argument. Make and display the table T whose (counter) diagonal sums form the product of the coefficients ec i. Similar remarks apply to the hyperbolic and circular functions treated in Sections 2G.sinh) 0. and he may therefore miss the fact that all is based only on the bare definitions given in Sections 2E and 2F. [ t00 is 1*1 A4 A5
t01+t10 is (1*_1)+(1*1) ]
Use the scheme of A3 on larger subtables of T to check further elements of the polynomial product.Chapter 6 Properties of Functions 67
The final expression uses the scaling conjunction of Section I of Chapter 2. t01. Experimentation
Hyperbolics.T by t00. However.H.7 and eca i. t10. [ T=: (ec */ eca) i. write explicit expressions for them. cosh=: 6&o. [Consider the functions f=: ^*^ and g=: ^@+: beginning by applying them to arguments such as f"0 i. load'plot' plot (cosh. Then verify that t00 and t01+t10 agree with the first two elements of the product polynomial given in the text. One hyperbolic may be plotted against the other as follows:
sinh=: 5&o. and t11. 5]
B. In the foregoing discussion we have used simple observations (such as the probable reciprocity of growth and decay) to motivate experiments that led to the statement and proof of significant identities. it seems better to adopt commonly used names at the outset. We may now conclude that the function ^ AM r describes growth at any rate. Thus:
. Repeat the exercises of this section for other relations between functions that might be known to you.5 and g"0 i.1*i:21
The resulting plot suggests a hyperbola satisfying the equation 1= (sqr x)-(sqr y). and f3 for the functions. and that negative values of r subsume the case of decay.7.
54308 1 1.71828 7.1 @sinh * cosh)) ((2"0 * cosh * sinh)-(2"0 * sinh * cosh)) (2"0 * ((cosh * sinh) .*:@sinh) (*:@cos + *:@sin) is is 1 1
See Section K of Chapter 2 for justification of the steps in the proof:
(*:@cosh . each is either odd or even.7622 1. The Exponential Family
We have now shown how the growth.
D.1 sin d.70 Calculus ^ EVEN x 10.(sinh * cosh))) 0"0
The circular case differs only in the values for the derivatives:
cos d. and hyperbolic functions can be expressed in terms of the single exponential function ^ :
.1 .54308 3.0179 _3.0179 (^EVEN x)+(^ODD x) 0.
C.*:@sinh) d.1752 3.1
is -@sin is cos
C1
Write and test a proof of the fact that the sum of the squares of the functions 1&o.367879 1 2.0497871 0.*:@sinh d.1) ((*: d.0677 ^ ODD x _10. is 1.62686 _1. Proofs
We will now use the definitions of the hyperbolic and circular functions to establish the two main conjectures of Section B:
(*:@cosh .0677 3. decay.38906 20.62686 10. and 2&o. they yield 1 for any argument:
(sinh = sinh ODD) (cosh = cosh EVEN) (sin = sin ODD) (sinh = ^ ODD) (cosh = ^ EVEN) (cos = cos EVEN)
B1 Repeat Exercises A2-A5 with modifications appropriate to the circular and hyperbolic functions.135335 0. that is.7622 10.1 @cosh*sinh)-(*: d.1752 0 1. The following functions are all tautologies.0855
Since the coefficients that define the hyperbolic and circular functions each have zeros in alternate positions.1 (*:@cosh d.
the measurement of threesided figures. We will illustrate some of them below by tautologies. that is two pi.a and the complementary angle 0. the cosine of a is the horizontal (or x) coordinate of the point whose arc is a. so may a three-sided figure be characterized as trigonal or triangular. Moreover. given by pi=: o. because they can be defined in terms of the coordinates of a point on a unit circle (with radius 1 and centre at the origin) as functions of the length of arc to the point. the coordinates of the end points of arcs of lengths 1p1 and 0. 1.
sin a 1 a
cos a
Figure F1 Taken together with these remarks. The sine and cosine are also called circular functions.
. Trigonometric Functions
Just as a five-sided (or five-angled) figure may be characterized either as pentagonal or pentangular. measured counter-clockwise from the reference point with coordinates 1 0. and the ratio of the circumference of a circle to its diameter is called pi.Chapter 6 Properties of Functions 73 5 ^ 4+3 78125 (5^4)*(5^3) 78125
E1
Comment on the question of whether the equivalence of */n#m and m^n holds for the case n=:0. This section concerns the equivalence of the functions sin and cos (that have been defined only by differential equations) and the corresponding trigonometric functions sine and cosine. or by the constant 1p1.
F. the supplementary angle 1p1&. The circular functions therefore have the period 2p1. As illustrated in Figure F1. as illustrated for the first of them:
S=: 1&o. and the sine of a is the vertical coordinate. The first of these words suggests the etymology of trigonometry.a are found by moving clockwise from these points.5p1&. the properties of the circle make evident a number of useful properties of the sine and cosine.5p1 are _1 0 and 0 1. The length of arc is also called the angle. each of which can be tested by enclosing it in parentheses and applying it to an argument.
Similarly for the cosine:
(C@+:@-: = <:@+:@*:@C@-:) (C = <:@+:@*:@C@-:) (C@-: =&| %:@-:@>:@C)
Tautologies may be re-expressed in terms of arguments i and x as illustrated below for S@+ and C@+:
i=:0.C) (S@-: =&| (+:@*: I)@(1"0 .1 (S i+x) = ((S i)*(C x)) + ((C i)*(S x)) (C i+x) = ((C i)*(C x)) .= *&C + *&S
Since a S@+ a is equivalent to (the monadic) S@+:. Using the results of Section 2A.*:@S
The theorem of Pythagoras can be used to obtain two further forms of the identity for C@+: :
C@+: = -.C))
The last two tautologies above compare magnitudes (=&|) because the square root yields only the positive of the two possible roots.+:@*:@S@-:) (C = 1"0 .@+:@*:@S C@+: = <:@+:@*:@C
An identity for the sine of the half angle may be obtained as follows:
(C@+:@-: = 1"0 .C)) (S@-: =&| %:@-:@(1"0 .= (S@[ * C@]) .*&S C@.Chapter 6 Properties of Functions 75
The following sum formulas for the sine and cosine are well-known in trigonometry:
S@+ = (S@[ * C@]) + (C@[ * S@]) S@.+:@*:@S@-:) (+:@*:@S@-: = 1"0 . we may obtain the following identities for the double angle:
S@+: = +:@(S * C) C@+: = *:@C .((S i)*(S x))
Derivatives. we may express the secant slope of the sine function at the points x and i+x as follows:
.(C@[ * S@]) C@+ = *&C .
the magnitude of the area of the sector with arc length (angle in radians) i lies between the areas of the triangles OSC and OST. Multiplying by 2 and dividing by S i gives the relative sizes C i and i%S i and %C i . In the unit circle of Figure F2. and its length is called the tangent of i. Its value (S%C) i follows from the ratios in the similar triangles. Hence. it will be necessary to obtain limiting values of the ratios (S i)%i and (1-C i)%i. the ratio i%S i lies between C i and %C i. Finally. both of which are 1 if i=: 0.76 Calculus ((S i+x)-(S x))%i
Using the sum formula for the sine we obtain the following equivalent expressions:
(((S i)*(C x)) + ((C i)*(S x)) .(S x)*((1-C i)%i) ((C x)*((S i)%i)) . also 1. the desired limiting ratio (S i)%i is the reciprocal. the lengths of the relevant sides are as shown below: OC
C i
CS
S i
OS
1
ST
(S%C) i
S
i O C T
Figure F2
ST is the tangent to the circle. The values of the cited areas are therefore -:@(S*C) i and -:@i and -:@(S%C) i .(S x))%i (((S i)*(C x)) + (S x)*(<:C i))%i (((S i)*(C x))%i) . Moreover.(S x)*((1-C i)%i)
To obtain the derivative of S from this secant slope. for
. The limiting value of (1-C i)%i is given by the identity +:@*:@S@-: = :
(1-C i)% i (+: *: S i%2) % i (*: S i%2) % (i%2) 1"0-C.
7. For example. and 6.b for various scalar values of a and b and comment on the results. by +/ . and a vector r that is normal to each of two vectors p and q is said to be normal to the plane defined by them. and the vector perpendicular or normal to this plane is an important derivative called the curl of the vector function. Dot and Cross Products
As illustrated in Section 3E."0 and perm=: +/ . Substituting these limiting values in the expression for the secant slope ((C x)*((S i)%i)) .and cos=:2&o. and illustrates their properties. the values of S and sin and of C and cos agree at the argument 0. somewhat more generally. The angle between two vectors is defined analogously. and we see that the relations between S and C and their derivatives are the same as those between sin and cos and their derivatives. * ]
G. namely:
((C x)*(1)) . and the limit of S i%2 is 0. that is. Thus:
a=: 1 2 3 +/a*b 16 [ b=: 4 3 2
. The angle between two rays from the origin is defined as the length of arc between their intersections with a circle of unit radius centred at the origin. where a and b are scalars. [ f is a tautology recognizable as
(sin(a+b))=((sin a)*(cos b))+((cos a)*(sin b))]
F2
Define other tautologies known from trigonometry in the form used in F1.(S x)*(0) C x
Similar analysis shows that the derivative of C is -@S. the vectors in this matrix lie in a plane. The dot product may be defined by +/@* or. Similar notions apply in three dimensions. including the dot or scalar product and the cross or vector product. hence the limit of (1-C i)%i is their product. in Sections 6. 6.. 0. F1 Define f=:sin@(+/) = perm@:sc and sc=:1 2&o. for example.Chapter 6 Properties of Functions 77 ((S i%2)%(i%2)) * (S i%2)
The limit of the first factor has been shown to be 1. Again we will leave interpretations to the reader. We will now present a number of results needed in its definition. Proofs of these properties may be found in high-school level texts as. then evaluate f a. in the sense that it is normal to every vector of the form (a*p)+(b*q). Moreover. * . The remainder of this section defines the dot and cross products. the vector derivative of the function */\ yields a matrix result. or 45 degrees. the angle between the vectors 3 3 and 0 2 is 1r4p1 (that is. and will defer comment on them to exercises.8. [ Consider the use of det=: -/ . * and sin=:1&o. If the angle between two vectors is 1r2p1 radians (90 degrees). one-fourth of pi) radians.(S x)*((1-C i)%i) we obtain the expression for the derivative of the sine. they are said to be perpendicular or normal.12 of Coleman et al [11].
* m _20
G1
Experiment with the dot and cross products.2@>@indices ]skm=: *: . when applied to skew arrays of odd order (having an odd number of items) it is self-inverse. it applies to arrays other than vectors. beginning with vectors in 2-space (that is with two elements) for which the results are obvious.2474 (L a cross b) % */ L a.: |: i. Normals
We now use the function e introduced in Section 3E to define a function norm that is a generalization of the cross product. Moreover.:) a sin b 0. cos) b 1
The cross product is not commutative The product of the sine of the angle between the vectors with the product of their lengths The sine of the angle The sine function
The following expressions suggest interpretations of the dot and cross products that will be pursued in exercises:
c=: 4 1 2 c dot a cross b _20 m=: c. and produces a result that is normal to its argument. Sketch the rays defined by the vectors.:b 0.:b m 4 1 2 1 2 3 4 3 2 -/ . 3 3 0 _4 _16 4 0 _12 16 12 0 ]v=: -: +/ +/ skm * e #skm _12 16 _4
A skew matrix
.607831 sin=: L@cross % */@(L@. Thus:
indices=:{@(] # <@i.) Result is called an "e-system" by McConnell [4] e=:C.Chapter 6 Properties of Functions 79 cross=: -/@(1 _1&rot@[ * _1 1&rot@]) a cross b _5 10 _5 (a. Continue with other vectors in 2-space and in 3-space.:b) dot a cross b 0 0 b cross a 5 _10 5 L a cross b 12.607831 a +/@:*:@(sin .a.
H. showing their intersection with the unit circle (or sphere).!.
and (because h is small) as an approximation to the tangent slope of the exponential.31511j1. for the case of the discontinuous integer part function <. the function h S ^ has the behaviour expected of a secant slope:
^ y=: 2j3 _7. various interpretations of a particular function definition are possible (as in vol=: */ and cost=: */).)' f=: ^ h=: 1e_8 sf=: h S f sf x=: 1 2. We will illustrate this point by three examples. @ (] -&y. If the spacing h is complex.71828
define and use the function sf. -&x. either by suggesting too little or too much. depending upon the background of the reader. sf can be helpfully interpreted as the secant slope of the exponential with spacing h. However. and any one of them may be either helpful or confusing. Thus:
h S <. Example 1. The sentences:
S=: 2 : '%&x. this interpretation would be misleading because its "tangent slope" at the point 1 is infinite. x 1e8
Example 2. Introduction
As remarked in Section 3A.83
Chapter
7
Interpretations and Applications
A. Moreover.04274 h=: 1e_6j1e_8 h S ^ y
. A helpful interpretation may also be misleading.
sin)@*"0
may be interpreted as the "Position of a car .
B.019998 r S + y 1 _1 i S + y
The problem arises because the conjugate is not an analytic function.. Example 3.31511j1.04274
Again the interpretation of the function h S f as an approximation to the tangent slope is valid. Because the phrase involved a derivative.
. the phrase f D. of course. the expression in quotes could be considered as an application of the circular functions. and could be posed as a word problem requiring as its solution a definition of the function f. the point a.04274 (r=: 1e_6j0) S ^ y _7. For example.1&path=:(cos.31511j1. there are an infinity of different paths through complex numbers from complex a to complex b. the (continuous) conjugate function + shows unusual behaviour:
h S + y 0. This observation leads to the more difficult. and although there is a clearly defined "path" through real numbers between a pair of real numbers a and b. Section 2D interprets the integral of a function f as a function that gives the area under the graph of f from a point a (that is.31511j1... Conversely.04274 (i=: 0j1e_8) S ^ y _7.f a on the graph of f) to a second point b. Similarly. but how should we visualize the area under a function that gives a complex result? It is. possible to interpret the integral as a complex result whose real and imaginary parts are the areas under the real and imaginary parts of f. but highly useful. or as word problems in math. However. then the function:
f=:0.84 Calculus _7. the corresponding word problem would be considered as an application of the calculus. This interpretation is helpful for real-valued functions. notion of integration along a prescribed path (called a line or contour integral). and sin=:1&o. A clear and simple discussion of this matter may be found in Churchill [12].1 radians per second". a notion not hinted at by the interpretation of integration as the area under a curve.1 may be interpreted as the velocity of the car whose position is prescribed by f. respectively. if cos=:2&o. However. the beginning and end points may themselves be complex. moving on a circular path at an angular velocity of 0. Applications and Word Problems
What we have treated as interpretations of functions may also be viewed as applications of math.9998j_0.
1_23.1_22.4 _4. and f is a specific polynomial whose (tightly) formatted results:
(fmt=: 5.1_12.7_11.6_17.1 1.2 _6.5 0.1 1.1_12. or from other calculus texts.2 _26.1 over the same interval illustrates the obvious fact that the derivative is zero at an extremum (minimum or maximum):
fmt f d. The method used is sometimes called the bisection method.5 1.4 _1.1 0. A graph of the derivative f d.7 _8.2_25.3 1.3 27.Chapter 7 Interpretations And Applications 85
Just as a reader's background will determine whether a given interpretation is helpful or harmful in grasping new concepts in the calculus.5_17.9_12. Moreover.6 1. then p.4 _9.8 1.0 0. Thus:
m=: +/ % # bis=: 1 : '2&{.0_24.8 _2.1_10.8 1.1 x 1.0 _11.0 2.9_21.5_11.6 _4.9 0. so will it determine the utility of word problems.4 _7.9 1.2 and a minimum near 4.4_10.7 0.) ])' Interval that bounds a root of f f y=: 1 4 1.9_24.6 10 0.4 7.3 _2.9_14. the interval is repeatedly halved in length by using the midpoint (that is.1 2.2 _5.4_19.8 _8.6 _4.0_23.7_26.6 0.1 _0.6_25.9.9_12.7 _6.8 1.9 _5.4_11.9) near 1.6 _0.8_11.8 2.3_16.5 1
One step of the bisection method
.0_11.
C. Since we are concerned only with real roots we will define a simple adverb for determining the value of a root in a specified interval. ] #~ m ~:&(*@x. a graph of the function over the interval from 0 to 4 shows their location more precisely.9 1.6_11.2 _8.9 1.1_12.0 31. and encourage the reader to choose further applications from any field of interest.0
suggest that it has a (local) maximum (of 1. is a polynomial in terms of coefficients.4 1.0 _9.4 _9.1 _1.9_20.3_26.8 10.1_26.4 1.5 0.4_14.8 1.6_10. Extrema and Inflection Points
If f=: (c=: 0 1 2.6_25.0 _10. where the function values at the ends of the interval must differ in sign.0 _5.1*>:i.@(m .1 _7.5 13.1_15.9 1.9 _3.8 23.4 16.5 19.7_18.9 _8.3 _5.1 _7.5 _1.1_22.0
We may therefore determine the location of an extremum by determining the roots (arguments where the function value is zero) of the derivative function. the mean) together with that endpoint for which the function value differs in sign.3 0.4 1.6 _3.4_11.0 _6.1 5..2 _2.0 _1.0_25.5 _2 0. We will limit our treatment of interpretations and applications to a few examples.2 _9.7 1.1&":) f x=: 0.1 2.2 0.0 _21.75 _20 f bis y 2.25)&p.7_12.0 3.0 _0.6 _2.6 1.
g x=: 3 intersecting the axis at a point nx. [
f droot + _0. As a consequence.60
Draw a tangent at the point x.1 x. The derivative of the function can be used in a method that normally converges much faster.
D. The function g=: (]-1:)*(]-2:) has roots at 1 and 2.25)&p. but a maximum. d. The length x-nx is the run that produces the rise g x with the slope g d. Newton's Method
Although the bisection method is certain to converge to a root when applied to an interval for which the function values at the endpoints differ in sign.x.@ in the definition of bis? [ C3 Remove the phrase and try f bis 1 3 ]
For various coefficients c. make tables or graphs of the derivative c&p.1 x) 2. D to determine intervals bounding roots.
Used in Section C
. ] C2 What is the purpose of 2&{.33333 g nx 0. this convergence is normally very slow.0001
It is not a minimum.0 and note that nx is a much better approximation to the nearby root at 2 than is x. nx=:x-(g x) % (g d.1 x) is a better approximation to the root at 2. Thus:
x=:3 g=: (]-1:)*(]-2:) g x 2 ]nx=:x-(g x) % (g d. using an adverb N as follows:
N=:(1 : '] .Chapter 7 Interpretations And Applications 87
C1
Test the assertion that droot is a local minimum of f .g y=: 1r20*i.0001 0 0. and use them with bis to determine extrema of the polynomial c&p.444444
A root can be determined by repeated application of this process. % x.1') (^:_) f=: (c=: 0 1 2.5 _2 0. although convergence is assured only if the initial guess is "sufficiently near" the root. as shown by its graph:
plot y.
This expression produces the vector derivative with respect to each of the approximate roots.2 7 8.15625
In this form it is clear that the vector of residuals produced by x. Thus:
d=: cfr 1 2 2j3 4 2j_4 ]roots=: (norm d) K begin d 4 2j3 2j_4 2 1 /:~roots 1 2j3 2j_4 2 4
Sorted roots
The definition of the adverb k (for a single step of Kerner) can be revised to give an alternative equivalent adverb by replacing the division (%) by matrix division (%. Thus:
ak=: 1 : ']-x.&p.28125 4.
F."1))@(-/~))' c ak b 2. and removing the phrase (<0 1)&|:@ that extracts the diagonal of the matrix produced by the subsequent phrase. ((1&(*/\. . which must all be reduced to zero) is divided by the matrix produced by the expression to the right of %. E1 E2 Find all roots of the functions used in Section C. %. Define some polynomials that have complex roots. For example:
MD=: ("2) (D.@<:@#) 1 3 3 1 0 0j1 0j1.5 1 6 3 1 4 2 7 8 5 1 6 det m _2
The determinant is a function of rank 2 that produces a rank 0 result. its derivative is therefore a rank 2 function that produces a rank 2 result. * yields the determinant of a square matrix argument. * ]m=: >3 1 4.09375 2. For example:
det=: -/ . 1). and use Kerner's method to find all roots. These matters are left for exploration by the reader. like the analogous case of the direct calculation of the derivative in the adverb NP it is a direct calculation of the derivative without explicit use of the vector derivative adverb VD=: ("1) (D.41421
For example.1)
.46875 3. Determinant and Permanent
The function -/ .).90 Calculus
(begin=: %:@-@i.&p. (the values of the function applied to the putative roots. the coefficients d=: cfr 1 2 2j3 4 2j_4 define a polynomial with two complex roots.
the complementary minor of any element of a matrix is the matrix obtained by deleting the row and column in which the element lies.Chapter 7 Interpretations And Applications 91 det MD m 34 28 _33 _2 _2 2 _20 _16 19
This result can be checked by examining the evaluation of the determinant as the alternating sum of the elements of any one column. each weighted by the determinant of its respective complementary minor.)"2 ^:2 box minors m box minors alph box^:2 minors^:2 alph
[The function minors produces the complementary minors of its argument. *) 350 per MD m 50 52 37 10 38 8 36 32 23 m
F1
Read the following sentences and try to state the meanings of the functions defined and the exact results they produce. Then enter the expressions (and any related expressions that you might find helpful) and again try to state their meanings and results. Corresponding results can be obtained for the permanent. agreeing with the leading element of the derivative. the matrix occupying the remaining rows and columns. 4 4 box=: <"2 minors=: 1&(|:\.1 sqm) % (det minors sqm) ((+/ .:1 6.1 sqm (det D.*@minors)sqm
.1)%+/ . defined by the function +/ . For example. the complementary minor of the leading element of m is the matrix m00=: 7 8. For example:
(per=: +/ .*.*D. whose determinant is 34.
alph=: 4 4$ 'abcdefghijklmnop' m=: i. the derivative with respect to any given element is its weighting factor.] F2 Enter and then comment upon the following sentences:
sqm=: *:m det minors sqm det D.
this agrees with the solution f used at the outset.@(%&180p_1).*&4r360)"0 CH 0 1 90 180 360 0 1 0 0.000609111 0.0111111 _0. and indeed to the osculating (kissing) plane that touches the helix at the point given by CH.0003042 0 _2. provides interpretations of the vector calculus that should prove understandable to anyone with an elementary knowledge of coordinate geometry.1249e_5 _0.0111111 _0. as developed by Eisenhart in his book of that title [13]. The sine function 1&o. the following are solutions:
u=: (s+t)%2"0 v=: (s-t)%0j2"0
The cosine function 2&o.0111111 _0. We will provide a glimpse of his development.000304571 0 _1.3163e_6 _0.0174427 0.@(%&180p_1).0631e_5 _0.
Since u is equivalent to the cosine function.
J. and experiment with similar differential equations.000304432 0 _1.0111111 1 0 1 0 _1 2 _2.00121748 0.0111111 x 0 0 0 _5.44921e_16 1 4 D=: ("0) (D. In particular. 1) x=:0 1 2 3 4 CH D x 0.0174533 0. any linear combination of the basic solutions s and t is also a solution.000304602 0.59424e_5 _0.0174506 0. Differential Geometry
The differential geometry of curves and surfaces.0174294 0.2&o.Chapter 7 Interpretations And Applications 95
Moreover.0174524 0.0174108 CH D D
0 0. beginning with a function which Eisenhart calls a circular helix. with a rise of 4 units per revolution:
CH=:(1&o. their derivatives produced by CH D D are the directions of the binormals.999848 0.
. I1 Enter the expressions of this section. The binormal is perpendicular to the tangent.000303875 0
The derivatives produced by CH D in the expression above are the directions of the tangents to the helix. The following defines a circular helix in terms of an argument in degrees.000913435 0.0111111 _0.
and exponential. Replace the constant multiple function for the last component by other functions.@[ * 1&|. 2. To compute the directions of the principal normal we must determine a vector perpendicular to two other vectors. all of which can be modelled by a paper tube. the use of a sheet of transparent plastic will make visible successive laps of the helix.(_1&|. Consequently the definition of a function dfl to give degrees from length is given by:
dfl=: %&((%: +/ *: 4 360) % 360)
and the function CH@dfl defines the helix in terms of its own length. square. From the foregoing discussion of the paper tube model it is clear that the length of the helix corresponding to 360 degrees is the length of the hypotenuse of the triangle with sides 360 and 4. Hold a third needle in the direction of the principal normal.@]) . to produce a helix on an elliptical cylinder. and compare with the computed results. such as the square root. In the case of the helix defined by CH. A drawing to scale can be made by marking the point of overlap on the paper. This can be
. it is possible to choose an argument that is intrinsic to the curve. and again compare with the computed results. Puncture a thin sheet of flat cardboard and hang it on the binormal needle to approximate the osculating plane. 3. or the following simpler vector product function:
vp=: (1&|. An accurate rendering of a helix can be made by drawing a sloping straight line on a sheet of paper and rolling it on the tube. 4. Puncture the tube to hold the needle in the direction of the binormal. and it is clear that the choice of the argument to describe a curve is rather arbitrary. As Eisenhart points out. For this we can use the skew array used in Section 6I. Multiply the functions for the first two elements by constants a and b respectively. * q 0 0 b +/ . 2.@[ * _1&|.96 Calculus
These matters may be made more concrete by drawing the helix on a mailing tube or other circular cylinder. It is possible to modify the definition of the function CH to produce more complex curves. For example: 1. and drawing the straight line with a rise of 4 units and a run of the length of the circumference. Use a nail or knitting needle to approximate the tangent at one of the points where its directions have been computed. Finally. it is easy to determine the relation between the path length and the degree argument.@]) a=: 1 2 3 [ b=: 7 5 2 ]q=: a vp b _11 19 _9 a +/ . unrolling it. Then proceed as follows: 1. namely the length along its path. * q
Although we used degree arguments for the function CH we could have used radians. which lies in the osculating plane perpendicular to the tangent.
since the difference (f I x+h)-(f I x)is approximately the area of the rectangle with base h and altitude f x. Moreover. then the area is approximately the sum of the areas of the rectangles with the common base h and the altitudes f h*i. Approximate Integrals
Section M of Chapter 2 developed a method for obtaining the integral or anti-derivative of a polynomial.Chapter 7 Interpretations And Applications 97
modelled by removing the cardboard core from the cylinder and flattening it somewhat to form an approximate ellipse. leading to methods known by names such as Simpson's Rule.512 1 1.n (4: %~ ^&4) y
.216 0. if the area under the curve is broken into n rectangles each of width x%n.096 5. and Section N outlined a method for approximating the integral of any function by summing the function values over a grid of points to approximate the area under the graph of the function.
f
Figure C1
x
x+h
Figure C1 can also be used to suggest a way of approximating the function AREA=: f I.
K.n 0 0.728 2. the approximation approaches equality for small h. Better approximations to the integral can be obtained by weighting the function values.064 0.n. We will here develop methods for producing these weights.008 0. and use them in the definition of an adverb (to be called I) such that f I x yields the area under the graph of f from 0 to x. The fact that the derivative of f I equals f can be seen in Figure C1.744 4. the secant slope of the function f I is approximately f.832 +/h*cube h*i. For example:
h=: y % n=: 10 [ y=: 2 cube=: ^&3 cube h*i.
75 4. using g groups of 1+2*k points each:
n=: (g=: 4) * 2 * (k=: 1) ]h=: n %~ x=: 5 0.1 +/h*w*^&2 grid +/h*w*^&4 grid 41.875 2.1 1 4 2 4 2 4 2 4 1 w=: 3%~ 1.(4 2 $~ <: 2*g).0.24 4
The approximation can be improved by taking a larger number of points. each group being fitted by a polynomial of degree 2*k.375 5 1.25 1.102
. For example. and since each altitude other than the first and last enter into two trapezoids.6667 625. For example:
w=: 3%~1 4 1 h=: (x=: 5)%(n=:2) ]grid=: h*i.5 5 f=: ^&2 w*f grid 0 8. resulting in weights of the form 3%~1 4 2 4 2 4 2 4 1. For example.5.125 3.98 Calculus 3. linear approximations to the function between grid points.6667 +/h*w* ^&4 grid 651.04
The trapezoids provide.042
Exact integral of ^&2 Exact result is 625
Better approximations are given by several groups of three points.5 .(4 2 $~ <: 2*g).625 1. n+1 0 2.0. but it can also be improved by using the areas of the trapezoids of altitudes f h*k and f h*k+1 (and including the point h*n).5 +/h*w*cube h*i. n+1 4.5 1 1 1 1 1 1 1 1 1 0. the integration produced is exact.33333 +/h*w* f grid 41.33333 8.5.(1 #~ n-1). the case k=: 1 provides fitting by a polynomial of degree 2 (a parabola) and a consequent weighting of 3%~1 4 1 for the three points. n+1 0 0. If the function to be fitted is itself a polynomial of degree two or less.625 ]grid=: h*i. Thus:
]w=: 0.(1 #~ n1). much better approximations to the integral can be obtained by using groups of 1+2*k points.5 3. this is equivalent to multiplying the altitudes by the weights w=: 0. Since the area of each trapezoid is its base times the average of its altitudes. in effect.5 0.
_1 x 0.780924 4 * (2 EW 2) ai cir 1 3.8 (x^5)%5 0.6 204.66667 9 21.8 (cir=:0&o.6 204.) 1 3.1 3. Areas and Volumes
The integral of a function may be interpreted as the area under its graph.14132 o.9 104858
K1 Use the integral adverb I to determine the area under the square root function up to various points.1 102.7 104854 (x^10) % 10 0.866025
Weights give exact results for integral of fourth power
is the altitude of a unit circle Approximation to area under cir (area of quadrant) Approximation to pi
For use in exercises and in the treatment of interpretations in Section L.2 6.Chapter 7 Interpretations And Applications 101 0.5 1 1 0. we will define the adverb I in terms of the weights 4 EW 4.9 104858 ^&9 d.1237 4*(20 EW 3) ai (0&o. they enclose an area.333333 2.4 5904. K2 Since the graphs of the square and the square root intersect at 0 and 1.0999966 102.
L.4 48.1 102. is %:@(1"0-*:) and cir 0.14159 0&o.3333 (1 EW 2) ai (^&4) x 0. Determine its size. four groups of a polynomial approximation of order eight:
I=: (4 EW 4) ai ^&9 I x=: 1 2 3 4 0. we will use the adverb I defined in the preceding section. For example:
(0&o.)0 0.) I 1
Approximate area of quadrant of circle
.4 5904. [ (%:I-*:I) 1 or (%:-*:)I 1 ] K3 Experiment with the expression (f . To approximate integrals.f I D) x for various functions f and arguments x.397 5904.4 48. that is.866025 0 (2 EW 2) ai cir 1 0.2 6.
333317 2.37717 28.102 Calculus 0.14159 12.3583 1.13963
Approximation to pi
*: I x=: 1 2 3 4 0.00105389 cade I x 1.66654 8. The volume is therefore called a volume of revolution.@*:@] " 0 ca x 3.5423 1.333333 2. as the volume of a pyramid.784908 4 * (0&o.5664 28.425168 0. the equivalent function ^&3 % 3"0 is a well-known expression for the volume of a pyramid.99956 21.273 67.0575403 0.00778723 0.2743 50. For example:
cade=: ca@^@cade x 0.66667 9 21.0174
h*x
x
Figure L1 By drawing a figure analogous to Figure L1. Similarly for a function that defines the area of a circle in terms of its radius:
ca=: o.04715 8.2655 ca I x 1.) I 1 3.3323 (^&3 % 3"0) x 0. Alternatively.56746 1. it can be interpreted as the volume of a three-dimensional solid as illustrated in Figure L1. Functions other than ] (the 45-degree line) can be used to generate volumes of revolution. it may be seen that the cone whose volume is determined by ca I can be generated by revolving the 45-degree line through the origin about the axis. that is. In particular.3333
The foregoing integral of the square function can be interpreted as the area under its graph.57123
Area of circle whose radius is the decaying exponential Volume of revolution of the decaying exponential
.
a 59. if p t gives the position at time t of a body suspended on a spring or rubber band. whose value at 0 is infinite. More specifically. p=: (a*sin)+(b*cos).99978 -/f I b.2. where a and b are constant functions. and their rates of change (that is.2) must be zero. or mere observation of everyday phenomena. the area approximated is the area over the same interval from 0 to y. This relation can be simplified to 0: = p . which is itself a simple linear function of the position p. and whose value at b-a is %b.9967
However. For example.9967
f I a=: 2 3. and its third derivative (jerk). then the acceleration of the body (p d. the position of a body as a function of time is related to its first derivative (velocity).25
The integral of the reciprocal from 2 to 4 The natural log of 2
L1
Use integration to determine the areas and volumes of various geometrical figures. derivatives). If position is measured from the rest position (where the body rests after motion stops) this linear function is simply multiplication by a constant function c determined by the elasticity of the spring. where the constant function m is the mass of the body. more generally. this approach will not work for a function such as %. In other words. its second derivative (acceleration). In such a case we may use the related function %@(+&a).693147
g b-a 0.2) is proportional to the force exerted by the spring. Physical Experiments
Simple experiments.
. (c*p)-(m*p d. whose value at 0 is %a. can provide a host of problems for which simple application of the calculus provides solutions and significant insights. and c*p must be equal and opposite to m*p d.
M.2. The reason is that phenomena are commonly governed by simple relations between the functions that describe them.9965 (f I b) -(f I a) 59. including cones and other volumes of revolution.c2 * p d.693163 ^. The area under f from a to b can be determined as a simple difference.5 g I b-a 0. 2 0. where c2 is the constant function defined by c2=: m%c. The function p is therefore (as seen in Section I) the sine function. For example:
f=: ^&3 f I b=: 4 63.Chapter 7 Interpretations And Applications 103
Because the expression f I y applies the function f to points ranging from 0 to y. or. Thus:
g=: %@(+&a) g 0 0.
and enjoys the same form of electrical oscillation.e.
Coordinate geometry also provides problems amenable to the calculus. unlike the sine and cosine functions which continue with undiminished amplitude. a solution of such a linear differential equation is given by ^@r. then ^r may also be written as (^x)*(^j. c=: (1&o. The use of a heavier fluid would increase the damping. e=: (a*1&o. 1)+(f*p D.. In other words. resulting in non-oscillating solutions in terms of the hyperbolic functions sinh and cosh. For example. and raise the following question: Could the body be completely damped. The performance of actual experiments might also lead one to watch for other phenomena governed by differential equations of the same form. As seen in Section I. the differential equation:
0: = (d*p)+(e*p D. The amount of electrical charge remaining in a capacitor draining through a resistor (used in circuits for introducing a time delay). of course.
. The amount of water remaining in a can at a time t following the puncture of its bottom by a nail.1 is the current (whose value determines the voltage drop across the resistor).2&o. Similarly. most of us could perform the corresponding "thought experiment" and so avoid the effort of an actual experiment. y. y).) gives the coordinates of an ellipse. then q d. if the function q describes the quantity of electrical charge in a capacitor whose terminals are connected through a resistor and a coil.). and the gradient c D. because it commonly leads to the consideration of interesting related problems. 2)
provides a more accurate relation. For example. For example. a positive value of the factor f (the coefficient of p d. The difference is due to resistance (from friction with the air and internal friction in the rubber band) which is approximately proportional to the velocity. Because oscillations similar to those described above are such a familiar sight.)"0 is a rank 1 0 function that gives the coordinates of a circle. However. Other systems concerning motion suggest themselves for actual or thought experiments: * * * The voltage generated by a coil rotating in a magnetic field.2 is its rate of change (which determines the voltage drop across the coil). showing that the position function is a product of a decay function (^x) and a periodic function (^j. Such a positive factor cannot. the charge q satisfies the same form of differential equation that describes mechanical vibrations. However. In other words. 1 gives the slope of its tangent. coming to rest with no oscillation whatever? The answer is that no value of the decay factor ^x could completely mask the oscillatory factor ^j.104 Calculus
This result is only an approximation.(b*2&o.. direct observations of the effect of greater damping can result from immersing the suspended body in a pail of water. y. where r is a (usually complex) root of the polynomial (d. If r=: x+j. since a body oscillating in this manner will finally come to rest. and q d.f)&p.2) will provide real roots r. be realized in the experiment described. y) like the solution to the simpler case in which the (resistance) constant e was zero. the performance of actual experiments is salutary.
they would prove unsatisfactory in a text devoted to physics: they ignore the matter of relating the coefficients in the differential equations to the actual physical measurements (Does mass mean the same as weight? In what system of units are they expressed?). On the other hand.Chapter 7 Interpretations And Applications 105
If we are indeed surrounded by phenomena so clearly and simply described by the calculus. Although the brief treatments of mechanical and electrical vibrations given here may provide significant insights into their solutions. a superficial treatment that does not lead the student far enough to actually produce significant new results is likely to leave her uninterested. would make difficult its use by a student in some other discipline looking only for guidance in calculus. Textbook pictures of suspension bridges with encouraging but unhelpful remarks that calculus can be used to analyze the form assumed by the cables. 2.
. and they ignore the practicality of the computations required. 4. although essential in a physics text. 3. they ignore questions concerning the goodness of the approximation to the actual physical system. for which we will now essay some answers: 1. The use of scalar notation makes it difficult to reach the interesting results of the vector calculus in an introductory course. are more likely to discourage than stimulate a student. The treatment of such matters. Emphasis on rigorous analysis of limits in an introductory course tends to obscure the many interesting aspects of the calculus which can be enjoyed and applied without it. why is it that so many students forced into calculus fail to see any point to the study? This is an important question.
.
Kline states (page 167): But there is a deeper reason. like Newton's colleagues at the time of his development of what came to be the calculus. which can lead to false ideologies and even to destruction... of reason is far more beneficial than blind trust. . have serious qualms about the validity of assuming a quantity r to be non-zero and then..". and to learn the techniques for assuring that they are being properly observed. or are there important lessons to be learned from the centuries-long effort to put the calculus on a "firm" foundation? If so. a more mature student familiar with the use of rigorous axiomatic and deductive methods would. In other words. Up to about 1500. the fact that the derivative so determined leads to consistent and powerful results would only tend to confirm a faith in the validity of the arguments.107
Chapter
8
Analysis
A. and then setting r to zero in the resulting expression) might appear not only persuasive but conclusive. On the other hand. mathematicians were [now] contributing concepts rather than abstracting ideas from the real world. at a convenient point in the argument.
. As Morris Kline says in the preface to his Mathematics: The Loss of Certainty [14]: But intellectually oriented people must be fully aware of the powers of the tools at their disposal. Moreover. Introduction
To a math student conversant only with high-school algebra and trigonometry. the concepts of mathematics were immediate realizations of or abstractions from experience. Should a student interested primarily in the practical results of the calculus dismiss such qualms as pedantic "logic-chopping". as well as the capabilities. Concerning "This history of the illogical development [of the calculus] .. asserting it to be zero. and how may they be approached? The important lesson is to appreciate the limitations of the methods employed. what are they. A subtle change in the nature of mathematics had been unconsciously made by the masters. Recognition of the limitations. the arguments used in Section 1E to determine the exact derivative of the cube (dividing the rise in the function value by the run r.
however small. and perhaps to supplement it with Lakatos' equally readable account of the interplay between proof and refutation in mathematics.01 _0. and equally ingenious refutations. Figure B1 provides a graphic picture of the role of the frame function: d=: a fr e specifies the half-width of a frame such that the horizontal boundary lines at e and -e are not crossed by the graph of g-h within the frame.0001 a+i 4 2 3.1 2.0001 0.
B.01 0. Limits
The function h=: (*: . and applied to series in Section D. Thus:
g=: 6"0 h=: (*:-9"0) % (]-3"0) a=: 3 h a 0 ]i=: . as x approaches a.001 0.1 0. one constructs functions to contradict the conclusions of our predecessors and one will never be able to apply them for any other purpose.1 0.9999 h a+i 7 5 6.1 5. for any positive value e.0001 5. or limit. a student should be aware of the fact that weird and difficult functions sometimes brought into presentations of the calculus are included primarily because of their historical role as refutations.9 3.01 0.0001
We might therefore say that h x approaches a limiting value.3"0) applied to the argument a=: 3 yields the meaningless result of zero divided by zero. provided that y differs from a by no more than d. The central concept required to analyze derivatives is the limit.99 6. the expression e>:|(g h) y is true for any y such that (|y-a) <: a fr e.001 2. it is introduced in Section B.001 0.01 5. We make a more precise definition of limit as follows: The function h has the limit g at a if there is a frame function fr such that for any positive value of e.(+.5) 1 _1 0.99 3. Students are urged to read it in full.0001 _0.01 2.9999 |(g-h) a+i 1 1 0.999 6. In other words. The words of Poincare (quoted by Kline on page 194) are worth remembering: When earlier. In this case the limit is the constant function 6"0. even though it differs from h a.9 6.108 Calculus
Chapter VII of Kline provides a brief and readable overview of ingenious attempts to put the calculus on a firm basis. there is a value d=: a fr e such that h y differs from g y by no more than e.001 5.1 0. On the other hand.999 3.0001 2.1 _0. a list of arguments that differ from a by successively smaller amounts appear to be approaching the limiting value g=:6"0. on the contrary. Today.001 _0.001 0. new functions were introduced.-)"0 (10^-i.01 0.9"0) % (] . the purpose was to apply them. In particular.
.
x
We will now obtain a simple upper bound for the magnitude of the difference (that is. For example. generates coefficients for a polynomial that approximates its own derivative.(x-a)^4 (0 0 0 4 0 p. x (1 _4 6 * a^3 2 1) p.x)-(4 w -a)p. beginning with the final expression above. For example: e=: 0. Continuity
Informally we say that a function f is continuous in an interval if its graph over the interval can be drawn without lifting the pen. Since the coefficients produced by ec decrease rapidly in magnitude (the 20th element is %!19. |(g-a&h) x). the largest term is a^1
The
final
expression
provides
the
a=: e % (6*+/|x^0 1 2). and the growth function (exponential) is defined as the limiting value for an infinite number of terms. but not in an interval that contains integers. to avoid length problems)
x=:5 |(1 _4 6 * a^3 2 1) p. x a%~((a*0 0 0 4 0)p. x)-(4 w -a)p.
D. we define a function f to be continuous in an interval if it possesses a limit at every point in the interval. approximately 8e_18). x | +/1 _4 6*(a^3 2 1)*x^i. the integer part function <.1 to 0. x)-a%~(0 0 0 0 1 p.Chapter 8 Analysis 111 (4*x^3)-a %~ (x^4) .x)-(0 0 0 0 1 p. Formally. would converge to a
.000806451
basis the
for a frame function: if magnitude of the difference
C. as in x=: 5.3) +/1 4 6*(a^3 2 1)*|x^0 1 2 +/6*(a^3 2 1)*|x^0 1 2 +/6*a*|x^0 1 2 6*a*+/|x^0 1 2 a* (6*+/|x^0 1 2)
Magnitude of (g-a&h) x Polynomial as sum of terms Sum of mags>:mag of sum
a is non-negative
For a<1. and continuing with a sequence of expressions that are greater than or equal to it: (If the expressions are to be entered. then |(g-a&h) x will not exceed e.x a%~(1 _4 6 * a^ 4 3 2) p. is continuous in the interval from 0.001 a=: e % (6*+/|x^0 1 2) |(g-a&h) x 0. it seemed reasonable to assume that the polynomial (ec i. Convergence of Series
The exponential coefficients function ec=:%@!.n)&p. x should be set to a scalar value.3 +/(|1 _4 6)*(|a^3 2 1)*(|x^i.9.
the series %@>:@i. If at a given term t in a series the remaining terms are decreasing in such a manner that the magnitudes of the ratios between each pair of successive terms are all less than some value r less than 1.112 Calculus
limit for large n even when applied to large arguments.-r) */ r S n 1 0 0 0 0 0 0 0 0 0 _59049 -r^10 _59049 +/dsums _59048 (1-r) * r T n _59048
If the magnitude of r is less than 1. It might seem that the sum of a series whose successive terms approach zero would necessarily approach a limiting value. which has a fixed ratio r. We will now examine more carefully the conditions under which a sum of such a series approaches a limit.-r) */ r S n 1 3 9 27 81 243 729 2187 6561 19683 _3 _9 _27 _81 _243 _729 _2187 _6561 _19683 _59049 ]dsums=:+//.n. the sum of the entire series therefore approaches a limit. the value of r^n in the numerator of r T n approaches zero for large n.166667
. However. For example:
S=: [ ^ i. For example:
ec=:%@! j=: 4 ec j-0 1 0. n provides a counter example. since (by considering sums over successive groups of 2^i. and the numerator itself therefore approaches 1. then the magnitude of the sum of the terms after t is less than the magnitude of t%(1-r). k elements) it is easy to show that its sum can be made as large as desired. The expression ec j-0 1 gives a pair of successive coefficients of the polynomial approximation to the exponential. This can be illustrated by the series r^i. and has a sum equal to (1-r^n) % (1-r). consequently.@] T=: (1"0-^)%(1"0-[) r=: 3 n=: 10 r S n 1 3 9 27 81 243 729 2187 6561 19683 +/ r S n 29524 r T n 29524
A proof of the equivalence of T and the sum over S can be based on the patterns observed in the following:
(1. if this quantity can be shown to approach 0. and %/ec j-0 1 gives their ratio. the result of r T n approaches %(1-r) for large n.(1.0416667 0.
)`(2&o. At some point this ratio becomes less than 1.n)&p. We will illustrate this by first developing a series approximation to the arctangent.540302 0.)`(3&o.841471 0. applied to x is x times this.Chapter 8 Analysis 113 %/ec j-0 1 0.1r6p1. Derivative of the tangent 2.1r4p1..25 %j 0.866025 cos x 0.866025 0.) sin x 0. one-quarter pi):
]x=: 1. after removing the alternate zero coefficients.57735 1 1.55741 0.523599 0.785398 1. Express the derivative as the limit of a polynomial 5. Derivative of the inverse tangent 3. that is.57735 1 1. and the series for the exponential therefore converges.73205 (sin % cos) x 1. the inverse tangent _3o.0472 D=:("0) (D. Similar proofs of convergence can be made for the series for the circular and hyperbolic sines and cosines. namely. Another generally useful proof of convergence can be made for certain series by establishing upper and lower bounds for the series.0472 '`sin cos tan arctan'=: (1&o. x%j.1) tan D Definition of tan (sin % cos) D (sin%cos)*(sin D%sin)-(cos D%cos) θ7§2K tan*(cos%sin)-(-@sin%cos) §2K tan * %@tan +tan
. Apply the polynomial to the argument 1 to get a series whose sum approximates the arctangent of 1 (that is.785398 1.1r3p1 1 0. Express the derivative as a polynomial in the tangent 4.73205 INV=: ^:_1 tan INV tan x 1 0.707107 0.)`(_3&o. The development proceeds in the following steps: 1.5 0. This method applies if the elements alternate in sign and decrease in magnitude.707107 0.5 tan x 1.25
The ratio of the corresponding terms of the polynomial (ec i.523599 0.55741 0. Integrate the polynomial 6.
76046 0.804601 0.785398 0.Chapter 8 Analysis 115 0.
1r4p1 .754268 0.808079 0. Second column (sums of even number of terms) are increasing lower bounds of limit.744012 0.866667 0. +/gaor 1000 0.834921 0.72381 0. [Group pairs of successive elements to form a sum of positive or negative terms]
.820935 0.813091 0.785148
D1 D2
Test the derivations in this section by enclosing a sentence in parens and applying it to an argument.764601 0. as in (1: + *:@tan) x Prove that a decreasing alternating series can be bounded as illustrated.767564
of terms) are decreasing upper bounds of limit.
.
117
Appendix Topics in Elementary Math
A. the sums. x) * (c p.: c) p. and a sum of monomials is called a polynomial. It is important for many reasons. In particular. x
"Sum" of coefficients Sum polynomial Product of polynomials "Product" of coefficients Product polynomial
. it is easily expressed in terms of sums. and it is closed under a number of operations. x) 2 12 36 80 150 b +/@.@(*/) c) p. denoted by p. the items of its list left argument being called the coefficients of the polynomial:
pol=: +/@([ * ] ^ i. that is. products. and integral powers. x) + (c p.: c 2 5 4 1 (b +/@. x 2 12 36 80 150 (b p.@#@[) " 1 0
For example:
c=: 1 2 3 [ x=: 0 1 2 3 4 c pol x 1 6 17 34 57 1 3 3 1 pol x 1 8 27 64 125
The polynomial may therefore be viewed as a weighted sum of powers. products.@(*/) c 1 5 10 10 5 1 (b +//. For example:
x=: 0 1 2 3 4 [ b=: 1 2 1 [ c=: 1 3 3 1 Sum of polynomials (b p. We now define a polynomial function. it can be used to approximate almost any function of practical interest. Polynomials
An atomic constant multiplied by an integer power (as in a"0 * ^&n) is called a monomial. It is important enough to be treated as a primitive. and integrals of polynomials are themselves polynomials. . derivatives. the weights being specified by the coefficients. x) 1 32 243 1024 3125 b +//.
D x 2&|. f -y equals -f y. Define a function hat to give Heron's area of a triangle.*' and use it to display the results of Exercises G2. using VD=: ("1)(D.
[ AP5
rFd=: %&180@o.1) .1) x=: 1 2 3 4 5 |. [AREA=: -:@(0&{ * 1&{ * 1&o. f -y equals f y. as in show |. and explain the (near) zero result in the final element. and then enter them to validate your predictions:
D=: ("1) (D. and verify that the results agree with those given in the text.5 1 . Plots of even and odd functions show their graphic properties: the graph of an even function is "reflected" in the vertical axis. ])"1
]
AP8
Define a function bc such that bc n yields the binomial coefficients of order n.@rFd@{:)"1]
AP6
Experiment with the vector derivative of the triangle area function of Exercise G5. D x 3 1 0 2 &{ D x +/\ D x +/\.
]
Define a function AREA such that AREA v yields the area of a triangle with two sides of lengths 0{v and 1{v and with an angle of 2{v degrees between them. whose areas are easily computed. 0. [
hat=: %:@(*/)@(-:@(+/) . try the case hat VD 3 4 5. and the odd part in the origin.
Exercises AP1 AP2 Enter the expressions of this section. Define a function rFd to produce radians from degrees. a function tbc such that tbc n yields a table of all binomial coefficients up to
. D x . In particular. and compare rFd 90 180 with
o. and experiment with its vector derivative hat VD. [AREA VD 2 3 90]
AP7
Heron's formula for the area of a triangle is the square root of the product of the semiperimeter with itself less zero and less each of the three sides.122 Calculus
For an even function. Predict the results of each of the following sentences.0: . for an odd function. D x
AP3 AP4
Define show=: {&'. Test it on triangles such as 2 3 90 and 2 3 30.
using the argument x=:3 1 4 1 6 [
L = i. Also experiment with matrices of complex numbers and with the use of the matrix inverse and matrix product functions upon them.Appendix 123
order n. hyperbolic sine and hyperbolic cosine. Test it on the linear functions L=:|.@>: ! ]
tbc=: !/~ @ (i. 4 is equivalent to x^n+1 for various values of n. and a function tabc for the corresponding alternating binomial coefficients.
AP10 Write an expression to yield the matrix m such that mp&m is equivalent to a given linear function L. cosine."1 and L=:3&A. # x ]
AP11 Experiment with the use of various functions on imaginary and complex numbers. x=: i.@tbc
]
AP9
Test the assertion that (bc n) p."1. the sine.@>:) tabc=: %.
. including the exponential. [
bc=: i. | 677.169 | 1 |
Edexcel iGCSE Maths
Edexcel International GCSE Maths
The iGCSE Maths course is harder than normal GCSEs. It involves some topics that are normally studied at AS level, such as differentiation, sets and functions. The exam can be taken in the Summer or in January. There are two papers, each of 2 hours counting for 50% each. The exam can be taken at Foundation tier (maximum grade C) or higher tier.
Exam dates for 2014
Course content
See the specification (link above) for more details
Number
Use numerical skills in a purely mathematical way and in real-life situations.
Algebra
Use letters as equivalent to numbers and as variables.
Understand the distinction between expressions, equations and formulae.
Use algebra to set up and solve problems.
Demonstrate manipulative skills.
Construct and use graphs.
Geometry
Use properties of angles.
Understand a range of transformations.
Work within the metric system.
Understand ideas of space and shape.
Use ruler, compasses and protractor appropriately | 677.169 | 1 |
Teaching Mathematics: A Sourcebook of AIDS, Activities, and Strategies
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Teaching Mathematics: A Sourcebook of AIDS, Activities, and Strategies
Max A. Sobel
3rdPublisher: Pearson Custom PublishingFormat: Paperback
This text addresses the art of teaching mathematics while also providing specific aids and activities in arithmetic, geometry, algebra, and probability and statistics for use in the classroom. The authors pay close attention to the role, importance, methods, and techniques of motivation. They present ideas that will generate attention, interest, and surprise among students, and will thus foster creative thinking. The material is based on talks given by Sobel and Maletsky at professional meetings, as well as the actual application of their ideas in undergraduate and graduate classes they taught. Additionally, many laboratory and discovery activities have been used by the authors in teaching junior and senior high school math classes | 677.169 | 1 |
Functions of a complex variable,: With applications,
So you will learn things again in new ways, and gain a powerful new set of tools. Knowledge of such Riemannian concepts as the Levi-Civita connection and curvature will be helpful, but not essential. Also, I received the product very quickly. Curves and surfaces were explored without ever giving a precise definition of what they really are (precise in the modern sense). In addition to its contribution to mathematical foundations and to computer science, mathematical logic and its methods have also led to the solution of a number of important problems in other fields of mathematics such as number theory and analysis.
Pages: 140
Publisher: Interscience Publishers; 2nd edition (1943)
ISBN: B0007F549A
Curvature and Homology
In classical algebraic geometry, the algebra is the ring of polynomials, and the geometry is the set of zeros of polynomials, called an algebraic variety Elliptic Operators and Compact read here read here. Dover edition (first published by Dover in 1988), paperback, 240 pp., ISBN 0486656098. Lots of interesting examples, problems, historical notes, and hard-to-find references (refers to original foreign language sources). Thorpe, John A., Elementary Topics in Differential Geometry, Springer-Verlag, 1979, hardcover, 253 pp., ISBN 0387903577 Symplectic Invariants and Hamiltonian Dynamics (Modern Birkhäuser Classics) Symplectic Invariants and Hamiltonian. Observe that however you do this, at every point of your sheet of paper there will always be a direction along which perfectly straight lines exist Tensor Algebra and Tensor download pdf tiny-themovie.com. Mathematical visualization of problems from differential geometry. This web page gives an equation for the usual immerson (from Ian Stewart, Game, Set and Math, Viking Penguin, New York, 1991), as well as one-part parametrizations for the usual immersion (from T. Nordstrand, The Famed Klein Bottle, ) and for the figure eight form (from Alfred Gray's book, 1997) epub. The treatment of these themes blends the descriptive with the axiomatic. The authors present the results of their development of a theory of the geometry of differential equations, focusing especially on Lagrangians and Poincare-Cartan forms , cited: Elementary Topics in download here Enough examples have been provided to give the student a clear grasp of the theory , e.g. Integral Geometry and Inverse Problems for Kinetic Equations (Inverse and Ill-Posed Problems) download for free. Thanks to everyone who came along and made it a fabulous event. We have a winner! @biancapascall with the first Clifford torus! pdf. The simplest way to perform the trick is to take a rope that is 12 units long, make a knot 3 units from one end and another 5 units from the other end, and then knot the ends together to form a loop. However, the Egyptian scribes have not left us instructions about these procedures, much less any hint that they knew how to generalize them to obtain the Pythagorean theorem: the square on the line opposite the right angle equals the sum of the squares on the other two sides Differential Forms and download here
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To fix this, set the correct time and date on your computer , source: Elementary Topics in download for free Beginning in the 19th century, various mathematicians substituted alternatives to Euclid's parallel postulate, which, in its modern form, reads, "given a line and a point not on the line, it is possible to draw exactly one line through the given point parallel to the line." They hoped to show that the alternatives were logically impossible epub$ online. Central Point: There exists on each generator of a general ruled surface a special point, called the central point of the generator Differential Geometry and read for free Differential Geometry and Physics:. Past speakers at these events include Keenan Crane, Fernando de Goes, Etienne Vouga, Mathieu Desbrun, and Peter Schröder. These notes grew out of a Caltech course on discrete differential geometry (DDG) over the past few years , e.g. Reduction of Nonlinear Control Systems: A Differential Geometric Approach (Mathematics and Its Applications) Reduction of Nonlinear Control Systems:. The MSRI Computing Group uses another horoball diagram as their logo. Thomas Banchoff animates the Hopf fibration. Gallery of interactive on-line geometry. The Geometry Center's collection includes programs for generating Penrose tilings, making periodic drawings a la Escher in the Euclidean and hyperbolic planes, playing pinball in negatively curved spaces, viewing 3d objects, exploring the space of angle geometries, and visualizing Riemann surfaces Comprehensive Introduction to Differential Geometry (Volumes 1 and 2) ccc.vectorchurch.com.
In these examples, there is similarity of the corresponding small elements. When this relation holds, the mapping is said to be conformal. Note: An isometric mapping preserves both distances and the angles, whereas a conformal mapping just preserves angles. A one- one correspondence of P (u, v) on S and Hence, if u= constant and v= constant are isothermic, any other isothermic system mapping of the surface on the plane Modeling of Curves and Surfaces with MATLAB® (Springer Undergraduate Texts in Mathematics and Technology) Now, on the other side of the stone, on the other face and in another language, we have the crisis and the possible death of mathematics in itself. Given then a proof to explicate as one would a text. And, first of all, the proof, doubtless the oldest in history, the one which Aristotle will call reduction to the absurd The Geometry of Physics: An download pdf The Geometry of Physics: An. The uniqueness of this text in combining geometric topology and differential geometry lies in its unifying thread: the notion of a surface. With numerous illustrations, exercises and examples, the student comes to understand the relationship between modern axiomatic approach and geometric intuition Lectures on the Differential read here . This course is an introduction to smooth manifolds and basic differential geometry , e.g. Selected Papers II read for free langleyrealestatesearch.com. If you require any further information or help, please visit our support pages: The present book grew out of notes written for a course by the same name taught by the author during in 2005 Differential Manifolds (Dover read pdf read pdf. By page 18 the author uses these terms without defining them: Differentiable Manifold,semigroup, Riemannian Metric, Topological Space, Hilbert Space, the " " notation, vector space, and Boolean Algebra. Fortunately for me, I have a fairly extensive math education, and self-studied Functional Analysis, so I wasn't thrown for a loop;but for many others -- brace yourselves! 1) Here is a quote: "The collection of all open sets in any metric space is called the topology associated with the space." Topics discussed are; the basis of differential topology and combinatorial topology, the link between differential geometry and topology, Riemanian geometry (Levi-Civita connext Families of Conformally download here download here. There was earlier scattered work by Euler, Listing (who coined the word "topology"), Mobius and his band, Riemann, Klein, and Betti. Indeed, even as early as 1679, Leibniz indicated the desirability of creating a geometry of the topological type ref.: Tensor Calculus and Analytical Dynamics (Engineering Mathematics) download here. It also helped to construct graphs which are Dirac isospectral. The matrix is also valuable for doing computations in geometry. Today, one can with a dozen lines of computer algebra system code produce the cohomology groups for any graph. The Dirac operator also allows to to see the graph theoretical Gauss-Bonnet-Chern theorem as an example of a discrete index theorem. [November 4, 2012] The Lusternik-Schnirelmann theorem for graphs [ ArXiv ] Plane Networks and their download here | 677.169 | 1 |
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Alpha Mathematics
Alpha Mathematics | Grades 1-5
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Alpha Math consists of two volumes for each grade that have distinctive and rich content. The content is built on the Common Core State Standards. The instructions and practices are designed in accordance with the learning domains of Bloom's taxonomy to help build a deep, strong and innovative foundation in mathematics.
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giving to the student complete familiarity
with
all
the essentials of the subject. but "cases" that are taught only on account of tradition.
and ingenuity
while the cultivation of the student's reasoning power is neglected.
owing
has certain distinctive features."
this book.
All
practical
teachers
know how few
students understand and
appreciate the more difficult parts of the theory. and conse-
.
All unnecessary methods
and "cases" are
omitted. etc. in order to make every example a
social
case of a memorized method. not only taxes a student's memory unduly but in variably leads to mechanical modes of study. Typical in this respect is the
treatment of factoring in
many
text-books
In this book
all
methods which are of
and which are applied in advanced work are given. are omitted..
"
While
in
many
respects
similar to the author's
to its peculiar aim. The entire study of algebra becomes a mechanical application of memorized
rules.
omissions serve not only practical but distinctly pedagogic " cases " ends.
however.
specially
2. short-cuts that solve only examples
real value. chief
:
among
These
which are the following
1.PREFACE
IN
this
book the attempt
while
still
is
made
to shorten the usual course
in algebra.
Elementary Algebra.
Such a large number of methods. All parts of the theory whicJi are beyond the comprehension
of
the student or wliicli are logically
unsound are
omitted. Until recently the tendency was to multiply as far as possible.
manufactured for this purpose.
as quadratic equations and
graphs.
enable students
who can devote only a minimum
This arrangement will of time to
algebra to study those subjects which are of such importance for further work. " The book is designed to meet the requirements for admis-
sion to our best universities
and
colleges. etc. there has been placed at the end of the book a collection of exercises which contains an abundance
of
more
difficult
work. This made it necessary to introduce the theory of proportions
.
differ
With very few
from those
exceptions
all
the exer
cises in this
book
in the
"Elementary Alge-
bra". in particular the
requirements of the College Entrance Examination Board.vi
PREFACE
quently hardly ever emphasize the theoretical aspect of alge bra. all proofs for the sign age
of the product of
of the binomial
3.
may
be used to supplement the other.
TJie exercises are slightly simpler than in the larger look. hence either book
4. The presenwill be found to be tation of problems as given in Chapter
V
quite a departure from the customary way of treating the subject. and it is hoped that this treatment will materially diminish the difficulty of this topic for young students. are placed early in the course. Moreover. all elementary proofs theorem for fractional exponents. For the more ambitious student. however.
In regard
to
some other features of the book.
The best way to introduce a beginner to a new topic is to offer Lim a large number of simple exercises. the following
may
be quoted from the author's "Elementary Algebra":
which
"Particular care has been bestowed upon those chapters in the customary courses offer the greatest difficulties to
the beginner.
Topics of practical importance. especially problems and factoring.
two negative numbers. e.g. a great deal of the theory offered in the avertext-book is logically unsound .
" Graphical methods have not only a great practical value. such examples.
but they unquestionably furnish a very good antidote against 'the tendency of school algebra to degenerate into a mechanical application of
memorized
rules.
is
based principally upon the alge-
. elementary way. and hence the student is more easily led to do the work by rote
than when the arrangement
braic aspect of the problem. based upon statistical abstracts.'
This topic has been preit is
sented in a simple.
McKinley
than one that gives him the number of Henry's marbles.
to solve a
It is
undoubtedly more interesting for a student
problem that results in the height of Mt. are
frequently arranged in sets that are algebraically uniform.
nobody would find the length Etna by such a method.
By studying proportions during the first year's work. and
of the
hoped that some
modes of representation given
will be considered im-
provements upon the prevailing methods. and commercial are numerous.
and they usually involve difficult numerical calculations."
Applications taken from geometry. Moreover.
viz. an innovation which seems to mark a distinct gain from the pedagogical point
of view.PREFACE
vii
and graphical methods into the first year's work. in
"
geometry
. The entire work in graphical methods has been so arranged that teachers who wish
a shorter course
may omit
these chapters.
while in the usual course proportions
are studied a long time after their principal application. the student will be able to utilize this knowledge where it is most
needed.
of the Mississippi or the height of Mt. But on the other hand very few of such applied examples are
genuine applications of algebra. physics. but the true study of algebra has not been sacrificed in order to make an impressive display of sham
life
applications.
is such problems involves as a rule the teaching of physics by the
teacher of algebra.
William P.
NEW
YORK.
. however.
ARTHUR SCHULTZE.
pupil's knowlso small that an extensive use of
The average
Hence the
field of
suitable for secondary school
tations.
April.
genuine applications of elementary algebra work seems to have certain limi-
but within these limits the author has attempted to
give as
many
The author
for
simple applied examples as possible. desires to acknowledge his indebtedness to Mr. 1910.
edge of physics.viii
PREFACE
problems relating to physics often
offer
It is true that
a field
for genuine applications of algebra. Manguse for the careful reading of the proofs and
many
valuable suggestions.
27. a = 4.
28.
Express in algebraic symbols 31. 30. a = 3.
23. geometry. 6 = 7.
35. 10-14
The
representation of numbers by letters makes it posvery briefly and accurately some of the principles of arithmetic.
6 = 4. = 3.
a =3.
Read the expressions
of Exs. 6. a =4. physics. and other sciences. 6 = 5.
12 cr6
-f-
6 a6 2
6s.6 . and the area of the
is
triangle
S
square feet (or squares of other units selected).
.
Six
2
.
sible to state
Ex.
w
cube plus three times the quantity a minus
plus 6 multiplied
6.
The quantity a
6
2 by the quantity a
minus
36. 25. a
a=3. of this exercise?
What kind of expressions are Exs.
:
6.6 -f c) (6
a
+ c).
26.
34.
6.
a = 3.
24. 6 = 5. 33.
then
8 = \ V(a + 6 + c) (a 4.
30. 6 = 6. a = 2.
and
If the three sides of a triangle contain respectively c feet (or other units of length).
22. 6 = 2. 38. 6 = 6. 6 = 3. 2-6 of the exercise.
Twice a3 diminished by 5 times the square root of the quantity a minus 6 square.
37.12
17
&
*
ELEMENTS OF ALGEBRA
18
'
8
Find the numerical value of 8 a3
21. if
:
a = 2.
Six times the square of a minus three times the cube of Eight x cube minus four x square plus y square. a = 4.
a.c) (a . Six times a plus 4 times
32.
29. 6 = 1.
6=2.
b 14.g. and c
13
and
15
=
=
=
. if v
:
a. 14. then a 13.16 centimeters per second. and 13 inches.16
1
= 84. c. 13.
the area of the triangle equals
feet.) Assuming g
. Find the height of the tree. if v = 30 miles per hour.INTRODUCTION
E. 15 therefore feet.
S = | V(13-hl4-fl5)(13H-14-15)(T3-14-i-15)(14-13-f-15)
= V42-12-14.
2.
84 square
EXERCISE
1. 12.
4.e. and 5 feet.
By
using the formula
find the area of a triangle
whose
sides are respectively
(a) 3. d. and 15 feet.
(c) 4. How far does a body fall from a state of rest in T ^7 of a (c)
A
second ?
3.
=
(a)
How
far does
a body fall from a state of rest in 2
seconds ?
(b)
*
stone dropped from the top of a tree reached the ground in 2-J. An electric car in 40 seconds.
(b) 5.
i. count the resistance of the atmosphere.
9
distance s passed over by a body moving with the uniform velocity v in the time t is represented by the formula
The
Find the distance passed over by A snail in 100 seconds. the three sides of a triangle are respectively 13. if v . b. A train in 4 hours.seconds. A body falling from a state of rest passes in t seconds 2 over a space S (This formula does not take into ac^gt 32 feet.
. A carrier pigeon in 10 minutes. if v = 50 meters per second 5000 feet per minute.
This number cannot be expressed exactly.) Find the surface of a sphere whose diameter equals
(a)
7.
$ = 3.
ELEMENTS OF ALGEBRA
If
the radius of a circle
etc.
of this formula
:
The The
interest on interest
$800
for 4 years at
ty%.
to Centigrade readings:
(b)
Change the following readings
(a)
122 F.
(c)
8000 miles.).
If the diameter of a sphere equals d units of length.
diameter of a sphere equals d
feet.
(c)
10
feet. If cated on the Fahrenheit scale.
6
Find the volume of a sphere whose diameter equals:
(b)
3
feet. the
3.
:
8000 miles.
2 inches.14
4.14
square meters.
fo
If
i
represents the simple interest of
i
p
dollars at r
in
n
years. then
=p
n
*
r
%>
or
Find by means
(a)
(b)
6.
(c)
5 miles.
is
H
2
units of length (inches.14d (square units).).
If the
(b) 1 inch.
(c)
5
F.
on $ 500 for 2 years at 4 %.
meters. the equivalent reading C on the Centigrade scale may be found by the formula
F
C
y
= f(F-32).
~
7n
cubic feet. and the value given above is only an
surface
$=
2
approximation.
the area
etc.
5.
denotes the number of degrees of temperature indi8.
then the
volume
V=
(a) 10 feet. Find the area of a circle whose radius is
It
(b)
(a) 10 meters.14 is frequently denoted by the Greek letter TT. (The number 3.
32 F.
.
square units (square inches.
or positive and negative numbers.
While
in arithmetic the
word sum
refers only to
the
result obtained
by adding positive numbers. Or in the symbols of algebra
$4) =
Similarly. In algebra.$6) + (-
$4) = (-
$10). SUBTRACTION. Thus a gain of $ 2 is considered the sum of a gain of $ 6 and a loss of $ 4. the fact that a loss of
loss of
+ $2. however. or that
and
(+6) + (+4) = +
16
10.CHAPTER
II
ADDITION.
of
$6 and a gain
$4
equals a
$2 may be
represented thus
In a corresponding manner we have for a loss of $6 and a
of
loss
$4
(. but we cannot add a gain of $0 and a loss of $4. in algebra this word includes also the results obtained by adding negative. In arithmetic we add a gain of $ 6 and a gain of $ 4.
Since similar operations with different units always produce analogous results. AND PARENTHESES
ADDITION OF MONOMIALS
31. we call the aggregate value of a gain of 6 and a loss of 4 the sum of the two.
. we define the sum of two numbers in such a way that these results become general.
ELEMENTS OF ALGEBRA
These considerations lead to the following principle
:
If two numbers have the same sign. 12. if :
a
a
= 2. add their absolute values if they have opposite signs. 10.
EXERCISE
Find the sum
of:
10
Find the values
17. find the numerical values of a + b
-f c-j-c?.
21. 4.
23. the average of 4 and 8
The average The average
of 2.
The average
of two
numbers
is
average of three numbers average of n numbers is the
is one half their sum. = 5.
. 24. 23-26. the one third their sum. d = 0.
33. + -12.16
32.
4
is
3 J.
l-f(-2).
-
0.
(_
In Exs.
is 0. c =
4.
22.
Thus. and the sum of the numbers divided by n. 5. subtract their absolute values and
. '.3.
(always) prefix the sign of the greater. is 2.
+ (-9).
5. c = = 5.
of:
20.
19.
18.
(-17)
15
+ (-14).
6
6
= 3.
of 2.
d = 5.
Find the average of the following temperatures 27 F.ADDITION. $500 loss. 12. 7 yards. affected
by the same exponents. Find the average gain per year of a merchant. 3. 34.
.
:
and
1. 09. . 5 and
12.
= 22. \\ Add 2 a. 37.
Similar or like terms are terms which have the same
literal factors.
42.
. 1.
are similar terms.
:
34.
40. 32.3.5. & = 15.
13. 6. and 3 F. 0. $3000 gain.
:
48.4.
25.13.
5 a2 &
6 ax^y and
7 ax'2 y. . 38.
31.
29.
6.
which are not
similar.
2. 55.
35. if his yearly gain or loss during 6 years was $ 5000 gain. c = 0.. = -13... SUBTRACTION. -11 (Centigrade). 66. 6.
4
F. $7000 gain. Find the average temperature of New York by taking the average of the following monthly averages 30. 10. 7 a.
&
28.
-'
1?
a
26. 7 a.
30.
'
Find the average of the following
34.
AND PARENTHESES
d = l.
33. and 3 a. 10.
:
Find the average temperature of Irkutsk by taking the average of the following monthly temperatures 12. 32.7.
3 and 25.7.
.
.
and
-8
F. 41.
Dissimilar or unlike terms are terms
4 a2 6c and o
4 a2 6c2 are dissimilar terms.
2. 43. or
and
. 74. $1000 loss.5. }/ Add 2 a.
= -23. c=14. 10. and 3 a. 60.
or 16
Va + b
and
2Vo"+~&. 39. -4. and 4. .
36. and 3 yards.
27. 72.
^
'
37.
sets of
numbers:
13. and $4500 gain.
d=
3.
.
2
.
-f
4 a2.
The sum The sum
of a of a
Dissimilar terms cannot be united into a single term.
EXERCISE
Add:
1. or
a
6.
Vm
-f. and 4 ac2
is
a
2 a&
-|-
4 ac2.
10. b wider sense than in arithmetic.
12(a-f b)
12.
11
-2 a +3a -4o
2.18
35.
5 a2
.
sum of two such terms can only be them with the -f. in algebra it may be considered b. Algebraic sum.
2 a&. The indicated by connecting and a 2 and
a
is
is
-f-
a2
.
14
.
12
2 wp2 .
-3a
. either the difference of a and b or the sum of a and
The sum
of
a.13 rap
25 rap 2.
9(a-f-6).
13.ii.
b
a
-f (
6).
1
\
-f-
7 a 2 frc
Find the sum of
9.
5Vm + w.
ab
7
c
2
dn
6.
11.
12Vm-f-n.
.
ELEMENTS OF ALGEBRA
The sum
of 3
of
two similar terms
x2
is
is
another similar term.sign.
The sum
x 2 and
f
x2 .
5l
3(a-f-6).
In algebra the word sum is used in a 36.
12
13
b sx
xY xY 7 #y
7. While in arithmetic a denotes a difference only.
2(a-f &). 7 rap2.
:
2 a2.
+ 6 af
.
This gives by the same method.
(-
6)
-(-
= . In addition.
5
is
2. the algebraic sum and one of the two numbers is
The algebraic sum is given. In subtraction.
3 gives
3)
The number which added
Hence.3. may be stated number added to 3 will give 5? To subtract from a the number b means to find the number which added to b gives a.
From
5 subtract
to
The number which added
Hence.
State the other practical examples which show that the number is equal to the addition of a
40.
From
5 subtract
to
. and the
required number the difference. from What 3.
7. Subtraction is the inverse of addition.
41.
if
x
Ex. a-b =
x.
may
be stated in a
:
5 take form e.
3.
+b
3.
From
5 subtract
+ 3.
The
results of the preceding examples could be obtained
by the following
Principle.
Ex. The student should perform mentally the operation of chang8 2 6 from 6 a 2 fc. Or in symbols. change the sign
of the subtrahend and
add. the other number is required. Therefore any example in subtraction
different
.
1.
Ex. called the minvend.2.
NOTE. and their algebraic sum is required. ing the sign of the subtrahend thus to subtract 6 a 2 6 and 8 a 2 6 and find the sum of change mentally the sign of
.
2.
3 gives 5
is
evidently 8.
AND PARENTHESES
23
subtraction of a negative
positive number. SUBTRACTION.g.
To
subtract.
6
-(-3) = 8.
a.
. two numbers are given. the given number the subtrahend.ADDITION.
I. The beginner will find it most convenient
at every step to
remove only those parentheses which contain
(7 a
no others.
Hence
the
it is
sign
may
obvious that parentheses preceded by the -f or be removed or inserted according to the fol:
lowing principles
44.c.
.
45. may be written as follows:
a
-f ( 4.6 b -f (.
AND PARENTHESES
27
SIGNS OF AGGREGATION
43.
& -f
c.
If there is no sign before the first term within a paren*
-f-
thesis.b c = a a
&
-f-
-f.2 b .a
-f-
= 4a
sss
7a
12
06
6.
6
o+(
a
+ c) = a =a 6 c) ( 4-.& c
additions
and sub-
+ d) = a + b
c
+ d. SUBTRACTION.
A
moved
w
may be resign of aggregation preceded by the sign inserted provided the sign of evei'y term inclosed is
E. If we wish to remove several signs of aggregation. A sign of aggregation preceded by the sign -f may be removed or inserted without changing the sign of any term.
(b
c)
a
=a
6 4-
c.
4a-{(7a + 6&)-[-6&-f(-2&.a~^~6)]} = 4 a -{7 a 6 b -[. Ex.ADDITION.
II.g. the sign
is
understood.
a+(b-c) = a +b . one occurring within the other.
tractions
By using the signs of aggregation.
66
2&-a + 6
4a
Answer.c.
changed. we may begin either at the innermost or outermost.a^6)]
-
}
.
46. Simplify 4 a f
+ 5&)-[-6& +(-25.
z
+ d.
The sum
of tKe squares of a
and
b. )X
6.
7.
12. The minuend is always the of the two numbers mentioned. SUBTRACTION.
5 a2
2.
2m-n + 2q-3t.
10. The product of the sum and the difference
of
m and n.
2.
In each of the following expressions inclose the last three in a parenthesis preceded by the minus sign
:
-27i2 -3^ 2 + 4r/.
y
-f-
8
.
The product The product
m
and
n.
4 xy
7 x* 4-9 x + 2.4 y* .
II.
13.7-fa.
6 diminished
.
a-\-l>
>
c
+
d.
EXERCISE
AND PARENTHESES
16
29
In each of the following expressions inclose the last three terms in a parenthesis
:
1.
3.
terms
5.
The sum^)f
m
and
n.
'
NOTE. 9.
7.
5^2
_ r .
6.ADDITION.
Nine times the square of the sum of a and by the product of a and b.
m
x
2
4.
difference of the cubes of n and m.
4.
p + q + r-s.
8.
The sum
of the fourth powers of a of
and
6.
EXERCISES
IN"
ALGEBRAIC EXPRESSION
17
:
EXERCISE
Write the following expressions
I. .
The
difference of a
and
6.
3.
The The
difference of the cubes of
m
and
n.
and the subtrahend the second.
m and n.
5.
Three times the product of the squares of The cube of the product of m and n.
first.
of the cubes of
m and
n.
The square of the difference of a and b.2 tf .1.
dif-
of the squares of
a and
b increased
by the
square root of
15.)
.
The sum
The
of a
and
b multiplied
b is equal to the difference of
by the difference of a and a 2 and b 2
.
18.
difference of the cubes of a
and
b divided
by the
difference of
a and
6.
a plus the prod-
uct of a and
s
plus the square of
-19. (Let a and b represent the numbers.
x cube minus quantity 2 x2 minus 6 x plus
The sum
of the cubes of a.
d.
6 is equal to the square of
b. 16.
and
c
divided by the
ference of a and
Write algebraically the following statements:
V 17.30
14.
b. The difference of the squares of two numbers divided by the difference of the numbers is equal to the sum of the two numbers. 6.
ELEMENTS OF ALGEBRA
The sum
x.
and forces produced at by 3 Ib.
If the
two loads balance. what force
31
is
produced by tak(
ing away 5 weights from
B ? What therefore is
5)
x(
3) ?
. weights. is 5 x ( 3) ?
7. weight at B ?
If the
addition of five 3
plication example. weights at A ? Express this as a multibalance.
A
A
A
1.
3. what force is produced by the Ib.
By what sign is an upward pull at A represented ? What is the sign of a 3 Ib. weight at A ? What is the sign of a 3 Ib. 2.
two loads balance. If the two loads balance.
force is produced
therefore.
5.CHAPTER
III
MULTIPLICATION
MULTIPLICATION OF ALGEBRAIC NUMBERS
EXERCISE
18
In the annexed diagram of a balance. is
by
taking away 5 weights from
A?
5
X 3?
6.
If the two loads
what
What. let us consider the and JB.
4. what force is produced by the addition of 5 weights at B ? What. therefore. applied at let us indicate a downward pull at by a positive sign.
however. (-5)X4. times is just as meaningless as to fire a gun
tion
7
Consequently we have to define the meaning of a multiplicaif the multiplier is negative.
Thus.
(-
9)
x (-
11) ?
State a rule by which the sign of the product of
two
fac-
tors can be obtained. such as given in the preceding exercise. a result that would not be obtained by other assumptions.
4x(-3)=-12. Practical examples^
it
however. To take a number 7 times.
4
x(-8) = ~(4)-(4)-(4)=:-12.4)-(.
9
9. 5x(-4). and we may choose any definition that does not lead to contradictions.
(.
ELEMENTS OF ALGEBRA
If
the signs obtained by the
true.
examples were generally
method of the preceding what would be the values of
(
5x4. thus. or plied by 3.
the multiplier is a negative number.
Multiplication
by a negative
integer is a repeated
sub-
traction.32
8. Multiplication by a positive integer is a repeated addition. becomes meaningless
if
definition.
.
48. 4 multiplied by 3.9) x
11.
(
(. or
4x3 =
=
(_4) X
The preceding
3=(-4)+(-4)+(-4)=-12.
x
11.4)-(-4) = +
12.
In multiplying integers we have therefore four cases
trated
illus-
by the following examples
:
4x3 = 4-12.
NOTE. make venient to accept the following definition
:
con-
49. This definition has the additional advantage of leading to algenumbers which are identical with those for positive numbers. 4 multi44-44-4 12.4) x
braic laws for negative
~ 3> = -(. 9 x
(-
11).
2 a2 6 a8
2 a*
*
-
2"
a2
-7
60. however.1.3 a 2 + a8 .3 a 2 + a8
a
a = =-
I
1
=2
-f
2
a
4.
Ex.a6
=2
by numerical
Examples
in multiplication can be checked
substitution.4. Multiply 2 a . To multiply two polynomials. multiply each term of one by each term of the other and add the partial products thus formed.
2a-3b a-66 2 a .3
b
by a
5
b.
Multiply 2
+ a -a.
2.
Check. this method
tests only the values of the coefficients
and not the values of
the exponents. Since errors.
a2
+ a8 + 3 .
.M UL TIP LICA TION
37
58.
The most convenient way of adding the partial products is to place similar terms in columns. the student should
apply this test to every example. are far more likely to occur in the coefficients than anywhere else. If the polynomials to be multiplied contain several powers of the same letter.3 a
3
2
by 2 a
:
a2 + l. 1 being the most convenient value to be substituted for all letters.
If
Arranging according to ascending powers
2
a
.3 ab
2
2 a2
10 ab
-
13 ab
+ 15 6 2 + 15 6 2
Product.a6 4 a 8 + 5 a* . Since all powers of 1 are 1. the work becomes simpler and more symmetrical by arranging these expressions according to either ascending or descending powers. as illustrated in the following example
:
Ex.a
.
59.
the product of two binomials whose corresponding terms are similar is equal to the product of the first two
terms.
that the square of each term is while the product of the terms may have plus always positive.
The middle term
or
Wxy-12xy
Hence in general.
2
2
+ 2) (10 4-3).
7%e square of a polynomial is equal to the sum of the squares of each term increased by twice the product of each term with each that follows it.
65. plus the product of the
EXERCISE
Multiply by inspection
1. (4s + y)(3-2y). (3m + 2)(m-l).&
+ c) = a + tf + c
.
2
(2x y
(6
2
2
+ z )(ary + 2z ).
(5a-4)(4a-l). (100 + 3)(100 + 4).
(x
i-
5
2
ft
x 2 -3 6 s).
8.
11.
or
The student should note
minus
signs.42
ELEMENTS OF ALGEBRA
of the result is obtained
product of 5 x
follows:
by adding the These products are frequently called the cross products. ) (2
of a polynomial. plus the
last terms. and are represented as
2 y and 4y 3 x.
4.
3. 14.
5.
The square
2
(a 4.
(2a-3)(a + 2).
sum of the
cross products. 9.
2 2 2 2 (2a 6 -7)(a & +
5). 13.
(5a6-4)(5a&-3).
6.
2
(2m-3)(3m + 2).
:
25
2.
7.
.
2
10.
((5a?
(10
12.-f
2 a&
-f
2 ac
+ 2 &chence
it
is
an equation
of
condition.
81. An identity is an equation of the letters involved.
=11.
83.r
-f9
= 20
is
true only
when
a.
(rt+6)(a-ft)
=
2
-
b'
2
.
. in the equation 2 x 0.
in
Thus x
12 satisfies the equation x
+
1
13. ond member or right side is that part which follows the sign of
equality. second member is x
+
4
x
9. is said to satisfy an equation.
A set of numbers which when substituted for the letters an equation produce equal values of the two members.
y
=
7 satisfy
the equation x
y
=
13.
ber
equation is employed to discover an unknown num(frequently denoted by x.CHAPTER V
LINEAR EQUATIONS AND PROBLEMS
79. y y or z) from its relation to
63
An
known numbers.
.
x
20.
which
is
true for all values
a2 6 2 no matter what values we assign to a Thus. The sign of identity sometimes used is = thus we may write
.
82.
the
80. (a + ft) (a b) and b.
. An equation of condition is an equation which is true only for certain values of the letters involved.
The
first
member
or left side of an equation
is
that part
The secof the equation which precedes the sign of equality. An equation of condition is usually called an equation.
the
first
member
is
2 x
+
4.
Thus.
54
84.2.
3. the products are equal.
86.
5.
.
the divisor equals zero.
the
known quan
x) (x -f 4)
tities are
=
.
expressed in arithmetical numbers
literal
is
as (7
equation
is
one in which at least one of the
known
quantities as x -f a letters
88. called axioms
1.
a.
Transposition of terms.
(Axiom
2)
the term a has been transposed from the left to thQ
right
member by changing
its
sign. an^ unknown quantity which satisfies the equation is
a root of the equation. the
sums are
equal.
90.
ELEMENTS OF ALGEBRA
If
value of the
an equation contains only one unknown quantity. the quotients are equal. 87.
E.
= bx
expressed by a letter or a combination of
c. the remainders are
equal.e.
but 4 does not equal
5.
Like powers or
like roots
of equals are equal.
.
9
is
a root of the equation 2 y
+2=
is
20.
A numerical
equation is one in
which
all
.
If equals be subtracted from equals.
If equals be added
to equals.
85. A
2
a.
A
linear equation or
which when reduced
first
to its simplest
an equation of the first degree is one form contains only the
as
9ie
power of the unknown quantity.
If equals be multiplied by equals.
To
solve
an equation
to find its roots.
A
term may be transposed from
its sign.g.
NOTE.
If equals be divided by equals.
one member to another by changing
x + a=.
4.
Axiom
4
is
not true
if
0x4
= 0x5.b.
fol-
A
linear equation is also called a simple equation.
89.
2. x
I. Consider the equation b Subtracting a from both members.
2
= 6#-f7.
The process
of solving equations depends upon the
:
lowing principles.
3.
6.
10. 15.
1.
33
2.
one part equals
is 10.
a. so that one part
The
difference between
is s. greater one is g.
Find the greater one.
7.
5.
EXERCISE
1. Hence 6 a must be added
to a to give
5.
What number divided by 3 will give the quotient a? ? What is the dividend if the divisor is 7 and the quotient
?
.
is
a?
2
is
c?. so that one part Divide a into two parts. and the smaller one
parts.
Ex.
Divide a into two parts. one yard will cost
100
-dollars.
x
-f-
y yards cost $ 100
. one yard will cost -
Hence
if
x
-f
y yards cost $ 100. 11. so that
of c ?
is
p.58
Ex.
two numbers
and the and the
2
Find the greater one.
If 7
2.
13. is d. Divide 100 into two
12. is b.
4.
9. 6.
find the cost of one yard
14. or 12 717.
smaller one
16. The difference between two numbers Find the smaller one.
$> 100 yards cost one hundred dollars.
find the
has ra dollars.
?/
31.
A
feet wide.
How many years
A
older than
is
B?
old.
If
B
gave
A
6
25.LINEAR EQUATIONS AND PROBLEMS
18. and B has n dollars.
A
dollars.
34.
32.
How many
cents
has he ?
27.
22.
26.
and 4
floor of a room that is 3 feet shorter wider than the one mentioned in Ex.
20. The greatest of three consecutive the other two.
How many
cents are in d dollars ? in x dimes ?
A has
a
dollars.
Find the area of the Find the area of the
feet
floor of
a room that
is
and 3
30. 28.
How many
cents had he left ?
28.
and
c cents.
A man
had a
dollars.
Find
35.
feet wider than the one
mentioned in Ex.
59
What must
The
be subtracted from 2 b to give a?
is a.
numbers
is x. square feet are there in the area of the floor ?
How many
2 feet longer
29. b dimes.
sum
If A's age is x years.
24.
is
A A
is
# years
old.
and
B
is
y years old. and B's age is y years 28. 33. A room is x feet long and y feet wide. rectangular field is x feet long and the length of a fence surrounding the field. find the of their ages 6 years hence.
and spent
5 cents.
19. Find the sum of their ages
5 years ago.
Find
21.
y years
How
old was he 5 years ago ?
How
old will he be 10 years hence ?
23. amount each will then have.
48. What fraction of the cistern will be second by the two pipes together ?
44.
A
cistern can be filled
in
alone
fills it
by two pipes.
a.
The two
digits of a
number
are x and
y. What fraction of the cistern will be filled by one pipe in one minute ?
42.
of 4. If a man walks 3 miles per hour. If a man walks n miles in 4 hours.
Find the
number.
b
To express in algebraic symbols the sentence: " a exceeds much as b exceeds 9.
how many
how many
miles will
he walk in n hours
38.60
ELEMENTS OF ALGEBRA
wil\
36.
c
a
b
=
-
9.
-46. he walk each hour ?
39.
m is the
denominator.
% % %
of 100
of
x.
per
Find 5 Find 6
45.
as
a exceeds
b
by as much
as c exceeds 9.
How many
x years ago
miles does a train
move
in
t
hours at the
rate of x miles per hour ?
41.
miles does
will
If a man walks r miles per hour. and "by as much as" Hence we have means equals (=)
95.
How
old
is
he
now ?
by a pipe in x minutes.50. how many miles he walk in n hours ?
37.
49. find the fraction.
.
of m. Find a
47.
The numerator
If
of a fraction exceeds the denominator
by
3. The first pipe x minutes. and the second pipe alone fills it in
filled
y minutes.
-.
Find
a.
If a
man walks
?
r miles per hour.
A
was 20 years
old." we have to consider that in this by statement "exceeds" means minus ( ).
A
cistern
is
filled
43. in how many hours he walk n miles ?
40.
Find x
% %
of 1000.
of a increased
much
8.
The product
of the
is
diminished by 90 b divided by 7.
In
many
word
There are usually several different ways of expressing a symbolical statement in words. etc.LINEAR EQUATIONS AND PROBLEMS
Similarly.
of a and 10 equals 2
c. thus:
a
b
= c may
be expressed as follows
difference between a
:
The
and
b is
c.
-80.
by one third of b equals 100.
third of x equals
difference of x
The
and y increased by 7 equals
a.
The double
as
7.
5.
80.
c.
3.
8
-b ) + 80 = a
.
EXERCISE
The The double The sum
One
34
:
Express the following sentences as equations
1.
cases it is possible to translate a sentence word by in algebraic symbols in other cases the sentence has to be changed to obtain the symbols.
of x increased by 10 equals
x.
equal to the
sum and the difference of a and b sum of the squares of a and
gives the
Twenty subtracted from 2 a
a.
c.
The
excess of a over b
is c. Four times the difference of a and b exceeds c by as
d exceeds
9.
=
2
2
a3
(a
-
80.
a exceeds b by c. the difference of the squares of a
61
and
b increased
-}-
a2
i<5
-
b'
2
'
by 80 equals the excess of a over
80
Or. 2.
double of a
is
10.
4.
9.
same
result as 7
subtracted from
. a is greater than b by b is smaller than a by
c.
6.
amounts. of 30 dollars. In 3 years A will be twice as old as B. pays to C $100. a third sum of 2 x + 1 dollars.
->.
18. B's.
6
%
of m B's. the
sum
and C's
money
(d)
(e)
will be $ 12.
#is5%of450.
is
If A's age is 2 x.
first
00
x % of the
equals one tenth of the third sum. and
(a)
(6)
A
If
has $ 5 more than B.
A
If
and
B
B together have $ 200 less than C. express in algebraic
3x
:
10.*(/)
(g)
(Ji)
Three years ago the sum of A's and B's ages was 50. they have equal
of A's.62
10.
12. sum equals $20. (d) In 10 years A will be n years old. (e) In 3 years A will be as old as B is now. and C have respectively 2 a. and C's ages will be 100.
17. the first sum equals 6 % of the third sura.
a. B.
ELEMENTS OF ALGEBRA
Nine
is
as
much below a
13. they have equal amounts.
.
as 17
is
is
above
a.
m is x %
of n.
(a)
(b)
(c)
A is twice as old as B. A is 4 years older than
Five years ago A was x years old.000. B's age
20.
A
gains
$20 and B
loses
$40.
symbols
B.
and C's age
4
a.
a. 16. In 10 years the sum of A a second sum.
5x
A sum of money consists of x dollars.
(c)
If each
man
gains $500..
11. express in algebraic symbols
:
-700. 14.
x
is
100
x%
is
of 700.
x
4-
If A. 3 1200 dollars.
50
is
x % of
15.
be 30
.
A
will
Check. number of
yards.
Simplifying. Find A's present age. the required
.
Transposing. 2. x = 20.
6 years ago he was 10
.
Ex.
number by x (or another letter) and express the yiven sentence as an equation.
1.
3z-40:r:40-z. x= 15.
Uniting.
Check.
x+16 = 3(3-5). the
.
Ex. In 15 years A will be three times as old as he was 5 years ago.
Let x = the number.
In 15 years
10. be three times as old as he was 5 years ago.
NOTE.
3
x
+
16
=
x
x
(x
-
p)
Or. The student should note that x stands for the number of and similarly in other examples for number of dollars. verbal statement (1)
(1) In 15 years
A
will
may be expressed in symbols (2). exceeds 40 by as much as 40 exceeds the no. x + 15 = 3 x
3x 16
15. The solution of the equation (jives the value of the unknown number. 4 x = 80. In order to solve them.
equation is the sentence written in alyebraic shorthand.
Three times a certain number exceeds 40 by as Find the number.
Let x
The
(2)
= A's present age.
3 x or 60 exceeds 40
+ x = 40 + 40. etc.LINEAR EQUATIONS AND PROBLEMS
63
PROBLEMS LEADING TO SIMPLE EQUATIONS
The simplest kind of problems contain only one unknown number.
.
by 20
40 exceeds 20 by 20.
Uniting.
Transposing. = x x
3x
-40
3x
40-
Or. denote the unknown
96. Three times a certain no.
much
as 40 exceeds the number.
number. 15.
Dividing.
-23 =-30. The equation can frequently be written by translating the sentence word by word into algebraic symbols in fact.
but
30
=3
x
years. Write the sentence in algebraic symbols.
Forty years hence
his present age.2.
Six years hence a
12 years ago. 14.
to
42 gives a
sum
equal to 7 times the
original
6.
ELEMENTS OF ALGEBRA
56
is
what per cent
of 120 ?
= number
of per cent.
35
What number added
to twice itself gives a
sum
of
39?
44. then the
problem expressed in symbols
W
or.
Dividing.64
Ex.
300
56.
Hence
40
= 46f.
3.
Find the number.
How
old
is
man will be he now ?
twice as old as he was
9.
A will
be three times as old as to-da3r
. by as much as 135 ft. Four times the length of the Suez Canal exceeds 180 miles by twice the length of the canal.
Find
8. A train moving at uniform rate runs in 5 hours 90 miles more than in 2 hours.
What number
7
%
of
350?
Ten times the width of the Brooklyn Bridge exceeds 800 ft.
47 diminished by three times a certain number equals 2.
14 50
is
is
4
what per cent of 500 ? % of what number?
is
12.
.
120. How long is the Suez
Canal?
10.
twice the number plus
7.
5.
Find the width of the Brooklyn Bridge.
EXERCISE
1.
A
number added
number.
Let x
3.
4.
Find the number whose double exceeds 30 by as much
as 24 exceeds the number. % of
120.
Find the number whose double increased by 14 equals Find the number whose double exceeds 40 by 10.
Uldbe
66
| x x
5(5 is
=
-*-.
13. exceeds the width of the bridge. How many miles per hour does it run ?
. 11. Find the number.
The other verbal statement.
F
8.
is
the equation. Vermont's population increased by 180.
numbers (usually the smaller one) by
and use one of the
given verbal statements to express the other unknown number in terms of x. and
as
15. The sum of the two numbers is 14. then dollars has each ? many
have equal amounts of money. If A gains A have three times as much
16.
Ex. Maine's population increased by 510.
How many
dol-
A has
A
to
$40.
One number exceeds another by
:
and their sum
is
Find the numbers. two verbal statements must be given. If the first farm contained twice as many acres as
A man
number
of acres. Ill the simpler examples these two
lems they are only implied.
statements are given directly. and
B
has $00.
times as
much
as A.
1.
the second one. If a problem contains two unknown quantities. and Maine had then twice as many inhabitants as Vermont.
. which gives the value of
8.
x.
65
A
and
B
$200. while in the more complex probWe denote one of the unknown
x. In 1800 the population of Maine equaled that of Vermont. During the following 90 years. Find
the population of Maine in 1800. B will have lars has A now?
17.
how many
acres did he wish to
buy
?
19.
make A's money
equal to 4 times B's
money
wishes to purchase a farm containing a certain He found one farm which contained 30 acres too many.
A
and
B
have equal amounts of money.000.
How many dollars
must
?
B
give to
18.LINEAR EQUATIONS AND PROBLEMS
15.000.
B
How
will loses $100.
14.
97.
The problem consists of two statements I. written in algebraic
symbols. One number exceeds the other one by II. and another which lacked 25 acres of the required number.
five
If
A
gives
B
$200.
26
= B's number of marbles after the exchange.
<
Transposing. = 3. to
Use the simpler statement.
. 2. Let
x
3x
express one
many as A. 25 marbles to B.
x
x =14
8.
has three times as many marbles as B. the greater number. A has three times as many marbles as B.
Dividing.
. the sum of the two numbers is 14. the smaller number.
Another method for solving this problem is to express one unknown quantity in terms of the other by means of statement II viz.
A will lose.
Then.
Let
x
14
I
the smaller number.
If
A gives
are
:
A
If
II. B will have twice as many as A.
x
3x
4-
and
B
will gain.
2x
a?
x
-j-
= 6. unknown quantity
in
Then.
o\
(o?-f 8)
Simplifying.
x
= 8.
= B's number of marbles.
.
which leads
ot
Ex.
Statement
x
in
=
the larger number.
The two statements
I. I.
expressed symbols is (14 x) course to the same answer as the first method.
If
we
select the first one.
To
express statement II in algebraic symbols. although in general the simpler one should be selected.
Uniting. 26 = A's number of marbles after the exchange.66
ELEMENTS OF ALGEBRA
Either statement may be used to express one unknown number in terms of the other. consider that by the
exchange
Hence.
+
a-
-f
-f
8
= 14. A gives B 25 marbles. B will have twice as
viz.
/
.
terms of the other.
and
Let x
= the
Then x -+. = 14.
in algebraic
-i
symbols produces
#4a.= The second statement written
the equation ^
smaller number. = A's number of marbles. 8 = 11. 8 the greater number.
Check.25 = 20. Dividing.
Selecting the cent as the denomination (in order to avoid fractions).75.240.
dollars
and dimes
is
$3. the number of half dollars. etc.
. the number of dimes.
differ
differ
and the greater and their sum
times
Two numbers
by
60.
w'3. the
price.10.
Dividing. their sum
+ +
10 x 10 x
is
EXERCISE
36
is five
v
v. of dollars to the number of cents.
Two numbers
the smaller. The number of coins II.
The sum of two numbers is 42.10.
67
x
-f
25 25
Transposing.
*
98. Find the numbers. B's number of marbles. x = the number of half dollars.
6 dimes
= 60
= 310.
is 70. x = 6.
by 44. 40 x .
The numbers which appear
in the equation should
always
be expressed in the same denomination.
Uniting.
Let
11
= the number of dimes.550 -f 310. 6 half dollars = 260 cents.
50 x
Transposing.
50(11 660 50 x
-)+ 10 x = 310. How many are there of each ?
The two statements are I.
Find
the numbers. 45 .
6 times the smaller. Eleven coins.
greater
is
. and the Find the numbers. 11 x = 5.
Uniting.
x x
+
= 2(3 x = 6x
25
25). then..LINEAR EQUATIONS AND PROBLEMS
Therefore. 3 x = 45.. Check. consisting of half dollars and dimes.
Simplifying. 15 + 25 = 40.
2. have a value of $3.5 x .
x
from
I. but 40 = 2 x 20.
Never add the number number of yards to their
Ex. x = 15.
1. A's number of marbles.
60.
(Statement II)
Qx
. cents..10. .$3.
50.
*
'
. The value of the half
:
is 11.
we
express the statement II in algebraic symbols. 3.
Simplifying.
and twice the altitude of Mt.
United
States. the night in
Copenhagen
lasts 10 hours
longer than the day.
9.68
4. Find their ages. How many volcanoes are
in the
8.
Find
Find two consecutive numbers whose sum equals 157. and B's age is as below 30 as A's age is above 40.
?
Two
vessels contain together 9 pints. and four times the former equals five times the latter.
What
is
the altitude of each
mountain
12. How many inches are in each part ?
15. and in 5 years A's age will be three times B's.
and twice the greater exceeds Find the numbers.
ELEMENTS OF ALGEBRA
One number
is
six
times another number.
of volcanoes in
Mexico exceeds the number
of volcanoes in the United States by 2.
3 shall be equal to the other increased by
10.000
feet.
Two numbers
The number
differ
by
39.
How many
hours does the day last ?
. Mount Everest is 9000 feet higher than Mt.
A's age is four times B's.
On December
21.
cubic foot of iron weighs three times as much as a If 4 cubic feet of aluminum and
Ibs.
6.
it
If the smaller
one contained 11 pints more. Everest by 11.
find the
weight of a cubic
Divide 20 into two parts. McKinley.. McKinley exceeds the altitude of
Mt.
would contain three times as
pints does each contain ?
much
13.
tnree times the smaller by 65.
5. the number. Twice 14. and the
greater increased by five times the smaller equals 22.
7.
and in Mexico
?
A
cubic foot of aluminum. What are their ages ?
is
A A
much
line 60 inches long is divided into two parts.
2 cubic feet of iron weigh 1600 foot of each substance. the larger part exceeds five times the smaller part by 15 inches.
as the larger one. one of which increased by
9.
How many
14 years older than B.
11.
they would have 3. B. Tf it should be difficult to express the selected verbal state-
ment
directly in algebraical symbols. III. or 66 exceeds 58 by 8. II.
Ex. If A and B each gave $5 to C.
the
the
number
of dollars of dollars of dollars
A
B
C
has. then
three times the
money by
I. then three times the sum of A's and B's money would exceed C's money by as much as A had originally. The solution gives
:
3x
80
Check.
number
of dollars of dollars
B
C
had.
A
and B each gave $ 5
respectively.
I. number of dollars A had. original amount.
If
4x
= 24.
8(8
+ 19)
to C. and 68. try to obtain
it
by a
series of successive steps.
.
69
If a
verbal statements must be given. and the other
of x
problem contains three unknown quantities.
are
:
C's
The three statements
A. The third
verbal statement produces the equation.
1.
first
According to
3 x
number
number
and according
to
80
4
x
=
the
express statement III by algebraical symbols. let us consider the words ** if A and B each gave $ 5 to C. and B has three as A. and C together have $80.LINEAR EQUATIONS AND PROBLEMS
99. 4 x = number of dollars C had after receiving $10.
has.
Let
x
II. B. bers is denoted by x. 19.
times as
much
as
A.
sum of A's and B's money would exceed much as A had originally. = number of dollars B had after giving $5."
To
x
8x
90
= number of dollars A had after giving $5.
number
had. If A and B each gave $5 to C. = 48.
has.
5
5
Expressing in symbols Three times the sum of A's and B's money exceeds C's money by A's 3 x ( x _5 + 3z-5) (90-4z) = x. three One of the unknown num-
two are expressed in terms by means of two of the verbal statements. B has three times as much as A. x = 8. and C together have $80.
A
and the number of sheep was twice as large as the number How many animals of each kind did he buy ?
of horses and cows together.
and. = the number of dollars spent for horses. + 35 x 4. 9 -5 = 4 . the third five times the first. 2 (2 x -f 4) or 4 x
Therefore.
85 (x 15 (4 x
I
+ 4)
+
8)
= the number of sheep. each horse costing $ 90. 4 x -f 8 = 28. and the sum of the
. = the number of dollars spent for cows. according to III.
x
-j-
= the
number of horses.
The
I.
Uniting.140 + (50 x x 120 = 185. The number of cows exceeds the number of horses by 4. x -f 4 = 9. 90
may
be written.
first. number of sheep. according to II. number of horses.
+ 35 (x +-4) -f 15(4z-f 8) = 1185. 28 2 (9 5). number of cows.
The total cost equals $1185. and each sheep $ 15.
x 35
-f
+
=
+
EXERCISE
1.
+
8
90 x
and.70
ELEMENTS OF ALGEBRA
man spent $1185 in buying horses.
1 1
Check.
Find three numbers such that the second is twice the 2.
first
the third exceeds the second by and third is 20. The number of sheep is equal to twice tho number of horses and
x 4
the
cows together. and 28 sheep would cost 6 x 90 -f 9 + 316 420 = 1185. cows.
III. number of cows. = the number of dollars spent for sheep
Hence statement
90 x
Simplifying.
37
Find three numbers such that the second is twice the first.
x
Transposing.
28 x 15 or 450
5 horses. x = 5. 2.
Dividing. 90 x -f 35 x + GO x = 140 20 + 1185. 185 a = 925. 9 cows. The number of cows exceeded the number of horses by
4. and the difference between the third and the second is 15
2. each cow $ 35. and Ex.
three statements are
:
IT. Let
then. sheep.
13. and the third part exceeds the second by 10.
A
12.
"Find three
is 4. what is the length of each?
has 3.
-
4.
is five
numbers such that the sum of the first two times the first. If twice
The sum
the third side.000 more inhabitants than Philaand Berlin has 1.
first. what is the population of each city ?
8. increased by three times the second side. men.000 more than Philadelphia (Census 1905).
v
-
Divide 25 into three parts such that the second part first. In a room there were three times as many children as If the number of women.
twice the
6. the copper.
what are the three angles ?
10. The gold.
first.
v
.
A
is
Five years ago the What are their ages ?
C.
twice as old as B. women. and children together was 37.LINEAR EQUATIONS AND PROBLEMS
3. and the third exceeds the
is
second by
5. equals 49 inches. and is 5 years younger than sum of B's and C's ages was 25 years.
7.
9.
If the second angle of a triangle is 20 larger than the and the third is 20 more than the sum of the second and
first. the second one is one inch longer than the first. how many children
were present ?
x
11.000.
the
first
Find three consecutive numbers such that the sum of and twice the last equals 22. and the sum of the first and third is 36.
The
three angles of any triangle are together equal to
180.
the third
2.
Find three consecutive numbers whose sum equals
63.000.
71
the
Find three numbers such that the second is 4 less than the third is three times the second.
New York
delphia.
and
of the three sides of a triangle is 28 inches. and 2 more men than women. and the pig iron produced in one year (1906) in the United States represented together a value
. If the population of New York is twice that of Berlin.
and quantities
area.000.
and Massachusetts has one more than California and Colorado If the three states together have 31 electoral votes. together.e. number of miles A
x
x
walks.
start at the same hour from two towns 27 miles walks at the rate of 4 miles per hour.
Find the value of each.
how many
100.
Hence
Simplifying.
14. = 5.
. then x 2 = number of hours B walks.
The copper had twice
the value of the gold. or time. 3 and 4. statement "A and B walk from two towns 27 miles apart until they meet " means the sum of the distances walked by A and B equals 27 miles. it is frequently advantageous to arrange the quantities in a systematic manner.000 more than that
the copper. number of hours.
B
many
miles does
A
walk
?
Explanation. and distance.
= 35.
A and B
apart.000. First fill in all the numbers given directly. but stops 2 hours on the way.
7
Uniting.g.
3z + 4a:-8 = 27. of 3 or 4 different kinds. i. Let x = number of hours A walks. speed. After how many hours will they meet and how
E. and A walks at the rate of 3 miles per hour without stopping. we obtain 3 a. width. Since in uniform motion the distance is always the product of
rate
and time. such as length.
California has twice as
many
electoral votes as Colorado.000.
of
arid the value of the iron
was $300.
Dividing. 8 x = 15.
has each state
?
If the example contains Arrangement of Problems.72
of
ELEMENTS OF ALGEBRA
$ 750.
3x
+
4 (x
2)
=
27. and 4 (x But the 2) for the last column.
How many pounds of each kind did he buy ?
8. were increased by 3 yards.
invested at 5 %. how much did each cost per yard ?
6. and the cost
of silk
of the auto-
and 30 yards of cloth cost together much per yard as the cloth. and in order to raise the required sum each of the remaining men had to pay one dollar more.
Find the share of each. but as two of them were unable to pay their share. A man bought 6 Ibs.
Twenty men subscribed equal amounts
of
to raise a certain
money. and the sum Find the length of their areas is equal to 390 square yards.
Six persons bought an automobile.
A
sets out later
two hours
B
. and its width decreased
by 2 yards. paid 24 ^ per pound and for the rest he paid 35 ^ per pound.
sions of the field.
A sum
?
invested at 4 %.
Ten yards
$
42. but four men failed to pay their shares. and a second sum.
A
If its length
rectangular field is 2 yards longer than it is wide. The second is 5 yards longer than the first. of coffee for $ 1.
If the silk cost three times as
For a part he 7. and how far will each then have traveled ?
9. How much did each man subscribe ?
sum
walking at the rate of 3 miles per hour.55. twice as large. and follows on horseback traveling at the rate of 5 miles per hour.
as a
4.
A
of each. What are the
two sums
5.
mobile. the area would remain the same. together bring $ 78 interest.
3.
2.
sum $ 50
larger invested at 4
brings the same interest Find the first sum. After how many hours will B overtake A.74
ELEMENTS OF ALGEBRA
EXERCISE
38
rectangular field is 10 yards and another 12 yards wide.
1.
Find the dimen-
A
certain
sum
invested at 5
%
%. each of the others had to pay
$ 100 more.
walking at the same time in the same If A walks at the rate
of 2
far
miles per hour. how must B walk before he overtakes A ?
walking at the rate of 3 miles per hour. and B at the rate of 3 miles per hour. Albany and travels toward New York at the rate of 30 miles per hour without stopping.
A
sets out
two hours
later
B
starts
New York to Albany is 142 miles. traveling by coach in the opposite direction at the rate of 6 miles per hour. how many miles from New York will they meet?
X
12.LINEAR EQUATIONS AND PROBLEMS
v
75
10. but
A has
a start of 2 miles.
The
distance from
If a train starts at
. After how many hours.will they be 36 miles apart ?
11. and from the same point. and another train starts at the same time from New York traveling at the rate of 41 miles an hour.
A
and
B
set out
direction.
The prime
factors of 10 a*b are 2. if it contains
no other
factors (except itself
and unity)
otherwise
.
it is
composite.
a2
to 6. at this
6
2
. if.
104.
a-
+
2 ab
+ 4 c2
.CHAPTER
VI
FACTORING
101.
a factor of a 2
A
factor is said to be prime. 6.
+ 62
is
integral with respect to a. a. it contains no indicated root of this letter
. if it does contain
some indicated root of
.
\-
V&
is
a
rational with respect to
and
irrational with respect
102. but fractional with respect
103. consider
105.
The
factors of
an algebraic expression are the quantities
will give the expression.
76
.
which multiplied together
are considered factors. 5. An expression is integral with respect to a letter.
-f-
db
6
to b.
we
shall not. as.
An
expression
is integral
and rational with respect
and rational.
irrational.
vV
.
a.
expression is rational with respect to a letter.
An
after simplifying.
J Although Va'
In the present chapter only integral and rational expressions
b~
X
V
<2
Ir
a2
b'
2
2
?>
.
this letter.
stage of the work. if this letter does not occur in any denominator. if it is integral to all letters contained in it.
110.
Ex.9 x if + 12 xy\
2
The
greatest factor
common
2
to all terms
flcy*
is
8
2
xy'
.
109.9 x2 y 8 + 12
3 xy
-f
by
3
xy\
and the quotient
But.
Factor G ofy 2
. 2.
107.
77
Factoring
is
into its factors.
2 4 x + 3) is factored if written (x' would not be factored if written x(x and not a product.3 sy + 4 y8). .
55.
2.
in the
form
4)
+3.FACTORING
106.
factors of 12
&V
is
are 3. or that a
=
6)
(a
= a .
.
1.
it
follows
that a 2
.
TYPE
I.9 x2^ + 12 sy* = 3 Z2/2 (2 #2 . since (a + 6) (a 2 IP factored.)
Ex. x. for this result is a sum. 2.
it fol-
lows that every method of multiplication will produce a method
of factoring.
y.3 6a + 1).
x.
Since factoring
the inverse of multiplication.62
can be
&).
or
Factoring examples may be checked by multiplication by numerical substitution. 8) (s-1).
POLYNOMIALS ALL OF WHOSE TERMS CONTAIN A COMMON FACTOR
(
mx + my+ mz~m(x+y + z).
Divide
6
a% .
E.
Factor
14 a*
W-
21 a 2 6 4 c2
+ 7 a2 6
2
c2
7
a2 6 2 c 2 (2 a 2
.
01.g.
The factors
of a
monomial can be obtained by inspection
2
The prime
108.
An
the process of separating an expression expression is factored if written in the
form of a product.62 + &)(a 2
. dividend
is
2 x2
4
2
1/
.
?/.
It (a.
Hence
6 aty 2
= divisor x quotient.
a).
2.
3.11 a + 30. but only in a limited number
of ways as a product of two numbers.
as p.
EXERCISE
Besolve into prime factors :
40
4.
m -5m + 6. or 11 and 7 have a sum equal to 4.
5.11) (a
+
7).
determine whether
In solving any factoring example.6 = 20. of this type. it is advisable to consider the factors of q first.FACTORING
Ex.
If
30 and whose
sum
is
11 are
5
a2
11
a = 1.30 = (a . the student should first all terms contain a common monomial factor. the two numbers
have opposite
signs. If q is positive.4 . a 2 .
11 a2 and whose sum The numbers whose product is and a.
2
6.
.
79
Factor a2
-4 x .
tfa2 -
3.
and the greater one has the same sign
Not every trinomial
Ex.G) = . Hence z6 -? oty+12 if= (x -3 y)(x*-4 y ).
Ex.
.
Factor a2
.
Therefore
Check.11.1 1 a
tf
a 4. + 30 = 20.4 x .11 a
2
.
11
7.5) (a 6). the two numbers have both the same sign as p.
Since a number can be represented in an infinite number of ways as the sum of two numbers.
Ex.1 afy 8 The two numbers whose product is equal to 12 yp and whose sum equals 3 8 7 y are -4 y* and -3 y*.
Factor
+ 10 ax .5) (a .77 =
(a. 2 11 a?=(x + 11 a) (a.
+
112..
or
77
l. If q is negative.
4.
but of these only
a:
Hence
2
. Hence fc -f 10 ax
is
10 a are 11 a
-
12 /.
. can be factored. Factor x? .
is
The two numbers whose product and -6. however.
We may consider
1. or 7 11. and (a .
77 as the product of 1 77.
and that they must be negative. 64 may be considered the
:
product of the following combinations of numbers 1 x 54. none of the binomial factors can contain a monomial factor.83 x
-f-
54.e-5
V A
x-1
3xl \/ /\
is
3
a. If py? -\-qx-\-r does not contain any monomial factor.5 . X x 18. or
G
114. 27 x 2. 9 x 6. exchange the
signs of the second terms of the factors. Hence only 1 x 54 and 2 x 27 need
be considered.1).
.31 x
Evidently the
last
2
V A
6. the signs of the second terms are minus.17 x
2o?-l
V A
5
-
13 a
combination
the correct one. viz. and r is negative. 2.5) (2 x . 54 x 1. and after a little practice the student possible should be able to find the proper factors of simple trinomials
In actual work
at the first trial.
The work may be shortened by the
:
follow-
ing considerations
1. then the second terms of
have opposite signs. 18 x 3. 2 x 27.
3.
a. we have to reject every combination of factors of 54 whose first factor contains a 3.
sible
13 x
negative. but the opposite sign. 3 x and x.
Factor 3 x 2
.
all
it is not always necessary to write down combinations. Since the first term of the first factor (3 x) contains a 3. If p is poxiliw. the second terms of the factors have same sign as q.
If
the factors
a combination should give a sum of cross products.FACTORING
If
81
we consider that the
factors of -f 5
as
must have
is
:
like signs. all pos-
combinations are contained in the following
6x-l
x-5 .13 x + 5 = (3 x .
Ex.
.
The
and
factors of the first term consist of one pair only.
11 x
2x.
the
If p and r are positive. 6 x 9. which has the same absolute value as the term qx.
5.
are prime can be found by inspection.
33
2
7
3
22 3 2
. of (a
and (a
+
fc)
(a
4
is
(a
+ 6)
2
. F.
6.
3. C.
24
s
. C.
F. of 6 sfyz.
II
2
. The H. of
:
48
4.
the algebraic factor of highest degree common expressions to these expressions thus a 6 is the II.
13 aty
39 afyV. C.CHAPTER
VII
HIGHEST COMMON FACTOR AND LOWEST COMMON MULTIPLE
HIGHEST COMMON FACTOR
120.
C. and GO aty 8 is 6 aty.
Two
common
factor except unity
The H. If the expressions have numerical coefficients.
8
. F. of
aW.
-
23 3
.
+
8
ft)
and
cfiW is
2
a 2 /) 2
ft)
.
5
s
7
2
5. and prefix it as a coefficient to H.
2.
.
2
2
.
The student should note
H. C. of the algebraic expressions. C. F.
expressions which have no are prime to one another. of a 7 and a e b 7
.
121.
89
. F.
EXERCISE
Find the H.
C. F. F.
The
highest
is
common
factor (IT.
5
2
3
. C. of
two or more monomials whose factors
.
C. The H. find by arithmetic the greatest common factor of the coefficients. F.) of
two or more
.
15
aW. of a 4 and a 2 b is a2
The H. aW.
3
.
54
-
32
.
5
7
34 2s
. C. Thus the H. F.
122. F. 12 tfifz.
25
W. is the lowest
that the power of each factor in the power in which that factor occurs in any
of the given expressions.
M. of tfy and xy*. C.
M. C.
= (a -f
last
2
&)'
is
(a
-
6) .
NOTE. C.
&)
2
M. C. of as -&2 a2 + 2a&-f b\ and 6-a. C.
of 12(a
+
ft)
and (a
+ &)*( -
is
12(a
+ &)( .
M of the algebraic expressions. M.
C.
a^c8
3
. ory is the L.
. C.(a + &) 2 (a
have the same absolute value.
Find the L.
etc.
M. find by arithmetic their least common multiple and prefix it as a coefficient to the L. If the expressions have a numerical coefficient.
2
The The
L.
Common
125. thus.
A
common
remainder.
of 4 a 2 6 2 and 4 a 4
-4 a 68
2
. 60
x^y'
2
.
126.LOWEST COMMON MULTIPLE
91
LOWEST COMMON MULTIPLE
multiple of two or more expressions is an which can be divided by each of them without a expression
124.
2. L. M. M.
The
L.
4 a 2 &2
_
Hence.6 3 ).
1. C.
128. To find the L. Obviously the power of each factor in the L. M.6)2. C. is equal to the highest power in which it occurs in any of the
given expressions.
.
of 3
aW.
Ex. two lowest common multiples. of several expressions which are not completely factored.
6
c6 is
C a*b*c*.M.
Find the L. M.
Hence the L.
127.
2 multiples of 3 x
and 6 y are 30 xz y. each set of expressions has
In example ft).
=4 a2 62 (a2 .
The
lowest
common
multiple (L.C.
300 z 2 y. resolve each expression into prime factors and apply the method for monomials. but opposite
. L. of the
general.
Ex.M.C.) of
two or more
expressions is the common multiple of lowest degree. which
also
signs. C.
etc. Thus. If both terms of a fraction are multiplied or divided by the same number) the value of the fraction is not altered.
Reduce
~-
to its lowest terms.
a?.
and
i
x mx = my y
terms
A
1. thus -
is identical
with a
divisor b the denominator.
an indicated quotient.
Remove
tor. however.
fraction
is
in its lowest
when
its
numerator
and
its
denominator have no
common
factors.
131. the value of a fraction is not altered by multiplying or dividing both its numerator and its denominator by the same number. as 8.
rni
Thus
132.
The dividend a is called the numerator and the The numerator and the denominator
are the terms of the fraction.CHAPTER
VIII
FRACTIONS
REDUCTION OF FRACTIONS
129. but we
In arithmetic.
TT
Hence
24
2 z = --
3x
.
successively all
2
j/' .ry ^
by
their H.
and denominators are considered.
Ex. F.
All operations with fractions in algebra are identical
with the corresponding operations in arithmetic.
common
6
2
divisors of
numerator and denomina-
and z 8
(or divide the terms
.
a b
= ma
mb
.
130.
A
-f-
fraction is
b. C. only positive integral numerators shall assume that the
all
arithmetic principles are generally true for
algebraic numbers.
the product of two fractions is the product of their numerators divided by the product of their denominators.
1).
and the terms of
***. we have
(a
+ 3) (a -8) (-!)'
NOTE.
-
by 4
6' .M.-1^22
' .
1.
and
Tb reduce fractions to their lowest common denominator.
M.
To reduce
to a fraction with the
denominator 12 a3 6 2 x2 numerator
^lA^L O r 2 a 3
'
and denominator must be multiplied by
Similarly. and 6rar 3 a? kalr
.
Reduce -^-.3)O -
Dividing this by each denominator.
multiply each quotient by the corresponding numerator.by 3 ^
A
2
' .
Ex
-
Reduce
to their lowest
common
denominator.M.3) (-!)'
=
. and
135.r
2
2
.
mon
T denominator. C.
we may extend this method
to integral expressions.C.
by any quantity without altering the value of the fraction. of the denominators for the common denominator.
.
TheL.
we have
the quotients (x
1).
^
to their lowest
com-
The
L. Divide the L. C.C.
2>
.~16
(a
+ 3) (x.
.
3 a\ and 4
aW
is
12 afo 2 x2 .
.
Since a
(z
-6 + 3)(s-3)O-l)'
6a. =(z
(x
+ 3)(z.
we have
-M^.D. take the L.
-
of
//-*
2
. by the denominator of each fraction. multiplying the terms of
22
.
Multiplying these quotients by the corresponding numerators and writing the results over the common denominator.96
134.
and
(a-
8). we may use the same process as in arithmetic for reducing fractions to the lowest
common
denominator.
ELEMENTS OF 'ALGEBRA
Reduction of fractions to equal fractions of lowest common Since the terms of a fraction may be multiplied
denominator.
Ex.
+
3).
or.
2.
2
a
Ex.g. Common factors in the numerators and the denominators should be canceled before performing the multiplication.
we may extend any
e.
!.
Since -
= a.)
Ex. multiply the
142.
-x
b
c
=
numerator by
To multiply a fraction by an
that integer. each
numerator and denomi-
nator has to be factored.
Simplify 1 J
The
expreeaion
=8
6
.102
ELEMENTS OF ALGEBRA
MULTIPLICATION OF FRACTIONS
140.
fractions to integral numbers. Fractions are multiplied by taking the product of tht numerators for the numerator. (In
order to cancel
common
factors.
expressed in symbols:
c
a
_ac b'd~bd'
principle proved for
b
141. and the product of the denominators for the
denominator.
integer.
F J Simplify
.
Divide X-n?/
.
144.
1.104
ELEMENTS OF ALGEBRA
DIVISION OF FRACTIONS
143.
The reciprocal of ?
Hence the
:
+*
x
is
1
+ + * = _*_. and the principle of division follows
may
be expressed as
145.
The The
reciprocal of a
is
a
1
-f-
reciprocal of J
is
|
|.
* x* -f xy 2
by
x*y
+y
x'
2
3
s^jf\ =
x'
2
x*
. To divide an expression by a fraction.
The
reciprocal of a
number
is
the quotient obtained by
dividing 1
by that number. To divide an expression by a fraction. invert the divisor and multiply it by the dividend.
8
multiply
the
Ex. x a + b
obtained by inverting
reciprocal of a fraction
is
the fraction. expression by the reciprocal of the fraction
. Integral or mixed divisors should be expressed in fractional form before dividing. :
a 4-1
a-b
* See page 272.
1.180..20
C.
C
is
the circumference of a circle whose radius
R.
100
C. 2. When between 3 and 4 o'clock are the hands of
a clock together
?
is
At
3 o'clock the hour hand
15 minute spaces ahead of the minute
:
hand.
. = 16^. A can do a piece of work in 3 days and B in 2 days.
~^ = 15
11 x
'
!i^=15.
is
36.
12.
of minute spaces the
hour hand moves
Therefore x
~ = the number of minute spaces the minute hand moves
more than the hour hand.
Multiplying by
DividingFind
R in terms of C and
TT..
Ex. = the number of minute spaces the minute hand moves
over.
. Ex. hence the question would be formulated After how many minutes has the minute hand moved 15 spaces more than the hour hand ?
Let then
x x
= the required number of minutes after 3 o'clock.
and
12
= the number
over.114
35.
A would do
each day ^ and
B
j.minutes after x=
^
of
3 o'clock.
x
Or
Uniting. 2 3
.
PROBLEMS LEADING TO FRACTIONAL AND LITERAL EQUATIONS 152.
then
= 2 TT#. In how many days can both do it working together ?
If
we denote
then
/-
the required
number
by
1.
days by x and the piece of work while in x days they would do
respectively
ff
~ and and hence the sentence written in algebraic symbols ^.
3. u The accommodation train needs 4 hours more than the express train. the required
number
of days.
Solving.
Explanation
:
If
x
is
the rate of the accommodation train.
32
x
= |. what is
the rate of the express train
?
180
Therefore.
= the
x
part of the
work both do
one day.
But
in
uniform motion Time
=
Distance
.
in
Then
Therefore.
the rate of the express train. The speed of an express train is $ of the speed of an If the accommodation train needs 4 accommodation train. then
Ox
j
5
a
Rate Hence the rates can be expressed.
Ex.
Clearing.
180
Transposing." gives the equation /I). 4x = 80. hours more than the express train to travel 180 miles.
= 100 + 4 x.
fx
xx*
=
152
+4
(1)
Hence
=
36
= rate
of express train."
:
Let
x -
= the
required
number
of days. and the statement.
or 1J.
How much money had he at first?
12
left
After spending ^ of his
^ of his money and $15.
Find A's
8.
is oO.
its
Find the number whose fourth part exceeds part by 3.
Two numbers
differ
l to s of the smaller. How
did the
much money
man
leave ?
11.
by 3. and of the father's age. which was $4000.
of his present age.
A man left ^ of his property to his wife.
ceeds the smaller by
4.
is
equal
7. a man had How much money had he
at first?
.
3.
Find two consecutive numbers such that
9.
and 9
feet above water.
money and $10.
ex-
What
5.
fifth
Two numbers
differ
2. to his son. and found that he had \ of his original fortune left.
Find a number whose third and fourth parts added
together
2.116
ELEMENTS OF ALGEBRA
EXERCISE
60
1.
Twenty years ago A's age was |
age. one half of What is the length
of the post ?
10
ter.
make
21. and J of the greater Find the numbers. -|
Find their present ages. to his daughand the remainder. A man lost f of his fortune and $500.
9
its
A
post
is
a fifth of
its
length in water.
J-
of the greater
increased by ^ of the smaller equals
6. and one half the greater Find the numbers.
The sum
10 years hence the son's age will be
of the ages of a father and his son is 50.
length in the ground.
are the
The sum of two numbers numbers ?
and one
is
^ of the other.
by 6.
investments.
air.
At what time between 4 and
(
5 o'clock are the hands of
a clock together?
16.
and losing
1-*-
ounces when weighed in water?
do a piece of work in 3 days.)
22. and after traveling 150 miles overtakes the accommodation train. 1.
How much money
$500?
4%. and B In how many days can both do it working together
in
?
12 days. A can do a piece of work in 4 clays.
152.
A has invested capital
at
more
4%. Ex. what is the rate of the express train? 152.
A man
has invested
J-
of his
money
at
the remainder at
6%.
A can
A
can do a piece of work in 2 days. If the accommodation train needs 1 hour more than the express train to travel 120 miles. Ex.FRACTIONAL AND LITERAL EQUATIONS
13. In how many days can both do it working together ? ( 152. what is the
14.
after
rate of the latter ?
15. and has he invested if
his animal interest therefrom is
19. If the rate of the express train is -f of the rate of the accommodation train. ^ at 5%. 3.)
At what time between 7 and 8
o'clock are the
hands of
a clock together ?
17.
117
The speed
of an accommodation train
is
f of the speed
of an express train. An ounce of gold when weighed in water loses -fa of an How many ounce. ounces of gold and silver are there in a mixed mass weighing
20 ounces in
21. and an ounce of silver -fa of an ounce.) (
An express train starts from a certain station two hours an accommodation train.
?
In
how many days can both do
working together
23. Ex. 2. and B in 4 days.
. and
it
B in 6 days.
at 4J % and P> has invested $ 5000 They both derive the same income from their How much money has each invested ?
20.
At what time between 7 and
8 o'clock are the hands of
?
a clock in a straight line and opposite
18.
Then
ft
i.
they can both do
in 2 days. Ex.
is 42. n x
Solving. therefore.
we
obtain the equation
m m
-.
make
it
m
6
A can do this work in 6 days Q = 2.
.009 918.
ELEMENTS OF ALGEBRA
The
last three questions
and their solutions differ only two given numbers. In how
in the numerical values of the
:
many days
If
can both do
we
let
x
= the
it working together ? required number of days.
To
and
find the numerical answer. Hence. by taking for these numerical values two general algebraic numbers. A in 6.
26. Answers to numerical questions of this kind may then be found by numerical substitution. 3. e. Find three consecutive numbers whose sum equals m. .=
m
-f-
n
it
Therefore both working together can do
in
mn
-f-
n
days. 2.
B in 5 The problem to be solved. is A can do a piece of work in m days and B in n days.414.
Find three consecutive numbers whose sum
Find three consecutive numbers whose sum
last
:
The
two examples are
special cases of the following
problem 27. and n = 3. B in 16. if
B
in 3 days. and apply the
method of
170.
25. A in 4.g. m and n. is 57.= -. Find the numbers if m = 24 30. B in 12. A in 6.118
153. B in 30.e.
6
I
3
Solve the following problems
24. it is possible to solve all examples of this type by one example.
.
Find the side of the square. After how many hours do they rate of n miles per hour. 2 miles per hour.
.
119
Find two consecutive numbers the difference of whose
is 11.
34.
meet. (c) 16.
same hour from two towns. 3 miles per hour.721. 3J miles per hour.
4J-
miles per hour. If each side of a square were increased by 1 foot. d miles the first traveling at the rate of m.
The
one:
31.001. solve the following ones Find two consecutive numbers the difference of whose squares
:
find the smaller number.000. (d) 1. and how many miles does each travel ?
32. two pipes together ? Find the numerical answer. respectively. the second at the apart.
by two pipes in m and n minutes In how many minutes can it be filled by the respectively.
is ?n
.
squares
29. the rate of the
first.
A cistern can
be
filled
(c)
6 and 3 hours. After how many hours do they meet. (b) 35 miles.
and
the rate of the second are. if m and n are. and how many miles does each travel ? Solve the problem if the distance. the area would be increased by 19 square feet. (a) 20 and 5 minutes. 5 miles per hour.
squares
30.FRACTIONAL AND LITERAL EQUATIONS
28. (b) 149. (b) 8 and 56 minutes. 2 miles per hour.
33. the
Two men start at the same time from two towns.
:
(c)
64 miles. respectively (a) 60 miles. 88 one traveling 3 miles per hour.
Find two consecutive numbers -the difference of whose
is 21.
last three
examples are special cases of the following
The
difference of the squares of
two consecutive numbers
By using the result of this problem.
is (a)
51.
Two men
start at the
first
miles
apart. and the second 5 miles per hour.
The
first
156.
b
is
a
Since a ratio
a
fraction.
Thus the
written a
:
ratio of a
b
is
.
1.g. the denominator
The
the
157. 158."
we may
write
a
:
b
= 6.
b.
terms are multiplied or divided by the same number. b is the consequent.
Simplify the ratio 21 3|.
The
ratio of
first
dividing the
two numbers number by the
and
:
is
the quotient obtained by
second.
Ex. 6 12 = .CHAPTER X
RATIO AND PROPORTION
11ATTO
154.
:
A somewhat shorter way
would be to multiply each term by
120
6. a ratio
is
not changed
etc.
A
ratio
is
used to compare the magnitude of two
is
numbers.
is
numerator of any fraction
consequent.
b. instead of writing
6 times as large as
?>.
the symbol
being a sign of division. the second
term the consequent.
The
ratio -
is
the inverse of the ratio -.
term of a ratio
a
the
is
is
the antecedent.or a *
b
The
ratio is also frequently
(In most European countries this symbol is employed as the usual sign of division.
antecedent.
:
:
155.
E. all principles
relating
to
fractions
if its
may
be af)plied to ratios. etc.
.5. the antecedent.
In the ratio a
:
ft.) The ratio of 12 3 equals 4.
" a Thus.
proportional between a
and
c.
1.
4|-:5f
:
5.
9.
In the proportion a b
:
=
b
:
c.
$24: $8.
Simplify the following ratios
7. either mean the mean proportional between the first and the last terms.
8^-
hours.
159.
extremes.
16.
5 f hours
:
2.
17.
two
|
ratios.
7|:4 T T
4
terms.
11.
b is the
mean
b.
Transform the following
unity
15.
27 06: 18 a6. the second
and fourth terms of a proportion are the and third terms are the means.
3:4.
61
:
ratios
72:18.
12.
18.
10.
3
8.
62:16.
:
is
If the means of a proportion are equal.
and
c
is
the third proportional to a and
. and c. The last term d is the fourth proportional to a. and the last term the third proportional to the first and second
161.
A
proportion
is
a statement expressing the equality of
proportions.
:
a-y
.
:
ratios so that the antecedents equal
16:64. b.RATIO
Ex.
3:1}.
= |or:6=c:(Z are
The
first
160. a and d are the extremes.
16 x*y
64 x*y
:
24 48
xif.
:
1.
equal
2.
7f:6J.
term
is
the fourth proportional to the
:
In the proportion a b = c c?. b and c the means.
6.
16a2 :24a&.
4. The last
first three.
3.
:
:
directly proportional
may say.
ccm.
ELEMENTS OF ALGEBRA
Quantities of one kind are said to be directly proper
tional to quantities of another kind. and we
divide both
members by
we have
?^~ E. if the ratio of any two of the first kind.
Clearing of fractions. of a proportion. then G ccm. " we " NOTE. are
: : :
inversely proportional.
pro-
portional. If the product of two numbers is equal to the product of two other numbers^ either pair may be made the means. then 8 men can do it in 3 days. = 30 grams 45 grams. and the
other pair the extremes.122
162. of iron weigh .) b = Vac.
t/ie
product of the means
b
is
equal
to the
Let
a
:
=c
:
d.'*
Quantities of one kind are said to be inversely proportional to quantities of another kind.
ad =
be. Instead of u
If 4
or 4 ccm. or 8 equals the inverse ratio of 4 3. Hence the weight of a mass of iron is proportional to its volume.
q~~ n
.
If
(Converse of
nq.30 grams. Hence the number of men required to do some work.
2
165.
164.
If 6 men can do a piece of work in 4 days. In any proportion product of the extremes.
!-.e. is equal to the ratio of the corresponding two
of the other kind. of iron weigh 45 grams.
:
c. 6 ccm.)
mn = pq.
163. and the time necessary to do it. i.__(163. a b
:
bettveen two
numbers
is
equal to
the square root
Let the proportion be
Then Hence
6
=b = ac.
163. if the ratio of any two of the first kind is equal \o the inverse ratio of the corresponding two of
the other kind. 3 4.
The mean proportional
of their product.
briefly.
the area of the larger? the same.126
54. The number of men (m) is inversely proportional to the number of days (d) required to do a certain piece of work. the volume of a
The temperature remaining
body of gas inversely proportional to the pressure. and
the time necessary for it.
What
will be the
volume
if
the pressure
is
12 pounds per square inch ?
.
and the area of the smaller
is
8 square inches.
(d)
The sum
of
money producing $60
interest at
5%. (c) The volume of a body of gas (V) is
circles are to
each
inversely propor-
tional to the pressure (P). and the time.
57. State whether the quantities mentioned below are directly or inversely proportional (a) The number of yards of a certain kind of silk.
A
line 11 inches long
on a certain
22 miles.
and the speed
of the train.
(b)
The time a
The length
train needs to travel 10 miles.
ELEMENTS OF ALGEBEA
State the following propositions as proportions : T (7 and T) of equal altitudes are to each.inches long represents
map corresponds to how many miles ?
The
their radii.
areas of circles are proportional to the squares of If the radii of two circles are to each other as
circle is
4
:
7. and the
:
total cost.
what
58. and the area
of the rectangle.
1
(6) The circumferences (C and C ) of two other as their radii (R and A").
56.
the squares of their radii
(e)
55.
A
line 7^. (e) The distance traveled by a train moving at a uniform rate.
(d)
The
areas
(A and
A') of two circles are to each other as
(R and R'). othei
(a) Triangles
as their basis (b
and
b').
(c)
of a rectangle of constant width. under a pressure of 15 pounds per square inch has a volume of
gas
is
A
16 cubic
feet.
. 4 inches long. = the second number.
Divide 108 into two parts which are to each other
7. 11 x -f 7 x = 108.
:
Ex.
11
x
x
7
Ex.
x=2. 11 x = 66 is the first number so that
Find^K7and BO. When a problem requires the finding of two numbers which are to each other as m n.
Therefore
7
=
14
= AC.
Then
Hence
BG = 5 x.RATIO AND PROPORTION
69.
2 x
Or
=
4.
Hence
or
Therefore
Hence
and
= the first number. AB = 2 x.
as 11
Let
then
:
1. it is advisable to represent these unknown numbers by mx and nx.
127
The number
is
of miles one can see from an elevation of
very nearly the mean proportional between h and the diameter of the earth (8000 miles). x = 6. produced to a point C.000
168.
Let
A
B
AC=1x.
is
A line AB. 18 x = 108.
4
'
r
i
1
(AC): (BO) =7: 5. 2. 7 x = 42 is the second number.
How many
7.
:
Divide 39 in the ratio 1
:
5. 6.
What
are the parts ?
5. 9.
7. How many ounces of copper and zinc are in 10 ounces of brass ?
6.
Divide 20 in the ratio 1 m.
11.
13. Water consists of one part of hydrogen and 8 parts of
If the total surface of the earth
oxygen. The three sides of a triangle are respectively a. 12.
The
total area of land is to the total area of
is
water as
7 18.000.
How many
grams of hydrogen are contained in 100
:
grams
10.
14.
:
197. Brass is an alloy consisting of two parts of copper and one part of zinc.
Gunmetal
tin. and c inches. and 15 inches.
12.128
ELEMENTS OF ALGEBRA
EXERCISE
63
1. If c is divided in the ratio of the other two. How many gen. cubic feet of oxygen are there in a room whose volume is 4500
:
cubic feet?
8.
m
in the ratio x:
y
%
three sides of a triangle are 11. How
The
long are the parts ? 15.
of water?
Divide 10 in the ratio a
b.
:
Divide a in the ratio 3
Divide
:
7.
A line 24 inches
long
is
divided in the ratio 3
5. 2.
3. and the longest is divided in the ratio of the other two.)
000 square miles.
Divide 44 in the ratio 2
Divide 45 in the ratio 3
:
9.
:
4. what are
its
parts ?
(For additional examples see page 279. find the number of square miles of land and of water.
y
(3)
these
unknown numbers can be found.
2 y = . From (3) it follows y 10 x and since
by the same values of x and
to be satisfied
y. y =
5
/0 \ (2)
of values.
Hence. if
. the equation is satisfied by an infinite number of sets Such an equation is called indeterminate.
If
satisfied
degree containing two or more by any number of values of
2oj-3y =
6.-L
x
If
If
= 0.e.
The
root of (4)
if
K
129
. is x = 7. y = 1.y=--|. values of x and y.
An
equation of the
first
unknown numbers can be the unknown quantities. there is only one solution. etc.
However.
the equations have the two values of
y must be equal.-. =.
if
there
is
different relation
between x and
*
given another equation. expressing a y. x = 1.
a?
(1)
then
I.CHAPTER XI
SIMULTANEOUS LINEAR EQUATIONS
169. such as
+ = 10.
Hence
2s -5
o
= 10 _ ^
(4)
= 3. which substituted in (2) gives y both equations are to be satisfied by the same Therefore.
By By
Addition or Subtraction.
viz.
4y
. 3.
Therefore. 6 and 4 x y not simultaneous.
E.
A
system of two simultaneous equations containing two
quantities is solved by combining them so as to obtain
unknown
one equation containing only one
173.
of elimination
most frequently used
II.X. the last set inconsistent.130
170.
ELEMENTS OF ALGEBRA
A
system
of simultaneous equations is
tions that can be satisfied
a group of equa by the same values of the unknown
numbers.
x
-H
2y
satisfied
6 and 7 x 3y = by the values x = I.
172.
Solve
-y=6x
6x
-f
Multiply (1) by
2.
6x
.26. Independent equations are equations representing different relations between the unknown quantities such equations
.
cannot be reduced to the same form. The first set of equations is also called consistent. same relation.
= .
~ 50.
(3)
(4)
Multiply (2) by
-
Subtract (4) from (3).
unknown
quantity.3 y = 80.
ELIMINATION BY ADDITION OR SUBTRACTION
175.24.
Substitution.
The process of combining several equations so as make one unknown quantity disappear is called elimination.
21 y
. for they cannot be satisfied by any value of x and y. 26 y = 60. y
I
171. and 3 x + 3 y =. for they are 2 y = 6 are But 2 x 2.
174. for they express the x -f y 10. Any set of values satisfying 5 x + 6 y = 60 will also satisfy the equation 3 x -f.
to
The two methods
I. y = 2. 30 can be reduced to the same form -f 5 y Hence they are not independent.
are simultaneous equations.
z + x = 2 n.
(1)
100s
+ lOy + z + 396 = 100* + 10y + x.
Let
x
y z
= the
the digit in the hundreds' place.
unknown quantity by
every verbal statement as an equation. 1 digit in the tens place.
Check. the number. and to express
In complex examples. + z = 2p.
and Then
100
+
10 y
+z-
the digit in the units' place.
The
three statements of the problem can
now be
readily expressed in
. and if 396 be added to the number.SIMULTANEOUS LINEAR EQUATIONS
143
x
29.
.
+2+
6
= 8. however.)
it is advisable to represent a different letter. as many verbal statements as there are unknown quantities.
to express
it is difficult
two of the required
digits in
terms
hence we employ 3
letters for the three
unknown
quantities. Problems involving several unknown quantities must contain. The digit in the tens' place is | of the sum of the other two digits. either directly or implied.
+
396
= 521.
# 4. y
31.
M=i.y
125
(3)
The solution of these equations gives x Hence the required number is 125.
. Simple examples of this
kind can usually be solved by equations involving only one
unknown
every
quantity. 1.
2
= 1(1+6).
2
= 6.
y
*
z
30.
Find the number.
1
=
2.
= 2 m. the first and the last digits
will be interchanged.2/
2/
PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS
183.
Ex.
symbols:
x
+
y
+z-
8.
(
99.
x
:
z
=1
:
2.
Obviously
of the other
.
=
l. The sum of three digits of a number is 8.
(1) (2)
12.
xy
a:
2y 4y
2.
Ex.
.
By
expressing the two statements in symbols.
4
x
= 24.
direction.
2. the fraction is reduced to | and if both numerator and denominator of the reciprocal of the fraction be dimin-
ished by one.
ELEMENTS OF ALGE13KA
If both numerator and denominator of a fraction be
.
x
y
= the = the
x
denominator
.
Or
(4)-2x(3). From (3)
Hence xy
Check. 5_
_4_
A.
Find the
fraction. and C travel from the same place in the same B starts 2 hours after A and travels one mile per hour faster than A.
+
I
2
(1)
and
These equations give x
Check.
x 3
= 24.
3+1 5+1
4_2.
increased by one. the fraction
Let and
then y
is
reduced to
nurn orator. y = 3.
= the
fraction.
=
Hence the
fraction
is
f. who travels 2 miles an hour faster than B.
Since the three
men
traveled the
same
distance.
3. = 8. the distance traveled by A.
2. B. C.
3
xand y
I
1
(2)
5.
6
x 4
= 24. x 3x-4y = 12.
we
obtain. starts 2 hours after B and overtakes A at the same How many miles has A then traveled? instant as B.144
Ex.
(3) C4)
=
24 miles.
8
= xy + x xy = xy -f 3 x 2 y = 2.
If
9 be added to the number. and the second one increased by 5 equals twice
number.
number by
the
first
3.
6.
the
number
(See Ex.}.
If the
numerator of a fraction be trebled.
1.
to
L
<>
Find the
If the
numerator and the denominator of a fraction be If 1 be subtracted from increased by 3. the last two digits are interchanged.
part of their difference equals
4. Find the number. Four times a certain number increased by three times another number equals 33. the Find the fraction.
Half the sum of two numbers equals 4. to the number the digits will be interchanged. and the fourth 3.
Five times a certain number exceeds three times another 11.
If 4 be
Tf 3 be
is J.
?
What
9.
The sum
18
is is
and
if
added
of the digits of a number of two figures is 6.
Find the numbers. and
its
denomi-
nator diminished by one. and four times the first digit exceeds the second digit by 3.
added to the numerator of a fraction. if its numerator and its denominator are increased by 1. and twice the numerator What is the fracincreased by the denominator equals 15. and the two digits exceeds the third digit by 3. If the denominator be doubled. Find the fraction. the fraction equals . Find the numbers.SIMULTANEOUS LINEAR EQUATIONS
EXERCISE
70
145
1. its value added to the denominator. both terms.
.
7. Find the numbers. the value of the fraction is fa.
If 27 is
10. Find the number. and the second increased by 2 equals three times the first.
The sum
of the first
sum
of the three digits of a number is 9. the fraction is reduced
fraction.)
added to a number of two digits. and the numerator increased by 4. the digits will be interchanged.
tion ?
8.
2.
183.
5. it is reduced to J. fraction is reduced to \-. A fraction is reduced to J.
bringing a total yearly interest of $530. the rate of interest?
18. respectively ?
16. Two cubic centimeters of gold and three cubic centimeters of silver weigh
together 69 J. the annual interest would be $ 195.
much money
is
invested at
A sum
of
money
at simple interest
amounted
in 6 years
to $8000. and B's age is \ the sum of A's and C's ages.grams. Ten years ago the sum of their ages was 90.
Ten years ago A was B was as
as old as
B
is
old as
will be 5 years hence .
.
and
money and
17. How 6 %. If the rates of interwere exchanged. now.
A
sum
of $10.
partly at
5 %. and the 5% investment brings $15 more interest than the 4 % investment. What was the amount of each investment ?
A man
%
5%. and in 5 years to $1125.
14. Three cubic centimeters of gold and two cubic centimeters of silver weigh together 78 grains. Find the weight of one cubic centimeter of gold and one cubic centimeter of silver. and 4 %.146
ELEMENTS OF ALGEBRA
11. What was the amount of each investment ?
15. the rate of interest ?
What was
the
sum
of
A sum
of
money
at simple interest
amounted
in 2 years
to $090. Find
the rates of interest. Twice A's age exceeds the sum of B's and C's ages by 30. If the sum of
how
old
is
each
now ?
at
invested $ 5000. partly at 5% and partly at 4%.
19. 12.
and 5 years ago
their ages is 55. a part at 6 and the remainder bringing a total yearly interest of $260. A man invested $750.000
is
partly invested at
6%.
in 8 years to $8500. 5 %. and partly at 4 %.
What was
the
sum and
rates
est
The sums of $1500 and $2000 are invested at different and their annual interest is $ 190.
13. Find their present ages. and The 6 investment brings $ 70 more interest than the 5
%
%
4%
investments together.
24. cows.
On
/).
. How many did he sell
of each if the total
number
of animals
was 24?
21.
. what is
that
=
OF. then AD = AF. and F. but if A would double his pace. BE. and $15 for each sheep. B find angles a. If angle ABC = GO angle BAG = 50. c.
BC=7. BD = HE. and their difference by GO . and AC = 5 inches.SIMULTANEOUS LINEAR EQUATIONS
147
20.
1
NOTE. The number of sheep was twice the number of horses and cows together. If one angle exceeds the sum of the other two by 20.
25. $ 50 for each cow. and F. receiving $ 100 for each horse. are taken so
ABC. three
AD = AF. and sheep. and e.
the three sides of a triangle E. angle c = angle d. for $ 740.
In the annexed diagram angle a = angle b.
23.
It takes
A two hours
longer
24 miles.
triangle
Tf
AD. and F '(see diagram). and angle e angle/. ED = BE.
the length of
NOTE. and GE = CF.
Find their
rates of walking. andCL4 = 8. and CE If AB = G inches.
A
r
^
A
circle is inscribed in triangle
sides in D.
points. and CF?
is
a circle
inscribed in the
7<7.
E.
An C touch ing the sides in D.
is
the center of the circum-
scribed circle. BC = 7 inches. A farmer sold a number of horses. and angle BCA = 70. The sum of the 3 angles of a triangle is 180.
Find the parts of the
ABC touching the three sides if AB = 9. he would walk it in two hours less than
than
to travel
B
B. what are the angles of the triangle ?
22.
respectively.
(2.
is the abscissa. B. PM. Abscissas measured to the riyht of the origin.
or its equal
OM. and PJ/_L XX'. then
the position of point is determined if the lengths
of
P
P3f and
185. and whose ordinate is usually denoted by (X ?/). and point the origin..
(2.
(3.
The
abscissa
is
usually denoted by
line XX' is called the jr-axis. (7. jr.
-3).
is
The point whose abscissa is a.
It'
Location of a point.
the ordinate of point P.
2). and PN _L YY'. hence
The
coordinates lying in opposite directions are negative. and ordinates abore the x-axis are considered positive .
?/.
PN are given.
.
186.
PN.
lines
PM
the
and P^V are
coordinates
called
point P.
two fixed straight lines XX' and YY' meet in at right angles. and
respectively represented
Dare
and
by
(3 7 4). the ordinate by ?/.CHAPTER
XII*
GRAPHIC REPRESENTATION OF FUNCTIONS AND
EQUATIONS
184. and r or its equal OA is
.
first
3). YY' they-axis. Thus the points A.
* This chapter
may
be omitted on a
148
reading.
The
of
Coordinates.
(-4.
. (-2. Graphic constructions are greatly facilitated by the use of cross-section paper.
What
is
the locus of
(a?. (-4.2).
0). (0.
71
2).4).
two variable quantities are so related that
changes of the one bring about definite changes of the other. 0).e.
Where do Where do
Where do
all
points
lie
whose ordinates
tfqual
4?
9.
4)
and
(4. which of its coordinates
known ?
13.(!.1).
whose coordinates are given
NOTE. -2).
4.
2.
(-5. 11.3).
(-1. i.)
EXERCISE
1.
What
are the coordinates of the origin ? If
187. -!). 1). (4.
(-3.
=3?
is
If a point lies in the avaxis.
6.
-2).
What
Draw
is
the distance of the point
(3. paper ruled with two sets of equidistant and parallel linos intersecting at right angles. -4). -3).
and measure
their
distance.
6.
1).
the quadrilateral whose vertices are respectively
(4.
2J-).GRAPHIC REPRESENTATION OF FUNCTIONS
The
is
149
process of locating a point called plotting the point.
Plot the points
(6.
8.
Plot the points:
(4.and(l.
12. 3). (0.
Graphs. (4.
3.
(4. 3).
Plot the points: (-4.
Draw
the triangle whose vertices are respectively
(-l.
.
all all
points
points
lie
lie
whose abscissas equal zero ?
whose ordinates equal zero?
y) if y
10.
4).
Plot the points
:
(0. the mutual dependence of the two quantities may be represented
either by a table or
by a diagram. (4. 0). 0). (See diagram on page 151.
4)
from the
origin ?
7.
ically
each representing a temperature at a certain date. or the curved line the temperature. C.. Thus the first table produces 12 points. may be represented graphby making each number in one column the abscissa.
ure the ordinate of F. in like manner the average temperatures for every value of the time. and the corresponding number in the adjacent column the ordinate of a point. however.
A graphic
and
it
impresses upon the eye
all
the peculiarities of
the changes better and quicker than any numerical compilations. we obtain an uninterrupted sequence
etc.150
ELEMENTS OF ALGEBRA
tables represent the average temperature
Thus the following
of
New
volumes
1
Y'ork City of a certain
to 8 pounds.
ABCN
y
the so-called graph of
To
15
find
from the diagram the temperature on June
to be 15
.
By representing
of points.
15.
we meas1
.
188. B.
may be found
on Jan. D.
representation does not allow the same accuracy of results as a numerical table. A. but it indicates in a given space a great many more
facts than a table. 10
.
Thus the average temperature on May
on April 20.
1.
.
from January 1 to December 1. and the amount of gas subjected to pressures from
pound
The same data.
:
72
find approximate answers to the following
Determine the average temperature of New York City on (a) May 1. (c) January 15. Whenever a clear Daily papers represent ecpnoniical facts graphically. the graph
is
applied. (b) July 15.
EXERCISE
From the diagram
questions
1. uses them. the merchant.
.
physician. the
matics. as the prices and production of commodities. the rise and fall of wages. and to deduce general laws therefrom. etc. (d) November 20. concise representation of a
number
of numerical data
is
required. The engineer.
is
10. from what date to what date would it extend ?
If
.
1 ?
does
the
temperature
increase from
11.
How
much. At what date is the average temperature highest the highest average temperature?
?
What What
is
4..?
is
is
the average temperature of
New York
6. 1?
11
0. ?
-
3. (d) 9 0. (1)
10
C.
June
July
During what month does the temperature increase most
?
rapidly
12.
During what month does the temperature change least?
14. At what date is the average temperature lowest? the lowest average temperature ?
5.
Which month
is
is
the coldest of the year?
Which month
the hottest of the year?
16..
When
What
is
the temperature equal to the yearly average of
the average temperature from Sept.152
2. 1 to Oct. ?
9. (freezing
point) ?
7.
on
1 to
the
average.
How much warmer
1 ?
on the average
is it
on July 1 than
on
May
17.
15.
During what months
above 18 C.
ELEMENTS OF ALGEKRA
At what date
(a) G
or dates
is
New York
is
C.
During what month does the temperature decrease most
rapidly ?
13.
When
the average temperature below
C.
is
ture
we would denote the time during which the temperaabove the yearly average of 11 as the warm season.. (c)
the average temperature oi 1 C.
From what
date to what date does the temperature
increase (on the average)?
8.
NOTE.
a temperature chart of a patient.
Represent graphically the populations
:
(in
hundred thou-
sands) of the following states
22.
153
1?
When is the average temperature the same as on April
Use the graphs of the following examples for the solution of concrete numerical examples.09 yards. Construct a diagram containing the graphs of the mean temperatures of the following three cities (in degrees Fahren-
heit)
:
21.GRAPHIC REPRESENTATION OF FUNCTIONS
18.
20.
Draw
. One meter equals 1.
19. in a similar manner as the temperature graph was applied in examples 1-18. transformation of meters into yards.
Hour
Temperature
.
Draw
a
graph for the
23. From the table on page 150 draw a graph representing the volumes of a certain body of gas under varying pressures.
binding.
from
R
Represent graphically the = to R = 8 inches. the value of a of this quantity will change. etc.
x*
x
19.
function
If the value of a quantity changes.
ELEMENTS OF ALGEBRA
If
C
2
is
the circumference of a circle whose radius
is J2.50.) On the same diagram represent the selling price of the books.
+7
If
will
respec-
assume the values 7. x
7 to 9. if he sells 0.
Show
graphically the cost of the
REPRESENTATION OF FUNCTIONS OF ONE VARIABLE
189.
4.
.
2
is
called
x
2 xy
+ 7 is a function of x. gas. 1 to 1200 copies.
29.5
grams.. etc. 3. 2 8 y' + 3 y is a function of x and
y.
A
10 wheels a day.
190. 2 x -f 7 gradually from 1 to 2. Represent graphically the cost of butter from 5 pounds if 1 pound cost $.
26.154
24.
(Assume ir~
all circles
>2
2
. and $.
28.50..
then
C
irJl.. amount to $8.50 per copy
(Let 100 copies = about \.)
T
circumferences of
25. 9. if 1 cubic centimeter of iron weighs 7.
3. if each copy sells for $1.
to
27.
books from
for printing.
to 20 Represent graphically the weight of iron from cubic centimeters.inch.
x increases will change gradually from
13.g. the daily average expenses for rent.
If
dealer in bicycles gains $2 on every wheel he sells.
The
initial cost of
cost of manufacturing a certain book consists of the $800 for making the plates. An expression involving one or several letters a function of these letters. e. represent his daily gain (or loss). Represent graphically the distances traveled by a train in 3 hours at a rate of 20 miles per hour. if x assumes
successively the
tively
values
1. 2 .
2.
(-
2. (1. hence
various values of x
The values of a function for the be given in the form of a numerical table.
is
A
constant
a quantity whose value does not change in the
same discussion.
3
(0.
Graph
of a function. be also represented by a graph.
values of x2
nates are the corresponding i. however. and (3. 4).GRAPHIC REPRESENTATION OF FUNCTIONS
191.
a*. as
1. construct
'. plot points which
lie
between those constructed above. to con struct the graph x of x 2 construct a series of -3 points whose abscissas rep2 resent X) and whose ordi1
tions
.
9).g.2 x
may
4 from x
=
4.
Draw the graph of x2 -f.
.
(1^. 3 50. E.
may
.
may.
it is
In the example of the preceding article.e.1).0). The values of func192. If a more exact diagram
is
required. x a variable.
155
-A
variable is a quantity
whose value changes in the
same
discussion.
etc. to
x = 4.1).
2
(-1. (2.
-J).
1
the points (-3.
To
obtain the values of the functions for the various values of
the
following arrangement
be found convenient
:
.
2). while 7 is a constant.
Ex.
is
supposed to change. for x=l. 4).
9). 2.
Q-. Thus the table on page 1G4 gives the values of the functions x 2 x3 and Vsr.
and join
the
points in order.
If
If
Locating
ing
by a
3) and (4.
7
.
(To avoid
very large ordinatcs. (-2. 2 4 and if y = x -f.
(-3.
if
/*
4
>
1i >
>
?/
=
193. straight line produces the required graph. and joining in order produces
the graph
ABC.20). 4).-. hence two points are sufficient for the construction
of these graphs.
4J..
Ex.
A
Y'
function of the
first
degree is an integral
rational function
involving only
the
power of the variable.
194. etc... y = 6. the scale unit of the ordinatcs is taken smaller than that of the x. and join(0.2 x
.
Draw
y
z x
the graph of
= 2x-3.
r
*/
+*
01
.4).
.
2. j/=-3.
= 0. = 4.156
ELEMENTS OF ALGEBRA
Locating the points(
4. the function
is
frequently represented
by a single letter.
It can be
proved that the
graph is a straight
of a function of the first degree
line.. as y. or ax + b -f c are funclirst
tions of the first degree. Thus in the above example.)
For brevity.
rf
71
. Thus 4x + 7. (4. -1). 5).
25.
y=
formula graphically.24 or x =
P and
Q. then
y = .
.) scale are
expressed in
degrees of the Centigrade (C.
From
grade equal to
(c)
the diagram find the number of degrees of centi-1 F.
to Fahrenheit readings
:
Change
10
C. i.e. then
cXj
where
c is a constant. 32 FShow
any convenient number). Represent 26. it is evidently possible Thus to find to find graphically the real roots of an equation..
that the graph of two variables that are directly proportional is a straight line passing through the origin (assume
for c
27. Therefore x = 1.
A body
moving with a uniform
t
velocity of 3 yards per
second moves in
this
seconds a distance d
=3
1. 14 F.
that
graph with the o>axis...158
24. the abscissas of 3. what values of x make the function x2 + 2x 4 = (see 192). 9 F. If two variables x and y are directly proportional.
ELEMENTS OF ALGEBRA
Degrees of the Fahrenheit
(F. we have to measure the abscissas of the intersection of the
195.where x
c is
a constant..
1
C.24..) scale by the formula
(a)
Draw
the graph of
C = f (F-32)
from
to
(b)
4 F F=l.
if c
Draw
the locus of this equation
= 12.
Ex.
NOTE.
If
x
=
0.
(f
.
Hence. y = -l.1.
?/
=4
AB.
Ex.
?/. because their graphs are straight lines.
.
if
y
=
is
0. locate points
(0.
T
.
199.e.
i.
unknown
quantities. 2).
Hence
if
if
x
x
-
2.
If the given equation is of the we can usually locate two
y.
Thus
If
in
points without solving the equation for the preceding example:
3x
s
.
Hence we may
join (0.
3x
_
4
.
fc
= 3. solve for
?/. Draw the locus of 4 x + 3 y = 12.
1)
and
0).2 y ~ 2. = 0.
4)
and
(2.
first
degree. that can be reduced
Thus
to represent
x
-
-
-L^-
\
x
=2
-
graphically.160
ELEMENTS OF ALGEBRA
GRAPHIC SOLUTION OF EQUATIONS INVOLVING TWO UNKNOWN QUANTITIES
198.
represent graphically equations of the form y function of x ( 1D2). and join the required graph. 0). we can construct the graph or locus of any
Since
we can
=
equation involving two
to the above form.
== 2.
and joining by a straight
line. Represent graphically
Solving for
y ='-"JJ y.
y=
A
and construct
x
(
-
graphically.
Graph
of
equations involving two
unknown
quantities.
4) and
them by
straight line
AB
(3.
y y
2.
X'-2
Locating the points
(2. Equations of the first degree are called linear equations.
produces the
7*
required locus.2.
To
find the roots of
the system.
and CD.15. we obtain the roots.
parallel have only one point of intersection.
3.
203.
equation
x=
By measuring
3. The roots of two simultaneous equations are represented by the coordinates of the point (or points) at which their
graphs intersect.
Since two straight lines which are not coincident nor simultaneous
Ex. viz.
202.
AB
y
= .
The
every
coordinates
of
point in satisfy the equation
(1).
AB
but only one point
in
AB
also satisfies
(2).
Graphical solution of a linear system. linear equations have only one pair of roots.
201.
By
the
method
of
the preceding article construct the graphs
AB
and
and
CD
of
(1)
(2) respectively.GRAPHIC REPRESENTATION OF FUNCTIONS
161
200.57. P.
(2)
.
Solve graphically the equations
:
(1)
\x-y-\. the point of intersection of the coordinate of P.1=0. The coordinates of every point of the graph satisfy the given equation. and every set of real values of x and y satisfying the given equation is represented by a point in
the locus.
0). 3). we of the
+
y*
= 25.
2 equation x
3). 2.
2. 4.
P
graphs meet in two and $.
-
4.
4.e.
intersection.
4.9.
1. Since the two
-
we obtain DE.
4.
Locating two points of equation (2).
In general.
Inconsistent equations.
(1)
(2)
cannot be satisfied by the same values of x and y.0.
Locating the
points
(5. etc. (-2.
Solving (1) for y. parallel graphs indicate inconsistent equations. This is clearly shown by the graphs of (1) arid (2).
5.
4.
4. 0. 3. (4.
3. construct CD the locus of (2)
of intersection. e.
obtain the graph (a circle)
AB C
joining. and
. 1.
4.y~ Therefore. the point
we
obtain
Ex.
4.162
ELEMENTS OF ALGEBRA
graph.
y equals
3.. the graph
of
points
roots.5. i. (-4.
3.
Measuring the coordinates
of P.
Solve graphically the
:
fol-
lowing system
= =
25.
There can be no point of
and hence no
roots.
(1)
(2)
-C. if x equals
respectively
0. 0) and (0.
and
+ 3).g.
x2
. there are two pairs of By measuring the coordinates of
:
P and Q we find
204. and joining by a
straight line. AB the locus of (1). which consist of a
pair of parallel lines.
Using the method of the preceding para.
3x
2 y = -6.5. 5.
V25
5.
. = 0.
The equations
2
4
= 0. they are inconsistent. 0.
or
-3
for
(usually written
3)
.
tity
.
109
.CHAPTER XIV
EVOLUTION
213.
Every odd root of a quantity has
same sign as
and
2
the
quantity. called real
numbers.
= x means
= 6-. it is evidently impossible to express an even root of a negative quantity by Such roots are called imaginary the usual system of numbers. which can be simplified no further. Since even powers can never be negative.
4
4
. and ( v/o* = a. or y
~
3.
27
=y
means
r'
=
27.
Evolution
it is
is the operation of finding a root of a quan the inverse of involution.
V
\/P
214.
V9 = +
3.
numbers. for distinction.
Thus
V^I is an imaginary number.
a)
4
= a4
.
or x
&4 .
\/a
=
x means x n
=
y
?>
a.
for (-f 3) 2
(
3)
equal
0. etc. for (+ a) = a \/32 = 2.
It follows
from the law of signs
in evolution that
:
Any
even root of a positive.
215.
\/"^27=-3.
2. (_3) = -27. quantity
may
the
be either 2wsitive
or negative.
1. and
all
other numbers are.
however.
2ab
.e. it is not known whether the given
expression is a perfect square.2 ab + b
. i.72 aW + 81 &
4
.
second term 2ab by the double of
by dividing the the so-called trial divisor.
a2
+ & + c + 2 a& .2 ac .
#2
a2
-
16. let us consider the relation of a -f. a -f.
12.
ELEMENTS OF ALGEBEA
4a2 -44a?> + 121V2 4a
s
.
.
2
.2 &c.
15.
The
term
a'
first
2
.
10.
term a of the root
is
the square root of the
first
The second term
of the root can be obtained
a. and b (2 a -f b).
The work may be arranged
2
:
a 2 + 2 ab
+ W \a + b
.>
13.
2 2
218.
14. the that 2 ab -f b 2
=
we have then to consider sum of trial divisor 2 a.
a-\-b
is
the root
if
In most cases.
8
.172
7.
and
b.
multiplied by b must give the last two terms of the
as follows
square.b 2 2 to its square. In order to find a general method for extracting the square root of a polynomial.
2
49a 8 16 a 4
9.
+ 6 + 4a&.
the given expression is a perfect square.
mV-14m??2)-f 49. 11.
The square
.
2. the required root
(4
a'2
8a
+
2}.
4 x2
3
?/
8 is
the required square foot. the first term of the answer.
is
As
there
is
no remainder. 6 a. and consider Hence the their sum one term.
.
*/''
. 8 a 2 .
.24 a + 4 -12 a + 25 a8
s
.
1.24 afy* -f 9 tf.
Explanation.
. 1. 8 a 2
-
12 a
+4
a
-f 2.
Ex. 8 /-.
\
24 a 3
4-f
a2
10 a 2
Second remainder.EVOLUTION
Ex.
First trial divisor. 8 a 2
2. We find the first two terms of the root by the method used in Ex.
219.
Arrange the expression according to descending powers root of 10 x 4 is 4 # 2 the lirst term of the root.
by division we
term of the
root.
173
x*
Extract the square root of 1G
16x4
10 x*
__
.
of x. The process of the preceding article can be extended to polynomials of more than three terms.
10 a 4
8
a. 2 Subtracting the square of 4x' from the trinomial gives the remainder '24 x'2 + y.
and so
forth.
Arranging according to descending powers of
10 a
4
a.
.
First complete divisor. By doubling 4x'2 we obtain 8x2 the trial divisor. 8 a 2 Second complete divisor.
. As there is no remainder.
-
24 a
3
+
25 a 2
-
12 a
+4
Square of 4 a First remainder. we obtain the next term of the root 3 y 3 which has to be added to 2 the trial divisor.
double of this term
find the next
is
the
new
trial divisor. Second trial divisor. 24# 2 y 3 by the trial divisor Dividing the first term of the remainder.
Extract the square root of
16 a 4
. Multiply the complete divisor Sx' 3y 3 by Sy 8 and subtract the product from the remainder.
the preceding explanation it follows that the root has two digits. the integral part of the square root of a number less than 100 has one figure.
1.
Ex.000. of a number between 100 and 10.176.
= 80.EVOLUTION
220. Hence the root is 80 plus an unknown number. the first of which is 9 the square root of 21'06'81 has three digits. Hence if we divide the digits of the number into groups.
As
8
x 168
=
1344. two figures.
The
is
trial divisor
=
160. the square root of 7744 equals 88. and the square root of the greatest square in
units.
7744 80 6400
1
+8
160
+ 8 = 168
1344
1344
Since a
2 a
Explanation. then the number of groups is equal to the number of digits in the square root.
square root of arithmetical numbers can be found to the one used for algebraic
Since the square root of 100 is 10..
Find the square root of 7744000 is 100. beginning at the
and each group contains two digits (except the last. and the complete divisor
168. of 10.
2.
a 2 = 6400. Thus the square root of 96'04' two digits. etc. the first of which is 4. the first of which is 8. which may contain one or two).
175
The
by a method very similar
expressions.000 is 1000. and we may apply the method used in algebraic process. of 1.
From
A
will
show the
comparison of the algebraical and arithmetical method given below identity of the methods.
the
consists of
group is the first digit in the root. etc.1344. Therefore 6 = 8. and the first remainder is.
first
.000.
Ex.
Find the square root of 524.
0961
are
'.
Roots of common fractions are extracted either by divid-
ing the root of the numerator by the root of the denominator.
places.
EXERCISE
Extract the square roots of
:
82
.688
4
45 2 70
2 25
508
4064
6168 41)600
41344
2256
222.70
6.10.
3. annex a cipher.GO'61.1T6
221.
ELEMENTS OF ALGEKRA
In marking
off groups in a number which has decimal begin at the decimal point.
Find the square root of
6/. or by transforming the common fraction into a decimal.
we must
Thus the groups
1'67'24.
The groups
of 16724.
in .7 to three decimal places.1 are
Ex. and if the righthand group contains only one digit.
12.
If the hypotenuse
whose angles
a
units of length.
r.
and the sum
The
sides of
two square
fields are as
3
:
5.
:
6.
solve for d.
If a 2 4.
The
two numbers
(See
is
2
:
3. its area contains
=a
2
-f-
b2
.
9
&
-{-
c#
a
x
+a
and
c.
and they con-
tain together 30G square feet.
2.
22
a.
2
.
Find the numbers.
26. 25.b 2 If s
If
=c
. and the two other sides respectively
c
2
contains
c
a and b units. The sides of two square fields are as 7 2.180
on
__!_:L
ELEMENTS OF ALGEBRA
a.
solve for
r.
is
one of
_____
b
The side right angle. and the first exceeds the second by 405 square yards.
27.
If s
= 4 Trr
'
2
.
2
:
3.
2a
-f-
1
23. opposite the right angle is called the hypotenuse (c in the diagram). is 5(5.
Find the side
of each field.
EXERCISE
1. 24.
.
If 2
-f 2 b*
= 4w
2
-f c
sol ve for
m.
3.
and their product
:
150.
A
right triangle is a triangle.
4.
Find
is
the number.
If
G=m m
g
.
.
84
is
Find a positive number which
equal to
its
reciprocal
(
144).
find a in terms of 6
.
28.
108.
29. Find the side of each field.
2
.
solve for v.
'
4.)
of their squares
5.
If 22
= ~^-.
228.
may
be considered one half of a
rec-
square units.
Three numbers are to each other as 1 Find the numbers. then
Since such a triangle
tangle.
A
number multiplied by
ratio of
its fifth
part equals 45.
=
a
2
2
(' 2
solve for solve for
= Trr
.
the radius of a sphere whose surface equals
If the radius of a sphere is r.
and the two smaller
11.
9. we have
of
or
m = |.
.
add
(|)
Hence
2
. passes in t seconds 2 over a space s yt Assuming g 32 feet.
The area $
/S
of a circle
2
.
Find the
sides. and the third side is 15 inches.
The area
:
sides are as 3
4.
The hypotenuse
of a right triangle is 2.)
COMPLETE QUADRATIC EQUATIONS
229.
sides.
24.
The following
ex-
ample
illustrates the
method
or
of solving a complete quadratic
equation by completing the square.
.
the formula
= Trr
whose radius equals r is found by Find the radius of circle whose area S
equals (a) 154 square inches.
radii are as 3
14.
Method
of completing the
square.
-J-
=
12. The hypotenuse of a right triangle is to one side as 13:12.
Two
circles together contain
:
3850 square
feet. To find this term. its surface
(Assume
ir
=
2 .
member can be made a complete square by adding 7 x with another term.
make x2
Evidently 7 takes the place 7x a complete square
to
to
which corresponds
m
2
.)
13.
8. in how many seconds will a body fall (a) G4 feet.
7r
(Assume
and their
=
2 7
2
.
of a right triangle Find these sides. and the
other two sides are as 3
4. Find the unknown sides and the area. A body falling from a state of rest.2
7
.
181
The hypotenuse
of a right triangle
:
is
35 inches.
Find the
radii.
4.
Solve
Transposing. x* 7 x=
10.
Find these
10.
is
and the other
two
sides are equal.QUADRATIC EQUATIONS
7.7 x -f 10 = 0.
.
8 = 4 wr2 Find 440 square yards. let us compare x 2
The
left
the perfect square x2
2
mx -f m
to
2
. (b) 100 feet?
=
.
2m. (b) 44 square feet.
article. -\-bx-\.184
ELEMENTS OF ALGEBRA
45
46.
49.
= 12.
Solving this equation we obtain
by the method of the preceding
2a
The
roots of
substituting the values of a.
2
Every quadratic equation can be
reduced to the general form.
. and c in the general answer.
2x
3
4.
ao.c
= 0.
o^
or
-}-
3 ax == 4 a9
7 wr
.
Solution
by formula.
231.
any quadratic equation may be obtained by 6.
=8
r/io?.
=0.
x
la
48.
-5.QUADRATIC EQUATIONS
Form
51.
area
A
a perimeter of 380
rectangular field has an area of 8400 square feet and Find the dimensions of the field.
56.
7.
Divide CO into two parts whose product
is 875.1.9.
1.
.
2. Find the sides.
58. but frequently the conditions of the problem exclude negative or fractional answers.
Find the number.
54.
two numbers is 4.
The sum
of the squares of
two consecutive numbers
85.
Find
the numbers. Find the number.
8. The
11. and the difference Find the numbers.
57.
88
its reciprocal
A
number increased by three times
equals
6J.
of their reciprocals is
4.
number by 10.0.3.3.
3.
1.
-2.
EXERCISE
1.
-4.3.
is
Find two numbers whose product
288.
Problems involving quadratics have
lems of this type have only one solution.
What
are the
numbers
of
?
is
The product
two consecutive numbers
210.
:
3.
-2. and equals 190 square inches.
The
difference of
|.
and whose
product
9. -2.
Twenty-nine times a number exceeds the square of the 190.
2.
and whose sum
is
is 36.0.2.
55.
PROBLEMS INVOLVING QUADRATICS
in general two answers. and consequently many prob-
235.
5.
3.
Find two numbers whose difference
is 40.
Find a number which exceeds
its
square by
is
-|. feet.
6.
189
the equations whose roots are
53.
G.
-2.
its
sides of a rectangle differ by 9 inches.
0.
52.
dollars. one of which sails two miles per hour faster than the other.
.190
12.
15.
vessel sail ?
How many
miles per hour did the faster
If 20. At what rates do
the steamers travel ?
18. he had paid 2 ^ more for each apple. and the slower reaches its destination one day
before the other. If a train had traveled 10 miles an hour faster.
Two steamers
and
is
of 420 miles.
14.
19. and lost as many per cent Find the cost of the watch. a distance One steamer travels half a mile faster than the two hours less on the journey.
and gained as many per Find the cost of the horse.10. exceeds its widtK AD by 119 feet.
c equals 221
Find
AB and AD.
A man
A man
sold a
as the watch cost dollars. he would have received two horses more for the same money. What did he pay for each
apple ?
A man bought a certain number of horses for $1200.
The diagonal
:
tangle as 5 4.
of a rectangle is to the length of the recthe area of the figure is 96 square inches. Find the rate
of the train.
other. What did he pay for
21. A man bought a certain number of apples for $ 2.
13.
If he
each horse ?
.
17.
ply between the same two ports. had paid $ 20 less for each horse.
as the
16. sold a horse for $144. and lost as many per cent Find the cost of the watch.
watch cost
sold a watch for $ 21. Two vessels.
ELEMENTS OF ALGEBRA
The length
1
B
AB of a rectangle.
A man
cent as the horse cost dollars. ABCD. start together on voyages of 1152 and 720 miles respectively. he would have received 12 apples less for the same money. and Find the sides of the rectangle. and the line BD joining
two opposite
vertices (called "diagonal")
feet. it would have needed two hours less to travel 120 miles.
watch for $ 24.
A needs 8 days more than B to do a certain piece of work. If the area of the walk is equal to the area of the plot. Find and CB.
237. so that the rectangle.
24. The number of eggs which can be bought for $ 1 is equal to the number of cents which 4 eggs cost. a point taken.I) -4(aj*-l)
2
= 9. and working together.
1.
By formula. how wide is the walk ?
23. as
0.
Find the side of an equilateral triangle whose altitude
equals 3 inches. 30 feet long and 20 feet wide. the two men can do it in 3 days.
Find
TT r (Area of a circle . 23 inches long.
A rectangular
A
circular basin is surrounded
is
-
by a path 5
feet wide. and the unknown factor of one of these terms is the square of the unknown factor of the
other.
27.
EQUATIONS IN THE QUADRATIC FORM An equation is said to be in the quadratic form
if it
contains only two unknown terms.QUADRATIC EQUATIONS
22.
.
^-3^ = 7.
=9
Therefore
x
=
\/8
= 2. is surrounded by a walk of uniform width. contains B 78 square inches. Equations in the quadratic form can be solved by the methods used for quadratics.
of the area of the basin.
(tf. In how many days can B do the work ?
=
26.
B
AB
AB
-2
191
grass plot.
Solve
^-9^ + 8 =
**
0.
and the area of the path
the radius of the basin.) 25. How many eggs can be bought for $ 1 ?
236.
Ex. constructed with and CB as sides.
is
On the prolongation of a line AC.
or x
= \/l = 1.
such as 2*. we let these quantities be what they must be if the exponent law of multiplication is generally true. ~ a m -f. hence. instead of giving a formal definition of fractional and negative exponents.
m
IV.
the direct consequence of the defiand third are consequences
FRACTIONAL AND NEGATIVE EXPONENTS
243.
no
Fractional and negative exponents. that a
an
= a m+n
.*
III. very important that all exponents should be governed by the same laws.
II.
Then the law
of involution.
(ab)
.
The
first
of these laws
is
nition of power.
244.a" = a m n
mn .
for all values
1
of
m and n. and
.
We assume.
we may choose
for such
symbols any definition that
is
con-
venient for other work. while the second of the first.
must be
*The symbol
smaller than.
>
m therefore. however.
It is. The following four fundamental laws for positive integral exponents have been developed in preceding chapters
:
I.CHAPTER XVI
THE THEORY OF EXPONENTS
242. (a m ) w
. 4~ 3 have meaning according to the original definition of power. (a ) s=a m = aw bm
a
.
provided
w > n.
= a""
<
.
a m a" = a m+t1 ."
means "is greater than"
195
similarly
means "is
.
ml.
Assuming these two
8*.
To
find the
meaning
of
a fractional exponent.
(bed)*.
m$.
. 25.
28. at.
etc.
a\
26.
n 2 a. or zero exponent
equal
x.
245.
a?*.
30.
^=(a^)
3*
3
.
Hence
Or
Therefore
Similarly.
Write the following expressions as radicals :
22. as. (xy$.196
ELEMENTS OF ALGEBRA
true for positive integral values of n. 4~ . 3*.
Let
x
is
The operation which makes the fractional exponent disappear evidently the raising of both members to the third power.
disappear.
-
we
find
a?
Hence we
define a* to be the qth root of of.
= a.
23.
a*.
'&M
A
27.
we
try to discover the
let the
meaning of
In every case we
unknown quantity
and apply to both members of the equation that operation which makes the negative.
31.
e.
29.g.
0?=-^. a . since the raising to a positive integral power is only a repeated multiplication.
24. fractional.
laws.
a8 a
2
=
1
1
.
ELEMENTS OF ALGEBRA
To
find the
meaning
of a negative exponent. Factors
may
be transferred
from
the
numerator
to
the
denominator of a fraction.
each
is
The
fact that a
if
=
we
It loses its singularity
1 sometimes appears peculiar to beginners.
by changing the sign of
NOTE.
Let
x=
or".
etc. or
the exponent.
.
a
a
a
= =
a a a
a1
1
a.
cr n.g.
an x = a.2
=
a2
. in which
obtained from the preceding one by dividing both
members by
a.
Or
a"#
= l. consider the following equations. e.
Multiplying both members by
a".
vice versa.
248.198
247.
2.
1 Multiply 3 or
+x
5 by 2 x
x.
1.
lix
=
2x-l
=+1
Ex.202
ELEMENTS OF ALGEBRA
32.
40.
V ra
4/
3
-\/m
33.
If
powers of
a?. the term which does not contain x may be considered as a term containing #.
Divide
by
^
2a
3 qfo
4.
6
35. The
252.
we wish to arrange terms according to descending we have to remember that.
powers of x arranged are
:
Ex.2 d
.
1.
34.
Arrange in descending powers of
Check.
Ex.
43.
51.
V3 .
52.
49.
Va
-v/a. a
VS
-f-
a?Vy
= -\/ -
x*y
this
Since surds of different orders can be reduced to surds of
the same order.
47.214
42.
268.
a fraction.
ELEMENTS OF ALGEHRA
(3V5-5V3)
S
.V5) ( V3 + 2 VS).
60.
it
more convenient to multiply dividend and divisor by a factor which makes the divisor rational.
(2
45.
44.
is
1
2.
.
E.
Ex. the quotient of the surds
is
If.
(5V2+V10)(2V5-1).
53.
48.
(3V3-2Vo)(2V3+V5).
-v/a
-
DIVISION OF RADICALS
267.y.
(5V7-2V2)(2VT-7V2).
(V50-f 3Vl2)-4-V2==
however.
46.
all
monomial surds may be divided by
method.
(3V5-2V3)(2V3-V3). Monomial surdn of the same order may be divided by multiplying the quotient of the coefficients by the quotient of the
surd factors.
Divide
VII by v7.
by V7.73205
we
simplify
JL-V^l
V3
*>
^>
division
Either quotient equals . To show that expressions with rational denominators are simpler than those with irrational denominators.
is
Since \/8
12 Vil
=
2 V*2..
3. the rationalizing factor
x
' g
\/2. arithTo find.
VTL_Vll '
~~"
\/7_V77
.57735. called rationalizing the
the following examples
:
215
divisor. the by 3 is much easier to perform than the division by
1. we have
V3
But
if
1. e.
Divide 4 v^a by
is
rationalizing factor
evidently \/Tb
hence.
we have
to multiply
In order to make the divisor (V?) rational.
Divide 12 V5
+ 4V5 by V.
Hence
in arithmetical
work
it
is
always best to
rationalize the denominators before dividing. however.
.
The
2.
/~
}
Ex.
.
.73205. is illustrated
by
Ex.RADICALS
This method.
1.
+ 4\/5 _ 12v 3 + 4\/5 V8 V8
V2 V2
269. metical problems afford the best illustrations.g.by the usual arithmetical method. Evidently.
4\/3~a'
36
Ex.
y n is divisible by x -f ?/.
2. if w is odd. if n is even.
2 8 (3 a )
+8=
+
288.
Factor 27 a* -f
27 a 6
8.xy +/).
ar
+p=
z6
e.
xn -f. For substituting y for x."
.
We may
6
n 6 either a difference of two squares or a
dif-
* The symbol
means " and so forth to.
2
Ex.
actual division
n.
It
y is
not divisible by
287.
Ex.
The
difference of
two even powers should always be
considered as a difference of two squares.
it
follows from the Factoi
xn y n is always divisible by x y.
286.
Two
special cases of the preceding propositions are of
viz.
Factor
consider
m
m
6
n9
.
By
we obtain the other
factors.
1.230
285.
-
y
5
=
(x
-
can readily be seen that #n -f either x + y or x y. xn y n y n y n = 0. if n For ( y) n -f y n = 0.
ELEMENTS OF ALGEBRA
positive integer.
2.g. If n is a Theorem that
1.
and have
for
any positive integral value of
If n
is
odd.
x* -f-/
= (x +/)O .
:
importance.
is
odd.
By making x
any * assigned
zero.e.e.
x
-f 2.
is satisfied
by any number.
(1)
is
an
identity. equation.
ELEMENTS OF ALGEBRA
Interpretation of ?
e.i
solving
a problem
the result
or oo indicates that the
all
problem has no solution.
and becomes infinitely small.
TO^UU"
sufficiently small.
(1)
= 0.
1.
oo is
= QQ. however
x approaches the value
be-
comes
infinitely large.
I. cancel. or infinitesimal) This result is usually written
:
305.000
a.
.
The
~~f
fraction .
306.242
303.
and
.
ToU"
^-100 a.g. the answer is indeterminate. or that x may equal any finite number.
(a:
Then
Simplifying.
creases. without exception.
Let
2.
be the numbers.
1.
the
If in an equation
terms containing
unknown quantity
cancel.
i.
as
+ l.can be
If
It is
made
larger than
number.x'2 2 x =
1.
Hence such an equation
identity. while the
remaining terms do not
cancelj the root is infinity.increases
if
x
de-
x
creases.
.decreases
X
if
called infinity.
of the second exceeds the product of the first
Find three consecutive numbers such that the square and third by 1.
Or.
= 10.
The
solution
x
=-
indicates that the problem
is indeter-
If all terms of an minate.
(1).
great.
+
I)
2
x2
'
-f
2x
+
1
-x(x + 2)= . it
is
an
Ex.
Hence any number will satisfy equation the given problem is indeterminate.
i.
Interpretation of
QO
The
fraction
if
x
x
inis
infinitely large.
customary to represent this result
by the equation ~
The symbol
304.
The area of a
nal 41 feet.
is 6.
and the sum of
(
228. and
its
The diagonal
is
is
perimeter
11. Find these sides. Find the numbers. 12. The volumes of two cubes differ by 98 cubic centimeters.
255 and the sum of
5.
10. Find the side of each square. 148 feet of fence are required. But if the length is increased by 10 inches and
12. equals 4 inches.
the
The mean proportional between two numbers sum of their squares is 328. Find two numbers whose product whose squares is 514.244
3.
of a right triangle is 73. 190.)
53 yards. Find the dimensions of the
field. and the side of one increased by the side of the other e. and the edge of one exceeds the edge of the other by 2 centimeters. increased by the edge of the other. and the edge of one.
rectangle is 360 square Find the lengths of the sides.
ELEMENTS OF ALGEBRA
The
difference between
is
of their squares
325.
9.
is
is
17 and the
sum
4. the area becomes
-f%
of
the original area.
103.
Find the
sides.
and the diago(Ex. The sum of the areas of two squares is 208 square feet.
13.
14. To inclose a rectangular field 1225 square feet in area.
8.
Find the
sides of the rectangle.
6.
146 yards.
Find the other two
sides. Find the edge of each cube. p.
of a rectangular field
feet.
and
is
The area of a rectangle remains unaltered if its length increased by 20 inches while its breadth is diminished by 10 inches.quals 20 feet. Two cubes together contain 30| cubic inches.)
The area
of a right triangle is 210 square feet.
.
two numbers Find the numbers.
The hypotenuse
is
the other two sides
7.
is
the breadth
diminished by 20 inches. Find the edges.
and the
hypotenuse
is 37.
the quotient
is 2.SIMULTANEOUS QUADRATIC EQUATIONS
15.
differ by 8 inches.) (Area of circle
and
=
1
16.
and
if
the digits will be interchanged.
is
20 inches.
245
The sum of the radii of two circles is equal to 47 inches. and the equal to the surface of a sphere Find the radii.
The
radii of
two spheres
is
difference of their surfaces
whose radius = 47T#2. Find the radii.)
17.
. their areas are together equal to the area of a circle whose radius is 37 inches. irR *. Find the number.
by the product of 27 be added to the number. (Surface of sphere
If a
number
of
two
digits be divided
its digits.
The first is an ascending. a
11.
-4.
. to produce the 4th term.
.
. An arithmetic progression (A. is derived from the preceding by the addition of a constant number.
-f
..
to each
term produces the next term.
of a series are its successive numbers. a -f d. 15 is 9 -f. The progression is a.
a
+
d..
To
find the
nth term
/
of an A. 3 d must be added to a.
:
7. a
3d. progression. P.
added to each term to obtain the next one..
Since d
is
a
-f
3
d... a + 2 d.
a. P.
+
2 d.
17..) is a series.
The terms
ARITHMETIC PROGRESSION
308.
10.
309. P. 12.
of the following series is
3. to produce the 3d term.
series 9..CHAPTER XX
PROGRESSIONS
307.
to
A series
is
a succession of numbers formed according
some
fixed law.
Hence
/
= a + (n . 2 d must be added to a. 16.
The common
Thus each
difference is the
number which added
an A.7.1) d.
11. the second a descending.. 3. 19. the first
term a and
the
common difference d being given. except the first. ..11 246
(I)
Thus the 12th term of the
3
or 42. (n 1) d must be added to a.
The common differences are respectively 4. each term of which. to produce the nth term. and d..
5.
if
a = 5. d = 3.
Find the 10th term of the
series 17.
Find the 101th term of the
series 1..-.
-3.
series
.
(d) 1J. 8..
Or
Hence
Thus
from
(I)
= (+/).
= 99.3 a = -l.-.
Find the 12th term of the
-4.
Which
(6)
(c)
of the following series are in A..
2*=(a + Z) + (a + l) + (a + l)
2s = n
*
.PROGRESSIONS
310.-
(a
+ + (a +
l)
l)...
247
first
To
find the
sum s
19
of the first
n terms of an A.' cZ == . P. 6. 2J.
115.
2
EXERCISE
1. 3.. . P.
5.
Find the 7th term of the Find the 21st term
series
.8.
.
= I + 49 = *({ +
. a = 2. 7.
Find the nth term of the
series 2.. 3. 2.
-|.
. 21.
3..
7.
the last term
and the common difference d being given..
5.
6.
1.4.. 4.
. 2
sum
of the first 60
I
(II)
to find the
' '
odd numbers. 5.
= -2.
9.
.
8.
Adding. 6.
2. 3. the
term
a. ?
(a) 1.
1-J..
1. 19..16.
first
2
Write down the
(a)
(6)
(c)
6 terms of an A..
= a + (a
Reversing the order. 99) = 2600.. P. -10.. -7. -4^.
series 2. 5.
..
of the series 10.
Find the 5th term of the
4. -24..
9.
6
we have
Hence
. d
. 8.
(x +"l) 4.
+ 2-f-3 + 4 H
hlOO. 7. 7.
to 7 terms.
:
3. 11.
.
1+2+3+4H
Find the sum of the
first
n odd numbers.1 -f 3.
Sum
the following series
14. Jive quantities are involved.
33.
>
2-f
2. 1|.
1. In most problems relating to A.
. hence if any three of them are given.
\-n.
1. the other two may be found by the solution of the simultaneous equations
.5
H + i-f
-f-
to 10 terms. and for each than for the preceding one.
19.
to 20 terms. 11.
(i)
(ii)
. 2J.
$1
For boring a well 60 yards deep a contractor receives yard thereafter 10^ more How much does he receive all
together ?
^S5 A bookkeeper accepts a position at a yearly salary of $ 1000.
to 10 terms. strike
for the first yard.
16. 11.
23. striking hours only.
4.7 -f
to 12 terms.
to 8 terms. 12.
ELEMENTS OF ALGEBRA
last
term and the sum of the following series :
.
rf.
.
to 20 terms. 31.
1J. 16.
+ 3.
2.
Q^) How many times
in 12 hours ?
(&fi)
does a clock..
22.
'.
8.
11.
17.
to 15 terms.
. How much does he receive (a) in the 21st year (6) during the first 21 years ?
j
311.
6.
13.
-.
to 16 terms.(#
1
2) -f (x -f 3) H
to
a terms.
to 20 terms.
21.
15. P. 29.
20. 15.248
Find the
10.
.
15.
7.
3.
.
12.
18. and a yearly increase of $ 120.
.
. Find a and Given s = 44. n = 13. and all his savings in 5 years amounted to $ 6540.
7. = 52. Given a = |. n = 20.
has the series 82.
I
Find
I
in terms of a. Find a Given a = 7.
f
J 1 1
/
. Given a = . Given a = 4. Find n. Find d. Find d and Given a = 1700.
and
s.
12. 78. n
has the series
^
j
.
10. How much did he save the first month?
19.
T?
^.
A
$300
is
divided
among 6 persons
in such a
way
that each
person receives $ 10 did each receive ?
more than the preceding
one. s = 70.
m
and
n
2.
17.250
ELEMENTS OF ALGEBRA
EXERCISE
116
:
Find the arithmetic means between
1. 74.
8. d = 5.
y and #-f-5y. Find d.
13. n = 16.
I. of 5 terms
6. Given a = 1.
15.3. P.
= 16. = ^ 3 = 1.
f?
.
produced.
Between 4 and 8
insert 3 terms (arithmetic
is
means)
so
that an A. = 83.
How much
.
4. = 45. = 17.
a x
-f-
b
and a
b. s == 440.
16. n = 4. Find?.
man saved each month $2 more than in the pre 18. n = 17.
Between 10 and 6
insert 7 arithmetic
means
.
11.
6?
9.
3. = 1870.
a+
and
b
a
b
5.
How many terms How many terms
Given d = 3. Find w.
14. ceding one. n.
a?*2 To obtain the nth term a must evidently be multiplied by
. the first term a and
the ratios r being given.
or 81
315.
The progression is a.PROGRESSIONS
251
GEOMETRIC PROGRESSION
313. P. 36..
the following form 8
nf +
q(l-r")
1
r
.
2
arn
(2)
Subtracting (1) from
(2).g.
Therefore
Thus the sum
= ^ZlD.
s(r
1)
8
= ar"
7*
JL
a. P..
is
16(f)
4
.
Hence
Thus the 6th term
l
= ar
n~l
..
A geometric progression
first. is
it
(G.
.. To find the sum s of the first n terms term a and the ratio r being given. +1.
-2.
(I)
of the series 16.
and
To
find the
nth term
/ of
a G..)
is
a series each term of
which.
|. except the
multiplying
derived from the preceding one by by a constant number.
4-
(1)
.. or.
4.
The
314. P.. 36.. 24.
of a G.
E. 108.
r
n~ l
.
ratios are respectively 3. 24.arn ~ l .
NOTE. <zr .
. -I.
.
ar8
r. 12.
2 a. called the ratio. rs =
s
2
-. 36.
(II)
of the
8 =s
first
6 terms of the series 16.
g==
it is
convenient to write formula' (II) in
*. ar. the first
= a + ar -for ar -f ar Multiplying by r.
If
n
is less
:
than unity.
fl
lg[(i)
-l]
==
32(W -
1)
= 332 J.
...
4.
.
-.
288.
. whose and whose second term is 8. 4. 72. whose and whose common ratio is 4.._!=!>.
f. 144. Jive quantities are in.
Find the 6th term of the
series J.
10.. if any three of them are given. 18.
-fa. 3.
To
insert 5 geometric
means between 9 and 576. 9. + 5.
. 1.
+-f%9 %
.
first
term
is
125 and
whose common
..
(b) 1.5. ?
(c)
2.
Find the 7th term of the Find the 6th term of the
Find the 9th term of the
^.
. In most problems relating to G.
9.
is 16. the other two be found by the solution of the simultaneous equations :
may
(I)
/=<!/-'. 9.l.
series
.. 36..
8.72.
series 6.
0. P.. 25.
.
or
7.18.
..
first
term
4.
7.5.
series 5.252
ELEMENTS OF ALGEBRA
316. P.. P.
Evidently the total
number
of terms is 5
+ 2.
\
t
series
.
. 144. hence.
Write down the first 5 terms of a G.18.
(d) 5. 20. P.
676
t
Substituting in
= r6 = 64. 80.
Ex.
36.
117
Which
(a)
of the following series are in G.
4..
is 3.
Write down the first 6 terms of a G.
i 288.
Hence n
=
7.
.
Hence the
or
series is
0.
72.
EXERCISE
1.288.*. P.
volved .
2
term
3.
first
5.
144.
Find the 5th term of a G.. f..
|.
. I
= 670..
r^2.
a
=
I.
. 36.
.
(it.
.4. whose
.54.
l.
And the
required
means are
18. 676.6..
6.
576.
-fa.
series
Find the llth term of the Find the 7th term of the
ratio is
^.
.
.
z 2
-92.
if
each
increased 2 feet.
176. and 5 h.
. x*
185. side were one foot longer.
train.-36.
An
The two
express train runs 7 miles an hour faster than an ordinary trains run a certain distance in 4 h.
A
each
177.
same
result as the
number
diminished by
175.
3 gives the
same
result as the
numbet
multiplied by
Find the number.
power one of the two Find the power of each.
A
boy
is
father.
two boys is twice that of the younger. Find the age
5 years older than his sister
183. 4 a 2
y-y
-42. was three times that of the younger.
188.
A
the
boy
is
as old as his father
and
3 years
sum
of the ages of the three is 57 years.
The length
is
of a floor exceeds its width
by 2
feet.
7/
191.
180. 6 in each row the lowest row has 2 panes of glass in each window more than the middle row. z 2 + x . and the middle row has 4 panes in each window more than the upper row there are in all 168 panes of glass. Find the dimensions of the floor.
+x-
2.
2
2
+
a
_ no.56.
190.
-ll?/-102. and the father's present age is twice what the son will be 8 years
hence.
Find the number.266
173.
sister
.
+
a.
The age
of the elder of
it
three years ago of each.
13 a + 3. 189.
181.
179. the
sum
of the ages of all three is 51.
dimension
182.
is
What are their ages ? Two engines are together
more than the
of 80 horse
16 horse power
other.
father.
187.
3 gives the
174. Four years ago a father was three times as old as his son is now. aW + llab-2&. + 11 ~ 6.
What
is
the distance?
if
square grass plot would contain 73 square feet more Find the side of the plot.
178. How many are there in each window ?
.
and | as old as his Find the age of the
Resolve into prime factors
:
184.
respectively.
186.
.
ELEMENTS OF ALGEBRA
A A
number increased by
3. 12 m.
number divided by
3.
younger than his Find the age of
the father. 15 m. the ana of the floor will be increased 48 square feet. 10x 2 192. A house has 3 rows of windows.
.
18 be subtracted from the number. far did he walk all together ?
A
. the order of the digits will be inverted.a)(x b
b)
(x
b
~
)
412. How long is each road ?
423. Find the number of miles an hour that A and B each walk.
a
x
)
~
a
2 b
2
ar
a
IJ a.
mx ~
nx
(a
~
mx
nx
c
d
d
c)(:r
lfi:r
a
b)(x
.
hour. Find the number Tn 6 hours
.
-f
a
x
-f
x
-f c
1
1
a-b
b
x
415.
a
x
a
x
b
b
x
c
b
_a
b
-f
x
414.
-
a)
-2
6 2a.
(x
. and was out 5 hours.
A man
drives to a certain place at the rate of 8 miles an
Returning by a road 3 miles longer at the rate of 9 miles an hour.278
410. In a
if
and
422.(c rt
a)(x
-
b)
=
0.
421. and at the rate of 3^ miles an hour.
A
in 9 hours
B walks
11 miles
number of two digits the first digit is twice the second.c) .
down again
How
person walks up a hill at the rate of 2 miles an hour.(5 I2x
~r
l
a)
.
4x
a
a
2 c
6
Qx
3 x
c
419. 411.
420418 ~j-o. he takes 7 minutes longer than in going.
had each at first?
B
B
then has
J
as
much
spends } of his money and as A. A sum of money at simple interest amounts in 8 months to $260.
A
spends \ of his.
If 1
be added to the numerator of a fraction
it
if 1
be added to the denominator
it becomes equal becomes equal to ^. Find two numbers such that twice the greater exceeds the by 30.
.
486. If 31 years were added to the age of a father it would be also if one year were taken from the son's age
.
least
The sum
of three
numbers
is
is
21. Find the principal and the rate of
interest.
if
the
sum of
the digits be multiplied by
the digits will be inverted.
half the
The greatest exceeds the sum of the greatest and
480. There are two numbers the half of the greater of which exceeds the less by 2.
481. Find the numbers. and in 18 months to $2180. Find the numbers.
latter
would then be twice the
son's
A
and B together have $6000.
to
.
thrice that of his son
and added to the father's.
years.
by 4. and if each be increased by 5 the Find the fraction. Find their ages.
477. A sum of money at simple interest amounted in 10 months to $2100. and a fifth part of one brother's age that of the other. and 5 times the less exceeds the greater by 3.
483.
487.
479.
485. What is that fraction which becomes f when its numerator is doubled and its denominator is increased by 1. Find the
fraction. In a certain proper fraction the difference between the nu merator and the denominator is 12. the Find their ages.
A
number
consists of
two
digits
4. also a third of the greater exceeds half the less by 2. age. Find the sum and the rate of
interest. and in 20 months to $275.
whose difference
is
4.
Find
the number. and the other number least. fraction becomes equal to |. Of the ages of two brothers one exceeds half the other by 4 is equal to an eighth of 482. How much money
less
484.282
ELEMENTS OF ALGEBRA
476. and becomes when its denominator is doubled and its numerator increased by 4 ?
j|
478.
CD.
and CA=7.
. Throe numbers are such that the
A
the
first
and second equals
. L. An (escribed) and the prolongations of BA and BC in Find AD. A vessel can be filled by three pipes. if and L. and BE. and one overtakes the other in 6 hours. Tu what time will it be filled if all run
M
N
N
t
together?
529.REVIEW EXERCISE
285
525.
and third equals \\ the sum third equals \.
and B together can do a piece of work in 2 days. A boy is a years old his mother was I years old when he was born.
it
separately
?
531. 90. if the number be increased by Find the number. 37 pounds of tin lose 5 pounds. When weighed in water. they would have met in 2 hours.
527.
AC
in /).
E
533.
. N. and 23 pounds of lead lose 2 pounds.
sum of the reciprocals of of the reciprocals of the first of the reciprocals of the second and
the
sum
528. it is filled in 35 minutes. the first and second digits will change places. A number of three digits whose first and last digits are the same has 7 for the sum of its digits. M. In how many days can each alone do the same work?
526. (a) How many pounds of tin and lead are in a mixture weighing 120 pounds in air. A can do a piece of work in 12 days B and C together can do the same piece of work in 4 days A and C can do it in half the time in which B alone can do it.
touches
and
F respectively. AB=6. if L and Af in 20 minutes. How long will B and C take to do
.
530. Two persons start to travel from two stations 24 miles apart. In
circle
A ABC. Find the numbers.
532. Tf and run together. B and C and C and A in 4 days. What are their rates of
travel?
. BC = 5. If they had walked toward each other. in 28 minutes. and losing 14 pounds when weighed in water? (b) How many pounds of tin and lead are in an alloy weighing 220 pounds in air and 201 pounds in water ?
in 3 days. his father is half as old again as his mother was c years ago. Find the present ages of his father and mother.
.
550.
d.
How
is
t /
long will
I
take 11
men
2
t' .
b. The greatest value of the function.
Draw
the graphs of the following functions
:
538. Represent the following table graphically
TABLE OF POPULATION (IN MILLIONS) OF UNITED STATES. 2
-
x
-
x2
.
-
3 x. If
to
feet is the length of a
seconds.e.
to
do the work? pendulum. FRANCE. x 8
549. the function.10 marks. x 2 544.
The values of y.
540. i. 2 541
2.
x*. One dollar equals 4.
545.
547.
-
3 x.286
ELEMENTS OF ALGEBRA
:
534.
+
3.
c. z 2
-
x x
-
5. Draw the graph of y 2 and from the diagram determine
:
+
2 x
x*. The roots of the equation 2 + 2 x x z = 1. 3 x
539.
548. 536. x
2
+
x. then / = 3 and write
=
3.
-
7. AND BRITISH ISLES
535. x *-x
+
x
+
1.
. 2 x
+
5. x*
-
2
x.
from x
=
2 to x
= 4.
e. formation of dollars into marks. The values of x if y = 2.
543.
a. GERMANY.3
Draw
down
the time of swing for a
pendulum
of length
8 feet. the time of whose swing a graph for the formula from / =0
537.
546.
542. The value of x that produces the greatest value of y. 2|. if x = f 1.
723.
.l
+
8
-8
+
ft)'
(J)-*
(3|)*
+
(a
+
64-
+ i. If a pound of tea cost 30 J* more than a pound of coffee.
in value. 724. 721.
The
difference of the cubes of
two consecutive numbers
is
find them. What number exceeds its reciprocal by {$.44#2 + 121 = 0.292
709.
of a rectangle is 221 square feet and its perimeter Find the dimensions of the rectangle. In how many days can A build the wall?
718. paying $ 12 for the tea and $9 for the coffee. How shares did he buy ?
if
726. A man bought a certain number of shares in a company for
$375. 727.
714
2
*2
'
+
25
4
16
|
25 a2
711. **-13a: 2
710. if 1 more for 30/ would diminish
720.
722. Find two consecutive numbers whose product equals 600. what is the
price of the coffee per
pound ?
:
Find the numerical value of
728. and working together they can build it in 18 days. Find the altitude of an equilateral triangle whose side equals a.25 might have bought five more for the same money.
ELEMENTS OF ALGEBRA
+36 = 0. 12
-4*+
-
8. 725.
a:
713.
___ _ 2* -5 3*2-7
715. needs 15 days longer to build a wall than B. 717. Find the price of an apple. Find two numbers whose 719. Find four consecutive integers whose product is 7920. 16 x* . he
many
312?
he had waited a few days until each share had fallen $6.
729. 2n n 2 2 -f-2aar + a -5 = 0.
217
.
716. 3or
i
-16 .
sum is a and whose product equals J.
What two numbers
are those whose
sum
is
47 and product
A man
bought a certain number of pounds of tea and
10 pounds more of coffee.
The area
the price of 100 apples by $1.40 a 2* 2 + 9 a 4 = 0.
A
equals CO feet.
Find the sides of the rectangle. and 10 feet broader.
feet. (y + *) = .
two numbers Find the numbers.
is 20.
943. A plantation in rows consists of 10. The perimeter of a rectangle is 92 Find the area of the rectangle.
and the sum of
their areas 78$. feet.
. and B diminishes his as arrives at the winning post 2 minutes before B.square inches.
rate each
man
ran in the
first
heat. In the second
heat
A
. 152.
y(
934. there would have been 25 more trees in a row. The difference of two numbers cubes is 513. Find the side of each two
circles is
IT
square.
A
is
938.
944.
935.
942. and also contains 300 square feet. y(x + y + 2) = 133. The diagonal of a rectangle equals 17 feet. (y
(* + y)(y +*)= 50. *(* + #) =24.
and the
difference of
936. Find the length and breadth of the first rectangle.000 trees.
34
939.
The sum of two numbers Find the numbers.
the difference of their
The
is
difference of
their cubes
270.
Assuming
= -y.
+ z)=18.
is
3
.300
930.
two squares is 23 feet.
The sum
of the perimeters of
sum
of the areas of the squares is 16^f feet.
much and A then
Find at what
increases his speed 2 miles per hour. In the first heat B reaches the winning post 2 minutes before A. How many rows are there?
941.102.
931. z(* + y + 2) = 76. + z) =108.
the
The sum
of the perimeters of
sum
of their areas equals 617 square feet.
ELEMENTS OF ALGEBRA
(*+s)(* + y)=10. Find the numbers. s(y
932.
is 3. a second rec8 feet shorter.
2240.
two squares equals 140 feet. = ar(a? -f y + 2) + a)(* + y
933.
and the sum of their cubes
is
tangle
certain rectangle contains 300 square feet. and the Find the sides of the and
its
is
squares.
937. If each side was increased by 2 feet. The
sum
of the circumferences of
44 inches. the area of the new rectangle would equal 170 square feet.
diagonal
940. (3 + *)(ar + y + z) = 96.
find
the radii of the two circles. Tf there had been 20 less rows. A and B run a race round a two-mile course.
Find in what time both will do it. at Find the his rate of traveling. The diagonal of a rectangular is 476 yards.
P and
Q. A rectangular lawn whose length is 30 yards and breadth 20 yards is surrounded by a path of uniform width.
. A number consists of three digits whose sum is 14.
Find the number.
. the area lengths of the sides of the rectangle.
953.
950. The sum of the contents of two cubic blocks
the
of the heights of the blocks is 11 feet.
Two
starts
travelers. that
B
A
955. The area of a certain rectangle is equal to the area of a square side is 3 inches longer than one of the sides of the rectangle.
sum
Find an edge of
954. and that B. What is its area?
field is
182 yards. Find the width of the path if its area is 216 square yards. and
its
perim-
948.
A and
B.
unaltered. triangle is 6. its area will be increased 100 square feet.
is
407 cubic feet.
951. When
from
P
A
was found that they had together traveled 80 had passed through Q 4 hours before.
952. at
the same time
A
it
starts
and
B
from
Q
with the design to pass through Q. if its length is decreased 10 feet and its breadth increased 10 feet.
949. and if 594 be added to the number. If the breadth of the rectangle be decreased by 1 inch and its
is
length increased by 2 inches. The area of a certain rectangle is 2400 square feet. the difference in the lengths of the legs of the Find the legs of the triangle.
.
overtook
miles. Find its length and breadth. and the other 9 days longer to perform the work than if both worked together. distance between P and Q. Find the number.REVIEW EXERCISE
301
945. Find two numbers each of which
is
the square of the other. Two men can perform a piece of work in a certain time one takes 4 days longer. and travels in the same direction as A. each block. set out from two places. the digits are reversed. the square of the middle digit is equal to the product of the extreme digits. The square described on the hypotenuse of a right triangle is 180 square inches. was 9 hours' journey distant from P.
whose
946. A certain number exceeds the product of its two digits by 52 and exceeds twice the sum of its digits by 53.
Find the
eter
947.
P.3 '
Find the 8th
983. and the
common
difference.
987. Find the
first
term.
of n terms of an A.
of n terms of 7
+
9
+ 11+
is
is
40.
985. named Sheran.
989.001
4.2
.
303
979.
all
A
perfect number
is
a number which equals the sum
divisible.001
+
. Find four numbers in A. then this
sum multiplied by
(Euclid.)
the last term
the series
a perfect number.
first
984.-.---
:
+
9
-
-
V2
+
. such that the product of the and fourth may be 55.
0.
If
of
2
of
integers + 2 1 + 2'2
by which
is
it
is
the
sum
of
the series
2 n is prime. Insert 8 arithmetic means between
1
and
-. P. and the sum of the first nine terms is equal to the square of the sum of the first two. doubling the number for each successive square on the board.+ lY L V.
Find the number of grains which Sessa should have received.
"(. is 225.
to
n terms.REVIEW EXERCISE
978.
How many
sum
terms of 18
+
17
+
10
+
amount
.
to oo.
v/2
1
+ +
+
1
4
+
+
3>/2
to oo
+
+
. Find the sum of the series
988.
to 105?
981.
1.-..
What
2 a
value must a have so that the
sum
of
+
av/2
+
a
+
V2
+
. 5
11.
Find four perfect
numbers. 4 grains on the 3d. and so on. and of the second and third 03.
986.
. 2 grains on the 2d. The 21st term of an A..
980. Find the value of the infinite product 4
v'i
v7-!
v^5
.
to infinity
may
be 8?
.. P..1
+
2.04
+
. The Arabian Araphad reports that chess was invented by amusement of an Indian rajah.
Find
n.
992. The
term.. who rewarded the inventor by promising to place 1 grain of wheat on
Sessa for the
the 1st square of a chess-board.
990. Insert 22 arithmetic means between 8 and 54..01
3. The sum
982.
P. are
45 and 765
find the
numbers. are unequal.
1001. prove that they cannot be in A. Find (a) the sum of all circumferences. P. The side of an equilateral triangle equals 2.
998. The
fifth
term of a G.
is 4. c. In an equilateral triangle second circle touches the first circle and the sides AB and AC. (I) the sum of the perimeters of
all
squares.
1000. 1003. In a circle whose radius is 1 a square is inscribed. (6) the sum of the infinity.
512
996. are 28 and find the numbers. One of them travels uniformly 10 miles a day. Each stroke of the piston of an air
air contained in the receiver.304
ELEMENTS OF ALGEBRA
993. and so forth to infinity. the sides
of a third triangle equal the altitudes of the second.
. Two travelers start on the same road.
and the
fifth
term
is
8 times
the second . areas of all triangles. and so forth to Find (a) the sum of all perimeters. in this square a circle. inches. The sides of a second equilateral triangle equal the altitudes of the first.
ABC
A A
n
same
sides. (a) after 5 strokes. P. Under the conditions of the preceding example. after how strokes would the density of the air be xJn ^ ^ ne original
density ?
a circle is inscribed. Insert 3 geometric means between 2 and 162. Insert 4 geometric means between 243 and 32.
994. The other travels 8 miles the first day and After how increases this pace by \ mile a day each succeeding day.
997. in this circle a square.
The sum and sum
.
AB =
1004. (6) after n
What
strokes?
many
1002.
and
if
so forth
What
is
the
sum
of the areas of all circles. at the same time. P.
pump removes
J
of the
of air
is
fractions of the original amount contained in the receiver. ft. P. If a. and G. The sum and product of three numbers in G. find the series.
third circle touches the second circle and the
to infinity. many days will the latter overtake the former?
.
of squares of four
numbers
in
G. 995.
999.
To meet the requirements of the College Entrance Examination Board. physics.
i2mo. great many
work. Particular care has been bestowed upon those chapters which in the customary courses offer the greatest difficulties to the beginner.
Half
leather. and commercial life.
xiv+563
pages. given.
A
examples are taken from geometry.D.
Half leather. but none of the introduced illustrations is so complex as to require the expenditure of
time for the teaching of physics or geometry. but these few are treated so thoroughly and are illustrated by so many varied examples that the student will be much better prepared for further
The Exercises are superficial study of a great many cases. The more important subjects
tions. etc. 64-66 FIFTH AVBNTC.
which have been omitted from the body of the work Indeterminate Equahave been relegated to the Appendix.
THE MACMILLAN COMPANY
PUBLISHERS.
xi 4-
373 pages. proportions and graphical methods are introduced into the first year's course. which has been retained to serve as a basis for higher work.
$1.ELEMENTARY ALGEBRA
By ARTHUR SCHULTZE. without the sacrifice of scientific accuracy and thoroughness. comparatively few methods are heretofore. so that the Logarithms. than by the
.
not
The Advanced Algebra is an amplification of the Elementary. Ph. The author
has emphasized Graphical Methods more than is usual in text-books of this grade. but the work in the latter subject
has been so arranged that teachers
who
wish a shorter course
may omit
it
ADVANCED ALGEBRA
By ARTHUR SCHULTZE.10
The treatment of elementary algebra here is simple and practical. especially
duction into Problem
Work
is
very
much
Problems and Factoring. and the Summation of Series is here presented in a novel form. All subjects now required for admission by the College Entrance Examination Board
have been omitted from the present volume.
$1.
HEW TOSS
. The introsimpler and more natural than the
methods given
In Factoring. save Inequalities.25
lamo. book is a thoroughly practical and comprehensive text-book. very numerous and well graded there is a sufficient number of easy examples of each kind to enable the weakest students to do some work.
proportions and graphical methods are introduced into the first year's course.
THE MACMILLAN COMPANY
PUBLISHBSS.D. All subjects now required for admission by the College Entrance Examination Board
have been omitted from the present volume.
not
The Advanced Algebra is an amplification of the Elementary. The author
grade. physics.
$1. Ph. especially
duction into Problem
Work
is
very
much
Problems and Factoring. comparatively few methods are
given. so that the tions.
Half leather. The introsimpler and more natural than the
methods given heretofore.
has emphasized Graphical Methods more than is usual in text-books of this and the Summation of Series is here presented in a novel form.25
i2mo. To meet the requirements of the College Entrance Examination Board.
xi
-f-
373 pages.
$1. than by the superficial study of a great many cases. 64-66
7HTH
AVENUE.
12010. without
Particular care has been the sacrifice of scientific accuracy and thoroughness. great many
A
examples are taken from geometry. there is a sufficient number of easy examples of each kind to enable the weakest students to do some work. but the work in the latter subject
has been so arranged that teachers
who
wish a shorter course
may omit
it
ADVANCED ALGEBRA
By ARTHUR SCHULTZE.ELEMENTARY ALGEBRA
By ARTHUR Sen ULTZE. book is a thoroughly practical and comprehensive text-book. which has been retained to serve as a basis for higher work. and commercial life. Logarithms. but none of the introduced illustrations is so complex as to require the expenditure of
time for the teaching of physics or geometry. etc.
xiv+56a
pages.
HEW YOKE
.10
The treatment
of elementary algebra here
is
simple and practical. The Exercises are very numerous and well graded. The more important subjects
which have been omitted from the body of the work Indeterminate Equahave been relegated to the Appendix. bestowed upon those chapters which in the customary courses offer the greatest difficulties to the beginner.
In Factoring.
HatF leather. save Inequalities. but these few are treated so thoroughly and are illustrated by so many varied examples that the student will be much better prepared for further
work.
80 cents
This Geometry introduces the student systematically to the solution of geometrical exercises. $1. under the heading Remarks". at the
It
same
provides a course which stimulates him to do original time. The numerous and well-graded Exercises the complete book.
text-book in Geometry
more
direct
ositions
7.
SEVENOAK.
Algebraic Solution of Geometrical Exercises is treated in the Appendix to the Plane Geometry .
Cloth. Cloth.
$1.
and no attempt has been made
to present these solutions in such form
that they can be used as models for class-room work.
lines. Difficult Propare made somewhat? easier by applying simple Notation . 7 he
. Hints as to the manner of completing the work are inserted The Order 5. 10. more than 1200 in number in 2. The Analysis of Problems and of Theorems is more concrete and practical than in any other
distinct pedagogical value.
.
THE MACMILLAN COMPANY
PUBLISHERS.r and.
izmo. aoo pages.10
By ARTHUR
This key will be helpful to teachers who cannot give sufficient time to the Most solutions are merely outsolution of the exercises in the text-book.10
L..
i2mo.
xtt-t
PLANE GEOMETRY
Separate. SCHULTZE.
PLANE AND SOLID GEOMETRY
F. State: . iamo.
of Propositions has a
Propositions easily understood are given first and more difficult ones follow .
xii
+ 233 pages. 64-66 FIFTH AVENUE. 9. Many proofs are presented in a simpler and manner than in most text-books in Geometry 8.
wor. 6. guides him in putting forth his efforts to the best
advantage. Pains have been taken to give Excellent Figures
throughout the book. 4. Proofs that are special cases of general principles obtained from the Exercises are not given in detail.
KEY TO THE EXERCISES
in
Schultze and Sevenoak's Plane and Solid Geometry.
NEW YORK
. These are introduced from the beginning 3.
By ARTHUR SCHULTZE and
370 pages. Attention is invited to the following important features I. Ph.
Half
leather.D. Preliminary Propositions are presented in a simple manner . The Schultze and Sevenoak Geometry is in use in a large number of the leading schools of the country.
ments from which General Principles may be obtained are inserted in the " Exercises.
. $1.
causes of the inefficiency of mathematical teaching.
.
12mo.
. and Assistant Professor of Mathematics in New York University
of
Cloth.
THE MACMILLAN COMPANY
64-66 Fifth Avenue. of these theoretical views."
The treatment
treated are
:
is concrete and practical.
New York
DALLAS
CHICAGO
BOSTON
SAN FRANCISCO
ATLANTA
. " is to contribute towards book/ he says in the preface.
. enable him to
" The chief object of the speak with unusual authority.25
The
author's long
and successful experience as a teacher
of mathematics in secondary schools and his careful study of the subject from the pedagogical point of view. 370 pages.The Teaching
of
Mathematics
in
Secondary Schools
ARTHUR SCHULTZE
Formerly Head of the Department of Mathematics in the High School Commerce.
.
. Most teachers admit that mathematical instruction derives its importance from the mental training that it But in affords.
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. and not from the information that it imparts.
methods of teaching mathematics the first propositions in geometry the original exercise parallel lines methods of the circle attacking problems impossible constructions applied problems typical parts of algebra. New York City. making mathematical teaching less informational and more disciplinary. a great deal of mathematical spite
teaching
is
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informational.
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how
demonstrate. Typical topics the value and the aims of mathematical teach-
ing
.
The author's aim is to keep constantly before the
This book
pupil's mind the general movements in American history and their relative value in the development of our nation.40
is distinguished from a large number of American text-books in that its main theme is the development of history the nation.
An exhaustive system of marginal references. and a full index are provided.
THE MACMILLAN COMPANY
64-66 Fifth Avenue.AMERICAN HISTORY
For Use
fa
Secondary Schools
By ROSCOE LEWIS ASHLEY
Illustrated.
Cloth. All
smaller movements and single events are clearly grouped under these general movements. This book is up-to-date not only in its matter and method.
but in being fully illustrated with
many excellent maps.
$1. which put the main stress upon national development rather than upon military campaigns. which have been selected with great care and can be found in the average high school library. Studies and Questions at the end of each chapter take the place of the individual teacher's lesson plans.
photographs. diagrams.
diagrams. " This volume
etc. supply the student with plenty of historical
narrative on which to base the general statements and other classifications made in the text. Maps.
New York
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Complete School Math and English Educational Package
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Showing 1 to 30 of 91
4.5 Basis and Dimension
- Structure of Vector Spaces
Recognize bases in the vector spaces Rn, Pn,
and Mm,n.
Test a basis in a vector space
Find the dimension of a vector space
Definition of Basis
A set of vectors S = cfw_ v1, v2, , vn in a vector
spac
5.3 Orthonormal Bases:
Gram-Schmidt process
Show that a set of vectors is orthogonal
Represent a vector relative to an orthonormal
basis
Apply Gram-Schmidt orthonormalization
process
Review: Basis for a vector space V
S = cfw_ v1, v2, , vn is called a
Chapter 5 Inner Product Spaces
In this chapter we will enrich vector spaces
by adding another operation, called Inner
Product. Through it we will extend some
concepts in Rn, such as length, distance, angle,
orthogonality, to more general vector
spaces.
5.
Chapter 3 Determinants (Find Determinants)
3.1 Determinant of a Matrix
Find the determinant of a 2 x 2 matrix
Find the minors and cofactors of a matrix
Use expansion by cofactors to find the
determinant of a matrix
Find the determinant of a triangular
4.7 Coordinates and Change of Basis
Find a coordinate matrix relative to a basis
in Rn.
Find the transition matrix from basis B to
B in Rn.
Represent coordinates in general
n-dimensional spaces.
Coordinates Relative to a Basis
Let B = cfw_v1, v2, vn be
6.4 Transition Matrix and Similarity
Introduction to the problem
Let T: V V be a linear transformation. B and B
are two bases for V. A is the matrix for T relative
to B and A is the matrix for T relative to B
([T(v)]B = A[v]B and [T(v)]B = A[v]B )
Q. What
Math 207
Lesson 6: Computing determinants
Sections 3.1, 3.2
24 pts.
Show all work to receive full credit. Clearly indicate your answers. Erase or cross out all work you do not want considered.
Each problem is worth 4 points.
1. Determine all minors Mij an
Math 207
Lesson 8: Vector operations, vector spaces, and vector subspaces
Sections 4.1, 4.2, 4.3
32 pts.
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Each problem is worth 4 p
Math 207
Lesson 11: Kernel and range
Sections 6.1, 6.2
24 pts.
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Each problem is worth 4 points.
1. The transformation T : R2 R2 , d
Math 207
Lesson 4: Matrix inverses and LU factorization
Sections 2.3, 2.4
32 pts.
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Each problem is worth 4 points.
1. Use the formu
Math 207
Lesson 1: Solution sets of linear systems
Section 1.1
24 pts.
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Each problem is worth 4 points.
1. Graph the system of line
Recent Documents
Math 207
Lesson 2: Gaussian elimination
Section 1.2
24 pts.
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Each problem is worth 4 points.
1. Identify the elementary row operati
Math 207
Lesson 3: Matrix arithmetic
Sections 2.1, 2.2
32 pts.
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1 0
1. Compute 2A 3B for the matric
Math 207
Lesson 12: The matrix of a linear transformation
Sections 6.3
24 pts.
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1. Find the standar
Math 207
Lesson 13: Eigenvalues, eigenvectors, and diagonalizable matrices
Sections 7.1, 7.2
24 pts.
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Math 207
Lesson 9: Linear independence, bases, and rank
Sections 4.4, 4.5, 4.6
32 pts.
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1. Determin | 677.169 | 1 |
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Finite Element Beginnings
This Electronic Book, by engineer and teacher David Pintur, is an introduction to the principles
of the finite element method. If you use, or intend to use, existing finite element packages
but want a deeper theoretical understanding of the methodology, this book is ideal. Through
a variety of examples, you get a solid foundation for establishing finite element applications
so you can move on to more advanced programs. And, because it's based on Mathcad's
"live" math environment, every number, formula, and plot can be adapted to solve your
individual problems. You can change parameters and plots and watch Mathcad recalculate
answers right there in the book. Selecting interpolation functions, assembling stiffness
matrices, and solving for field variables are just some of the areas covered in this Electronic
Book.
Platform: Windows
Requires Mathcad 3.1 or higher, 5 MB hard disk space
Available for immediate download (size 3517782 bytes) or
ground shipment
Topics include: Historical Perspective of the Finite Element Method, Basic Concepts of Linear Elas-
ticity, the Principles of Minimum Potential Energy and Direct Method, Using Interpolation Concepts
in One and Two Dimensions, Mapped Elements, and much more.
The potential within a
unit circle, given by the
Laplace Equation, can be
determined using finite
element discretization.
Finite Element Beginnings
This Electronic Book is unique in that it merges the actual computer implementation with
the method's theoretical basis. The Mathcad environment allows the reader freedom to
experiment and explore the concepts which are introduced. As the title implies, Finite
Element Beginnings provides an introduction to the method which the reader can use as a
springboard to more advanced issues and applications.
Introduction
Definition and Basic Concepts
The Process of Discretization
Discrete Systems
Continuous Systems
Comparison to the Finite Difference Method
Seven Basic Steps of the Finite Element Method
Discretizing the Continuum
Selecting Interpolation Functions
Finding Element Equations
Assembling the Elements
Applying the Boundary Conditions
Solving the System of Equations
Making Additional Computations
Brief History of the Finite Element Method
The Discrete Approach: A Physical Interpretation
Introduction
A Simple Elastic Spring
TABLE OF CONTENTS (page 1 of 6)
Finite Element Beginnings
A System of Springs
Step 1: Discretize the Spring System
Step 2: Select Interpolation Functions
Step 3: Find the Element Properties
Step 4: Assemble the Elements
Step 5: Apply the Boundary Conditions
Step 6: Solve the System of Equations
Step 7: Additional Calculations
Assembling the Elements
An Example Finite Element Mesh
The Assembly Algorithm
Properties of the Assembled Stiffness Matrix
How to Treat Boundary Conditions
The Direct Method
The Payne and Irons Technique
Matrix Partitioning
A Discrete Finite Element Algorithm in One Dimension
Application to Other Discrete Systems
Truss Analysis
Element Stiffness Matrix in Global Coordinates
Stiffness Derivation Using Local Coordinates
A Finite Element Algorithm for Trusses in Two Dimensions
Truss Algorithm with Discussion
Truss Algorithm without Discussion
TABLE OF CONTENTS (page 2 of 6)
Finite Element Beginnings
Introduction to Finite Elements of Elastic Continua
Introduction
Continuity of Elements in a Continuum
Basic Concepts in Three Dimensional Linear Elasticity
The Displacement Field
Strain Components
Stress Components
Constitutive Laws
The Principle of Minimum Potential Energy
Plane Stress and Plane Strain
A Triangular Element in Plane Stress
The Direct Method for a Triangular Element
Interpolation of Displacement
Strain-Displacement Equation
Stress-Strain Relationship
Equivalent Forces for a Stress Field
The Stiffness Matrix
a) Summary of the Direct Method
The Energy Method for Elastic Elements
The Stiffness Matrix
How to Treat Surface Tractions
Final Remarks
Comparison of the Direct and Energy Methods for Plane Stress
A Finite Element Code for Plane Strain
Plane Stress Code With Discussion
Plane Stress Code Without Discussion
TABLE OF CONTENTS (page 3 of 6)
Finite Element Beginnings
Element Interpolation and Shape Functions
Introduction
The Essence of the Finite Element Method
Linear Interpolation in One Dimension
Piecewise Linear Interpolation
Higher-Order Polynomials in One Dimension
Quadratic Interpolation in One Dimension
Piecewise Quadratic Interpolation
Generalization to Higher Orders
Derivatives of Shape Functions
Linear Interpolation and Differentiation
Quadratic Interpolation and Differentiation
Continuity Requirements
Polynomials in Two Dimensions
A Linear Triangular Element
A Four Node Rectangular Element
A Specialized Rectangular Element
Shape Functions Using Normalized Coordinates
1-D Lagrangian Shape Functions
2-D Lagrangian Shape Functions
2-D Serendipity Shape Functions
Final Remarks
Mapped Elements
Introduction
TABLE OF CONTENTS (page 4 of 6)
Finite Element Beginnings
Mapping in One Dimension
Differentiation and Integration
a) Newton-Cotes Quadrature
b) Gauss Quadrature
c) Summary
Mapping in Two Dimensions
Evaluation of Element Equations
Transformation of Derivatives
The Area Integral and Numerical Integration
a) Integration of Mapped Quadratic Elements
Integration Along Element Boundaries
Shape Functions Along Element Boundaries
Reduction to One Dimension on Boundaries
Evaluating a Distributed Edge Load
Finite Element Code Using Isoparametric Plane Stress Elements
Linear Isoparametric Plane Stress Elements
Quadratic Isoparametric Plane Stress Elements
The Method of Weighted Residuals
Introduction
Overview of Residual Methods
Problem Definition
Approximate Solution Using Trial Functions
TABLE OF CONTENTS (page 5 of 6)
Finite Element Beginnings
a) Point Collocation
b) Subdomain Collocation
c) Galerkin's Method
Comparison of the Three Methods
Applying Galerkin's Method to Finite Elements
One Dimension: Integration by Parts
Finite Element Code in One Dimension
Two Dimensions: Green's Theorem
Finite Element Applications
Laplace's Equation in a Circular Disk
a) Linear Finite Element Code of Laplace's Equation
Laplace's Equation in a Rectangular Region
a) Quadratic Finite Element Code of Laplace's Equation
Concluding Remarks
References
Index
TABLE OF CONTENTS (page 6 of 6)
Finite Element Beginnings
Piecewise Linear Interpolation
In the previous section, a linear variation of the field variable valid over the domain of a
single element was defined. Now, a linear interpolation over a series of connected elements
is shown in the figure below:
Figure 1: Example of Finite Element Mesh
Suppose that the exact solution to the field variable is
and the field variable at each node corresponds to the exact solution:
Using the linear shape functions, N1 and N2, the field variable across the domain of all the
elements can be approximated in a piecewise manner, as demonstrated in the following
algorithm:
Shape functions
SAMPLE PAGE (page 1 of 3)
Finite Element Beginnings
Piecewise Linear Interpolation
Define x values along each element
Variation of field variable
Figure 2: Plot of Field Variable over Entire Domain
SAMPLE PAGE (page 2 of 3)
Figure 2 demonstrates a
piecewise linear interpolation
between the values of the field
variable at each node. It is
piecewise because the
distribution is calculated one
piece at a time, element by
element, and the value of f(x)
within the element is interpolated
linearly from the nodal values.
Finite Element Beginnings
Error Analysis
The relative error, fj,k, for the first three elements is redrawn:
Figure 3: Close-up of First Three Elements
Observe that the piecewise representation of the field variable over the physical domain
of the problem is exact at the node points while it is only an approximation across each
element domain. The error in the approximation is the shaded area between the exact
curve and the linear approximation. To increase the accuracy of the approximation, then
the following could be done:
1.Increase the number of elements over the domain.
2.Keep the same number of elements, but increase the order of the interpolation.
Increasing the Order of Interpolation
For rapidly varying functions, a linear approximation would require a large amount of
elements to accurately reflect the function's gradients.
SAMPLE PAGE (page 3 of 3)
Alternatively, higher-order elements could be formulated that
would also have higher-order interpolation functions and
more nodes per element.
The next section investigates higher-order interpolation
functions. | 677.169 | 1 |
How to learn college algebra in a day
If you know basic Arithmetic, you can learn PRACTICAL Algebra in 20 minutes because Algebra is easy. Please include your IP address in your email. You will need basic math skills, such as adding, subtracting, multiplying, and dividing. Review your basic math operations. Although it can be a little bit tricky, mastering these concepts is necessary to moving forward in math. There is no fast and simple way to pass college algebra. You must put in the work to understand the concepts and that takes time.
Around 7th grade, though, I got bored with my advanced math class and quit paying attention and quit trying. I tested out of all my high school maths (by the skin of my teeth) and never looked back. Quote:Official Course Description:MAT122LEC3 Credit(s)3 Period(s)Intermediat on how to learn algebra fast.Algebra is a common headache for many students. For someone who is not well grounded on the basics -- addition, subtraction, multiplication, division, exponents, ratios and fractions -- algebra can indeed be confusing and frustrating.
However, the beauty of algebra, as it is with math, is that it is completely logical and everything has a solution. It may be puzzling but solving the puzzle can be very gratifying and fulfilling. For someone who is not well grounded on the basics -- addition, subtraction, multiplicationmore, you can learn it, practice it and get better at it. Even as a baby you learn to count. Startingfrom that tiny age you will start to learn how to use building blocks howto count and then move on to drawing objects and figures. Mathematics is one of the first things you learn in life. Even asa baby you learn to count.
Starting from that tiny age you will startto learn how to use building blocks how to count and then move onto drawing objects and figures. All of these things are importantpreparation to doing algebra. The key to opportunityThese are the years of small beginnings until the day comes thatyou have to be abl. | 677.169 | 1 |
Integral geometry deals with the problem of determining functions by their integrals over given families of sets. These integrals de?ne the corresponding integraltransformandoneofthemainquestionsinintegralgeometryaskswhen this transform is injective. On the other hand, when we work with complex measures or forms, operators appear whose kernels are... more...
Provides a concise overview of the core undergraduate physics and applied mathematics curriculum for students and practitioners of science and engineering Fundamental Math and Physics for Scientists and Engineers summarizes college and university level physics together with the mathematics frequently encountered in engineering and physics calculations.... more...
Applied Finite Mathematics, Second Edition presents the fundamentals of finite mathematics in a style tailored for beginners, but at the same time covers the subject matter in sufficient depth so that the student can see a rich variety of realistic and relevant applications. Some applications of probability, game theory, and Markov chains are given.... more...
This new addition to the bestselling 20 Minutes a Day series helps students master the math essentials found on all levels of the COMPASS, ASSET, and ACCUPLACER exams. These exams, given by colleges country-wide, are taken by students to place them into the appropriate courses. Placing out of courses with a high score can mean credits (and money)... more...
This book provides an introduction to index numbers for statisticians, economists and numerate members of the public. It covers the essential basics, mixing theoretical aspects with practical techniques to give a balanced and accessible introduction to the subject. The concepts are illustrated by exploring the construction and use of the Consumer... more... | 677.169 | 1 |
Overview
This workshop briefly introduces participants to concepts in basic linear algebra and proceeds to discuss matrix computations and algorithms that underlie data science and computational engineering. Upon completion, participants will have an understanding of what is behind black box software packages and be able to make more informed decisions about what type of algorithm may be best for a given application. After familiarization with basic matrix and vector operations, the workshop discusses the foundational concepts on which many algorithms used in data mining, machine learning and deep learning are built. Examples include solving linear systems, eigenvalues and eigenvectors, and factorizations. Exercises help participants apply concepts to problems in optimization, machine learning and statistics.
Recommended background: knowledge of vector calculus.
Topics Include
Basics: matrices, vectors and fundamental operations: products, norms
Solving linear systems
Least squares
Eigenvalues and eigenvectors
Fundamental factorizations: QR, tall-skinny QR, SVD
To view other workshop descriptions, or to get general information about the 2016 ICME Summer Workshop Series, click here | 677.169 | 1 |
Mathematical Olympiad challenges by Titu Andreescu(
Book
) 43
editions published
between
2000
and
2009
in
English
and held by
653 WorldCat member
libraries
worldwide
"Mathematical Olympiad Challenges is a rich collection of problems put together by two experienced and well-known professors
and coaches of the U.S. International Mathematical Olympiad Team. Hundreds of beautiful, challenging, and instructive problems
for algebra, geometry, trigonometry, combinatorics, and number theory were selected from numerous mathematical competitions
and journals. An important feature of the work is the comprehensive background material provided with each grouping of problems."
"Aimed at motivated high school and beginning college students and instructors, this work can be used as a text for advanced
problem-solving courses, for self-study, or as a resource for teachers and students training for mathematical competitions
and for teacher professional development, seminars, and workshops."--Jacket
Mathematical miniatures by Svetoslav Savchev(
Book
) 14
editions published
between
2003
and
2014
in
English
and held by
395 WorldCat member
libraries
worldwide
102 combinatorial problems : from the training of the USA IMO team by Titu Andreescu(
Book
) 12
editions published
in
2003
in
English
and held by
365 WorldCat member
libraries
worldwide
"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."--Jacket
Mathematical Olympiad treasures by Titu Andreescu(
Book
) 30
editions published
between
2002
and
2012
in
English
and held by
347 WorldCat member
libraries
worldwide
Mathematical Olympiad Treasures aims at buiding a bridge between ordinary high school exercises and more sophisticated, intricate
and abstract concepts and problems in undergraduate mathematics. The book contains a stimulating collection of problems in
the subjects of geometry and trigonometry, algebra, number theory and combinatorics. While it may be considered a sequel to
Mathematical Olympiad Challenges, the focus of Treasures in on engaging a wider audience of undergraduates to think creatively
in applying techniques and strategies to problems in the real world. The problems are clustered by topic into self-contained
sections. Unlike Challenges, however, Treasures begins with elementary facts, followed by a number of carefully selected problems
and an extensive discussion of their solutions. This discussion then leads to more complicated and more intellectually challenging
problems, as well as their solutions. Throughout the book students are encouraged to express their ideas, conjectures, and
conclusions in writing. The goal is to help readers develop a host of new mathematical tools and strategies that will be useful
beyond the classroom and in a number of disciplines. -- from back cover
Complex numbers from A to--Z by Titu Andreescu(
Book
) 41
editions published
between
2004
and
2014
in
English and German
and held by
289 WorldCat member
libraries
worldwide
"The book reflects the unique experience of the authors. It distills a vast mathematical literature, most of which is unknown
to the western public, and captures the essence of an abundant problem culture. The target audience includes undergraduates,
high school students and their teachers, mathematical contestants (such as those training for Olympiads or the W.L. Putnam
Mathematical Competition) and their coaches, as well as anyone interested in essential mathematics."--Jacket
A path to combinatorics for undergraduates : counting strategies by Titu Andreescu(
Book
) 13
editions published
between
2003
and
2004
in
English
and held by
257 WorldCat member
libraries
worldwide
"This approach to combinatorics is centered around challenging examples, fully-worked solutions, and hundreds of problems
- many from Olympiads and other competitions, and many original to the authors. Each chapter highlights a particular aspect
of the subject and casts combinatorial concepts in the guise of questions, illustrations, and exercises that are designed
to encourage creativity, improve problem-solving techniques, and widen the reader's mathematical horizons." "The authors'
previous text, 102 Combinatorial Problems, makes a fine companion volume to the present work, which is ideal for Olympiad
participants and coaches, advanced high school students, undergraduates, and college instructors. The book's unusual problems
and examples will stimulate seasoned mathematicians as well. A Path to Combinatorics for Undergraduates is a lively introduction
not only to combinatorics, but also to mathematical ingenuity, rigor, and the joy of solving puzzles."--BOOK JACKET
104 number theory problems : from the training of the USA IMO team by Titu Andreescu(
Book
) 22
editions published
between
2007
and
2010
in
English and Japanese
and held by
245 WorldCat member
libraries
worldwide
Develops problem-solving skills needed to excel in mathematical contests and in mathematical research in number theory. This
book contains problems that encourage students to express their ideas in writing to explain how they conceive problems, what
conjectures they make, and what conclusions they reach
103 trigonometry problems : from the training of the USA IMO team by Titu Andreescu(
Book
) 20
editions published
between
2005
and
2010
in
English
and held by
227 WorldCat member
libraries
worldwide
103 Trigonometry Problems contains highly-selected problems and solutions used in the training and testing of the USA International
Mathematical Olympiad (IMO) team. Though many problems may initially appear impenetrable to the novice, most can be solved
using only elementary high school mathematics techniques. Key features: Gradual progression in problem difficulty builds and
strengthens mathematical skill and techniques ; Basic topics include trigonometric formulas and identities, their applications
in the geometry of the triangle, trigonometric equations and inequalities, and substitutions involving trigonometric functions
; Problem-solving tactics and strategies, along with practical test-taking techniques, provide in-depth enrichment and preparation
for possible participation in various mathematical competitions ; Comprehensive introduction (first chapter) to trigonometric
functions, their relations and functional properties, and their applications in the Euclidean plane and solid geometry expose
advanced students to college level material. -- from back cover
Geometric problems on maxima and minima by Titu Andreescu(
Book
) 23
editions published
between
2005
and
2006
in
English
and held by
218 WorldCat member
libraries
worldwide
Questions of maxima and minima have great practical significance, with applications to physics, engineering, and economics;
they have also given rise to theoretical advances, notably in calculus and optimization. Indeed, while most texts view the
study of extrema within the context of calculus, this carefully constructed problem book takes a uniquely intuitive approach
to the subject: it presents hundreds of extreme value problems, examples, and solutions primarily through Euclidean geometry.
Key features and topics: * Comprehensive selection of problems, including Greek geometry and optics, Newtonian mechanics,
isoperimetric problems, and open questions, such as Malfatti?s problem * Unified approach to the subject, with emphasis on
geometric, algebraic, analytic, and combinatorial reasoning * Presentation and application of classical inequalities, including
Cauchy--Schwarz and Minkowski?s Inequality; important results in real variable theory, such as the Intermediate Value Theorem;
and emphasis on simple but useful geometric concepts, including transformations, convexity, and symmetry * Clear solutions
to the problems, often accompanied by figures * Hundreds of exercises of varying difficulty, from straightforward to Olympiad-caliber
Written by a team of established mathematicians and teachers, this work draws on the authors? experience in the classroom
and as Olympiad coaches. By exposing readers to a wealth of creative problem-solving approaches, the text communicates not
only geometry but also algebra, calculus, and topology. Ideal for use at the junior and senior undergraduate level, as well
asin enrichment programs and Olympiad training for advanced high school students, this book?s breadth and depth will appeal
to a wide audience, from secondary school teachers and pupils to graduate students, professional mathematicians, and puzzle
enthusiasts. TOC:Preface * Methods for Solving Geometric Problems on Maxima and Minima * Selected Types of Geometric Problems
on Maxima and Minima * Miscellaneous Problems * Hints and Solutions * Glossary
Number theory : structures, examples, and problems by Titu Andreescu(
Book
) 22
editions published
in
2009
in
English
and held by
168 WorldCat member
libraries
worldwide
While the forefront of number theory is replete with sophisticated and famous open problems, at its foundation are basic,
elementary ideas that can stimulate and challenge beginning students. This introductory text focuses on a problem-solving
approach to the subject, situating each concept within the framework of an example for readers to solve
An introduction to diophantine equations : a problem-based approach by Titu Andreescu(
Book
) 22
editions published
between
2002
and
2010
in
English
and held by
145 WorldCat member
libraries
worldwide
"This problem-solving book is an introduction to the study of Diophantine equations, a class of equations in which only integer
solutions are allowed. The material is organized in two parts: Part I introduces the reader to elementary methods necessary
in solving Diophantine equations, such as the decomposition method, inequalities, the parametric method, modular arithmetic,
mathematical induction, Fermat's method of infinite descent, and the method of quadratic fields; Part II contains complete
solutions to all exercises in Part I. The presentation features some classical Diophantine equations, including linear, Pythagorean,
and some higher degree equations, as well as exponential Diophantine equations. Many of the selected exercises and problems
are original or are presented with original solutions. [This book] is intended for undergraduates, advanced high school students
and teachers, mathematical contest participants - including Olympiad and Putnam competitors - as well as readers interested
in essential mathematics. The work uniquely presents unconventional and non-routine examples, ideas, and techniques."--Back
cover
Problems in real analysis : advanced calculus on the real axis by Teodora-Liliana T Rădulescu(
Book
) 22
editions published
in
2009
in
English
and held by
108 WorldCat member
libraries
worldwide
"Problems in Real Analysis: Advanced Calculus on the Real Axis features a comprehensive collection of challenging problems
in mathematical analysis that aim to promote creative, non-standard techniques for solving problems. This self-contained text
offers a host of new mathematical tools and strategies Which develop a connection between analysis and other mathematical
disciplines, such as physics and engineering. A broad view of mathematics is presented throughout; the text is excellent for
the classroom or self-study. It is intended for undergraduate and graduate students in mathematics, as well as for researchers
engaged in the interplay between applied analysis, mathematical physics, and numerical analysis."--Jacket
Quadratic Diophantine equations by Titu Andreescu(
Book
) 14
editions published
between
2007
and
2015
in
English and Spanish
and held by
58 WorldCat member
libraries
worldwide
"This monograph treats the classical theory of quadratic Diophantine equations and guides the reader through the last two
decades of computational techniques and progress in the area. These new techniques combined with the latest increases in computational
power shed new light on important open problems. The authors motivate the study of quadratic Diophantine equations with excellent
examples, open problems, and applications. Moreover, the exposition aptly demonstrates many applications of results and techniques
from the study of Pell-type equations to other problems in number theory. The book is intended for advanced undergraduate
and graduate students as well as researchers. It challenges the reader to apply not only specific techniques and strategies,
but also to employ methods and tools from other areas of mathematics, such as algebra and analysis"--
Essential linear algebra with applications : a problem-solving approach by Titu Andreescu(
Book
) 17
editions published
between
2014
and
2016
in
English
and held by
57 WorldCat member
libraries
worldwide
Rooted in a pedagogically successful problem-solving approach to linear algebra, the present work fills a gap in the literature
that is sharply divided between elementary texts and books that are too advanced to appeal to a wide audience. It clearly
develops the theoretical foundations of vector spaces, linear equations, matrix algebra, eigenvectors, and orthogonality,
while simultaneously emphasizing applications and connections to fields such as biology, economics, computer graphics, electrical
engineering, cryptography, and political science. Ideal as an introduction to linear algebra, t | 677.169 | 1 |
...
Show More to help students reach different results. A variety of fundamental proofs demonstrate the basic steps in the construction of a proof and numerous examples illustrate the method and detail necessary to prove various kinds of theorems.Jumps right in with the needed vocabulary-gets students thinking like mathematicians from the beginningOffers a large variety of examples and problems with solutions for students to work through on their ownIncludes a collection of exercises without solutions to help instructors prepare assignmentsContains an extensive list of basic mathematical definitions and concepts needed in abstract mathematics | 677.169 | 1 |
Calculus I Advice
Showing 1 to 2 of 2
I recommend this course for those who are going into STEM related fields, because it is never hurts to better understand the mathematics and it will be helpful in the work force. An important part of life is to make calculations and predictions.
Course highlights:
The highlight of this course is better understanding mathematics, before I had a hard time comprehending the subject. It has taught me to gain an understanding of limits and integrals.
Hours per week:
12+ hours
Advice for students:
It is very important to practice, without practicing I would not have succeed in this course
Course Term:Spring 2017
Professor:Arthur Morris
Course Required?Yes
Course Tags:Math-heavyAlways Do the ReadingCompetitive Classmates
Feb 11, 2016
| Would highly recommend.
This class was tough.
Course Overview:
Mr. Williams teaches the class in a way where you enjoy learning. He will make jokes and keep things interesting throughout every lecture. He is a very caring man and a wonderful teacher; if a student is stuck on a problem in his class, he will help you get through it and will not judge you or think less of you for not being able to understand it right away.
Course highlights:
I felt like Calculus was going to be a death sentence for my GPA but I was able to pass the class with a B. The way Mr. Williams teaches will keep you engaged and interested throughout the entire semester.
Hours per week:
12+ hours
Advice for students:
If you are going to take Calculus, be prepared for a LOT of homework and a LOT of studying. Calculus has so many different formulas that need to be memorized that you really need to prepare yourself in order to pass the class. | 677.169 | 1 |
This activity has passed
About This Activity
Description
Teacher Referral Required to Sign Up for this Course. Students enrolled in this course did not meet criteria for placement into 9th grade Geometry. They have been recommended by their eighth grade math teacher because they have the potential to be successful in Geometry with more practice of prerequisite skills. Students will complete lessons and activities based on the 400 Level College and Career Readiness Skills. Skill bands included will be: basic operations and applications, probability, statistics, data analysis, number concepts and properties, expressions, equations and inequalities, graphical representations, properties of plane figures, and measurement. If students demonstrate proficiency of skills assessed throughout this course they will be placed into 9th grade Geometry this Fall | 677.169 | 1 |
ALGEBRA 2
REMEDIATION BY
STANDARD
Preparing for the FSA
MAFS.912.A-APR.1.1
UNDERSTAND THAT POLYNOMIALS FORM A SYSTEM ANALOGOUS TO THE
INTEGERS; NAMELY, THEY ARE CLOSED UNDER THE OPERATIONS OF
ADDITION, SUBTRACTION, AND MULTIPLICATION; ADD, SUBTRACT, AND
M
AP Calculus AB Advice
Showing 1 to 3 of 6
This is a great opportunity to get your upper level math courses started with extra attention, support, and extra time. This is considered a college course, you will be taking it in high school. Besides the extra semester to grab on the concepts, you will get one of the most prepared, intelligent, and understanding porfessor you will ever get the chance to get.
Course highlights:
I learned the basics of calculus and its applications to real life issues. The varios opportunities to get ahead and understand every concept allow you to have great discussions in class without the stress of not passing.
Hours per week:
6-8 hours
Advice for students:
Do your work and your readings, but most of all, ask questions. Take advantage of the conpetitive and the knwledgable prifessor. He always has time and patience to help you, make sure to ask for help.
Course Term:Fall 2015
Professor:collins
Course Tags:Math-heavyGo to Office HoursCompetitive Classmates
Jun 16, 2017
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
Calculus can be daunting, showing uo to class and doing what you have to will get you through the course with very little to figure out on your own. Lots of help and support.
Course highlights:
Lots of time to get help. Clear and easy instructions. Bonding and fun activities. Relevant and current information. I learned the basics of calculus and its applications to other fileds. Also, tips to get through college.
Hours per week:
6-8 hours
Advice for students:
Do your part and you will have fun and learn lots
Course Term:Fall 2015
Professor:collins
Course Tags:Math-heavyBackground Knowledge ExpectedGo to Office Hours
Feb 06, 2017
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
I would recommend this course because it expands your mind in the math aspect. Mr.Collins is a great teacher and you will understand his teaching skills and would have a blast learning from him.
Course highlights:
The highlights of this course is that you will understand the material giving to you if you put your mind and soul into it. I learned that you need to study hard, do the homework and listen if you plan to do well in this class.
Hours per week:
6-8 hours
Advice for students:
The advice i would give to students is that you need to study at least an hour every day, read the book material, and do the homework to be successful in this class. | 677.169 | 1 |
mPustakSubtract Mathematics Practice app that lets you learn with a lot of fun!Graphviewer scientific equation solver You can calculate the function inflection points
Multiple functions can be plotted in different colors at once
The intersection points of the different equations are calculated and can be shown | 677.169 | 1 |
Algebra: Working with Rational Exponents
Be sure that you have an application to open
this file type before downloading and/or purchasing.
854 KB|13 pages
Product Description
This Smart Notebook file can be used at the foundation for a lesson that introduces students to using rational exponents. Students learn to change from radical expressions to exponential expressions. This lesson is appropriate for Algebra I.
The zipped file contains a Smart Notebook file and a set of directions for using the Smart Notebook file. | 677.169 | 1 |
Edexcel GCSEMathematics Linked Pair Pilot
Here, you'll find everything you need to study for or to teach the Edexcel GCSE Mathematics Linked Pair Pilot, including key documents and the latest news.
Students of our Edexcel linked pair maths GCSEs will cover the rigorous core national curriculum programme of study while also gaining a broader grounding in both methods in mathematics and applications of mathematics.
Course materials
Support for
Why choose this specification?
The 'linked pair' pilot began in September 2010, and consists of two GCSEs:
Methods, which looks at the pure aspects of mathematics
Applications, which includes using mathematics in everyday contexts (including financial applications) and problem-solving in real-life scenarios.
The 'linked pair' pilot is worth two GCSEs, and students must enter for both for either to count in the performance measures.
Jointly, the linked pair GCSEs cover the rigorous core national curriculum programme of study, which is also assessed by the single GCSE. The pair, in addition, give a broader grounding in both methods in mathematics and applications of mathematics. | 677.169 | 1 |
From what I hear, a fair number of homeschooling parents worry about teaching algebra. Why? Because it's the first branch of math that they've forgotten — in whole or in part.
So, what 's a homeschooler to do?
OPTION 1: Find a curriculum that teaches algebra in a way that's both fun and clear.
OPTION 2: Get a tutor.
OPTION 2: Enroll your child at the local community college.
But each of these options comes with a problem. Finding a great algebra curriculum is difficult as most curricula are dull and don't explain concepts clearly. Getting a tutor who works well with your child can be difficult and expensive. And with the community college option, the school may not even offer the entry-level algebra course your child needs.
In this brief article I'd like to present a few options for algebra that you may not have considered, options that can really work.
First — and why does this sound like an ad? — you can get MY products, the Algebra Survival Guide and Algebra Survival Guide Workbook. The two books, used together, offer an Algebra 1 program that is fun, easy to use, and clear as a silver bell on Sunday morning.
I know this sounds like an infomercial, but keep in mind that I am genuinely enthusiastic about my products — because so many people have told me how much they like them. So please allow me to explain what makes these books perfect for the task at hand.
The Algebra Survival Guide is written in a Q&A format: question in the voice of a befuddled student, answer in the voice of a kindly tutor. As a result, reading the book is like stepping into a pleasant dialogue, not getting blasted by a boring lecture. Secondly the pages are designed in such a way that they present just one concept at a time. This format keeps students from feeling overwhelmed — a major problem in math education. To help students nail down the concepts and skills, the pages offer short sets of practice problems at the bottom, with all answers provided. Finally, the book is written in a conversational style, very much like the way a teenager might talk to a friend. Nothing fancy, and no textbook-ese. To round out the program, the complementary Algebra Survival Guide WORKBOOK provides thousands of additional practice problems, gently nudging students along the path from awareness to competence to mastery! In short, the Algebra Survival Guide and Workbook are inspired, cover to cover by the recognition that students are human beings, not math-learning droids, and they need to be treated as such.
Using the Guide and the Workbook, students will learn the following topics in a standard Algebra 1 curriculum: Properties of Numbers / Sets of Numbers / Positive and Negative Numbers / Order of Operations / Like Terms / Absolute Value / Exponents / Radicals / Factoring / Solving Equations / the Coordinate Plane / and (even the dreaded ) Word Problems.
What's more, the Algebra Survival Guide offers stories from everyday life that help students make sense of algebra's tricky abstractions. Examples: the book offers a tug-of-war story to help students grasp the rules for positive and negative numbers; a detective story to demonstrate the process of solving equations. As "icing on the cake," the book offers cartoons to bring algebra's abstractions to life, a, fold-out "Emergency Fact" poster to display algebra's key rules and formulas — and even a board game that turns practicing lessons into playful fun.
To check out the Algebra Survival Guide and Algebra Survival Guide Workbook, click on over to Amazon.com and read the glowing reviews, many written by homeschoolers.
Another approach to the algebra dilemma is to purchase a standard curriculum, and use the Algebra Survival products as a supplement. If you take this route, I recommend textbooks by McDougal Littel or Prentice Hall. I've found that these two companies' upper-level math textbooks are logically organized and well edited.
While these texts are well written, students do at times get confused by certain concepts. Not to fear, though. You can use the Algebra Survival Guide like an algebraic life-preserver. For example, suppose your child gets stuck in the chapter on rules of exponents. Switch for a while to the Algebra Survival Guide, turning to the exponents chapter. Once your child grasps the explanations in this book, switch back to the textbook to continue with the curriculum.
A third option is to do some online learning. There are many companies that now offer online instruction, and I have jumped onto that bandwagon too. I now offer live online instruction through SKYPE. Working this way, I either tutor students struggling with another curriculum, or teach students directly from my program. I can even do classes with a number of students. If either possibility appeals, shoot me an email — josh@SingingTurtle.com, — and I'll help you explore the possibilities.
No matter which path you pursue, there are a variety of ways to teach your children algebra without giving yourself a year-long headache. Please keep in mind that I, a published math author, am available to help homeschoolers in whatever way I can.
Josh Rappaport, former homeschooling father, lives, writes, and tutors in Santa Fe, New Mexico. Josh is the author of several math books, including the Card Game Roundup series and PreAlgebra Blastoff! In addition to the Algebra Survival Guide and Workbook. Josh tutors and teaches online — and also offers math workshops at homeschooling conferences. Josh can be reached at josh@SingingTurtle.com or by calling 505-690-2351. Or visit Josh through his blog: mathchat.wordpress.com
Go Green Now!
Subscribe to The Way Home E-newsletter & receive a free digital download | 677.169 | 1 |
MATH 0115 – Foundations of Mathematics I
(0 credits) 5 hours lecture, 1 hour tutorial
This is a credit-free upgrading course; special fees apply.
This course comprehensively covers the essential topics of algebra and the basic principles of geometry to an intermediate level. The course is designed to meet the needs of students who have a minimal background in algebra or whose proficiency may have declined during years away from formal education. Successful completion of this course prepares students for Mathematics 0130 or Mathematics 0132, where algebraic skills are utilized to study more advanced topics. | 677.169 | 1 |
Learn the concepts and methods of linear algebra, and how to use them to think about computational problems arising in computer science. Coursework includes building on the concepts to write small programs and run them on real data.
Featuring recorded lectures from the Harvard School of Engineering and Applied Sciences course Computer Science 20, this course covers widely applicable mathematical tools for computer science, including topics from logic, set theory, combinatorics, number theory, probability theory, and graph theory. It includes practice in reasoning formally and proving theorems. Students meet twice a week via web conference to solve problems collaboratively.
In recent years calligraphers have felt an increased demand for two unique skills: modern flourishesthat are elegant (without looking like they came out of the eighteenth century) and digitized calligraphythat | 677.169 | 1 |
Be sure that you have an application to open
this file type before downloading and/or purchasing.
1 MB|3 (including 1 cover and 1 answer key)
Product Description
How confident the students are when evaluating the explicit formula of a geometric sequence? Here is a fun way to check for students' understanding in "Finding a term of a geometric sequence given the explicit formula" .
An explicit formula is provided and students are asked to find a specific term using the formula. Some explicit formulas contain fractions. It is really up to the teacher whether students could use a calculator (and to what extend) or work the questions manually. In the preview, please find examples of similar questions. The subscript of "a" will range from 1 to 11.
This maze focuses purely on evaluating explicit formulas for a particular term in a geometric sequence. It uses the notation "a with a number subscript".
There are 15 possible sequences provided in each of the maze. From start to end, the student will be able to answer 13 questions out of the 15 provided to get to the end of the maze | 677.169 | 1 |
Scientific calculator Calculator simple interface but with all the scientific functions without advertising
Uniconver Unit converter and volume & area calculator for various geometrical shapesClickr Clickr is a simple application that allows you to count things like people, cars or even sheep just by tapping on the screen or pressing the trackball
Scientific Calculator AD A scientific calculator is a type of electronic calculator, usually but not always handheld, designed to calculate problems in science, engineering, and mathematics | 677.169 | 1 |
Calculus Homework Help
Calculus is a fundamental math discipline that consists of two parts: differential calculus and integral calculus. This is the main part of math education today. Calculus is a key to another, even more complex branches of mathematics that study functions and limits, and in wide sense are called mathematical analysis. Calculus is widely applied in science, economics and engineering, it is able to solve a big variety of problems which can't be solved with only algebra methods. Differential calculus is used in calculations related to velocity and acceleration, slope of the curve and optimization. On the other hand, integral calculus is applied in calculations of areas, volumes, curve lengths, centers of gravity, work and pressure (in physics). Calculus is applied to get better representation of nature of space, time and motion.
You have to know large set of formulas to be able to solve calculus assignment, have skills of their application and understand material in details. You also need several examples of solution of different types of problems to be capable to complete similar exercises yourself. Every method used in solution must be clear for you. Calculus homework help provided by our service contains step-by-step explanation of the solution, so you will understand everything and will be able to explain it to your classmates! Deep understanding of the subject of calculus will help you with math studying in future, develop your mind and allow to apply your knowledge in practice. Assignment4Student expert team is ready to provide calculus homework help you get at school or university. Our help is always qualified and done in time! You are welcome to ask any questions concerning your task or solution we provide.
This subject is very important in every discipline you study or field you work. Mathematical formalism experienced in calculus is the basis of your further studying. Our tutors will solve calculus assignment for you with as much details and explanations as you need. You just need to create new order or write us an email.
You can see more details about how we provide calculus homework help on Home and How It Works pages. | 677.169 | 1 |
List of resources to fully understand Mathematics
Mathematics is a very essential subject that will always be part of our lives, you'll always need to understand some basics to move forward.
Through my studies and still going, this list has helped me a lot to understand new things. These are the only sites you need, no more searching. Hope it helps you.
Khan Academy - A video site that teaches you everything and step by step, all is generalized, not much examples
Patrick JMT - A video site with some tutorials but the main focus is on known problems, that are solved
Math is fun - From the first glance it will look like it's made for kids, I could say not but it's made in mind so that even a kid understands it. It's not easy to find those advanced tutorials you're looking for, just google "mathisfun.com keyword"
Wolfram Alpha - An online calculator with amazing features - it does almost everything and for extended functionality it's like $3-$5 a month
Matlab - A very huge and costly software but a one that also is very good for generating graphs and doing advanced calculations, you need to learn how to code (it's easy) but it pays off if you're doing a lot of reports and data processing, you know where to find it without paying | 677.169 | 1 |
Using Sine and Cosine Graphs
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this file type before downloading and/or purchasing.
209 KB|2 pages
Product Description
This worksheet has students use the sine and cosine graphs to answer some questions. A calculator is NOT to be used to complete this worksheet. Teaching activities for the sine and cosine wave are available at The Math Machine. | 677.169 | 1 |
Grade 8 Math: Functions
This topic covers an introduction to functions. This series of three lessons introduces the student to a definition of a function. Many definitions related to functions are given.
This topic also covers whether a given graph represents a function or a relationship between two variables.
Functions: Definitions
This lesson introduces the student to functions. The instruction defines the domain and range of a function, as well. Then, the student learns the definition for function. They also learn that relationships between numbers that are not functions are called relations.
Functions: Notation
In this lesson the student learns the standard f(x) notation for a function. They also learn the terms independent and dependent variable.
Functions: Vertical Line Test
In this lesson, the student learns about the vertical line test. The lesson explains that the vertical line test is used to determine whether a graph of a two variable relationship is a function or is just a relation. | 677.169 | 1 |
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Sophie
is hungry and goes off in search of
food. Since Sophie is not very smart, she
moves in a straight line. Let be the distance
Sophie has traveled at time
Section
8.2.2
Eulers Method
Leonhard Euler ("Oiler) , 15 April 1707 18 September
1783, was a pioneering Swiss mathematician and physicist.
He made important discoveries in fields as diverse as
infinitesimal calculus and graph theory. He also introduced
mu
History of Theories of Evil
Thomas Hobbes
Declared that human life was solitary, poor, nasty, brutish, and short, that the natural
state of man was beast like.
Civilization recuses humanity
Jean-Jacques Rousseau
Believed that man is born naturally good
Beach Park
Block 3 Homographs and Homophones Assessment
Name
Question #9:
Write the homograph that could fill both blanks.
The white _ spread its wings and _ down
from the roof to the ground.
Question #1:
Write the homograph that could fill both blanks.
T
POPULATION EFFECT ON THE ENVIROMENT AND CHALLENGES
Population growth and distribution have significant roles to play in the sustainability
of the world's vast resources. Not only the number of people, but also the lifestyle,
consumption patterns, and regi
Economics
Writing Assignment: Prioritizing Economic Goals
Top Scores (A-B) - The top scoring students will do the following:
Provide a paragraph of at least five sentences that describes the economic goal.
Provide two logical reasons that support their
Mehwish Butt
214246995
ANTH 1120E
Rhiannon Mosher
March 9, 2016
Question: How has cultural change affected the settlement of the global movement of people,
ideas, goods, and money around the world in peoples everyday lives?
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Unformatted text preview: 5E-10(pp 622-631) 1/18/06 9:18 AM Page 622 CHAPTER 10
By analyzing pairs of differential equations we gain
insight into population
cycles of predators and
prey, such as the Canada
lynx and snowshoe hare. W
150 100 R
3000 W R
W 50 120
2000
80 0 1000 2000 3000 R
1000
40 0 t¡ t™ t£ Differential Equations t 5E-10(pp 622-631) 1/18/06 9:18 AM Page 623 Perhaps the most important of all the applications of calculus is to differential equations. When physical scientists
or social scientists use calculus, more often than not it is to
analyze a differential equation that has arisen in the process of modeling some phenomenon that they are studying.
Although it is often impossible to find an explicit formula for the solution of a differential equation, we will see that graphical and numerical approaches provide the
needed information. |||| 10.1 Modeling with Differential Equations |||| Now is a good time to read (or reread) the
discussion of mathematical modeling on page 25. In describing the process of modeling in Section 1.2, we talked about formulating a mathematical model of a real-world problem either through intuitive reasoning about the phenomenon or from a physical law based on evidence from experiments. The mathematical
model often takes the form of a differential equation, that is, an equation that contains an
unknown function and some of its derivatives. This is not surprising because in a realworld problem we often notice that changes occur and we want to predict future behavior
on the basis of how current values change. Let's begin by examining several examples of
how differential equations arise when we model physical phenomena. Models of Population Growth
One model for the growth of a population is based on the assumption that the population
grows at a rate proportional to the size of the population. That is a reasonable assumption
for a population of bacteria or animals under ideal conditions (unlimited environment, adequate nutrition, absence of predators, immunity from disease).
Let's identify and name the variables in this model:
t time ͑the independent variable͒
P the number of individuals in the population ͑the dependent variable͒
The rate of growth of the population is the derivative dP͞dt. So our assumption that the
rate of growth of the population is proportional to the population size is written as the
equation
1 dP
kP
dt where k is the proportionality constant. Equation 1 is our first model for population
growth; it is a differential equation because it contains an unknown function P and its
derivative dP͞dt.
Having formulated a model, let's look at its consequences. If we rule out a population
of 0, then P͑t͒ Ͼ 0 for all t. So, if k Ͼ 0, then Equation 1 shows that PЈ͑t͒ Ͼ 0 for all t.
This means that the population is always increasing. In fact, as P͑t͒ increases, Equation 1
shows that dP͞dt becomes larger. In other words, the growth rate increases as the population increases. 623 5E-10(pp 622-631) 624 ❙❙❙❙ 1/18/06 9:18 AM Page 624 CHAPTER 10 DIFFERENTIAL EQUATIONS P Let's try to think of a solution of Equation 1. This equation asks us to find a function
whose derivative is a constant multiple of itself. We know that exponential functions have
that property. In fact, if we let P͑t͒ Ce kt, then
PЈ͑t͒ C͑ke kt ͒ k͑Ce kt ͒ kP͑t͒
t Thus, any exponential function of the form P͑t͒ Ce kt is a solution of Equation 1. When
we study this equation in detail in Section 10.4, we will see that there is no other solution.
Allowing C to vary through all the real numbers, we get the family of solutions
P͑t͒ Ce kt whose graphs are shown in Figure 1. But populations have only positive
values and so we are interested only in the solutions with C Ͼ 0. And we are probably concerned only with values of t greater than the initial time t 0. Figure 2 shows the physically meaningful solutions. Putting t 0, we get P͑0͒ Ce k͑0͒ C, so the constant C
turns out to be the initial population, P͑0͒.
Equation 1 is appropriate for modeling population growth under ideal conditions, but
we have to recognize that a more realistic model must reflect the fact that a given environment has limited resources. Many populations start by increasing in an exponential
manner, but the population levels off when it approaches its carrying capacity K (or
decreases toward K if it ever exceeds K). For a model to take into account both trends, we
make two assumptions: FIGURE 1 The family of solutions of dP/dt=kP
P 0 t ■ dP
Ϸ kP if P is small (Initially, the growth rate is proportional to P.)
dt ■ dP
Ͻ 0 if P Ͼ K (P decreases if it ever exceeds K.)
dt FIGURE 2 The family of solutions P(t)=Ce kt
with C>0 and t˘0 A simple expression that incorporates both assumptions is given by the equation 2 P P =K equilibrium
solutions
P =0
0 FIGURE 3 Solutions of the logistic equation t ͩ ͪ dP
P
kP 1 Ϫ
dt
K Notice that if P is small compared with K, then P͞K is close to 0 and so dP͞dt Ϸ kP. If
P Ͼ K , then 1 Ϫ P͞K is negative and so dP͞dt Ͻ 0.
Equation 2 is called the logistic differential equation and was proposed by the Dutch
mathematical biologist Pierre-François Verhulst in the 1840s as a model for world population growth. We will develop techniques that enable us to find explicit solutions of the
logistic equation in Section 10.5, but for now we can deduce qualitative characteristics of
the solutions directly from Equation 2. We first observe that the constant functions
P͑t͒ 0 and P͑t͒ K are solutions because, in either case, one of the factors on the right
side of Equation 2 is zero. (This certainly makes physical sense: If the population is ever
either 0 or at the carrying capacity, it stays that way.) These two constant solutions are
called equilibrium solutions.
If the initial population P͑0͒ lies between 0 and K, then the right side of Equation 2 is
positive and the population
decreases. Notice that, in either case, if the population approaches the carrying capacity
͑P l K͒, then dP͞dt l 0, which means the population levels off. So we expect that the
solutions of the logistic differential equation have graphs that look something like the ones
in Figure 3. Notice that the graphs move away from the equilibrium solution P 0 and
move toward the equilibrium solution P K . 5E-10(pp 622-631) 1/18/06 9:18 AM Page 625 SECTION 10.1 MODELING WITH DIFFERENTIAL EQUATIONS ❙❙❙❙ 625 A Model for the Motion of a Spring
Let's now look at an example of a model from the physical sciences. We consider the
motion of an object with mass m at the end of a vertical spring (as in Figure 4). In Section 6.4 we discussed Hooke's Law, which says that if the spring is stretched (or compressed) x units from its natural length, then it exerts a force that is proportional to x:
m equilibrium
position restoring force Ϫkx 0 x m where k is a positive constant (called the spring constant). If we ignore any external resisting forces (due to air resistance or friction) then, by Newton's Second Law (force equals
mass times acceleration), we have x FIGURE 4 m 3 d 2x
Ϫkx
dt 2 This is an example of what is called a second-order differential equation because it
involves second derivatives. Let's see what we can guess about the form of the solution
directly from the equation. We can rewrite Equation 3 in the form
d 2x
k
x
2 Ϫ
dt
m
which says that the second derivative of x is proportional to x but has the opposite sign. We
know two functions with this property, the sine and cosine functions. In fact, it turns
out that all solutions of Equation 3 can be written as combinations of certain sine
and cosine functions (see Exercise 3). This is not surprising; we expect the spring to oscillate about its equilibrium position and so it is natural to think that trigonometric functions
are involved. General Differential Equations
In general, a differential equation is an equation that contains an unknown function and
one or more of its derivatives. The order of a differential equation is the order of the highest derivative that occurs in the equation. Thus, Equations 1 and 2 are first-order equations
and Equation 3 is a second-order equation. In all three of those equations the independent
variable is called t and represents time, but in general the independent variable doesn't
have to represent time. For example, when we consider the differential equation
4 yЈ xy it is understood that y is an unknown function of x.
A function f is called a solution of a differential equation if the equation is satisfied
when y f ͑x͒ and its derivatives are substituted into the equation. Thus, f is a solution of
Equation 4 if
f Ј͑x͒ xf ͑x͒
for all values of x in some interval.
When we are asked to solve a differential equation we are expected to find all possible
solutions of the equation. We have already solved some particularly simple differential 5E-10(pp 622-631) 626 ❙❙❙❙ 1/18/06 9:18 AM Page 626 CHAPTER 10 DIFFERENTIAL EQUATIONS equations, namely, those of the form
yЈ f ͑x͒
For instance, we know that the general solution of the differential equation
yЈ x 3
is given by
y x4
ϩC
4 where C is an arbitrary constant.
But, in general, solving a differential equation is not an easy matter. There is no systematic technique that enables us to solve all differential equations. In Section 10.2, however, we will see how to draw rough graphs of solutions even when we have no explicit
formula. We will also learn how to find numerical approximations to solutions.
EXAMPLE 1 Show that every member of the family of functions y 1 ϩ ce t
1 Ϫ ce t is a solution of the differential equation yЈ 1 ͑y 2 Ϫ 1͒.
2
SOLUTION We use the Quotient Rule to differentiate the expression for y: yЈ |||| Figure 5 shows graphs of seven members of
the family in Example 1. The differential equation
shows that if y Ϸ Ϯ1, then yЈ Ϸ 0. That is
borne out by the flatness of the graphs near
y 1 and y Ϫ1.
5 ͑y 2 Ϫ 1͒ 5
_5 FIGURE 5 ce t Ϫ c 2e 2t ϩ ce t ϩ c 2e 2t
2ce t
͑1 Ϫ ce t ͒2
͑1 Ϫ ce t ͒2 The right side of the differential equation becomes
1
2 _5 ͑1 Ϫ ce t ͒͑ce t ͒ Ϫ ͑1 ϩ ce t ͒͑Ϫce t ͒
͑1 Ϫ ce t ͒2 1
2 ͫͩ 1 ϩ ce t
1 Ϫ ce t ͪ ͬ ͫ
2 Ϫ1 1
2 ͑1 ϩ ce t ͒2 Ϫ ͑1 Ϫ ce t ͒2
͑1 Ϫ ce t ͒2 ͬ 1
4ce t
2ce t
2 ͑1 Ϫ ce t ͒2
͑1 Ϫ ce t ͒2 Therefore, for every value of c, the given function is a solution of the differential
equation.
When applying differential equations, we are usually not as interested in finding a family of solutions (the general solution) as we are in finding a solution that satisfies some
additional requirement. In many physical problems we need to find the particular solution
that satisfies a condition of the form y͑t0 ͒ y0 . This is called an initial condition, and the
problem of finding a solution of the differential equation that satisfies the initial condition
is called an initial-value problem.
Geometrically, when we impose an initial condition, we look at the family of solution
curves and pick the one that passes through the point ͑t0 , y0 ͒. Physically, this corresponds
to measuring the state of a system at time t0 and using the solution of the initial-value problem to predict the future behavior of the system. 5E-10(pp 622-631) 1/18/06 9:18 AM Page 627 SECTION 10.1 MODELING WITH DIFFERENTIAL EQUATIONS ❙❙❙❙ 627 EXAMPLE 2 Find a solution of the differential equation yЈ 2 ͑y 2 Ϫ 1͒ that satisfies the
1 initial condition y͑0͒ 2. SOLUTION Substituting the values t 0 and y 2 into the formula y 1 ϩ ce t
1 Ϫ ce t from Example 1, we get
2 1 ϩ ce 0
1ϩc
0
1 Ϫ ce
1Ϫc Solving this equation for c, we get 2 Ϫ 2c 1 ϩ c, which gives c 1 . So the solution
3
of the initial-value problem is
y |||| 10.1 Exercises
(d) Find a solution of the differential equation yЈ xy that satisfies the initial condition y͑1͒ 2. 1. Show that y x Ϫ x Ϫ1 is a solution of the differential equation xyЈ ϩ y 2x. 2. Verify that y sin x cos x Ϫ cos x is a solution of the initial- 7. (a) What can you say about a solution of the equation yЈ Ϫy 2 just by looking at the differential equation?
(b) Verify that all members of the family y 1͑͞x ϩ C ͒ are
solutions of the equation in part (a).
(c) Can you think of a solution of the differential equation
yЈ Ϫy 2 that is not a member of the family in part (b)?
(d) Find a solution of the initial-value problem value problem
yЈ ϩ ͑tan x͒y cos2 x y͑0͒ Ϫ1 on the interval Ϫ͞2 Ͻ x Ͻ ͞2.
3. (a) For what nonzero values of k does the function y sin kt satisfy the differential equation yЉ ϩ 9y 0 ?
(b) For those values of k, verify that every member of the family of functions yЈ Ϫy 2 is also a solution.
4. For what values of r does the function y e rt satisfy the differential equation y Љ ϩ yЈ Ϫ 6y 0? 5. Which of the following functions are solutions of the differen- tial equation y Љ ϩ 2yЈ ϩ y 0?
(a) y e t
(b) y e Ϫt
(c) y te Ϫt
(d) y t 2e Ϫt
6. (a) Show that every member of the family of functions
2 y Ce x ͞2 is a solution of the differential equation yЈ xy.
(b) Illustrate part (a) by graphing several members of the family of solutions on a common screen.
(c) Find a solution of the differential equation yЈ xy that satisfies the initial condition y͑0͒ 5. y͑0͒ 0.5 8. (a) What can you say about the graph of a solution of the equa- y A sin kt ϩ B cos kt ; 1 ϩ 1et
3 ϩ et
3
1 Ϫ 1et
3 Ϫ et
3 ; tion yЈ xy 3 when x is close to 0? What if x is large?
(b) Verify that all members of the family y ͑c Ϫ x 2 ͒Ϫ1͞2 are
solutions of the differential equation yЈ xy 3.
(c) Graph several members of the family of solutions on a
common screen. Do the graphs confirm what you predicted
in part (a)?
(d) Find a solution of the initial-value problem
yЈ xy 3 y͑0͒ 2 9. A population is modeled by the differential equation ͩ dP
P
1.2P 1 Ϫ
dt
4200 ͪ (a) For what values of P is the population increasing?
(b) For what values of P is the population decreasing?
(c) What are the equilibrium solutions? 5E-10(pp 622-631) 628 ❙❙❙❙ 1/18/06 9:18 AM Page 628 CHAPTER 10 DIFFERENTIAL EQUATIONS 10. A function y͑t͒ satisfies the differential equation 13. Psychologists interested in learning theory study learning dy
y 4 Ϫ 6y 3 ϩ 5y 2
dt
(a) What are the constant solutions of the equation?
(b) For what values of y is y increasing?
(c) For what values of y is y decreasing?
11. Explain why the functions with the given graphs can't be solu- tions of the differential equation
dy
e t͑ y Ϫ 1͒2
dt
(a) y dP
k͑M Ϫ P͒
dt (b) y 1 t 1 t 12. The function with the given graph is a solution of one of the following differential equations. Decide which is the correct
equation and justify your answer.
y 0 A. yЈ 1 ϩ xy |||| 10.2 B. yЈ Ϫ2xy k a positive constant is a reasonable model for learning.
(c) Make a rough sketch of a possible solution of this differential equation. 1 1 curves. A learning curve is the graph of a function P͑t͒, the
performance of someone learning a skill as a function of the
training time t. The derivative dP͞dt represents the rate at
which performance improves.
(a) When do you think P increases most rapidly? What
happens to dP͞dt as t increases? Explain.
(b) If M is the maximum level of performance of which the
learner is capable, explain why the differential equation x C. yЈ 1 Ϫ 2xy 14. Suppose you have just poured a cup of freshly brewed coffee with temperature 95ЊC in a room where the temperature
is 20ЊC.
(a) When do you think the coffee cools most quickly? What
happens to the rate of cooling as time goes by? Explain.
(b) Newton's Law of Cooling states that the rate of cooling of
an object is proportional to the temperature difference
between the object and its surroundings, provided that this
difference is not too large. Write a differential equation that
expresses Newton's Law of Cooling for this particular situation. What is the initial condition? In view of your answer
to part (a), do you think this differential equation is an
appropriate model for cooling?
(c) Make a rough sketch of the graph of the solution of the
initial-value problem in part (b). Direction Fields and Euler's Method
Unfortunately, it's impossible to solve most differential equations in the sense of obtaining an explicit formula for the solution. In this section we show that, despite the absence
of an explicit solution, we can still learn a lot about the solution through a graphical
approach (direction fields) or a numerical approach (Euler's method). Direction Fields
Suppose we are asked to sketch the graph of the solution of the initial-value problem
yЈ x ϩ y y͑0͒ 1 We don't know a formula for the solution, so how can we possibly sketch its graph? Let's
think about what the differential equation means. The equation yЈ x ϩ y tells us that the
slope at any point ͑x, y͒ on the graph (called the solution curve) is equal to the sum of the 5E-10(pp 622-631) 1/18/06 9:18 AM Page 629 SECTION 10.2 DIRECTION FIELDS AND EULER'S METHOD ❙❙❙❙ 629 x- and y-coordinates of the point (see Figure 1). In particular, because the curve passes
through the point ͑0, 1͒, its slope there must be 0 ϩ 1 1. So a small portion of the solution curve near the point ͑0, 1͒ looks like a short line segment through ͑0, 1͒ with slope 1
(see Figure 2).
y y Slope at
(¤, fi) is
¤+fi. Slope at
(⁄, ›) is
⁄+›. (0, 1) 0 x Slope at (0, 1)
is 0+1=1. 0 x FIGURE 1 FIGURE 2 A solution of yª=x+y Beginning of the solution curve through (0, 1) As a guide to sketching the rest of the curve, let's draw short line segments at a number of points ͑x, y͒ with slope x ϩ y. The result is called a direction field and is shown in
Figure 3. For instance, the line segment at the point ͑1, 2͒ has slope 1 ϩ 2 3. The direction field allows us to visualize the general shape of the solution curves by indicating the
direction in which the curves proceed at each point.
y y (0, 1)
0 1 2 x 0 1 2 FIGURE 3 FIGURE 4 Direction field for yª=x+y x The solution curve through (0, 1) Now we can sketch the solution curve through the point ͑0, 1͒ by following the direction field as in Figure 4. Notice that we have drawn the curve so that it is parallel to nearby
line segments.
In general, suppose we have a first-order differential equation of the form
yЈ F͑x, y͒
where F͑x, y͒ is some expression in x and y. The differential equation says that the slope
of a solution curve at a point ͑x, y͒ on the curve is F͑x, y͒. If we draw short line segments
with slope F͑x, y͒ at several points ͑x, y͒, the result is called a direction field (or slope
field). These line segments indicate the direction in which a solution curve is heading, so
the direction field helps us visualize the general shape of these curves. 5E-10(pp 622-631) 630 ❙❙❙❙ 1/18/06 9:18 AM Page 630 CHAPTER 10 DIFFERENTIAL EQUATIONS y EXAMPLE 1 2 (a) Sketch the direction field for the differential equation yЈ x 2 ϩ y 2 Ϫ 1.
(b) Use part (a) to sketch the solution curve that passes through the origin. 1 SOLUTION
_2 _1 0 1 2 (a) We start by computing the slope at several points in the following chart: x -1 yЈ x ϩ y Ϫ 1
2 FIGURE 5 2 1 0 1 2 0 1 2 Ϫ2 Ϫ1 0 1 2 ... 0 0 0 0 1 1 1 1 1 ... 3 0 Ϫ1 0 3 4 1 0 1 4 ... Now we draw short line segments with these slopes at these points. The result is the
direction field shown in Figure 5.
(b) We start at the origin and move to the right in the direction of the line segment
(which has slope Ϫ1 ). We continue to draw the solution curve so that it moves parallel
to the nearby line segments. The resulting solution curve is shown in Figure 6. Returning
to the origin, we draw the solution curve to the left as well. y _1 2 Ϫ1 0 y _2 _2 Ϫ2 x x The more line segments we draw in a direction field, the clearer the picture becomes.
Of course, it's tedious to compute slopes and draw line segments for a huge number of
points by hand, but computers are well suited for this task. Figure 7 shows a more detailed,
computer-drawn direction field for the differential equation in Example 1. It enables us to
draw, with reasonable accuracy, the solution curves shown in Figure 8 with y-intercepts
Ϫ2, Ϫ1, 0, 1, and 2. -1 _2 FIGURE 6 3 3 Module 10.2A shows direction fields and
solution curves for a variety of differential equations. _3 3 _3 _3 _3 FIGURE 7
R E L switch
FIGURE 9 3 FIGURE 8 Now let's see how direction fields give insight into physical situations. The simple electric circuit shown in Figure 9 contains an electromotive force (usually a battery or generator) that produces a voltage of E͑t͒ volts (V) and a current of I͑t͒ amperes (A) at time t.
The circuit also contains a resistor with a resistance of R ohms ( ⍀ ) and an inductor with
an inductance of L henries (H).
Ohm's Law gives the drop in voltage due to the resistor as RI.
1 L dI
ϩ RI E͑t͒
dt which is a first-order differential equation that models the current I at time t . 5E-10(pp 622-631) 1/18/06 9:18 AM Page 631 SECTION 10.2 DIRECTION FIELDS AND EULER'S METHOD ❙❙❙❙ 631 EXAMPLE 2 Suppose that in the simple circuit of Figure 9 the resistance is 12 ⍀, the
inductance is 4 H, and a battery gives a constant voltage of 60 V.
(a) Draw a direction field for Equation 1 with these values.
(b) What can you say about the limiting value of the current?
(c) Identify any equilibrium solutions.
(d) If the switch is closed when t 0 so the current starts with I͑0͒ 0, use the direction field to sketch the solution curve.
SOLUTION (a) If we put L 4, R 12, and E͑t͒ 60 in Equation 1, we get
4 dI
ϩ 12I 60
dt or dI
15 Ϫ 3I
dt The direction field for this differential equation is shown in Figure 10.
I
6 4 2 0 1 2 3 t FIGURE 10 (b) It appears from the direction field that all solutions approach the value 5 A, that is,
lim I͑t͒ 5 tlϱ (c) It appears that the constant function I͑t͒ 5 is an equilibrium solution. Indeed, we
can verify this directly from the differential equation. If I͑t͒ 5, then the left side is
dI͞dt 0 and the right side is 15 Ϫ 3͑5͒ 0.
(d) We use the direction field to sketch the solution curve that passes through ͑0, 0͒, as
shown in red in Figure 11. dI
15 Ϫ 3I
dt I
6 4 2 0 FIGURE 11 1 2 3 t 5E-10(pp 632-641) 632 ❙❙❙❙ 1/18/06 9:20 AM Page 632 CHAPTER 10 DIFFERENTIAL EQUATIONS Notice from Figure 10 that the line segments along any horizontal line are parallel. That
is because the independent variable t does not occur on the right side of the equation
IЈ 15 Ϫ 3I . In general, a differential equation of the form
yЈ f ͑y͒
in which the independent variable is missing from the right side, is called autonomous.
For such an equation, the slopes corresponding to two different points with the same
y-coordinate must be equal. This means that if we know one solution to an autonomous
differential equation, then we can obtain infinitely many others just by shifting the graph
of the known solution to the right or left. In Figure 11 we have shown the solutions
that result from shifting the solution curve of Example 2 one and two time units (namely,
seconds) to the right. They correspond to closing the switch when t 1 or t 2. Euler's Method
The basic idea behind direction fields can be used to find numerical approximations to
solutions of differential equations. We illustrate the method on the initial-value problem
that we used to introduce direction fields:
yЈ x ϩ y
y solution curve 1 y=L(x) 0 FIGURE 12 First Euler approximation 1 x y͑0͒ 1 The differential equation tells us that yЈ͑0͒ 0 ϩ 1 1, so the solution curve has slope
1 at the point ͑0, 1͒. As a first approximation to the solution we could use the linear
approximation L͑x͒ x ϩ 1. In other words, we could use the tangent line at ͑0, 1͒ as a
rough approximation to the solution curve (see Figure 12).
Euler's idea was to improve on this approximation by proceeding only a short distance
along this tangent line and then making a midcourse correction by changing direction as
indicated by the direction field. Figure 13 shows what happens if we start out along the
tangent line but stop when x 0.5. (This horizontal distance traveled is called the step
size.) Since L͑0.5͒ 1.5, we have y͑0.5͒ Ϸ 1.5 and we take ͑0.5, 1.5͒ as the starting point
for a new line segment. The differential equation tells us that yЈ͑0.5͒ 0.5 ϩ 1.5 2, so
we use the linear function
y 1.5 ϩ 2͑x Ϫ 0.5͒ 2x ϩ 0.5
as an approximation to the solution for x Ͼ 0.5 (the orange segment in Figure 13). If we
decrease the step size from 0.5 to 0.25, we get the better Euler approximation shown in
Figure 14.
y 1
0 y 1 1.5
0.5 1 x 0 0.25 1 x FIGURE 13 FIGURE 14 Euler approximation with step size 0.5 Euler approximation with step size 0.25 In general, Euler's method says to start at the point given by the initial value and proceed in the direction indicated by the direction field. Stop after a short time, look at the
slope at the new location, and proceed in that direction. Keep stopping and changing direc- 5E-10(pp 632-641) 1/18/06 9:20 AM Page 633 SECTION 10.2 DIRECTION FIELDS AND EULER'S METHOD y slope=F(x¸, y¸)
(⁄, ›) hF(x¸, y¸)
h
y¸ 0 ⁄ x FIGURE 15 633 tion according to the direction field. Euler's method does not produce the exact solution to
an initial-value problem—it gives approximations. But by decreasing the step size (and
therefore increasing the number of midcourse corrections), we obtain successively better
approximations to the exact solution. (Compare Figures 12, 13, and 14.)
For the general first-order initial-value problem yЈ F͑x, y͒, y͑x 0͒ y0 , our aim is to
find approximate values for the solution at equally spaced numbers x 0 , x 1 x 0 ϩ h,
x 2 x 1 ϩ h, . . . , where h is the step size. The differential equation tells us that the slope
at ͑x 0 , y0 ͒ is yЈ F͑x 0 , y0 ͒, so Figure 15 shows that the approximate value of the solution
when x x 1 is
y1 y0 ϩ hF͑x 0 , y0 ͒
Similarly, x¸ ❙❙❙❙ y2 y1 ϩ hF͑x 1, y1 ͒ In general, yn ynϪ1 ϩ hF͑x nϪ1, ynϪ1 ͒ EXAMPLE 3 Use Euler's method with step size 0.1 to construct a table of approximate values for the solution of the initial-value problem
yЈ x ϩ y y͑0͒ 1 SOLUTION We are given that h 0.1, x 0 0, y0 1, and F͑x, y͒ x ϩ y. So we have y1 y0 ϩ hF͑x 0 , y0 ͒ 1 ϩ 0.1͑0 ϩ 1͒ 1.1
y2 y1 ϩ hF͑x 1, y1 ͒ 1.1 ϩ 0.1͑0.1 ϩ 1.1͒ 1.22
y3 y2 ϩ hF͑x 2 , y2 ͒ 1.22 ϩ 0.1͑0.2 ϩ 1.22͒ 1.362
Module 10.2B shows how Euler's method
works numerically and visually for a
variety of differential equations and
step sizes. This means that if y͑x͒ is the exact solution, then y͑0.3͒ Ϸ 1.362.
Proceeding with similar calculations, we get the values in the table:
n xn yn n xn yn 1
2
3
4
5 0.1
0.2
0.3
0.4
0.5 1.100000
1.220000
1.362000
1.528200
1.721020 6
7
8
9
10 0.6
0.7
0.8
0.9
1.0 1.943122
2.197434
2.487178
2.815895
3.187485 For a more accurate table of values in Example 3 we could decrease the step size. But
for a large number of small steps the amount of computation is considerable and so we
need to program a calculator or computer to carry out these calculations. The following
table shows the results of applying Euler's method with decreasing step size to the initialvalue problem of Example 3.
Step size
|||| Computer software packages that produce
numerical approximations to solutions of
differential equations use methods that are
refinements of Euler's method. Although Euler's
method is simple and not as accurate, it is the
basic idea on which the more accurate methods
are based. Euler estimate of y͑0.5͒ Euler estimate of y͑1͒ 0.500
0.250
0.100
0.050
0.020
0.010
0.005
0.001 1.500000
1.625000
1.721020
1.757789
1.781212
1.789264
1.793337
1.796619 2.500000
2.882813
3.187485
3.306595
3.383176
3.409628
3.423034
3.433848 5E-10(pp 632-641) 634 ❙❙❙❙ 1/18/06 9:20 AM Page 634 CHAPTER 10 DIFFERENTIAL EQUATIONS Notice that the Euler estimates in the table seem to be approaching limits, namely, the
true values of y͑0.5͒ and y͑1͒. Figure 16 shows graphs of the Euler approximations with
step sizes 0.5, 0.25, 0.1, 0.05, 0.02, 0.01, and 0.005. They are approaching the exact solution curve as the step size h approaches 0.
y 1 FIGURE 16 Euler approximations
approaching the exact solution 0 0.5 1 x EXAMPLE 4 In Example 2 we discussed a simple electric circuit with resistance
12 ⍀, inductance 4 H, and a battery with voltage 60 V. If the switch is closed when
t 0, we modeled the current I at time t by the initial-value problem dI
15 Ϫ 3I
dt I͑0͒ 0 Estimate the current in the circuit half a second after the switch is closed.
SOLUTION We use Euler's method with F͑t, I͒ 15 Ϫ 3I, t0 0, I0 0, and step size h 0.1 second: I1 0 ϩ 0.1͑15 Ϫ 3 ؒ 0͒ 1.5
I2 1.5 ϩ 0.1͑15 Ϫ 3 ؒ 1.5͒ 2.55
I3 2.55 ϩ 0.1͑15 Ϫ 3 ؒ 2.55͒ 3.285
I4 3.285 ϩ 0.1͑15 Ϫ 3 ؒ 3.285͒ 3.7995
I5 3.7995 ϩ 0.1͑15 Ϫ 3 ؒ 3.7995͒ 4.15965
So the current after 0.5 s is
I͑0.5͒ Ϸ 4.16 A 5E-10(pp 632-641) 1/18/06 9:20 AM Page 635 SECTION 10.2 DIRECTION FIELDS AND EULER'S METHOD |||| 10.2 ❙❙❙❙ 635 Exercises 1. A direction field for the differential equation yЈ y (1 Ϫ 4 y 2)
1 I (iii) y͑0͒ Ϫ3 II y y 2 is shown.
(a) Sketch the graphs of the solutions that satisfy the given
initial conditions.
(i) y͑0͒ 1
(ii) y͑0͒ Ϫ1
(iv) y͑0͒ 3 2 2x _2 2x _2 (b) Find all the equilibrium solutions.
y _2 _2 3 III IV y y 2 2 2 1 _3 _2 _1 0 1 2 3 x 2x _2 2x _2 _1 _2 _2 _3 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 7. Use the direction field labeled I (for Exercises 3–6) to 2. A direction field for the differential equation yЈ x sin y is sketch the graphs of the solutions that satisfy the given
initial conditions.
(a) y͑0͒ 1
(b) y͑0͒ 0
(c) y͑0͒ Ϫ1 shown.
(a) Sketch the graphs of the solutions that satisfy the given
initial conditions.
(i) y͑0͒ 1
(ii) y͑0͒ 2
(iii) y͑0͒
(iv) y͑0͒ 4 ■ _2 8. Repeat Exercise 7 for the direction field labeled III.
9–10 |||| Sketch a direction field for the differential equation. Then
use it to sketch three solution curves. (v) y͑0͒ 5 9. yЈ 1 ϩ y (b) Find all the equilibrium solutions.
■ y ■ ■ 10. yЈ x 2 Ϫ y 2
■ ■ ■ ■ ■ ■ ■ ■ ■ 11–14 |||| Sketch the direction field of the differential equation.
Then use it to sketch a solution curve that passes through the
given point. 5 4 11. yЈ y Ϫ 2x, ͑1, 0͒ 13. yЈ y ϩ x y, 3 ■ ■ ■ 12. yЈ 1 Ϫ x y,
14. yЈ x Ϫ x y, ͑0, 1͒
■ ■ ■ ͑0, 0͒
͑1, 0͒ ■ ■ ■ ■ ■ ■ 2 CAS
1 _3 _2 _1 0 1 2 3 x 15–16 |||| Use a computer algebra system to draw a direction field
for the given differential equation. Get a printout and sketch on it
the solution curve that passes through ͑0, 1͒. Then use the CAS to
draw the solution curve and compare it with your sketch. 15. yЈ y sin 2x
■ 3–6 |||| Match the differential equation with its direction field
(labeled I–IV). Give reasons for your answer. 3. yЈ y Ϫ 1
5. yЈ y Ϫ x
2 4. yЈ y Ϫ x
2 6. yЈ y 3 Ϫ x 3 CAS ■ ■ 16. yЈ sin͑x ϩ y͒
■ ■ ■ ■ ■ ■ ■ ■ ■ 17. Use a computer algebra system to draw a direction field for the differential equation yЈ y 3 Ϫ 4y. Get a printout and sketch
on it solutions that satisfy the initial condition y͑0͒ c for
various values of c. For what values of c does lim t l ϱ y͑t͒
exist? What are the possible values for this limit? 5E-10(pp 632-641) 636 ❙❙❙❙ 1/18/06 9:20 AM Page 636 CHAPTER 10 DIFFERENTIAL EQUATIONS 22. Use Euler's method with step size 0.2 to estimate y͑1͒, where 18. Make a rough sketch of a direction field for the autonomous y͑x͒ is the solution of the initial-value problem yЈ 1 Ϫ x y,
y͑0͒ 0. differential equation yЈ f ͑ y͒, where the graph of f is as
shown. How does the limiting behavior of solutions depend
on the value of y͑0͒? 23. Use Euler's method with step size 0.1 to estimate y͑0.5͒, where y͑x͒ is the solution of the initial-value problem yЈ y ϩ x y,
y͑0͒ 1. f(y) 24. (a) Use Euler's method with step size 0.2 to estimate y͑1.4͒, _2 _1 0 1 2 where y͑x͒ is the solution of the initial-value problem
yЈ x Ϫ x y, y͑1͒ 0.
(b) Repeat part (a) with step size 0.1. y ; 25. (a) Program a calculator or computer to use Euler's method to
compute y͑1͒, where y͑x͒ is the solution of the initial-value
problem
19. (a) Use Euler's method with each of the following step sizes to estimate the value of y͑0.4͒, where y is the solution of the
initial-value problem yЈ y, y͑0͒ 1.
(i) h 0.4
(ii) h 0.2
(iii) h 0.1
(b) We know that the exact solution of the initial-value
problem in part (a) is y e x. Draw, as accurately as you
can, the graph of y e x, 0 ഛ x ഛ 0.4, together with the
Euler approximations using the step sizes in part (a).
(Your sketches should resemble Figures 12, 13, and 14.)
Use your sketches to decide whether your estimates in
part (a) are underestimates or overestimates.
(c) The error in Euler's method is the difference between
the exact value and the approximate value. Find the errors
made in part (a) in using Euler's method to estimate the
true value of y͑0.4͒, namely e 0.4. What happens to the error
each time the step size is halved?
20. A direction field for a differential equation is shown. Draw, with a ruler, the graphs of the Euler approximations to the
solution curve that passes through the origin. Use step sizes
h 1 and h 0.5. Will the Euler estimates be underestimates
or overestimates? Explain.
y dy
ϩ 3x 2 y 6x 2
dx
(i) h 1
(iii) h 0.01 y͑0͒ 3 (ii) h 0.1
(iv) h 0.001
3 (b) Verify that y 2 ϩ eϪx is the exact solution of the differential equation.
(c) Find the errors in using Euler's method to compute y͑1͒
with the step sizes in part (a). What happens to the error
when the step size is divided by 10?
CAS 26. (a) Program your computer algebra system, using Euler's method with step size 0.01, to calculate y͑2͒, where y
is the solution of the initial-value problem
yЈ x 3 Ϫ y 3 y͑0͒ 1 (b) Check your work by using the CAS to draw the solution
curve.
27. The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of C farads (F), and a resistor with
a resistance of R ohms (⍀). The voltage drop across the capacitor is Q͞C, where Q is the charge (in coulombs), so in this case
Kirchhoff's Law gives 2 RI ϩ Q
E͑t͒
C But I dQ͞dt, so we have
R
1 1
dQ
ϩ
Q E͑t͒
dt
C Suppose the resistance is 5 ⍀, the capacitance is 0.05 F, and a
battery gives a constant voltage of 60 V.
(a) Draw a direction field for this differential equation.
(b) What is the limiting value of the charge?
C
0 1 2 x 21. Use Euler's method with step size 0.5 to compute the approxi- mate y-values y1, y2 , y3 , and y4 of the solution of the initialvalue problem yЈ y Ϫ 2x, y͑1͒ 0. E R 5E-10(pp 632-641) 1/18/06 9:20 AM Page 637 SECTION 10.3 SEPARABLE EQUATIONS (c) Is there an equilibrium solution?
(d) If the initial charge is Q͑0͒ 0 C, use the direction field to
sketch the solution curve.
(e) If the initial charge is Q͑0͒ 0 C, use Euler's method with
step size 0.1 to estimate the charge after half a second. 637 at a rate of 1ЊC per minute when its temperature is 70ЊC.
(a) What does the differential equation become in this case?
(b) Sketch a direction field and use it to sketch the solution
curve for the initial-value problem. What is the limiting
value of the temperature?
(c) Use Euler's method with step size h 2 minutes to
estimate the temperature of the coffee after 10 minutes. 28. In Exercise 14 in Section 10.1 we considered a 95ЊC cup of cof- fee in a 20ЊC room. Suppose it is known that the coffee cools |||| 10.3 ❙❙❙❙ Separable Equations
We have looked at first-order differential equations from a geometric point of view (direction fields) and from a numerical point of view (Euler's method). What about the symbolic
point of view? It would be nice to have an explicit formula for a solution of a differential
equation. Unfortunately, that is not always possible. But in this section we examine a certain type of differential equation that can be solved explicitly.
A separable equation is a first-order differential equation in which the expression for
dy͞dx can be factored as a function of x times a function of y. In other words, it can be
written in the form
dy
t͑x͒f ͑y͒
dx
The name separable comes from the fact that the expression on the right side can be "separated" into a function of x and a function of y. Equivalently, if f ͑y͒ 0, we could write
dy
t͑x͒
dx
h͑y͒ 1 where h͑y͒ 1͞f ͑y͒. To solve this equation we rewrite it in the differential form
h͑y͒ dy t͑x͒ dx
|||| The technique for solving separable differential equations was first used by James Bernoulli
(in 1690) in solving a problem about pendulums
and by Leibniz (in a letter to Huygens in 1691).
John Bernoulli explained the general method in a
paper published in 1694. so that all y's are on one side of the equation and all x's are on the other side. Then we integrate both sides of the equation: y h͑y͒ dy y t͑x͒ dx 2 Equation 2 defines y implicitly as a function of x. In some cases we may be able to solve
for y in terms of x.
We use the Chain Rule to justify this procedure: If h and t satisfy (2), then
d
dx so d
dy ͩy and
Thus, Equation 1 is satisfied. ͩy ͪ ͩy h͑y͒ dy ͪ h͑y͒ dy h͑y͒ d
dx dy
t͑x͒
dx
dy
t͑x͒
dx ͪ t͑x͒ dx 5E-10(pp 632-641) 638 ❙❙❙❙ 1/18/06 9:20 AM Page 638 CHAPTER 10 DIFFERENTIAL EQUATIONS EXAMPLE 1 dy
x2
2.
dx
y
(b) Find the solution of this equation that satisfies the initial condition y͑0͒ 2.
(a) Solve the differential equation |||| Figure 1 shows graphs of several members
of the family of solutions of the differential
equation in Example 1. The solution of the initialvalue problem in part (b) is shown in red. SOLUTION (a) We write the equation in terms of differentials and integrate both sides:
y 2 dy x 2 dx 3 _3 3 2 dy y x 2 dx 1
3 yy y3 1 x3 ϩ C
3 where C is an arbitrary constant. (We could have used a constant C1 on the left side and
another constant C 2 on the right side. But then we could combine these constants by
writing C C 2 Ϫ C1.)
Solving for y, we get _3 3
y sx 3 ϩ 3C FIGURE 1 We could leave the solution like this or we could write it in the form
3
y sx 3 ϩ K where K 3C. (Since C is an arbitrary constant, so is K .)
3
(b) If we put x 0 in the general solution in part (a), we get y͑0͒ sK . To satisfy the
3
initial condition y͑0͒ 2, we must have sK 2 and so K 8.
Thus, the solution of the initial-value problem is
3
y sx 3 ϩ 8 |||| Some computer algebra systems can plot
curves defined by implicit equations. Figure 2
shows the graphs of several members of the
family of solutions of the differential equation
in Example 2. As we look at the curves from left
to right, the values of C are 3, 2, 1, 0, Ϫ1, Ϫ2,
and Ϫ3. EXAMPLE 2 Solve the differential equation SOLUTION Writing the equation in differential form and integrating both sides, we have ͑2y ϩ cos y͒dy 6x 2 dx y ͑2y ϩ cos y͒dy y 6x 4 3
_2 2 _4 FIGURE 2 dy
6x 2
.
dx
2y ϩ cos y 2 dx y 2 ϩ sin y 2x 3 ϩ C where C is a constant. Equation 3 gives the general solution implicitly. In this case it's
impossible to solve the equation to express y explicitly as a function of x.
EXAMPLE 3 Solve the equation yЈ x 2 y.
SOLUTION First we rewrite the equation using Leibniz notation: dy
x2y
dx 5E-10(pp 632-641) 1/18/06 9:20 AM Page 639 SECTION 10.3 SEPARABLE EQUATIONS |||| If a solution y is a function that satisfies
y͑x͒ 0 for some x, it follows from a
uniqueness theorem for solutions of differential
equations that y͑x͒ 0 for all x. If y ❙❙❙❙ 639 0, we can rewrite it in differential notation and integrate:
dy
x 2 dx
y y y 0 dy
y x 2 dx
y Խ Խ ln y x3
ϩC
3 This equation defines y implicitly as a function of x. But in this case we can solve
explicitly for y as follows: ԽyԽ e Խ Խ e ͑x ͞3͒ϩC e Ce x ͞3
3 ln y 3 3 y Ϯe Ce x ͞3 so We can easily verify that the function y 0 is also a solution of the given differential
equation. So we can write the general solution in the form
3 y Ae x ͞3
where A is an arbitrary constant ( A e C, or A Ϫe C, or A 0).
y
6
4 |||| Figure 3 shows a direction field for the differential equation in Example 3. Compare it with
Figure 4, in which we use the equation
3
y Ae x / 3 to graph solutions for several values
of A. If you use the direction field to sketch
solution curves with y-intercepts 5, 2, 1, Ϫ1,
and Ϫ2, they will resemble the curves in
Figure 4. 6
2 _2 _1 0 1 2 x
_2 _2
_4
_6 _6 FIGURE 4 FIGURE 3
R E FIGURE 5 EXAMPLE 4 In Section 10.2 we modeled the current I͑t͒ in the electric circuit shown in
Figure 5 by the differential equation
L switch 2 L dI
ϩ RI E͑t͒
dt Find an expression for the current in a circuit where the resistance is 12 ⍀, the inductance is 4 H, a battery gives a constant voltage of 60 V, and the switch is turned on when
t 0. What is the limiting value of the current?
SOLUTION With L 4, R 12, and E͑t͒ 60, the equation becomes 4 dI
ϩ 12I 60
dt 5E-10(pp 632-641) 640 ❙❙❙❙ 1/18/06 9:20 AM Page 640 CHAPTER 10 DIFFERENTIAL EQUATIONS dI
15 Ϫ 3I
dt or
and the initial-value problem is dI
15 Ϫ 3I
dt I͑0͒ 0 We recognize this equation as being separable, and we solve it as follows:
dI y 15 Ϫ 3I y dt ͑15 Ϫ 3I 0͒ Խ
Խ
Խ 15 Ϫ 3I Խ e Ϫ 1 ln 15 Ϫ 3I t ϩ C
3
|||| Figure 6 shows how the solution in
Example 4 (the current) approaches its limiting
value. Comparison with Figure 11 in Section 10.2
shows that we were able to draw a fairly accurate solution curve from the direction field. Ϫ3͑tϩC͒ 15 Ϫ 3I ϮeϪ3CeϪ3t AeϪ3t
I 5 Ϫ 1 AeϪ3t
3 6 1
Since I͑0͒ 0, we have 5 Ϫ 3 A 0, so A 15 and the solution is y=5 I͑t͒ 5 Ϫ 5eϪ3t
The limiting current, in amperes, is
0 lim I͑t͒ lim ͑5 Ϫ 5eϪ3t ͒ 5 Ϫ 5 lim eϪ3t 2.5 tlϱ tlϱ tlϱ 5Ϫ05 FIGURE 6 Orthogonal Trajectories
An orthogonal trajectory of a family of curves is a curve that intersects each curve of the
family orthogonally, that is, at right angles (see Figure 7). For instance, each member of
the family y mx of straight lines through the origin is an orthogonal trajectory of the
family x 2 ϩ y 2 r 2 of concentric circles with center the origin (see Figure 8). We say that
the two families are orthogonal trajectories of each other.
y x orthogonal
trajectory
FIGURE 7 FIGURE 8 EXAMPLE 5 Find the orthogonal trajectories of the family of curves x ky 2, where k is an arbitrary constant.
SOLUTION The curves x ky 2 form a family of parabolas whose axis of symmetry is the x-axis. The first step is to find a single differential equation that is satisfied by all 5E-10(pp 632-641) 1/18/06 9:20 AM Page 641 SECTION 10.3 SEPARABLE EQUATIONS ❙❙❙❙ 641 members of the family. If we differentiate x ky 2, we get
1 2ky dy
dx dy
1
dx
2ky or This differential equation depends on k, but we need an equation that is valid for all
values of k simultaneously. To eliminate k we note that, from the equation of the given
general parabola x ky 2, we have k x͞y 2 and so the differential equation can be
written as
dy
1
dx
2ky 1
x
2 2 y
y dy
y
dx
2x or This means that the slope of the tangent line at any point ͑x, y͒ on one of the parabolas is
yЈ y͑͞2x͒. On an orthogonal trajectory the slope of the tangent line must be the negative reciprocal of this slope. Therefore, the orthogonal trajectories must satisfy the differential equation
dy
2x
Ϫ
dx
y y This differential equation is separable, and we solve it as follows: y y dy Ϫy 2x dx
y2
Ϫx 2 ϩ C
2 x x2 ϩ 4 FIGURE 9 y2
C
2 where C is an arbitrary positive constant. Thus, the orthogonal trajectories are the family
of ellipses given by Equation 4 and sketched in Figure 9.
Orthogonal trajectories occur in various branches of physics. For example, in an electrostatic field the lines of force are orthogonal to the lines of constant potential. Also, the
streamlines in aerodynamics are orthogonal trajectories of the velocity-equipotential
curves. Mixing Problems
A typical mixing problem involves a tank of fixed capacity filled with a thoroughly mixed
solution of some substance, such as salt. A solution of a given concentration enters the tank
at a fixed rate and the mixture, thoroughly stirred, leaves at a fixed rate, which may differ
from the entering rate. If y͑t͒ denotes the amount of substance in the tank at time t, then
yЈ͑t͒ is the rate at which the substance is being added minus the rate at which it is being
removed. The mathematical description of this situation often leads to a first-order separable differential equation. We can use the same type of reasoning to model a variety of
phenomena: chemical reactions, discharge of pollutants into a lake, injection of a drug into
the bloodstream. 5E-10(pp 642-651) 642 ❙❙❙❙ 1/18/06 9:21 AM Page 642 CHAPTER 10 DIFFERENTIAL EQUATIONS EXAMPLE 6 A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that contains 0.03 kg of salt per liter of water enters the tank at a rate of 25 L͞min. The solution
is kept thoroughly mixed and drains from the tank at the same rate. How much salt
remains in the tank after half an hour?
SOLUTION Let y͑t͒ be the amount of salt (in kilograms) after t minutes. We are given that y͑0͒ 20 and we want to find y͑30͒. We do this by finding a differential equation satisfied by y͑t͒. Note that dy͞dt is the rate of change of the amount of salt, so
dy
͑rate in͒ Ϫ ͑rate out͒
dt 5 where (rate in) is the rate at which salt enters the tank and (rate out) is the rate at which
salt leaves the tank. We have ͩ rate in 0.03 kg
L ͪͩ 25 L
min ͪ 0.75 kg
min The tank always contains 5000 L of liquid, so the concentration at time t is y͑t͒͞5000
(measured in kilograms per liter). Since the brine flows out at a rate of 25 L͞min, we
have
rate out ͩ y͑t͒ kg
5000 L ͪͩ 25 L
min ͪ y͑t͒ kg
200 min Thus, from Equation 5 we get
dy
y͑t͒
150 Ϫ y͑t͒
0.75 Ϫ
dt
200
200
Solving this separable differential equation, we obtain
dy y 150 Ϫ y y Խ Խ Ϫln 150 Ϫ y |||| Figure 10 shows the graph of the function
y͑t͒ of Example 6. Notice that, as time goes by,
the amount of salt approaches 150 kg. dt
200 t
ϩC
200 Since y͑0͒ 20, we have Ϫln 130 C, so Խ t
Ϫ ln 130
200 Խ Ϫln 150 Ϫ y y Խ 150 Ϫ y Խ 130e Ϫt͞200 150 Therefore 100 Since y͑t͒ is continuous and y͑0͒ 20 and the right side is never 0, we deduce that
150 y͑t͒ is always positive. Thus, 150 Ϫ y 150 Ϫ y and so Խ 50 0 FIGURE 10 Խ y͑t͒ 150 Ϫ 130eϪt͞200
200 400 t The amount of salt after 30 min is
y͑30͒ 150 Ϫ 130eϪ30͞200 Ϸ 38.1 kg 5E-10(pp 642-651) 1/18/06 9:21 AM Page 643 SECTION 10.3 SEPARABLE EQUATIONS |||| 10.3
1–10 1. |||| Solve the differential equation.
2. 5. ͑1 ϩ tan y͒yЈ x ϩ 1 7. dy
te t
dt
y s1 ϩ y 2 9. du
2 ϩ 2u ϩ t ϩ tu
dt 8. yЈ ■ ■ 10.
■ members of the family of solutions (if your CAS does implicit
plots). How does the solution curve change as the constant C
varies?
CAS xy
2 ln y dz
ϩ e tϩz 0
dt ■ ■ ■ ■ ■ ■ y͑1͒ 0 12. dy
y cos x
,
dx
1 ϩ y2 y͑0͒ 1 15. du
2t ϩ sec 2t
,
dt
2u 16. dy
te y,
dt ■ ■ y͑0͒ 0 ■ ■ ■ ■ ■ ■ ■ |||| Find the orthogonal trajectories of the family of curves.
Use a graphing device to draw several members of each family on
a common screen. 28. x 2 Ϫ y 2 k
30. y keϪx ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 32. In Exercise 28 in Section 10.2 we discussed a differential equa- u͑0͒ Ϫ5 tion that models the temperature of a 95ЊC cup of coffee in a
20ЊC room. Solve the differential equation to find an expression
for the temperature of the coffee at time t. y͑1͒ 0 33. In Exercise 13 in Section 10.1 we formulated a model for learning in the form of the differential equation 0 Ͻ x Ͻ ͞2 y͑1͒ Ϫ1
■ ■ ■ ■ ■ ■ ■ ■ 19. Find an equation of the curve that satisfies dy͞dx 4x y and whose y-intercept is 7.
20. Find an equation of the curve that passes through the point ͑1, 1͒ and whose slope at ͑x, y͒ is y 2͞x 3.
21. (a) Solve the differential equation yЈ 2x s1 Ϫ y 2. (b) Solve the initial-value problem yЈ 2x s1 Ϫ y 2, y͑0͒ 0,
and graph the solution.
(c) Does the initial-value problem yЈ 2x s1 Ϫ y 2, y͑0͒ 2,
have a solution? Explain. Ϫy
; 22. Solve the equation e yЈ ϩ cos x 0 and graph several mem- bers of the family of solutions. How does the solution curve
change as the constant C varies?
CAS ■ to find an expression for the charge at time t. Find the limiting
value of the charge. 3 ; ■ 31. Solve the initial-value problem in Exercise 27 in Section 10.2 P͑1͒ 2 ■ ■ 29. y ͑x ϩ k͒Ϫ1 17. yЈ tan x a ϩ y, y͑͞3͒ a, ■ ■ ; 27–30 ■ dP
sPt,
dt 18. x yЈ ϩ y y 2, ■ 26. yЈ x 2͞y 27. y kx 2 13. x cos x ͑2y ϩ e 3y ͒yЈ,
14. |||| 25. yЈ 1͞y |||| Find the solution of the differential equation that satisfies
the given initial condition. dy
y 2 ϩ 1,
dx 25–26 (a) Use a computer algebra system to draw a direction field
for the differential equation. Get a printout and use it to sketch
some solution curves without solving the differential equation.
(b) Solve the differential equation.
(c) Use the CAS to draw several members of the family of solutions
obtained in part (b). Compare with the curves from part (a). 11–18 11. 24. Solve the equation yЈ x sx 2 ϩ 1͑͞ ye y ͒ and graph several dy
e 2x
dx
4y 3 du
1 ϩ sr
6.
dr
1 ϩ su 2 ■ CAS 4. yЈ y 2 sin x 3. ͑x 2 ϩ 1͒yЈ xy ■ 643 Exercises y
dy
dx
x ■ ❙❙❙❙ 23. Solve the initial-value problem yЈ ͑sin x͒͞sin y, y͑0͒ ͞2, and graph the solution (if your CAS does implicit plots). dP
k͑M Ϫ P͒
dt
where P͑t͒ measures the performance of someone learning a
skill after a training time t, M is the maximum level of performance, and k is a positive constant. Solve this differential
equation to find an expression for P͑t͒. What is the limit of this
expression?
34. In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C:
A ϩ B l C. The law of mass action states that the rate
of reaction is proportional to the product of the concentrations
of A and B:
d ͓C͔
k ͓A͔͓B͔
dt
(See Example 4 in Section 3.4.) Thus, if the initial concentrations are ͓A͔ a moles͞L and ͓B͔ b moles͞L and we write 5E-10(pp 642-651) 644 ❙❙❙❙ 1/18/06 9:21 AM Page 644 CHAPTER 10 DIFFERENTIAL EQUATIONS x ͓C͔, then we have
dx
k͑a Ϫ x͒͑b Ϫ x͒
dt
CAS (a) Assuming that a b, find x as a function of t. Use the fact
that the initial concentration of C is 0.
(b) Find x ͑t͒ assuming that a b. How does this expression
for x ͑t͒ simplify if it is known that ͓C͔ a͞2 after
20 seconds?
35. In contrast to the situation of Exercise 34, experiments show that the reaction H 2 ϩ Br 2 l 2HBr satisfies the rate law
d ͓HBr͔
k ͓H 2 ͔͓Br 2 ͔ 1͞2
dt
and so for this reaction the differential equation becomes
dx
k͑a Ϫ x͒͑b Ϫ x͒1͞2
dt
where x ͓HBr͔ and a and b are the initial concentrations of
hydrogen and bromine.
(a) Find x as a function of t in the case where a b. Use the
fact that x͑0͒ 0.
(b) If a Ͼ b, find t as a function of x. [Hint: In performing the
integration, make the substitution u sb Ϫ x.]
36. A sphere with radius 1 m has temperature 15ЊC. It lies inside a concentric sphere with radius 2 m and temperature 25ЊC. The
temperature T ͑r͒ at a distance r from the common center of the
spheres satisfies the differential equation
d 2T
2 dT
0
ϩ
dr 2
r dr
If we let S dT͞dr, then S satisfies a first-order differential
equation. Solve it to find an expression for the temperature T ͑r͒
between the spheres. 37. A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is
converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at that
time. Thus, a model for the concentration C C͑t͒ of the glucose solution in the bloodstream is
dC
r Ϫ kC
dt
where k is a positive constant.
(a) Suppose that the concentration at time t 0 is C0.
Determine the concentration at any time t by solving the
differential equation.
(b) Assuming that C0 Ͻ r͞k, find lim t l ϱ C͑t͒ and interpret
your answer.
38. A certain small country has $10 billion in paper currency in circulation, and each day $50 million comes into the country's
banks. The government decides to introduce new currency by
having the banks replace old bills with new ones whenever old currency comes into the banks. Let x x ͑t͒ denote the amount
of new currency in circulation at time t, with x ͑0͒ 0.
(a) Formulate a mathematical model in the form of an initialvalue problem that represents the "flow" of the new
currency into circulation.
(b) Solve the initial-value problem found in part (a).
(c) How long will it take for the new bills to account for 90%
of the currency in circulation?
39. A tank contains 1000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 L͞min. The solution
is kept thoroughly mixed and drains from the tank at the same
rate. How much salt is in the tank (a) after t minutes and
(b) after 20 minutes?
40. A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of
5 L͞min. Brine that contains 0.04 kg of salt per liter of water
enters the tank at a rate of 10 L͞min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L͞min.
How much salt is in the tank (a) after t minutes and (b) after
one hour?
41. When a raindrop falls, it increases in size and so its mass at time t is a function of t, m͑t͒. The rate of growth of the mass is
km͑t͒ for some positive constant k. When we apply Newton's
Law of Motion to the raindrop, we get ͑mv͒Ј tm, where v is
the velocity of the raindrop (directed downward) and t is the
acceleration due to gravity. The terminal velocity of the
raindrop is lim t l ϱ v͑t͒. Find an expression for the terminal
velocity in terms of t and k.
42. An object of mass m is moving horizontally through a medium which resists the motion with a force that is a function of the
velocity; that is,
m d 2s
dv
m
f ͑v͒
dt 2
dt where v v͑t͒ and s s͑t͒ represent the velocity and position
of the object at time t, respectively. For example, think of a
boat moving through the water.
(a) Suppose that the resisting force is proportional to the
velocity, that is, f ͑v͒ Ϫk v, k a positive constant. (This
model is appropriate for small values of v.) Let v͑0͒ v0
and s͑0͒ s0 be the initial values of v and s. Determine v
and s at any time t . What is the total distance that the
object travels from time t 0?
(b) For larger values of v a better model is obtained by supposing that the resisting force is proportional to the square
of the velocity, that is, f ͑v͒ Ϫk v 2, k Ͼ 0. (This model
was first proposed by Newton.) Let v0 and s0 be the initial
values of v and s. Determine v and s at any time t . What is
the total distance that the object travels in this case?
43. Let A͑t͒ be the area of a tissue culture at time t and let M be the final area of the tissue when growth is complete. Most cell
divisions occur on the periphery of the tissue and the number
of cells on the periphery is proportional to sA͑t͒. So a reason- 5E-10(pp 642-651) 1/18/06 9:21 AM Page 645 APPLIED PROJECT HOW FAST DOES A TANK DRAIN? CAS able model for the growth of tissue is obtained by assuming
that the rate of growth of the area is jointly proportional to
sA͑t͒ and M Ϫ A͑t͒.
(a) Formulate a differential equation and use it to show that
the tissue grows fastest when A͑t͒ M͞3.
(b) Solve the differential equation to find an expression
for A͑t͒. Use a computer algebra system to perform the
integration. m dv
mtR 2
Ϫ
dt
͑x ϩ R͒2 (a) Suppose a rocket is fired vertically upward with an initial
velocity v0. Let h be the maximum height above the surface
reached by the object. Show that gravitational force on an object of mass m that has been
projected vertically upward from Earth's surface is
mtR 2
͑x ϩ R͒2 where x x͑t͒ is the object's distance above the surface at
time t , R is Earth's radius, and t is the acceleration due to
gravity. Also, by Newton's Second Law, F ma m ͑dv͞dt͒ 645 and so 44. According to Newton's Law of Universal Gravitation, the F ❙❙❙❙ v0 ͱ 2tRh
Rϩh [Hint: By the Chain Rule, m ͑dv͞dt͒ mv ͑dv͞dx͒.]
(b) Calculate ve lim h l ϱ v0. This limit is called the escape
velocity for Earth.
(c) Use R 3960 mi and t 32 ft͞s2 to calculate ve in feet
per second and in miles per second. APPLIED PROJECT
How Fast Does a Tank Drain?
If water (or other liquid) drains from a tank, we expect that the flow will be greatest at first
(when the water depth is greatest) and will gradually decrease as the water level decreases. But
we need a more precise mathematical description of how the flow decreases in order to answer
the kinds of questions that engineers ask: How long does it take for a tank to drain completely?
How much water should a tank hold in order to guarantee a certain minimum water pressure for
a sprinkler system?
Let h͑t͒ and V͑t͒ be the height and volume of water in a tank at time t. If water leaks through
a hole with area a at the bottom of the tank, then Torricelli's Law says that
dV
Ϫa s2th
dt 1 where t is the acceleration due to gravity. So the rate at which water flows from the tank is proportional to the square root of the water height.
1. (a) Suppose the tank is cylindrical with height 6 ft and radius 2 ft and the hole is circular with radius 1 in. If we take t 32 ft͞s2, show that y satisfies the differential equation
1
dh
Ϫ
sh
dt
72
(b) Solve this equation to find the height of the water at time t, assuming the tank is full at
time t 0.
(c) How long will it take for the water to drain completely?
2. Because of the rotation and viscosity of the liquid, the theoretical model given by Equation 1 isn't quite accurate. Instead, the model
2 dh
ksh
dt 5E-10(pp 642-651) 646 ❙❙❙❙ 1/18/06 9:21 AM Page 646 CHAPTER 10 DIFFERENTIAL EQUATIONS |||| This part of the project is best done as a
classroom demonstration or as a group project
with three students in each group: a timekeeper
to call out seconds, a bottle keeper to estimate
the height every 10 seconds, and a record keeper
to record these values. is often used and the constant k (which depends on the physical properties of the liquid) is
determined from data concerning the draining of the tank.
(a) Suppose that a hole is drilled in the side of a cylindrical bottle and the height h of the
water (above the hole) decreases from 10 cm to 3 cm in 68 seconds. Use Equation 2 to
find an expression for h͑t͒. Evaluate h͑t͒ for t 10, 20, 30, 40, 50, 60.
(b) Drill a 4-mm hole near the bottom of the cylindrical part of a two-liter plastic soft-drink
bottle. Attach a strip of masking tape marked in centimeters from 0 to 10, with 0 corresponding to the top of the hole. With one finger over the hole, fill the bottle with water
to the 10-cm mark. Then take your finger off the hole and record the values of h͑t͒ for
t 10, 20, 30, 40, 50, 60 seconds. (You will probably find that it takes 68 seconds for
the level to decrease to h 3 cm.) Compare your data with the values of h͑t͒ from
part (a). How well did the model predict the actual values?
3. In many parts of the world, the water for sprinkler systems in large hotels and hospitals is supplied by gravity from cylindrical tanks on or near the roofs of the buildings. Suppose
such a tank has radius 10 ft and the diameter of the outlet is 2.5 inches. An engineer has to
guarantee that the water pressure will be at least 2160 lb͞ft 2 for a period of 10 minutes.
(When a fire happens, the electrical system might fail and it could take up to 10 minutes for
the emergency generator and fire pump to be activated.) What height should the engineer
specify for the tank in order to make such a guarantee? (Use the fact that the water pressure
at a depth of d feet is P 62.5d. See Section 9.3.)
4. Not all water tanks are shaped like cylinders. Suppose a tank has cross-sectional area A͑h͒ at height h. Then the volume of water up to height h is V x0h A͑u͒ du and so the Fundamental
Theorem of Calculus gives dV͞dh A͑h͒. It follows that
dV
dV dh
dh
A͑h͒
dt
dh dt
dt
and so Torricelli's Law becomes
A͑h͒ dh
Ϫa s2th
dt (a) Suppose the tank has the shape of a sphere with radius 2 m and is initially half full of
water. If the radius of the circular hole is 1 cm and we take t 10 m͞s2, show that h
satisfies the differential equation
͑4h Ϫ h 2 ͒ dh
Ϫ0.0001 s20h
dt (b) How long will it take for the water to drain completely? APPLIED PROJECT
Which Is Faster, Going Up or Coming Down?
Suppose you throw a ball into the air. Do you think it takes longer to reach its maximum
height or to fall back to Earth from its maximum height? We will solve the problem in this project but, before getting started, think about that situation and make a guess based on your physical intuition.
1. A ball with mass m is projected vertically upward from Earth's surface with a positive initial
velocity v0. We assume the forces acting on the ball are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude
p v͑t͒ , where p is a positive constant and v͑t͒ is the velocity of the ball at time t. In both Խ Խ 5E-10(pp 642-651) 1/18/06 9:21 AM Page 647 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY |||| In modeling force due to air resistance,
various functions have been used, depending
on the physical characteristics and speed of
the ball. Here we use a linear model, Ϫpv,
but a quadratic model (Ϫpv 2 on the way up
and pv 2 on the way down) is another possibility for higher speeds (see Exercise 42 in
Section 10.3). For a golf ball, experiments
have shown that a good model is Ϫpv 1.3
going up and p v 1.3 coming down. But no
matter which force function Ϫf ͑v͒ is used
[where f ͑v͒ Ͼ 0 for v Ͼ 0 and f ͑v͒ Ͻ 0
for v Ͻ 0], the answer to the question
remains the same. See F. Brauer, "What
Goes Up Must Come Down, Eventually,"
Amer. Math. Monthly 108 (2001),
pp. 437–440. ❙❙❙❙ 647 the ascent and the descent, the total force acting on the ball is Ϫpv Ϫ mt. [During ascent,
v͑t͒ is positive and the resistance acts downward; during descent, v͑t͒ is negative and the
resistance acts upward.] So, by Newton's Second Law, the equation of motion is
mvЈ Ϫpv Ϫ mt
Solve this differential equation to show that the velocity is Խ Խ v͑t͒ ͩ ͪ mt Ϫpt͞m
mt
e
Ϫ
p
p v0 ϩ 2. Show that the height of the ball, until it hits the ground, is ͩ y͑t͒ v0 ϩ mt
p ͪ m
mtt
͑1 Ϫ eϪpt͞m ͒ Ϫ
p
p 3. Let t1 be the time that the ball takes to reach its maximum height. Show that t1 ͩ m
mt ϩ pv0
ln
p
mt ͪ Find this time for a ball with mass 1 kg and initial velocity 20 m͞s. Assume the air
1
resistance is 10 of the speed. ; 4. Let t2 be the time at which the ball falls back to Earth. For the particular ball in Problem 3,
estimate t2 by using a graph of the height function y͑t͒. Which is faster, going up or coming
down?
5. In general, it's not easy to find t2 because it's impossible to solve the equation y͑t͒ 0 explicitly. We can, however, use an indirect method to determine whether ascent or descent
is faster; we determine whether y͑2t1 ͒ is positive or negative. Show that
y͑2t1 ͒ m 2t
p2 ͩ xϪ ͪ 1
Ϫ 2 ln x
x where x e pt1͞m. Then show that x Ͼ 1 and the function
f ͑x͒ x Ϫ 1
Ϫ 2 ln x
x is increasing for x Ͼ 1. Use this result to decide whether y͑2t1 ͒ is positive or negative. What
can you conclude? Is ascent or descent faster? |||| 10.4 Exponential Growth and Decay
One of the models for population growth that we considered in Section 10.1 was based
on the assumption that the population grows at a rate proportional to the size of the
population:
dP
kP
dt
Is that a reasonable assumption? Suppose we have a population (of bacteria, for instance) 5E-10(pp 642-651) 648 ❙❙❙❙ 1/18/06 9:21 AM Page 648 CHAPTER 10 DIFFERENTIAL EQUATIONS with size P 1000 and at a certain time it is growing at a rate of PЈ 300 bacteria per
hour. Now let's take another 1000 bacteria of the same type and put them with the first population. Each half of the new population was growing at a rate of 300 bacteria per hour.
We would expect the total population of 2000 to increase at a rate of 600 bacteria per
hour initially (provided there's enough room and nutrition). So if we double the size, we
double the growth rate. In general, it seems reasonable that the growth rate should be proportional to the size.
The same assumption applies in other situations as well. In nuclear physics, the mass
of a radioactive substance decays at a rate proportional to the mass. In chemistry, the rate
of a unimolecular first-order reaction is proportional to the concentration of the substance.
In finance, the value of a savings account with continuously compounded interest increases
at a rate proportional to that value.
In general, if y͑t͒ is the value of a quantity y at time t and if the rate of change of y with
respect to t is proportional to its size y͑t͒ at any time, then
dy
ky
dt 1 where k is a constant. Equation 1 is sometimes called the law of natural growth (if k Ͼ 0)
or the law of natural decay (if k Ͻ 0). Because it is a separable differential equation we
can solve it by the methods of Section 10.3: y dy
y k dt
y Խ Խ
ԽyԽ e ln y kt ϩ C
ktϩC e Ce kt y Ae kt
where A ( Ϯe C or 0) is an arbitrary constant. To see the significance of the constant A,
we observe that
y͑0͒ Ae k ؒ 0 A
Therefore, A is the initial value of the function.
Because Equation 1 occurs so frequently in nature, we summarize what we have just
proved for future use.
2 The solution of the initial-value problem
dy
ky
dt y͑0͒ y0 y͑t͒ y0 e kt is Population Growth
What is the significance of the proportionality constant k? In the context of population
growth, we can write
3 dP
kP
dt or 1 dP
k
P dt 5E-10(pp 642-651) 1/18/06 9:21 AM Page 649 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 649 The quantity
1 dP
P dt
is the growth rate divided by the population size; it is called the relative growth rate.
According to (3), instead of saying "the growth rate is proportional to population size" we
could say "the relative growth rate is constant." Then (2) says that a population with constant relative growth rate must grow exponentially. Notice that the relative growth rate k
appears as the coefficient of t in the exponential function y0 e kt. For instance, if
dP
0.02P
dt
and t is measured in years, then the relative growth rate is k 0.02 and the population
grows at a rate of 2% per year. If the population at time 0 is P0 , then the expression for the
population is
P͑t͒ P0 e 0.02t
TABLE 1 Year Population
(millions) 1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000 1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080 EXAMPLE 1 Assuming that the growth rate is proportional to population size, use the data
in Table 1 to model the population of the world in the 20th century. What is the relative
growth rate? How well does the model fit the data?
SOLUTION We measure the time t in years and let t 0 in the year 1900. We measure the
population P͑t͒ in millions of people. Then the initial condition is P͑0͒ 1650. We are
assuming that the growth rate is proportional to population size, so the initial-value problem is dP
kP
dt P͑0͒ 1650 From (2) we know that the solution is
P͑t͒ 1650e kt
One way to estimate the relative growth rate k is to use the fact that the population in
1910 was 1750 million. Therefore
P͑10͒ 1650e k͑10͒ 1750
We solve this equation for k:
e 10k
k 1750
1650
1
1750
ln
Ϸ 0.005884
10
1650 Thus, the relative growth rate is about 0.6% per year and the model becomes
P͑t͒ 1650e 0.005884t 5E-10(pp 642-651) 650 ❙❙❙❙ 1/18/06 9:21 AM Page 650 CHAPTER 10 DIFFERENTIAL EQUATIONS Table 2 and Figure 1 allow us to compare the predictions of this model with the actual
data. You can see that the predictions become quite inaccurate after about 30 years and
they underestimate by a factor of more than 2 in 2000.
TABLE 2 P Year Model Population 1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000 1650
1750
1856
1969
2088
2214
2349
2491
2642
2802
2972 1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080 6000 Population
(in millions)
P=1650e 0.005884t 20 40 60 80 100 t Years since 1900
FIGURE 1 A possible model for world population growth Another possibility for estimating k would be to use the given population for 1950,
for instance, instead of 1910. Then
P͑50͒ 1650e 50k 2560 |||| In Sections 7.2 and 7.4* we modeled the
same data with an exponential function, but
there we used the method of least squares. k 1
2560
ln
Ϸ 0.0087846
50
1650 The estimate for the relative growth rate is now 0.88% per year and the model is
P͑t͒ 1650e 0.0087846t
The predictions with this second model are shown in Table 3 and Figure 2. This exponential model is more accurate over a longer period of time, but it too lags behind reality
in recent years.
TABLE 3 P Year Model Population 1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000 1650
1802
1967
2148
2345
2560
2795
3052
3332
3638
3972 1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080 6000 Population
(in millions) P=1650e 0.0087846t 20 40 60 80 Years since 1900
FIGURE 2 Another model for world population growth 100 t 5E-10(pp 642-651) 1/18/06 9:21 AM Page 651 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 651 EXAMPLE 2 Use the data in Table 1 to model the population of the world in the second
half of the 20th century. Use the model to estimate the population in 1993 and to predict
the population in the year 2010.
SOLUTION Here we let t 0 in the year 1950. Then the initial-value problem is dP
kP
dt P͑0͒ 2560 and the solution is
P͑t͒ 2560e kt
Let's estimate k by using the population in 1960:
P͑10͒ 2560e 10k 3040
k 1
3040
ln
Ϸ 0.017185
10
2560 The relative growth rate is about 1.7% per year and the model is
P͑t͒ 2560e 0.017185t
We estimate that the world population in 1993 was
P͑43͒ 2560e 0.017185͑43͒ Ϸ 5360 million
The model predicts that the population in 2010 will be
P͑60͒ 2560e 0.017185͑60͒ Ϸ 7179 million
The graph in Figure 3 shows that the model is fairly accurate to date, so the estimate for
1993 is quite reliable. But the prediction for 2010 is riskier.
P
6000 P=2560e 0.017185t Population
(in millions) FIGURE 3 A model for world population growth
in the second half of the 20th century 20 Years since 1950 40 t 5E-10(pp 652-661) 652 ❙❙❙❙ 1/18/06 9:23 AM Page 652 CHAPTER 10 DIFFERENTIAL EQUATIONS Radioactive Decay
Radioactive substances decay by spontaneously emitting radiation. If m͑t͒ is the mass
remaining from an initial mass m0 of the substance after time t, then the relative decay rate
Ϫ 1 dm
m dt has been found experimentally to be constant. (Since dm͞dt is negative, the relative decay
rate is positive.) It follows that
dm
km
dt
where k is a negative constant. In other words, radioactive substances decay at a rate proportional to the remaining mass. This means that we can use (2) to show that the mass
decays exponentially:
m͑t͒ m0 e kt
Physicists express the rate of decay in terms of half-life, the time required for half of
any given quantity to decay.
EXAMPLE 3 The half-life of radium-226 ( .226 Ra) is 1590 years.
88 (a) A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of .226 Ra
88
that remains after t years.
(b) Find the mass after 1000 years correct to the nearest milligram.
(c) When will the mass be reduced to 30 mg?
SOLUTION (a) Let m͑t͒ be the mass of radium-226 (in milligrams) that remains after t years. Then
dm͞dt km and y͑0͒ 100, so (2) gives
m͑t͒ m͑0͒e kt 100e kt
In order to determine the value of k, we use the fact that y͑1590͒ 2 ͑100͒. Thus
1 100e 1590k 50 so e 1590k 1
2 and
1590k ln 1 Ϫln 2
2
kϪ ln 2
1590 m͑t͒ 100eϪ͑ln 2͞1590͒t Therefore We could use the fact that e ln 2 2 to write the expression for m͑t͒ in the alternative
form
m͑t͒ 100 ϫ 2 Ϫt͞1590
(b) The mass after 1000 years is
m͑1000͒ 100eϪ͑ln 2͞1590͒1000 Ϸ 65 mg 5E-10(pp 652-661) 1/18/06 9:23 AM Page 653 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 653 (c) We want to find the value of t such that m͑t͒ 30, that is,
100eϪ͑ln 2͞1590͒t 30 or eϪ͑ln 2͞1590͒t 0.3 We solve this equation for t by taking the natural logarithm of both sides:
150 Ϫ ln 2
t ln 0.3
1590 m=100e _(ln 2)t/1590 t Ϫ1590 Thus
m=30
0 FIGURE 4 4000 ln 0.3
Ϸ 2762 years
ln 2 As a check on our work in Example 3, we use a graphing device to draw the graph of
m͑t͒ in Figure 4 together with the horizontal line m 30. These curves intersect when
t Ϸ 2800, and this agrees with the answer to part (c). Newton's Law of Cooling
Newton's Law of Cooling states that the rate of cooling of an object is proportional to
the temperature difference between the object and its surroundings, provided that this
difference is not too large. (This law also applies to warming.) If we let T͑t͒ be the temperature of the object at time t and Ts be the temperature of the surroundings, then we
can formulate Newton's Law of Cooling as a differential equation:
dT
k͑T Ϫ Ts͒
dt
where k is a constant. We could solve this equation as a separable differential equation
by the method of Section 10.3, but an easier method is to make the change of variable
y͑t͒ T͑t͒ Ϫ Ts . Because Ts is constant, we have yЈ͑t͒ TЈ͑t͒ and so the equation
becomes
dy
ky
dt
We can then use (2) to find an expression for y, from which we can find T .
EXAMPLE 4 A bottle of soda pop at room temperature (72Њ F) is placed in a refrigerator
where the temperature is 44Њ F. After half an hour the soda pop has cooled to 61Њ F.
(a) What is the temperature of the soda pop after another half hour?
(b) How long does it take for the soda pop to cool to 50Њ F?
SOLUTION (a) Let T͑t͒ be the temperature of the soda after t minutes. The surrounding temperature
is Ts 44Њ F, so Newton's Law of Cooling states that
dT
k͑T Ϫ 44)
dt
If we let y T Ϫ 44, then y͑0͒ T͑0͒ Ϫ 44 72 Ϫ 44 28, so y is a solution of the
initial-value problem
dy
y͑0͒ 28
ky
dt 5E-10(pp 652-661) 654 ❙❙❙❙ 1/18/06 9:23 AM Page 654 CHAPTER 10 DIFFERENTIAL EQUATIONS and by (2) we have
y͑t͒ y͑0͒e kt 28e kt
We are given that T͑30͒ 61, so y͑30͒ 61 Ϫ 44 17 and
28e 30k 17 e 30k 17
28 Taking logarithms, we have
k ln ( 17 )
28
Ϸ Ϫ0.01663
30 Thus
y͑t͒ 28e Ϫ0.01663t
T͑t͒ 44 ϩ 28e Ϫ0.01663t
T͑60͒ 44 ϩ 28e Ϫ0.01663͑60͒ Ϸ 54.3
So after another half hour the pop has cooled to about 54Њ F.
(b) We have T͑t͒ 50 when
44 ϩ 28e Ϫ0.01663t 50
6
e Ϫ0.01663t 28 T
72 t 44 6
ln ( 28 )
Ϸ 92.6
Ϫ0.01663 The pop cools to 50Њ F after about 1 hour 33 minutes.
Notice that in Example 4, we have 0 FIGURE 5 30 60 90 t lim T͑t͒ lim ͑44 ϩ 28e Ϫ0.01663t ͒ 44 ϩ 28 ؒ 0 44 tlϱ tlϱ which is to be expected. The graph of the temperature function is shown in Figure 5. Continuously Compounded Interest
EXAMPLE 5 If $1000 is invested at 6% interest, compounded annually, then after 1 year
the investment is worth $1000͑1.06͒ $1060, after 2 years it's worth
$͓1000͑1.06͔͒1.06 $1123.60, and after t years it's worth $1000͑1.06͒t. In general, if
an amount A0 is invested at an interest rate r ͑r 0.06 in this example), then after
t years it's worth A0͑1 ϩ r͒t. Usually, however, interest is compounded more frequently,
say, n times a year. Then in each compounding period the interest rate is r͞n and there
are nt compounding periods in t years, so the value of the investment is ͩ ͪ A0 1 ϩ r
n nt 5E-10(pp 652-661) 1/18/06 9:23 AM Page 655 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 655 For instance, after 3 years at 6% interest a $1000 investment will be worth
$1000͑1.06͒3 $1191.02 with annual compounding $1000͑1.03͒6 $1194.05 with semiannual compounding $1000͑1.015͒12 $1195.62
$1000͑1.005͒36 $1196.68 ͩ with quarterly compounding
with monthly compounding $1000 1 ϩ 0.06
365 ͪ 365 ؒ 3 $1197.20 with daily compounding You can see that the interest paid increases as the number of compounding periods ͑n͒
increases. If we let n l ϱ, then we will be compounding the interest continuously and
the value of the investment will be ͩ ͪ ͫͩ ͪ ͬ
ͫ ͩ ͪͬ
ͫ ͩ ͪͬ
nt r
n A͑t͒ lim A0 1 ϩ
nlϱ lim A0
nlϱ A0 lim 1ϩ r
n A0 lim 1ϩ 1
m nlϱ mlϱ 1ϩ n͞r n͞r rt rt m r
n rt (where m n͞r) But the limit in this expression is equal to the number e (see Equation 7.4.9 or 7.4*.9).
So with continuous compounding of interest at interest rate r, the amount after t years is
A͑t͒ A0 e rt
If we differentiate this equation, we get
dA
rA0 e rt rA͑t͒
dt
which says that, with continuous compounding of interest, the rate of increase of an
investment is proportional to its size.
Returning to the example of $1000 invested for 3 years at 6% interest, we see that
with continuous compounding of interest the value of the investment will be
A͑3͒ $1000e ͑0.06͒3
$1000e 0.18 $1197.22
Notice how close this is to the amount we calculated for daily compounding, $1197.20.
But the amount is easier to compute if we use continuous compounding. 5E-10(pp 652-661) 656 ❙❙❙❙ 1/18/06 9:23 AM Page 656 CHAPTER 10 DIFFERENTIAL EQUATIONS |||| 10.4 Exercises 1. A population of protozoa develops with a constant relative Year 2. A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth
medium divides into two cells every 20 minutes. The initial
population of a culture is 60 cells.
(a) Find the relative growth rate.
(b) Find an expression for the number of cells after t hours.
(c) Find the number of cells after 8 hours.
(d) Find the rate of growth after 8 hours.
(e) When will the population reach 20,000 cells?
3. A bacteria culture starts with 500 bacteria and grows at a rate proportional to its size. After 3 hours there are 8000 bacteria.
(a) Find an expression for the number of bacteria after t hours.
(b) Find the number of bacteria after 4 hours.
(c) Find the rate of growth after 4 hours.
(d) When will the population reach 30,000? ; Population Year Population 1900
1910
1920
1930
1940
1950 growth rate of 0.7944 per member per day. On day zero the
population consists of two members. Find the population size
after six days. 76
92
106
123
131
150 1960
1970
1980
1990
2000 179
203
227
250
275 (a) Use the exponential model and the census figures for 1900
and 1910 to predict the population in 2000. Compare with
the actual figure and try to explain the discrepancy.
(b) Use the exponential model and the census figures for 1980
and 1990 to predict the population in 2000. Compare with
the actual population. Then use this model to predict the
population in the years 2010 and 2020.
(c) Graph both of the exponential functions in parts (a) and (b)
together with a plot of the actual population. Are these
models reasonable ones?
7. Experiments show that if the chemical reaction N2O5 l 2NO 2 ϩ 1 O 2
2 4. A bacteria culture grows with constant relative growth rate. After 2 hours there are 600 bacteria and after 8 hours the count
is 75,000.
(a) Find the initial population.
(b) Find an expression for the population after t hours.
(c) Find the number of cells after 5 hours.
(d) Find the rate of growth after 5 hours.
(e) When will the population reach 200,000?
5. The table gives estimates of the world population, in millions, from 1750 to 2000:
Year Population Year 790
980
1260 1900
1950
2000 1650
2560
6080 Ϫ d͓N2O5͔
0.0005͓N2O5͔
dt (See Example 4 in Section 3.4.)
(a) Find an expression for the concentration ͓N2O5͔ after
t seconds if the initial concentration is C.
(b) How long will the reaction take to reduce the concentration
of N2O5 to 90% of its original value? Population 1750
1800
1850 takes place at 45ЊC, the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows: (a) Use the exponential model and the population figures for
1750 and 1800 to predict the world population in 1900 and
1950. Compare with the actual figures.
(b) Use the exponential model and the population figures for
1850 and 1900 to predict the world population in 1950.
Compare with the actual population.
(c) Use the exponential model and the population figures for
1900 and 1950 to predict the world population in 2000.
Compare with the actual population and try to explain the
discrepancy.
6. The table gives the population of the United States, in millions, for the years 1900–2000. 8. Bismuth-210 has a half-life of 5.0 days. (a) A sample originally has a mass of 800 mg. Find a formula
for the mass remaining after t days.
(b) Find the mass remaining after 30 days.
(c) When is the mass reduced to 1 mg?
(d) Sketch the graph of the mass function.
9. The half-life of cesium-137 is 30 years. Suppose we have a 100-mg sample.
(a) Find the mass that remains after t years.
(b) How much of the sample remains after 100 years?
(c) After how long will only 1 mg remain?
10. After 3 days a sample of radon-222 decayed to 58% of its orig- inal amount.
(a) What is the half-life of radon-222?
(b) How long would it take the sample to decay to 10% of its
original amount? 5E-10(pp 652-661) 1/18/06 9:23 AM Page 657 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY 11. Scientists can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper
atmosphere by cosmic rays converts nitrogen to a radioactive
isotope of carbon, 14 C, with a half-life of about 5730 years.
Vegetation absorbs carbon dioxide through the atmosphere and
animal life assimilates 14 C through food chains. When a plant
or animal dies, it stops replacing its carbon and the amount of
14
C begins to decrease through radioactive decay. Therefore,
the level of radioactivity must also decay exponentially. A
parchment fragment was discovered that had about 74% as
much 14 C radioactivity as does plant material on Earth today.
Estimate the age of the parchment.
12. A curve passes through the point ͑0, 5͒ and has the property that the slope of the curve at every point P is twice the
y-coordinate of P. What is the equation of the curve?
13. A roast turkey is taken from an oven when its temperature has reached 185ЊF and is placed on a table in a room where the
temperature is 75ЊF.
(a) If the temperature of the turkey is 150ЊF after half an hour,
what is the temperature after 45 min?
(b) When will the turkey have cooled to 100ЊF?
14. A thermometer is taken from a room where the temperature is 20ЊC to the outdoors, where the temperature is 5ЊC. After one
minute the thermometer reads 12ЊC.
(a) What will the reading on the thermometer be after one
more minute?
(b) When will the thermometer read 6ЊC?
15. When a cold drink is taken from a refrigerator, its temperature is 5ЊC. After 25 minutes in a 20ЊC room its temperature has
increased to 10ЊC.
(a) What is the temperature of the drink after 50 minutes?
(b) When will its temperature be 15ЊC?
16. A freshly brewed cup of coffee has temperature 95ЊC in a 20ЊC room. When its temperature is 70ЊC, it is cooling at a rate of
1ЊC per minute. When does this occur?
17. The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is
constant. At 15ЊC the pressure is 101.3 kPa at sea level and
87.14 kPa at h 1000 m.
(a) What is the pressure at an altitude of 3000 m?
(b) What is the pressure at the top of Mount McKinley, at an
altitude of 6187 m?
18. (a) If $500 is borrowed at 14% interest, find the amounts due at the end of 2 years if the interest is compounded
(i) annually, (ii) quarterly, (iii) monthly, (iv) daily,
(v) hourly, and (vi) continuously. ; ❙❙❙❙ 657 (b) Suppose $500 is borrowed and the interest is compounded
continuously. If A͑t͒ is the amount due after t years, where
0 ഛ t ഛ 2, graph A͑t͒ for each of the interest rates 14%,
10%, and 6% on a common screen.
19. (a) If $3000 is invested at 5% interest, find the value of the investment at the end of 5 years if the interest is compounded (i) annually, (ii) semiannually, (iii) monthly,
(iv) weekly, (v) daily, and (vi) continuously.
(b) If A͑t͒ is the amount of the investment at time t for the case
of continuous compounding, write a differential equation
and an initial condition satisfied by A͑t͒.
20. (a) How long will it take an investment to double in value if the interest rate is 6% compounded continuously?
(b) What is the equivalent annual interest rate?
21. Consider a population P P͑t͒ with constant relative birth and death rates ␣ and , respectively, and a constant emigration
rate m, where ␣, , and m are positive constants. Assume that
␣ Ͼ . Then the rate of change of the population at time t is
modeled by the differential equation
dP
kP Ϫ m
dt where k ␣ Ϫ  (a) Find the solution of this equation that satisfies the initial
condition P͑0͒ P0.
(b) What condition on m will lead to an exponential expansion
of the population?
(c) What condition on m will result in a constant population?
A population decline?
(d) In 1847, the population of Ireland was about 8 million and
the difference between the relative birth and death rates was
1.6% of the population. Because of the potato famine in the
1840s and 1850s, about 210,000 inhabitants per year emigrated from Ireland. Was the population expanding or
declining at that time?
22. Let c be a positive number. A differential equation of the form dy
ky 1ϩc
dt
where k is a positive constant, is called a doomsday equation
because the exponent in the expression ky 1ϩc is larger than that
for natural growth (that is, ky).
(a) Determine the solution that satisfies the initial condition
y͑0͒ y0.
(b) Show that there is a finite time t T (doomsday) such that
lim t l T Ϫ y͑t͒ ϱ.
(c) An especially prolific breed of rabbits has the growth term
ky 1.01. If 2 such rabbits breed initially and the warren has
16 rabbits after three months, then when is doomsday? 5E-10(pp 652-661) 658 ❙❙❙❙ 1/18/06 9:23 AM Page 658 CHAPTER 10 DIFFERENTIAL EQUATIONS APPLIED PROJECT
Calculus and Baseball
In this project we explore three of the many applications of calculus to baseball. The physical
interactions of the game, especially the collision of ball and bat, are quite complex and their
models are discussed in detail in a book by Robert Adair, The Physics of Baseball, 3d ed.
(New York: HarperPerennial, 2002).
1. It may surprise you to learn that the collision of baseball and bat lasts only about a thou- sandth of a second. Here we calculate the average force on the bat during this collision by
first computing the change in the ball's momentum.
The momentum p of an object is the product of its mass m and its velocity v, that is,
p mv. Suppose an object, moving along a straight line, is acted on by a force F F͑t͒
that is a continuous function of time.
(a) Show that the change in momentum over a time interval ͓t0 , t1 ͔ is equal to the integral of
F from t0 to t1; that is, show that
t1 Batter's box An overhead view of the position of a
baseball bat, shown every fiftieth of
a second during a typical swing.
(Adapted from The Physics of Baseball) p͑t1 ͒ Ϫ p͑t0 ͒ y F͑t͒ dt
t0 This integral is called the impulse of the force over the time interval.
(b) A pitcher throws a 90-mi͞h fastball to a batter, who hits a line drive directly back
to the pitcher. The ball is in contact with the bat for 0.001 s and leaves the bat with
velocity 110 mi͞h. A baseball weighs 5 oz and, in U.S. Customary units, its mass is
measured in slugs: m w͞t where t 32 ft͞s 2.
(i) Find the change in the ball's momentum.
(ii) Find the average force on the bat.
2. In this problem we calculate the work required for a pitcher to throw a 90-mi͞h fastball by first considering kinetic energy.
The kinetic energy K of an object of mass m and velocity v is given by K 1 mv 2. Sup2
pose an object of mass m, moving in a straight line, is acted on by a force F F͑s͒ that
depends on its position s. According to Newton's Second Law
F͑s͒ ma m dv
dt where a and v denote the acceleration and velocity of the object.
(a) Show that the work done in moving the object from a position s0 to a position s1 is equal
to the change in the object's kinetic energy; that is, show that
s1 2
2
W y F͑s͒ ds 1 mv1 Ϫ 1 mv 0
2
2
s0 where v0 v͑s0 ͒ and v1 v͑s1 ͒ are the velocities of the object at the positions s0 and s1.
Hint: By the Chain Rule,
dv
dv ds
dv
m
m
mv
dt
ds dt
ds
(b) How many foot-pounds of work does it take to throw a baseball at a speed of
90 mi͞h?
3. (a) An outfielder fields a baseball 280 ft away from home plate and throws it directly to the
catcher with an initial velocity of 100 ft͞s. Assume that the velocity v͑t͒ of the ball after
t seconds satisfies the differential equation dv͞dt Ϫv͞10 because of air resistance. How long does it take for the ball to reach home plate? (Ignore any vertical motion of
the ball.)
(b) The manager of the team wonders whether the ball will reach home plate sooner if it
is relayed by an infielder. The shortstop can position himself directly between the outfielder and home plate, catch the ball thrown by the outfielder, turn, and throw the ball to 5E-10(pp 652-661) 1/18/06 9:23 AM Page 659 SECTION 10.5 THE LOGISTIC EQUATION ; |||| 10.5 ❙❙❙❙ 659 the catcher with an initial velocity of 105 ft͞s. The manager clocks the relay time of the
shortstop (catching, turning, throwing) at half a second. How far from home plate should
the shortstop position himself to minimize the total time for the ball to reach the plate?
Should the manager encourage a direct throw or a relayed throw? What if the shortstop
can throw at 115 ft͞s?
(c) For what throwing velocity of the shortstop does a relayed throw take the same time as a
direct throw? The Logistic Equation
In this section we discuss in detail a model for population growth, the logistic model, that
is more sophisticated than exponential growth. In doing so we use all the tools at our disposal—direction fields and Euler's method from Section 10.2 and the explicit solution of
separable differential equations from Section 10.3. In the exercises we investigate other
possible models for population growth, some of which take into account harvesting and
seasonal growth. The Logistic Model
As we discussed in Section 10.1, a population often increases exponentially in its early
stages but levels off eventually and approaches its carrying capacity because of limited
resources. If P͑t͒ is the size of the population at time t, we assume that
dP
Ϸ kP
dt if P is small This says that the growth rate is initially close to being proportional to size. In other words,
the relative growth rate is almost constant when the population is small. But we also want
to reflect the fact that the relative growth rate decreases as the population P increases and
becomes negative if P ever exceeds its carrying capacity K, the maximum population that
the environment is capable of sustaining in the long run. The simplest expression for the
relative growth rate that incorporates these assumptions is ͩ ͪ 1 dP
P
k 1Ϫ
P dt
K Multiplying by P, we obtain the model for population growth known as the logistic differential equation: 1 ͩ ͪ dP
P
kP 1 Ϫ
dt
K Notice from Equation 1 that if P is small compared with K, then P͞K is close to 0 and so
dP͞dt Ϸ kP. However, if P l K (the population approaches its carrying capacity), then
P͞K l 1, so dP͞dt l 0. We can deduce information about whether solutions increase or
decrease directly from Equation 1. If the population P lies between 0 and K, then the right
side of the equation is positive
and the population decreases. 5E-10(pp 652-661) 660 ❙❙❙❙ 1/18/06 9:23 AM Page 660 CHAPTER 10 DIFFERENTIAL EQUATIONS Direction Fields
Let's start our more detailed analysis of the logistic differential equation by looking at a
direction field.
EXAMPLE 1 Draw a direction field for the logistic equation with k 0.08 and carrying
capacity K 1000. What can you deduce about the solutions?
SOLUTION In this case the logistic differential equation is ͩ dP
P
0.08P 1 Ϫ
dt
1000 ͪ A direction field for this equation is shown in Figure 1. We show only the first quadrant
because negative populations aren't meaningful and we are interested only in what happens after t 0.
P
1400
1200
1000
800
600
400
200 FIGURE 1 Direction field for the logistic
equation in Example 1 0 20 40 60 80 t The logistic equation is autonomous (dP͞dt depends only on P, not on t), so the
slopes are the same along any horizontal line. As expected, the slopes are positive for
0 Ͻ P Ͻ 1000 and negative for P Ͼ 1000.
The slopes are small when P is close to 0 or 1000 (the carrying capacity). Notice that
the solutions move away from the equilibrium solution P 0 and move toward the
equilibrium solution P 1000.
In Figure 2 we use the direction field to sketch solution curves with initial populations
P͑0͒ 100, P͑0͒ 400, and P͑0͒ 1300. Notice that solution curves that start below
P 1000 are increasing and those that start above P 1000 are decreasing. The slopes
are greatest when P Ϸ 500 and, therefore, the solution curves that start below P 1000
have inflection points when P Ϸ 500. In fact we can prove that all solution curves that
start below P 500 have an inflection point when P is exactly 500 (see Exercise 9).
P
1400
1200
1000
800
600
400
200 FIGURE 2 Solution curves for the logistic
equation in Example 1 0 20 40 60 80 t 5E-10(pp 652-661) 1/18/06 9:23 AM Page 661 SECTION 10.5 THE LOGISTIC EQUATION ❙❙❙❙ 661 Euler's Method
Next let's use Euler's method to obtain numerical estimates for solutions of the logistic differential equation at specific times.
EXAMPLE 2 Use Euler's method with step sizes 20, 10, 5, 1, and 0.1 to estimate the population sizes P͑40͒ and P͑80͒, where P is the solution of the initial-value problem ͩ dP
P
0.08P 1 Ϫ
dt
1000 ͪ P͑0͒ 100 SOLUTION With step size h 20, t0 0, P0 100, and ͩ F͑t, P͒ 0.08P 1 Ϫ P
1000 ͪ we get, using the notation of Section 10.2,
P1 100 ϩ 20F͑0, 100͒ 244
P2 244 ϩ 20F͑20, 244͒ Ϸ 539.14
P3 539.14 ϩ 20F͑40, 539.14͒ Ϸ 936.69
P4 936.69 ϩ 20F͑60, 936.69͒ Ϸ 1031.57
Thus, our estimates for the population sizes at times t 40 and t 80 are
P͑40͒ Ϸ 539 P͑80͒ Ϸ 1032 For smaller step sizes we need to program a calculator or computer. The table gives
the results.
Step size Euler estimate of P͑40͒ Euler estimate of P͑80͒ 20
10
5
1
0.1 539
647
695
725
731 1032
997
991
986
985 Figure 3 shows a graph of the Euler approximations with step sizes h 10 and h 1.
We see that the Euler approximation with h 1 looks very much like the lower solution
curve that we drew using a direction field in Figure 2.
P
1000 h=1
h=10 FIGURE 3 Euler approximations of the
solution curve in Example 2 0 20 40 60 80 t 5E-10(pp 662-671) 662 ❙❙❙❙ 1/18/06 9:25 AM Page 662 CHAPTER 10 DIFFERENTIAL EQUATIONS The Analytic Solution
The logistic equation (1) is separable and so we can solve it explicitly using the method of
Section 10.3. Since ͩ ͪ dP
P
kP 1 Ϫ
dt
K
we have
dP y P͑1 Ϫ P͞K͒ y k dt 2 To evaluate the integral on the left side, we write
1
K
P͑1 Ϫ P͞K͒
P͑K Ϫ P͒
Using partial fractions (see Section 8.4), we get
K
1
1
ϩ
P͑K Ϫ P͒
P
KϪP
This enables us to rewrite Equation 2: y ͩ 1
1
ϩ
P
KϪP Խ Խ ͪ Խ dP y k dt Խ ln P Ϫ ln K Ϫ P kt ϩ C
ln Ϳ
Ϳ Ϳ
Ϳ KϪP
Ϫkt Ϫ C
P
KϪP
eϪktϪC eϪCeϪkt
P
KϪP
AeϪkt
P 3 where A ϮeϪC. Solving Equation 3 for P, we get
K
Ϫ 1 AeϪkt
P
so P ? P
1
K
1 ϩ AeϪkt K
1 ϩ AeϪkt We find the value of A by putting t 0 in Equation 3. If t 0, then P P0 (the initial
population), so
K Ϫ P0
Ae 0 A
P0 5E-10(pp 662-671) 1/18/06 9:25 AM Page 663 SECTION 10.5 THE LOGISTIC EQUATION ❙❙❙❙ 663 Thus, the solution to the logistic equation is
P͑t͒ 4 K
1 ϩ AeϪkt K Ϫ P0
P0 where A Using the expression for P͑t͒ in Equation 4, we see that
lim P͑t͒ K tlϱ which is to be expected.
EXAMPLE 3 Write the solution of the initial-value problem ͩ dP
P
0.08P 1 Ϫ
dt
1000 ͪ P͑0͒ 100 and use it to find the population sizes P͑40͒ and P͑80͒. At what time does the population
reach 900?
SOLUTION The differential equation is a logistic equation with k 0.08, carrying capacity K 1000, and initial population P0 100. So Equation 4 gives the population at
time t as P͑t͒ 1000
1 ϩ AeϪ0.08t where A P͑t͒ Thus 1000 Ϫ 100
9
100 1000
1 ϩ 9eϪ0.08t So the population sizes when t 40 and 80 are |||| Compare these values with the Euler
estimates from Example 2:
P͑40͒ Ϸ 731
P͑80͒ Ϸ 985 P͑40͒ 1000
Ϸ 731.6
1 ϩ 9eϪ3.2 P͑80͒ 1000
Ϸ 985.3
1 ϩ 9eϪ6.4 The population reaches 900 when
1000
900
1 ϩ 9eϪ0.08t
Solving this equation for t, we get
|||| Compare the solution curve in Figure 4 with
the lowest solution curve we drew from the
direction field in Figure 2.
1000 P= FIGURE 4 1
eϪ0.08t 81
1
Ϫ0.08t ln 81 Ϫln 81 P=900 0 1 ϩ 9eϪ0.08t 10
9 t 1000
1+9e _0.08t
80 ln 81
Ϸ 54.9
0.08 So the population reaches 900 when t is approximately 55. As a check on our work, we
graph the population curve in Figure 4 and observe where it intersects the line P 900.
The cursor indicates that t Ϸ 55. 5E-10(pp 662-671) 664 ❙❙❙❙ 1/18/06 9:25 AM Page 664 CHAPTER 10 DIFFERENTIAL EQUATIONS Comparison of the Natural Growth and Logistic Models
In the 1930s the biologist G. F. Gause conducted an experiment with the protozoan Paramecium and used a logistic equation to model his data. The table gives his daily count
of the population of protozoa. He estimated the initial relative growth rate to be 0.7944 and
the carrying capacity to be 64.
t (days) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 P (observed) 2 3 22 16 39 52 54 47 50 76 69 51 57 70 53 59 57 EXAMPLE 4 Find the exponential and logistic models for Gause's data. Compare the
predicted values with the observed values and comment on the fit.
SOLUTION Given the relative growth rate k 0.7944 and the initial population P0 2, the
exponential model is P͑t͒ P0 e kt 2e 0.7944t
Gause used the same value of k for his logistic model. [This is reasonable because
P0 2 is small compared with the carrying capacity (K 64). The equation
1 dP
P0 dt Ϳ ͩ
t0 k 1Ϫ 2
64 ͪ Ϸk shows that the value of k for the logistic model is very close to the value for the exponential model.]
Then the solution of the logistic equation in Equation 4 gives
P͑t͒
A where K
64
Ϫkt
1 ϩ Ae
1 ϩ AeϪ0.7944t
K Ϫ P0
64 Ϫ 2
31
P0
2 P͑t͒ So 64
1 ϩ 31e Ϫ0.7944t We use these equations to calculate the predicted values (rounded to the nearest integer)
and compare them in the table.
t (days) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 P (observed) 2 3 22 16 39 52 54 47 50 76 69 51 57 70 53 59 57 P (logistic model) 2 4 9 17 28 40 51 57 61 62 63 64 64 64 64 64 64 P (exponential model) 2 4 10 22 48 106 ... We notice from the table and from the graph in Figure 5 that for the first three or four
days the exponential model gives results comparable to those of the more sophisticated
logistic model. For t ജ 5, however, the exponential model is hopelessly inaccurate, but
the logistic model fits the observations reasonably well. 5E-10(pp 662-671) 1/18/06 9:25 AM Page 665 SECTION 10.5 THE LOGISTIC EQUATION ❙❙❙❙ 665 P P=2e0.7944t
60
40 P=
20 64
1+31e _0.7944t FIGURE 5
0 The exponential and logistic
models for the Paramecium data 4 8 16 t 12 Other Models for Population Growth
The Law of Natural Growth and the logistic differential equation are not the only equations that have been proposed to model population growth. In Exercise 14 we look at
the Gompertz growth function and in Exercises 15 and 16 we investigate seasonalgrowth models.
Two of the other models are modifications of the logistic model. The differential
equation
dP
P
kP 1 Ϫ
Ϫc
dt
K ͩ ͪ has been used to model populations that are subject to "harvesting" of one sort or another.
(Think of a population of fish being caught at a constant rate). This equation is explored
in Exercises 11 and 12.
For some species there is a minimum population level m below which the species tends
to become extinct. (Adults may not be able to find suitable mates.) Such populations have
been modeled by the differential equation ͩ ͪͩ ͪ dP
P
kP 1 Ϫ
dt
K 1Ϫ m
P where the extra factor, 1 Ϫ m͞P, takes into account the consequences of a sparse population (see Exercise 13). |||| 10.5 Exercises 1. Suppose that a population develops according to the logistic equation
dP
0.05P Ϫ 0.0005P 2
dt
where t is measured in weeks.
(a) What is the carrying capacity? What is the value of k ?
(b) A direction field for this equation is shown at the right.
Where are the slopes close to 0? Where are they largest?
Which solutions are increasing? Which solutions are
decreasing?
(c) Use the direction field to sketch solutions for initial populations of 20, 40, 60, 80, 120, and 140. What do these P
150 100 50 0 20 40 60 t 5E-10(pp 662-671) 666 ❙❙❙❙ 1/18/06 9:25 AM Page 666 CHAPTER 10 DIFFERENTIAL EQUATIONS solutions have in common? How do they differ? Which
solutions have inflection points? At what population levels
do they occur?
(d) What are the equilibrium solutions? How are the other solutions related to these solutions? ; 2. Suppose that a population grows according to a logistic model
with carrying capacity 6000 and k 0.0015 per year.
(a) Write the logistic differential equation for these data.
(b) Draw a direction field (either by hand or with a computer
algebra system). What does it tell you about the solution
curves?
(c) Use the direction field to sketch the solution curves for
initial populations of 1000, 2000, 4000, and 8000. What
can you say about the concavity of these curves? What is
the significance of the inflection points?
(d) Program a calculator or computer to use Euler's method
with step size h 1 to estimate the population after
50 years if the initial population is 1000.
(e) If the initial population is 1000, write a formula for the
population after t years. Use it to find the population after
50 years and compare with your estimate in part (d).
(f) Graph the solution in part (e) and compare with the
solution curve you sketched in part (c).
3. The Pacific halibut fishery has been modeled by the differential equation ͩ ͪ y
dy
ky 1 Ϫ
dt
K 5. The population of the world was about 5.3 billion in 1990. Birth rates in the 1990s ranged from 35 to 40 million per year
and death rates ranged from 15 to 20 million per year. Let's
assume that the carrying capacity for world population is
100 billion.
(a) Write the logistic differential equation for these data.
(Because the initial population is small compared to the
carrying capacity, you can take k to be an estimate of the
initial relative growth rate.)
(b) Use the logistic model to estimate the world population in
the year 2000 and compare with the actual population of
6.1 billion.
(c) Use the logistic model to predict the world population in
the years 2100 and 2500.
(d) What are your predictions if the carrying capacity is
50 billion?
6. (a) Make a guess as to the carrying capacity for the U.S. population. Use it and the fact that the population was
250 million in 1990 to formulate a logistic model for the
U.S. population.
(b) Determine the value of k in your model by using the fact
that the population in 2000 was 275 million.
(c) Use your model to predict the U.S. population in the years
2100 and 2200.
(d) Use your model to predict the year in which the U.S. population will exceed 300 million.
7. One model for the spread of a rumor is that the rate of spread where y͑t͒ is the biomass (the total mass of the members of
the population) in kilograms at time t (measured in years), the
carrying capacity is estimated to be K 8 ϫ 10 7 kg, and
k 0.71 per year.
(a) If y͑0͒ 2 ϫ 10 7 kg, find the biomass a year later.
(b) How long will it take for the biomass to reach 4 ϫ 10 7 kg?
4. The table gives the number of yeast cells in a new laboratory culture.
Time (hours) Yeast cells Time (hours) Yeast cells 0
2
4
6
8 18
39
80
171
336 10
12
14
16
18 509
597
640
664
672 (a) Plot the data and use the plot to estimate the carrying
capacity for the yeast population.
(b) Use the data to estimate the initial relative growth rate.
(c) Find both an exponential model and a logistic model for
these data.
(d) Compare the predicted values with the observed values,
both in a table and with graphs. Comment on how well
your models fit the data.
(e) Use your logistic model to estimate the number of yeast
cells after 7 hours. is proportional to the product of the fraction y of the population
who have heard the rumor and the fraction who have not heard
the rumor.
(a) Write a differential equation that is satisfied by y.
(b) Solve the differential equation.
(c) A small town has 1000 inhabitants. At 8 A.M., 80 people
have heard a rumor. By noon half the town has heard it. At
what time will 90% of the population have heard the rumor?
8. Biologists stocked a lake with 400 fish and estimated the carrying capacity (the maximal population for the fish of that
species in that lake) to be 10,000. The number of fish tripled in
the first year.
(a) Assuming that the size of the fish population satisfies the
logistic equation, find an expression for the size of the population after t years.
(b) How long will it take for the population to increase
to 5000?
9. (a) Show that if P satisfies the logistic equation (1), then ͩ ͪͩ P
d 2P
k 2P 1 Ϫ
dt 2
K 1Ϫ 2P
K ͪ (b) Deduce that a population grows fastest when it reaches half
its carrying capacity. ; 10. For a fixed value of K (say K 10), the family of logistic
functions given by Equation 4 depends on the initial value P0
and the proportionality constant k. Graph several members of 5E-10(pp 662-671) 1/18/06 9:25 AM Page 667 SECTION 10.5 THE LOGISTIC EQUATION 11. Let's modify the logistic differential equation of Example 1 as ͩ dP
P
0.08P 1 Ϫ
dt
1000 CAS CAS ͪ 14. Another model for a growth function for a limited population is given by the Gompertz function, which is a solution of the
differential equation Ϫ 15 ͩͪ K
dP
P
c ln
dt
P (a) Suppose P͑t͒ represents a fish population at time t, where t
is measured in weeks. Explain the meaning of the term Ϫ15.
(b) Draw a direction field for this differential equation.
(c) What are the equilibrium solutions?
(d) Use the direction field to sketch several solution curves.
Describe what happens to the fish population for various
initial populations.
(e) Solve this differential equation explicitly, either by using
partial fractions or with a computer algebra system. Use the
initial populations 200 and 300. Graph the solutions and
compare with your sketches in part (d). where c is a constant and K is the carrying capacity.
(a) Solve this differential equation.
(b) Compute lim t l ϱ P͑t͒.
(c) Graph the Gompertz growth function for K 1000,
P0 100, and c 0.05, and compare it with the logistic
function in Example 3. What are the similarities? What are
the differences?
(d) We know from Exercise 9 that the logistic function grows
fastest when P K͞2. Use the Gompertz differential equation to show that the Gompertz function grows fastest
when P K͞e. 12. Consider the differential equation ͩ P
dP
0.08P 1 Ϫ
dt
1000 ͪ 15. In a seasonal-growth model, a periodic function of time is Ϫc as a model for a fish population, where t is measured in weeks
and c is a constant.
(a) Use a CAS to draw direction fields for various values of c.
(b) From your direction fields in part (a), determine the values
of c for which there is at least one equilibrium solution. For
what values of c does the fish population always die out?
(c) Use the differential equation to prove what you discovered
graphically in part (b).
(d) What would you recommend for a limit to the weekly catch
of this fish population? introduced to account for seasonal variations in the rate of
growth. Such variations could, for example, be caused by
seasonal changes in the availability of food.
(a) Find the solution of the seasonal-growth model
dP
kP cos͑rt Ϫ ͒
dt ; 1Ϫ m
P (a) Use the differential equation to show that any solution is
increasing if m Ͻ P Ͻ K and decreasing if 0 Ͻ P Ͻ m.
(b) For the case where k 0.08, K 1000, and m 200,
draw a direction field and use it to sketch several solution
curves. Describe what happens to the population for various
initial populations. What are the equilibrium solutions?
(c) Solve the differential equation explicitly, either by using
partial fractions or with a computer algebra system. Use the
initial population P0 . where k, r, and are positive constants.
(b) By graphing the solution for several values of k, r, and ,
explain how the values of k, r, and affect the solution.
What can you say about lim t l ϱ P͑t͒?
follows: some species there is a minimum population m such that the
species will become extinct if the size of the population falls
below m. This condition can be incorporated into the logistic
equation by introducing the factor ͑1 Ϫ m͞P͒. Thus, the modified logistic model is given by the differential equation ͩ ͪͩ ͪ P͑0͒ P0 16. Suppose we alter the differential equation in Exercise 15 as 13. There is considerable evidence to support the theory that for P
dP
kP 1 Ϫ
dt
K 667 (d) Use the solution in part (c) to show that if P0 Ͻ m, then the
species will become extinct. [Hint: Show that the numerator in your expression for P͑t͒ is 0 for some value of t.] this family. How does the graph change when P0 varies? How
does it change when k varies?
follows: ❙❙❙❙ dP
kP cos 2͑rt Ϫ ͒
dt ; P͑0͒ P0 (a) Solve this differential equation with the help of a table of
integrals or a CAS.
(b) Graph the solution for several values of k, r, and . How do
the values of k, r, and affect the solution? What can you
say about lim t l ϱ P͑t͒ in this case?
17. Graphs of logistic functions (Figures 2 and 4) look suspiciously similar to the graph of the hyperbolic tangent function
(Figure 3 in Section 7.6). Explain the similarity by showing
that the logistic function given by Equation 4 can be written as [ P͑t͒ 1 K 1 ϩ tanh ( 1 k͑ t Ϫ c͒)
2
2 where c ͑ln A͒͞ k. Thus, the logistic function is really just a
shifted hyperbolic tangent. 5E-10(pp 662-671) 668 ❙❙❙❙ 1/18/06 9:25 AM Page 668 CHAPTER 10 DIFFERENTIAL EQUATIONS |||| 10.6 Linear Equations
A first-order linear differential equation is one that can be put into the form
dy
ϩ P͑x͒y Q͑x͒
dx 1 where P and Q are continuous functions on a given interval. This type of equation occurs
frequently in various sciences, as we will see.
An example of a linear equation is xyЈ ϩ y 2x because, for x 0, it can be written
in the form
yЈ ϩ 2 1
y2
x Notice that this differential equation is not separable because it's impossible to factor the
expression for yЈ as a function of x times a function of y. But we can still solve the equation by noticing, by the Product Rule, that
xyЈ ϩ y ͑xy͒Ј
and so we can rewrite the equation as
͑xy͒Ј 2x
If we now integrate both sides of this equation, we get
xy x 2 ϩ C or C
x yxϩ If we had been given the differential equation in the form of Equation 2, we would have
had to take the preliminary step of multiplying each side of the equation by x.
It turns out that every first-order linear differential equation can be solved in a similar
fashion by multiplying both sides of Equation 1 by a suitable function I͑x͒ called an
integrating factor. We try to find I so that the left side of Equation 1, when multiplied by
I͑x͒, becomes the derivative of the product I͑x͒y:
3 I͑x͒͑yЈ ϩ P͑x͒y͒ ͑I͑x͒y͒Ј If we can find such a function I , then Equation 1 becomes
͑I͑x͒y͒Ј I͑x͒Q͑x͒
Integrating both sides, we would have
I͑x͒y y I͑x͒Q͑x͒ dx ϩ C
so the solution would be
4 y͑x͒ 1
I͑x͒ ͫy ͬ I͑x͒Q͑x͒ dx ϩ C 5E-10(pp 662-671) 1/18/06 9:26 AM Page 669 SECTION 10.6 LINEAR EQUATIONS ❙❙❙❙ 669 To find such an I, we expand Equation 3 and cancel terms:
I͑x͒yЈ ϩ I͑x͒P͑x͒y ͑I͑x͒y͒Ј IЈ͑x͒y ϩ I͑x͒yЈ
I͑x͒P͑x͒ IЈ͑x͒
This is a separable differential equation for I , which we solve as follows: y dI
y P͑x͒ dx
I ԽԽ ln I y P͑x͒ dx
I Ae x P͑x͒ dx
where A Ϯe C. We are looking for a particular integrating factor, not the most general
one, so we take A 1 and use
I͑x͒ e x P͑x͒ dx 5 Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where I
is given by Equation 5. Instead of memorizing this formula, however, we just remember
the form of the integrating factor.
To solve the linear differential equation yЈ ϩ P͑x͒y Q͑x͒, multiply both sides by
the integrating factor I͑x͒ e x P͑x͒ dx and integrate both sides. EXAMPLE 1 Solve the differential equation dy
ϩ 3x 2 y 6x 2.
dx SOLUTION The given equation is linear since it has the form of Equation 1 with P͑x͒ 3x 2 and Q͑x͒ 6x 2. An integrating factor is
I͑x͒ e x 3x 2 dx ex 3 3 Multiplying both sides of the differential equation by e x , we get
|||| Figure 1 shows the graphs of several members of the family of solutions in Example 1.
Notice that they all approach 2 as x l ϱ. ex 6
C=2 3 dy
3
3
ϩ 3x 2e x y 6x 2e x
dx
d x3
3
͑e y͒ 6x 2e x
dx or C=1
C=0 Integrating both sides, we have C=_1
_1.5 1.8
C=_2 3 3 3 e x y y 6x 2e x dx 2e x ϩ C _3 FIGURE 1 y 2 ϩ CeϪx 3 5E-10(pp 662-671) 670 ❙❙❙❙ 1/18/06 9:26 AM Page 670 CHAPTER 10 DIFFERENTIAL EQUATIONS EXAMPLE 2 Find the solution of the initial-value problem x 2 yЈ ϩ xy 1 xϾ0 y͑1͒ 2 SOLUTION We must first divide both sides by the coefficient of yЈ to put the differential
equation into standard form:
6 yЈ ϩ 1
1
y 2
x
x xϾ0 The integrating factor is
I͑x͒ e x ͑1͞x͒ dx e ln x x
Multiplication of Equation 6 by x gives
xyЈ ϩ y 5 1
x 1
dx ln x ϩ C
x y and so ͑xy͒Ј or xy y Then
|||| The solution of the initial-value problem in
Example 2 is shown in Figure 2. 1
x ln x ϩ C
x Since y͑1͒ 2, we have (1, 2)
0 2 4 ln 1 ϩ C
C
1 Therefore, the solution to the initial-value problem is
_5 ln x ϩ 2
x y FIGURE 2 EXAMPLE 3 Solve yЈ ϩ 2xy 1.
SOLUTION The given equation is in the standard form for a linear equation. Multiplying by
the integrating factor
|||| Even though the solutions of the differential
equation in Example 3 are expressed in terms of
an integral, they can still be graphed by a computer algebra system (Figure 3). e x 2x dx e x
we get 2 2 e x yЈ ϩ 2xe x y e x
2 (e x y)Ј e x or 2.5
C=2 2 2 2 2 e x y y e x dx ϩ C Therefore
_2.5 2 2.5
2 C=_2 Recall from Section 8.5 that x e x dx can't be expressed in terms of elementary functions.
Nonetheless, it's a perfectly good function and we can leave the answer as _2.5 FIGURE 3 y eϪx 2 ye x2 dx ϩ CeϪx 2 5E-10(pp 662-671) 1/18/06 9:26 AM Page 671 SECTION 10.6 LINEAR EQUATIONS ❙❙❙❙ 671 Another way of writing the solution is
y eϪx 2 y x 0 2 e t dt ϩ CeϪx 2 (Any number can be chosen for the lower limit of integration.) Application to Electric Circuits
R E L switch In Section 10.2 we considered the simple electric circuit shown in Figure 4: An electromotive force (usually a battery or generator) produces a voltage of E͑t͒ volts (V) and a current of I͑t͒ amperes (A) at time t . The circuit also contains a resistor with a resistance of
R ohms (⍀) and an inductor with an inductance of L henries (H).
Ohm's Law gives the drop in voltage due to the resistor as RI . FIGURE 4 L 7 dI
ϩ RI E͑t͒
dt which is a first-order linear differential equation. The solution gives the current I at time t .
EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is 12 ⍀ and the
inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closed
when t 0 so the current starts with I͑0͒ 0, find (a) I͑t͒, (b) the current after 1 s, and
(c) the limiting value of the current.
SOLUTION
|||| The differential equation in Example 4 is
both linear and separable, so an alternative
method is to solve it as a separable equation
(Example 4 in Section 10.3). If we replace the
battery by a generator, however, we get an equation that is linear but not separable (Example 5). (a) If we put L 4, R 12, and E͑t͒ 60 in Equation 7, we obtain the initial-value
problem or dI
ϩ 12I 60
dt I͑0͒ 0 dI
ϩ 3I 15
dt 4 I͑0͒ 0 Multiplying by the integrating factor e x 3 dt e 3t, we get
e 3t dI
ϩ 3e 3tI 15e 3t
dt
d 3t
͑e I͒ 15e 3t
dt
e 3tI y 15e 3t dt 5e 3t ϩ C
I͑t͒ 5 ϩ CeϪ3t Since I͑0͒ 0, we have 5 ϩ C 0, so C Ϫ5 and
I͑t͒ 5͑1 Ϫ eϪ3t ͒ 5E-10(pp 672-681) ❙❙❙❙ 672 1/18/06 5:16 PM Page 672 CHAPTER 10 DIFFERENTIAL EQUATIONS |||| Figure 5 shows how the current in Example 4
approaches its limiting value. (b) After 1 second the current is
I͑1͒ 5͑1 Ϫ eϪ3 ͒ Ϸ 4.75 A 6 lim I͑t͒ lim 5͑1 Ϫ eϪ3t ͒ (c) y=5 tlϱ tlϱ 5 Ϫ 5 lim eϪ3t
tlϱ 5Ϫ05
2.5 0 EXAMPLE 5 Suppose that the resistance and inductance remain as in Example 4
but, instead of the battery, we use a generator that produces a variable voltage of
E͑t͒ 60 sin 30t volts. Find I͑t͒. FIGURE 5 SOLUTION This time the differential equation becomes 4 dI
ϩ 12I 60 sin 30t
dt dI
ϩ 3I 15 sin 30t
dt or The same integrating factor e 3t gives
d 3t
dI
͑e I͒ e 3t
ϩ 3e 3tI 15e 3t sin 30t
dt
dt |||| Figure 6 shows the graph of the current
when the battery is replaced by a generator. Using Formula 98 in the Table of Integrals, we have 2 e 3tI y 15e 3t sin 30t dt 15
0 e 3t
͑3 sin 30t Ϫ 30 cos 30t͒ ϩ C
909 5
I 101 ͑sin 30t Ϫ 10 cos 30t͒ ϩ CeϪ3t 2.5 Since I͑0͒ 0, we get
50
Ϫ 101 ϩ C 0 _2 |||| 10.6
1–4 5
50
I͑t͒ 101 ͑sin 30t Ϫ 10 cos 30t͒ ϩ 101 eϪ3t so FIGURE 6 Exercises Determine whether the differential equation is linear. |||| 1. yЈ ϩ e x y x 2 y 2 2. y ϩ sin x x 3yЈ 3. xyЈ ϩ ln x Ϫ x 2 y 0 12. 4. yЈ ϩ cos y tan x ■ ■ 5–14 |||| ■ ■ ■ ■ ■ ■ ■ ■ 13. ͑1 ϩ t͒
■ Solve the differential equation. 5. yЈ ϩ 2y 2e x 6. yЈ x ϩ 5y 7. xyЈ Ϫ 2y x 2 8. x 2 yЈ ϩ 2xy cos 2 x 9. xyЈ ϩ y sx
10. 1 ϩ xy xyЈ
11. dy
ϩ 2xy x 2
dx dy
x sin 2x ϩ y tan x,
dx Ϫ͞2 Ͻ x Ͻ ͞2 du
ϩ u 1 ϩ t, t Ͼ 0
dt ■ 14. t ln t
■ ■ 15–20 dr
ϩ r te t
dt
■ |||| 17. ■ ■ ■ Solve the initial-value problem. 15. yЈ x ϩ y,
16. t ■ y͑0͒ 2 dy
ϩ 2y t 3,
dt t Ͼ 0, y͑1͒ 0 dv
2
Ϫ 2tv 3t 2e t , v͑0͒ 5
dt ■ ■ ■ ■ ■ 5E-10(pp 672-681) 1/18/06 5:17 PM Page 673 SECTION 10.6 LINEAR EQUATIONS 18. 2xyЈ ϩ y 6x, x Ͼ 0, 20. x
■ dy
y
Ϫ
x,
dx
xϩ1
■ ; 21–22 ■ ■ y͑4͒ 20 y͑1͒ 0, ■ ■ ■ xϾ0
■ RI ϩ
■ ■ ■ But I dQ͞dt (see Example 3 in Section 3.4), so we have Solve the differential equation and use a graphing calculator or computer to graph several members of the family of
solutions. How does the solution curve change as C varies? R xϾ0 22. yЈ ϩ ͑cos x͒y cos x
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 23. A Bernoulli differential equation (named after James 31. Let P͑t͒ be the performance level of someone learning a skill Observe that, if n 0 or 1, the Bernoulli equation is linear.
For other values of n, show that the substitution u y 1Ϫn transforms the Bernoulli equation into the linear equation
du
ϩ ͑1 Ϫ n͒P͑x͒u ͑1 Ϫ n͒Q͑x͒
dx
Use the method of Exercise 23 to solve the differential 25. yЈ ϩ as a reasonable model for learning, where k is a positive
constant. Solve it as a linear differential equation and use your
solution to graph the learning curve. y3
2
y 2
x
x 32. Two new workers were hired for an assembly line. Jim 26. yЈ ϩ y xy 3
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 27. In the circuit shown in Figure 4, a battery supplies a constant voltage of 40 V, the inductance is 2 H, the resistance is 10 ⍀,
and I͑0͒ 0.
(a) Find I͑t͒.
(b) Find the current after 0.1 s.
28. In the circuit shown in Figure 4, a generator supplies a voltage ; as a function of the training time t. The graph of P is called a
learning curve. In Exercise 13 in Section 10.1 we proposed the
differential equation
dP
k͓M Ϫ P͑t͔͒
dt equation.
24. xyЈ ϩ y Ϫxy 2 Suppose the resistance is 5 ⍀, the capacitance is 0.05 F, a
battery gives a constant voltage of 60 V, and the initial charge
is Q͑0͒ 0 C. Find the charge and the current at time t.
and E͑t͒ 10 sin 60t. Find the charge and the current at time t. dy
ϩ P͑x͒y Q͑x͒y n
dx |||| 1
dQ
ϩ
Q E͑t͒
dt
C 30. In the circuit of Exercise 29, R 2 ⍀, C 0.01 F, Q͑0͒ 0, Bernoulli) is of the form 24–26 Q
E͑t͒
C ■ |||| 21. xyЈ ϩ y x cos x, 673 capacitor is Q͞C, where Q is the charge (in coulombs), so in
this case Kirchhoff's Law gives y͑͒ 0 19. xyЈ y ϩ x sin x,
2 ❙❙❙❙ of E͑t͒ 40 sin 60t volts, the inductance is 1 H, the resistance
is 20 ⍀, and I͑0͒ 1 A.
(a) Find I͑t͒.
(b) Find the current after 0.1 s.
(c) Use a graphing device to draw the graph of the current
function.
29. The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of C farads (F), and a resistor
with a resistance of R ohms (⍀). The voltage drop across the processed 25 units during the first hour and 45 units during the
second hour. Mark processed 35 units during the first hour and
50 units the second hour. Using the model of Exercise 31 and
assuming that P͑0͒ 0, estimate the maximum number of
units per hour that each worker is capable of processing.
33. In Section 10.3 we looked at mixing problems in which the volume of fluid remained constant and saw that such problems
give rise to separable equations. (See Example 6 in that
section.) If the rates of flow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable.
A tank contains 100 L of water. A solution with a salt concentration of 0.4 kg͞L is added at a rate of 5 L͞min. The
solution is kept mixed and is drained from the tank at a rate
of 3 L͞min. If y͑t͒ is the amount of salt (in kilograms) after
t minutes, show that y satisfies the differential equation
dy
3y
2Ϫ
dt
100 ϩ 2t C Solve this equation and find the concentration after 20 minutes.
E R 34. A tank with a capacity of 400 L is full of a mixture of water and chlorine with a concentration of 0.05 g of chlorine per liter.
In order to reduce the concentration of chlorine, fresh water is
pumped into the tank at a rate of 4 L͞s. The mixture is kept 5E-10(pp 672-681) 674 ❙❙❙❙ 1/18/06 5:17 PM Page 674 CHAPTER 10 DIFFERENTIAL EQUATIONS stirred and is pumped out at a rate of 10 L͞s. Find the amount
of chlorine in the tank as a function of time. (a) Solve this as a linear equation to show that
mt
v
͑1 Ϫ eϪct͞m ͒
c 35. An object with mass m is dropped from rest and we assume that the air resistance is proportional to the speed of the object.
If s͑t͒ is the distance dropped after t seconds, then the speed is
v sЈ͑t͒ and the acceleration is a vЈ͑t͒. If t is the acceleration
due to gravity, then the downward force on the object is
mt Ϫ cv, where c is a positive constant, and Newton's Second
Law gives
m |||| 10.7 dv
mt Ϫ cv
dt (b) What is the limiting velocity?
(c) Find the distance the object has fallen after t seconds.
36. If we ignore air resistance, we can conclude that heavier objects fall no faster than lighter objects. But if we take air
resistance into account, our conclusion changes. Use the
expression for the velocity of a falling object in Exercise 35(a)
to find dv͞dm and show that heavier objects do fall faster than
lighter ones. Predator-Prey Systems
We have looked at a variety of models for the growth of a single species that lives alone in
an environment. In this section we consider more realistic models that take into account
the interaction of two species in the same habitat. We will see that these models take the
form of a pair of linked differential equations.
We first consider the situation in which one species, called the prey, has an ample food
supply and the second species, called the predators, feeds on the prey. Examples of prey
and predators include rabbits and wolves in an isolated forest, food fish and sharks, aphids
and ladybugs, and bacteria and amoebas. Our model will have two dependent variables and
both are functions of time. We let R͑t͒ be the number of prey (using R for rabbits) and W͑t͒
be the number of predators (with W for wolves) at time t.
In the absence of predators, the ample food supply would support exponential growth
of the prey, that is,
dR
kR
dt where k is a positive constant In the absence of prey, we assume that the predator population would decline at a rate proportional to itself, that is,
dW
ϪrW
dt where r is a positive constant With both species present, however, we assume that the principal cause of death among the
prey is being eaten by a predator, and the birth and survival rates of the predators depend
on their available food supply, namely, the prey. We also assume that the two species
encounter each other at a rate that is proportional to both populations and is therefore proportional to the product RW. (The more there are of either population, the more encounters there are likely to be.) A system of two differential equations that incorporates these
assumptions is as follows:
W represents the predator.
R represents the prey. 1 dR
kR Ϫ aRW
dt dW
ϪrW ϩ bRW
dt where k, r, a, and b are positive constants. Notice that the term ϪaRW decreases the natural growth rate of the prey and the term bRW increases the natural growth rate of the
predators. 5E-10(pp 672-681) 1/18/06 5:17 PM Page 675 SECTION 10.7 PREDATOR-PREY SYSTEMS |||| The Lotka-Volterra equations were proposed
as a model to explain the variations in the shark
and food-fish populations in the Adriatic Sea
by the Italian mathematician Vito Volterra
(1860–1940). ❙❙❙❙ 675 The equations in (1) are known as the predator-prey equations, or the Lotka-Volterra
equations. A solution of this system of equations is a pair of functions R͑t͒ and W͑t͒ that
describe the populations of prey and predator as functions of time. Because the system is
coupled (R and W occur in both equations), we can't solve one equation and then the other;
we have to solve them simultaneously. Unfortunately, it is usually impossible to find
explicit formulas for R and W as functions of t. We can, however, use graphical methods
to analyze the equations.
EXAMPLE 1 Suppose that populations of rabbits and wolves are described by the LotkaVolterra equations (1) with k 0.08, a 0.001, r 0.02, and b 0.00002. The time t
is measured in months.
(a) Find the constant solutions (called the equilibrium solutions) and interpret
the answer.
(b) Use the system of differential equations to find an expression for dW͞dR.
(c) Draw a direction field for the resulting differential equation in the RW-plane. Then
use that direction field to sketch some solution curves.
(d) Suppose that, at some point in time, there are 1000 rabbits and 40 wolves. Draw the
corresponding solution curve and use it to describe the changes in both population levels.
(e) Use part (d) to make sketches of R and W as functions of t.
SOLUTION (a) With the given values of k, a, r, and b, the Lotka-Volterra equations become
dR
0.08R Ϫ 0.001RW
dt
dW
Ϫ0.02W ϩ 0.00002RW
dt
Both R and W will be constant if both derivatives are 0, that is,
RЈ R͑0.08 Ϫ 0.001W͒ 0
WЈ W͑Ϫ0.02 ϩ 0.00002R͒ 0
One solution is given by R 0 and W 0. (This makes sense: If there are no rabbits or
wolves, the populations are certainly not going to increase.) The other constant solution is
W 0.08
80
0.001 R 0.02
1000
0.00002 So the equilibrium populations consist of 80 wolves and 1000 rabbits. This means that
1000 rabbits are just enough to support a constant wolf population of 80. There are neither too many wolves (which would result in fewer rabbits) nor too few wolves (which
would result in more rabbits).
(b) We use the Chain Rule to eliminate t:
dW
dW dR
dt
dR dt so dW
dW
dt
Ϫ0.02W ϩ 0.00002RW
dR
dR
0.08R Ϫ 0.001RW
dt 5E-10(pp 672-681) 676 ❙❙❙❙ 1/18/06 5:17 PM Page 676 CHAPTER 10 DIFFERENTIAL EQUATIONS (c) If we think of W as a function of R, we have the differential equation
dW
Ϫ0.02W ϩ 0.00002RW
dR
0.08R Ϫ 0.001RW
We draw the direction field for this differential equation in Figure 1 and we use it to
sketch several solution curves in Figure 2. If we move along a solution curve, we
observe how the relationship between R and W changes as time passes. Notice that the
curves appear to be closed in the sense that if we travel along a curve, we always return
to the same point. Notice also that the point (1000, 80) is inside all the solution curves.
That point is called an equilibrium point because it corresponds to the equilibrium solution R 1000, W 80.
W W 150 150 100 100 50 50 0 FIGURE 1 1000 2000 0 3000 R FIGURE 2 Direction field for the predator-prey system 3000 R 2000 1000 Phase portrait of the system When we represent solutions of a system of differential equations as in Figure 2, we
refer to the RW-plane as the phase plane, and we call the solution curves phase trajectories. So a phase trajectory is a path traced out by solutions ͑R, W͒ as time goes by. A
phase portrait consists of equilibrium points and typical phase trajectories, as shown in
Figure 2.
(d) Starting with 1000 rabbits and 40 wolves corresponds to drawing the solution curve
through the point P0(1000, 40). Figure 3 shows this phase trajectory with the direction
field removed. Starting at the point P0 at time t 0 and letting t increase, do we move
clockwise or counterclockwise around the phase trajectory? If we put R 1000 and
W P™
140
120
100
80 P£ P¡ 60
40 P¸ (1000, 40) 20 FIGURE 3 Phase trajectory through (1000, 40) 0 500 1000 1500 2000 2500 3000 R 5E-10(pp 672-681) 1/18/06 5:18 PM Page 677 SECTION 10.7 PREDATOR-PREY SYSTEMS ❙❙❙❙ 677 W 40 in the first differential equation, we get
dR
0.08͑1000͒ Ϫ 0.001͑1000͒͑40͒ 80 Ϫ 40 40
dt
Since dR͞dt Ͼ 0, we conclude that R is increasing at P0 and so we move counterclockwise around the phase trajectory.
We see that at P0 there aren't enough wolves to maintain a balance between the populations, so the rabbit population increases. That results in more wolves and eventually
there are so many wolves that the rabbits have a hard time avoiding them. So the number
of rabbits begins to decline (at P1 , where we estimate that R reaches its maximum population of about 2800). This means that at some later time the wolf population starts to
fall (at P2 , where R 1000 and W Ϸ 140). But this benefits the rabbits, so their population later starts to increase (at P3 , where W 80 and R Ϸ 210). As a consequence, the wolf population eventually starts to increase as well. This happens when the
populations return to their initial values of R 1000 and W 40, and the entire cycle
begins again.
(e) From the description in part (d) of how the rabbit and wolf populations rise and fall,
we can sketch the graphs of R͑t͒ and W͑t͒. Suppose the points P1 , P2 , and P3 in Figure 3
are reached at times t1 , t2 , and t3 . Then we can sketch graphs of R and W as in Figure 4.
R W
140 2500 120
2000 100 1500 80
60 1000 40
500
0 20
t¡ t™ 0 t t£ t¡ t™ t t£ FIGURE 4 Graphs of the rabbit and wolf
populations as functions of time To make the graphs easier to compare, we draw the graphs on the same axes but with
different scales for R and W, as in Figure 5. Notice that the rabbits reach their maximum
populations about a quarter of a cycle before the wolves.
R
3000 W R
W 120
Number 2000
of
rabbits 80 Number
of
wolves 1000
40 FIGURE 5 Comparison of the rabbit
and wolf populations 0 t¡ t™ t£ t 5E-10(pp 672-681) 678 ❙❙❙❙ 1/18/06 5:18 PM Page 678 CHAPTER 10 DIFFERENTIAL EQUATIONS An important part of the modeling process, as we discussed in Section 1.2, is to interpret our mathematical conclusions as real-world predictions and to test the predictions
against real data. The Hudson's Bay Company, which started trading in animal furs in
Canada in 1670, has kept records that date back to the 1840s. Figure 6 shows graphs of the
number of pelts of the snowshoe hare and its predator, the Canada lynx, traded by the company over a 90-year period. You can see that the coupled oscillations in the hare and lynx
populations predicted by the Lotka-Volterra model do actually occur and the period of
these cycles is roughly 10 years.
160 hare
120 9 lynx
Thousands 80
of hares 6 Thousands
of lynx 40 3 FIGURE 6 Relative abundance of hare and lynx
from Hudson's Bay Company records 0 1850 1875 1900 1925 Although the relatively simple Lotka-Volterra model has had some success in explaining and predicting coupled populations, more sophisticated models have also been proposed. One way to modify the Lotka-Volterra equations is to assume that, in the absence
of predators, the prey grow according to a logistic model with carrying capacity K. Then
the Lotka-Volterra equations (1) are replaced by the system of differential equations ͩ ͪ dR
R
kR 1 Ϫ
dt
K Ϫ aRW dW
ϪrW ϩ bRW
dt This model is investigated in Exercises 9 and 10.
Models have also been proposed to describe and predict population levels of two
species that compete for the same resources or cooperate for mutual benefit. Such models
are explored in Exercise 2. |||| 10.7 Exercises 1. For each predator-prey system, determine which of the vari- ables, x or y, represents the prey population and which
represents the predator population. Is the growth of the prey
restricted just by the predators or by other factors as well? Do
the predators feed only on the prey or do they have additional
food sources? Explain.
dx
(a)
Ϫ0.05x ϩ 0.0001xy
dt
dy
0.1y Ϫ 0.005xy
dt
(b) dx
0.2x Ϫ 0.0002x 2 Ϫ 0.006xy
dt
dy
Ϫ0.015y ϩ 0.00008xy
dt 2. Each system of differential equations is a model for two species that either compete for the same resources or cooperate
for mutual benefit (flowering plants and insect pollinators, for
instance). Decide whether each system describes competition
or cooperation and explain why it is a reasonable model. (Ask
yourself what effect an increase in one species has on the
growth rate of the other.)
dx
(a)
0.12x Ϫ 0.0006x 2 ϩ 0.00001xy
dt
dy
0.08x ϩ 0.00004xy
dt
dx
(b)
0.15x Ϫ 0.0002x 2 Ϫ 0.0006xy
dt
dy
0.2y Ϫ 0.00008y 2 Ϫ 0.0002xy
dt 5E-10(pp 672-681) 1/18/06 5:18 PM Page 679 ❙❙❙❙ SECTION 10.7 PREDATOR-PREY SYSTEMS
|||| A phase trajectory is shown for populations of rabbits ͑R͒
and foxes ͑F͒.
(a) Describe how each population changes as time goes by.
(b) Use your description to make a rough sketch of the graphs of R
and F as functions of time. y 6. 3–4 species 1 1200
1000
800 F 600 300 3. 679 400 species 2 200
0 200
■ ■ ■ ■ 10
■ ■ 15
■ ■ ■ t
■ ■ 7. In Example 1(b) we showed that the rabbit and wolf popula- t=0 100 ■ 5 tions satisfy the differential equation
dW
Ϫ0.02W ϩ 0.00002RW
dR
0.08R Ϫ 0.001RW 0 800 400 1200 1600 R 2000 By solving this separable differential equation, show that
R 0.02W 0.08
e
e 0.00002R 0.001W 4. F where C is a constant.
It is impossible to solve this equation for W as an explicit
function of R (or vice versa). If you have a computer algebra
system that graphs implicitly defined curves, use this equation
and your CAS to draw the solution curve that passes through
the point ͑1000, 40͒ and compare with Figure 3. t=0 160 C 120 8. Populations of aphids and ladybugs are modeled by the 80 equations
dA
2A Ϫ 0.01AL
dt 40 0 ■ ■ 400 ■ ■ 800 ■ ■ 1200 ■ ■ R 1600 ■ ■ ■ ■ 5–6 |||| Graphs of populations of two species are shown. Use them
to sketch the corresponding phase trajectory. 5. y dL
Ϫ0.5L ϩ 0.0001AL
dt
(a) Find the equilibrium solutions and explain their
significance.
(b) Find an expression for dL͞dA.
(c) The direction field for the differential equation in part (b) is
shown. Use it to sketch a phase portrait. What do the phase
trajectories have in common?
L species 1 200 400 species 2 300
100 200
100 0 1 t 0 5000 10000 15000 A 5E-10(pp 672-681) ❙❙❙❙ 680 1/18/06 5:18 PM Page 680 CHAPTER 10 DIFFERENTIAL EQUATIONS (d) Suppose that at time t 0 there are 1000 aphids and
200 ladybugs. Draw the corresponding phase trajectory and
use it to describe how both populations change.
(e) Use part (d) to make rough sketches of the aphid and ladybug populations as functions of t. How are the graphs
related to each other? (b) Find all the equilibrium solutions and explain their
significance.
(c) The figure shows the phase trajectory that starts at the point
͑1000, 40͒. Describe what eventually happens to the rabbit
and wolf populations.
(d) Sketch graphs of the rabbit and wolf populations as functions of time. 9. In Example 1 we used Lotka-Volterra equations to model popu- lations of rabbits and wolves. Let's modify those equations as
follows:
dR
0.08R͑1 Ϫ 0.0002R͒ Ϫ 0.001RW
dt CAS 10. In Exercise 8 we modeled populations of aphids and ladybugs with a Lotka-Volterra system. Suppose we modify those equations as follows:
dA
2A͑1 Ϫ 0.0001A͒ Ϫ 0.01AL
dt dW
Ϫ0.02W ϩ 0.00002RW
dt
(a) According to these equations, what happens to the rabbit
population in the absence of wolves? dL
Ϫ0.5L ϩ 0.0001AL
dt W (a) In the absence of ladybugs, what does the model predict
about the aphids?
(b) Find the equilibrium solutions.
(c) Find an expression for dL͞dA.
(d) Use a computer algebra system to draw a direction field for
the differential equation in part (c). Then use the direction
field to sketch a phase portrait. What do the phase trajectories have in common?
(e) Suppose that at time t 0 there are 1000 aphids and
200 ladybugs. Draw the corresponding phase trajectory and
use it to describe how both populations change.
(f) Use part (e) to make rough sketches of the aphid and
ladybug populations as functions of t. How are the graphs
related to each other? 70 60 50 40 30
600 |||| 800 1000 1200 1400 1600 10 Review R ■ CONCEPT CHECK 1. (a) What is a differential equation? (b) What is the order of a differential equation?
(c) What is an initial condition?
2. What can you say about the solutions of the equation yЈ x 2 ϩ y 2 just by looking at the differential equation?
3. What is a direction field for the differential equation yЈ F͑x, y͒?
4. Explain how Euler's method works.
5. What is a separable differential equation? How do you solve it?
6. What is a first-order linear differential equation? How do you solve it? ■ 7. (a) Write a differential equation that expresses the law of natural growth. What does it say in terms of relative
growth rate?
(b) Under what circumstances is this an appropriate model for
population growth?
(c) What are the solutions of this equation?
8. (a) Write the logistic equation. (b) Under what circumstances is this an appropriate model for
population growth?
9. (a) Write Lotka-Volterra equations to model populations of food fish ͑F͒ and sharks ͑S͒.
(b) What do these equations say about each population in the
absence of the other? 5E-10(pp 672-681) 1/18/06 5:19 PM Page 681 CHAPTER 10 REVIEW ■ TRUE-FALSE QUIZ ❙❙❙❙ 681 ■ 5. The equation e x yЈ y is linear. Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement. 6. The equation yЈ ϩ xy e y is linear. 1. All solutions of the differential equation yЈ Ϫ1 Ϫ y 4 are 7. If y is the solution of the initial-value problem decreasing functions. ͩ ͪ 2. The function f ͑x͒ ͑ln x͒͞x is a solution of the differential dy
y
2y 1 Ϫ
dt
5 equation x 2 yЈ ϩ xy 1. 3. The equation yЈ x ϩ y is separable.
4. The equation yЈ 3y Ϫ 2x ϩ 6xy Ϫ 1 is separable. y͑0͒ 1 then lim t l ϱ y 5. ■ EXERCISES ■ y
3 1. (a) A direction field for the differential equation yЈ y͑ y Ϫ 2͒͑ y Ϫ 4͒ is shown. Sketch the graphs of the
solutions that satisfy the given initial conditions.
(i) y͑0͒ Ϫ0.3
(ii) y͑0͒ 1
(iii) y͑0͒ 3
(iv) y͑0͒ 4.3
(b) If the initial condition is y͑0͒ c, for what values of c is
lim t l ϱ y͑t͒ finite? What are the equilibrium solutions? 2
1 _3 y
6 _2 0 _1 1 3 x 2 _1
_2 4 _3 2 0 1 2 (b) Use Euler's method with step size 0.1 to estimate y͑0.3͒
where y͑x͒ is the solution of the initial-value problem in
part (a). Compare with your estimate from part (a).
(c) On what lines are the centers of the horizontal line
segments of the direction field in part (a) located? What
happens when a solution curve crosses these lines? x 4. (a) Use Euler's method with step size 0.2 to estimate y͑0.4͒, 2. (a) Sketch a direction field for the differential equation where y͑x͒ is the solution of the initial-value problem yЈ x͞y. Then use it to sketch the four solutions that
satisfy the initial conditions y͑0͒ 1, y͑0͒ Ϫ1, y͑2͒ 1,
and y͑Ϫ2͒ 1.
(b) Check your work in part (a) by solving the differential
equation explicitly. What type of curve is each solution
curve?
3. (a) A direction field for the differential equation yЈ x 2 Ϫ y 2 yЈ 2xy 2 (b) Repeat part (a) with step size 0.1.
(c) Find the exact solution of the differential equation and
compare the value at 0.4 with the approximations in
parts (a) and (b).
5–8 is shown. Sketch the solution of the initial-value problem
yЈ x Ϫ y
2 2 Solve the differential equation. |||| 5. yЈ xeϪsin x Ϫ y cos x y͑0͒ 1 Use your graph to estimate the value of y͑0.3͒. y͑0͒ 1 6. 8. x 2 yЈ Ϫ y 2 x 3e Ϫ1͞x 7. ͑3y 2 ϩ 2y͒yЈ x cos x
■ ■ ■ ■ ■ dx
1 Ϫ t ϩ x Ϫ tx
dt ■ ■ ■ ■ ■ ■ ■ 5E-10(pp 682-683) 1/18/06 9:30 AM Page 682 682 ❙❙❙❙ CHAPTER 10 DIFFERENTIAL EQUATIONS 9–11 |||| Solve the initial-value problem. 9. xyyЈ ln x, y͑1͒ 2 10. 1 ϩ x 2xyyЈ, x Ͼ 0, 11. yЈ ϩ y sxeϪx,
■ ■ ■ 19. The von Bertalanffy growth model is used to predict the length ■ y͑1͒ Ϫ2 y͑0͒ 3
■ ■ ■ ■ ■ ■ ■ ■ x
; 12. Solve the initial-value problem 2yyЈ xe , y͑0͒ 1, and graph the solution.
13–14 |||| ■ ■ 20. A tank contains 100 L of pure water. Brine that contains Find the orthogonal trajectories of the family of curves. ■ 0.1 kg of salt per liter enters the tank at a rate of 10 L͞min.
The solution is kept thoroughly mixed and drains from the tank
at the same rate. How much salt is in the tank after 6 minutes? k
14. y
1 ϩ x2 13. kx 2 ϩ y 2 1
■ ■ ■ ■ ■ ■ ■ ■ L͑t͒ of a fish over a period of time. If L ϱ is the largest length
for a species, then the hypothesis is that the rate of growth in
length is proportional to L ϱ Ϫ L , the length yet to be achieved.
(a) Formulate and solve a differential equation to find an
expression for L͑t͒.
(b) For the North Sea haddock it has been determined that
L ϱ 53 cm, L͑0͒ 10 cm, and the constant of proportionality is 0.2. What does the expression for L͑t͒ become
with these data? ■ 15. A bacteria culture starts with 1000 bacteria and the growth rate is proportional to the number of bacteria. After 2 hours the
population is 9000.
(a) Find an expression for the number of bacteria after t hours.
(b) Find the number of bacteria after 3 h.
(c) Find the rate of growth after 3 h.
(d) How long does it take for the number of bacteria to double?
16. An isotope of strontium, 90Sr, has a half-life of 25 years. (a) Find the mass of 90Sr that remains from a sample of 18 mg
after t years.
(b) How long would it take for the mass to decay to 2 mg?
17. Let C͑t͒ be the concentration of a drug in the bloodstream. As the body eliminates the drug, C͑t͒ decreases at a rate that is
proportional to the amount of the drug that is present at the
time. Thus, CЈ͑t͒ ϪkC͑t͒, where k is a positive number
called the elimination constant of the drug.
(a) If C0 is the concentration at time t 0, find the concentration at time t.
(b) If the body eliminates half the drug in 30 h, how long does
it take to eliminate 90% of the drug?
18. (a) The population of the world was 5.28 billion in 1990 and 6.07 billion in 2000. Find an exponential model for these
data and use the model to predict the world population in
the year 2020.
(b) According to the model in part (a), when will the world
population exceed 10 billion?
(c) Use the data in part (a) to find a logistic model for the population. Assume a carrying capacity of 100 billion. Then
use the logistic model to predict the population in 2020.
Compare with your prediction from the exponential model.
(d) According to the logistic model, when will the world population exceed 10 billion? Compare with your prediction in
part (b). 21. One model for the spread of an epidemic is that the rate of spread is jointly proportional to the number of infected people
and the number of uninfected people. In an isolated town of
5000 inhabitants, 160 people have a disease at the beginning of
the week and 1200 have it at the end of the week. How long
does it take for 80% of the population to become infected?
22. The Brentano-Stevens Law in psychology models the way that a subject reacts to a stimulus. It states that if R represents the
reaction to an amount S of stimulus, then the relative rates of
increase are proportional:
k dS
1 dR
R dt
S dt
where k is a positive constant. Find R as a function of S.
23. The transport of a substance across a capillary wall in lung physiology has been modeled by the differential equation
R
dh
Ϫ
dt
V ͩ ͪ
h
kϩh where h is the hormone concentration in the bloodstream, t is
time, R is the maximum transport rate, V is the volume of the
capillary, and k is a positive constant that measures the affinity
between the hormones and the enzymes that assist the process.
Solve this differential equation to find a relationship between h
and t.
24. Populations of birds and insects are modeled by the equations dx
0.4x Ϫ 0.002xy
dt
dy
Ϫ0.2y ϩ 0.000008xy
dt
(a) Which of the variables, x or y, represents the bird population and which represents the insect population? Explain. 5E-10(pp 682-683) 1/18/06 9:30 AM Page 683 CHAPTER 10 REVIEW (b) Find the equilibrium solutions and explain their
significance.
(c) Find an expression for dy͞dx.
(d) The direction field for the differential equation in part (c) is
shown. Use it to sketch the phase trajectory corresponding
to initial populations of 100 birds and 40,000 insects. Then
use the phase trajectory to describe how both populations
change. ❙❙❙❙ 683 y
260
240
220
200
180
160
140
120
100 y
400 10000 15000 35000 45000 x (d) Sketch graphs of the bird and insect populations as
functions of time. 300 26. Barbara weighs 60 kg and is on a diet of 1600 calories per day, 200
100 0 25000 20000 40000 60000 x of which 850 are used automatically by basal metabolism. She
spends about 15 cal͞kg͞day times her weight doing exercise. If
1 kg of fat contains 10,000 cal and we assume that the storage
of calories in the form of fat is 100% efficient, formulate a differential equation and solve it to find her weight as a function
of time. Does her weight ultimately approach an equilibrium
weight?
27. When a flexible cable of uniform density is suspended between (e) Use part (d) to make rough sketches of the bird and insect
populations as functions of time. How are these graphs
related to each other?
25. Suppose the model of Exercise 24 is replaced by the equations dx
0.4x ͑1 Ϫ 0.000005x͒ Ϫ 0.002xy
dt
dy
Ϫ0.2y ϩ 0.000008xy
dt two fixed points and hangs of its own weight, the shape
y f ͑x͒ of the cable must satisfy a differential equation of the
form
dy 2
d 2y
1ϩ
2 k
dx
dx ͱ ͩ ͪ where k is a positive constant. Consider the cable shown in the
figure.
(a) Let z dy͞dx in the differential equation. Solve the resulting first-order differential equation (in z ), and then integrate
to find y.
(b) Determine the length of the cable.
y (a) According to these equations, what happens to the insect
population in the absence of birds?
(b) Find the equilibrium solutions and explain their
significance.
(c) The figure shows the phase trajectory that starts with
100 birds and 40,000 insects. Describe what eventually
happens to the bird and insect populations. (b, h) (_b, h) (0, a)
_b 0 b x 5E-10(pp 684-685) 1/18/06 9:30 AM Page 684 PROBLEMS
PLUS 1. Find all functions f such that f Ј is continuous and
x [ f ͑x͒] 2 100 ϩ y ͕[ f ͑t͒] 2 ϩ [ f Ј͑t͒] 2 ͖ dt
0 for all real x 2. A student forgot the Product Rule for differentiation and made the mistake of thinking that ͑ ft͒Ј f ЈtЈ. However, he was lucky and got the correct answer. The function f that he
2
used was f ͑x͒ e x and the domain of his problem was the interval ( 1 , ϱ). What was the
2
function t ?
3. Let f be a function with the property that f ͑0͒ 1, f Ј͑0͒ 1, and f ͑a ϩ b͒ f ͑a͒ f ͑b͒ for all real numbers a and b. Show that f Ј͑x͒ f ͑x͒ for all x and deduce that f ͑x͒ e x. 4. Find all functions f that satisfy the equation ͩy ͪͩy f ͑x͒ dx ͪ 1
dx Ϫ1
f ͑x͒ 5. A peach pie is taken out of the oven at 5:00 P.M. At that time it is piping hot: 100ЊC. At 5:10 P.M. its temperature is 80ЊC; at 5:20 P.M. it is 65ЊC. What is the temperature of the
room? 6. Snow began to fall during the morning of February 2 and continued steadily into the after- noon. At noon a snowplow began removing snow from a road at a constant rate. The plow
traveled 6 km from noon to 1 P.M. but only 3 km from 1 P.M. to 2 P.M. When did the snow
begin to fall? [Hints: To get started, let t be the time measured in hours after noon; let x ͑t͒
be the distance traveled by the plow at time t; then the speed of the plow is dx͞dt. Let b be
the number of hours before noon that it began to snow. Find an expression for the height
of the snow at time t. Then use the given information that the rate of removal R (in m3͞h)
is constant.]
y 7. A dog sees a rabbit running in a straight line across an open field and gives chase. In a rectan- gular coordinate system (as shown in the figure), assume:
(i) The rabbit is at the origin and the dog is at the point ͑L, 0͒ at the instant the dog first
sees the rabbit.
(ii) The rabbit runs up the y-axis and the dog always runs straight for the rabbit.
(iii) The dog runs at the same speed as the rabbit.
(a) Show that the dog's path is the graph of the function y f ͑x͒, where y satisfies the differential equation (x, y) x
0 FIGURE FOR PROBLEM 7 (L, 0) x d 2y
dx 2 ͱ ͩ ͪ
1ϩ dy
dx 2 (b) Determine the solution of the equation in part (a) that satisfies the initial conditions
y yЈ 0 when x L. [Hint: Let z dy͞dx in the differential equation and solve the
resulting first-order equation to find z; then integrate z to find y.]
(c) Does the dog ever catch the rabbit?
8. (a) Suppose that the dog in Problem 7 runs twice as fast as the rabbit. Find a differential equation for the path of the dog. Then solve it to find the point where the dog catches the
rabbit.
(b) Suppose the dog runs half as fast as the rabbit. How close does the dog get to the rabbit?
What are their positions when they are closest? 684 5E-10(pp 684-685) 1/18/06 9:30 AM Page 685 9. A planning engineer for a new alum plant must present some estimates to his company regard- ing the capacity of a silo designed to contain bauxite ore until it is processed into alum. The
ore resembles pink talcum powder and is poured from a conveyor at the top of the silo. The
silo is a cylinder 100 ft high with a radius of 200 ft. The conveyor carries 60,000 ft 3 and
͞h
the ore maintains a conical shape whose radius is 1.5 times its height.
(a) If, at a certain time t, the pile is 60 ft high, how long will it take for the pile to reach the
top of the silo?
(b) Management wants to know how much room will be left in the floor area of the silo when
the pile is 60 ft high. How fast is the floor area of the pile growing at that height?
(c) Suppose a loader starts removing the ore at the rate of 20,000 ft 3 when the height of
͞h
the pile reaches 90 ft. Suppose, also, that the pile continues to maintain its shape. How
long will it take for the pile to reach the top of the silo under these conditions?
10. Find the curve that passes through the point ͑3, 2͒ and has the property that if the tangent line is drawn at any point P on the curve, then the part of the tangent line that lies in the first
quadrant is bisected at P.
11. Recall that the normal line to a curve at a point P on the curve is the line that passes through P and is perpendicular to the tangent line at P. Find the curve that passes through the point
͑3, 2͒ and has the property that if the normal line is drawn at any point on the curve, then the
y-intercept of the normal line is always 6.
12. Find all curves with the property that if the normal line is drawn at any point P on the curve, then the part of the normal line between P and the x-axis is bisected by the y-axis. 685 ...
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This note was uploaded on 02/08/2010 for the course M 340L taught by Professor Lay during the Spring '10 term at École Normale Supérieure. | 677.169 | 1 |
GCSE Mathematics: Revision and Practice: Higher: Homework Book
Synopsis
GCSE Mathematics: Revision and Practice: Higher: Homework Book by David Rayner
This book has been specifically written for the new two-tier GCSE specification for first examination in 2008. The book is targeted at the Higher tier GCSE, and it comprises units organised clearly into homeworks designed to support the use of the Higher Students' Book in the same series. Each unit offers: *Homework exercises directly related to the Students' book section *Exam-style homeworks to build confidence and practice key techniques *Routine practice to reinforce key skills and ideas. *Full answers in the accompanying answer book It forms part of a suite of two homework books at GCSE, in which the other book caters for the Foundation tier. | 677.169 | 1 |
List of Free Online Algebra Courses and Lessons
At m you can study math online - for free. If you are studying at high school or are preparing for college this is the perfect place for you. We have decided to divide our material into four math courses: Pre-algebra, Algebra 1, Algebra 2 and Geometry - on these pages you will find all the theory needed to complete your.
To start taking math video lessons online - choose in the right navigation bar above. Under each and every lesson you will find a corresponding math video lesson. Good luck!
The rule for adding terms. Subtracting a negative number. 1 4. Multiplying and dividing signed numbers, the Rule of Signs. 1 5. Reciprocals and zero. The definition of reciprocals. The definition of division.
The general form. Parallel and perpendicular lines. The point-slope formula. The two-point formula. 35. Simultaneous linear equations The method of addition. The method of substitution. Cramer's Rule: The method of determinants. Three equations in three unknowns.
Rules for 0. 1 6. Some rules of algebra, the rule of symmetry. Commutative rules. Inverses. Two rules for equations. 1 7. Removing grouping symbols, parentheses. Brackets. Braces. The relationship of b a to a b.
24. Equations with fractions Clearing of fractions. 25. Word problems that lead to equations with fractions The whole is equal to the sum of the parts. Same time problem: Upstream-downstream. Total time problem.
Home 1 1. Algebraic expressions, the four operations and their signs. The function of parentheses. Terms versus factors. Powers and exponents. The order of operations. Evaluating algebraic expressions. 1 2. Signed numbers: Positive and negative, the absolute value and the algebraic sign. | 677.169 | 1 |
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Summary
What knowledge of mathematics do secondary school math teachers need to facilitate understanding, competency, and interest in mathematics for all of their students? This unique text and resource bridges the gap between the mathematics learned in college and the mathematics taught in secondary schools. Written in an informal, clear, and interactive learner-centered style, it is designed to help pre-service and in-service teachers gain the mathematical insight they need to engage their students in learning mathematics in a multifaceted way that is interesting, developmental, connected, deep, understandable, and often, surprising and entertaining.Features include Launch questions at the beginning of each section, Student Learning Opportunities, Questions from the Classroom, and highlighted themes throughout to aid readers in becoming teachers who have great "MATH-N-SIGHT":M Multiple Approaches/RepresentationsA Applications to Real LifeT TechnologyH HistoryN Nature of Mathematics: Reasoning and ProofS Solving ProblemsI Interlinking Concepts: ConnectionsG Grade LevelsH Honing of Mathematical SkillsT Typical ErrorsThis text is ideally suited for a capstone mathematics course in a secondary mathematics certification program, is appropriate for any methods or mathematics course for pre- or in-service secondary mathematics teachers, and is a valuable resource for classroom teachers. | 677.169 | 1 |
I am going to high school now. As math has always been my problem area , I purchased the course books in advance. I am plan studying a couple of chapters before the classes begin . Any kind of pointers would be much appreciated that could help me to start studying 6th grade distributive property worksheetsworksheets myself.
How about some more particulars about your problem with 6th grade distributive property worksheetsworksheets? I might be able to suggest . If you are not able to get a good guidance or some one to sit and sort out your problem or if if it is too expensive , then there might be another solution to your problem. There are some good algebra software that you can explore. I checked them out myself. It came across to me as fine as any tutor can be. I would choose Algebrator for the kind of answer that you are in search of . What is nice about it is that it hand holds you to the solutions rather than just providing the answer. Why not try it out?
I allow my son to use that program Algebrator because I believe it can effectively assist him in his algebra problems. It's been a long time since they first used that program and it did not only help him short-term but I noticed it helped in improving his solving capabilities. The software helped him how to solve rather than helped them just to answer. It's fantastic!
Algebrator is one of the best tools that would give you all the fundamentals of 6th grade distributive property worksheetsworksheets. The exceptional training offered by the Algebrator on mixed numbers, multiplying fractions, linear equations and fractional exponents is second to none. I have checked out 3-4 home tutoring mathematics software and I found this to be extra ordinary. The Algebrator not only offers you the primary principles but also helps you in clearing any challenging Remedial Algebra question with ease. The quick formula reference that comes with Algebrator is very detailed and has almost every formula relating to Algebra 2.
Algebrator is a very simple software and is certainly worth a try. You will also find several interesting stuff there. I use it as reference software for my math problems and can swear that it has made learning math more enjoyable.
Thanks a million for the elaborate information. We will surely check this out. Hope we get our assignments completed with the aid of Algebrator. If we have any technical queries with respect to its usage , we would definitely come back to you again. | 677.169 | 1 |
books.google.com - Off,... Geometry
Basic Geometry
Off, the authors deduce plane Euclidean geometry by utilizing only five fundamental postulates. Incorporation of the system of real numbers in three of the five postulates of this geometry gives these assumptions great breadth and power. They lead the reader at once to the heart of geometry. It is because of the underlying power, simplicity, and compactness of this geometry that the authors called the book Basic Geometry. The book is designed for a one-year course in plane geometry. For advanced students, the authors incorporated certain material from three-dimensional and so-called modern geometry. A rich variety of exercises as well as many illustrations applying the abstract geometrical concepts to real life provide an excellent source of teaching material. | 677.169 | 1 |
Mathematics For Engineers Vol 2
ISBN 9788125930846
ISBN-10
8125930841
Binding
Paperback
Number of Pages
276 Pages
Language
(English)
Subject
Biography: science, technology & engineering
This is a sequel to Mathematics for Engineers Volume 1 by the same authors. Like its predecessor, this book also contains a large number of solved examples for students to internalize the concepts. With this provision it is expected that aspiring engineers will not only be able to master the concepts, but also learn the techniques of solving any kind of mathematical problems.
As the book has gradually evolved from the lectures delivered by the authors and their colleagues over the years, authors have been able to use their experiences to design it so that even the mediocre students are able to understand complex concepts, and study with ease and minimum assistance from the teachers. | 677.169 | 1 |
Math Teacher Has Right Formula
`He`s More Of A Friend Than A Teacher` To His Students
The 34 students in Alan Wojtach`s 8th-grade algebra class at Cary Junior High School are learning about solving simultaneous equations, or SSE, as he calls it.
They have an idea of what it means, he says. They discussed the concept the day before and, if they`ve done their homework, they have worked some SSE problems on their own.
`` `Solving simultaneous equations`-say that five times fast,`` Wojtach jokes with the class.
He draws a graph on the chalkboard and writes two equations beside it, one on top of the other:
y (LT) x + 1
y (GT) - x - 5
``Bill, how do I graph these?`` Wojtach asks. ``Should it be a solid line or a dotted line?``
``A dotted line,`` Bill says.
Wojtach draws a diagonal dotted line and diagonally shades in an area below it.
(A dotted line means that the points are not part of the solution; when an equation has a less-than or greater-than sign in it, you use a dotted line. Wojtach says. An equation with a greater-than-or-equal-to, or a less-than-or- equal-to sign takes a solid line.)
``How do we do the next one?`` he continues. ``Should this one be solid or dotted?``
``Dotted,`` another student responds.
Wojtach, looking pleased at the correct answers, draws in another diagonal dotted line and then shades in the area above it.
An area in the middle on the right side of the graph has been shaded twice, where the lines have intersected.
``Any number (on the graph) that got shaded both times will make those equations true at the same time, and that`s what simultaneous means,`` Wojtach explains.
He continues working problem after problem, filling the chalkboard over and over with equations until his students appear to understand what solving simultaneous equations is all about.
They have grasped the concept ``very easily,`` he says. Algebra is usually taught in the first year of high school, but this 8th-grade class is learning it now, he says.
Teaching advanced math to younger students has been a trend in recent years, partly because of modern technology, Wojtach says. As students become more familiar with calculators and computers, they are better equipped to handle more difficult math, he says.
``They want to learn the higher skills,`` he says. ``That eagerness to learn math is what I really like.
``I like (teaching) junior high (students). They`re young enough that you can tell them corny jokes, and they moan and groan. But they`re mature enough that you can get into some really good discussions in math. I like the dichotomy.``
Although he has also taught science during his 11 years at the school, Wojtach, who describes his age as ``greater than 30,`` now teaches only math to 6th and 8th graders. In addition to the advanced class, he teaches four levels of 6th-grade classes and an 8th-grade pre-algebra class.
``I like science. But I prefer math. I really like it, and I have an aptitude for it,`` he says. ``It seems to come easy to me. I`m very logical. I like the logic of (math); I like the structure. I`m a very ordered person, and math is very ordered.``
Having a love of the subject and understanding it well enough to relay it to students, who may each comprehend it differently, are critical factors in teaching it well, Wojtach says. ``I try to explain things to students in the easiest ways possible.``
Some of the students in the advanced class say that he does that successfully.
Dan Bauchiero, 14, says Wojtach is a good teacher ``because he makes things fun for the class. He teaches things very well. He`s more of a friend than a teacher.``
Ryan Frank, 14, says he had Wojtach in 6th grade and was glad to have him again. ``Mr. Wojtach is an exceptional teacher,`` he says. ``He understands students very well, and he realizes our problems with math. (He) teaches math to us thoroughly. He doesn`t rush through . . . the assignments.``
Principal Robert Lipinski agrees that Wojtach ``will do whatever he can to help the kids succeed.`` He has high expectations for his students, but he makes himself available to help them so they can reach those expectations, Lipinski says.
``My goal . . . is for them is to understand math and use that to become productive adults,`` Wojtach says. The idea is ``to have the students become better problem-solvers,`` not just in computations, he says, but in problems that can be used more in day-to-day life, such as how to estimate the area of a room if you need to paint it.
However, learning basic math skills, such as adding, subtracting, multiplying and dividing, never can be replaced, Wojtach says. Calculators are important, and ``I do use them in class,`` he says, but he adds that students can`t rely on them to replace learning. They are allowed to use calculators only after they have mastered the basic math skills, he says. | 677.169 | 1 |
...
Show More generous, but not excessive, dose of numerical analysis. Topics are introduced on a need to know basis to concisely illustrate the practical implementation of a variety of algorithms and demystify seemingly esoteric numerical methods. Algorithms that can be explained without too much elaboration and can be implemented within a few dozen lines of computer code are discussed in detail, and computer programs in Fortran, C++, and Matlab are provided. Algorithms whose underlying theories require long, elaborate explanations are discussed at the level of first principles, and references for further information are given. The book uses numerous schematic illustrations to demonstrate concepts and facilitate their understanding by providing readers with a helpful interplay between ideas and visual images. Real-world examples drawn from various branches of science and engineering are presented. Updated information on computer technology and numerical methods is included, many new and some original topics are introduced. Additional solved and unsolved problems are | 677.169 | 1 |
MATH Algebra 2 Advice
Showing 1 to 3 of 5
It is a required course, so students must take it. It is a challenging class where students must individually solve the given problems.
Course highlights:
I learned about logs (and natural logs), parabolas, factoring, systems of equations, and other math concepts.
Hours per week:
6-8 hours
Advice for students:
Students should spend their efforts into homework in order to be prepared for the class.
Course Term:Winter 2017
Professor:Mrs. Falcone
Course Tags:Math-heavyMany Small AssignmentsCompetitive Classmates
Apr 27, 2017
| No strong feelings either way.
This class was tough.
Course Overview:
It is a challenging class which provides individual thinking in order to figure out the practices given by the teacher.
Course highlights:
I learned about logs (and natural logs), parabolas, factoring, and other math concepts.
Hours per week:
6-8 hours
Advice for students:
I would be prepared to study during the weekends in order to understand the concepts better.
Course Term:Spring 2017
Professor:Mrs. Falcone
Course Required?Yes
Course Tags:Math-heavyMany Small AssignmentsCompetitive Classmates
Nov 30, 2016
| Would highly recommend.
Not too easy. Not too difficult.
Course Overview:
I would most certainly recommend it. I have struggled with math all my life and I have never had a teacher more understanding and willing to help. She guided me when I was lost, corrected me when I was wrong, and reassured me when I had almost given up. She is a sweet and kind lady dedicated to mathematics with a deep understanding and love for the material.
Course highlights:
I learned essential basic mathematical skills to prepare me for future courses. Although the course was challenging in it's own material, it is the foundation for numerous fields of mathematics, so pay attention!
Hours per week:
3-5 hours
Advice for students:
Pay attention. As obvious of a tip this is, it's true. Everything that you need to know will be said at one time or another, and she will often allude to it if she is planning to incorporate it into the next upcoming test! | 677.169 | 1 |
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Cours algebre matrice pdf
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Every night on both cours algebre matrice pdf local and national cours algebre matrice pdf news there is a story about someone having their identity cours algebre matrice pdf, yet cours algebre matrice pdf one ever mentions a solution to the problem. | 677.169 | 1 |
For introductory courses in Differential Equations.This text provides the conceptual development and geometric visualization of a modern differential equations course while maintaining the solid foundation of algebraic techniques that are still essential to science and engineering students. It reflects the new excitement in differential equations as the availability of technical computing environments likeMaple, Mathematica, and MATLAB reshape the role and applications of the discipline. New technology has motivated a shift in emphasis from traditional, manual methods to both qualitative and computer-based methods that render accessible a wider range of realistic applications. With this in mind, the text augments core skills with conceptual perspectives that students will need for the effective use of differential equations in their subsequent work and study.
"synopsis" may belong to another edition of this title.
From the Publisher:
Reflecting the shift in emphasis from traditional methods to new computer-based methods, this text -- in classic Edwards and Penney style -- focuses on the mathematical modeling of real-world phenomena as the goal and constant motivation for the study of differential equations. It offers a fresh computational flavor in figures, examples, problems, and projects throughout, and features a broad range of real-world applications. The book also covers topics not found in other similar texts.
From the Back Cover:
This book reflects the new excitement in Elementary Differential Equations as the availability of specialized scientific computing environments and software systems continues to reshape the role and applications of the discipline. The new approach found in the Second Edition employs tools like Maple, Mathematica, and MATLAB to shift in emphasis from traditional manual methods to new computer-based methods that open a wider range of realistic applications. Coverage begins and ends with discussions of mathematical modeling of real-world phenomena, with a fresh new computational flavor evident in figures, examples, problems, and projects throughout the text. | 677.169 | 1 |
ISBN-13: 9780821893968 Algebraic Geometry has been at the center of much of mathematics for hundreds of years. It is not an easy field to break into, despite its humble beginnings in the study of circles, ellipses, hyperbolas, and parabolas. This text consists of a series of exercises, plus some background information and explanations, starting with conics and ending with sheaves and cohomology. The first chapter on conics is appropriate for first-year college students (and many high school students). Chapter 2 leads the reader to an understanding of the basics of cubic curves, while Chapter 3 introduces higher degree curves. Both chapters are appropriate for people who have taken multivariable calculus and linear algebra. Chapters 4 and 5 introduce geometric objects of higher dimension than curves. Abstract algebra now plays a critical role, making a first course in abstract algebra necessary from this point on. The last chapter is on sheaves and cohomology, providing a hint of current work in algebraic | 677.169 | 1 |
exemplification in mathematics education - Science, Technology ... The word example is used in mathematics education in a wide variety of ways. ... ground' between exercises and worked examples, for instance when a teacher ... mathematical object in question, rather than with qualities of the object itself. ..... ...
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Model solving in mathematical programming by H.P. Williams, Wiley ... solution of real problems receive very little atten- tion, having been dealt with by
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GMAT Algebra and Equation Solving
In this module, you'll learn all of the concepts, techniques and strategies needed to answer algebra and equation-solving questions on the GMAT. In particular, this module covers:
Simplifying algebraic expressions
Expanding expressions
3 types of factoring
Basic equations
Equations with fractions
Quadratic equations
Systems of equations
Equations with square roots
Equations with exponents
Equations with absolute value
Strange operations
3 types of inequalities
The coordinate plane
Equations of lines
Graphs of quadratics ALGEBRA AND EQUATION SOLVING VIDEOS
1.
Lesson:
Introduction to Algebra (5:25)
2.
Lesson:
Simplifying Expressions (5:44)
3.
Question:
Evaluate Expression with 37's (2:12)
4.
Lesson:
Expanding Expressions (7:21)
5.
Lesson:
FOIL Method for Expanding (3:55)
6.
Question:
Product with Roots (1:55)
7.
Lesson:
Factoring – Greatest Common Factor (5:48)
8.
Lesson:
Factoring - Difference of Squares (4:19)
9.
Lesson:
Factoring - Quadratics (6:39)
10.
Lesson:
Combining Factoring Techniques (3:46)
11.
Lesson:
Simplifying Rational Expressions (6:51)
12.
Question:
Simplify a Rational Expression (1:30)
13.
Question:
Tricky Simplification (2:49)
14.
Question:
353 squared (0:45)
15.
Lesson:
Basic Equation Solving (5:05)
16.
Lesson:
Eliminating Fractions (7:38)
17.
Lesson:
Quadratic Equations (8:44)
18.
Question:
Rational Sum of Solutions (1:43)
19.
Lesson:
2 Equations with 2 Unknowns (9:36)
20.
Question:
System Question (2:43)
21.
Lesson:
Number of Solutions to a System (3:37)
22.
Lesson:
3 Equations with 3 Unknowns (2:28)
23.
Lesson:
Equations with Square Roots (8:20)
24.
Lesson:
Equations and Powers (4:37)
25.
Lesson:
Absolute Value Equations (3:46)
26.
Question:
Absolute Value Question (3:47)
27.
Lesson:
Function Notation (5:01)
28.
Question:
Functions f and g (1:32)
29.
Lesson:
Strange Operators (2:22)
30.
Question:
Integers Between x and y (1:21)
31.
Lesson:
Working with Formulas (4:26)
32.
Lesson:
Inequalities - Part I (8:41)
33.
Question:
Is 5-2x Greater than -1? (2:34)
34.
Question:
2x over y (3:19)
35.
Lesson:
Inequalities - Part II (5:54)
36.
Question:
Find Integer x (2:39)
37.
Question:
v-w (2:01)
38.
Lesson:
Inequalities and Absolute Value (6:17)
39.
Lesson:
Quadratic Inequalities (5:21)
40.
Question:
Quadratic and Absolute Value (3:03)
41.
Lesson:
The Something Method (1:42)
42.
Question:
1/10 and 1/2 (2:36)
43.
Lesson:
The Coordinate Plane (3:39)
44.
Question:
Which Quadrant? (2:48)
45.
Lesson:
Distance Between Two Points (3:30)
46.
Lesson:
Graphing Lines (8:12)
47.
Lesson:
Vertical and Horizontal Lines (2:54)
48.
Lesson:
Slope (5:04)
49.
Lesson:
x-intercepts and y-intercepts (2:57)
50.
Question:
Passing Through Quadrant II (2:07)
51.
Lesson:
Slope y-intercept Form (2:59)
52.
Question:
Variable Slope (1:08)
53.
Lesson:
Writing Equations (3:29)
54.
Lesson:
Graphs of Quadratics (1:29)
55.
Question:
Which Equation is Best? (1:09)
56.
Question:
Is x Less than 0? (1:46)
57.
Question:
Expansion with Roots (1:41)
58.
Question:
Functions f and g (1:19)
59.
Question:
Does xy = Square of x? (2:23)
60.
Question:
(x-y)/xy (0:45)
61.
Question:
x/2 and so on (1:50)
62.
Question:
Is 2x less than y+1? (1:43)
63.
Question:
f(4)= (1:06)
64.
Question:
Sum of Solutions (2:11)
65.
Question:
Value of x in a System (1:35)
66.
Question:
Line k (1:24)
67.
Question:
Sum of Squares (0:56)
68.
Question:
px+qy+r (2:43)
69.
Question:
Distance Between Intercepts (2:58)
70.
Question:
Powers of 2 (1:24)
71.
Question:
Absolute Solution Sum (1:53)
72.
Question:
Intersection of Lines A and B (3:19)
73.
Question:
Value of x+2y (1:54)
Total viewing time for the GMAT Algebra and Equation Solving module: 254 minutes.
Additional Practice
NOTE: In each of the above video lessons, you'll find a Related Resources box containing several practice questions that are specifically related to the concepts covered in that lesson (e.g., the lesson on inequalities). Conversely, the practice questions below cover a variety of topics under the umbrella of GMAT Algebra and Equation Solving, 2015), you can enter your responses directly into our downloadable Improvement Chart. This will help track your progress and identify areas of weakness | 677.169 | 1 |
Need an answer to a mathematical question? Wolfram|Alpha is a big online
calculator and dictionary all in one. Instead of searching the web, you can
do dynamic computations based on a vast collection of built-in data,
algorithms, and methods. | 677.169 | 1 |
Unit Goals:
- The students will use the methods of substitution and graphing to solve systems of equations. - The students will use definitions and properties to perform operations on matrices. - The students will use matrices to solve systems of equations, find areas of triangles, and write coded messages.
Unit Objectives:
- When given systems of equations, the students will be able to solve the systems using substitution and verify this solution by graphing the equations.
- When given systems of linear equations in two variables, the student will be able to solve each system using the method of elimination.
- Given the graphs of systems of equations in two variables, the students will be able to interpret the graphs and determine the number of solutions and the consistency of the equations.
- Given pairs of matrices, the students will be able to determine whether or not the matrices are equal.
- Students should be able to perform operations on matrices such as, adding and subtracting matrices and multiplying the matrices by scalars.
- Given problems containing 2x2, square and triangular matrices, the students will be able to describe the process of and find the determinant of each matrix..
- Given problems containing square matrices, the student will be able to define and find the minors and cofactors of each of the matrix.
- When given triangles, the students will be able to find the matrix that corresponds to each triangle, find the determinant and use it to find the area of the triangle.
- Given systems of equations in matrix form, students should be able to use Cramer's Rule to solve each of the systems.
- When given matrices, students will be able use the matrices to encode and decode messages and use the determinants of the matrices to decide whether or not points are collinear.
- When given scenarios in the form of a word problem, the student will be able to set up the problem using systems of equations and matrices and solve for the unknown variables.
Unit Assessments:
- The students will be assessed by a Jeopardy game at the end of the lesson. Each student in each group will be expected to answer one question on their own. The work is not expected to be perfect, but the student should have a good understanding of the process and steps of the substitution method.
- As the students complete the elimination and graphing worksheet, I will walk around the room observing the student's work and progress. The students must show all work, graphs and if necessary explain the reasoning behind their answers. Students will be chosen to put work on board or answer from their desk at random. Grading will be based on accuracy on the worksheet, as well as the student's effort to complete the problems and participation in class. The grading scale will be according to that listed with the worksheet. - The students will be assessed on their completion and accuracy of the worksheet. This worksheet will be given at the end of class and will be treated like a quiz. It will assess the student's knowledge of the content and determine how much time is spent reviewing this material at the beginning of the next class. The grading will be as is listed at the bottom of the quiz. - The students will generate definitions and possible lesson name by studying matrices and the research topics. They should show understanding of determinants, minors and cofactors for bother square, 2x2 and triangular matrices. For assessment, they will be asked to take a short comprehension quiz to determine their understanding of the section and content.
- The student will show mastery of the concepts in this unit by creating a cumulative unit project. The project will consist of one page per concept in which the students are to list steps and provide examples for each of the concepts. The goal will be for the students to create a booklet containing the most important things from the chapter that they could use to teach other student's this unit. The students will add things to the project as they are completed. The grading will be according the rubric for the project. | 677.169 | 1 |
GRE Subject Mathematics
The test consists of approximately 66 multiple-choice questions drawn from courses commonly offered at the undergraduate level.
Approximately 50 percent of the questions involve calculus and its applications — subject matter that can be assumed to be common to the backgrounds of almost all mathematics majors.
About 25 percent of the questions in the test are in elementary algebra, linear algebra, abstract algebra and number theory. The remaining questions deal with other areas of mathematics currently studied by undergraduates in many institutions.
The following content descriptions may assist students in preparing for the test. The percents given are estimates; actual percents will vary somewhat from one edition of the test to another.
CALCULUS — 50%
Material learned in the usual sequence of elementary calculus courses — differential and integral calculus of one and of several variables — includes calculus-based applications and connections with coordinate geometry, trigonometry, differential equations and other branches of mathematics.
ALGEBRA — 25%
Elementary algebra: basic algebraic techniques and manipulations acquired in high school and used throughout mathematics
Other topics: general topology, geometry, complex variables, probability and statistics, and numerical analysis
The above descriptions of topics covered in the test should not be considered exhaustive; it is necessary to understand many other related concepts. Prospective test takers should be aware that questions requiring no more than a good precalculus background may be quite challenging; such questions can be among the most difficult questions on the test. In general, the questions are intended not only to test recall of information but also to assess test takers' understanding of fundamental concepts and the ability to apply those concepts in various situations | 677.169 | 1 |
PRENTICE HALL MATH ALGEBRA 1 STUDENT EDITION book download
10-2 Cell Division Mitosis Spindle forming Prophase Prophase is the first and longest phase of mitosis. The centrioles separate and take up positions on Centromere Chromosomes opposite sides of (paired chromatids) the nucleus. Slide 18 of 38 End Show Copyright Pearson Prentice Hall
10-2 Cell Division Mitosis Spindle forming The centrioles lie in a region called the centrosome. The centrosome helps to organize the spindle, a fanlike microtubule structure that helps Centromere separate the Chromosomes chromosomes. (paired chromatids) Slide 19 of 38 End Show Copyright Pearson Prentice Hall
The sequence for the Prentice Hall High School Math series is intertwined and interchangeable with some elements of the Middle School courses. Students using the Prentice Hall Middle School Math Courses can choose to do Course 1 and Course 2, and then they have the option of doing Course 3 or Algebra Readiness. After that, students will then do Algebra 1, Geometry, and Algebra 2. ~ Donna
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Prentice Hall is comprised mostly of double rooms. All rooms in Prentice Hall are carpeted and air-conditioned. All beds use twin-extra long sheets. Microfridges are provided by Residence Services. The loft-able furniture present in the rooms provides a multitude of furniture arrangement options. To learn more about what the lofts are and how they can be used, visit our 'Loft Instructions' page by The first floor is accessible to students with mobility impairments. An ADA approved washer, dryers, and restroom facilities are provided.
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10-2 Cell Division Mitosis Centriole Metaphase The second phase of mitosis is metaphase. The chromosomes line up across the center of the cell. Microtubules connect the centromere of each Spindle Slide 22 of 38 chromosome to the End Show Copyright Pearson Prentice Hall poles of the spindle.
Prentice Hall Biology Chapter 3
10-2 Cell Division Cell Division In eukaryotes, cell division occurs in two major stages. The first stage, division of the cell nucleus, is called mitosis. The second stage, division of the cell cytoplasm, is called cytokinesis. Slide 3 of 38 End Show Copyright Pearson Prentice Hall | 677.169 | 1 |
Over the past few years there has been much discussion
of the quality and quantity of calculus instruction that should be required
as part of an undergraduate education in mathematics. We have heard plans
to trim down the calculus instruction to a leaner form and to add more
computer interaction, graphics, and applications to make it livelier. Others
want the calculus course used as a pump to push more students on to advanced
scientific studies, not as a filter to screen out students who don't match
the profile for an elite few preparing for a career in mathematics or physics.
Certainly it is unhelpful to blame inadequate student preparation for the
bulk of the problems in calculus courses today.
It is time to recognize the obvious: The students taking
calculus courses today have backgrounds, needs, and goals different from
those of students of 20 years ago. We need to adjust in a meaningful way
to the legitimate demands for change. If we look critically at what most
textbooks and our own lectures are saying to students, we may be a little
less likely to defend the current courses as presenting calculus as a triumph
of coherent human thought.
Who are we trying to kid? The malaise in the calculus
curriculum, which is manifest in the frustrations and distress of our students,
has a rather simple source: What we teach does not make sense as a coherent
body of knowledge. We give our students a little of this and that; here
an application to graphing, there an application to physical science. There
are very few common threads to hold together the fabric of the calculus
we now present. And even when those threads are present, we don't pay enough
attention to them for our students to notice their importance.
For a moment or two let's face some hard questions.
What percentage of the students in our calculus courses are
going on to take more mathematics?
Of those who complete their formal mathematics training with
calculus, how many will ever use the things that we emphasize in our courses?
For those who believe that calculus is or should be considered
a "liberal arts" course, what in our courses has value as a liberal art?
Of the students who proceed beyond calculus, how many would
characterize it as a superb example of the human ability to unite conceptually
an enormous collection of problems and techniques?
Whatever the audience, do the students attain a basic comprehension
of the conceptual nature of the calculus and its applications?
There may be excellent reasons for what we teach in the first-year
calculus curriculum, reasons that are both persuasive and sensible; but
they are not explicit in our courses today, nor is it clear that we are
aware of them and are treating them implicitly in our instruction. What
sense is there in placing a treatment of the differential between the algebraic
rules for differentiation and the chain rule? After spending so much time
on using calculus to help in graphing, why do we practically ignore the
calculus in discussing the conics? Why (except perhaps for some of the
trigonometric functions) do we treat the transcendental functions
after
the
fundamental theorem of calculus?
Does your text or your classroom presentation have a response
to these questions? It's not an adequate response to say the reasons lie
in some later course. For most of the students, there will not be any later
course; and calculus is very likely the last chance for mathematics to
make sense to them.
The key to solving the "calculus problem" will not be
discovered by wandering through a maze of applications and computer graphics.
Nor will it be found in a move toward "harder" problems and more rigor.
We
must try to makesense
in our calculus instruction. The topics we discuss should make sense both
internally and in context, to ourselves as instructors and to our students
as learners. This criterion will provide the knife for cutting and the
thread for reassembling the calculus curriculum of the next 40 to 50 years.
If a topic is sensibly organized by itself and sensibly placed with regard
to the other topics, then it should remain a part of the course. But if
it fails to make sense locally or globally, it needs careful reassessment
and revision.
In reviewing the calculus for sensibility, we will benefit
from agreeing on some basic principles. Both instructor and student will
find it easier to make sense of a topic if they can refer to two or three
principal themes that are developed in calculus. I suggest here three themes
that I believe are particularly significant for reviewing and revising
the calculus curriculum, namely,
differential equations,
estimation, and
mathematical models.
After some illustrations of how these themes serve as reviewing
stan dards, I will leave it for you to judge their utility in testing the
calculus syllabus for sensibility.
The first theme is that everything
in a calculus course can be related to the study of differential equations.
Consider
how this theme can affect the understanding of the derivative form of the
fundamental theorem of calculus. What was formerly treated as a very useful
tool for evaluating definite integrals can be viewed also as a theorem
about the existence (and uniqueness) of a solution to a differential equation.
Or take the treatment of the differential for estimating the value of a
function; this technique makes much more sense when followed up with a
discussion of using Euler's method to estimate particular solutions to
a differential equation with an initial condition.
The treatment of the differential also ties in well with
the second theme, that estimation is valuable for both numerical
and conceptual development.Another
example of this theme assisting in the organization of the course is in
the treatment of infinite series. Infinite series can be thought of as
the continuation of efforts to estimate function values with polynomials
(as in Taylor's Theorem) and thereby serve to estimate some difficult definite
integrals as well as solve some differential equations.
The final theme I suggest is the importance
of models as sources for concepts and interpretations as well as for applications.Although
practically every treatment of the differential presents a graphical interpretation
of this tool, none to my knowledge gives the simple and very convincing
position-velocity interpretation: that is, if you know the velocity and
position of a moving object at a certain time and you wish to estimate
the change in position for a short change in time, you should multiply
the velocity by the time change.
The modeling theme also could provide a more sensible
approach for the study of the transcendental functions. Rather than treating
them abstractly, with applications placed almost as afterthoughts, it might
make more sense to discuss first some (differential equation) models that
these functions solve. Indeed, the models often provide meaningful questions
that lead to many of the most important properties of these functions.
I think those are enough examples to give you a sense
of these themes. Now, go forth, and try them on your courses. And may your
solutions be sensible to yourselves and your students. | 677.169 | 1 |
Mathematics for Class XI is a reference book by R. D. Sharma for students of class XI following the syllabus issued by the Central Board of Secondary Education.
Summary Of The Book
Mathematics for Class XI is a reference book aimed at helping students of Class XI in their efforts to prepare for class exams as well as to provide a good foundation in the preparation for competitive exams.
The chapters in this book provide a description of concepts covered in Class XI, along with a multitude of problems for each chapter. The book has an algorithmic approach and comes with clear explanations of the theory supplemented by illustrations, making concepts easy to understand and remember. The book also contains concise summaries of concepts and formulae at the end of each chapter, making it a good book to have handy for revisions just before the exam.
For students looking to learn by practice, this book is a great source for problems. It comes with many exercises, and the problems listed are graded and come with the solved answers. The book also contains a number of exercises with unsolved problems that the student may use to further sharpen his/her skills.
Mathematics for Class XI addresses the entire syllabus for Class XI. Some topics covered in the book include trigonometric equations, functions, ratios, and graphs of trigonometric functions, sine and cosine formulae and their applications, mathematical induction, and transformation formulae.
With detailed explanations of the theory and a wealth of solved and unsolved problems, this book makes for a reliable source book for Class XI students to help in their efforts of doing well in their competitive exams.
About R. D. Sharma
R. D. Sharma is an Indian author and teacher. He is a widely published author of Mathematics textbooks and reference books.
Some other books by R. D. Sharma include Mathematics for Class XII, Mathematics for Class 12 (Set of 2 Volumes), Objective Mathematics For IIT-JEE, AIEEE and All Other Engineering Entrance Examinations, and Mathematics Class-VII.
R. D. Sharma has a Doctorate in Mathematics, and completed his Bachelor's and Master's degrees, with Honours and double gold medals, from the University of Rajasthan. He is presently with the Directorate of Technical Education, Delhi, serving as the Head of Department, Science and Humanities.
Firstly,,, i'm a bit unhappy because the the delivery time period was two-three days and 1 got it after four days. But it is not a very big problem. talking about the book,, it is one of a kind,, the most basic book for IIT basics + school exams. The price offered by flipkart is the best, as always. Thank you flipkart.
Helpful for conceptual clarity on doubts.Problem solving skills would get a boost if you work out all exercises.I would recommend each CBSE Class XI student must possess this mathematical gem.Definitely you would derive pleasure out of it.
R.D. Sharma is undoubtedly the best book for 11th grade mathematics that you could find anywhere. I started using R.D. SHARMA in class 10 and to be honest,I have been blown away ever since. If you want to build confidence in the subject of mathematics,look no further,because what you see in front of you is probably the best book available in the market. It's CBSE based,though but that's not much of an issue since ISC and CBSE Boards do not differ much in their Math syllabi,so to say. Moreover...
R.D.Sharma's Sir's books are always helpful.The way Dr.R.D.Sharma starts off with introducing the chapter and explaining the examples so clearly and the exercise followed by the examples are really useful.The exercise problems are slightly different from the examples but going through the examples would be of great use to solve the exercise ones. If we work out every exercise problems, understanding the concept becomes very easy. I am working as a lecturer in Engineering College and even toda...
Features 1) Number of examples problem based on each concepts 2) makes concepts easy to understand and have practical and application type of sums 3)Good quality of questions in exercises 4)MCQs at d end of each chapter 5)Summary contains impt formulas,definations and revalant points.
A very nice book for Boards preparation. The book was very easy to solve as I had the concepts beforehand. But if you are looking for something that gives you an edge over others, well try out some JEE books. But for boards, its enough
Firstly, the book is awesome... books written by dr r.d. sharma proves extremely useful for the students. This book contains a large number of well graded solved examples and many illustrative examples and problems are also included in it. The book has some unique features which fulfills the needs of students. the book is a fine blend of NCERT , CBSE and other competitive-examination based problems. students of different boards would also find it extremely helpful. Lastly, i would like to say...
If U Get More Than 90% In Mathematics , This Is The Book For U Man !! Otherwise Stay Away....................................................................................................................... | 677.169 | 1 |
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