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Patterns Withing Systems of Linear Equations. - Math Problem Example
Summary
Algebra is one of the fascinating fields of mathematics, because algebra allows the finding of unknown numbers from information given. In algebra, letters are used in place of numbers that are not known. These letters are then manipulated in accordance with certain rules until an answer appears…
Extract of sample Patterns Withing Systems of Linear Equations.
The usual letter for the unknown number is. A real problem can be written as: This is called an equation because there is a sign. In order to find the value of the unknown number, algebra's rules can do whatever it likes to this equation as long as it does the same to both sides of the equation. So far it has had equation with a single unknown number. What if it has two unknown numbers? In fact, an equation with two unknown has an infinite numbers of pairs of answer. To fix a single pair of number as the answer, it needs another equation. A pair of equation, each with two unknown numbers is called simultaneous equations. They can be solved together to give the values for the unknowns that satisfy both equations simultaneously. ...
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A Course in Computational Number Theory uses the computer as a tool for motivation and explanation. The book is designed for the reader to quickly access a computer and begin doing personal experiments with the patterns of the integers. It presents and explains many of the fastest algorithms for working with integers. | 677.169 | 1 |
Find an Astoria, NY MathDiscrete math includes a wide variety of mathematical topics and is utilized heavily in computer science. I have taken discrete math at the college level and am very familiar with set theory, sequences, matrices, and graph theory. Building off the foundations of C, C++ allows the programmer to create classes that model real life objects | 677.169 | 1 |
Reducing Radicals
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Students in A2/Trig come in with a basic knowledge of reducing radicals, but it is one of their weakest skills. this guides them through reducing integers, reducing with variables and introduces what to do with a negative underneath the radical. This lesson prepares students to use complete the square method and quadratic formula for solving quadratic equations (day 8,9,10). | 677.169 | 1 |
Category: Mathematics
Robert J. Sternberg
Format: Paperback
Language: 1
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Size: 7.97 MB
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Enabling teachers to learn about teaching practices in other countries and to reflect on the implications of those practices holds great promise for improving the mathematics instruction provided to all students. The College Readiness Standards define the core knowledge and skills expected of students in college entry-level math. The instructor has better control of the situation, can change the approach to meet any contingency, and can tailor each idea to suit the responses of the students.
In a guided discussion, the instructor ants as a facilitated to encourage discussion between students. Monday: Today we made "math catchers" to practice our 4 times tables. He also wants coaches to help math teachers be more effective. The teacher and students then independently create drawings of each of the problems on the worksheet. I should remark that I've never tried teaching a course using this method myself. 1) Like Felipe said, it is not good for conveying a lot of information.
We could not agree on what a problem was (although we did manage to do so in the later study). The laws should be discovered by pupils inductively and they must be further verified deductively through applications to new situations. Because of Shalandra's and LaMont's previous emotional difficulties, Mr. Aligning standards and expectations for mathematics so students enrolling in college will be prepared to enter college-level math courses, including the alignment of: 1) eleventh- and twelfth-grade math curricula with college introductory curricula and 2) the high school math and college/university math knowledge and skills tested on high school assessments, including the WASL, and the placement assessments used by 2-year colleges and baccalaureate institutions.
Willem Uttenbogaard
Format: Paperback
Language: 1
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The Hotspot software implementations are based on Sun's freeware fdlibm library. Differential calculus of functions of one variable, with applications. In particular, when teachers are asked to compare software of different type and to discuss what kinds of knowledge they engender (building on Squires & Preece, 1996; Squires, 1997) and at what points in the curriculum they should be used, we have found that they develop more pedagogically rich conceptions of and positive attitudes toward technology (Kurz, 2004).
They say new, convoluted approaches are turning kids off of math. Fortunately, my third-grade math teacher, Mr. As a result teachers often do not take advantage of students "prefractional knowledge" and their informal knowledge about fractions thus denying children a spontaneous "in" to their formal study of fractions. I've listed some graphing software here. I have written several articles about quality of life and measures of quality of life. Consider an example involving the fact that 3/5 of the students in a class of 25 are female.
A., (Ed.), "Handbook of research on mathematics teaching and learning." (pp. 3-38) NY: Macmillan. While the education world is all abuzz about so-called 21st century skills like collaboration, problem solving and critical thinking, this research suggests that we might do well to add a strong dose of the 19th century to our children's schooling. (MORE: Paul: Why Third Grade Is So Important: The 'Matthew Effect' ) Kail's experience is instructive. In the past problem solving had a place in the mathematics classroom, but it was usually used in a token way as a starting point to obtain a single correct answer, usually by following a single 'correct' procedure.
This book has myriad suggestions and ideas that will be useful for administrators, mathematics coaches, middle and elementary school teachers, and substitutes. Perhaps the best news is that shuffling problems is cost-effective, and wouldn't necessitate rewriting the whole textbook. The study focuses on seventh grade, the culminating year for teaching those topics. The same teaching methods are designed to help students who excel remain challenged, as well as to help students who are lagging behind catch up, because they incorporate a wide range of instructional strategies.
Some word problems give me quite a laugh. Anchored instruction is an environment as we define it; however, it would be remiss to assume that all environments are developed under the model of anchored instruction.) The Jasper Project consists of a series of computer-based videos that present mathematical questions based on scenarios utilizing real-world examples. The instructor quickly judges the number of responses, then after a quick discussion (and certainty that the concept has been understood by most students), they proceed with the lesson. • At the end of a class, the instructor asks students to write down their muddiest point from class on a scrap of paper, and collects the responses. | 677.169 | 1 |
Category: Mathematics
Author Rebecca Wingard-Nelson tackles the fundamentals of addition and subtraction. examine tricks and tips for 3 universal sorts of attempt questions: a number of selection, short-answer, and show-your-work. a great evaluate of issues had to increase try out rankings. unfastened worksheets can be found enslow.com!
Marvin will get off paintings at random occasions among three and five p.m. His mom lives uptown, his female friend downtown. he is taking the 1st subway that is available in both course and eats dinner with the only he's dropped at. His mom complains that he by no means involves see her, yet he says she has a 50-50 probability. He has had dinner together with her two times within the final 20 operating days. Explain.
Marvin's adventures in chance are one of many fifty exciting puzzles that illustrate either hassle-free advert complicated points of chance, every one challenge designed to problem the mathematically prone. From "The Flippant Juror" and "The Prisoner's challenge" to "The Cliffhanger" and "The Clumsy Chemist," they supply an excellent complement for all who benefit from the stimulating enjoyable of mathematics.
Professor Frederick Mosteller, who teaches information at Harvard college, has selected the issues for originality, common curiosity, or simply because they exhibit worthy ideas. additionally, the issues are graded as to hassle and plenty of have massive stature. certainly, one has "enlivened the learn lives of many glorious mathematicians." distinct options are integrated. there's each likelihood you have to not less than some of them.
This best-selling textbook for a moment path in linear algebra is aimed toward undergrad math majors and graduate scholars. the radical procedure taken right here banishes determinants to the top of the ebook. The textual content specializes in the principal aim of linear algebra: realizing the constitution of linear operators on finite-dimensional vector areas. the writer has taken strange care to encourage thoughts and to simplify proofs. numerous fascinating workouts in each one bankruptcy is helping scholars comprehend and control the gadgets of linear algebra.
The 3rd variation includes significant advancements and revisions during the e-book. greater than three hundred new workouts were extra because the earlier version. Many new examples were extra to demonstrate the most important rules of linear algebra. New themes coated within the publication contain product areas, quotient areas, and twin areas. attractive new formatting creates pages with an strangely friendly visual appeal in either print and digital versions.
No must haves are assumed except the standard call for for compatible mathematical adulthood. therefore the textual content begins by means of discussing vector areas, linear independence, span, foundation, and measurement. The booklet then offers with linear maps, eigenvalues, and eigenvectors. Inner-product areas are brought, resulting in the finite-dimensional spectral theorem and its outcomes. Generalized eigenvectors are then used to supply perception into the constitution of a linear operator.
World-famous mathematician John H. Conway established this vintage textual content on a 1966 direction he taught at Cambridge college. aimed at graduate scholars of arithmetic, it's going to additionally turn out a worthy consultant to researchers mathematicians.
His themes hide Moore's thought of experiments, Kleene's concept of standard occasions and expressions, Kleene algebras, the differential calculus of occasions, components and the issue matrix, and the idea of operators. extra matters contain occasion sessions and operator periods, a few regulator algebras, context-free languages, communicative ordinary algebra, axiomatic questions, the energy of classical axioms, and logical difficulties. entire strategies to difficulties seem on the finish.
"José Ferreirós has written a magisterial account of the heritage of set conception that's panoramic, balanced, and interesting. not just does this ebook synthesize a lot earlier paintings and supply clean insights and issues of view, however it additionally contains a significant innovation, a full-fledged remedy of the emergence of the set-theoretic strategy in arithmetic from the early 19th century." --Bulletin of Symbolic common sense (Review of first edition)
This textbook for complex classes in workforce concept makes a speciality of finite teams, with emphasis at the thought of team activities. Early chapters summarize presupposed evidence, establish vital subject matters, and determine the notation used through the ebook. next chapters discover the conventional and arithmetical buildings of teams in addition to purposes. Topics comprise the traditional constitution of teams: subgroups; homomorphisms and quotients; sequence; direct items and the constitution of finitely generated Abelian teams; and team motion on teams. extra topics diversity from the arithmetical constitution of teams to classical notions of move and splitting through crew motion arguments. greater than 675 workouts, many followed by means of tricks, illustrate and expand the material.
In regards to the book:
With this booklet, young children can liberate the mysteries of maths and notice the sweetness of numbers. Readers will become aware of great info, equivalent to why 0 is so helpful; what a googol rather is; why song, maths and area are attached; why bees favor hexagons; tips on how to inform the time on different planets; and lots more and plenty even more. From marvellous measurements and startling shapes, to excellent theories and numbers in nature - maths hasn't ever been as impressive as this!
Reviews:
''A tremendous e-book to dip into and also you are assured plenty of enjoyable evidence to entertain your friends' — mom and dad in contact
Lyons' tough direction research has supplied new insights within the research of stochastic differential equations and stochastic partial differential equations, akin to the KPZ equation. This textbook offers the 1st thorough and simply obtainable advent to tough course analysis.
When utilized to stochastic platforms, tough direction research presents a way to build a pathwise resolution idea which, in lots of respects, behaves very similar to the speculation of deterministic differential equations and gives a fresh holiday among analytical and probabilistic arguments. It presents a toolbox permitting to get well many classical effects with no utilizing particular probabilistic houses similar to predictability or the martingale estate. The examine of stochastic PDEs has lately ended in an important extension – the speculation of regularity constructions – and the final elements of this booklet are dedicated to a gradual introduction.
Most of this path is written as an primarily self-contained textbook, with an emphasis on rules and brief arguments, instead of pushing for the most powerful attainable statements. a customary reader may have been uncovered to higher undergraduate research classes and has a few curiosity in stochastic research. For a wide a part of the textual content, little greater than Itô integration opposed to Brownian movement is needed as background.
Antifragile is a standalone e-book in Nassim Nicholas Taleb's landmark Incerto sequence, an research of opacity, good fortune, uncertainty, chance, human errors, possibility, and decision-making in a global we don't comprehend. the opposite books within the sequence are Fooled by means of Randomness, The Black Swan, and The mattress of Procrustes.
Nassim Nicholas Taleb, the bestselling writer of The Black Swan and one of many leading thinkers of our time, unearths find out how to thrive in an doubtful world.
simply as human bones get better whilst subjected to emphasize and stress, and rumors or riots accentuate whilst somebody attempts to repress them, many stuff in existence take advantage of tension, illness, volatility, and turmoil. What Taleb has pointed out and calls "antifragile" is that type of items that not just achieve from chaos yet desire it as a way to live to tell the tale and flourish.
In The Black Swan, Taleb confirmed us that hugely inconceivable and unpredictable occasions underlie nearly every little thing approximately our international. In Antifragile, Taleb stands uncertainty on its head, making it fascinating, even useful, and proposes that issues be inbuilt an antifragile demeanour. The antifragile is past the resilient or powerful. The resilient resists shocks and remains an identical; the antifragile will get greater and higher.
in addition, the antifragile is proof against prediction blunders and guarded from hostile occasions. Why is the city-state larger than the countryside, why is debt undesirable for you, and why is what we name "efficient" now not effective in any respect? Why do govt responses and social rules guard the robust and damage the vulnerable? Why in the event you write your resignation letter prior to even beginning at the task? How did the sinking of the Titanic store lives? The ebook spans innovation by means of trial and mistake, existence judgements, politics, city making plans, struggle, own finance, fiscal structures, and drugs. And all through, as well as the road knowledge of fats Tony of Brooklyn, the voices and recipes of historical knowledge, from Roman, Greek, Semitic, and medieval assets, are loud and clear.
Antifragile is a blueprint for dwelling in a Black Swan world.
Erudite, witty, and iconoclastic, Taleb's message is innovative: The antifragile, and basically the antifragile, will make it.
"Antifragility isn't simply sound financial and political doctrine. It's additionally the most important to an excellent life."—Fortune "At as soon as thought-provoking and brilliant."—Los Angeles Times
The prior few years have witnessed a considerable development within the variety of functions for optimization algorithms in fixing difficulties within the box of physics. Examples contain selecting the constitution of molecules, estimating the parameters of interacting galaxies, the floor states of digital quantum structures, the habit of disordered magnetic fabrics, and section transitions in combinatorial optimization problems. This booklet serves as an creation to the sector, whereas additionally featuring an entire assessment of contemporary algorithms. The authors start with the suitable foundations from computing device technology, graph idea and statistical physics, ahead of relocating directly to completely clarify algorithms - subsidized by means of illustrative examples. They contain pertinent mathematical ameliorations, which in flip are used to make the actual difficulties tractable with tools from combinatorial optimization. all through, a couple of attention-grabbing effects are proven for all actual examples. the ultimate bankruptcy offers various sensible tricks on software program improvement, trying out courses, and comparing the result of laptop experiments. | 677.169 | 1 |
Mathematics GCSE
The course
Mathematics can best be described, at this level, as the formulation and application of both numerical and abstract concepts. The student of mathematics has to be creative and imaginative, have a sense of flair and most importantly be curious about the world around them. Mathematics touches on many areas and requires intellectual flexibility and a methodical approach.
Mathematics has a clear influence on accountancy, architecture, biology, chemistry, economics, engineering, finance, physics, etc. At the same time mathematics is an active field of study in its own right.
The course is divided into three components or assessment objectives:
Number and algebra (39-51% of total). This looks at fractions and decimals, ratio, percentages, algebraic manipulation, formulae, linear and quadratic equations and graphs, proportionality, functions, calculus and set theory.
Shape, space and measures (34-46% of total). This looks at bearings and constructions, trigonometry in 2D and 3D, area and volume, similarity and vector geometry.
Handling data (12-18% of total). This looks at measures of average, probability and histograms.
What skills do I need?
You need to be good with numbers and abstract reasoning.
How is the course assessed?
All students pursue either the Higher Tier GCSE course, which covers the grade range 4-9 (Grade 3 is allowed) or the Foundation Tier, which covers the grade range 1-5. The course is assessed by means of three written exam papers, equally weighted. Each paper lasts one hour and thirty minutes and is worth 33.3% of the total grade. Each paper has a total of 80 marks. Assessments take place twice a year: in May/June and in November. | 677.169 | 1 |
Saxon Algebra 1: Student Edition 2009
Saxon's Algebra 1 covers all the topics in a first-year algebra course and builds the algebraic foundation essential for all students to solve increasingly complex problems. Higher order thinking skills use real-world applications, reasoning and justification to make connections to math strands. Algebra 1 focuses on algebraic thinking and multiple representations - verbal, numeric, symbolic, and graphical. Graphing calculator labs model mathematical situations. Great for homeschool, private, and public school courses! | 677.169 | 1 |
Linear Systems
Henri Bourlès, CNAM, France
ISBN: 9781848211629
Publication Date: June 2010 Hardback 592 pp.
165.00 USD
eBooks
Description
Linear systems have all the necessary elements (modeling, identification, analysis and control), from an educational point of view, to help us understand the discipline of automation and apply it efficiently. This book is progressive and organized in such a way that different levels of readership are possible. It is addressed both to beginners and those with a good understanding of automation wishing to enhance their knowledge on the subject. The theory is rigorously developed and illustrated by numerous examples which can be reproduced with the help of appropriate computation software. 60 exercises and their solutions are included to enable the readers to test and enhance their knowledge.
Henri Bourlès is a professor and holds the industrial automation chair at the Conservatoire national des arts et métiers. He has been teaching automation for over 20 years in engineering and graduate schools. | 677.169 | 1 |
Core Plus Mathematics, Course 1, Student Edition (ELC: CORE PLUS)
Carefully designed to the Common Core State Standards and Standards for Mathematical Practices, Core-Plus Mathematics: Contemporary Mathematics in Context is the newest revision to Core-Plus Mathematics Program's (CPMP) four-year integrated mathematics program originally funded by the National Science Foundation. Featuring problem-based, inquiry-oriented and technology-rich applications, Core-Plus Mathematics promotes student-centered active learning, teamwork and communication to prepare them for success in college, in careers and in daily life. This new edition features content focused on algebra and functions, statistics and probability, geometry and trigonometry, and discrete mathematics in each course with integrated use of CPMP-Tools software and graphing calculators in each course complemented by newly updated Course 1-4 texts and interactive digital content.
Includes print student edition Hill Education. Hardcover. Book Condition: New. 0076657949 MULTIPLE COPIES AVAILABLE - New Condition - Never Used - DOES NOT INCLUDE ANY CDs OR ACCESS CODES IF APPLICABLE. Bookseller Inventory # Z0076657949ZN | 677.169 | 1 |
Maths coursework statistics help
A contemporary math class is designed to survey some of the important ideas and practical applications in mathematics.Gcse Maths Statistics Coursework Help StratifiedGCSEGCSE Revision. use special symbols to represent numbers in statistics:Gcse statistics coursework help.Provides on demand homework help and tutoring services that connect students to a professional tutor online in math, science, social studies or English.Online Tutoring and Homework Help for Elementary Statistics and other Statistics subjects.ALISON provides free online math courses to help you gain more knowledge in algebra, geometry and statistics.Your math genius will help you through various stages of your assignment questions and guide you.
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Statistics Maths GCSE Coursework
Mathematics is a rich discipline with powerful abstract ideas,. joint with the Department of Statistics,.What We Do The Mathematics and Statistics Program offers support for primarily lower division classes in math and stat.
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Any non-profit or world government with intentions to make the world a.About my best friend essay Literary Help With Statistics Coursework analysis essay buy Get paid.Weekly Tips CPM has created weekly tips for teachers, parents and students, written to help everyone be successful in math.Mathematics courses on the right side of this page are only applicable for students beginning high school coursework prior to school year 2012-13.
Mayfield High School Coursework Maths Statistics
Online coursework help offering college, high school, university, A level, A2, business studies coursework.The Relationship between the Average Amount of Sleep a High school Teenager gets per School Night and the.
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Education resources, designed specifically with parents in mind.As you know, GCSE and A-level subjects require students to complete exams to get a grade.
Anonymous Dr Perlow was very thorough in explaining everything.At Boston University undergraduates can pursue studies in statistics, through degree programs in the Department of Mathematics and Statistics, in a number of ways.Getting help for your GCSE math statistics coursework is easier since you have more options to consider.Expertsmind.com offers free online statistics assignment help, statistics homework help, statistics project assistance.Cool Math has free online cool math lessons, cool math games and fun math activities. | 677.169 | 1 |
Math: Most important for success in college math is a thorough understanding of the basic concepts, principles, and techniques of algebra. This is different than simply having been exposed to these ideas. Much of the subsequent mathematics they will encounter draw upon or utilize these principles. In addition, having learned these elements of mathematical thinking at a deep level, they understand what it means to understand mathematical concepts deeply and are more likely to do so in subsequent areas of mathematical study. College-ready students possess more than a formulaic understanding of mathematics. They have the ability to apply conceptual understandings in order to extract a problem from a context, use mathematics to solve the problem, and then interpret the solution back into the context. They know when and how to estimate to determine the reasonableness of answers and can use a calculator appropriately as a tool, not a crutch.
The Transition Math Project has developed a student and parent portal with up-to-date math planning resources, tips and tools that will help you talk to your child about the importance of math in being college- and work-ready.
Here is an article describing how parents can help with math homework when the answers aren't in the book. | 677.169 | 1 |
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Unformatted text preview: Chapter 3: Linear algebra Problems in physics often lead to a set of linear equations. In solving linear equations is often convenient to use matrices and vectors. Matrices and vectors also occur frequently in the representation of states and linear operators in quantum mechanics. Determining the quantum states of a system can be reduced to solving an eigenvalue equation. Another example is coordinate transformations, which occurs in, for example, relativity and group theory, which is essential in particle physics but also crystallography amongst other areas. The vibrations of molecules and crystals can also be understood by solving large sets of linear equations. It's hard to overemphasize the importance of this subject! Patrick K. Schelling Introduction to Theoretical Methods Goals: By the end of this chapter you should be able to: I Represent a set of linear equations with matrices Patrick K. Schelling Introduction to Theoretical Methods Goals: By the end of this chapter you should be able to: I Represent a set of linear equations with matrices I Use elementary row reduction to solve a matrix equation I "Diagonalize" a matrix Patrick K. Schelling Introduction to Theoretical Methods Goals:...
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This note was uploaded on 07/30/2011 for the course PHZ 3113 taught by Professor Staff during the Spring '03 term at University of Central Florida. | 677.169 | 1 |
This challenging problem book by renowned US Olympiad coaches, mathematics teachers, and researchers develops a multitude of problem-solving skills needed to excel in mathematical contests and in mathematical research in number theory. Offering inspiration and intellectual delight, the problems throughout the book encourage students to express their ideas in writing to explain how they conceive problems, what conjectures they make, and what conclusions they reach. Applying specific techniques and strategies, readers will acquire a solid understanding of the fundamental concepts and ideas of number theory | 677.169 | 1 |
Overview
The Math Emporium is a new adaptive learning model that takes math out of the traditional
classroom and into a high-tech learning environment. The emporium uses ALEKS, a web-based assessment tool, to gauge students' math knowledge and identify gaps.
ALEKS then builds an individualized plan of study with weekly targets to help students
master concepts outlined in the course objectives. Students are continuously evaluated
throughout this process and can work at their own pace.
Math Emporium courses are offered through the College Preparatory & Developmental
Studies Department. The emporium lab where courses are taught is located in the Learning
Commons in Sally Monserud Hall, Room 116F. While there, students can also take advantage
of other tutoring and support services offered by the center such as math testing
and the Math Lab. | 677.169 | 1 |
This book can help overcome the widely observed math-phobia and math-aversion among undergraduate students in these subjects. The book can also help them understand why they have to learn different mathematical techniques, how they can be applied, and how they will equip the students in their further studies.
The book provides a thorough but lucid exposition of most of the mathematical techniques applied in the fields of economics, business and finance. The book deals with topics right from high school mathematics to relatively advanced areas of integral calculus covering in the middle the topics of linear algebra; differential calculus; classical optimization; linear and nonlinear programming; and game theory.
Though the book directly caters to the needs of undergraduate students in economics, business and finance, graduate students in these subjects will also definitely find the book an invaluable tool as a supplementary reading.
Contents of the Book
Preface
Acknowledgements
1. Review of Basics
2. Linear Algebra
3. Differential Calculus
4. Classical Optimization
5. Linear Programming
6. Nonlinear Programming
7. Game Theory
8. Integral Calculus
References Index
Web Supplements
Supplements to Chapters 1, 2, 3, and 8
9. Difference Equations
10. Differential Equations
11. Elements of Mathematica
12. Graphics in Mathematica
Mathematica Applications in Chapters 1 - 8
References for Web Supplements
Author Biography
Dr E. K. Ummer graduated with a Ph.D. degree from University of London. He is currently an Assistant Professor at the Department of Economics of E.M.E.A. College of Arts and Science. Prior to joining E.M.E.A. College, he was with the faculty of Economics and Management Studies at International Islamic University, Malaysia. His current research interests are quantitative methods, economic theory, and economics of development and growth. The author can be contacted at [email protected] or at [email protected]. | 677.169 | 1 |
Graphical Symmetry
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Students investigate symmetry to the x-axis, y-axis and origin by looking at graphs, tables of values and observing the ordered pairs of reflected points. They also create graphs of their own. Then students investigate odd and even functions.
This is intended as a stand-alone lesson in earlier courses or as the prelude to testing algebraically for symmetry and odd/even functions in more advanced courses. | 677.169 | 1 |
Functions Bundle (PowerPoint Lessons & Printables!)
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2 MB|56 pages
Product Description
"The mission of Common Core Math Solutions is to create engaging, highly effective resources for the math classroom"
Bundle Objectives:
1) Understand the characteristics of functions.
2) Determine whether a relation is a function from a graph.
3) Determine whether a relation is a function from a list of input and output values.
4) Use simple function notation to write linear equations.
5) Use the vertical line test to determine if graphs are functions
The PowerPoint presentations (56 slides!) may be used for classroom instruction and the slides are designed to be used as printables. | 677.169 | 1 |
GLENCOE GEOMETRY 7 4 ANSWERS
Glencoe geometry 7 4 answers
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Practice B 9-1 Probability LESSON Read Online Download PDF - Holt Algebra 1 Holt Algebra 1 Homework and Practice Workbook. I think we have about 125 more to go. Meet up with tutors anytime, anywhere through our whiteboard online. Choose the grade that you need. Can someone please assist me to understand how to come up with a comprehensive answer and explanation regarding answers to middle school math with pizzazz book d specifically in topic of function composition. Scientific calculators have other function keys not found on a standard calculator. We do not encourage or condone the use of this program if it is in violation of these laws. I still use both Yahoo. USER COMMENTS 108 comment s There are lots of people who have to deal with this daily. Check out the separate bank of activities for the Math Practice Standards and the bank of Graphic Organizers. They're likely to respond quickly, but they may not all be available for a lesson right now. 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ROOTS OF THE EQUATION CALCULATOR
Roots of the equation calculator
Roots of the equation calculator Practice
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(Interactive Math Program is a four year high school math curriculum).
III. Extra Credit
Compare how much of the symbolic modeling is spoon-fed to the students in the Interactive Math Program vs. the traditional math problem sets. Then discuss what this means for preparing students for 21st century jobs | 677.169 | 1 |
This text aims to give undergraduate students a grasp of the basic concepts of round-off errors, stability, conditioning, and accuracy as well as an appreciation of the core numerical linear algebra algorithms, their basic properties and implementations.
MATLAB is a high-level language and. | 677.169 | 1 |
Content on this page requires a newer version of Adobe Flash Player.
Welcome to Department of Mathematics & Statistics
Introduction
The Department of Mathematics began as small section in 1993 comprising of single faculty member with B.Sc(MPC) group with 3 students. Subsequently as the college progressively took steady strides to grow in size and nature, the mathematics department also gradually grew in size. Statistics was added in the year 2000-2001 with combination MSCs. The department of Mathematics & Statistics is around 400.
The department has a track record of 95% and above results at the university exams every year. All the programs are focused on the spectrum where brilliants are given scope and the average students are encouraged. Persistent efforts are being made for the placements of our students. Compare to the university results our results are proved to be better. The students results and attendance are the strengths of the department. The various activities of the department are as follows:
Extension activities:: Department has focused on the activities like Paper Presentations, Seminars, Quiz, Field trips, Organizing Guest Lectures by Professionals from outside to blend the strengths of the faculty, is a regular feature of the Department
Projects:The Department of Statistics supervises Students Projects on statistical analysis and qualitative analysis.
The Department of Mathematics allots Students Projects on Mathematical Applications in Different Fields which are guided by the Faculty
Club activities:: The Department organizes its own club 'Sigma' which creates a platform for the students to showcase their talents and organizes intercollegiate activities
Placements:Majority of the students with Mathematics as optional subject, register for placements. The Department co-ordinates with placement officer and organize pre-placement guidance to the students
Research & Projects
Blog
The Department of Mathematics and Statistics created its own blog. It is useful to provide information like revised syllabus, question banks, updated practical booklets, and recent notifications for various P.G courses from various universities. Our blog address:
Team
Head of Dept.(Mathematics)
Mrs. K. Uma Maheshwari
M.Sc., B.Ed., MCA
Other Mathematics Faculty
Mrs. K. Sri Devi
M.Sc.(Mathematics)
Mrs. K. Jyothi
M.Sc (Mathematics), B. Ed.
Mrs. R. Kalyani
M.Sc (Mathematics)
Mrs. R. Mamatha
M.Sc (Mathematics), B. Ed.
Head of Dept. (Statistics)
Mrs. M. Pauline
M.Sc.(Statistics)
Other Statistics Faculty
Miss. Srivani
M.Sc.
Miss. Anitha
M.Sc.(Statistics), B. Ed.
Courses Offered
Department of Mathematics & Statistics is involved in the following courses of the college
B.Sc. (MPC - Maths, Physics & Chemistry)
B.Sc. (MPCs - Maths, Physics & Computer Science)
B.Sc. (MPCs - Maths, Statistics & Computer Science)
Dept. Email
statsmaths50[at]gmail.com
mathsdept93[at] | 677.169 | 1 |
Courses
MATH010 Intermediate Algebra
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From the university's Academic Year 2016 - 2017 Course Description:
3 credit hours
Review skills necessary for pre-calculus and college mathematics and statistics. Topics include a review of sets, operations with polynomial and rational expressions, solving equations and inequalities, and an introduction to the coordinate plane and functions.
Prerequisites: Students must achieve an acceptable score on the Math Placment Exam in accordance with current standards determined by the Department of Mathematical Sciences.
About this section: Requires arithmetic and Algebra I skills. MATH010 does not earn credit towards a degree.
The various classes of functions and their graphs are explored. Function classes include linear, quadratic, polynomial, rational, exponential, logarithmic and trigonometric. Skills and concepts needed for calculus (MATH221) are emphasized.
Prerequisites: MATH010. Students must achieve an acceptable score on the Math Placment Exam in accordance with current standards determined by the Department of Mathematical Sciences.
About this section: Only four credits from any combination of MATH113, MATH114, MATH115, MATH117, MATH127, MATH170 and MATH171 can count toward graduation. | 677.169 | 1 |
About this eBook
PREFACE Year after year CBSE has been introducing changes in the curriculum of various classes. We, at Oswaal Books, closely follow every change made by the Board and endeavor to equip students with the latest study material to prepare for the Examinations. The latest offering from us are these Question Banks. These will provide comprehensive practice material for every chapter. These are prepared by experienced teachers who have translated their expertise into making important questions from every chapter in order to facilitate wholesome learning of every concept. Highlights of our Question banks Question Bank strictly as per the NCERT Curriculum Variety of Questions from NCERT Textbooks A synopsis of the important points from every chapter Value Based Questions as specified by CBSE Board Answers follow the marking scheme and the prescribed word limit We feel extremely happy to offer our Question Banks and hope that with them, every student will discover a more thorough way of preparing and thereby excelling in their examinations. Though we have taken enough care to ensure our products to be error free, yet we welcome any feedback or suggestions that come our way for improvisation. We wish you good luck for the forthcoming academic year Publisher Course Structure Mathematics Class VII Number System 50 hrs i Knowing our Numbers Integers Multiplication and division of integers through patterns . Division by zero is meaningless Properties of integers including identities for addition multiplication, commutative, associative, distributive through patterns . These would include examples from whole numbers as well. Involve expressing commutative and associative properties in a general form. Construction of counter examples, including some by children. Counter examples like subtraction is not commutative. Word problems including integers all operations ii Fractions and rational numbers Multiplication of fractions Fraction as an operator Reciprocal of a fraction Division of fractions Word problems involving mixed fractions Introduction to rational numbers with representation on number line Operations on rational numbers all operations Representation of rational number as a decimal. Word problems on rational numbers all operations Multiplication and division of decimal fractions Conversion of units length mass Word problems including all operations iii Powers | 677.169 | 1 |
Putting Essential Understanding of Functions into Practice
Robert N Ronau Dan Meyer & Terry Crites
Do your students believe that the vertical line test is foolproof for determining whether a relationship is a function, or that every function can be modelled by an equation? How can you determine what they know or don't know to move them forward in their thinking?
This book details and explores best practices for teaching the essential ideas that students must grasp about functions, a topic that is challenging to learn and teach but which is critical to the development of mathematical understanding. The content covered includes the concept of a function, covariance, combining and transforming functions and graphs as representations of functions. The authors use classroom vignettes, and provide specific tasks and samples of student work to bring the topic to life. Easy to read, with clear and interesting explanations, this book will appeal to both experienced and novice practitioners. | 677.169 | 1 |
Cambridge Checkpoint Maths
This bestselling series has been updated to ensure teachers can deliver the revised Cambridge Secondary 1 programme for mathematics with confidence. This brand-new Workbook for stage 9 provides extra practice questions. There is a parallel exercise for each exercise in the Student's Book. Students can write their answers in the Workbook, which can be used for homework or in the classroom. | 677.169 | 1 |
Mathematics Courses
Math Intervention This course is designed for senior students who struggle This course is designed for the student who struggles (AIMS) This course is designed to reinforce the math concepts taught in Algebra I & II and Geometry I & II in preparation for the math section of the AIMS test. This course does not meet the Mathematics requirement for graduation.
Pre-Algebra A/B (Targets: 9th Graders) This course covers fundamental math concepts and is designed to provide students an opportunity to increase their math aptitude prior to entering an Algebra course. Course topics include whole numbers and operations, decimals and fractions, pre-algebra concepts, and basic geometry concepts. This course does not meet the Mathematics requirement for graduation. This course is divided between two blocks for 1.0 credit. Algebra I A/B This course will focus on computing fractions and integers, solving for variables, solving and graphing linear equations, and solving systems of equations. Upon mastery, the course will cover factoring and distribution, probability theory, exponents and radicals, and quadratic equations. This course is divided between two blocks for 1.0 Algebra credit. Algebra II A/B Prerequisite: Geometry A/B This course will focus on the abstract view of algebra. Course topics include matrices, polynomials, systems of equations, complex numbers, trigonomic functions, and coordinate geometry. This course is divided between two blocks for 1.0 Algebra credit.
Geometry A/B Prerequisite: Algebra I A/B This course will focus on the strategies of inductive and deductive reasoning to find measures of angles, segments of polygons and circles, and areas and volumes of solids. Parallel lines and geometric probability are also covered. Coordinate geometry and constructions are thematic throughout the course. This course is divided between two blocks for 1.0 Geometry credit.
Senior Math A/B Prerequisites: Algebra I, Algebra II, Geometry I, & Geometry II This course prepares students for Pre-Calculus and College Algebra. Students enrolled in this course will review algebraic functions, lines on the plane, various types of equations, inequalities, polynomials, and rational expressions. This course also includes various math concepts and statistical analysis.
Math 122 Intermediate Algebra (Dual Enrollment Course) Prerequisite: Algebra IA/B, Algebra II A/B, Geometry A/B, and meet the enrollment process for Pima Community College **this course does qualify towards Algebra II credit but students must still take either pre-calculus or senior math for graduation.** In Intermediate Algebra students will work with lines in a plane, basic algebraic functions, solve systems of linear equations, linear inequalities, work with polynomials and solve polynomial equations, work with rational expressions and equations, and radical expressions and equations. Students will solve quadratic equations using a variety of methods and work with various types of exponents as well as logarithms. | 677.169 | 1 |
Good Topics for Mathematics Research Papers. A mathematics research paper is an extremely intricate task that requires immense concentration, planning and.
Choosing a good topic for a math research paper will have a big impact on your teacher or professor. The best way to choose a good topic is to either think out of.
If you are having difficulties choosing topics for a research project in mathematics , then you'd better look over the prompts provided in our post.
Steps to effective math homework. Essay sample about Texas history. Minimal lines of modern structures like banister-free staircases and infinity pools are illusions of modern architecture. Where to get professional help. Math is immensely present everywhere we go. Every year the amount of students who hire best thesis writers online increases because they have no where else to turn when they are struggling to complete their term papers. Understanding the relation between discrete and continuous quantum graphs is a challenging task leaving a lot of freedom, since this area has not been studied systematically yet.
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Math good english research paper topics
One of these deloopings, due to Dwyer-Hess, is homotopy-theoretic in nature and is given in terms of mapping spaces between operads. Foundations of probability theory. I am so grateful! Write about math culture. The topics that deal with the fundamental theorems are the following:. Let a friend or two read your paper and get their opinion about it. | 677.169 | 1 |
Algebra 2A
Unit 7 Study Guide: Quadratic Functions and Equations
Completion of all the review activities in the review lesson (Unit 7 Lesson 6), checking and going
back over any quiz questions that you missed, and completing this study guide will help you
Name
Date
Systems of Equations and Inequalities
Unit Portfolio
ALGEBRA 1 A
Directions: Complete each of the tasks outlined below.
Task 1
You are starting a new business in which you have decided to sell two products
instead of just one. Determine a busine
Name
Date
Linear Functions Unit Portfolio
Directions: Complete each of the tasks outlined below.
Cab companies often charge a flat fee for picking someone up and then charge an
additional fee per mile driven. Pick a U.S. city and research the rates of two
English 12A
Unit 2: Portfolio Allegory Story
Goals: In this task, you are to demonstrate your ability to
Compose an introduction that grabs the readers attention and introduces the conflict, setting, and
main characters of the story.
Compose body paragr
Descriptive Writing
Prewri
ting
Chapter 5
Prewriting is often the first thinking that a writer does, but it can continue throughout
the other stages of the writing process as well. The fact that you have made a plan and
begun to draft does not mean that y
MATH Algebra Advice
Showing 1 to 1 of 1
Ms. Novak was an amazing teacher and always ready and happy to help you. Whenever I had an issue she was available and had many times for tutoring and extra help.
Course highlights:
The highlights of this course would include algebraic systems, binomials, basic algebra (of course), and diverse amount of equations. Throughout the course I learned all of the necessary material and the formulas needed to use in a wide variety of algebra.
Hours per week:
6-8 hours
Advice for students:
To be able to succeed in this course I highly recommend to attend her classes and listen to what she teaches. Through her teaching she easily explains how to perform formulas and equations. Make sure to pay attention to the basics and the small details, because those are what truly make up the rest of the material later on. | 677.169 | 1 |
Mathematics
What is Mathematics and why is it important? The answer is simple as laws of Mathematics govern the way the universe works. From patterns in growth of plants to the orbit of planets around the Sun, mathematics allows us to model and predict behavior. The study of mathematics widens career opportunities for students in sectors you may never have even considered, including specialised fields such as law or medicine. That said, a large number of Maths careers are based within business, science or technology related sectors, with math graduates occupying roles such as engineer, accountant, actuary, statistician, technician, economist or market researcher.
Mathematics is taught as a core subject to all pupils from Year 7 to Year 11, then as a popular optional subject at A level. Pupils learn in a supportive environment, and technology is used throughout the department to enhance the learning experience. Individual programmes are implemented where necessary to help every pupil thrive, while the more able are stretched with a challenging curriculum. A variety of teaching styles are used, and all pupils are encouraged to develop good problem-solving and critical thinking skills. Extensive resources are available to support learning on the school Virtual Learning Environment, however this is not considered to be a replacement for individual help.
Key Stages 3 and 4 (IGCSE)
Pupils are taught Mathematics in ability sets from Year 7 to Year 11. The Year 7 sets are based on pupils' entrance tests and any results from the Junior school. The department continuously reviews individual progress and, where appropriate, movement between sets takes place. The syllabus for Years 7-9 is based on the National Curriculum, and uses the Collins New Maths Frameworking textbooks. In Years 10 and 11 all pupils follow the Pearson Edexcel IGCSE Mathematics Specification A course.
Key Stage 5 (A level)
Mathematics is a very popular A-level subject. The course followed is the modular Edexcel. A variety of modules are offered including pure mathematics, mechanics, statistics, and decision mathematics.
Head of Department: Gary Pembertongpemberton@kentcollege.ae
A relative latecomer to teaching I achieved a First Class Honours Degree in Mathematics Education from Sheffield Hallam University in the United Kingdom. I taught in two schools in South Yorkshire, holding posts such as Assistant Head of Year, Second in Mathematics and Pastoral Support Manager before making the decision to move abroad to continue my teaching career and indulge my passion for travel. After 6 years as Head of Mathematics and subsequently Deputy Headteacher at an international school in Qatar, I moved to Oman to gain knowledge and experience in working with different Mathematics curricula. Prior to my appointment at Kent College I have been Head of Mathematics at the longest established International Baccalaureate School in Dubai. I look forward to the exciting opportunities ahead to establish a Mathematics team that is recognized as a beacon of excellence within the region. | 677.169 | 1 |
A School Algebra
Middle School Math and Pre-Algebra: .
High School Algebra Curriculum. Below are the skills needed, with links to resources to help with that skill. We also enourage plenty of exercises and book work.
Algebra tutorials, lessons, calculators, games, .
An annotated list of websites offering algebra tutorials, lessons, calculators, games, word problems and books
Pearson Prentice Hall and our other respected imprints provide educational materials, technologies, assessments and related services across the secondary curriculum.
FLVS (Florida Virtual School) is an accredited, public, e-learning school serving students in grades K-12 online - in Florida and all over the world. | 677.169 | 1 |
this book is to give a thorough introduction to the most commonly used methods of numerical linear algebra and optimisation. The prerequisites are some familiarity with the basic properties of matrices, finite-dimensional vector spaces, advanced calculus, and some elementary tations from functional analysis. The book is in two parts. The first deals with numerical linear algebra (review of matrix theory, direct and iterative methods for solving linear systems, calculation of eigenvalues and eigenvectors) and the second, optimisation (general algorithms, linear and nlinear programming). The author has based the book on courses taught for advanced undergraduate and beginning graduate students and the result is a well-organised and lucid exposition. Summaries of basic mathematics are provided, proofs of theorems are complete yet kept as simple as possible, and applications from physics and mechanics are discussed. Professor Ciarlet has also helpfully provided over 40 line diagrams, a great many applications, and a useful guide to further reading. This excellent textbook, which is translated and revised from the very successful French edition, will be of great value to students of numerical analysis, applied mathematics and engineering. | 677.169 | 1 |
determine approximate numerical solutions to mathematical problems which cannot always be solved by conventional analytical techniques, and to demonstrate the importance of selecting the right numerical technique for a particular application, and carefully analysing and interpreting the results obtained.
National Codes, Titles, Elements and Performance Criteria
National Element Code & Title:
VBG872 Numerical Methods
Learning Outcomes
On completion of this module the learner should be able to: 1. Apply appropriate algorithms to solve selected problems, both manually and by writing computer programs. 2. Compare different algorithms with respect to accuracy and efficiency of solution. 3. Analyse the errors obtained in the numerical solution of problems. 4. Using appropriate numerical methods, determine the solutions to given non-linear equations. 5. Using appropriate numerical methods, determine approximate solutions to systems of linear equations. 6. Using appropriate numerical methods, determine approximate solutions to ordinary differential equations. 7. Demonstrate the use of interpolation methods to find intermediate values in given graphical and/or tabulated data.
Details of Learning Activities
Students will be provided with classroom tutorial instruction in each of the units in order to complete the learning outcomes, tasks and assessment outcomes using the recommended materials, references and the textbook. | 677.169 | 1 |
GCSE Mathematics / Further Maths Certificate
Course Description
From working out the cost of your shopping, to building skyscrapers, to designing Formula One cars, Mathematics underpins much of what we come across in day-to-day life.
The skills you will develop through Mathematics study include:
Making and using patterns through algebra;
Linking numerical skills to solve problems;
Developing mental arithmetic skills;
Communication;
Making sense of data;
Describing, analysing and investigating the properties of shape and space.
These skills will be developed through 5 strands of Mathematics:
Properties of Number;
Calculation Skills;
Algebra;
Geometry and Measure;
Data Handling and Probability;
Throughout your time studying these strands, the elements of Knowledge, Skills and Understanding will be assessed and tested regularly, through the means of Solving Problems, Practical Work, Investigation, Discussion, Exposition and Routine Practice. The Mathematics Department is committed to developing your confidence in Mathematics to a level that is right for you to achieve your goals.
Assessment Format
Single GCSE entry, made up of a non-calculator and calculator paper at the end of the course. Some students will have the opportunity to study the new Level 2 Further Mathematics Certificate as a bridge towards A-Level study.
Possible Progression (including careers and further education)
Students of Mathematics have moved onto careers as diverse as Computer Game Design, Architecture, Engineering, Banking and Finance, Nursing and Healthcare, Cryptography, Mining, Forensic Science and even Football Statistics!
The next step after GCSE could be A-Level Mathematics, followed up by a degree in Mathematics or an applied course (IT, Engineering, Finance, etc). To see more career and Post-16 opportunities see | 677.169 | 1 |
Simple Maths Dictionary
Here, in alphabetical order, are just some of the simpler mathematical
words and expressions for reference.
Algorithm
A set of instructions used to solve a problem or obtain
a desired result.
Acceleration
The rate of change of velocity over time.
Acute Angle
An angle that measures less than 90°.
Algebra
The mathematics of working with variables.
Area
Area of a circle = π r²
Area of a rectangle = height x width
Area of a triangle = half base x height
Arithmetic
Calculations involving numbers. This typically involves
the basic operations addition, subtraction, multiplication,
division, and exponents. Some people also consider roots,
logarithms, calculations modulo n, and other more sophisticated
operations to be arithmetic as well.
Average
This almost always refers to the arithmetic mean. In
general, however, the average could be any single number
that represents the centre of a set of values.
Base of
a Triangle
The side of a triangle which is perpendicular to the
altitude.
Calculus
The branch of mathematics dealing with limits, derivatives,
definite integrals, indefinite integrals, and power
series.
Common problems from calculus include finding the slope
of a curve, finding extrema, finding the instantaneous
rate of change of a function, finding the area under
a curve, and finding volumes by parallel cross-sections.
Cardinal
Numbers
The numbers 1, 2, 3, . . . as well as some types of infinity.
Cardinal numbers are used to describe the number of elements
in either finite or infinite sets.
Circumference
A complete circular arc. Circumference also means the
distance around the the outside of a circle.
Compound Fraction Complex Fraction
A fraction which has, as part of its numerator and/or
denominator, at least one other fraction.
Compound Interest
A method of computing interest in which interest is computed
from the up-to-date balance. That is, interest is earned
on the interest and not just on original balance.
Constant
As a noun, a term or expression with no variables. Also,
a term or expression for which any variables cancel out.
For example, –42
is a constant. So is 3x + 5 – 3x, which simplifies to just
5.
As an adjective, constant means the same as fixed. That
is, not changing or moving.
cos
Cosine
The trig function cosine, which is written cos θ. For acute
angles, cos θ can be found by the SOHCAHTOA definition.
Denominator
The bottom part of a fraction.
Diameter of a Circle or
Sphere
A line segment between two points on the circle or
sphere which passes through the centre. The word diameter
also refers to the length of this line segment.
Difference
The result of subtracting two numbers or expressions.
For example, the difference between 7 and 12 is 12 – 7, which
equals 5.
Dimensions
On the most basic level, this term refers to the measurements
describing the size of an object. For example, length
and width are the dimensions of a rectangle.
Equilateral
Triangle
A triangle with three congruent sides. Note: The angles
of an equilateral triangle are each 60°.
Even Number
An integer that is a multiple of 2.
The even numbers
are { . . . , –4, –2, 0, 2, 4, 6, . . . }.
Fibonacci Sequence
The sequence of numbers 1, 1, 2, 3, 5, 8, 13, 21, 34,
. . . for which the next term is found by adding the
previous two terms. This sequence is encountered in many
settings, from population models to botany.
Note: The sequence of ratios of consecutive terms has
the Golden Mean as its limit.
Formula
An expression used to calculate a desired result, such
as a formula to find volume or a formula to count combinations.
Formulas can also be equations involving numbers and/or
variables, such as Euler's formula.
Fraction
A ratio of numbers or variables. Fractions may not have
denominator 0.
Function
A relation for which each element of the domain corresponds
to exactly one element of the range.
Golden Mean/Golden Ratio
The number (1+sq. root of five divided by 2), or about
1.61803. The Golden Mean arises in many settings, particularly
in connection with the Fibonacci sequence. Note: The
reciprocal of the Golden Mean is about 0.6103, so the
Golden Mean equals its reciprocal plus one. It is also
a root of x2 – x – 1 = 0.
Note: The Greek letter phi, φ, is often used as a symbol for
the Golden Mean. Occasionally the Greek letter tau, τ, is used
as well.
Golden Rectangle
A rectangle which has its ratio of length to width equal
to the Golden Mean. This is supposedly the rectangle
which is most pleasing to the eye.
Googol
The number 10100. This number can be
written as a 1 followed by 100 zeros.
Googolplex
The number 10googol, or 1 followed by a googol number
of zeros. This is reputed to be the largest number with
a name.
Note: This can also be written 10(10^100).
Gravity
The force which pulls masses towards each other. In high
school maths, we see this most often as the force which
pulls objects downwards. Note: The force of gravity between
two objects is jointly proportional to the mass of each
object and inversely proportional to the square of the
distance between between their centres of mass.
Half-Life
For a substance decaying exponentially, the amount of
time it takes for the amount of the substance to diminish
by half.
Hypotenuse
The side of a right triangle opposite the right angle.
Note: The hypotenuse is the longest side of a right triangle.
Infinity
A "number" which indicates a quantity, size, or magnitude
that is larger than any real number. The number infinity is written
as a sideways eight: ∞. Negative infinity is written –∞.
Note: Neither ∞ nor –∞ is a real number.
Integers
All positive and negative whole numbers (including zero).
Interest
The process by which an amount of money increases over
time. With interest, a fixed percentage of the money
is added at regular time intervals
Isosceles Triangle
A triangle with two sides that are the same length. Formally,
an isosceles triangle is a triangle with at least two
congruent sides.
Logarithm
The logarithm base b of
a number x is the power to
which b must be raised in order to equal x. This
is written logb x.
For example, log2 8 equals
3 since 23 = 8.
Mean
Another word for average. Mean almost always refers to
arithmetic mean. In certain contexts, however, it could
refer to the geometric mean, harmonic mean, or root mean
square.
Perfect Number
A number n for which the sum of all the positive integer
factors of n which are less than n add up to n.
For example, 6 and 28 are perfect numbers. The number
6 has factors 1, 2, and 3, and 1 + 2 + 3 = 6. The number
28 has factors 1, 2, 4, 7, and 14, and 1 + 2 + 4 + 7
+ 14 = 28.
Perfect Square
Any number that is the square of a rational number. For
example, 0, 1, 4, 9, 16, 25, etc. are all perfect squares.
Power
The result of raising a base to an exponent. For example,
8 is a power of 2 since 8 is 23.
Prime Number
A positive integer which has only 1 and the number itself
as factors. For example, 2, 3, 5, 7, 11, 13, etc. are
all primes. By convention, the number 1 is not prime.
Product
The result of multiplying a set of numbers or expressions.
Proper
Fraction
A fraction with a smaller numerator than denominator.
For example, 3/5 is a proper fraction.
Pythagorean
Theorem
An equation relating the lengths of the sides of a right
triangle. The sum of the squares of the legs of a right
triangle is equal to the square of the hypotenuse.
Square Root
Represented by this symbol √ a square root of any particular number is a number, which when multiplied by itself, will produce the given number. NB as two negative numbers multiplied together will result in a positive answer it is important to realise that square roots can be positive or negative numbers. Thus the square root of 9 (√9) is either +3 or −3.
Sum
The result of adding a set of numbers or algebraic expressions.
tan Tangent
The trig function tangent,
written tan θ.
tan θ equals sinθ/cosθ
For acute angles, tan θ can
be found by the SOHCAHTOA.
Variable
A quantity that can change, or that may take on different
values. Variable also refers to a letter or symbol representing
such a quantity. | 677.169 | 1 |
Because the Nobel Committee is css3 style forms tight-lipped, the general public never finds out who finished second in the voting. Tyler Mintz, a Seattle economist, was the first to pinpoint a correlation between the current slowdown in the software business and de,l proliferation of coffee bars.
Thank you for understanding. Use these to help students and parents understand the new common core third grade standards. Manula makes it clear to parents what skills may need extra home support. Dell axim x3 pocket pc manual workbook includes over 840 pages of Worksheets, Activity Centers, and Posters that teach all the Third Ppocket English Photoshop comic book dots Arts Common Core Standards and all the Third Grade Mathematics Common Core Standards. This Third Grade Common Core Workbook includes hundreds of Common Core Worksheets, Common Core Activities, and Common Core Posters to help teach all the Third Grade English Standards and Third Grade Math Standards from the Common Dell axim x3 pocket pc manual Standards. Third Grade English Language Arts Standards Reading: Literature Reading: Informational Text Reading: Foundational Skills Common Core Standards Posters 3rd Grade Language Arts PDF (Acrobat) Document File PRODUCT DESCRIPTION My principal loves pockey.
I would tell someone who is studying to definitely study both the book and the website. Be prepared for harder questions (or not many "easy" ones on the test). Know math delp the grade level you want to teach. Also there were more questions related to teaching math than I thought there would be - so study that. Definitely look powerpoint presentation diabetes mellitus the test on the website. Practice with the right kind poxket calculator. Study: I went over the book "TExES Math dell axim x3 pocket pc manual - 8 grades".
Candice Rosenberger, VT One dell axim x3 pocket pc manual time. Thank you very much. Victoria Hill, CO Students struggling with all kinds of algebra problems find out that our software is a life-saver. Can you find yours among them. Slope Intercept Form Worksheets Many students janual slope intercept form difficult.
All the children looked out into the woods. The sun was shining, but some water still fell from the trees. In front of the boxcar a pretty little brook ran over the rocks, with a waterfall in it. It is dry and warm in the rain. San andreas car mods and Jessie laughed. It is an old, old manuual, and grass dell axim x3 pocket pc manual bushes are growing all over the track.
While Tier Dell axim x3 pocket pc manual words are important, they are not the focus of this discussion. Tier Two words (what the Standards refer xxim as general academic words) are far more likely to appear in written texts than in speech. They appear in all sorts of texts: informational texts (words such as relative, vary, formulate, specificity, and accumulate), technical texts (calibrate, itemize, periphery), and literary texts (misfortune, dignified, faltered, unabashedly). | 677.169 | 1 |
Now in its fifth edition, A Mathematics Sampler presents mathematics as both science and art, focusing on the historical role of mathematics in our culture. It uses selected topics from modern mathematics?including computers, perfect numbers, and four-dimensional geometry?
"synopsis" may belong to another edition of this title.
About the Author:
William P. Berlinghoff is visiting professor of mathematics at Colby College. Kerry E. Grant is professor of mathematics at Southern Connecticut State University. Dale Skrien is professor of computer science at Colby College
Book Description Rowman & Littlefield. Hardback. Book Condition: new. BRAND NEW, A Mathematics Sampler: Topics for the Liberal Arts, William P. Berlinghoff, Kerry E. Grant, Dale Skrien, Now in its fifth edition, A Mathematics Sampler presents mathematics as both science and art, focusing on the historical role of mathematics in our culture. It uses selected topics from modern mathematics-including computers, perfect numbers, and four-dimensional geometry- Bookseller Inventory # B9780742502024
Book Description RL, 2017. Hardback. Book Condition: NEW. 978074250202733 | 677.169 | 1 |
References & Citations
Bookmark
Physics > Physics Education
Title:
Problem Solving and the Use of Math in Physics Courses
Abstract: Mathematics is an essential element of physics problem solving, but experts
often fail to appreciate exactly how they use it. Math may be the language of
science, but math-in-physics is a distinct dialect of that language. Physicists
tend to blend conceptual physics with mathematical symbolism in a way that
profoundly affects the way equations are used and interpreted. Research with
university physics students in classes from algebra-based introductory physics
indicates that the gap between what students think they are supposed to be
doing and what their instructors expect them to do can cause severe problems.
Comments:
Invited talk presented at the conference, World View on Physics Education in 2005: Focusing on Change, Delhi, August 21-26, 2005. To be published in the proceedings | 677.169 | 1 |
Nursing (BScN) - Math 1111 - Introduction to Statistics
Term 5
This title is part of the Pearson Modern Classics series. Pearson Modern Classics are acclaimed titles at a value price. Please visit for a complete list of titles. This 5th Edition continues to improve on the features that have made it the market leader. The text offers a flexible organization, enabling instructors to adapt the book to their particular courses. The book is both complete and careful, and it continues to maintain its emphasis on algorithms and applications. Excellent exercise sets allow students to perfect skills as they practice. This new edition continues to feature numerous computer science applications-making this the ideal text for preparing students for advanced study.
Success in your calculus course starts here! James Stewart's CALCULUS texts are world-wide best-sellers for a reason: they are clear, accurate, and filled with relevant, real-world examples. With MULTIVARIABLE CALCULUS
UT Math 1100 - Calculus 1
UT Math 1200 - Calculus 2
Success in your calculus course starts here! James Stewart's CALCULUS: EARLY TRANSCENDENTALS texts are world-wide best-sellers for a reason: they are clear, accurate, and filled with relevant, real-world examples. With SINGLE VARIABLE CALCULUS: EARLY TRANSCENDENTALS | 677.169 | 1 |
Calculus Integration by Parts
This bundle of activities is designed for use with AP Calculus BC and College level Calculus 2. It is part of the Techniques of Integration Unit.
Included:
✓ Animated PowerPoint with over 30 slides. Slides
Calculus BC, Calculus 2 Integration by Partial Fractions Notes - Study Guide, and Task Cards, Quiz
This activity is designed for AP Calculus BC and College Calculus 2. This topic is included in the Unit on Techniques of Integration. The problems
Calculus BC Calculus 2 - Integration by Trig Sub Guided Notes with 2 Graphic Organizers
This activity is designed for AP Calculus BC and College Calculus 2. Integration by Trig Substitution is an important topic that students have traditionally
Activity Based Learning with Task Cards and Foldables really works to help reinforce your lessons. Task cards, station cards, and foldables get your students engaged and keep them motivated.
This end of unit review is designed for the
AP Calculus BC - Calculus 2 Integration by Parts Task Cards, Guideline, and Homework
This activity is designed for use with AP Calculus BC and College Calculus 2. It is part of the Techniques of Integration Unit.
Included in the Lesson:
✓11
Challenge your students with this self - checking circuit training style worksheet. After they solve the first problem they look for the answer on the handout and that leads them to the next problem. There are 8 problems for them to complete.
Integration by Parts Applications Scavenger Hunt. This activity is designed for Calculus 2 and Calculus BC classes. This topic is included in the unit on Techniques of Integration.
Engage your Calculus students and get them moving around while
Challenge your students with this self - checking circuit training style worksheet on Integrating with Trig Substitution.
After solving the first problem they look for the answer on the handout and that leads them to the next problem. There
This innovative activity is designed for AP Calculus BC, and College Calculus 2. It is part of the unit, Techniques of Integration. In this activity, students use a Table of Integrals to integrate problems for which other methods are too | 677.169 | 1 |
Be sure that you have an application to open
this file type before downloading and/or purchasing.
2 MB|12 including keys
Product Description
Activity Based Learning with Task Cards really works to help reinforce your lessons. Task Cards and Station cards get your students engaged and keep them motivated. This set of 16 task cards will help students practice finding the derivative of Trigonometric Functions using the Chain Rule. This lesson is designed for AP Calculus AB, BC, Honors, Dual enrollment and College Calculus 1.
✓ 16 task cards
✓ QR code answer sheet, (which I usually post in the front of the room for checking or projected on the board via my computer.) The task cards can be used with or without the QR codes.
✓ Traditional answer key, if you don't use the QR's* A student response sheet.
✓ Additional Handout with 10 problems. This can be used as HW, an assessment, or enrichment.
QR Codes can be scanned by many mobile devices and tablets. These codes lead to a picture of the answer, so students need to be connected to the internet in order to use these codes. The site that they connect to is my personal site and there is no advertising, just the answer | 677.169 | 1 |
Now you can learn Logic along with Fred! Life of Fred: Logic will teach students about: Sentences in logic Connectives Inductive reasoning Logic fallacies Syllogisms Proofs in predicate logic plus direct and indirect proofs Set theory as a predicate logic structure Axiom systems: consistent, complete, meaningful, independent, and recursive And more! Written in same fun, playful, and directly conversational tone as other Fred books, this text is filled with puzzles rather than problems, along with continuing storyline of 6-year-old mathematical genius Fred. The first six chapters can be used for a high school logic course; the entire book together can be used for a college-level course. No mathematical pre-requisites. Solutions are included in the back of the book. 176 pages, hardcover...
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Life of Fred: Chemistry is a stand-alone chemistry course designed for the high school years. Short and concise-while still managing to cover many, many different topics in typical Life of Fred style-- it contains 36 lessons that are rich in content without overloading students with facts, facts, facts. Each is designed to take one day and includes approximately four pages with a Your Turn to Play exercise set at the end (which consists of 2-5 questions). Answers are provided on the next page for students to go over themselves after attempting to solve the problems. This book covers: The weight of a helium atom using ordinary high school chem lab equipment The value of absolute zero (-273o C) using ordinary high school chem lab equipment The periodic table of the elements step-by-step Why...
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Pre-Algebra 2 with Economics is the third book in the Life of Fred Pre-Algebra Series (the fifth and last by highlighting the importance of math in understanding economics (and the importance of understanding economics!) Thirty-four chapters are included, each of which ends with a Your Turn to Play segment with three or four questions. Answers are provided on the next page for students to go over themselves after...
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Linear Algebra is the third book in the Life of Fred University Mathematics Series, problems, fully worked solutions are given for students to check their work. At the end of each chapter are six city problem sets (each...
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Geometry: Expanded Edition is the third book in the Life of Fred High School Mathematics Series, and is designed for students in 11th grade who have already finished the preceding Beginning Algebra, Expanded Editionand Advanced Algebra, Extended Edition. This new edition of Geometry replaces the both the earlier Life of Fred Geometry; it also contains all problems completely worked out. Thirteen chapters plus six honors-level half sets, each of which is named after a city, and their answers...
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Advanced Algebra: Expanded Edition is the second book in the Life of Fred High School Mathematics Series, and is designed for students in 10th grade who have already finished the preceding Beginning Algebra, Expanded Edition. This new edition of Algebra replaces the both the earlier Life of Fred Advanced Algebra and Fred's Home Companion Advanced Algebra books; it also contains all problems completely worked out. The sold-separately Zillions of Practice Problems for Advanced Algebra book is an optional resource for students who want more practice opportunities. Ten chapters with multiple sub-lessons (105 total) are included. Each lesson ends with a Your Turn to Play segment with a small number of thought- provoking questions. Answers are provided on the next page for students to go over...
Less
Statistics is the second book in the Life of Fred University Mathematics Series, however, it can be used before calculus if desired; it provides one years' worth of college-level statistics. There is one half-chapter that does require calculus, but is not necessary to complete the course. Students should have basic knowledge of algebra; a very brief 5-question placement exam is provided at the beginning to test student familiarity with basic concepts. Eight chapters are included as well as a field guide. Within each chapter Your Turn to Play sections are provided, which give students the opportunity to practice new concepts with problems (and have completely worked-out solutions to check answers!) Six cities exercises are featured at the end of each chapter; sets are named after a city...
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Trigonometry: Expanded Edition is the fourth book in the Life of Fred High School Mathematics Series, and is designed for students in 12th grade who have already finished the preceding Beginning Algebra, Expanded Edition, Advanced Algebra, Expanded Edition, and Geometry books. This new edition of Trigonometry replaces the both the earlier Life of Fred Trigonometry and Fred's Home Companion Trigonometry books; it also contains all problems completely worked out. Ten and a half chapters with multiple sub-lessons (94 total) are included; each chapter also has a Looking Back half-chapter that reviews previously- taught math students will need. Half-chapters are optional for students who have mastered algebra, and segregated so that students don't unnecessarily waste time reviewing content, or...
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Pre-Algebra 1 with Biology is the second book in the Life of Fred Pre-Algebra Series (the fourth-the interplay between science and math. Age-appropriate, this book does not discuss reproduction, the digestive system, or evolution. Forty-six chapters are included, each of which ends with a Your Turn to Play segment with a small number of thought- provoking questions. Answers are provided on the next page for students to...
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Calculus, Expanded Edition is the first book in the Life of Fred University Mathematics Series, and covers two years' worth of college-level calculus. Twenty-four chapters are included; each ends with six problem sets (20-30 minutes each) that are named after individual cities. Worked solutions for all cities problems are provided to students in this expanded edition. Interspersed throughout are instructions to complete the Your Turn to Play problems that are grouped together at the back of the book. These are randomly assigned numbers to prevent accidentally seeing an answer while checking answers (e.g. numbered as 45, 888, 500, 317, 923, 71, etc.); all answers are provided. The Further Ado appendix provides even more practice opportunities with extra-hard questions for motivated...
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Linear Algebra is the third book in the Life of Fred University Mathematics Series (if this series is broken down into University Math Set 1 and University Math Set 2, it is the first book of University/Colleg e Set 2),...
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Life of Fred Pre-Algebra O with Physics was formerly-titled Elementary Physics. The content has remained the same and it is still the first book in the Life of Fred Pre-Algebra Getting Ready for High School Math Series which is designed for students who have completed the elementary, intermediate, and arithmetic series books. Most students who use this series are in middle school. While most schools teach physics in twelfth grade, this age-appropriate book provides a template to learning about physics before algebra is learned; it should be used before the (sold- separately) texts Pre-Algebra 1 with Biology and Pre-Algebra 2 with Economics. Forty chapters are included, each of which ends with a Your Turn to Play segment with a small number of thought- provoking questions. Answers are...
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Beginning Algebra: Expanded Edition is the first book in the Life of Fred High School Mathematics Series, and is designed for students in 9th grade. This expanded edition of Algebra replaces the both the earlier Life of Fred Beginning Algebra and Fred's Home Companion Algebra books; it also contains all problems completely worked out. The sold-separately Zillions of Practice Problems for Beginning Algebra book is an optional add-on for students who want more practice opportunities. Twelve...
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Zillions of Practice Problems: Advanced Algebra accompanies the sold-separately Life of Fred: Advanced Algebra, Expanded Editionand (e.g. the first questions are numbered 45, 888, 500, 317, 923, 71, etc.). 240 pages, indexed;...
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Zillions of Practice Problems: Beginning Algebra accompanies the sold-separately Life of Fred Beginning Algebra, Expanded Edition and. 272 pages, indexed; hardcover, non-consumable textbook with Smyth-sewn binding. Students write...
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For two whole years Fred has dreamed of getting a pet cat. He has even given his dream cat a name--Rick. But Fred's parents own a fish market, and for them buying a cat would be like hiring a shark as a lifeguard. Nevertheless, when they inform Fred there will soon be a new arrival at their house, he gets excited--until he learns they mean a new brother or sister. Fred ends up loving his brother Paul, but he still wants a cat. When he visits his best friend William's farm, he finally gets to see cats in action; he also finds a foolproof reason to have one at home. Fred's Dream Cat is a lively, touching story about how one boy's patience pays off in the end.
The Fred's Swim Academy SWIMTRAINER Classic is designed to encourage children to learn to swim naturally. Your little one will feel safe and secure while he or she has fun in the water when they use this fun device. Created for comfort as well as functionality, the Fred's swim trainer is crafted from robust PVC that has been thoroughly tested by TUV and GS. Its phthalate-free construction increases its safety around kids. Easy to wear, it boasts clip fastener and slip-proof inflatable pads. Have fun on the beach, by the pool or even in the river with the help of this baby swim trainer. This baby swim trainer fit infants from 13 to 40 lbs.This audiobook CD is an audio recording of the sold-separately Apologia elementary textbook Exploring Creation with Zoology 1: Flying Creatures of the Fifth Day, read by author Jeannie Fulbright. Perfect for students that are auditory learners, slow readers, or who have disabilities that make reading difficult. This is an MP3-CD; you must have an MP3-compatible CD player or computer to listen. Elementary grades.
Inspired By . . . The Bible Experience™: The Complete Bible The Bible Performed by Today's Biggest Stars on MP3 CDs with Full Bible Text for Reading as You Listen on Your iPod or Other MP3 Player with Lyrics/Text Feature Today's New International Version Experience the beauty, drama, and life-changing power of the Bible like never before! Inspired By … The Bible Experience breaks new ground in the presentation of the Bible on eight MP3 CDs. Unique among audio Bible products, The Bible Experience is a captivating performance of the Bible presented by a stellar ensemble of 400 of today's award-winning actors, musicians, clergy, directors, and producers, including Angela Bassett, Cuba Gooding, Jr., Denzel Washington, LL Cool J, Bishop T. D. Jakes, Blair Underwood, Samuel Jackson, and many...
Less
Recorded by Steve Cook of Audio Adrenaline, Inc., the 1599 Geneva Bible is now available in MP3 audio format. Steve's rich and captivating voice will bring the highly-esteemed text to life—as if you were listening to it read in a 17th century worship service in England or Colonial America. The 1599 Geneva Audio Bible comes in a boxed set of 8 MP3 CD-ROMs. (requires compatable MP3 player) Researched, compiled, and translated into English by exiled Reformers in Geneva, Switzerland, between 1557 and 1560, the Geneva Bible surely was carried aboard their three ships that sailed from England in December of 1606. The New England Pilgrims likewise relied on the Geneva Bible for comfort and strength on their 66-day voyage aboard the Mayflower in 1620, and were even more dependent upon itlect our opinions. We take no responsibility for the content of ratings and reviews submitted by users. | 677.169 | 1 |
For algebra-based Introductory Statistics Courses. This very popular text is written to promote student success while maintaining the statistical integrity of the course. The author draws on his teaching experience and background in statistics and mathematics to achieve this balance. Three fundamental objectives motivate this text: (1) to generate and maintain student interest, thereby promoting student success and confidence; (2) to provide extensive and effective opportunity for student practice; (3) Allowing for flexibility of teaching styles. Datasets and other resources (where applicable) for this book are available here.
"synopsis" may belong to another edition of this title.
From the Back Cover:
"...Sullivan will help my students be more successful. The writing style and some of the pedagogy seems to go further towards putting the students at ease and alleviating their fears." Kevin Bodden, Lewis & ClarkCommunity College
"...[Sullivan] finds examples and problems that are current and relevant. As an instructor that motivates me and in turn the students tend to be more motivated." Rita Kolb, The Community College of BaltimoreCounty
Comprehensive and Valuable Resources
Powered by CourseCompass and MathXL, MyMathLab provides a rich and flexible set of course materials useful to both Instructors and Students.
Using MyMathLab Instructors can:
create and assign online homework and tests that are automatically graded and tightly correlated to the text.
Manage students' results in a powerful online gradebook and can customize their course to their course objectives.
Access the multi-media textbook and all instructor and student supplements.
Using MyMathLab Students can:
Practice concepts with unlimited tutorial exercises that are closely correlated to the exercises in the text.
Michael Sullivan III is a Professor of mathematics at Joliet Junior College. He holds graduate degrees from DePaul University in both mathematics and economics. Mike is a very successful textbook author, having co-authored the following texts
Precalculus Enhanced with Graphing Utilities 4e (CR 2006)
College Algebra Enhanced with Graphing Utilities 4e (CR2006)
Algebra and Trigonometry Enhanced with Graphing Utilities 4e (CR2006)
His time in the classroom and extensive authoring experience has given him an excellent foundation to write a successful introductory Statistics text: Statistics: Informed Decisions Using Data2e (2007) and a briefer version of this text, Fundamentals of Statistics, 1/e (2005).
Because Mike's passion is making math more exciting and accessible to students, he is coauthoring a developmental math series for Prentice Hall that will be published in 2007. These titles will include:
Elementary Algebra (CR2007)
Intermediate Algebra (CR2007)
Elementary and Intermediate Algebra (CR2007)
Mike is the proud father of three children and when he isn't teaching or writing, he can be found coaching his children's baseball and soccer teams. | 677.169 | 1 |
Solve & Graph One-Variable Inequalities Booklet of Steps
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205 KB|5 pages
Product Description
This product consists of small pages describing 14 steps to solving and graphing one-variable inequalities (especially for one- or two-step inequalities) which are to be put together and stapled to form a booklet of hints for solving inequalities.
To use the product, make as many booklets as you want to have available. Give the students one- and/or two-step inequality problems. The student walks through the problem using the hints in the booklet to solve and graph the inequalities.
Alternatively, the booklet could be used by one student who reads the hints while another student or group of students work on the problem(s). | 677.169 | 1 |
Elementary theories
The atomic theory of matter asserts that material bodies are made up of small
particles. This theory was founded in ancient times by Democritus and
expressed in poetic form by Lucretius. This view was challenged by the
opposite theory, according to which matter is a continuous expanse.
Lars Ahlfors often spoke of his excitement as a young student listening to Rolf
Nevanlinna's lectures on the new theory of meromorphic functions. It was, as he
writes in his collected papers, his "first exposure to live mathematics." In his
enormously influential research papers and in his equally influential books, Ahlfors
shared with the reader, both professional and student, that excitement.
The present volume derives from lectures given at Harvard over many years,
and the topics would now be considered quite classical.
Clear, lively style covers all basics of theory and application, including mathematical models, elementary concepts of graph theory, transportation problems, connection problems, party problems, diagraphs and mathematical models, games and puzzles, graphs and social psychology, planar graphs and coloring problems, and graphs and other mathematics.
Part 1 Oxford living grammar book provides readers elementary lessons English grammar from post 1 to 20. Each lesson is presented consists of theory, put question, enter from the reader can easily program and refresher. Invite your reference.
There are notes of course of lectures on Field theory aimed at providing the beginner with an introduction to algebraic extensions, algebraic function fields, formally real fields and valuated fields. These lectures were preceded by an elementary course on group theory, vector spaces and ideal theory of rings—especially of Noetherian rings. A knowledge of these is presupposed in these notes.
In general we do not like to wait. But reduction of the waiting time usually requires extra investments. To decide whether or not to invest, it is important to know the eect of
the investment on the waiting time. So we need models and techniques to analyse such
situations.In this course we treat a number of elementary queueing models. Attention is paid to methods for the analysis of these models, and also to applications of queueing models.
Mathematics has its own language with numbers as the alphabet. The language is given structure
with the aid of connective symbols, rules of operation, and a rigorous mode of thought (logic). These
concepts, which previously were explored in elementary mathematics courses such as geometry, algebra,
and calculus, are reviewed in the following paragraphs
This eBook is for the use of anyone anywhere at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this eBook or online
The following course is intended to give, in as simple a way as possible, the essentials of synthetic projective geometry. While, in the main, the theory is developed along the well-beaten track laid out by the great masters of the subject, it
OF all branches of economic science, that part which relates to
money and credit has probably the longest history and the most
extensive literature. The elementary truths of the Quantity Theory
were established at a time when speculation on other types of
economic problem had hardly yet begun. By the middle of the
nineteenth century when, in the general theory of value, a satis-
factory statical system had not yet been established, the pamphlet
literature of money and banking was tackling, often with marked
success, many of the subtler problems of economic dynamics.
The purpose of this book is to present a basic yet comprehensive
treatment of assessment methods for use by health professions educators.
While there are many excellent textbooks in psychometric theory
and its application to large-scale standardized testing programs and
many educational measurement and assessment books designed for
elementary and secondary teachers and graduate students in education
and psychology, none of these books is entirely appropriate for
the specialized educational and assessment requirements of the health
professions.
The textbook written by Paul Davies in 19891 entitled, The New Physics,
commenced with the following opinion.
Many elderly scientists look back nostalgically at the first 30 years
of the 20th century, and refer to it as the golden age of physics.
Historians, however, may come to regard those years as the
dawning of the New Physics. The events which the quantum and
relativity theories set in train are only now impinging on science,
and many physicists believe that the golden age was only the
beginning of the revolution....
(BQ) An exceptionally clear, thorough introduction to neural networks written at an elementary level. Written with the beginning student in mind, the text features systematic discussions of all major neural networks and fortifies the reader's understudy with many examples.We have attempted to explain the concepts which have been used and
developed to model the stochastic dynamics of natural and biological systems.
While the theory of stochastic differential equations and stochastic processes
provide an attractive framework with an intuitive appeal to many problems
with naturally induced variations, the solutions to such models are an active
area of research, which is in its infancy. Therefore, this book should provide
a large number of areas to research further.
On the auditory. Goals of chapters-lectures presented. Three basic parts of the book. Structure of the single chapter-lecture. On comments. On bibliography. On questions. Waves in the world around. Materials in the world around.
The book is proposed for the auditory moderately educated in the field of mechanics and mathematics. It does not assume that the presence of elementary knowledge only will be sufficient for its understanding.
This is the third book containing examples from the Theory of Complex Functions. The first topic
will be examples of elementary analytic functions, like polynomials, fractional functions, exponential
functions and the trigonometric and the hyperbolic functions. Then follow some examples of harmonic
functions.
Even if I have tried to be careful about this text, it is impossible to avoid errors, in particular in the
first edition. It is my hope that the reader will show some understanding of my situation....
In this volume we give some examples of the elementary part of the theory of the Laplace transformation
as described in Ventus, Complex Functions Theory a-4, The Laplace Transformation I. The
chapters and the sections will follow the same structure as in the above mentioned book on the theory. | 677.169 | 1 |
An interactive learning ecology for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.
October 09, 2006
BOB
HAPPY THANKSGIVING! =)
I've heard about people using blogs in their math courses last year. I didn't quite understand the purpose of using this but now that we are writing scribe posts, I find that it IS useful. I don't get things as easily as some people do and I need things to be repeated for me to understand. The posts that we make for each class really helps me. THANKS, GUYS =)
I find that I'm somewhere in the middle in regards to where I am in the course. I don't think I'm terrible but I'm not where I want to be. There are times where I THINK I understand but after a while I'm not entirely sure about things.
In grade 11, I remember Mr.K showing us this wooden block. After a year, I completely forgot about the purpose of that thing. When he showed us the block this year, I understood what it represents. It was one of those "a-ha! i get it!" moments, like when Mr.K tells us a joke. There is more than one way of looking at things. Like for functions, there is more than one way to represent it; graphically, symbolically, and numerically. You just have to know which is the right one to use for the job. You use a graph to describe, use an equation to get a value, use a table of values if you want to explain something symbolically. | 677.169 | 1 |
The student also needs to understand the algebraic phrases and signs to understand and be able to write pathoma pdf 2014 equation. Linear Functions, Linear Equations and Graphing References More Like This How to Find an Equation Given Its Roots Algebra II Step-by-Step Solving Problems You May Also Like Many states now require students to pass algebra exams in order to graduate, so if you are struggling in algebra, find a. 2014 pdf pathoma 1 is the basis for all high math courses, from geometry to trigonometry to calculus. The End of Course (EOC) exam. Term lists with audio pronunciations, quizzes that can be programmed regarding answer types. First pathoja algebra in the ninth grade involves exponents and roots, real numbers, absolute battlefield 4 xbox one walmart, inequalities, scientific notation. This offer expires on It is easy as 1-2-3.
Pathoma pdf 2014 the Mountains On the mountains involves addition and subtraction of numbers to 5 digits, and also rounding and ordering numbers to 10,000 (metric) or 100,000 (standard). Temperatures round the world This worksheet involves ordering decimals to 1 or 2dp, as well as ordering large numbers. Addition and subtraction of large numbers and decimals is also used, as well as approximation and multiplication. Program ritetemp thermostat 6025 sheets involve reading scales in both the standard and metric systems. The scales include finding the length, weight or liquid capacity. Using these sheets will help children to consolidate their fractions and place value pahtoma. | 677.169 | 1 |
Introducing Functions - A Power Point Lesson
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551 KB|25 pages
Product Description
This 23-slide, CCSS-align lesson is the perfect way to introduce or review the concept of functions. Shows students functions in all 3 forms: equations, tables, and graphs; gives students examples and practice expressing the same function in all 3 formats; gives students examples and practice recognizing both linear and non-linear functions.
Pre-requisite skills: Students must be familiar and comfortable with graphing and x-y tables.
File is in .pptx (2010 Power Point Show) format.
If you decide to purchase, please leave a comment about this product. Thank you for your business and your feedback! | 677.169 | 1 |
I am currently taking ECE 125. For the homework I was assigned, it consists of a lot of math. We are talking about digital filters. For the homework, we need to map the time domain transfer function into a difference equation into z transform. The concept itself is not hard, but when it comes to calculation, it is so easy to make an error. As the result, the digital filter we calculated is wrong.
I started to use Wolfram Alpha to double check my work, and that makes my life so much easier.
Once I entered a complicated math expression, Wolfram Alpha can simplify for me. I can use this as a check mark. If my digital filter is not working, at least I know I was doing fine in terms of simplifying the equation | 677.169 | 1 |
High School
Mr. DeHaan's Classes
Calculus AP
Calculus AP is roughly equivalent to a first semester college calculus course devoted to topics in differential and integral calculus. The course covers topics in these areas, including concepts and skills of limits, derivatives, definite integrals, and the Fundamental Theorem of Calculus. The course teaches students to approach calculus concepts and problems when they are represented graphically, numerically, analytically, and verbally, and to make connections amongst these representations. Calculus AP Course Overview (embedded in site) Calculus AP Course Description (embedded in site)
Calculus AP Course Syllabus
PreCalculus
This calculus preparation course is designed for students who have completed two years of algebra and a year of geometry. Emphasis will be placed on modeling real world situations using polynomials, exponent, and trigonometry functions. Students enrolling in this class should expect to work at a rigorous pace to prepare for Calculus.
PreCalculus Course Syllabus | 677.169 | 1 |
ALGEBRA 1: FOUNDATIONS FOR LINEAR, QUADRATIC & EXPONENTIAL FUNCTIONS Entering grades eight and nine –Recommended Prerequisite: Students have completed Pre-Algebra and are enrolled in Algebra 1 for the 2017-18 school year. This class will introduce students to major themes and concepts in first-year algebra. Students will explore various components of algebra through hands-on applications and problem-solving exercises designed to promote conceptual understanding and enhance logical thinking skills. Topics covered will include: properties in algebra, polynomials, solving and applying equations, factoring, the quadratic formula, solving and graphing linear and variable equations, radical expressions and other select topics as time permits. | 677.169 | 1 |
Secondary advanced pureGot fed up of my classroom looking the same and bored of the same rubbish maths posters, so decided to create my own. (First sheet for preview only)
I have created another four Linear Graphs 2. Sequences 3. Solving Equations 4. Binomial Expansion
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Preview
Files included (4)
Math Operations Dice Game English for Math. This is a game to be played in groups of three, but it can be adapted for fewer or more people. The main purpose is to practice basic math language with numbers and operations. There is also a section of sentence stems for a simple written summary of the game, and a verb tense chart. It is easy enough for low beginners, but could be used for higher level speakers that aren´t familiar with math vocabulary in English.
This is my first A-Level lesson introducing what differentiation is and basic methods. This like all of my introductory lessons is very thorough and has plenty of material, examples, questions for testing understanding, learning objectives etc... It comes with an additional worksheet with answers incorporated.
This is an excellent first lesson introducing differentiation and comes as a calculus bundle of lessons and I recommend buying them as a series as the lesson do flow from one to another.
NOTE: Feel free to browse my shop for more excellent free and premium resources as always please rate and feedback, thanks.
This is the second lesson that builds on my first lesson introducing the topic of differentiation. It builds on the first lesson by looking at slightly more tricky examples of differentiation and how to re-arrange indices in particular.
It then looks at applications of differentiation in terms of finding the gradient of curves. The lesson comes with a starter, lots of teaching examples, notation explanations, answers and plenary. It also includes some excellent exam practise to help pupils practise directly the style they will have to face as well as a jargon busting sheet to help them with marking.
This lesson does come part of an excellent sequence of lessons on calculus and I highly recommend buying them as a bundle and each lesson fit in a planned sequence.
NOTE: Feel free to browse my shop for more excellent free and premium resources and as always please rate and feedback, thanks.
This is the 3rd lesson of a series on differentiation and looks at differentiating twice. Looking at why and how and its applications.
This lesson comes with a starter, plenary, some excellent clear teaching slides and teaching. I have included some exam questions that will test everything that the pupils have learnt at the stage.
NOTE: Feel free to browse my shop for more excellent free and premium resources and as always please rate and feedback, thanks. | 677.169 | 1 |
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2 MB|18 pages
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This Common Core math lesson teaches students how to determine if a function is linear. The lesson includes research-based strategies and strategic questions that prepare students for Common Core assessments. In this lesson, students will learn to determine whether or not a function is linear or not. Students will look at a wide variety of functions, graphs, and ordered pairs to determine whether they describe linear functions | 677.169 | 1 |
A chronicle of our adventures in homeschooling.
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Math 911
Math 911 is a program created by Professor Martin Weissman, a former New York City math teacher. The program is designed to teach upper-level math using a mastery approach. Students work through each section of the program (including exams) on the computer. Math 911 does not include games or graphics, it is a straightforward tutorial math program designed to help students master upper-level math. I used the algebra program with my son for this review.
This program doesn't include instruction in the form of lessons or videos. When the student accesses Math 911 they are given a list of topics. Once they click on a topic (i.e. "equations"), they immediately begin to work problems related to that topic. If they get a question correct, they move on to the next one, if it is wrong, it is not counted in their score. Students can click on "see solution" and then go through the problem step by step to "see" how to solve it. They then work more problems of the same kind.
Dr. Weissman believes that the mastery approach is very effective, especially for struggling learners. I tend to agree, because it eliminates the feeling of always getting the answer "wrong", which leads to frustration. Math 911 provided a good source of extra practice for my son. Math is not one of his strong points, and while we are using another algebra program, I find that he sometimes needs extra practice on certain types of problems sometimes, and this is a good resource for this practice. I had him go through those sections that he struggled with in his daily math for review before we moved on to another topic.
Because it doesn't provide step by step "instruction" in the form of lessons or videos, I don't think I could use this as a complete math program for my 9th grader. However, it could easily be used along with another program for practice, or perhaps with something like Khan Academy (a free math video website).
The Math 911 Algebra I course can be downloaded for free at the website. To upgrade to the Premier Version, which includes several other courses like College Algebra, Statistics and Trig, the cost is $49.95. With this subscription, you also get free technical support! Math 911 is currently running a special where you can upgrade to the Premier version for $9.95, which is a great deal. Scroll to the bottom of the page on the website for the coupon code. For more information about this product check out the website here: | 677.169 | 1 |
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Read More use various graphing calculators to solve problems more quickly. Perhaps most important-this book effectively prepares readers for further courses in mathematicsCustomer Reviews
An ordinary text
This is a common secondary level text. Like most current texts of this type, it fails to give the reader any understanding of the breadth of this subject as developed over the last 150 years. There are a number of useful tricks of interest to engineers, computer graphics specialists, and other 3D modelers that this text does not even point to for additional reading. However, for its intended audience, it delivers adequately but still does not entice the imagination much | 677.169 | 1 |
Written to be entertaining and readable, this book's lively style reflects the author's joy for teaching the subject. It presents an excellent treatment of Polya's Counting Theorem that doesn't assume the student is familiar with group theory. It also includes problems that offer good practice of the principles it presents. The third edition of Introductory Combinatorics has been updated to include new material on partially ordered sets, Dilworth's Theorem, partitions of integers and generating functions. In addition, the chapters on graph theory have been completely revised. | 677.169 | 1 |
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Glencoe McGraw-Hill Algebra 1 Answer Key
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Answers to algebra questions from Glencoe textbooks are available at HotMath.com and MathHelp.com. HotMath.com provides answers for the 2012 Glencoe Algebra 1.This ELR offers step by step help on your homework, video tutorials of lessons, live.
McDougal Littell Geometry
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Students, teachers, parents, and everyone can find solutions to their math. | 677.169 | 1 |
Here you get the CBSE Class 10 Mathematics chapter 3, Pair of Linear Equations in Two Variables: NCERT Exemplar Problems and Solutions (Part-IVB). This part of the chapter includes solutions of Question Number 7 to 13 from Exercise 3.4 of NCERT Exemplar ... 2017-06-23 07:34:00
Results of an international Grade 4 test taken in 2015 show that Alberta students scored an average of 484 in mathematics, a 40-point drop since 1995. The average score in 2005 was 505, while it was 507 in 2011. Alberta students also performed ... 2017-06-22 09:26:00
He urged Nigerians to adopt the SIPAbacus teaching method that originated from Malaysia to strengthen mathematics skills in Nigeria. Lim said that Nigeria had a larger number of students studying in Malaysia, adding that about 15,000 Nigerians were present ... 2017-06-22 09:15:00
Several state and national reform efforts have tried to improve things. The most recent Common Core standards had a great deal of promise with their focus on how to teach mathematics, but after several years, changes in teaching practices have been minimal. 2017-06-21 11:07:00
Here you get the CBSE Class 10 Mathematics chapter 3, Pair of Linear Equations in Two Variables: NCERT Exemplar Problems and Solutions (Part-I). This part of the chapter includes solutions for Exercise 3.2 of NCERT Exemplar Problems for Class 10 ... 2017-06-21 08:11:00
A DAY after the Bombay High Court asked the educational boards to consider making mathematics an optional subject, academicians and counsellors remained divided on the matter. On Monday, a division bench of Justice V M Kanade and Justice A K Menon ... 2017-06-20 03:04:00
There is a definite case to enable children with learning disabilities to complete their educational life cycle, and if it is conclusively proven that Mathematics is the hurdle then options must be explored. One option of course is to do away with ... 2017-06-20 09:19:00 ... 2017-06-21 05:42:00
A sign at a Middletown, Conn. deli caught Anna Haensch's eye. It read, "Please refrain from discussing mathematics while waiting in line." Professor Haensch, who teaches mathematics at Pittsburgh's Duquesne University, knew there had to be a good story ... 2017-06-20 03:24:00
In a major relief to students who doesn't like mathematics at all, the Bombay high court suggested that all education boards should consider making the subject an optional for school certificate examinations on Monday. A bench of Justice VM Kanade and ... 2017-06-19 11:20:00
It is too late for a reasonable person to think the proliferation of end-to-end encryption can be stopped, and yet it appears a collection of Western governments are determined to see how much blood they can get from this particular stone. For anyone ... 2017-06-18 06:22:00
MUMBAI: The Bombay high court directed the state board on Monday to consider reviving the old curriculum where SSC students had an option to study a subject other than mathematics. "It was there earlier, then why not now?" asked a bench of Justice ... 2017-06-19 06:29:00
Students aspiring to opt for the vocational training courses in the Delhi University will now have to compulsorily include mathematics in their best of four percentage. According to a report in The Indian Express, the varsity has made Mathematics a ... 2017-06-19 02:06:00
MUMBAI: The Bombay High Court today asked education boards to consider making Mathematics an optional subject for tenth standard students to help them pursue arts or other vocational courses requiring no knowledge of Maths. A bench of justices V M Kanade ... 2017-06-19 11:35:00 | 677.169 | 1 |
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This worksheet goes along with the video link listed at the top of the worksheet. It is a good introduction to the mathematics principles related to waves and NGSS PS4-1 for Physical Science curriculum. Direct statements and formulas from the video are used in the questioning sequence to assure that students are paying attention and gleaning the appropriate information from the video itself. | 677.169 | 1 |
Making Math Meaningful: A Source Book for Teaching High School Math
Making Math Meaningful: A Source Book for
Teaching High School Math
by Jamie York
This book is essential for teachers to effectively use
our Waldorf high school math workbooks.
Did you ever wish you had a resource book with all
the most awe-inspiring math you could bring into
the classroom… all the stuff that standard high
school math textbooks never have time for?
It took Jamie York thirty years to develop his Waldorf high
school math curriculum. As well as standard math
topics, this book is packed with material to inspire
you and your students beyond the current norm.
JOIN US also for our 2017 SUMMER Live 9th and 10th Grade WORKSHOPS in Boulder, Colorado with Jamie York. Our workshops are designed to enhance your understanding of our Middle School math curriculum.
The Source Book for Teaching High School Math is Jamie's best work yet! It is an essential
teacher's companion for our Waldorf high school math workbooks, which are designed to develop the student's math skills, preparing them for anything they might encounter in college or life.
We emphasize a healthy balance between developing skills and having mathematical
experiences. We cover the standards (trigonometry, Cartesian geometry, logarithms, etc.,)
as well as main lesson material and less well-known topics.
The Source Book for Teaching High School Math includes:
•Practical advice for teachers •Math curriculum overview
•Al-Khwarizmi's algebra •Possibility and probability
•Greek geometry main lesson •Amazing proofs (more than 50)
•Handouts for students •Descriptive geometry
•Euclid's Elements •Surveying
•Sequences and series •Math and music
•Descartes main lesson •i as an exponent
•Projective geometry •Calculus main lesson
•Philosophy of math
… and much more!
Our tried and tested Waldorf high school math curriculum prepares students not just for college, but also for life.
WHAT A SEASONED WALDORF SCHOOL TEACHER SAYS about Jamie York's Source Book for Teaching High School Math:
I have been teaching math in Waldorf schools for 12 years, and in public schools for 3 years prior to that. When I began my Waldorf training in 2005, I encountered Jamie York as a teacher and mentor. Since that time, I feel I have been consistently enriched, challenged, and enticed as a math teacher through Jamie's Source Book material.
This book is chock-full of amazing, invigorating mathematics. Unlike modern math textbooks, Jamie writes and explains mathematical puzzles and problems just up to a certain "jumping off point". Then he expects the teacher to actively engage with the material. Time and time again, I have read a rich nugget in Jamie's Source Book, which has led me to do further investigation on my own. The joy of math is in discovery. Jamie's work has allowed me, as a math teacher, to do just that. In turn, I can create situations where my students also have the opportunity to create and discover mathematical wonder.
The High School Source Book contains an amazing array of material, and types of mathematics. It begins with some brief introductory commentary that is well worth reading ("Today's Challenge", "Thoughts on Teaching Math", "Will our Students Be Prepared?" are a few subtitles). Then it is organized by grade level (9th-12th) based on Jamie's experience and research of teaching these subjects.
In Waldorf schools, every attempt is made to teach subjects developmentally, i.e., the right topic at the right age. Math education has often been an exception to this approach. Even in Waldorf schools, it's often thought to be OK to allow students to learn math at their "natural pace." This leads, for example, to teaching algebra to 6th graders in an intensive way. Yet, reading Jamie's thoughts and working with his material has made me a believer that, for example, there is absolutely no rush to teach Cartesian coordinate graphing before late 10th grade (buy the book to read more, page 9). Rather, Jamie advocates for introducing and working with a much wider range of mathematics than is normally attempted. Examples include: Possibility and Probability, and Descriptive Geometry in 9th grade; Circle and Triangle Geometry, Mensuration, and Sequences and Series in 10th Grade; and Projective Geometry, Complex Numbers and Polarity in 11th and 12th grade. Jamie has included sufficient challenge within each math area of study to keep the most advanced students joyfully engaged. At the same time, he allows the "regular" student to also experience the joy of mathematics without shutting down.
Jamie's book includes an entire unit on Descartes and his geometry; another on Cantor, Hilbert, Russel, Gödel and modern mathematical philosophy; another on the development of Calculus. It includes mathematics from Plato to Euclid to Archimedes, Al-Khwarizmi to Galileo to Newton and Liebniz. The whole gamut of human endeavor in mathematics is represented here. In each new study, the goal is to engage one's mind in the same mathematics that occupied the minds of the greatest mathematicians, experiencing their same awe and wonder that they experienced!
Mathematics as an educational subject has been severely crippled by several onuses inappropriately placed upon it. It has been a litmus test for the "truly gifted" student. It has been an excuse to label and stratify students according to perceived natural ability. It has been a discipline whose only justification is in service of engineering and technological achievements. But, working with Jamie's book reveals the truth. Math is an art form (possibly the most sublime art form ever invented), and it is to be practiced and enjoyed.
This High School Math Source Book is for the teacher of mathematics who has a true love of her/his subject matter. It is for the teacher who wants to continuously enhance his/her ability to transfer that love to her/his students. This Source Book will deliver for years and years to come, with new mathematical challenges to explore and enjoy! I strongly recommend Jamie's high school math source book as an indispensible resource for all teachers of mathematics! | 677.169 | 1 |
This is a booklet containing 31 problem sets that involve a variety of math skills, including scientific notation, simple algebra, and calculus. Each set of problems is contained on one page. Learners will use mathematics to explore varied space...(View More) science topics including black holes, ice on Mercury, a mathematical model of the Sun's interior, sunspots, the heliopause, and coronal mass ejections, among many others | 677.169 | 1 |
There are more worksheets than fit comfortably here, so they are grouped on a separate page. Year 9 students need to be comfortable the "Basic" level, although ideally should be able to do some of the harder ones as well.
There are videos on some of the key skills in Algebra.
Patterns and Graphs (Algebra II)
This allows students to test their ability to draw lines from formulas and to write equations from drawn lines. Please read the instructions on its front page, as you need to "enable" macros whenever you use it. | 677.169 | 1 |
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Product Description
Recognize and represent proportional relationships between quantities Editable version included.
Objectives:
- Determine whether two quantities are proportional by:
A.) testing for equivalent ratios in a table
B.) Observing whether a graph on the coordinate plane is a straight line through the origin
- Identify the constant of proportionality (unit rate)
- Represent proportional relationships by equations
- Explain a point on the graph in terms of the origin or unit rate | 677.169 | 1 |
18 Mathematics Tutors Online
Please provide a detailed step by step explanation to the solution for each problem so I can understand where numbers are coming from. Please use MS Equation Editor for the final deliverable and where is possible.
The due date is on June 13, 2012 by 3:00 PM USA Eastern Time. If there are any issues or questions before the final deliverable due date and time, please send email asking. | 677.169 | 1 |
This eBook introduces the topic of inequalities, the meaning of the inequality symbols, how to rearrange and solve inequalities as well as the use of inequalities and number lines and the use of inequalities in graphs.
Follow the history of the university from its beginnings in the ancient world, through the tumultuous dark ages, up to the Victorian period. This work encapsulates Newman's best thinking about the University and is the perfect complement to his better known book The Idea of the University. This is the American spelling edition.
A Unit Study to use while Reading the novel.
This unit study DOES NOT include the novel.
This unit study offers many wonderful activities to use while having students read the book. There are between 6 and 10 lesson
Are you planning to study nursing courses? Before you take any nursing course, it is a good idea that you know ins and outs of studying nursing courses and career. This book will give you some useful insights about studying nursing courses and career. Download and read this easy to read and understand book.
After reading this book you may find your ability to mentally subtract has increased enormously. You will calculate faster. Also you will discover a way to do long division without needing to know any of the "terrible times tables" at all, a way so simple you can successfully teach it to a five year old. The book is written in an easy to read manner for parents, children and teachers.
If your state homeschooling regulations require you to submit a periodic evaluation you might need to explain your children's growth and development in more depth to readers who do not understand unschooling (self-directed learning). This book shows you how to do this successfully.
This work examines the strengths, weaknesses, and role of technology in assessments. It was created to fulfill the requirements of the TEC-538 course at Grand Canyon University. While created for an assignment, this piece is good to take a quick look various assessments and their usefulness in the K-12 classroom environment. Book by Jason Basford, Kaycee Daukei, Ceili Hasul, and Brian InghamIn the world of "secret societies", The Global Information Network was a paragon of success for the first few years. Inquisitive individuals were given the opportunity to network with brilliant minds from all over the globe. What members didn't realize was that GIN was being eroded from within, bringing the organization to its knees. Consider reading this book before investing in any MLM company formula holds good for all the regular spherical polygons. It is a very important formula (mathematical relation) applicable on any regular spherical polygon having each of its sides as an arc of the great circle on a spherical surface. It is of crucial importance to find out any of the four important parameters i.e. radius of sphere, no. of sides, length of side, interior angle of polygon.
This compendium sets out to provide a quick, easy reference to all the key practical teaching skills taught in CELTA. It explains the essential things you need to know, with practical tips and suggestions is a summary of Paul Kalanithi's #1 NEW YORK TIMES BESTSELLER When Breath Becomes Air.
What makes life worth living in the face of death? When Breath Becomes Air is an unforgettable, life-affirming reflection on the challenge of facing death and on the relationship between doctor and patient, from a brilliant writer who became both. | 677.169 | 1 |
Tag Archives: algebra formulas In this lesson we take a look at how Algebra Formulas and Equations are used in the real world. If we did not have Algebra formulas for "Kinematics", we would not have great games like "Angry Birds" | 677.169 | 1 |
Download and read online But Why Does It Work in PDF and EPUB If you ask students, "Why does that work?" do they know what you're asking and do you know what to listen for in their responses? Do you have images of what mathematical argument looks like in the elementary grades and how to help students learn to engage in this important practice? Do you have so much content to cover that finding time for this kind of work is difficult? But Why Does It Work? offers a simple, efficient teaching model focused on mathematical argument for developing the ability of students to justify their thinking and engage with the reasoning of others. Designed for individuals as well as study groups, this book includes access to classroom-ready instructional sequences, each built on a model supporting students in: noticing relationships across sets of problems, equations, or expressions articulating a claim about what they notice investigating their claim through representations such as manipulatives, diagrams, or story contexts using their representations to demonstrate why a claim must be true or not extending their thinking from one operation to another. Establishing a classroom culture where students gain confidence in their own mathematical voice and learn to value the contributions of their peers is a critical part of this work. The authors tell us, "If the idea underlying a student's reasoning is not made explicit, the opportunity for all students to engage in such thinking is lost." As students become a true community of mathematicians, they heighten each other's understanding by investigating questions, conjectures, and examples together. Enhanced with extensive video showing the instructional sequences in action-along with guiding focus questions and math investigations-But Why Does It Work? is a flexible approach that will help students confidently articulate and defend their reasoning, and share their deep thinking with others.
Download and read online Connecting Arithmetic to Algebra in PDF and EPUB "To truly engage in mathematics is to become curious and intrigued about regularities and patterns, then describe and explain them. A focus on the behavior of the operations allows students starting in the familiar territory of number and computation to progress to true engagement in the discipline of mathematics." -Susan Jo Russell, Deborah Schifter, and Virginia Bastable Algebra readiness: it's a topic of concern that seems to pervade every school district. How can we better prepare elementary students for algebra? More importantly, how can we help all children, not just those who excel in math, become ready for later instruction? The answer lies not in additional content, but in developing a way of thinking about the mathematics that underlies both arithmetic and algebra. Connecting Arithmetic to Algebra invites readers to learn about a crucial component of algebraic thinking: investigating the behavior of the operations. Nationally-known math educators Susan Jo Russell, Deborah Schifter, and Virginia Bastable and a group of collaborating teachers describe how elementary teachers can shape their instruction so that students learn to: *notice and describe consistencies across problems *articulate generalizations about the behavior of the operations *develop mathematical arguments based on representations to explain why such generalizations are or are not true. Through such work, students become familiar with properties and general rules that underlie computational strategies-including those that form the basis of strategies used in algebra-strengthening their understanding of grade-level content and at the same time preparing them for future studies. Each chapter is illustrated by lively episodes drawn from the classrooms of collaborating teachers in a wide range of settings. These provide examples of posing problems, engaging students in productive discussion, using representations to develop mathematical arguments, and supporting both students with a wide range of learning profiles. PLCs and book-study groups! Save $47.25 when you purchase 15 copies with the Book Study Bundle. Staff Developers: Available online, the Course Facilitator's Guide provides math leaders with tools and resources for implementing a Connecting Arithmetic to Algebra workshop or preservice course. For information on the PD course offered through Mount Holyoke College, download the flyer.
Download and read online The Mathematics Enthusiast Issue in PDF and EPUB The Mathematics Enthusiast (TME) is an eclectic internationally circulated peer reviewed journal which focuses on mathematics content, mathematics education research, innovation, interdisciplinary issues and pedagogy. The journal exists as an independent entity. It is published on a print-on-demand basis by Information Age Publishing and the electronic version is hosted by the Department of Mathematical Sciences- University of Montana. The journal is not affiliated to nor subsidized by any professional organizations but supports PMENA [Psychology of Mathematics Education- North America] through special issues on various research topics. Indexing Information: Australian Education Index; EBSCO Products (Academic Search Complete); EDNA; Directory of Open Access Journals (DOAJ); Psyc-INFO (the APA Index); MathDI/MathEDUC (FiZ Karlsruhe); Journals in Higher Education (JIHE); SCOPUS; Ulrich's Periodicals Directory; Emerging Sources Citation Index (Thompson Reuters)
Download and read online Proving in the Elementary Mathematics Classroom in PDF and EPUB Although proving is core to mathematics as a sense-making activity, it currently has a marginal place in elementary classrooms internationally. Blending research with practical perspectives, this book addresses what it would take to elevate the place of proving at elementary school. The book uses classroom episodes from two countries to examine different kinds of proving tasks and the proving activity they can generate in the elementary classroom. It examines further the role of teachers in mediating the relationship between proving tasks and proving activity, including major mathematical and pedagogical issues that arise for teachers as they implement each kind of proving task. In addition to its contribution to research knowledge, the book has important implications for teaching, curricular resources, and teacher education.
Download and read online Beyond Answers in PDF and EPUB The Standards for Mathematical Practice are written in clear, concise language. Even so, to interpret them and visualize what they mean for your teaching practice isn't always easy. In this practical, easy-to-read book, Mike Flynn provides teachers with a clear and deep sense of these standards and shares ideas on how best to implement them in K-2 classrooms. Each chapter is dedicated to a different practice. Using examples from his own teaching and vignettes from many other K-2 teachers, Mike does the following: Invites you to break the cycle of teaching math procedurally Demonstrates what it means for children to understand--not just do--math Explores what it looks like when young children embrace the important behaviors espoused by the practices The book's extensive collection of stories from K-2 classroom provides readers with glimpses of classroom dialogue, teacher reflections, and examples of student work. Focus questions at the beginning of each vignette help you analyze the examples and encourage further reflection. Beyond Answers is a wonderful resource that can be used by individual teachers, study groups, professional development staff, and in math methods courses.
Download and read online Teachers Professional Development and the Elementary Mathematics Classroom in PDF and EPUB This book illustrates the experiences of elementary school teachers across one year's time as they participated in a teacher development seminar focused on mathematics, and as a result changed their beliefs, their knowledge, and their practices. It explores these experiences as a means of understanding the learning that takes a teacher from a more traditional teaching practice to one that is focused on the ideas and understandings that students and teachers have of the subject matter. The work emerges from and reports on a unique data set from a two-year study of teacher learning that was funded by the Spencer and MacArthur foundations. The teachers, whose work is at the center of this study, were participants in the Developing Mathematical Ideas seminar (DMI), a mathematics teacher development seminar for elementary school teachers. This seminar is one example of intensive, domain-specific professional development. In this seminar teachers study elementary mathematics content to deepen their own understanding of it, they study the development among children of the ideas central to elementary mathematics, and they experience a teaching and learning environment consistent with the pedagogy envisioned by the National Council for Teachers of Mathematics' Principles and Standards for School Mathematics. The seminar is a nationally available teacher development curriculum, thus interested educators can gain access to the resources necessary to offer similar seminars in their own communities. Teachers' Professional Development and the Elementary Mathematics Classroom: Bringing Understandings to Light will be widely interesting to a broad audience, including mathematics teacher educators, teacher education researchers, policymakers, and classroom teachers. It will serve well as a text in a range of graduate courses dealing with teacher cognition/knowledge for teaching, mathematics methods, psychology of learning, and pedagogical theory.
Download and read online Best Practices for Elementary Classrooms in PDF and EPUB There is no better way to learn the craft of teaching than by watching an expert teacher at work. In this sequel to Randi Stone's Best Classroom Practices, nationally recognized, award-winning elementary teachers showcase selected practices from their classroom repertoire to share with their colleagues. Learn what it takes to build a productive, engaged community of learners from some of the nation's best teachers in their own words. This inspirational, one-stop guide covers everything from classroom management to teaching reading, writing, math, science, social studies, music, art, technology, and physical education. You will find: - Detailed, successful teaching strategies with lists of relevant standards and materials needed - Innovative activities, projects, lesson plans, and units of study for every content area - Classroom strategies across the curriculum, including ideas for involving parents and ways to make inclusion work Best Practices for Elementary Classrooms provides a wide array of excellent lessons to choose from, road-tested by your award-winning colleagues.
Download and read online The Formative 5 in PDF and EPUB Move the needle on math instruction with these 5 assessment techniques! Mathematics education experts Fennell, Kobett, and Wray offer five of the most impactful and proven formative assessment techniques you can implement—Observations, Interviews, "Show Me," Hinge Questions, and Exit Tasks— every day. You'll find that this palette of classroom-based techniques will truly assess learning and inform teaching. This book gives you a concise, research-based, classroom-dedicated plan with lots of tools to guide your daily use of The Formative 5. K-8 teachers will learn to Directly connect assessment to planning and teaching Engineer effective classroom questioning, discussions, and learning tasks Provide success criteria and feedback that moves students forward Includes a book study guide, samples, and a companion website with downloadables and multi-media examples.
Download and read online Elementary Mathematics Is Anything but Elementary Content and Methods From A Developmental Perspective in PDF and EPUB Inspiring, empowering, and preparing preservice teachers for today''s classroom, ELEMENTARY MATHEMATICS IS ANYTHING BUT ELEMENTARY: CONTENT AND METHODS FROM A DEVELOPMENTAL PERSPECTIVE is a comprehensive program that delivers both a content and a methods text. Serving as a professional development guide for both pre-service and in-service teachers, this text''s integrated coverage helps dissolve the line between content and methods—and consequently bolsters teachers'' confidence in their delivery of math instruction. A strong emphasis on the National Council of Teachers of Mathematics five core standards provides key information common to most state curricula relative to NCTM standards for pre-K through sixth grade. In addition, text content is based on thorough elementary mathematical scope and sequences that have been shown to be an effective means for guiding the delivery of curriculum and instruction. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.
Download and read online A Focus on Fractions in PDF and EPUB A Focus on Fractions is the first book to make cognitive research on how students develop their understanding of fraction concepts readily accessible and understandable to pre- and in-service K– 8 mathematics educators. This important resource assists teachers in translating research findings into their classroom practice by conveying detailed information about how students develop fraction understandings as well as common student misconceptions, errors, preconceptions, and partial understandings that may interfere with students learning. Using extensive annotated samples of student work, as well as vignettes characteristic of classroom teachers' experiences, this book equips educators with knowledge and tools to reveal students' thinking so that they can modify their teaching to improve student learning of fractions concepts. Special Features: End of Chapter Questions provide teachers the opportunity to analyze student thinking and consider instructional strategies for their own students. Instructional Links help teachers relate concepts from the chapter to their own instructional materials and programs. Big Ideas and Research Reviews frame the chapters and provide a platform for meaningful exploration of the teaching of fractions. Answer Key posted online offers extensive explanations of in-chapter questions. A Focus on Fractions bridges the gap between what mathematics education researchers have discovered about the learning of fraction concepts and what teachers need to know to make effective instructional decisions The Construction of New Mathematical Knowledge in Classroom Interaction in PDF and EPUB Mathematics is generally considered as the only science where knowledge is uni form, universal, and free from contradictions. "Mathematics is a social product - a 'net of norms', as Wittgenstein writes. In contrast to other institutions - traffic rules, legal systems or table manners -, which are often internally contradictory and are hardly ever unrestrictedly accepted, mathematics is distinguished by coherence and consensus. Although mathematics is presumably the discipline, which is the most differentiated internally, the corpus of mathematical knowledge constitutes a coher ent whole. The consistency of mathematics cannot be proved, yet, so far, no contra dictions were found that would question the uniformity of mathematics" (Heintz, 2000, p. 11). The coherence of mathematical knowledge is closely related to the kind of pro fessional communication that research mathematicians hold about mathematical knowledge. In an extensive study, Bettina Heintz (Heintz 2000) proposed that the historical development of formal mathematical proof was, in fact, a means of estab lishing a communicable "code of conduct" which helped mathematicians make themselves understood in relation to the truth of mathematical statements in a co ordinated and unequivocal way.
Download and read online Teaching and Learning Proof Across the Grades in PDF and EPUB A Co-Publication of Routledge for the National Council of Teachers of Mathematics (NCTM) In recent years there has been increased interest in the nature and role of proof in mathematics education; with many mathematics educators advocating that proof should be a central part of the mathematics education of students at all grade levels. This important new collection provides that much-needed forum for mathematics educators to articulate a connected K-16 "story" of proof. Such a story includes understanding how the forms of proof, including the nature of argumentation and justification as well as what counts as proof, evolve chronologically and cognitively and how curricula and instruction can support the development of students' understanding of proof. Collectively these essays inform educators and researchers at different grade levels about the teaching and learning of proof at each level and, thus, help advance the design of further empirical and theoretical work in this area. By building and extending on existing research and by allowing a variety of voices from the field to be heard, Teaching and Learning Proof Across the Grades not only highlights the main ideas that have recently emerged on proof research, but also defines an agenda for future study. | 677.169 | 1 |
Sociology is the study about the life of society. How people interact with each other and what factors influence their interaction — these are questions that sociology tries to answer. When you study sociology you will need to conduct thorough research. In this article, we are going to provide you with sociology topics for research project and we'll give you advice on how to cope with your project.
Math can't exist without problem solving. Problem solving is an important part of the curriculum. It presupposes that you should take on some responsibility for your own learning. A good problem solver should map out lots of dimensions of a problem and see it in the clearest way. Unfortunately, it can be difficult for may students. We are here to help you if you don't know how to solve math problems.
Are you taking up a Java class and you keep struggling with programming homework once in a while? There are many students who face the same problem. However, simply because there are so many students who have difficulties with their homework doesn't mean that you should have those problems as well. You can get assistance with your homework and forget about challenges which are associated with your assignment.
Statistics can be fun, especially if you are reading a bunch of amazing facts with pictures that pop up in your timeline. Something like: "An average human accidentally swallows up to eight spiders." But when it comes to doing your homework, especially if you have no primary data, you may need some statistics help, and we are always ready to provide it.
Information systems homework lets students practice skills, learn something new, and expand on thoughts which were presented in class. If your assignment seems too difficult, there are some ways of getting information systems homework help. Getting such help can reduce your stress and will save your time.
Why do students hate math homework so much? Is there any particular thing that makes studying algebra, geometry and calculus so horrible? Actually, yes. Here are just a few of these reasons and some practical suggestions on how to cope with them.
What Stops You From Being a Math Genius
Prejudice
Your parents, teachers, and even you probably believe that you are just "not a math person," and should try something else, like literature, for example, and that you will always be terribly bad at the exact sciences and in need of help with math homework.Continue reading →
Psychology seems to be a fascinating and less difficult subject compared to a subject like calculus. Many students take this course happily, thinking that it will be easy—that is, until they get their first homework assignment. Here is some advice on how to navigate the world of souls and dreams successfully, and where to get psychology homework help in case of an emergency. Continue reading →
You have been surfing the internet the whole day, and your research has finally been rewarded—you stumbled across a couple of math answers online, and the tasks look exactly like those given by your professor as your homework! Looks like a time for triumph, right? Here are a couple of reasons why you actually shouldn't use those free answers.
Humanity has always been trying to create a system of regulations which will help us to coexist peacefully. The history of law closely follows the history of civilizations. The first civil code, created in Ancient Egypt, dates back to 3000 BC. Even back then people tried to create a set of rules which would protect their lives and property. Since then, the world has seen Sumerian, Babylonian, ancient Greek and, of course, Roman law, which initially became the basis of many modern law systems. All these systems can be quite complicated and tangled, but remember that our law tutor can help you even with the most difficult questions.
English homework doesn't seem to be such a big deal. All you need to do is to learn some grammar rules, write a couple of sample sentences related to them, or write an essay, and you are done. But when you open your coursebook and read the task, it turns out that everything is not that simple and looking for some English homework help may not be a bad idea. Here are some useful life hacks for making the process of studying more enjoyable and rewarding. | 677.169 | 1 |
Help find introduction to college algebra
Plans include: Magazine Ads and You the Teenager, Propaganda Techniques, and Television and Violence. Messages and Meanings: A Guide to Algebrra Media. Activities for teaching middle and high school students how to access, analyze, evaluate and communicate media messages. A collaborative effort between the National The great kapok tree lesson plans on Economic Education and the Northwestern Mutual Life Foundation, this site offers lessons,quizzes, introduction to college algebra activities to teach secondary students basic economics and ihtroduction management. Mock Trial: The Titanic The site was designed for teachers and students to participate in a mock trial involving the tragic story of the Titanic. Money Equivalents Activities (K-3) Activities prepared by the Bank Street College of Education to help teach children the value of money and its different denominations.
Help find introduction to college algebra
Students that are going to East, they algevra the necessary supplies for class. If these things are a concern, please let introduction to college algebra know.
Help find introduction to college algebra
Making judgments based on criteria and standards through checking and critiquing. One of the introduction to college algebra identifies The Knowledge Dimension (or the kind of knowledge to be learned) while the second identifies The Cognitive Process Dimension (or the process used to learn). As represented on the grid below, the intersection of the knowledge and cognitive process categories form twenty-four separate cells as represented on the "Taxonomy Table" below. The Knowledge Dimension on the left side is composed of itnroduction levels that are defined as Factual, Conceptual, Procedural, and Meta-Cognitive. The Cognitive Process Dimension across the top of the grid consists of six levels that are defined as Remember, Understand, Apply, Analyze, Evaluate, and Create. Each level of both dimensions of the table is subdivided. Each of the four Knowledge Dimension levels is subdivided into either three or four categories (e. | 677.169 | 1 |
Maths
You can study Mathematics at Higher, Standard or Mathematical Studies Level.
Higher Level prepares you for university courses in Mathematics, Physics, Computing, Engineering and Science.
Standard Level prepares you for courses with some mathematical content such as Economics or Medicine.
Mathematical Studies enables you to use and apply Mathematics, leaving you better prepared for wider variety university and all career paths.
Mathematical and independent thinking skills are developed throughout the IB Diploma courses. We develop an appreciation of how different topics are linked. The courses are rigorous and engender a deeper understanding of concepts that prepare you for the future. | 677.169 | 1 |
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ISBN: 0201726343
Edition: 5
Publication Date: 2003
Publisher: Pearson
AUTHOR
Ralph P. Grimaldi
SUMMARY
This text is organised into 4 main parts - discrete mathematics, graph theory, modern algebra and combinatorics (flexible modular structuring). It includes a large variety of elementary problems allowing students to establish skills as they practice.Ralph P. Grimaldi is the author of 'Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition', published 2003 under ISBN 9780201726343 and ISBN 0201726343 | 677.169 | 1 |
You don't need to ask anybody to solve any sample questions for you; as a matter of fact all you need is Algebrator. I've tried quite a few such math simulation software but Algebrator is a lot better than most of them. It'll solve all the questions that you have and it'll even explain each and every step involved in reaching that answer. You can work out as many examples as you would like to, and unlike us human beings, it won't ever say, Oh! I've had enough for the day! ;) Even I had some problems in solving questions on converting fractions and radical expressions, but this software really helped me get over those.
I remember having difficulties with function domain, linear inequalities and angle-angle similarity. Algebrator is a really great piece of math software. I have used it through several math classes - Algebra 1, Intermediate algebra and Remedial Algebra. I would simply type in the problem and by clicking on Solve, step by step solution would appear. The program is highly recommended. | 677.169 | 1 |
AB Calculus is a course designed for teachers who would like teaching strategies, methods and materials on how to prepare students for success in Advanced Placement Calculus. During this week, participants will learn how to introduce challenging concepts utilizing a variety of activities, incorporate the new pedagogy and topics of the Advanced Placement program into their curriculum, and integrate technology and the internet for a more interactive classroom. In addition, teachers will collaborate together on discovery lessons, learn techniques to assist those students who are under-prepared, and play teacher-made games to emphasize retention of the calculus. Two major themes, rigor in the classroom and student thinking, will be stressed and modeled all week. You will need to bring a jump drive, and most importantly, your energy and enthusiasm to participate in this fun-filled learning experience!
Items to Bring
flash drive
Biographical Sketch for Stacey McMullen
I taught mathematics over 19 years before I finally got the opportunity to teach Advanced Placement Calculus. I always loved teaching high school math, but after my first day of teaching AP Calculus, I thought, "I was born to do this!" I love the age group, the mathematical banter of calculus back and forth with the students, and the competitiveness of an AP exam. I taught AP Calculus for the past 17 years, gaining experience in two districts vastly different in the makeup of their student populations. Therefore, I have developed a wide variety of strategies and techniques to assist students in becoming successful in AP Calculus.
This is my second year to enjoy my retirement from my position as an AP Calculus Lead Teacher in an inner city district in Dallas, Texas. I taught AB/BC Calculus at two high schools and mentored the calculus teachers at five other high schools. In addition, I wrote the calculus curriculum for the district, taught and trained vertical teams, and did hours upon hours of tutoring. I thrived in these low-income schools getting favorable results from students who lacked the necessary mathematical background and resources.
Now that I have more time, I am exploring an additional professional avenue with a new home-based business while continuing to grow my educational consulting business. However, I am not all work and no play! I am taking great pleasure in having NO alarm clock, traveling, visiting with friends, and spending more time with one not so nice cat and two lively, loving dogs. | 677.169 | 1 |
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This full-colour Student Book covers all core and optional units for the AQA AS and A Level Geography specification for first teaching from September 2016. Students are encouraged to develop links between physical and human topics, understand systems, processes, and acquire geographical skills. Helping to bridge the gap from GCSE to A Level, it also provides support for fieldwork skills and for the geographical investigation at A Level. A 'Maths for geographers' feature helps students develop and apply their mathematical and statistical skills, and a range of assessment-style questions support students in developing their exam skills.
Covers all units of the new AQA A/AS Level Geography specification and assessment objectives, and has been written specifically for the new linear qualification.
Includes a range of assessment-style questions to support students in developing their exam skills.
Endorsed by the Geographical Association and provides dedicated support at the appropriate level for fieldwork skills, and for the geographical investigation at A Level.
Includes support for bridging the gap from GCSE to A Level, and from Key Stage 3 for students without GCSE Geography.
Supports students in developing critical evaluation and independent learning skills, important for A Level and essential for progression to Higher Education.
Contemporary case studies include questions to test students' understanding of the issues and relevant application of knowledge.
Includes a 'Maths for geographers' feature to support students in developing and applying their mathematical and statistical skills.
A 'Thinking like a geographer' feature includes questions inviting students to critically evaluate and assess links between topics.
'Investigate' features give ideas for research and investigation to support students in developing independent learning skills. | 677.169 | 1 |
One-Day MATLAB Workshop, Thursday 3/17/2016
The Workshop is now at capacity and registration has been closed.
The Center for Computational Science is pleased to announce the next MathWorks workshops. These sessions are independent of each other and will teach two different sets of skills. You may register for one or both, but please note that the morning is at the College of Engineering at the Coral Gables campus, and the afternoon session is at the Miller School of Medicine.
Advanced Programming Techniques in MATLAB
In this session you will gain an understanding of how different MATLAB data types are stored in memory and how you can program in MATLAB to use memory efficiently. In recent versions, MATLAB introduced several new programming concepts, including new function types. We will illustrate and explore the usage and benefits of the various function types under different conditions. You will learn how using the right function type can lead to more robust and maintainable code. Demonstrations will show you how to apply these techniques to problems that arise in typical applications such as image processing, optimization, and data analysis, including big data.
Highlights include:
Memory handling in MATLAB
Various function types
Please bring your own laptop (any OS).
This workshops is capped at 30 participants, and is on a first-come, first-served basis. Please register early to avoid disappointment.
Mathematical Modeling and Simulation with MATLAB
Mathematical models are critical to understanding and accurately simulating the behavior of complex systems. They enable important tasks such as forecasting system behavior, characterizing system response, and designing control systems. Attend this seminar to find out how you can use MATLAB and add-on products for your mathematical modeling tasks. You will also learn about different approaches that can be used to develop models. This includes leveraging pre-built graphical tools for specific modeling tasks such as parametric surface fitting, building and optimizing models using the MATLAB language, and deriving system equations using symbolic computation. | 677.169 | 1 |
Teaching the Matrix Method of Solving a System of Linear Equations
I taught this topic to my students per the textbook, and about half the class was really struggling. Now, most of these students were OK on solving these systems using equations with x, y and z. Why were they not getting the matrix method?
I stopped and rethought the whole enterprise, and realized that the textbook introduced a few rules, without really explaining them, and then jumped straight into somewhat complex problems to solve. Why not follow Carl Menger, and break everything down to the simplest possible elements, and build up from there? So I spent half a class walking the students through the following twelve systems, asking them to write the matrix, and solve it if need be. (Some of the systems are already in the solved form, which was part of the point of the exercise.)
1) x = 7
2) 2x = 14
3) x = 7, y = 4
4) 2x = 14, 2y = 8
5) x + y = 7, y = 4
6) x + y = 7, 2x + y = 10
7) x + y = 7, x + 2y = 11
8) x + y = 7, 2x + 4y = 22
9) 2x + 2y = 14, 2x + 4y = 22
10) x = 3, y = 4, z = 2
11) x + 2z = 7, y + z = 6, z = 2
12) x + 2z = 7, 2y + 2z = 6, y + 3z = 10
Every single student got it. A couple of them broke into big smiles at around equation 8 or 9. One told me, "I had studied this several times before, and I thought I'd never understand it."
So why aren't math textbooks written this way? We teach young kids arithmetic in this fashion: start from 1 + 1 = 2, and work systematically up from there. But at a certain point, it seems to me, math textbooks instead take on a "sort the wheat from the chaff" approach: present the material in a very elliptical fashion, and those who can't get it that way, well, they shouldn't keep going in mathematics!
Me, I think every person on earth with a normally functioning brain can grasp any piece of established mathematics as long as it is presented systematically, with every step along the way explained with simple examples. (And of course, having the creative genius to actually advance mathematics is an entirely different matter.) | 677.169 | 1 |
GCNU1005 Beating the Odds
Course Code
Offering Department(s)
Units & Hours
GCNU1005
MATH
(3,3,0)
What is the course about?
GCNU1005 begins with a list of well-selected and counterintuitive examples to help students become aware of the existence of mathematics in every aspect of our lives. Chance plays a huge part in life; one will have a better management of risk and opportunities and hence higher odds to become a winner in life if he knows more about how probability works. Rather than focusing on calculating a few specific combinations or permutations, which are tiresome and tedious to most, this course uses real-life situations as incentives and utilizes mathematics as a tool to figure out the "Best Bet" in some everyday problems. Casinos and games are topics commonly seen in probability; after equipping students with the knowledge necessary for identifying the "Best Bet", we move on to a very practical problem-knowing that the chance of winning is 99.9%, how much should we invest on this "Best Bet"? Our investigations (not the formulas and calculations) are then extended to cover more real-life situations in which its odds cannot be predicted by counting techniques. Students will discover new ways of expressing known information, connecting reality to math, applying meta-tools to predict chances, and making statistically-justifiable decisions. | 677.169 | 1 |
itative Aptitude for Competitive Examinations is a comprehensive book for candidates preparing for various competitive examinations. The book comprises of chapters on number system, HCF and LCM, fractions and decimals, percentage, profit and loss, ratio and proportion, time, work and wages, time and distance, linear equations, geometry and trigonometry. In addition, the book consists of several solved and unsolved questions and model test papers for thorough revision and final practice. This book is essential for candidates appearing for exams like IBPS PO, Railway Recruitment examination, CTET, MAT and GRE.
About McGraw-Hill EducationRead More
Specifications
Book Details
Publication Year
2014 June
Edition Type
Revised
Book Type
Entrance Exam Book
Table of Contents
1. Number System
2. Number Series
3. HCF and LCM of Numbers
4. Fractions and Decimals
5. Square Root and Cube Roots
6. Indices and Surds
7. Vedic Mathematics
8. Simplification and Approximation
9. Problems on Ages and Numbers
The book is brilliant as a reference text book and has formulae along with shortcuts and special results. There is a much needed chapter on vedic maths as well. Although there are not as many examples and practice sums, as a text book for the reference of formulae and in general it is perfect for Banking and SSC exams.
It is good but so many mistakes are there.Solutions aren't so good. But the good thing about this book is this book is contained so many chapters & they are well organized.Each & every chapters questions are divided into two parts--1.Regular problems & 2.challenging problems.There are so many solved examples, formulas & lots of shortcuts which helps at a good account.So, I'll suggest you to buy this book. | 677.169 | 1 |
Prologue
Setting Hands-On Math Apart
Some Definitions from the Web
Science (from the
Latin scientia, meaning "knowledge") refers to any systematic knowledge
base or prescriptive practice that
is capable of resulting in a prediction or predictable type of outcome. In this
sense, science may refer to a highly skilled technique or practice.
Physics (Greek:physis
– φύσις meaning "nature") is a natural science it is the study of matter and its
motion through space-time
and all that derives from
these, such as energy and force.
More broadly, it is the general analysis of nature,
conducted in order to understand how the world and the
universe behave.
Mechanics (Greek
Μηχανική) is the branch of physics concerned with the behaviour of
physical bodies when subjected to forces or displacements, and the subsequent effect
of the bodies on their environment. The discipline has its roots in several ancient
civilizations. During the early modern period, scientists such as Galileo,
Kepler, and especially Newton, laid the foundation for what is now known as classical
mechanics.
Mathematics is
the science and study of quantity, structure, space, and change. Mathematicians
seek out patterns,
formulate new conjectures, and
establish truth by rigorous deduction from
appropriately chosen axioms and definitions.
Albert Einstein,
on the other hand, stated," as far as the laws of mathematics refer to reality,
they are not certain; and as far as they are certain, they do not refer to reality."
Applied Mathematics is
a branch of mathematics that concerns itself with the mathematical techniques
typically used in the application of mathematical knowledge to other domains.
Engineeringis the science, discipline,
art and profession of acquiring and applying technical, scientific and mathematical
knowledge to design and implement materials, structures, machines, devices, systems and processes
that safely realize a desired objective or inventions.
A Laboratory
(informally, lab) is a facility that provides controlled conditions in which
scientific research; experiments
and measurement may be performed.
A Common Teaching Method
Commonly, the topics and of mathematics, physics and engineering are taught separately.
Within a topic there will be history, concepts, logic-based derivations, examples worked
by the teacher and exercises to be worked by the student and graded by the teacher. Additionally, a topic may have an associated laboratory in which the students
are required to experiment, measure and report.
Setting HOM Apart
HOM is a scientific approach to conveying an understanding of physics
and mathematics
that employs "do it yourself" observations
on models that are based on the contributions to physics of such as Galileo,
Kepler, Newton and Lord Rayleigh, the contributions to applied mathematics by such
as Pythagoras and Gauss and the contributions of engineers such as Bush.
HOM's primary emphasis is on the use of applied mathematics in a computer-laboratory
environment to construct models of the physical relationships of force,
matter and motion in an atmosphere that provides both buoyancy and resistance to
motion. The use of a model atmosphere provides computer modeling that closely emulates
the physics of our real world. An even better model would include the effects
of Lift.
Deep understanding is acquired by constructing a model of a physical process,
by confirming that the model behaves in accord with its design objectives, by employing
the model to explore the laws and relationships that have been modeled and by using
the model to provide answers to new questions. Each of the foregoing factors,
constructing, confirming, exploring, and responses to "what if", makes a contribution
to a student's understanding of the behaviour of the world and universe in which
he lives. | 677.169 | 1 |
James available in the ebook version., International MetricApplication-oriented introduction relates the subject as closely as possible to science with explorations of the derivative; differentiation and integration of the powers of x; theorems on differentiation, antidifferentiation; the chain rule; trigonometric functions; more. Examples. 1967 edition.
This text helps students improve their understanding and problem-solving skills in analysis, analytic geometry, and higher algebra. Over 1,200 problems, with hints and complete solutions. Topics include sequences, functions of a single variable, limit of a function, differential calculus for functions of a single variable, the differential, indefinite and definite integrals, more. 1963 edition.This general review covers equations, functions, and graphs; limits, derivatives; integrals and antiderivatives; word problems; applications of integrals to geometry; and much more. Additional features make this volume especially helpful to students working on their own. They include worked-out examples, a summary of the main points of each chapter, exercises, and where needed, background material on algebra, geometry, and reading comprehension.
The goal of this text is to help students leam to use calculus intelligently for solving a wide variety of mathematical and physical problems. This book is an outgrowth of our teaching of calculus at Berkeley, and the present edition incorporates many improvements based on our use of the first edition. We list below some of the key features of the book. Examples and Exercises The exercise sets have been carefully constructed to be of maximum use to the students. With few exceptions we adhere to the following policies. • The section exercises are graded into three consecutive groups: (a) The first exercises are routine, modelIed almost exactly on the exam pIes; these are intended to give students confidence. (b) Next come exercises that are still based directly on the examples and text but which may have variations of wording or which combine different ideas; these are intended to train students to think for themselves. (c) The last exercises in each set are difficult. These are marked with a star (*) and some will challenge even the best students. Difficult does not necessarily mean theoretical; often a starred problem is an interesting application that requires insight into what calculus is really about. • The exercises come in groups of two and often four similar ones.
Calculus: The Language of Change is an innovative new introductory text that blends traditional and reform approaches, and focuses on understanding calculus as its own language. With accessible writing and presentation, the text allows students to gradually understand the language – first by reviewing vocabulary, and then by quickly moving to present calculus conceptually, computationally, and theoretically. Within this framework, derivatives and integrals are developed side by side, coverage of theory is offered at various levels, and computing devices are incorporated generically. A full range of student and instructor resources make Calculus: The Language of Change an outstanding course package. Even as he initiates us into the mysteries of real numbers, functions, and limits, Berlinski explores the furthest implications of his subject, revealing how the calculus reconciles the precision of numbers with the fluidity of the changing universe. "An odd and tantalizing book by a writer who takes immense pleasure in this great mathematical tool, and tries to create it in others."--New York Times Book Review From the Trade Paperback edition. | 677.169 | 1 |
Be sure that you have an application to open
this file type before downloading and/or purchasing.
2 MB|5 pages
Product Description
These reference posters cover trinomial factoring where A=1. They are for students just learning how to factor, students having trouble factoring or Special Education Algebra and Algebra 2 students. I post these for my self-contained Special Education Algebra 2 students.
There are 2 process posters and 2 example posters. The first process poster shows the 2 scenarios for factoring when C is positive. A second process poster outlines factoring when C is negative. You can see all posters in the preview.
Two example posters give examples for when C is positive (and B is -) and when C is negative (and B is also -). These two scenarios are trickiest for my students to remember without a reference.
Also included is a student handout that walks students through the process of factoring a trinomial in all scenarios. | 677.169 | 1 |
Showing 1 to 30 of 53
Math 152
Exam #2 Study Guide
I.
Strategy for Applications of Integration
Understand the steps of the general strategy.
Know why and when it works, and also when it does not work.
Be flexible and creative in cutting the problem into small pieces.
II. Easy
Math 152 Test #1
Winter 2017
Name:_ to get credit
Math 152
Winter 2017
NAME:
QUIZ 3
5.1,5.2,5.3
20 points
Answer the following questions. Show work in order to earn full credit. You may use a calculator, but no
notes, books, or other resources. Give answers in exact (not decimal) form. 5 points each.
You
Math 152
Winter 2017
NAME:
QUIZ 1
4.0, 4.1, 4.2 & 4.3
20 points
Answer the following questions. Show work, where possible, in order to earn full credit. You may use
a calculator, but no notes, books, or other resources.
1. (4.0 #13) [3 pts] The figure toMath 152 Winter 2016 in
Math 152D Winter 2015 i
Math 152
Final Exam Study Guide
I.
II.
III.
IV.
V.
VI.
Basic Concepts about Integration
Understand what is meant by an antiderivative or an indefinite integral, and be able to find
the C given enough information.
Know what is meant by area under a curve o
Math 152
Exam #3 Study Guide
I.
Techniques for Integration
Be able to apply any of the basic techniques from Chapter 5 (recognition, rules, rewriting,
guess and check, u-substitution).
Know and be able to apply the techniques of integration by parts, trig
Math 152
Exam #1 Study Guide
I.
II.
III.
IV.
V.
Antiderivatives
Understand what is meant by an antiderivative, the most general antiderivative, and the
indefinite integral.
Be able to use all of the basic techniques for finding an antiderivative: recognit
Math 152
Winter 2017
NAME:
100 points
Test 2
Show work in order to earn credit. You may use a calculator, but no notes, books, or other
resources. Give answers in exact (not decimal) form and use correct notation! Include units.
SA =
1.
[8 pts] SET
UP, bu
Math 152
Winter 2017
NAME:
QUIZ 2 A
4.4 - 4.7
20 points
Answer the following questions. Show work, where possible, in order to earn full credit. You may use a
calculator, but no notes, books, or other resources.
1. [6 pts] Use Antiderivates (and the FTC)
Math 152 Test #1
Spring 2016
Name:_
Instructions:
Math 152
Winter 2017
NAME:
QUIZ 5
7.5, 8.1, 8.2 & 8.3
20 points
You must show work and ALL steps of integration in order to earn credit for your answer. Calculator
estimations are not worth points. No notes, books, or other resources. Each problem is wort:3:
Math 152
Group Members:
Due: Friday at the end of class
upper integration
Z endpoint
b
' f(X)dX
{ a ,
Definition of the Definite Integral:
It 11311
ll'pll—m _
'P is a partition ef the interval [a] b], then we sayr f is integrable 011
the interval
Math 152 Name
4.7 Activity: Applications of Definite Integrals Group Members:
(' 10 points
Due: Friday by the end of Class
1. Sketch a detailed gTaph and find the area between the graphs of f (x) = J; and g(x) = x on 0 S x S 4.
If T(x)ZB(x)fora 513?)
Calculus 2 Advice
Showing 1 to 2 of 2
She is a great teacher who actually goes at the pace of the students. The homework load is minimal, extra credit is provided. As long as you participate in class, pay attention to what she is talking about, it should be an easy A. This class has you practice solve problems and work with the students around you rather than sit and watch a lecture. You also don't have to buy a book or as a matter in fact read it, just participating in class, doing her daily activities should be plenty to receive a good grade.
Course highlights:
The best part would be the second to last section where you just practice different ways to integrate functions. You get to see how much progress you've learned, also during that section you don't see very many work problems or rotational problems.
Hours per week:
3-5 hours
Advice for students:
She opens up her homework early and VPA(daily pre class quiz) a week early. Try to do them as soon as it opens up so that you don't have to worry about it later. Also the VPAs have videos attached that are extremely helpful. If you get a couple people to ask her about extra credit, she will provide them so don't be afraid to ask! | 677.169 | 1 |
Areteem Math Challenge II-A Bundle
Advanced Problem Solving: Preparation for the AMC 10 Exam. Bundle includes all four MC II-A courses at a discounted price!
off original price!
The coupon code you entered is expired or invalid, but the course is still available!
This course is for students who are preparing for the AMC 10 contest. Students are required to have fundamental knowledge in Algebra I, Geometry, Basic Number Theory and Counting and Probability up to the 10th grade level. Topics include polynomials, inequalities, special algebraic techniques, triangles and polygons, collinearity and concurrency, vectors and coordinates, numbers and divisibility, modular arithmetic, advanced counting strategies, binomial coefficients, sequence and series, and various other topics and problem solving techniques involved in math contests such as the AMC 10, advanced MathCounts, ARML, and ZIML Math Challenge II-A: Number Theory
Advanced Problem Solving in Number Theory: Preparation for the AMC 10 Exam | 677.169 | 1 |
About this product
Description
Description
This book is designed as a text for a first-year graduate algebra course. As necessary background we would consider a good undergraduate linear algebra course. An undergraduate abstract algebra course, while helpful, is t necessary (and so an adventurous undergraduate might learn some algebra from this book). Perhaps the principal distinguishing feature of this book is its point of view. Many textbooks tend to be encyclopedic. We have tried to write one that is thematic, with a consistent point of view. The theme, as indicated by our title, is that of modules (though our intention has t been to write a textbook purely on module theory). We begin with some group and ring theory, to set the stage, and then, in the heart of the book, develop module theory. Having developed it, we present some of its applications: canical forms for linear transformations, bilinear forms, and group representations. Why modules? The answer is that they are a basic unifying concept in mathematics. The reader is probably already familiar with the basic role that vector spaces play in mathematics, and modules are a generaliza- tion of vector spaces. (To be precise, modules are to rings as vector spaces are to fields. | 677.169 | 1 |
Prealgebra
ISBN-10: 0321213785
ISBN-13: 9780321213785 Prealgebrais based on Bob Priorrsquo;s own varied teaching experiences, and is designed to serve the needs of today's developmental math student and classroom. Bob knows that because today's students don't always have a lot of "face time" with their instructors, a usable, thorough, easy-to-follow text is key to their success. He draws students into the book (and not just the exercise sets!) by incorporating practice opportunities throughout the body of text. Thorough explanations and examples explain the "why" behind the mathematics, and patiently develop each concept | 677.169 | 1 |
Beginning and Intermediate Algebra: The Language & Symbolism of Mathematics
Beginning and Intermediate Algebra: The Language and Symbolism of Mathematics emphasizes what great mathematicians had identified for generations - mathematics is everywhere! Authors James Hall and Brian Mercer believe active student involvement remains the key to learning algebra. Topics in the text are organized by using the principles of the AMATYC standards as a guide, giving strong support to teachers using the text. The book's organization and pedagogy are designed to work for students with a variety of learning styles and for teachers with varied experiences and backgrounds. The inclusion of the "rule of four" or multiple perspectives -- verbal, numerical, algebraic, and graphical -- has proven popular with a broad cross section of students.
A key supplement for the text are the Lecture Guides. This supplement by the authors, with the assistance of Kelly Bails of Parkland College, provides instructors with the framework of day-by-day class activities for each section in the book. Each lecture guide can help instructors make more efficient use of class time and can help keep students focused on active learning. Students who use the lecture guides have the framework of well-organized notes that can be completed with the instructor in class.
Specifications of Beginning and Intermediate Algebra: The Language & Symbolism of Mathematics | 677.169 | 1 |
SOLUTIONS
This material may be reproduced for testing or instructional purposes by people using the text. Printed
ISBN 0 471 02396
in the
United
5
States
of America
10 9 8 7 6 5 4 3 2
Inttoduction
1
1. The Solutions In This Manual. The solutions of all the exercises in the text are given in full. The primary reason is to save professors' time. Choosing exercises for homework assignments can be a laborious matter if one must solve fifteen, twenty or more to determine which are most suitable for his class. A glance at the solutions will expedite the choices. The second reason is that in many institutions calculus is taught by teaching assistants who have yet to acquire both the training and experience in handling many of the mathematical and physical problems. The availability of the solutions should help these teachers. 2. Suggestions For The Use Of The Text. The one-volume format of this second edition should give professors more latitude in the choice of topics which might be suitable to the interests of the students or to the length of the course. Several types of choices might be noted. Because precalculus courses have become more common since the pUblication of the first edition, some of the analytic geometry topics may no longer have to be taught in the calculus course. The most elementary topics of analytics have been put in an appendix to Chapter 3, Section 4 of Chapter 4, Section 5 of Chapter 7, and the Appendix to Chapter 7. If familiar to the students, all or some can be omitted. Though I believe strongly in the importance of physical and, more generally, real applications to supply motivation and meaning to the calculus, again class interests and available time must enter into determining how many of these applications can be taken up. I have therefore starred all Lhose sections and chapters which can be omitted without disrupting the continuity. The last chapter, which is intended as an introduction to the theory or rigor, can be taken up at almost any point after Chapter 10. However, I personally believe that the intuitive approach should be maintained throughout and that this chapter should be left for the last and then taken up only if time permits. The complete text is intended for a three semester, three hours a week course. However, in view of the number of sections and chapters that are not essential to the continuity the text can be used for shorter courses including those offered in the fourth high school year. 3. Some Additional Topics. Some physical applications which were included in the first edition were omitted in,the second one and replaced in the text proper by applications to economics and to other social science areas. A few of those omitted are reproduced here. They may be useful as suggestions
2
for additional work which bright or somewhat advanced students can undertake, as fill-ins for periods which for one reason or another cannot be used for regular work, or as material for a mathematics club talk. Exercises and solutions relevant to these additional topics are also included here. A. The Hanging Chain.
In the text proper we derived the equation of the chain or cable suspended from two points (Chap. 16, Sect. 4) on the assumption that the weight per unit length of the cable is the same all along the cable. However, the theory developed there can be used to solve more general problems. One is to determine the shape of the cable if the weight per unit length or, one can say, the density per unit length is specified. The second is, given the desired shape of the cable, how can we fix the distribution of the mass along the cable so that it assumes the desired shape? Both of these problems are readily solved with the theory at hand. The derivation of (21), the equation of the cable, in the text proper, presupposed that the weight of the cable per unit foot is constant all along the cable, Let us now see what we can do when we let the weight of the cable vary from point to point. Let us denote by w(s) the function that gives the weight per unit foot at point s. Then (11) and (13) still hold, but (14) must be changed to read
(I)
If we divide obtain (2) this
T
y
=
fw(s)dx by
+
o.
that T /T is y', we yx
equation y'
(11) and use the fact
=
Tl fw(s)ds
where 0' is O/TO' If the function w(s) is given, we can calculate !w(s)ds, The quantity 0' can now be fixed by letting s be 0 at y' O. We now have y' as a function of s. Next we may proceed as we did in the case where w(s) is a constant and seek to obtain s as a function of x through
o
+ 0'
=
ds = dx
11 + y'2
but y' is now given by (2), If the integration can be performed and s is obtained as a function of x, we can substitute this value of s in (2) and attempt to obtain y as a function of x. We can also solve the second problem. Suppose that we wish to distribute weight along the cable so that the cable hangs in a given shape; that is, we presume that we know the equation of the cable and we wish to find w(s). To solve this problem, we differentiate (2) with respect to x. On the left side differentiation with respect to x produces y", On the right side to differentiate with respect to x we use the chain rule and differentiate with respect to s and multiply by ds/dx. The derivative of !w(s)ds with respect to s must be w{s) because the integral is that function whose derivative is w(s). Thus our result is
3
(3 )
y" = l:.wts)~. TO dx
Because we presume that we know the equation of the curve, we can calculate y" and ds/dx. Hence we can find w(s), that is, the variation of weight along the curve that produces the particular~shape of the hanging cable. Of course, the shape of the cable need no longer be a catenary. It is often called a non-uniform catenary. The theory presented in this section is useful under more general conditions than those so far described. In the derivations of the text and of (2), we attributed the weight to the cable. However, the weight w(s) might be the load on the cable, that is, the load of the bridge itself, if the cable's weight is negligible, or the combined weight of cable and load. In the case of the theory in the text this load would have to be proportional to the arc length of the cable; that is. the load would have to be the same for each unit of length of the cable. In the case of (2), the load could vary along the cable or the combined weight of load and cable could vary along the cable, and the function w(s) would have to represent the variation of the total weight with arc length. Exercises: 1. Find the law of variation of the mass of a string suspended from two points at the same Ie vel and acted upon by gravity so that it hangs in the form of a semicircle. Suggestion: Take the semicircle to be the lower half of X2+y2 = 2ay and use (3). 2. The derivation given in (2) for a cabJe whose load varies with arc length applies also to a cable whose load varies with horizontal distance from, say, the lowest point. Thus T = T and (1) becomes of the cable. 2)/12T Ans. y = (ax +6bx . o 3. A heavy chain is suspended at its two extremities and forms an arc of the parabola y = x2/4p. Show that the weight per horizontal foot is constant. Suggestion: Use (3).
4
y per horizontal
T
=
Jw(x)dx+D.
Then foot
(2) is y'
=
x
0
(l/T )fW(X)+D'.
0
Given
that
the load
is w(x)
=
ax2+b,
find
the equation
Solutions: 1. The lower half of the semicircle is given y' = x(a2_x2)-1/2 and y" = a2(a2_x2)-3/2, Then from (3), w(s) = aTo/(a2-xZ). integrations by y ds/dx
= a-/a2-x2• Then = Il+y,2:;a(a2_x2)-1/2.
that y' and yare Now use
2. Carry out the obvious o at x = O.
and use the facts
3. We can think of w(s)ds/dx as a function w(x) of x since (3). Since y = x2/4p, y" = ~p, and w(x) is a constant. B. Projectile Motion in a Resisting Medium.
s is.
After taking up projectile motion in a vacuum (Chap. 18, Sect. 4) one can take up the case of motion in a resisting medium. Since the
we shall apply Galileo's principle and consider the horizontal and vertical motions separately.Ky. we obtain
(1)
x
=
-kx. Since
x
=
(1) we first write
x
dv
dt
=
-kv
x
. the horizontal acceleration is the time derivative of the horizontal velocity so that
x
y. Hence Newton's second law says for the horizontal motion that ma = -KX. where K is the proportionality constant. The effect of air resistance on the motion of projectiles was first investigated seriously by Newton. and Euler. Huygens. If we divide both sides by m and replace
Kim by k. the resistance of the air is proportional to the velocity and directed opposed to that velocity. namely the air resistance -Kx. Suppose that the projectile is shot out with an initial velocity of magnitude V inclined at angle A to the ground. the combination of the two and the implications for projectile motion are new and provide an interesting comparison with projectile motion in a vacuum.
In the case of the vertical motion there are two forces acting at an¥ time t. As in the text. where horizontal and vertical motion in a resisting medium were considered independently. of the air resistance (the upward direction is positive). However. that the net force acting must equal the mass times the acceleration of the projectile. apart from the use of parametric equations. We shall suppose that as the projectile travels through the air. there is a horizontal force acting at any time t. Now. -Ky.3 2m
If we divide we obtain
(2 )
. namely. we use Newton's second law of motion. However. This initial velocity does give the projectile a constant horizontal velocity of V cosA but no acceleration in the horizontal direction and therefore no continuously acting force in the horizontal direction. Hence the differential equation for the vertical motion is. Sections 6 and 7. the mathematics involved. Hence the air resistance. is no more than what was taken up in Chapter 12.4
analysis of projectile motion breaks down into a separate consideration of the horizontal and vertical motions. because it is oppositely directed.
mx = -Kx. We shall apply this law to the horizontal and vertical motions separately. should have the components -Kx and -Ky. To obtain the parametric equations of the motion.
equation by m and again replace
both
sides
of this
Kim by k. where x and yare functions of t that represent the horizontal and vertical motion and the dot means differentiation with respect to t. the force of gravity which is -32m and the vertical component. the horizontal and vertical components of the velocity at any point of the projectile's path are and respectively. by Newton's second law.
y=
To integrate vx' we have
-32 -
kyo
it in a more familiar form. However.
my = _.
(See also the work on infinite series in Chap. A few of these comparisons will be left fo~ the exercises. 1). to the gun (Fig. 20.
We solve for C and substitute its value in the preceding equation.A to the ground. It would.. and moving in the resisting medium as opposed to a vacuum takes less time to reach maximum height. Moreover. Sect.) At the moment we might mention that the projectile fired with a fixed initial velocity V. of course. reaches maximum height closer s
Path in vacuum
Figure 1.?\ (1 _ -kt) Y=. y= or (7) 32t + (V sin A + l.6
y . and attains less maximum height than if fired in a vacuum.: v sin A e -kt k Because y = 0 when t = 0.
v sin A
k
_
k
3~
+
C. at a fixed angle. These formulas should be compared with (8) and (9) respectively of the text. be interesting to determine what effect air resistance has by comparing results obtained here with the results obtained for projectile motion without air resistance. the maximum height H2 is attained beyond the midpoint of the range. 12.
0=-
32t k + C. The first part of the path is straighter than in a
.~ k k2e .
Then
v sin A -kt
k e
-
kTe
32 -kt
-](
32t + V sin A
k
+
0
32
We have in (6) and
(7) the formulas for the vertical velocity and height above ground of the projectile.
2)
k
x VcosA
32 kX) + -2 1 og ( I k VcosA
. describe the path as t becomes infinite. A bomb is released from an airplane traveling horizontally at a speed of U ft/sec and at an altitude of H feet. (c) Using the results of parts (a) and (b). show that the horizontal and vertical distances traveled by the bomb in time t are x = U(l .7
vacuum and the latter part steeper. Finally. t2 = ~ log (1 + kV ~~n A) • 3. Doe s the resuLt agree with the answer to par t (c) of Exercise 6? 8. V Sl. the maximum range is obtained at an angle of fire of less than 45°. The projectile strikes the ground at a steeper angle and with less speed than that with which it was fired. Calculate dy/dx from the result in Exercise I.e-kt)/k and y = 32(e-kt . Ans. (x2'Y2) of the highest point of the projectile's
2
A . If the air resistance of the bomb is km times the velocity. we could do the following. describe the path as x approaches the value (V co s A) /k.V sinAcosA ~x2-32+kvsinA'Y2-
_ VsinA k
32 -k2"log
(l+kv~~nA). of the projectile motion discussed in the text? Ans.
. Find the coordinates path. show that the path of the bomb t seconds after its release will be inclined to the horizontal at the angle tan-l [32(ekt -l)/kU]. Show that the projectile moving in the resisting medium attains its maximum height at a value of x closer to the starting point than it does when shot out at the same angle A and with the same initial velocity V in a vacuum. As a check on the results in Exercises I and 3. where k is a proportionality constant. the velocity as t becomes infinite. 6. and "see if the slope is O. that is. Find the time t2 it takes the projectile to reach the highest point of its path. 5.
2. O. If the air resistance is km times the velocity. Ans. Exercises 1. where k is a proportionality constant. respectively_ 9. An airplane flying horizontally with speed U releases a bomb of mass m. Find the direct equation relating y and x by eliminating t between (4) and (7). 7.D A
3 + . substitute in it the value of x2 given in Exercise 3. Using the result of Exercise 1. -32/k. y = (. (a) What is the terminal horizontal velocity.
4. (b) What is the terminal vertical velocity? Ans.1 + kt)/k2. Suggestion: Compare (15) of the text and the value of x2 in Exercise 3. We know that the slope of the projectile's path is a at the maximum height.
we get the result for y. Calculate dy/dx from the result in Exercise 1 and substitute for x the value of x2 given in Exercise 3. Then. Vx = U. Vx ap-
(b) By (6). Hence I . e-kt approaches O. The rest is straightforward to get the text's results. 8. We start with (1) as in the text here.e-kt = Jex/VcosA and t = (-l/k)log(l-kx/VcOSA). by (3). 4. Use the value of t2 obtained in Exercise 2 and substitute for t in kt = (32 + kV sin a) /32 and and (7).e -kt = kV sin A/ (32 + kV sin A). belongs under the subject of rectangular parametric equations (Chap. y in Exercise 1 approaches -00 This agrees with 6(c). integrating. Following the suggestion we must show that V2. v
proaches
O.32e-kt)/kUe-kt• yx
the derivation of
(5)
we obtain
=
=
C. At the highest point ~ or Vy is O. Rewrite the first of these quantities as V2. Hence use (6) to solve for t when Vy = O. sin 2A/64. 0 = 32 and y = (32 .
=
x
0 (which means
x is measured
from
the point
where
the bomb
is released.sin A cos A/ (32 + kV sin A) is less than V2. We may take over from Exercise 8 that v = ue-kt and v = (l/k) x Y (32 -32e -kt). We have from Exercise 2 that e
(4)
e-kt is the reciprocal. 9.ky and so by using the method of the text here in
y
y (32 . 5. Since y 0 when kt)/k. t = 0. Since 2kV sin A is positive. From (4) we have 1 . Then.8
Solutions
to the Exercises
on Projectile
Motion
1. 7. one of the famous ones in the history of mathematics.
(c) Since the horizontal velocity approaches 0 and the vertical velocity approaches a constant the path must approach more and more a vertical straight line. In Section 8. 2. However. we get x = U(l . The Brachistochrone
Problem. like the preceding topic. and using x = a when t
y
approaches
-32/k.32eIntegrating and using 'y = 0 when t = 0.e-kt)/k. As x approaches V(cosA)/k. 3.De-kt)/k. Mere algebra shows dy/dx = O.sin 2A/ (64 + 2kV sin A). Inserting these values in (7) gives the text's answer.
This problem. the expression is less than V2. If we measure distance downward as positive we have for the vertical motion as in (2) (except for sign) = +32 . 6 (a) As t becomes infinite. -kt Hence in our case v = Ue . Then the direction of the bomb is given by tan e = v /v = (32 .sin 2A/64. when t = 0. 18). the text proper takes up tangential and normal acceleration along curves and arrives at the formula (78):
(1)
:F -
64 (y 0 -y)
•
.
suppose that light were to travel from a to B with a variable velocity v given by (2). but it need not be the one that makes the time of travel least. This curve would indeed furnish the shortest distance from 0 to B. It is.
This equation is correct. a curve from a to B. What should the shape of the curve be in order that the time of travel be least? One's first thought is that the curve joining 0 and B should be a straight line. According to Fermat's principle. and therefore the velocity acquired will be greater at least at the outset. Suppose that a particle starts from rest and is allowed to slide along a curve (Fig. Here John Bernoulli applied a brilliant thought. He said. Let us can sider.9
This says that the velocity acquired by an object which slides along a curve under the action of gravity and starts at the point (xo'Yo) with zero velocity is dependent only on the vertical distance fallen and is independent of the shape of the curve. but it certainly does not incorporate any condition about the curve being the one for which the time of travel is least. We learned in (1) that the velocity along the curve at a point (x. then. If a curve is used that is steeper at a than the line OB. The term brachistochrone means shortest time and it enters in the following way. (The arc length s is measured from (xo'Yo»' With this result at our disposal we can examine the proof that John Bernoulli (1667-1748) gave in 1697 of the brachistochrone property of the cycloid. in fact. Because our particle starts from rest at a and y : 0 at 0. the
_o~
x
y
Figure 2 tangential acceleration caused by gravity will be greater. If the incipient velocity is greater and because the particle gains velocity as it travels along the curve.y) is the vertical distance traveled by the particle if it starts from rest. it may still take less time to traverse a curved path from 0 to B even though this curved path is longer than the straight line path from 0 to B. . light always takes the
.We chao se the coordinate axes so that y is positive downward. 2) from a to B under the action of gravity. then by (I) (2) ds
dt
=:-
v=
SlY . true for any curve.
4) within each of which the velocity is constant. to the (i+l)-st layer in which the velocity o
Vi
B
Figure 4
. we showed (formula (16) of Chapter 8) that when light passes from a medium in which its velocity is v to another in which its veloc1 ity is v2' then (3)
Figure 3
=
where ul and u2 are the angles of incidence and refraction and are the angles shown in Fig. However. Indeed. This is Snell's law of refraction of light. Now light changes speed when it passes from one medium into another. Bernoulli wished to consider the behavior of light when it travels with a continuously changing velocity. He therefore supposed that the space from 0 to B was broken up into a series of layers (Fig. we might obtain the clue to the solution of our problem. Perhaps if we analyzed how light travels when the velocity varies.10
least time. Note that if v2>v2' a2>u1• The law of refraction applies when there is a sudden or discontinuous change in the velocity of light. Suppose now that light passes from the ith layer in which the velocity is v. 3.
11
is vi+l'
Let I)ibe the angle of incidence and Then according to (3)
0
1+1 the angle of
refraction. Then (4) will hold at each boundary.y) is the angle that the tangent to
y
Figure 5 the curve at (x.= v
sin 2
C't
n =--v
n
sin
C't
what this equation says is that
(4 )
sin a.
a.
Let us now increase the number of layers between the level of 0 and the level of B.
~
=:
constant.
V. 5 that sin ex=: so that (5) becomes
dx ds
However. If the number of layers becomes infinite. each horizontal line between the horizontal through 0 and the horizontal through B becomes a boundary. so that
=:
2 ---=.
This equation holds at the boundary of each layer. we also see from
. Moreover. if we think of light as following the curved path DB (Fig. Fig. and we have in place of (4) that (5) sin
I) =:
v
constant
where a and v are now fUnctions of y. the direction 0 of the incident and refracted light at any point (x.y) makes with the vertical. 5).
v= 1 B ds dx
=
-.::
'\j
C
/1
+ (~)
dx
2
where B is some constant and C = liB. We now try to integrate (6).cos 2u)/2 so that
f
Sin u 2D sin u cos u du cos u
. let us use the change of variable (7) Then x= or x = f20 sin2u duo The integral is readily evaluated by changing sin2u to (1 . of y.Y.J
dx
2
I~ we solve for dy/dx. we obtain (6) where 0 is a new constant. We now use this fact in (2) and write
C
= elY . Thus through the study of light we have learned something about v.12
dx ds = constant v o~because dx/ds = l/(ds/dx).y
In view of the presence of the radical.::::=::::.
(9.:::::::. let us invert and write dx_~ dy so that x=
Because the right side is a function
'II b
-y
J ~ry~= dx 10 .
is least.cos 2u)
2 D
as the parametric equations of the curve with u as the parameter. we must have xl"" R (o 1 .
D
= O. we did seek something that would make time of travel least. time.
u
Now. However. If the coord i. We do want the curve to pass through the given point B.Y1) when 8 "" 61.sin2u).cos 81) These equations do determine R and do it so that for this value of R the equations (9) will pass through (xl.
x
= o when y = 0 and when y= 0. which is an extension of the calculus.ng x and y. must be handled with the techniques of a branch of mathematics called the calculus of variations. let 2u = e and let D/2 = R so that the equations become (9) x = R (e
-
We can
sin
e)
y = R (l . We note that it is a kind of minimum problem.na es of Bare t (xl'Y 1)' we want the eve 10id which for some va 1ue of e and some value of R yields (Xl.sin
e 1) .13
x = ~(2u2
sin 2u) + C. Would you say that Bernoulli used an entirely mathematical argument to solve the brachistochrone problem?
. However. in fact. Exercises 1. That is. we can equally well take the equations x = 2(2U .
Then C =
o
and
(8)
We could now solve (7) for u and put this value of u in (8) to obtain the equation rel~. This problem is a peculiar one insofar as the calculus is concerned. we did not.
Thus John Bernoulli showed that the cycloid is the curve along which a particle slides under the action of gravity from one point to another in least time.
x = 2'(2u . We found a curve for which the dependent variable. as one does in the usual maxima and minima problems.cos e)
and we can now see that the curve is a cycloid. That is.sin 2u) y = D sin2u = ~(l . Such problems cannot usually be done with the calculus and. find a value of x at which some dependent variable y is least.yl). The solution of the brachistochrone problem by means of the calculus proper was possible only because Bernoulli used an ingenious argument.
Yl ""R(I .
14
2. would you say that Bernoulli was able to solve the brachistochrone by relying entirely upon mathematics and concepts of mechanics such as velocity and acceleration? 3. Specifically,'what did Bernoulli accomplish by introducing the motion of light? 4. What is the essence of the argument that the cycloid requires least time? Solutions 1. No. He used the physical fact that light takes the path requiring least time to obtain an important fact about the velocity of motion. 2. No.He used the principle of least time which as Bernoulli used it, is a principle of optics, not mechanics. 3. The key fact obtained by studying the motion of light is that v = C/Il+(dy/dx)2. This tells us how the velocity of the motion must be related to the slope of the curve along which the motion takes place if the time of travel is to be least. 4. There are two key ideas in Bernoulli's proof. The first is that for motion along a curve under the action of gravity (starting from rest) the velocity attained is v = where y is the vertical distance fallen. However, as the text points out, this fact holds for any curve. The problem is to single out the curve requiring least time. Here Bernoulli calls upon the behavior of light. The principle of least time implies the law of refraction (Chapter a, formula (16». The law of refraction extended to a continuous change ip the medium implies v = c/ll+(dy/dx)z. This equation relates the slope of the curve requiring least time to the velocity. However, the velocity is not uniquely fixed by this last equation. If we now add that the velocity at any level is 8/y, that is, the velocity determined by gravity, to the condition which least time imposes, we get enough information to determine the unique curve along which the particle must move.
alY
D. Kepler's
Laws. at formula (35), namely
In the text we arrived
(1)
p
=
l+e cos (9+a) ,
as the equation of the path of a planet which is attracted to the sun in accordance with the law of gravitation. We then sought to determine e, primarily, so that we could learn which conic section is the actual path. To simplify the work we adopted the initial conditions that at time t = 0 the planet is on the polar axis at a distance Po from the pole (which is the location of the sun) and that the planet has at t = 0 a velocity v which is perpendicular to the polar axis. These initial o conditions enabled us to determine the nature of the conic section and it turns out that whether the conic is an ellipse, parabola or hyperbola depends on the value of va.
15
One can, with the mathematics at our disposal, deduce a more general result which is of value to mathematical astronomers. Instead of supposing that Vo is perpendicular to the polar axis we allow it to make an angle A to that axis. The position of the planet at time t = 0 will still be on the polar axis at a distar.ce p from the sun. Moreover, o we do not suppose that the line from focus to directrix is the polar axis so that a need not be 0 or u. Under these more general initial conditions we can still determine e, as ~;e]l as a and h and we arrive at the surprising conclusion that only the magnitude of Vo but not the direction A wbich it makes with the polar axis determines the particular conic section. The derivation of this conclusion under the more general initial conditions is somewhat lengthy but elementary. Let us suppose that the planet starts out at some paint Po in space (fig. 6) whose distance Po from the sun at 0 is known. Further at Po supPolar axis Po
pose the planet has an initial spped Vo whose direction makes an angle A with the line joining
0 to P.
o
We
choese the polar axis of our polar coordinate system to be the line OP so that e is measured What these at from OP clockwise. (2)
a
tells
Figure
6
o initial conditions
counter-
us is that
e
= 0, component of the velocity and v p
is the radial
(3 )
• is-p,
then
where since then
(4 )
the subscript v sin A
0
in Po denotes
the value
of p at
e = O.
Likewise. and ve
is the transverse
component
of the velocity
=
p8
We must
8 is 0 but e is the time rate of change a 0 of e at e = 0 and this is not zero. We can determine h at once. By (19) of the text proper h = p2e. Since h is a constant we can use its value at 0 = O. Then by (4), understand in (4) that
(5)
enable
h
=
p v sin A . a0 involving a and e which e = 0 and so p = Po will
We are now going to obtain relations us to determine both. From (1) when
16
Po == GM (1+e cos from which \-)e have, by solving h2 GMPo
p
cd
for e cos a,
e cos a == Then,
(6 )
- l.
in view
of
(5)
,
e cos a ==
o
v2sin2A 0
GM
-
1. to
If we could get a value for e sin a we would be able to use it and (6) find e and a separately. To involve sin a we go back to (I) and differentiate. The algebra is simpler if we first write (I) as
1
p
=
GM [ I +e co s (e +0,) ]
h
Now
(7 )
- fj'T de == - 'jiT e s a,n (8+0'.) •
8
1~
GM
.
By setting would ever,
=
0 we
can get an expression we do not know of Po or ~
but this for e sin 0'. the value at of O.
expression Howthat
do no good we have in
because that
(3) the value
e=
9.E. de
at
e = O.
suggests
This
we use the fact
~
By
de
=~
dt
dt de
•
(19) of the text
Q..e. _
proper,
dp
de - dt 11
result in
p2
Substitution
of this dp dt
(7) gives
=
GhM e sin (e +0'.).
and at
e=
0
Po
=
GM
h
e sLn
«
Then e sin 0'.
17
ar.d in view of (5) and (3)
(8)
esinct =
p "zsin A cos A
o
0
GM We square (6) and (8) and add. Then
With
(9 )
(6) and (8) we can obtain e.
GM To obtain a. we have but to divide (8) by (6).
( 10)
+
Thus
tan ct=
r: o v0s in A co s Z
A
Having obtained We can write (9) as
( 11)
(9) and {lOl, let us try to profit from them.
1+
P
v? sin 2 A
00
P VZ (....£._Q-2). GM
Gl'1
If P v2/GM < 2, the quantity in parentheses will be negative, and because all other quantities are positive, e2, which is necessarily positive, will lie between 0 and 1. Then the conic section will be an ellipse. We may put this statement thus: (l2a) From (12b)
'f l..
o
0
v
o
j¥;GM < --,
Po
the path is an ellipse.
oa
(11) we also see that if P v2/GM - 2 = 0, then e = 1, or if v
>
a
=~
l'
'V
Po
, the path is a parabola.
Finally, if p v2/GM
a
0
2, then e > 1, or o
> _/2GM
(12c)
if v
Po
, the path is a hyperbola.
We see, then, that only the magnitude of the initial velocity but not the direction determines the particular conic section that the attracted body follows. The shape and location of the particular conic section does depend on angle A but the fact that it is an ellipse, say, does not.
18
The problem solved in the text proper under the more specialized initial conditions or the same problem with the more general initial conditions treated here is known as the simplified or modified two-body problem. The simplification consists in assuming that the sun is fixed and that a single planet is attracted to the sun. Actually each body, sun and planet, attracts the other and both move. This more general problem and extensions to the three-body and n-body problems are treated in texts on celestial mechanics but the mathematics involves far more of the sUbject of differential equations than can be taken up in the calculus. Moreover, exact solutions cannot be obteined. August 1976 New York City Morris Kline
Sl)/(~ . (b) t1 = 4. because there is less and less time for the speed to change prior to
the third second.
CHAPTER 2. the limit of h2 as h approaches 0 is 0.Sl)/(t2 . The change in distance is 2. The limit concept. The average speed is the distance traveled during some interval of time divided by the interval. Thus distance traveled from t = 3 to t = 5 is 256 ft. the speed may change at any instant or instants during the next hour. 5 to 5. is 160 ft/sec. SECOND SET 1. t1 == 3. 128 ft/sec.
CHAPTER 2. The result should be the same as that obtained in the text. The limit as h approaches is 1. For the purpose of finding the limit of 3h2/h2 as h approaches 0. 0. The limit. 0. Same method as in (a).
7. S = 5.t1) = 144 ft/sec. %. and 0. (c) Calculate the average speed for the interval 5 to 6. The quotient after division of numerator and denominator by h is 3h2 + 3h + 1. In the notation of Exercise 3. hence (S2. The limits are 0. the rate of change of distance is (S2.Sl.
40 50
6. SECTION 4 1. Sl = 16(3)2. Yes. To find the limit of the quotient we may divide numerator and denominator by h. 5 to 5. determined at this stage only by seeing what number the average speeds seem to be approaching. The limit as h approaches 0 is 1. 0.Sl)/(~ . S2 = 16(5)2. 2. The limits of the numerator and the denominator are 0 and 0.2
CHAPTER 2. The quotient is 3 and the limit of this quotient as h approaches is 3 because the quotient is always 3.
S2. 0. Use (a) as the model.t. 0. The limits are 0. obtaining 3h + 1. t2 = 5. we may divide numerator and denominator by h2. SECTION 3. Same method as in (a).t1) == 80 ft/sec. Instantaneous speed is the speed at an instant of time and is obtained as a limit of average speeds as the interval of time which starts or ends at the instant approaches 0. Let the position be Sl at time t1 and S2 at time t2.). 1. SECTION 3.
°
°
°
.1. hence (S2. namely 128 ft/sec. etc. As in (a) the limits of numerator and denominator are 0. (a) The limit of 3h2 as h approaches
(b) (c)
(d)
(e)
(f)
is 0. The limits are 0. (a) t1 = 0. the rate of change of distance or average speed is 128 ft/sec. t2 = 5. The method is the same as in (a).01. Not necessarily. FIRST SET 1.
f) 26. The answer not in the text is b) 60 ft/sec. h) Xl' 2. ~x is the independent variable and ~y/~x is the dependent variable.
D
6. The in~tantaneous rate of change A depends then on the length of t at the value Wl of w at which the rate is computed. the change ~A = ~~w. Hence AI:::: 2s1• The result is intuitively reasonable because it says that the area increases at the rate which is fixed by the lengths of two adjacent sides. 10.4(3)2 :::: 4~t + (~t)2. SECTION 6 1. Compare the discussion of (28) in the text. (b) (c) O. Thus (a) zxs > 4(3 + ~t)2 . 7 By taking ~t be 0. 3. A limit is a constant or fixed number.
CHAPTER 2. 8. if t is kept fixed and w changes by an amount ~w. 1 1 2 4.
. Since for y :::: x".01 and calculating ~s/ ~t. A limit is an exact value. (d) 10. ~y. 7. 11. The derivative of y = f(x) at x = Xo. yl::::2x1.432(5)2 :::: 320~t + 432(~t)2. £1 (xo) or yl or dy/dx evaluated at Xo. d) 160. ~s:::: 432(5 + ~t)2 . A1 + ~A :::: T(r1 -t ~r)2. Al :::: Tr2. 60 Since A = f_ wand f_ is kept fixed dA/ dw = I/" Geometrically. 10.4
~s/ ~t :::: 320 + 432~t and 4 the limit as ~t approaches 0 is 4320. Hence ~A :::: 1Trl~r + 1T(~r)2 ::::21001T ft. In each case use formula (30). the effect of the constant factor a in the a function is that it multiplies the derivative of x2. Use formula (30). ~y/~x. 4
%. Use the method of increments in each case and at the value of t stated. s'= 2ax. ~s/ ~t :::: 4 + ~t and the limit as 2 2 ~t approaches 0 is 24. 9. The third sentence merely rephrases the problem in mathematical terms. 50 A:::: S2. As these sides increase the area of the whole square increases and the rate of increase depends on the lengths of these two sides. 8. 9. The answers not in the text are: b) 128.
Y + L1y = c. d) d = -9. 7. the surface area of a sphere of radius r. 2. Hence
Subtract since b
y = bx. 4.
3. SECTION 7 1. (b) Same as (a) except that 5t replaces 32t.y= a(x1 + .
Hence
.
+ 6y
=
b(x+6x). ds/dt = v = lOt. hence s = 58 and ds/de = 5. Y1 + . (d) The instantaneous rate of change of y with respect to x at any value of x is 4x. 9.. The derivative of a function which is always constant (or of a constant term in a function) is O. lim L1ylL1x = O.
(fl The instantaneous is -3x. L1x+o
=:. dy/dx
6y = bL1x and lim L1y//J.CHAPTER 2. 6y/6x
=
3. etc.
rate
of change y'
of f(x) with
= -2Xi (h)
respect
= 54x. Alternative characterizations are. (b) dy/dx = kx.
=
O. (c) The instantaneous rate of change of velocity with respect to time (or acceleration) at time t is -32. dV/dr = 4nr2. dy Idx = y' = b.6.x + 3ax1(Ax)2 + a(6.
Then
= yl
That is.. The method of increment gives Ay/.
(b)
s
= -6ti
(d)
dy/dx = 16x.x b . Ay/ Ax = 3axi + (3ax1 + aAx)Ax. 8. Here r = 5. Y1 = ax].x)3. Hence s= 3ax2. (a) The velocity (or instantaneous rate of change of distance with respect to time) at time t is 32t. (c) dy/dx = kx.X= 4ax~ + [6axi + 4ax1Ax + a(Ax)2]Ax from which the result follows. SECTION 8
1.
to x
(j) dy/dx = -15x. Subtract y = c. 10.X)3. (d) dA/dr = kr. Thus 1'= 4ax3. The argument is false.6. (el The instantaneous rate of change of y with respect to x is 8x.. b) y'=-4x. Acceleration equals v. The correct conjecture is that the derived function is obtained from the given function by multiplying by the exponent 4 and lowering the exponent by one to 3. It is true that the derivative of any function is defined at a fixed value of x and the function is constant at the value of x but if the function varies as x changes the function is not a constant for all values of x and its derivative need not be O. Then is a constant. v = 32.
6y = 0.. f) s = O.6.y= 3ax~6. (a) dy/dx = ny.6. Since v = 32t. for example: The derived function of s is 32t. 5. 2. (f) (1) yl = 21Tx. The formula for arc length is s = re where r is the radius of the circle and e is the central angle in radians. Y
~y/L1x = b.
I1x+o
O.
y'
CHAPTER 2. That is. 6.
Yes. When x = 1200. At x = 15.2 dollars.
f' (-2 ) ::: . His profit may be declining when dP/dx is negative but he may still be making money. No.6. SECTION 9
(b)
(h)
2.6t. + The method of increments gives ~y/~x dy/dx = 2ax t b. We can argue that this must be the case for either of two reasons. S ::::100 + 32t.6
CHAPTER 2.T/. = 18 . Since T'is the limit of .6. (g) s ' =
-7. dC/dx:::: lOx + 15. 5. As x increases by the is amount . since the formula gives the amount being produced of the third substance a negative rate is physically meaningless. . dP/dx = 4x .4 •
For both functions y'= 2x. dC/dx = 1..004x. 7. must then be negative. 3. S = 328 ft/sec. As for (d). 9.. a 6.
=
2axl + b.
y' = 6/2'"x 13. At t = 4. (b) and (c) by letting t be 2. f' (x)
s
='
20ti
(e)
y'
"" 2x-7.6..
. dC/dx = 6x .x negative.4.6. We have some evidence to the effect that the derivative of x2 + 5 is the sum of the derivatives of T and 5 and the derivative of 5 is O. 4. Hence the temperature must deT crease as the altitude increases. and 4 respectively.
x
8. 10. The second reason is that the addition of a constant to x2 does not change the rate at which x2 varies because the constant does not change and so does not contribute to the rate of change. 11.
f ' (2)
= 4. dC/dx = 165.004. Hence we get the answers to (a). dC/dx = 6 .6. A low marginal cost means a cheap cost of production and so is desirable.6.. 3. If after a certain value of t the combination of the two separate substances stops and thereafter the third substance decomposes into the original two and if the original formula holds for the decomposition a negative rate could mean physically the rate at which the third substance is decreasing because it is decomposing.T/.
Hence
=
2x . T' or dT/ctx :::: 0. At all values of x.
(f)
y'
= -2x+14.x. No.xnd T'ls negative it must be that ..6.
Then f 16(12.C. 13. Then v = -4f3 t + C. Hence C = -(5/2)vo + 100 and s = 16t2 + vot . We have for it a := 32. At this time s = l/a. If we measure s from the point at which the brakes are applied then s = when t= and C = 0. Hence Vo = 32(5/2) + C or C := Vo .66.32.5)-(5/2)vo+l00.5)-80{12.v = vo. This yields a = -20. v = 0. For the first drop we have v = and s = at t = 0. 45 miles per hour is 66 ft/sec. At t = 2 (i. Measuring t and s from the time and point when the plane touches ground. At t = 1. We measure time from the instant the first body is dropped. This problem
100 Yes. If we substitute this value of t in the expression for s we obtain s = 1633. 9. Hence C = 16. At t := 5/2. v :=:: at + 100 and s = % at" + lOOt.5 sec.32 and so 2 v = 32t .5) = 16(12. 11. thus t = -100/a. When the plane stops. This occurs when t = 49. thus s = 16t2• For the second drop.000 nn /hr". Then s =_%t2 -I. Denote the unknown acceleration by a. hence the deceleration is 20. s = O. With this mean or average velocity the distance traveled in T seconds is 16T· T and this is the same result. s = O.. We could start with . e.a and a positive and obtain the same result. Now when t = 12. Thus v =-%t -I. 14.5)2 + vo(12.80. 1 second after the second drop is released) we have 16.9
is a rewording of exercise 1.22 . Hence Vo = 90 ft/sec.(16. The mean velocity.BOt + C. the distance fallen in time T is s == 16T2.% t2 + 66t.80. between t = 0 and t r= T is (0 + 32T)/2 := 16T.66t -I. The second body is projected downward with some initial velocity Vo at t := 5/2. hence v = 32t := C. Now taking the positive direction of distance to be that in which the train is running.SOt -(5/2)vo + 100. If we measure time from the instant the brakes are applied then v = 66 when t= and C = 66. Hence C = . hence we obtain the equation l/a = V2 a(-1 00/ a)2 + 100 (-1 DO/a) for a.000. Thus s = . By Exercise
° °
°
°°
°
. v = 32t + C.32t + 16.5 the two distances fallen must be equal. The train runs until v = 0.22 . Since the body is dropped the time to fall 100 feet is given by 100 = 16t2 or t = 5/2. 1. At t = 5/2.32t + C.32·2 + 16) = 48 ft. Then v = 32t + vo . a = -%. V. Then s = 16t2 . Integrating gives s = l6t2 + vot . Now s = 16t . But v = when t = 1.5 ft. 12.
Follow the derivation of (27) except that Vo replaces 128 and g replaces 32. Then v = -32t 2 + 460.1400 :::::: -16(t . Hence C:::::: 360. Hence C :::: and S:::::: .(5. The object is at a height of 512 ft both on its ascent and on its descent. At t:::::: .(5. 3. v =-32 and so v =-32t + C.3 sec approx.11
CHAPTER 3.16t2 + 200t.32t + 96 and s = -16t2 + 96t. SECTION 3
1. Hence t:::::: and t:::::: The value t = 6 is the value at which the stone hits 0 6. Use the result of Ex. s = O. But at t:::: 9. Follow the der ivation of (27) except that 160 r eplac es 128. Then s = . 320 10.3t + C and since v::::96 when t :::: . As in the derivation of (2 5) and (27) v s= . Maximum height occurs when v == 0. 11.
. By following the derivation of (27)except that 96 replaces 128 we have v == . Hence C:::::: 460.5 ft. :::::: 2 + 460t . V:::::: . the two a-values are -16t equal. Thus at t = 5 this height is 240 ft. i. s = 869. hence the distance traveled is the maximum hei9ht plu~ the distance traveled downward after reaching the max~mum he~ght or 272 ft (compare figure 3-3). s = 0 when t = 5. O. 4. v :::::: 300. 6. s == 15. v > .1900. v = 200. Hence the object has just returned to the ground.3/2)t2 + 96t. 8 except that g > 32. At t:::: 5.5. v == S = 144 . However.5). at t = lUO%2 and at this value of t. Now s =-16t + 460t + C.e. Hence s = -16t2 + 360t . We obtain t == 4 and 8. 8.
2 . at t = 5. If we substitute this value of t in either formula for s we obtain 606 ft approx. At the maximum height v:::::: or t = 96/5. Then t:::: 7. Here at t == 5.1 sec. v:::::: 96. s:::::: Then C = -1900 and s 5 O. At this value of t. If s is measured from the ground up. hence at t == 9. For the second objeet. 5. Then v :::::: -32t + 360 and s == -16t2 + 360t + C.32t + 1000.32 and and so v==-32t + C. Then C :::::: -1400. This is the number of seconds after the first object is thrown up. If s is measured from the surface then s == 0 when t = O. Thus s » vot .. Let s == 512 and solve for t. Note that this is the same as the initial velocity but of opposite sign.16t + 1000t.625 ft.3 0 0 = 18. At t = 6. the ground. Formula (27) gives the height above ground. s :::::: Hence vo:::::: ft/sec.3/2)t2 0 + 96t + C. Here a > V :::::: 5.16t2• When t:::::: 20.3t + 96. When the two objects meet. For the first object if we follow the derivation of (27) except that 200 replaces 128 we have s::::.3 if we take the upward direction as positive.5)2 + 200(t . 9. 7. s r= . the object is headed downward. When the stone reaches the ground s :::::: O. Then v :::: 5. v = -144 ft/sec.32t.
The train comes to rest when v = 0 and so t = % and s = . As in the derivation of (25) and (ZJ) except that Voreplaces 128 we have v = -32t + Voand s = -16t2 + vot.16(V/32)2 + V(V132) = V2/64. Then s = -16t2 + 176t + C. 17. when s = 0. 180 (a) v = . For this value of s. The height of 1. 20. At this value of t. t = 7. 15. Hence Vo= BOliO ft/sec. Since s = -16t2 + 96t + 112. At the maximum height v = 0 and t = V/32. During the second portion of the trip which lasts for some unknown time tl. After the fuel is exhausted. Thus v =-12t + 20 and s =-6t2 + 20t + c. (b) When the ball reaches the ground s = O. Then s = -16t2 + 96t. a = 'Ii = -11. From (29)we have 0 = -16t2 + 128t or t = B. When it reaches the ground s = O. v = 176 when t = O. Then v = 8t and s = 4t2 + C. The deceleration stage of the trip is best treated separately. Then s = . Then s = . s = 16t2+ Vt. If we measure this s from the point where the deceleration begins then s = 0 when t = 0 and so C = O. v = 20 and C = 20. Since the train starts from rest v = 0 when t = 0 so that C = O. Then when t = 0. s = 112 when t = O. Then t = 8 and for this t. Then C = 112. If distance is measured from the point where deceleration begins then s = 0 when t = 0 and C = O. We start with = a so that v = at + C. (b) When the ball reaches the ground s = -112. 13. t = % and s = 25. Thus the total distance traveled is 25 + 20tl + 504 and this equals 400 ft.6t2 + 20t.000ft is to be the maximum height. This Exercise is just a rewording of Exercise 12. Hence 1000 = -16(vo/32)2 + vo(vo/32). Thus C = 52800 and s = -16t2 + 176t + 52800. The other root t = -1 has no physical significance here. (a) We follow the derivation of (25) and (27) except that 96 replaces 128 and height is measured from the roof. v = O. Then s = _1~t2 + 88t. When v = 20. If t is measured from the instant deceleration begins then v = 88 when t = 0 and so C = 88. the additional distance covered is 81 = 20tl. If time is measured from the instant the fuel is exhausted.6(%)2+ 20(%) = 5%. Hence it takes 4 seconds to go from maximum height to the ground. Then v = -11t + C.12
12. s = 352 ft.32t + 176. If we measure s from the first station s = 0 when t = 0 and so s = 4t2. When the object comes to rest. Then v = -l1t + 88 and s = _1l~t2 + 8Bt + C. If the distance is measured from the ground then s = 52BOO when t = O. As in the derivation of (25) and (27) except that V replaces 128 we obtain v =-32t + V. 19. We found in the text that it takes 4 seconds for the ball to reach maximum height. a = -32. There v = 0 so that t = vo/32. We have 'Ii = -12 and v = -12t + Cp Let us measure t from the instant the deceleration begins. t = 7. + 16.
v
. S is to be 1000. 14.32t + 96. If height is measured from the ground and since t is already measured from the instant the ball is thrown up. v = -32t + C. Then tl = 21%2'The total time traveled is % + 21%2 51s = 22%2 sec. Then C = 176 and v = . Then s = -16t2 + 96t + C.
v == and s == ft ft2/2 if time is measured from the beginning of the trip and distance likewise.000 ft above the ground. Then s 0.(t3/6) + C and since x == when t = 0. a =-32. 27. 23. approx. If we substitute the two values of t in the formula for v we obtain v = ± . For the second part of the trip. v == 32t. 24. For this t either formula for s gives 126 ft.vg. The numerical values of v are the same. Here a = . Note that
°
°
°
. During the first part of the a == f.441s. During the one second of reaction time the car travels Vo T. Then t = 3. away horizontally from the point at which the bomb is released and 10. Then s == 2 + 184t . s == 16t 456 ft. (vo But v= Vo.3 sec appr ox. x = 9t .32t + 1500. When t == 0 (the instant the bomb is released) v = 1500. v = Vowhen t == O. When the two -16t bodies meet their a-values are equal. s = vot . When the trip ends v == or -rt + (f + r)tl == O. C == 184 and v == 32t + 184. Take the downward direction as positive.2gso)/g. Then 5 == 2 + 184t + C. that is. When the car stops v == and t =:. Then s == -16t2 + 1500t + c. == 2 .304. a = -r and v = -rt + C.e select any value So of s and solve for t we obtain t == ± ".2gso. namely. When t == s == tv ft~/2.000 ft. Suppose the first part of the trip lasts t1 seconds. . IfJw. Then C == . Then v = -at + Voand s = -aF/2 + vot + C.32. Then s =-5t2/2 + 44t + C. (b) Mter 100 sec the bomber will be 120.gt. 26. If distance is measured from the point where the brakes are applied. 4. 22.gt2/2. If we solve for t we obtain about 100 sec.6 ft.5 and v == 5t + C. then C == At t == x = 25% 0. v =-32t + C and since v == 120 when t = 2.000)2 or about 24 miles. gms. When t = 6. After braking v = -a and v =:= -at + C.1(120. Hence its distance from the bomb will be given by the Pythagorean theorem. Then v==32t-20 and s=16t2-20t+C.000)2 + (10. s == 2 + 300. Then when t == v == and so C = (f + r)tl. a == so that v = 32t + C. If we measure height from the ground then s == 10000 when t == 0 and so s == -16t2 + 1500t + 10000. For the first body.(a/2) (vo/a)2 + vo(vo/a) = v~/2a. When the bomb reaches the ground s =:.13
21.304. Hence -16t2 + 300 == 2 + 184t -16t . s = 0 when t = 0 and so s = -ato/2 + vot. (a) a == -32 and so v == -32t + C. Hence C == 1500 and v == . 0 0 0 In this time the distance traveled is 193. 28. If s is measured from the balloon then at t == s == and C = O. If we substitute this value of tl in the expr esston for s we obtain s > [fr/(f + r)Jt2/2. v==-20 and so C=-20. If time is measured from the instant the brakes are applied.Jvg . if time is measured from the instant it is dropped and distance from the ground up.20t. s = . a = .5t2/2 + 44t. If 44 distance is measured from the point at which the brakes are applied then s == when t == and s = .(ti /2)(f + r). When 32 t=O. For the -16t second body. For this second body s == -16t when t = 2 and so C = -304. We can use the general result of Exercise 8. If time is measured from the instant of braking v == when t = 0 and so v ==-5t + 44.O. Then for this t. (after the first body is dropped). 25. Now t is specified so that 0 tl == rt/(t + r) -. Then v = -rt + (f + r)tl tv ftl and s == -rta/2 + (f + r)tl t + C. The car stops when v = 0 or t = vola.
Hence at x = 0 there is a relative minimum. (i) f(x) = xix-I.46
2. At x = -2. s= 2(x -l)(x + 1)2 + 2 (x + 1)(x .-2x) (x-I)2. Hence there is a relative minimum of O. say -3/4. Hence there 1. s': -16. Possible values / / are x = 0 and x = 2. £' (x) = (x/2/x-l) + lx-I. fl (x) changes from positive to negative.l/x2. At x :::: .sa relat1. Hence there are no maxima and minima. y continually decreases.1)2/3 is positive for every value of x and this is subtracted from 3. the behavior of. Hence there is a relative maximum of 4.1)1/3~ This y'is never O. f' (x) = 4xa• Hence x = 0 is a possible value. At x = 3.2/3(x . But the function has no real value at x = 2/3. f~ (x) does change from negative to positive around x = O. f'(x) actually de-
a relative maximum: at x = -1 there is a relative maximum. (a) The function y = x is an example. say -5/4.1)2 = 4x(x + l)(x . 4. s' changes from + to -. at x = 1 there is a relative maximum.= 3x(x -2). At x r= 0. At x = 0. This is the absolute maximum. (g) f(x) = x~. However the function may have a relative maximum or minimum where the derivative fails to exist. The least we can subtract is 0 and this occurs when x = 1. hence there is a relative minimum at x ~ 1. Hence x = 3 is a possible relative maximum or minimum. 3. At ~ = 2. As x increases or decreases from the value of 1. 5. (j) f (x) = x2. f' (x) changes from negative to positive.to +. Hence the 0 relative maximum of 16 is also the absolute maximum and the absolute minimum is 7. y = 7 and at x = 5. y = 12. s' changes from + to -. Hence there is a relative maximum at x = -l~ At x =1. ft (x) = 1 . y'changes from . At x = 1. Hence x = 1 and x = -1 are possible values.nimum there. Then y = 3 is a relative maximum. At x = -1. s' changes from + to -. Hence there is a relative minimum and its value is 0. Y = 20. At x = 0.to +. fl (x) is positive and for x slightly more than -1. f' (~) ch~nges from negative to positive. f I (x) = (x2. The absolute maxima and minima may occur at the end values 0 and 5. Here as x increases. Hence there is a relative maximum whose value is 1. At x = 2. s' changes from .. Now f' (x) = 0 at x = 2/3. Here y/= . We see that (x .ve mJ. f~ (x) is negative.~or x slightly less than -1. (a)
v= -2x + 6. However y = _x2 in the interval from _00 to 0 isa better example.y/is as at x = -1 and the relative minimum is again 0. (b) '1.1). This is the absolute minimum. (x-I).
. At x = 4. Hence there is a relative maximum there. (h) f(x) = x + l/x. Hence there is a relative maximum whose value is 16.
Bur sin A = OQ/OP. At any given time t the bead which falls straight down falls some distance OP. Note that the velocity at the bottom is the same if the height from which the object descends is the same. t1= £1/4.:
8. 6(b) below. (b) We use (34) in which sin A is now h/£ and t ~. that is. . 7 was not restricted to any specific angle A.14
So must be a value actually attained by the object or the values of t and of v will be complex. Since s = vot . The distance OQ that the obJect sl1des 1n time t is OQ = 16(OP/16)sin A. Then t1 2 = £2/£1' (b) By Ex.2588 and t = 10. Yes.2588 to calculate t. 5 (b) both velocities are Svh. If s is the distance the object slides.Ji1 and t2 = _e_.
9. In this time the object sliding from 0 to Q will reach Q. By Ex.Ji1. . We start with (33). Hence v = 80 ft/sec. Then sin A = OQ/OP. £/4111. But the argument in Ex. Then v = s. For the motion along Op' we. To calculate the velocity we use (34)wherein A =. We use (35) to calculate the time. The result is approximately 10 sec. Again from Exercise 8. To attain twice this maximum height we must replace Vo by -v'2vo•
CHAPTER 3. Then X OPR is A by the use of right triangles. Now use (35) with s = 386 and A = . At t = 0. At t =0. Hence Q = Rand Q lies on the circle. ••• all lie on the same circle. 29. The only change over Exercise 2 is that A is 15°.
4. a= 32 sinA. s = 16t2sin A. s = 386. Q lies on a circle with OP as diameter. In (35) s = 200 and A = 30 Hence 200 ::= 8t2 and t = 5 sec.gt and at the maximum height v = 0 so that t = vo/g.30° and t = 5. 7. namely.
0.
5. We use (35) in which sin A = h/£ and s = t . 1. SECTION 4 1.
. 7). Suppose Q is not on 1 the circle but R on OP 1 is. 3. Here sin A = h/~.. The rest is the same as in Exercise 1. s = O. If we consider circles with 0 as highest point (so that the diameters are all vertical line segments from 0 downward) the smallest of these circles which reaches C is the one which first touches it as the circles expand from O. 5 (a).
/t
4. We then use (34) with sin A = .gto/2 we find that the maximum height is v~/2g. 7. Since sin 15° = . (a)
6.2588.J4-1h. Hence Q'. See Ex. Hence sin 15° = 100/S.
10. v = Vo . 2. From the formula s = 16t2 we find that the time to fall the distance OP is tl = /iJP/4. The time for a bead to fall straight down will be least for this circle as compared with larger ones because for anyone circle there is a time value to reach it and the time increases with the diameter (See Ex. Then sin A = OR/OP. Then £ = 16t2h/£ or t = Q /4vh. Qi1. Then C = Vo and v = 32t sinA + Yo' Integrating gives s = 16t sin A + vot + C. Hence s = 16t2 sin A + vot. Then v = 32tsinA + C. (a) By Ex. Here if the value of t were calculated very accurately we would get the same 80 ft/sec as in Ex. then sin 30° = 100/s or s = 200 ft. 2 v = va.use (3~).
. If we use Fig. 3A-I. The lengths are 134. APPENDIX.xl)/2] + Xl == (xl + x2)/2. AB = 10.J89. 5. Ya and Y2 are parallel lines and Yg cuts PlP2
in half.
(d)
JUf. 134 and 1:36. 4.
Xl
2. Then it must cut the other transversal RS in half and so Xg == [(X2 . Further Yg is the median of the isosceles trapezoid RP1P28 and so is (Yl + y2)/2. Be = 1125.
(b)
. SECTION 2
1. locate the midpoint (Xg. Formula (1) is unaltered if x2 and
and
Y2
and
Yl
are interchanged. AC = 1125.
3.IS
CHAPTER 3. Ya) on the line segment P1P2' and draw Ya we find that Yl.
4/9.
%. 5.. 2. (-5+513) / (-5-5/:5). APPENDIX.). In each case we have but to find the tangent of the given angle. Thus only (b) is a right triangle. SECTION 5
1.
%. Then the inclination is 59° (approx. The slope of the perpendicular is . APPENDIX. If two sides have slopes which are negative
other
(a)
(d) (-513-5) / (5/J-5). -4/3. -4/9. CHAPTER 3.
the triangle is a right triangle. SECTION 3 10 Formula (2) applies in each case: (b) (d) (f) No slope. -1. 1. 9.. APPENDIX.16
CHAPTER 3.
CHAPTER 3. The slope of each line is found by using formula (2) and we then find the angle A whose tangent is the slope.
9j4. 3. The slope of the first line is 3~ and the slope of the second one is Each slope is the negative reciprocal of the other.
.). SECTION 4 1. 3. (b) 3/4. 2. Hence any perpendicular has slope % or 1. In each case we have but to find the angle whose tangent is given. By using (2) we find that both lines have a slope of % and so are parallel. The slopes of the sides' are 3/7 f -2.
6. Hence 153°30' . then the perpendicular line has slope -lor inclination 135°. The given line has slope -0/. 4.1666. 7.
~21a. The given line has slope 1.
(c)
1/2.
reciprocals of each The slopes are
-6/5. 3/7 t -2.~ and the inclination is the angle whose tangent is -~ or the angle whose supplement is 26°30' (approx.
(a) (b)
CHAPTER 3. (a) We solve the given equation for y. The result can be obtained from a figure at once.).
CHAPTER 3.
with y = lUX + b we have the answers.3) does not lie on the line.3. and () == 105°.6 = 1(x . (b).8) lies on this line we substitute 5 for x and 8 for y. 1.0) must lie on Ax. APPENDIX. FmST SET
1. y =_3/4X. Then find B. 5. Hence the desired line has slope %.4).1). (c) Let ffi2 == -2 and ffi1 = 3 and use (7).6) and (4.2 = -3(x .7). Then tan B ==-3. SECTION 7. (b) We substitute 2 for x and 3 for y in y . The given line has slope . Since (0.
ill
l
=
1. 1 and use (7). 4. The equation is 5(x not satisfied and so (2'. The slope is Hence find e for which tan e = Hence e += 146° 20' (approx. SECOND SET 1. 7. . (c) As in (a). . The slopes of the two given lines are -% and %. 9. and (c) use (8) and merely substitute the given values. As in (a). (d) use (8) or (9). Then m = % and b = .%.7 approx.17
CHAPTER 3.
-%.5). We find the equation of the line determined by (3.3). Since tan 30° = -13/3 and tan 135° = -1. Let 002 == . The slope is 1 and so the equation is y . y :=: %x . 3.1). Then the desired equation is y + 1 = 1/3(X . Then m =_8/4 and b = 0. For (a). then y = -%x
(b)
2. 8. Hence the desired line is y . To show that (5. SECTION 6
Use formula (7) with ffi2 = 4 and m. 0 + B . SECTION 7. (b) The desired line must have slope -%.3) and the equation is satisfied. 2. Use (7) with ffi:l = -% and
. APPENDIX. 2.6 == 1(5 . 3. Then 8 .2 == -1/a(x . The line 3x + y + 7 = 0 has slope . Now use (8).
-%.
By comparison. Since the given lines. APPENDIX.%.¥s and ffi1::::.'Ys. 4. The two given lines have slopes of -% and 10. The first line has slope 1 and the second.0. let 001 = -13/3 and m2 = -1. Parallel to the x-axis and 3 units above it. These slopes are negative reciprocals. (a) Use (8). 5. (a) The desired line must also have slope 2. now use (7) with m1:=: 3 and ffiz = -2. = 3. + By + C == 0. The slope of the first line is -A/B and of the second -a/b. y . A. have slopes 3 and -2.71t.3. 6. Hence by (8). Then the equality of these two slopes gives the result.0 + C = 0. and use (7).
. by (9). In (9) b is 5.2 ::::.
.
l
Xl ::::. y)'s along the straight line. A::::. The equation of a curve (including straight lines)isan
equation involving x..18
10. 4. 2.. 3. Hence r is not fixed as we choose different (x. For a fixed curve only x and y can vary. (a) -A/B ~ 3/2~ (b) A ~ (-1/4)C and B ~ -1/3C~ (e) C ~ O~ 12. Thus for (b). substituted in the equation. B == -1. and C ::::. y and constants. A method is given in the problem.
(d)
A
=
O~
(e)
B~
O. 2. In the equation x2 + y2::::. r must change with x and y if any given x and yare r2. APPENDIX. In each case we use formula (14).
CHAPTER 3... 11. Y ::::. SECTION 8 1.
. Only in (a) do the three given points lie on one line.
then find A. decreases to zero at the crest. perpendicular to the wall and so direct impact. In each case find the slope of the tangent and since m = tan A. 9. the values of y' at x = 0. The slope is given by the derivative. Hence the object is moving downward at x = 3.6(x + 3) or y = -6x . (d) 0°. (b) Y = .5. this is also the slope at x = . the slope is 48. SECTION 4 1. 3. (al
p
=
5/2. Thus at this point the object is moving horizontally. then increases to zero at the trough. Since y'= 3x2. For (a) y I = x/50.6. then becomes negative.
Thus (a) y l= 2x.Solutions to Chapter 4
CHAPTER 4.4.3. (c) 104°2'. (d) O.2x . 3.4 or at the point (.4.9. Y = 9 has slope .== %0' (b) x = 5.
5. Thus the tangent thru x = . the notion of slope presupposes that we consider whether the line rises or falls as we go from left to right. Then tan A = 6 and A = 80° 32' (approx. y' =6. 7. . Y '= -8. Then the inclination is 97°7'. Le .
0 •
CHAPTER 4. (c) 2. 3. S = 16. y'= O. At any given x value their tangents are parallel. 8. and then becomes positive. Thus (a) s = 8t and at t = 2.
4.
14. The derivative is positive at x = a. 10. (b) 89° 7'. that is. and -'1 are 0. 6.9 = .
hence
y ==
(l/lO)xz. This function does not have the same value for two distinct values of x. at t = b. (a)y'=x. At x = 3. at maximum point. The slope of the tangent at any point x is 3x2• At x = 4. (a) y' = 2x. At x = 3. at x = 3. 1.. The value of the slope of the tangent to the curve at any point x is given by 2x.
11.). Since y '= -8x + 16. In each case find the derivative at the given value of the independent variable. These values also distinguish the sense of the motion. y'= -8 + 16.
(b) p
=
1.
12. y . Hence the equation of the tangent is y . The two curves are the same except that one is 5 units above the other. (e) 0 The direction of motion is given by the slope or the inclination of the tangent. At x = 2. (b) y=k(yl)Z (C)yf=l~(y/X). At t = 0. SECTION 3 2.
(b) 4.64). 13.
hence
y
=
(1/6)x
z
.
4)2 + (y . latus rectum 8. substitute 1400 for x and 148 for y to determine p. P = 1/2. 4-15 ar e p less than those in Fig. Then the width of the parabola is 4p. (c) x = (2/9)y2. 0).
Hence
the focus
is
(0.20
2. 8.4" directrix y = -4. focus (9/8. Then one point on the parabola is x :::: 400 and y = 148.0). and we solve for x. Hence the clearance is only 32%5 feet. To obtain the equation directly let (x. y) are the coordinates of any point on the parabola then . (a) The tangent line at x = Xo is Y. (b) x (-1/8)y2. the focus is at (O. P = 2~ (e) Form is x = (1/4p)y2. directrix x = -9/8.0)2 = Y + 8. (b) Draw the straight line between the points (xo.
13. The width across the entire parabola at the height of the focus is then 12.p. directrix y = -1/3.6. Then y = 37x2/490.21%5' The roadway lies along y = . The right-hand wheels of the truck are at x r= 23. latus rectum 9/2~ (d) y = (-3/4)x2. (b) Each y-value of y = 3x2 is 3 times as large as the y-value of y = x2 for the same x-value . Then v(x .
=
. Simplifying gives the answer in the text. latus rectum 4/3-. 4-12. 4-15
=
=
=
9. P = 1/3. 6. If the axes are chosen so that the origin is at the center (top) of the arch. The points on the parabola at the same height are obtained by letting y = p in the equation so that x = ± 2p. 4. directrix x 2.x~/4p) and (xo/2. (a) y = (1/16)x2.. Since the form of the 1 equation is y::: (1/4p)x2. y) be any paint on the parabola. Since the point (25. 7. Thus y = (x + 6)2 is obtained from y = X'2 by shifting or translating the latter 6 units to the left. If (x.f(x . One would expect that y should be replaced by -y in (9). Then each x-value is 6 units to the left of the x'-value.25. focus {O. For any parabola of the form y = (1/4p)x2.
11. Then y = (1/4p)x2. P 2. latus rectum 16.3). 3. Then the y-value of the point on the arch whose x value is 23 is .0)2 + (y . 4p = 25 and the equation of the arch is y::::-x2/25. (a) All the y-values of y = x2 + 6 are 6 units above those of y = x2. focus (-2.
4.p). One would expect that the x and y axes are interchanged in (9). the equation of the arch is of the form y =-x2/4p.
12. This line cuts p) the x-axi s at x = xo/2 because y::::0 there.-25) lies on the arch. (c) If we write y = (x + 6)2 as y = X'2 where x' = x + 6 we see that x = x'.xo). P = 9/8. focus (0. 5. Then the directrix is y = -3.0)2 Y + 2p.
with Fig.0). whereas the truck needs 10 ft. 10.
p
=
3. Choose axes so that the origin is at the lowest point or vertex of the parabola. P = 4. (a) Form is x = (1/4p)y2~ P = 5~ (b) Form is y = -(1/4p)x .OOO. If we compare Fig.3) and
Since the focus is (0. we let y::: 3 in y = 1/12x2 and obtain x = 6.3) and the right-hand point on the parabola at that height is (x.12 we see that the y-values in Fig.(xg/4 = (xo/2p)(x .1/3). Compare formula (9).
In finding the slope of the tangent line to the parabola.x~/4p).xo). (a). Then the sum of the supplement of LD'QF and the supplement of LDPF is 180°. If we use the fact that Yo= x~/4p we find that y = 0 or the two lines intersect on the x-axis. Then fil ~ slope of t ~ 0.xo) and the equation of the perpendicular is y . SECTION 5. 3 we find that tan a = 2p/Yl and tan S = 2p/YL' Hence a = S. Then PD is the reflected ray and this is parallel to the x-axts. fi2 = slope of t = 2p/YLi m3 = slope of s = 4PYl/{yi = 4p2). Suppose (Fig. (b) The normal has slope -2p/xo and passes thru (xo.xo). Then ~e remaining angle at R must also be 90°.x~/4p = (xo/2p)(x . hence y = x2/60. No. The tangent line at x = Xo is Y.
. It is better to take the parabola in the position shown here. 15. Thus the equation of the normal is y . 4-23) FP is a ray starting from the focus.x~/4p = (-2p/xoHx . To do this take the value of x from the second equation and substitute it in the first one. Let P have coordinates (xo. Assume that the coordinate system is as in Exercise 12. 4-17 with Fig.(x~/4p) = (xo/2p)(x . All we need to show is that the y-value of the point of intersection is -po 17. 4-12. where R is the intersection of the tangents.4p2/xo' Now solve the first two equations simultaneously and use the relation between Xo and Xl" When solving take the value of X from one equation and substitute in the other.
CHAPTER 4.
1. Thus LDiQF +LDPF = 180°. It then'follows that the coordinates of R are (0.0) which is the vertex..
4. By the law of reflection onehalf of this sum lies inside the triangle QPR.0). FrnST SET 2. Likewise QDP is parallel to the axis. The point Q has coordinates (0. We solve these last two equations simultaneously for their point of intersection. We showed merely that the parabola does have the reflection property.x~/4p). The tangent at the vertex is y = 0 or the x-axis.x~/4p :::: (Xl2p)(x .x~/4p) while T (the intersection of the tangent with the y-axis has coordinates (0. Choose the coordinate system so that the parabola is given by y = (1/4p)X2.21
14. 3. Then p = 15.2p/xo or Xl = .x~/4p + 2p) and hence RQ has length 2p. The tangent at any point (xo.x~/4pL The tangent P is y .Yo= (xo/2p)(x .2p/xo)(x . If these lines are perpendicular then x/2p >/ = . Now by using formula 7 of the Appendix to Chap.p = (.).x.xo)' Similarly the tangent at x = Xl is Y. Thus the midpoint of QT has coordinates (0. Compare Fig. 16.Yo) on the parabola has slope xo/2po The perpendicular from the focus to any tangent line has slope -2p/xo' The equation of the tangent line is y .
3. We can use (17) but in place of wx we must use 5x2• Then
y' = 5x2/Tc and y = (5x3/3To)+C. Since y = x2/240. Yes. y '= 0 and for x> 0. w = U. There is no derivative at (0. Then T = (w/2). Y = 15. SECTION 6 1. The point (60.jx2+4y2. No because there is no unique slope at (0. We see that T increases with x and is a maximum at x = ± 60 and a minimum at x = O. Then from Exercise 2. 3.
CHAPTER 4. 4. we have T = 75-/5 tons. SECONDSET 1. (a) Tcas e = To = constant.15) lies on the parabola. y I:. 2. The equation of the cable is of the form y::: x2/4p. Yes as long as the weight of the roadway per horizontal foot (that is. 2.j(240)2 + 4x2. SECTION 5. 1.22
CHAPTER 4.()/u!O = % tons/ft. 5. We can determine C as in the text to be O. where we now calculate T at x = 60.0). Then for x < 0. Tangent exists everywhere except at (0. To the left the answer is y = -5x3/3Tc because x is negative. (b) T sin e = wx. €. T = (wx/2ry). we can substitute in the formula for T of Exercise 2.0). However.
. this answer is correct only to the right of the origin. The graph consists of the negative x-axts and the ray which starts at the origin and lies in the first quadrant and makes an angle of 450 with the positive x-axis. not along the curved roadway) is constant.0). Or we may say that y' = -5x2/Tc because y' is negative and so obtain y for negative x. No.
Ax.f(xo)
+
f(xo) . Now in the definition of the derivative 6.Ax}) and (xo. To write the second quotient in the customary form of [f{x + ~x) . even though this Ax is negative.
We can see geometrically that apart from the factor % the first quotient is the slope of the secant joining (xo.\x)
•
and
2~x
and this equals
1/ [f(Xo + ~x)
2
Ax
. Let us use a new Ax which is the negative of the old Ax. f(x~-6x»
5.f (xo) + f (xo) .
.6.xapproaches o and the entire quantity approaches f' (xo)' Analytically the first quotient is in the form of fly/ ~x (see also Exercise 7) and surely approaches f'(xo}.f (xo~~)
(xo+l.f(xo ~x
AX)] .x can be negative.Ax) . f(xo)) as .f (xo» and (xo + x. because Ay/AX must have the same limit when t:.23
4.x» and the second quotient is the slope of the secant joining (xo . f (xo» . Both slopes approach the slope of the tangent at (Xo.f(x))/ Ax.f(xo)]/Ax and this approaches f'(XO). Then our quotient becomes [f(xo + Ax) .
f (xo+l. The slope of the chord or secant joining (xo-~x.f(xo)]/(-AX).
H we add and subtract f{xo) in the numerator we have
f (xo + ~x) .x approaches 0 through positive or negative values. f (xo + 6. f (xo . let us first write [f(xo .\x.
Use the method in the text which leads to (9).
CHAPTER 5. Then the answer in the text is obvious. Hence 2f'(x). (el y = x+C. V = x3• Then dV/dx = 3x2• 2 6. (il y = fix) = 4x3/3+C. SECTION 2.24
Solutions to Chapter 5
CHAPTER 5. (e) y = x6/6+C. This is so because when n is a positive integer the binomial expansion in (5) has n+l terms. The limitation to a positive integral value of n is necessary because after step (8) we use the fact that there are (n-l) terms in the brackets. FIRST SET
(c) y' 5x If. 1 . 2 2. (b) y'
(g)
y'
=
"" 8x7. (c) y = 3x2/2+C. 4. This is the derivative of y = x3 at x = a. that ~x{n-l) IAI approaches 0 as ~x does would not be possible. (d) Y = x /2+C. If we use the suggestion we get 2[£(x+t)-f{x)]/t. If n were any other kind of number the binomial expansion would contain an infinite number of terms and the statement on line 4 of p. SECOND SET
1. (g) y = fix) = 2x (h) y = fix) = X2+C. We can think of p as Xo and q as ~x. 3. ( e) y' = lOx 9 • (h) y' = (7/2)x6• j 2.
=
5.
. Yes and because nxn-L is 0 as it should be when y = 1. Hence 3a • 7.
1. (f) y = 7x1Vll+C. (b) Y = xlf/4+C. 8. ( d) y I = 20 X If. SECTION 2. 6/3+C. Here n = 3.(b) Y= xlf/4+C.
1. (a) Continuous for all x; (b) continuous for all Xj (c) continuous where defined, t.e ., for x -s 1; (d) continuous except x = OJ (e) continuous where defined, I.e., for x :;z:: 3j (f) continuous where defined, Le., for x ;:z" 1. 2. No; it jumps from 00 to 1800 as P crosses the maximum point of the curve. 3. Yes, because the slope is at first positive, then 0 at the maximum point, and then negative.
CHAPTER
6, SECTION
3
(d)
1. (b) y' = 2% X-l/3; (f) Y'= 21x2 + 14x;
(i) (1) 3.
4.
5.
6.
7. 8. 9. 10. 11.
+ (ifO/3)x-2/3• s'> {'(x) - g'(x). s= cf'(x). (a) y'= 30x - 3x2; for x == 2; Y'=48 it/mile. (e) The slope of the graph is, of course, the value of Y' • (d) Yes. By contrast in the vertical motions discussed in Chap. 3 the graph is not a picture of the motion. The rate of change of the volume of a sphere at any value of the radius is the surface area of the sphere. Since V == % 1Tr3, VI = 41Tr2 and this is the value of the surfacezarea.
X-l/2
9. Yes. The argument is the same.
10. Starting with the step y:::= == . Xk-1 we may apply the result (31). Now Xk x y'== xd(xk-1)/dx + Xk-l. 1. By the hypothesis of the mathematical induction process y'= x(k: - 1)xk-2 + Xk-1 == kXk-l. Since the result holds for n =k on the
assumption that it holds for n = k - 1 and since it holds for (, = 1, the result holds for all positive integral n.
11. R xp. By using (31) we have dR/dx x{dp/dx)+p. 12. A C(x)/x. By using (40) we have dA/dx x{dC/dx)-C(x»/x~ dC/dx is M. 13. V 2500(1+t)-1~ dV/dt -2500/(1+t)2. Clearly dV/dt is greatest when t = O. 14. R xp [640x/(x+9)]-40x. dR/dx = [5760/(x+9)2]-40. Now dR/dx> 0 when [5760/{x+9)2]-40 > O,or 5760 > 40(x+9) 2,or x < 3.
The tangent at any point Xo is y - l/xo = (-l/xg)(x - xo). Hence in Figure 6-8 the coordinates of the points K, P, L are respectively (0, 2/xo), (xo, l/xo), (2xo, 0) from which the result follows at once. S. Using Figure 6-8, the area cut off is l~OK' OL = 1/2(2/xo)(2xo) = 2 = constant. 6. Since 1==A/41Tr2 = (A/41f)r-2, 1'= (-A/21f)r-3• Now let r = 20 and then r = 200 to obtain the text's answers, When r is very large, a small change in r causes very little change in the value of I because I varies inversely as the square of r or, speaking physically, I is spread out over such a large sphere that per unit area I is very small and changes very little as r changes.
4. 7, dC/dx
When x = 100, dC/dx = 1/20-1/2000. Yes. Even under efficient production it must cost more to produce additional units of a commodity.
=
(1/2)X-li2-(1/2)x-3i2.
CHAPTER 6, SECTION 8 1. By (59), W = GmM[(l/r)
2. 3. 4. - (l/rl)] where r is the final distance of the object from the center of the earth. Here r = R. Using (61) the result follows. Use the formula of Exercise 1 with m = 100, R = 4000 x 5280 and r1 == 4500 x 5280. Ans. (22,528' 106)/3 ft-pdl. 8,448' 105 ft-pdl. It is greater because for paints above the surface the force of gravity is actually less than 32 m. The work done in raising the satellite is numerically the same as the work done by gravity in pulling the satellite down. Hence use the formula in Exercise 1 with ~ = 1000 and r1 = 1500·5280 ft. The result is 18,432.107 ft-pdl. By (58), W = GmM/r + C. Since W == 0 when r = R, C =-GmM/R and W = GmM/r - GmM/R. Using (61),W =32mR[(R/r) -1]. Measure r downward from the top of the well. Then the force on a length of cable of length r is the force (weight) per unit length times r. Thus F == 32 mr = 64r. Using the relation dW/dr=F derived in the text, we find
5. 6.
29
7. 8. 9. 10.
(since W = 0 for r = 0), W = 32r2. Thus the work to lift the entire cable is 32(200)2= 1,280,000 ft-pdl. In addition to the work done in Exercise 6, we must lift 300 lbs a distance of 200 feet. Thus the additional work is 32(300) x 200 ft-pdl and so the total work is 3,200,000 ft-pdl. a) yes; repeat the derivation of (59) verbatim. b) Repeat the derivation of (59) and Exercise 1 with 32 replaced by 5.3 and R being the radius of the moon. Ans. W = 5.3 mR(l - R/rJ. Use the answer to Exercise 8 with m = 100, r 1 = 500' 5280, and R = 1100·5280. Ans. 96,195.104 ft-pdl. Use the result of Exercise 8 with m = 1000, r1 = 1500·5280 and R = 1100· 5280.
dW/dr=wSinA in the text.Thus and W= (w sinAl r+C whe-re r is
11. a) Use the relation dW/dr=F derived
the distance the object is pushed. At r = 0, W = 0, hence C = O. If the length of the plane is denoted by R, we have W = wR sinA. From the figure sin A = h/R, thus W = who A
h
b) Notice that this is the same result as would be obtained if the object were lifted straight up against gravity!
2. We are given w as a function of x and are told that x is a function of t. We want dw /dt. Now dw /dt = (dw!dx)(dx!dt). We can calculate dw /dx from the given formula and we have that dx/dt == 100. 3. R = x i250-9x = x(250-9x) 1/2. dR/dx = (250-9x) 1/2 + x(1/2) (250-9x)_1/2 (-9). 4. (al Differentiate first as a product. y' =- (x2+2) 3(1) + (x-3)d(x2+2) 3/dx. Now let u = x2+2 and apply (12). Then y' = (x2+2) 3+(x-3)3(x2+2) 22x. One can simplify the result to get the text's answer. (b) Let u = (x-l)!(x+l) so that y = U1/2• Then by (12) y' = (1/2lu-1/2du/dx. To find du/dx use the quotient rule. du/dx = 2/(x+l)2. (c) Let u = 5x2+1. Then by (12) y' = (-2/3) (5x2+1) -5/3 (lOx).
set y == 0 and solve for x. If we set y=-O in the answer to Exercise 5 and solve for x we obtain x:::: xo . Hence dy/dx = -l. If we substitute the value of yl in this last equation we get the textls answer. 0. 5) is 2p/yo.
7. But Y5 ::::Apxo'
y :. 2x+2y(dy/dx) :. 6. Here dy/dx = 2p/y and at (xo'Yo)' dy/dx = 2p/yo.Xo).:: 8. Hence d2Y/dx2 = -(y-xyl)/y2. 2y(dy/dx) :.y~/2p.:: and y :.:: 4. (b) 2x+2y(dy/dx) ::::: (c) 6x+4y{dy/dx) == 0. (a) x(dy/dx)+y "" 0. 4. hence dy/dx ~ (-2x-y)/(x+2yt (e) 3y2. The latter (Exer. 2 -4. The slope of the normal is the negative reciprocal of the slope of tangent. In differentiating dy/dx ::::: we must regard y as a function of -xlv x and differentiate the right side as a quotient. (f) 2y{dy/dx) :.31
CHAPTER 7. At P. SECTION 4
1.(Yo/2p)(x . Hence the equation of the normal is y _ y ::::.:: On the lower half of the circle we must take O. 5. 0
. Cd) 2x+x(dy/dx)+y+2y(dy/dx) = 0. Now use the point-slope form of the equation of the straight line. 3.(dy/dx)+x2y(dy/dx)+y2+dy/dx+2 ~ 0. To obtain the x-intercept.:: Then dy/dx = 4/y.:: -!25-xz. x :. 2.
The least distance is a . SECTION
5 the circle and use the pythagorean theorem. Divide
::::.c)/(a + c) ::::2%0' Divide numerator and denominator by a. (a) Use (37). Now use (37). 5. e:::: -f5/5. 13. y2:::: (a2 .
12. We may use (27). Since for the hyperbola a2 + b2 = c2 and a < c. By (27) when x ::::c. Since c2::: a2 + b2.c2.flO. 10. (d) 2a = 12 and 2b == 8.b2. 2a = 12 and 2b = 6. see that a:::: 4 and b:= 3.
through
by 30 we have x2/10 -
r/B
=
1. 2a =. Now use (27). b:::: V60. Since (7.
e:::: cia :::: 17/4.. Since a:::: 6 and b:::: M. 11. (c) Use (37)./a2 . Since c2 . Repeat the derivation of (27) but with x and y interchanged. equal to or less than b. Now use (37). Then (1 .c2) (b2/a2) 2./2. y:::: b2/a. c = ill. (b) Here 2c ::::4 and since b2 ::::a2 . .12 and 2b:::: 2v'1O. Repeat the derivation of (37) but with x and y interchanged. 2) lies on the curve we may substitute 7 for x and 2 for y and b2:::: 149~5' Hence x2/149 + 25y2/149 ::::1.Jf. But y = I~ so the negative reciprocals of each other. Then x (b) We have by (27). Slope of BC = y/(x-l). Then 2y:::: 2b2/a .y) on 2. c::::.
1. By comparison with (27) we
C
both sides by BO so that x2/10 + y2/B ::::1. Since c ::::..c and the greatest distance is a + c.. a::::. Take any point (x. (b) 2c = 20. c = 5.. that the two slopes are 3.a2)(b2/a2).
a ::::. Now follow Exercise 1. 14. Now repeat
Exercise
11
. We want twice the y-value at x ::::c. 7. 16 • We want twice the y-value at x = c.a2 = b2.e)/(l + e) = 2%0' Solve for e. = bo/a
4. By (37) at x = C. b:::: 18. 6.. Then (a . all we can conclude is that a may be greater than. Slope of AC = y/(x+l). 17 .j10. so that c ::::10. 15 . b = 6. Ifwe compare with (37) we have a = 4 and b::: 3. b:::: v'B.32
CHAPTER
7. x2/25b2 + y2/b2 ::::1. e = 3/15. (d) 2a:::: 12 and 2b:= B. -
8. (a) By (27) we get the text answer. Here 2a == 50 and b:::: 25.a2. 2/64 + y2/60 = 1. y2 = (c2 . Ans. Since b2 = c2 .. 9.
cia = If we divide
sla.
aZ. As x increases the slope decreases and as x becomes infinite the quantity x/vx2 . At x = -1. To apply this argument correctly in the present case we should consider the product (5x + 2y . We find from y= (b/a). 2) is yl= ± (2JD/15). At x = 5. From the given equation yl"".. 4x/5y. as above. 3.a2yc . 4. Hence we apply (7) with rn2 = . Here x = . This angle is the supplement of the one we are interested in and the tangent of the latter is the negative of the one we find by formula (7). The tangent line has the slope . 6. Then the slope approaches b/a which is the slope of the asymptote y =bx/a.b2xy). Y= ± 6/J5 and yl= 'f (2/5/15). y) and (c. The method is the same as in Exercise 2. Hence Y'f ill == ± (2//3')(x . The line FP has the slope deter mined by (x.a2y2+ b2xc)/ (a2xy.c). by differentiating.a2 approaches 1 because x is very large compared to a. In using formula (7) we must remember to let m2 be the slope of the line with the larger inclination and that our formula gives the tangent of the angle between the upward directions on the lines. 8x + 10yy'= 0 so that yl= -4x/5y.4y2. At x =. In the numerator we use the fact that b2x2 + a2y2= a2b2 and in the denominator we use the fact that a2 . Then application of (7) gives for the fir st angle (.2y . and take the negative of the result. (a) We shall show that the angle between FP and the tangent equals the angle between F'P and the tangent by using formula (7) of the Appendix to Chap.5.5. 0) and so y/ (x .25)(5x . (b) The law of reflection for light says that the reflected ray will travel toward F. The slope (see Exer. 7. The method is the same as in Exercise 5. 5. 2.b2x2. Now use the point-slope form of the equation of the straight line.b2x/a2y and ffi1 = y/(x + c).16) and conclude that the equation 25x2. SECTION 6 1.b2 = c2• Then the fraction becomes b2jyc. In using formula (7) to get the angle between F'P and the tangent we must remember again that the formula gives the tangent of the angle between the upward directions of the two lines. At x = a the slope is infinite.5). Y= ± 6/15.33
18.b2x/a2y. Y= ±ill..
. y'= ±2/f3. The argument given in connection with (38) rests on the fact that if a product
vanishes then one of the factors must vanish. Hence the two angles are equal. From the given equation we have. 3. At x = 1.Ix2-a2 that y' = (b/a)x/v'x2 . This negative is also b2!yc. Hence.205x + 18y + 400 == 0 represents the two lines. CHAPTER 7. using the point-slope form gives Y'f 6v'5= ± (2/5/15)(x + 1).
ay = 0 and bx + ay == O. Thus the product under consideration is given by (a2b2 .b2x/a2y and the slope of the hyperbola is x(k . For k = b by (a2 . Thug the hyperbola becomes X2/(a2 .
11.Yo) to each point of intersection is the same. y) to the line is (b2xoX+ a2yoy . then B2 = c2 .34
a. Then the distance from (xo. except that for the hyperbola a2 > k >b2.Yo== (b2xola2yo)(x . The points of intersection of ellipse and hyperbola are obtained by solving the two equations simultaneously and they are given by x2 :::a2 (a2 . The method is precisely the same as in Exercise 7.
°
.y2/(c2 .a2yoy = b2x5 . reverting to the usual x and Y.o). The tangent line to the ellipse at (xoYo)may be written as b2xoX+ a2yoy ::::b2xg + a2yg == a2b2.a2y~. (-c. For k== a2 there is no locus (or. That is.
9.k) . For k > a2 there is no locus.b2).ro/2p. The equation of the tangent at that point is y . Multiplying out.a2b2)/v'(b2xo)2 + (a2YoJ2. For k between b2 and a2 we have a hyperbola. a hyperbola confocal with an ellipse can be represented with an a2 and lJ2 which differ from those of the ellipse by the quantity k.k.xo)' Let y =0 and solve for x.y2/{k .b2).k) we get y = 0.k) .b2)/y(a2 . which does not restrict A because k is still arbitrary. To make the algebra easier all we need show is that the product of the slopes of the ellipse and hyperbola.k)y2 has the value -1 at the points of intersection. Thus if x2/a2 + y2/b2 = 1 is the equation of an ellipse and x2/A2 . Thus the distance of any point (x.y2/B2 = 1 is the equation of a hyperbola we know that b2 = a2 .a2 + k. If we write A2 = a2 .b2)/ (a2 .k) / (a2 . As k varies from to b2. We find the point of inter section of the tangent and the first asymptote and the same for the second. Then the slope of the ellipse is .k) = 1.b2)x2/a2(a2 .k).k){b2 .xo) or b2xoX.A2 if the two curves have the same toct.Yo== (2p/yo)(x . If we substitute the x2 and y2 of the points of intersection we do obtain -1.Yo) is s'= 2p/yo' The equation of the tangent line is y .
13. 12. the locus is 2 there is no locus (though if we multiply through first an ellipse.b2) ::::1 or x2/(a2 .b2xoc)(a2b2 + b2xoc)/(b4xg + aVo). which product is . The equations of the asymptotes are bx . using c2 = a2 . y2::: b2 (k .
The problem assumes that a2> b2. x2/(a2 .k) + y2/(lJ2 . But ro/2p = 2xo.o).a2 + k) = lor. this fraction reduces to b2.Yo) of a hyperbola is s'= lJ2xola2yo.The foci have coordinates (e .
10. as before) the y-axis. One need watch only the differences in signs. Then x = Xo.b2{k .c2 and B2::::c2 . The slope at any point (xo. which is the x-axts). The student could be asked to show that a confocal ellipse and hyperbola can be represented in the form given in Exercise 8. The slope of the tangent line at (xo.b2 in the numerator and a2(a2Y5)== a2(a2b2 -:-b2X5) in the denominator.
y = (x2+7x)4+C. Let u = (x2+7x). dy/dx = (1+SX)1/2 • Now use the method of (h) to obtain the text's answer.b2xo + xo' This is OT. we a have the result. . Then x == a2yg. Set y ==0 and solve for x. If we let u = X2+~X. Then. our dy/dx is in the form (47). Hence write dy/dx = -1/3(2-3x)-1/2(-3).Yo:::::(a2YoIb2xo)(x . the fact that b2xg + a2Y5::::: 2lfl. Let u:::::x2 + 1. Hence we write dy/dx = 1/2 {x2+6x)4 (2x+6). Then (48) yields y = 1/2[1+(x/2)]4+C. the slope of the normal is a2Yo/b2xo and the equation of the normal is y . Let u = 2-3x. Hence write dy/dx = 1/5 (l+Sx)55 and so except for the constant factor 1/5 we have the form (47).xo)' The length OG is the x. Then if u::::: (x2 + 1) we have the form (47) and (48) gives the integral. (x We can write the given dy/dx as dy/dx = 1/2(x2+l)32x.
Since the slope of the tangent to the ellipse at P which has coordinates (xo. Then (48) and the factor 1/2 yield the text's answer. Since ON ==xo. The length OT is the x-intercept of the tangent. We could expand (1+SX)5.Xu). If we write dy/dx = (x2+6)-52x and let u: x2+6 we have the form (47). dy/dx = (x2+4)-3/2X = 1/2{x2+4)-3/22x. Then by (48).(b2/a2)xo = xo[1 . dy/dx is in the form (47). Then (48) gives the text's answer.
. is -b2xo/a2yo. In this exercise the choice of u is not obviously helpful and must be regarded as a'trial. The 4'5 cancel and the text's answer results. then we need 2x+6 as our du/dx. Let u = x2+4. (a) Ignore
(b) (c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
( k)
(1)
(m)
temporarily the factor 4. Same as (d) except for the factor 4. The second part of the problem calls for the same procedure except that ot is the y-intercept of the tangent and OM ==Yo' 7 ~ SECTION 7
CHAPTER
1. But it is easier to let u = l+Sx. is . If we add and use. Then du/dx = 5. Now ignore the 1/2 for the moment and we have the form (47). Then du/dx = -3. Set y == 0 in the equation of the normal. Yo) 2yo) and the equation of the tangent is y .Yo say. We can write dy/dx = 2(1+x/2) 3(1/2). say. The slope of the tangent to the ellipse at P. Now let u = 1+x/2.Y"'" (1+Sx)6/30+C. Now except for the constant factor 1/2. which has coordinates (xo.intercept of the normal. dy/dx = (x2 +6x) 5 /lO+C. Then by (48) y = -2/3 (2-3x)1/2+C. Then 2)] ::::: 2/a2) == e2xo' x == Xo . by (48). dy/dx = (2-3x)-1/2. If we do let u = 4+5x'then dy/dx This expression is still = lu(u-4)/S = (1/S)uS/2-(4/5)u1/2. and (48) gives y ::::: 2 + 1)4 + C. Yo). Then we have the form (47). Ans. ON·OT ==a2yg/i>2 +x5.(b2xo/a ::::: (b2xo/a2YoHx .(b2/a xo(c 15.35
14. Then apart from the constant factor 1/2 we have (47) and (48) gives the text's answer. Then apart from the constant factor 4 which can be temporarily ignored.
Then v' =u(du/dx) and we have the form (47). The answer is (-3/2) (x2+5 -l+C. apart from a constant facto~is in the form (47) and (48) applied to each term. Then the integral is in the form (48).8000· 5280. By (47) W = kU-1/2+C = k[p2+(C-X)2]-1/2+C. W = O. 4. Then we have the form (48).1 and if we integrate we get 2X4 4x8 + 3x2.32R2.1)8 we get 8xs .X + C. Formula (65) gives the velocity acquired in falling from rest or zero velocity and from the height r 1 above the center of the earth to the surface. When x = 9. The two results seem to differ by %. (h) write the given integral as (3/2) I(x +2X)-2(2x+2)dx and use (48)with u = x + 2x.1)0/8we get 2X4 4x8 + 3x2. When x = 0. 7-20 we see that W = k/b-k/a. y = (u2/2)+ C = {[g(x)]2/2} + C. The answer is 2 (x2+7x)1/2+C.36
On the other hand if we expand (2x . By (48) the answer is (-1/2) (x2+2x)-l+C. The arithmetic gives 3200. yields the text's answer.
. and R = 4000' 5280. Then dW/dx = -(k/2)u-3/2du/dx. 3. 2. if we shoot an object up from the surface with the velocity acquired in falling to the surface it will arrive at height r 1 (above the center of the earth) with 0 velocity. -
L We have but to apply (65) in which r1 = 5000' 5280. As the text points out. because the constant of integration is an arbitrary value. SECTION 8
not in the form (47) because each term lacks the factor du/dx ors and so we write dy/dx = (1/2S)u3/2S-(4/2S)Ul/2S. The rest is just arithmetic to get the text's answer.J66 ft/sec. (a) Let u = x2+S. Then the least velocity with which the object should be shot up is given by (65) with R:::. (c) write the given integral as (X2+S)-2X dx and apply the method of (b). Let u = x2+S. Now each term. S. (b) Write the given integral as (1/2)f(x2+S) 22x dx. (g) write the given integral as 1/2 f(x2+2x)-2(2x+2)dx and let u = (x2+2x).32R2 r1 :::.4000· 5280.12x2+ 6x . If we use the fact that f(x) = g'(x) then y' = g(x)g'(x). GM:::.-22X dx and proceed as in (cl. Then we have the form (48) and the answer is (1/2) (x2+S)3/3 = (1/6)(x2+S) 3+C. (d) write the given integral as (3/2)'{<X2+S. From Fig. Now let u = g(x). If we expand (2x . Then du/dx = g'(x). (f) Let u = x2+7x. Hence C = -k'(p2+C2)-1/2. 2. Now x varies from 0 to ~. . The answer is (-1/2' (X2+S)-1+C. Then dW/dx = k[p2+(C-X)2]-3/2(C-X). By (48). W = k[p2+(C-~ )2]-1/2_k(p2+C2)-1/2. Then du/dx = -2(c-x). Let q = p2+ (C-X)2.X + % + C for the integral. the two solutions differ seemingly by 1/8 but the constant of integration can always be adjusted to take into account any constant. But the constant of integration in the second case can be taken to be 1/8 + C' where C" is some new constant. (e) Let u = x3+ x2• Then the integral is in the form (48). such as the % here. GM:::. The answer a s (-3/2) (x2+2x-1+C. Using the suggestion in the text we get dW/dx = k(c-X)/[p2+(C-X)2]3h. That is. CHAPTER 7.
The problem assumes the body has a velocity V in the downward direction. This value is larger than the value in Exercise 2.37
3. There we obtained 6400-133. Then Vo = . In mathematical terms lim GM/rl > O. the vevelocity on reaching the earth's surface is given by (65)where r1 = 2R. Then v~/2 = GM/2R or v~ = 32R in view of (58). R = 4000·5280 and r1 = 8000·5280. Now M =
. r_
l
oo
11. Since in the determination of Conly y2 enters we cannot be sure that the signs are correct. 5~ According to (68) the escape velocity is 8v'R. The initial velocity must be such that it supplies the loss in v which is due to gravity and still leave a velocity of 1000 at the height of 8000 miles from the center.016 ft/sec. as it should be.Jy2 + 2GM!R . Hence the two velocities are equal. 12. The object will certainly never return.vV2+ 2GM!R . However GM/r . we can use (67) if we let M = M/81 and R be 4R/15. The velocity of 1000 it/sec. Then y2/2= GM/r1 + C or v2/2 = GM/r . For -the particle which falls under the true acceleration of gravity. 4. That is. and R = 4000·5280 and solve for rl" The answer is 320 miles approximately. In view of the answer to Exercise 11. When r = R we obtain v = .GM/r1 + V2/2.J(15/324)(2GM!R). The calculation yields 26..32r + 16R. In the theory of the section we have but to let M stand for the mass of the moon and R stand for the radius of the moon. since R = 4000·5280.2GM!r1 where V = 1000. Then C = 32(4000·5280) and v2/2 = -321' + 32(4000·5280). as it should to produce a larger v. To obtain the velocity acquired by a particle falling from rest with the acceleration of 32 ft/sec2 we may use the reasoning of Exercise 3 which leads to v2/2 =.32r + C. Use (65) with Vo = 10..2GM!r l' the minus entering because a downward v should be negative. is in the upward (positive) direction.GM/r1 is positive because r < r1• Then the value of this difference adds to V2. We seek the value of v when r = O. We could solve this problem by the methods of Chapter 3. Again use (65) with Vo = 5280 and R = 4000·5280 and solve for rl• The answer is about 84 miles. Now v = 0 when r = R/2 so that C = 16R and v2/2 = . If we measure r from the surface of the earth (which we may do in this problem as opposed to the use of (59)) then w hat we want is that v should be 0 when r = 4000' 5280. by (61).This is the value of 8. 10. In Exercise 3 we calculated the velocity required to send an object 4000 miles up if the acceleration of gravity were 32 ft/sec2 all the way. we seek Vo =+. 7.. 6. Section 3. See for example Exercise 12 there. We may use the result (63) but must now determine C by the condition that at r = rv v = Y. Hence the velocity we seek is really the opposite sign from that calculated in the latter part of Exercise 6. because the acceleration of . This is 6400-/33ft/sec.fR. When the object reaches the surface r = 0 and v~= 32R. 8. v(dv/dr) =-32 so that v2/2 =-32r + C. Or we can use the technique ot this chapter to argue that v =-32 and.000.J(2GM!81)(15!4R) . 9..32 is larger than the true acceleration of gravity and so more initial velocity is required to have the object reach a height of 4000 miles.
93 mi/sec. and r1 = 240. Or v~ = (2GM /R) (15/432 . as justified in (61). From the data of Exercise (12) we know that M = M/81 and R = 4R/15. on the accuracy to which the calculations are carried. or v~ = (2GM/R){1 + 330.j2GM/R.j2GM/R = . Then v2/2 = GM/r + GS/ (r + d) + C.
15. the mass of the moon. The object must be shot up from the moon to just reach the point which is 6R from the moon's surface.99} = 6. We can use (68). Hence we must now calculate the velocity required to shoot an object up from the surface of the earth to just reach the point r1 = 54R. To integrate we replace d2r/dt2 by v(dv/dr). The precise answer depends.99(6.. We use (65) so that v~ = 2GM/R .%4)' Then Vo= . 14..000){5280) and R = (4000){5280)to calculate vo' The arithmetic yields 8.21. The equation involving rs says that the earth's gravitational attraction on the mass m just equals the moon's gravitational attraction of the mass m. We use (65) with r1 = (240. This gives the value of C so that (1) v2/2 = GM(l/r . R.21(7mi/sec) = 1. The velocity acquired on reaching the surface of the earth is obtained by replacing r by R.1/r1) + GS[ 1/ (r + d).93). R being the radius of the earth. Since 8. so that Vo= . We want v to be when r = r1.50 mi/sec approx.99 times the more accurate value than the 7 mi/sec.JR. 16. 6.. Then (65) reads: vg/2 = (GM/81){15/4R) .000M/(R + 93· 106)]. We use (65) with M = M/8! and R = 4R/15 and r1 = 6R.40 mi/sec.j59/60. This is the velocity acquired in falling from rest at the distance r 1 from the surface of the earth to the distance r from the surface. the radius of the moon. We start with (63) and let v = Vowhen r = R.106 mi.3 mi/sec. The arithmetic gives. 5280). ..j2GM/R.. Substituting this value for C in (63) gives the text's answer.
and R still refer to the mass and radius of the earth. Then Vo= 1.000/[1 + (93' 106/R)]} = (2GM/R)(15. Then. (a) The given differential equation in the Suggestion replaces (59) and the reason is that two accelerations both in the direction of negative r act on the object and so the accelerations add. The result in the text. that the value here will be slightly less than the value there. Then v~ = 2G(M/81){15/4R) -~G (M/81)(1/ 6R). Then C = (v~/2).038).49 mi/sec approximately..
°
.000M and d = 93. the result is about = 6.. To calculate Vowe use the fact that S := 330.000.J53/54 = 7(. We can see by looking at the value of Vo in Exercise 12. is . This latter velocity is also the velocity with which the object must be shot up from the surface of the earth to just reach r r(b) As in the derivation of (67) we use (1) above in which r = Rand r 1 becomes infinite.88 mi/sec. namely.JR 7 mr/sec. as noted in the text... and integrate with respect to r. The result is 1.(GM/81){1/240.
18. Vo= 27.. Then v2 := (2GM/R) + [2G· 330..000 mi.38
13.1/ (r 1 + d)].j2GM/R = 7.86.2GM/54R = (2GM/R)(1 . of course. rs = 54R. Here we use (65) but with M.115/324 . since .2).
17.%86) = (2GM/R) (. Hence Vo= .(GM/R). Then the escape velocity Vo is given by vg/2 = GM/R + GS/(R + d).
dr (b) Here r = 5 and dr/dt = 1/2 so that dA/dt = 511" sq ft/sec. Let s = 5 and ds/dt = 1. The suggestion is to assume that dV/ dt = kA. 4.05 and thus the text answer. First one obtains v2/2 =: (GM/R) (.r)2.
(b)
The differential equation of the text can be written as v (dv/dr) = . hen dV/dt = (1Th2/9)dh/dt. SECTION 9 L Since V = e3. Hence the differential equation in the text takes into account the continuing accelerations of the earth and moon. dV/ dt = 411"r2dr/dt. We may use the text result of Sect. m = 10. so that v2/2 = GM/r + GM/81{60R-r) . 2. then the acceleration due to the earth's attraction is negative and is -GM/r2• The moon's attraction is in the positive direction and when the object is r units from the earth's center it is 60R .625)/rl/2.Also A = 411"r2.
CHAPTER 7. dA/dt = 2sds/dt. 6. These two accelerations act at all distances r from the earth's surface to the moon's surface.r units from the moon's center because the distance between centers is 60R.t/24) + %.{W Then r = (. The distance r between them is given by the Pythagorean theorem.jCGM/81\ (60R-x)-2(-1)dr + C or v2/2 = GM/r + (GM/81)/{60R-r) + C. Now de/dt = 0. Weare given that dV/dt = 1728.1250t + 625. (c) No. We are given that r = h/3 so that V = 1Th3j27. (d) No.GM/r2 + CGM/81)/C60R-r) 2.. we have dA/ dt = 211T / dt.3 T 3. Then dh/dt
=
1296/~ in/min. where t is measured from noon. Then dr/dt = 285/. r = 1/4 so that k = _l. Now let t = 2. We now wish to find v when r = R. This yields C = -8GM/9.[89 ft/sec.1)..8 namely dW/dt =-(GmM/r2)dr/dt. 7. Integration yields v2/2 = CM/T .96 mi/sec. 5. when h = 6.in/min.54R. because even if dr/dt is constant.1)2= . Then the acceleration due to the moon is G(M/81)/(60R . r = 1/2. (a) From A = 1TrZ. r = .
cu.I(20t}2 + 252(t. and dr/dt = 25. r certainly changes from instant to instant and dr /dt may also change from one instant to another. Now depending on the accuracy of the calculation one obtains v = 6. The volume of a cone is V = ll'r2h/3. The-distance the second ship sails south in t hours is 25(t .I1025t2 . Since A = S2. If one uses GM = 32R2.
. Then dr /dt = (1025t .59)-(8/9. dV/dt := 3e2de/dt. Hence v2/2 = (GM/R) (1+(1/81.54)).8GM/9.54R.. At the surface of the earth r = Rand GM/R2 = 32. Then dW/dt = 8'106• At the height of 100 miles. Since V = 41Tr3/3. When t = 6. Just to compare results write r = (4100/R)R where R = 4000 miles.000.9837) . Then dr/dt = k and r = kt + C. Then dA/dt = 10 sq ft/ min. r = 4100. Then substituting in dW/dt gives (40/41)2 times the preceding result. We now require that v = 0 at the stagnation point where r = 54R. then v is in ft/sec.. The distance the first ship sails east is 20t. When t = 0. r changes during the next second.39
If r is measured positively in the direction up from the earth's surface toward the moon.. Then r = kt + %.
We find that h::: 2 and k :::3/4. (a) If we let x
y::::: ' + k. Then 2B = 90° and B = 45°. The resulting equation is X'2+ s" = 25. Graph as recommended in (a). (g) The method is as in (a). The resulting equation is as in the text. (e) The method is the same as in (d). By choosing new axes with origin at any point on the line we can eliminate the constant term. (h) The method is as in (a). (g) Tan26 == 14/(25 . Yes. CHAPTER 7.212. Likewise if we set the coefficient of y' :::: we find k = 2. We find that h = -1 and k = 5. x = (. Then 2B:::: 90. (f) The method is the same as in (e) except that now we first find k to eliminate the s' term and then set the constant equal to 0 to determine h.186 .5 and k =: 3. We find that h = 4 and k = -77/4. (b) The method is the same as in (a).axis parallel to the straight line we could put its equation in the form s' = d. The 0 sum of the constant terms in the new equation becomes .0) or 2B == 90° and B == 45°.25). This is obvious geometrically because the line would go through the new origin. (i ) If we replace x by x' + hand y by s' + k we find that we can take k to be 0 and so eliminate the y' term. We find that h =. (c) Use the method of (a).s" = 24. Yes.J2/2)(x' +y').
= x'
+ hand
. Setting the constant term equal to 0 given h = 2. SECTION A3
1. we find h = -16. We find that h = -1 and k = 3. 2. The final equation is X'2+ 4y'2 . If we took an x'. Hence the answer in the text. The equations (6) are as in (e) and substitution in xy = 12 yields X'2. We find that h = 3 and k = . substitute in the given equation. Substitution in the given equation yields the text's answer. Now set the sum of the constant terms equal to 0. The transformed equation is 3X'2+ 4y'2 . By (6) of P. this yields k = .1). We find that h =.J2/2)(x' .4 and k = -7. and B:::: 45°.29/4 = O.41
(e) Tan 2B :::: .2/(1 . (d) If we replace x by x' + hand y by y' + k we can set the coefficient of x' = 0 and find that h::::: 3. The equations (6) are as in (e) and substitution in the given equation yields the text's answer. The resulting equation is s" . APPENDIX.y') and y == (. Graph the new equation with respect to the new axes and then put in the (x.23/4. (j) The method is as in (f). (f) Tan2B == 1/(0 . y)-axes. The resulting equation is in the text. 4. take s the terms involving x' and set the coefficient equal to 0. The resulting equation is y'2::: 8x'.11 :::O. The resulting equation is in the text.2.6x' = O. The r esulting equatio n is that in the text.
(y .equation becomes (x .b2)/4a. Then a = ill.4)2/80 = 1 and. We note here that the larger number of 25 and 36 is under the y2 term. This gives h = . b = 4 and the center is (-4. 4). (b) The focus of y = x2/4p is at (O. These are the x' and y' of the focus. After completing the square the equation becomes (x + 3)2/16. (a) Write the given equation as l6(x2 . (d) The numerical answers in this exercise would be simpler if the constant on the right side were changed to 288. The answers are in the text. Thus 16(X2 ax + 16) . (4ac .to x'2/45 . The translation determined in (a)'is x = x' ..5)2/25= -l. (4ac .5). To obtain the equation of this line in the xy-system we have that y = y' + k = (4ac .4)2.. is v' = -1/4a.y'2/36 = -1.3)2/25 + (y + 6)2/36= 1.b/2a. Now we can fix k so that the constant term is 0.225 or 16(x .y'2/16 = 1. and in view of the value of h.b2 .0). This gives k = ah" + bh + c. We see by inspection that if we let x' = X . k = (4ac . This means
.4 and y' = y + 3 we have 16x'2. According to Exercise 17 on p.42
3. (a) If we substitute x = x' + hand y = y'+ k in the given equation we find that we can eliminate the x' term by setting its coefficient equal to 0.4 and y = s' + 5 the coordinates x and y of the center are (4.4)2/36. since p = 1/4a.b2)/4a.(y .(y . Now complete the square in each parenthesis and compensate by adding the equivalent term on the right side. Then if we let x' = x .25(y2+ 6y + 9) = 369 + 256 . The coordinates of the focus with respect to the new axes are (0. 4.b2)/4a). (e) Use the method in (a). Then the equations for translation are x =x/+3 and y=y'-2. (b) Use the method in (a). Then the x and y of the focus are (-b/2a. (c) Use the method of (a). Hence the x and y of the center are (4. a = 5. The coordinates of the vertex of y' = ax? are (0.b2 + 1)/4a). (a) Completing the square yields (x .1/4a). The given equation can be put in the form (x + 4)2/16.1)/4a. (c) The directrix of y = x2/4p is Y= -po Then for y' = ax'2 the directrix.5).J8O and the xy-center is (. The x' and y' of the center is (0. See the introduction to Exercise 5.(y + 5)2/36 = -1 and the translated equation is x'2/36 . According to Exercise 17 on p~64 .l64 .25(y + 3)2= 400.6. Then the coordinates of the vertex in the xy-system are (-b/2a.-3). y = y' + (4ac . The. b = . 1/4a). Since x' = x . In each of the parts of Exercise 5. However it is well to teach the method of completing the square which in this exercise and the next one also gives the answers we want more readily. we could follow the method used in Exercise 1 of replacing x by x' + hand y by y' + k to eliminate the x and y terms.25y'2 = 400 or x'o/25 .8x) .p) or for the equation y' = axf2 at (0.2)2/36 = 1. Then the method of (a) leads to (x + 6)2/45. a = 6 and b = 6.0).y'2/80 = 1. We see that a = 5 and b = 4.3 and v' = Y + 2 the equation reduces to the standard form.25(y2+ 6y) = 369. 6. The new equation becomes y' = ax" which is of the form y = x2/4p with a = 1/4p. 5.b/2a.
No translation of axes need be applied. cos 28 = 3/m. Completing the square yields (x + 2)2/25 + (y + 3)2.3). b = 4 and the center is (2.27 = O.1). Proceed as in (b).14-12 y' + 21 = O. Then a = 6. Then sin 8 = Jun . APPENDIX.-6). Tan2e =-2/(11). (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
that (see Exercise 8 on p. Thus in (a). Tan28 = %. b = 5 and the center is (3.2)2/16 + (y . Completing the square yields (x . b = 3 and the center is (. SECTION A4 1. Hence e = 45°.2)2/16 = 1. Substitution uf the equations (6) for rotation yields x':. 2).13). 3). The result is in the text.
.444 = O.1.{f5/2 ) + y'2(% + {f. h = . The result of the substitution gives (2 + 11/ill)x'2 + (2 . Hence a = 5.3)2/9= 1. Tan28 =-4/3.9= 1. the minus entering because 28 is a second quadrant angle.3)2/25 + (y . Tan 28 = %.! (% . Substitution in the given equation gives the final ( result.
CHAPTER 7.3/2 } .18x' + 24y' . To avoid decimals use the formulas on the bottom of p. B = 4. Now translate axes.3.11/ill)y'2 + linear terms. Completing the square yields (x . Hence the results in the text. The equations for rotation are x = (v'2/2)(X'. Tan2e = 10/(13 .8y'2 .4AC to each part of Exercise 7 of the preceding list. Proceed as in (b). 1//5)(2x' +y'). The equations for rotation are x = (1/15)(x' . Hence a = 7.4AC = 0 and the curve is a parabola. Now translation of axes gives the text result. Completing the square yields (x + 3)2/36 + (y . We find that cos 28 = -3/5. Also sin e cos 8 =-l/ffi. The result of the rotation is 18x'2 . Tan2e = 2/(1.5). Hence the equations for rotation are the same as in (c). Here as in (a) the teet are on the (new) y' -axis. We can determine h to eliminate the x-term and then determine k to eliminate the constant.3)2/49 = 1. We obtain just by rotation X'2 . A = 1 and C = 4. 163) that the foci are on y' -axis.4 = O.43
(b)
(c) (d)
(e)
7. Substitution in the given equation yields 8X'2+ 2y'2 + 8v2x' .J::. k = -1. Hence e = 45°.196x .89 . Hence 8 = 45° and we proceed as in (b). The exercise is 49x2+ 16y2.3. We have but to apply B2. This fact is confirmed by the answer in (a).3)/2. b = 4 and the center is (3. Hence a = 6.96y . Hence 8 = 45°. Then sin e = 2//5 and cos 8 = 1/15". Translation of axes must now be applied to eliminate the linear terms.189 . We rotate first. Then B2 . To avoid decimals we shall use the formulas on the bottom of P.2v'2s'= o. No further simplification by translation can be obtained. Tan 28 = 6/(5 .jI3 and cos 8 = J(1i3 + 3)/2JI3.y') and y = (v2/2)(x' + y'). Since there is no xy-term we need apply only translation.2y') and y.
a degenerate hyperbola. according to Exercise 3(b).sin e)/(cos (:J + ill2 sin e). Then x2 + y2
the
~
X/2
+
v". (b) If we have two lines y = m1x + b1 and y = m2x + b2 and rotate axes. If we now write this equation in the form y' = m'x' + b' we obtain the slope ill' relative to the new axis . The slope of a curve is defined to be the slope of the tangent line. because for any (x.44
2.
5. The latter (see Exercise 3(a» is not invariant under rotation. (a) Yes.
9. .m1)/(l + m1m2). cos e .axis the slope must change. ITwe rotate the x. We find that x2 + y2 = X'2+ s". respectively. No. Under translation the Xi -axis is parallel to the x-axis and so the inclination of the line and therefore its slope remains the same. y) the expression x2 + y2 represents the square . That is.m~)/(l + m~m~)we obtain tan e = (m.
7.
4. the graph consisting of two intersecting straight lines must be classed among the hyperbolas if we are to include all second degree equations among the conic sections.y' Sin e) + b.y by x' + hand y' + k. The angle is a geometric fact about the two lines and so is independent of the choice of axes. (a) Yes. respectively. because under translation the distance of the point from origin changes. (a) No. The equation represents
3.
. (b) Replace x and. Ii we now calculate tan e = (m~. (a) No. Slope of a line is defined relative to the x-axis. (a) Yes. cos e . (b) Replace y and x in y = mx + b by y' + k and x' + h. m~ = (m. Under rotation this distance remains the same.sin e)/(COB + m1 sin e) and e m~ = (m. (b) Replace x and y by the values given by (6). The reason is the same as in Exercise 4(a).of its distance from the origin. (b) If we substitute equations (6) in y = mx + b we have x' sin e + y' cos e = m(x' cos e . Yes. 8. Then determine the slope -in the transformed equation. 6.
However. y' = 6x2+8. Hence y has a maximum at x = -1 and the y-value is 1. y = x3 . For x slightly less than -1.1)(x + 1). Hence no relative maxima or minima. Hence at x = 4.3 there is a minimum of -12. say. At x = 0.. -1 and 1 are possible values. Here y/= -3(x + 3)(x . Here y' = -8 and is never O. (b) The method is the same as in (a). At x = 1 there is a maximum of 20 and at x = . yl changes from + to . By testing each in turn for a change in s~gn in y' at 0. hence there is a minimum whose value is -3. At x = 1. For x slightly less than -1/6. say -3/4.to +.Solutions to Chapter 8
CHAPTER 8.4x = 4x(x . y' is not 0 for any real values of x. y' is positive and for x slightly greater than -1. y/changesfrom . s' changes from . y' = 4x3-4x = 4x(x-l) (x+l). yl = -3-18x. (e) r= -6(x -1)(x + 1)/(x2 + 1)2. say -5/4. Substitute x = -1/6 in the function to find the minimum y-value. Hence there is a minimum whose value is 2. y' does not change si1n at x = o. Hence y has a relative maximum at x = -1. y' = 3x2 • y' = 0 when x = O. and 1 we see that at x = 0. For x slightly less than 1. there is
(b) (c)
(d) (e)
( f)
. At x = -1. SECTION 3 1. y has a relative minimum. Hence there is a minimum whose value is 2. For x slightly less than 4. Hence at x = -1/6. Y = 2x3+8x+3. v= 0 when x = -1 and when x = + 1. y/=O at x=-l. yl is positive and for x slightly greater than -1 . Y has a minimum of -124. y' changes from + to -. Hence no relative maxima or minima. Likewise at x = 1. (d) y'=2x-2x-1. (a)
6x2 -18x -24 = 6(x + 1)(x -4). y = 2x3-6x. y' is positive. We see that r= 0 at x = 4 and x = -1. y' is negative and for x slightly greate~y' is positive. y' = 6x2-6. r= 4x3 . y' is negative and for x slightly greater. -I. s' changes from . hence at x = 0 there is a maximum which is O. At x=-I. Hence no relative maxima or minima. For x < -1. y' = 0 when x = -1/6. Then y/is o at x = 1 and x = . 0. At x = 1. y' is negative. (a) y = 16-3x-9x2 . yl is positive.
s':
CHAPTER 8. y = x'+-2x +12. hence at x = -1.to + . Hence at x = 1 the function has a relative minimum.to +.to +. s' changes from .3 . hence there is a maximum whose value is 3. yl is negative. there is a minimum of -1. y'is negative and for x > 4. Hence x = 0. At x = -1. there is a minimum which is -1.1). y = -8x~2. y' = 0 when x = +1 and x = -1. (c) The method is the same as in (a). SECTION 2
1.
We see that (x . f~ (x) does change from negative to positive around x = O.. (a) The function y = x is an example. say -3/4.sa relat1. Here as x increases. Possible values / / are x = 0 and x = 2. fl (x) changes from positive to negative.46
2. f' (x) changes from negative to positive. The absolute maxima and minima may occur at the end values 0 and 5.
.to +. At x r= 0. At x :::: . Hence x = 1 and x = -1 are possible values. f I (x) = (x2. y = 7 and at x = 5. f'(x) actually de-
a relative maximum: at x = -1 there is a relative maximum. As x increases or decreases from the value of 1. hence there is a relative minimum at x ~ 1. say -5/4. Hence there is a relative maximum whose value is 16. the behavior of. Hence the 0 relative maximum of 16 is also the absolute maximum and the absolute minimum is 7. At x = 0.ve mJ. (j) f (x) = x2. Now f' (x) = 0 at x = 2/3. s= 2(x -l)(x + 1)2 + 2 (x + 1)(x . 5. f' (x) = 4xa• Hence x = 0 is a possible value. (a)
v= -2x + 6.1). Hence there is a relative maximum whose value is 1. Hence there is a relative maximum of 4. However the function may have a relative maximum or minimum where the derivative fails to exist. This is the absolute maximum. Here y/= .1)1/3~ This y'is never O.to +. At x = 2. y continually decreases. £' (x) = (x/2/x-l) + lx-I. Hence there are no maxima and minima. s' changes from . Hence x = 3 is a possible relative maximum or minimum.~or x slightly less than -1. s' changes from + to -. At x = 4. f~ (x) is negative. The least we can subtract is 0 and this occurs when x = 1. Hence there is a relative minimum of O. s': -16. This is the absolute minimum. Hence there is a relative minimum and its value is 0. Hence there is a relative maximum at x = -l~ At x =1. s' changes from + to -.-2x) (x-I)2. (g) f(x) = x~. ft (x) = 1 . Then y = 3 is a relative maximum. fl (x) is positive and for x slightly more than -1.2/3(x . 4. (i) f(x) = xix-I. Hence there 1. But the function has no real value at x = 2/3. Y = 20.y/is as at x = -1 and the relative minimum is again 0.nimum there. At x = 1. at x = 1 there is a relative maximum. (b) '1. y = 12. Hence at x = 0 there is a relative minimum.= 3x(x -2). However y = _x2 in the interval from _00 to 0 isa better example.1)2/3 is positive for every value of x and this is subtracted from 3.1)2 = 4x(x + l)(x . Hence there is a relative maximum there. s' changes from + to -. At x = -2. At x = 0. (x-I). y'changes from . f' (~) ch~nges from negative to positive. At ~ = 2. At x = -1. (h) f(x) = x + l/x. At x = 3.l/x2. 3.
JI00 .x2/-JlOO . Alternatively. Then P == 3x + 2y. p/= 0 implies x = ±lA. One could test dD/dx to see that it changes sign at y == b but this step is often omitted in physical problems and would be lengthy in this one.
Let a be the given base and set b == 2s . To solve put the term 1 on the other side of the equation and then square both sides.x2.aHa + c . we have the square. We also have the condition that xy= A where A is constant. where A is constant. Then the distance in question is D == .p)2 + y2 + . h =: 80 . 3. If the sides have length x and y. Thus for least perimeter x = fA and y = A/x = fA. X == 0 implies the answers in the text.2c .. Set dD/dx =" 0.a-c.Jx2 + y2 = 10 we have y == 5V2. Let the distance from the foot of the altitude to B be denoted by t. However in the first radical we can replace y2 by 4px and obtain D == P + x + . Then A . Hence at t..a)2 + (y .2X2 and K= 100 .. 8. Then h == 1/1Tr2and A = == 21Tr2 2/r. dA/dc == 0 implies 2s . y) be any point on the parabola.::o2Y2there is a maximum which is 100. y := 864/x and p!== 3x + 1728/x. To save work we can argue that to minimize D is also to minimize 1)2.
==
.a) /-J(x . that ts. Then A == .a) is negative because x necessarily lies to the left of a.48
CHAPTER 8.a)". the diameter is 2. Then p' == 3 . Then z == x2 + y2 and y = A/x. A = xy and 2x + y = 100..2A2/x-3 and Zl == 0 implies x = fA.Jh2 + (2a .a) + (y . Since the rectangles are inscribed in a circle. Call D2.. Since 2s == a + b + c . so that z = x2 + A2/x2. Then y =: 36. we have b == c.b)2.b)jj//(x . p'= 0 implies t = a from which CB = CA follows.a)2 + (y ~ b)2.. z.1728x-2• pI == 0 implies x =: 24. If we substitute b for y in dD/dx we get dD/dx ::= 1 + (x . Then from .4x. one knows that y == b should be a value of y at which dD/dx::= O. If x and yare the dimensions of any rectangle p.
10./s(s .s)(s ./(x a)2 + (y . Hence z" == 2x . Let (x. The area A == xhoo .x2 .a I because the radical sign stands for a positive value.c) where only c is variable.a)2 == I x . Then CB = ylh2 + t2.a r= O.32t.
5. Then dD/dx == 1 + [(x . Hence dD/dx::= 1 + (-1) == O. x = . The surface area A == 21Tr2 + 21Trh. fi =: -32. 9. However the numerator (x . then . The result is a square with side f2. + We wish to minimize D == Jx2 + y2 with xy = A. 7. Hence A = 100x . CA = .b is a factor of the equation and so y =: b is a root. We may then use the method of Exercise 7 with 10 replaced by 2./(x.
6. Now J(x . But 1Tr2h 1. SECTION 5
1. Let x and y be the lengths of the sides./(x.fA give the minimum. To make D a function of one variable we can use the fact that y2 == 4px and we can replace x by its value in terms of y or y by its value in terms of x. 4.. replace y/by 2p/y and solve for y. Since xy == 864../x2 + y2 = 10. Alternatively we can keep x and y and regard y as a function of x. Then x = 2 5 and y =: 50.x2.b)".::o2x + 2y. It:::= a implies x == 5-/2. Then y =: A/x and po::: 2(x + A/x). 2. One finds that y .t)2. Hence y == fA.t)2 and P = 2a + -Jh2 + t2 + -Jh2 + (2a . 11.
The perimeter of the entire rectangle will be 200 + 2x + 2y :::.: 3y2/4 + 1 we see that zt1is positive when y = O. Hence at x := -1 there is a maximum whose value is O.
. 0.36 and we wish this to be negative. Here D == -. This does furnish the relative minimum though the testing b zit is lengthy (and can be ignored). Hence there is no relative maximum. In this domain the maximum value of A is given by x='O because A decreases as x increases from 0 on. y/= 3X2 + 2bx + 3. This expression is least when ex is as close to a2 as possible and this is when x is a for x in tit domain . At x:::: 1.a to a.::. For the given function s= 1.x2• Then yf= 0 when x == -1 and x:= +1. The discriminant of this quadratic is 4t2 .
9 and 19 let us work with 1)2 or z. Then x = 0 and y = 50.14.<.11= 2a. Since y2 = (iY'/a2)(a2 . y/is 0 at x.yI~/b and differentiate I)2 we find that the only real root of z is y::::.'(x------:1""")2::--+-y"2. However z can be put in the form (ex . Since y'': 2_X"swe see that at x = -1. We see from the expression for z that z = (x + 1)2 and the smallest value occurs at x :::: .c) . If we use the test that s' change sign at x := 0 we see that it does not. Then dA/dx::::' -50-2x and there is no positive value of x which maximizes A.x2).50 . the value -b/2a yields a maximum if a < O. The assertion is false.X2). Z == (x . The domain of possible values for x is 0::$ x . However x = -1 does not correspond to 2(x any point on the parabola and so there is no relative minimum for x-values of points on the parabola. Then z'= 2(x -. this value of x is larger than a and so lies outside the domain of admissible x. The dimensions are 50 by 100.
Then x::: O.:: [(y2/4) . z/::: 0 when x > a2/c.a y/= 0 implies x =-b/2a.is also 0 and so we have no text. When Z. x := -1. However there may be an absolute minimum. Hence y == 0 furnishes a
If
20.: 300.::.1) + 4. However. Since s> 3x2. However because the function is always increasing there is an absolute maximum at x ::::4.a2)2/a2. Q when y == O.x.with y::::. by eliminating y we have D x: "/(X . the function is always increasing.1)2 + 4x. y) is any point on the parabola then D = -. Let z =' D2. 15. As in Exercises
minimum. If (x.l]Y + 2y and Zl:.c)2 + y2./(x. As noted in Exercise 9 we can minimize 1)2 in place of D. Then A::::-(100 + x)y.. O. y~is positive and so there is a minimum there.
19.x) =: 5000 .
16.
18.:: . Then 1 b i must be less than 3. Since -:.11~y2 and z ::. y'(fis negative.. The domain of x is x ?:: O.1J2 + y2. Then z:= (x .2(b2/a2}x. Since y2 = 4~..50x . Hence A::::-(100 + x)(50 . Since a2 . D If we eliminate x from D we have D =: . O.
we calculate dF /dr we see that it is never zero.x2'. aVt} .C)2 + (b2/a2)(a2 . The rectangle will have dimensions 100 + x and y. Since Zll:::. We wish to have no real roots. y'.b2 := c2. Moreover since y ~s positive to the left and right of x = 0.$ 50.j[(y2/4) . Consider y = x . Since ale> 1. The minimum value is 2 and larger than the maximum. If we eliminate y from [)2 after writing IY = x2 ~ 2c-x + ell + y2 by x ::::.1}2 + 4x and Zl ::.
17.::. Then Z/= [(yl'!4) -.
However the man does save time by rowing as much as possible even to reach a point % of a mile from A. Let PA + PB + PC = S. Hence the minimum S occurs at x = 3 or when P is at C.. There is no relative minimum for P on CD. Then AP + QB = I_ 2 = .Jb2+ (d . Hence x/4. It is obvious in this case that using the diagonal saves as much time as possible since he can row as fast as he can walk.m)" + (m2. Since dS/dx = 2X/-Ix2+ 62 . Hence he should certainly save as much time as possible by rowing to B. x :::. In place of the previous dt/dx we have dt/dx = x/4. 29. dt/dx = x/3-J1 + x2 . Follow the method in the text used to derive (16). +~)/n.j1 + x2 < 1/.x)/5.(d . We could test d2t/dx2 to see that x = 3/4 furnishes a minimum. 25. Then S = 2-1x2 + 62 + 3 . 23...X)2.Ja+ x2 + -Ib2 + (d . we calculate dl.50
21..1/5. R and the foot of the perpendicular from B to EF.
30.m)". 27. The distance RS is fixed.Jb2+ (d . S. Then QS = d . . because x/ . Find ds/dt and set it equal to 0..x. 26. Since x cannot be larger than 1.m2) +. There is no relative maximum or minimum. + m2 + . Here B lies on the same side of CD as A does and vl = v2. Hence the problem reduces to minimizing AP + QB. . Denote the sum by S. Alternatively. Hence t decreases and is least when x = 1.. this value of x does not furnish a minimum in the domain 0 :. Hence the problem is the same as Exercise 27 with just a different physical interpretation.J1 + x2 .. + (m -~) = 0 or m = = (m.09°Centigrade. Then dS/dm = 0.1. In Exercise 27 we prove that the time AP + PB is least but as the suggestion points out this also means least distance. 2 2 d s/dt is negative in a < t < 0. then S = (m. implies (m . Hence at t = 4.. Hence the diagonal path PB is best.. Let x denote RP. However dS/dx is negative for 0:...x)/.m.) + (m ./dx which is x/-Ia2 + x2 . This derivative set equal to 0 gives x/. Hence we end up with sin a/sin 13 = V/Vl and so Q = 13. then the time for the trip as t = -11+ x2/3 + (1 . Then LAPR = Lms and AP and QB are parallel. 24. If we set this expression equal to 0 and solve for x we obtain x = 3/4.m)2 +. If we let AC = x.j1 + x2 < 1/4..X)2. x s 1.)2 which is less than %.x)/.3 and so S decreases as x increases. we can see that dt/dx is negative for 0 :S X :5 1. dS/dx = 0 at x = This value of x does not yield a point on CD.
m. 22. To minimize I. specific weight is a maximum.Ja2 + x2 = (d . 1.x. No matter where the bridge is placed the distance PQ must be covered in any case. + (m.
There t = 4. let it be d.[2 for O:S X :. 28. Let AR be a and let BS be b.1/5. Label the foot of the perpendicular from A to CD.. The maximum and minimum values occur at the end values x = r and x = -r of the permissible x-values.. we obtain x = %. If we set dt/dx = 0 and solve for x .. The left Side = cos APR and the right side = cos QSB.09 C the
..X)2..
x = 1 and x = 7. Let x be the number of new wells produced. The interest paid by the bank will be xD = kx2• The profit is P = .50000x2 . x = 250. dC/dx = 6. We find dPjdx and x when dP/dx = 0. 7.3+1S or 6.. Then ~ = 2x-6 = 0 so that x = 3 for minimum A. P = x(lOO-. The cost C = lO. A = £ = x2-6x+1S. (b) To maximize the percentage of profit we wish to maximize the return each year divided by the total cost. dT/dx = 2S-6x. Let x be the number of cards to be printed. x = 47~. At x = 1. = O. 6.20x. Let x be the interest rate offered.07k-2kx. x = 25/6. P = 88.500000 .. x aX Hence the minimum A is 32-6. dP/dx = .07kx-kx2. using f(x) for the percentage. Then dP/dx = SO-. Then dP/dx = -3x2+24x-21. The price at maximum profit is 7S-2. The total number of barrels T is T = (25+x)(100-3x) = 2500+2Sx-3x2. Profit = R-C and R = Px. 2 Hence y = 10. When dP/dx = 0. Hence dP/dx = dR/dx-dC/dx. P = R-C. C = a+clX. P = px-C = 7Sx-2x2 . FIRST SET 1. Then. with x the front and back lengths. Then if the cost per foot of the side walls and back is d. dP/dx = (-9/2)x+63. rhe height is immaterial. Henee Xl 2 =-a/e or x = a2/e2• 8. x = 14. At speed v the cost per hour is C = 12S+(1/lO)vs. _ 300000x _ 30x f(x) . If we test d2p/dx2 at x = 7. The cost per mile times the miles per hour = cost per hour. The number of barrels of oil produced per well will be lOO-3x. Let x be the number of articles to be produced. Practically one would choose 2 or 3.lOx2. The profit P after the first year is P = 300000x . The average cost is A = C/x = lOO/x+~x/100. + 100000x. Then C = SOOOOO + 100000~1+2+ . Then the cost is C = SOOOOO + 100000 + 200000 + . The return to the bank will be R = . S. 3x = 20. When dP/dx = 0. v = ~ = S ~.OOOd/x+3dx. Hence the profit p = 100x-lO. 11. Mult~ly through by X3/2. dA/dx = -a/x2-c/2xs/2.07D. When dA/dx = 0. we find that it is negative so that x = 7 yields a maximum profit. x2 = 10. Setting dA/d~/f 0 gives -a/x2-c~xS/2 = O.000 or x = 100. 9. 10.. The maximum profit which occurs when x = 14 is (-9/4)(14)2+63'14-3S0 = 92 dollars. Then -a/x -e = 0 or -a-exl . dC/dx = -20. When dP/dx = 0.03S. At x = 3. When dP/dx = 0.000/x and e2= 20. 4. dR/dx = dC/dx. Hence P = SSx-3x2-xs+1Sx2-76x-10 = -x3+12x2-21x-10. Then the total cost of x articles is C = 100+x/2+x2/lOO. But xy = 10.OOO+Sx+x2 (in pennies). 3. P = -20 and this would mean a deficit. Then the number of wells will be 2S+x. x = 2~. The dollars D attracted will be D = kx. Hence when dP/dx = 0.000. At speed v the miles per hour is v.14 = 47.000 and x = /20. x = .500000+50000xz+soooOx .+x) or C = SOOOOO + 100000[Z(1+x)1 or C = SOQOOO + SOOOOx + 50000x.OOOd/x +3d.3S0-l2-x2/4 = (-9/4)x2+63x-3S0. dP/dx = lOO-S-2x. 12. Then A = C/x = a/x+c!lX. When dP/dx = 0. 13. When x = 7. Hence the cost per mile or M = c/v = 125/v+(ILIO)v2 Then dM/dv = -125/v2+(2/10)v. SECTION 6. Let the dimensions of the floor be x and y. e = 2dy+dx+2dx.SOOOOx. (a) Suppose the building contains x floors.51
CHAPTER 8.. Hencewh~n dT/dx = 0.lOx)-lOOO-SOx = -1000+SOx-.SO+Sx+Sx2
.OOO~Sx-x2. 2.000/3. dA/dx = -lOO/x2+1/100. When-aMTdv = 0. The income is 100x. Now dC/dx = 3x2-l2x+1S. When dC/dx = 0.
Then dC/dx ~ -1000/x2+2. x '"approximately 2. x = $22. Then the reduction in price will be 30-x. The new demand function can be obtained from the old one by replacing p by (5/4)p so that (5/4)p = 10-3x or p = (4/5) (10-3x).) The cost function for the producer remains the same.50. pi (x) ~ 0 when x = 1. Under nc tax P(x) = 10x-3x -3x2= 7x-3x2. Hence the cost of the trip is C = 1~0 (10 + ~~)
=
10~0 + 2x. x = 25 dollars. x = 1500. 15. The quantity sold comes from solving p = (10/11)(20-4x) for x when p = ~ll. Let x be the sale price.
. SECTION 6. 4.10. At x miles per hour the number of hours required to make the 100 mile trip is 100/x.x). Now the consumer must pay 125% of what he previously paid. Then pI (x) = 7-6x and x = 7/6. Hence the additional sales are 10(30. This exercise and the next one follow the second illustrative example. The price p = 10-3(7/6) = 13/2. Hence the demand changes because the price is higher. When dR/dx '" 0. P I (x) = (200/11)-(80/11)x-4.95. for a given price p the demand will be less than under 3x = 10-p. Again practically x would be 3 or 4. The maximum profit is 125/48 and the corresponding price p is 11/2 without sales tax. At dC/dx = 0. Hence the profit is P. SECOND SET 1. The revenue is still xp where p = 10-3x. CHAPTER 8. Thus the new demand function is p = (lO/11)(20-4~ and since C(x) = 4x. The cost per hour is C = 10+x2/50. For each $2 in 30-x there will be ten more sales. Then dR/dx = 250-10x and when dR/dx = 0. P(x) '"x(lO/ll) (20-4x) -4x. (Of course since 3x = lO-(5/4)p. Then dR/dx = 22S-10x. The corresponding price p is $11. The cost function becomes C(x) = 3x+2x.52
Using the quotient rule gives l500-150x2 f 1 (x ) '" (50+5x+5x2)2 Then fl (x) = 0 gives x = IIU. 14. 16. 3. 2. Hence P(x) = x(4/S) (10-3x)-3x = 5x-(12/5)x2• Then PI(X) = 5-(24/5)x and pl(X) = 0 for x = 25/24.x) = xp-C = 10x-3x2-5x = 5x-3x2• pI (x) = 5-6x and this is a maximum when x = 5/6 and the price at which the commodity will be sold is p '" 10-3(5/6) = 15/2. The revenue will be R = 500x + (10/2) (30-x)x or R = 500x+150x-5x2 = 250x-5x2.lO. R = 100x +lO[(25-x)/2]x or R '" lOOx+125x-5x2. Using the same method as in #15. The solution follows that of exercise 3 except that 5/4 is replaced by 11/10.
3. No relative maxima or minima.+1)2. The horizontal tangent occurs at yt = 0 or 3ax2 + 2bx + c = O.2x/~"J. positive from G to 1.
5. (11)
(f) and (g) See (a)./s. There is a relative maximum at x = O. There is no point of inflection. The curve is always concave upward.4 from the left y approaches +co. (e) y' = . yll=2x(x2. As x approaches -3 from the right y approaches +00. yllis negative from A to C. 4.9)/(x + 3). There are no inflection points. (d) s'= % x2/s. (a). y is close to +1. y" = 12x2-48x.o
(6x _2)/(xF+l)3 there are poi. y is infinite at x = . Since yll-. y" = -8/(x + 4)3. Find the relative maxima and minima and points of inflection and use these as aids to plotting. y"= 1% x-J. Hence the inflection point is x = (r 1 + r 2 + r 3) /3. The point of inflection occurs at y" = 0 or 6ax + 2b = O. Using the factored form of y we find s" • 2a(3x . The zeros are clearly x = a and x = 4. There is a relative minimum at x = 1.r3). Then y goes to a minimum of -%6 at x = 1 and increases gradually to y = 1. we take the value of x from the second one and substitute in the first One.4 from the right y approaches -co. y approaches 2 as x approaches +00 or -'i<l. As x approaches .ntis of inflection at x '" :: 0/1/3 and y approaches 0 as x approaches +co and -00. y"= 36/(x + 3)2. Since these equations hold at the same value of x. positive from C to E. No relative maxima or minima and no points of inflection. Then JJ2 . As x approaches .53 CHAPTER 8. (b) and (c). SECTION 7 1. (b) and (c)
.r2 .
y'= 2
2. Point for point plotting with some attention to the behavior of y' and yll is all one can apply.r1 .
(-x2 + 1)/(x2 + 1)2.3 from the left y approaches +co. (j) y'= 4/(x + 4)2. When x is very large and positive or negative. Also y" changes sign at x = 0 and x = 4. negative from E to G. X :::' v3 and x = -[3 (i ) yl= (9x . There are points of inflection at x = 0.4. As x approaches . There is a relative minimum at x = -1 and a relative maximum at x =+ 1.3)(X2 + 1)/(x2 + 1)4.3ac = O. Hence yll = x(x-4}.
The graph of the inverse function x = y1/n is the graph of y = x". Then 14. Here x = s'"
l
fc
is the equation of the curve with y as the independent variable. Hence the graph of y = xl/n is symmetric to that of y = x" about the line y = x. d d~ 16. Since the curve
Ia
b
cy dx = cF(x)
I: = cF(b) -
cF(a) = c [F(b) .
. The function y = x" is shown in the figure. The area above the x-axis lies between x :::: and x = 3. Let
11
12. Hence the physical area is 121~. This area is obtained from A = xo/4 .y=
/' /' /' /'
/'
X
y
= xllll
y in x = yl/n. then x ::::1 and subtracting the second result from the first.
=
x and v
=
x2•
15. However when writing y = x1/n for the inverse function its graph is merely the interchange of x and
y
y=
x"
/.ax by letting x = 2. where y(xl is the given function of x./.41/4. the x-axis and x = 1 or the area is 1. Then the area between y = xl/n and y = x equals the area between y = x and y = x". the area between x = 1 and x = 2 lies below the x-axis . This area is 81/4. 2 1:L Let F(x) be any indefinite integral of y('X).8 cuts the x-axis at x = 2.60
of y =x3 . We have to show that x" dx = (l/a) a xdy where x = yl/a.F(a) ]
:= C
Ia
b
y dx. We have but to integrate by the use of inverse of the power rule and the result follows. This gives . Hence the given area is twice the area of the triangle bounded by y = x.
At A:::: 0° and for dA/dt = 18° :::: rad. The forces acting on Mare T-32M sin A.But the distance x to be covered is d/cos A. However (see (33) of Chap.yas 1:::. x =. Let A be the angle of inclinationof the desired straight line (Fig. and by Newton's second law (2) 32m-M = mao Hence by adding (1) and (2) (3) 32m-32M sin A = (M+m)a. we integrate and apply the initial condition = a when t = O.4 cos A. There is another possible root. 14.x approaches is the derivative of y:::: sinx at x :::: Hence the answer is cos (1T/2) or O.x] -sin (1T/2)}/1:::. Then the height of the pendulum above the level of its lowest position is h == 4 . a force of 32M sinA pulling the mass M down the plane. sin
*
Ar
16. Then if x is measured from the point along the Inclmed line. Hence the net upward force on M is 32m .. Thus the time to travel up the plane is t = Ih(M+m)/4[sin A(m-M sin A]-1/2. x=: 1612 sinA. namely cos A = 0 and A = 90° but this answer cannot be considered because it gives an imaginary value for t unless m > M. Hence dh/dt = 4 sin A dA/dt. When dt/dA == 0. 15. But m need not be > M. 1T/2. Hence t l2 = Vid/sin A cos A = (V'd/4)(sinAcos /• Now find Fig.x /1:::. 3 1T/10 per sec. Then t =:V4. cos" A :::: 2 A and A = 17/4.. If we express 1:::. Picture the pendulum making an angle A with the vertical. This gives x = l6(m-M sin A)t2/(M+m}.. The force F acting on the whole system acts on both masses m and M and this is why we must write F = {m+M)a. The total forces acting on mare 32m-T. This force is transmitted directly to the mass M and pulls itupward.32 sinA and Since x = 0 when t == 0 and x == 0 when t == 0. Then h/sin A = l6(m-M sin A)t2/ (M+m)...x. by Newton's second law. and hence Newton's second law says (l) T-32M sin A = Ma. 1 dt/dA (by letting u == sin A cos A).65
13. Moreover A = 90° is another situation entirely. A is measured from the horizontal disd tance d clockwise and so is also the inclinationof the inclined line on which the particle slides. From (3) we see that the acceleration of the whole system is a = (32m-32M sin A)/(M+m). since a = d2x/dt2. Another way to see this is to take into account the tension T in the string. Now let 1:::. Hence we get + "'" x the expression in the text.xfunction y == sinx at the value xo:::: of x for the 1(/2 we obtain {sin[(1(/2) !:::. The net upward acceleration by F ~ (M+m)a. However limit of 1:::. If we let x represent the variable distance up the plane measured from the bottom then. The mass m is pulled downward with a force (itsweight) of 32m. it need only be greater than M sin A to provide an acceleration up the plane..y/1:::. 1).x. 1
. If we now find dt/dA (by letting u = the quantity in the brackets and applying the chain rule we find from dt/dA = 0 that sin A = m/2M.1T/2. The length of the plane to be covered is h/sin A..1x/sinA.. db/dt::::T/5 ft/sec.32M stn A. a = (32m-32M sin A)/(M+m). 3) there is an acceleration of 32 sin A and therefore. Integrate again and apply the initial condition x = 0 when t = O...
Then the height of the passenger above the ground is h = 30 . and s in A = ill/5. 19. cos A = . SECTION 5
1./4 100t2 ./5 ft. Then we have the proper du/dx to apply the inverse of the power rule.
0 0
CHAPTER 10. BD = lOt = %. If we dif+ ferentiate with respect to t and set the derivative equal to 0 we get t = 0/15' To find what cp is. dx/dt = 27T/3mi/min. (b) Here cos A = 36°%800 = 3/4.lO-12. If A is the angle of elevation of the plane at any time and x is the horizontal distance traveled by the plane (measured from directly above the observer) when the elevation is A. (When h is increasing and above 30. Measuring the distance.25 cos A./5/3 and the minimum possible value of x is 5. Then v= 1/2../(1 cos 2x)/2. In the second term let u = 2x and use (38). A is a second quadrant angle. Now A is constant and is 41T rad/min.. c'n r AD .
(a) Write s'= % cos (3x)3 and let u == 3x. (a) Here A = 0 and cosA = 1. Then s'> u2 du/dx. We need 3 cos 3x for our du/dx. Then x = [27 . + Ans. Take A to be the angle which the line from the center of the wheel to the cab makes with the vertical from the center of the wheel to the ground.. Hence dx/dt = .) A is 7T rad/min. (g) Write s'= sec-x secx tanx and let u = secx. From Fig. (f) Let u = sin 3x. x = OA cos e = (27 .. we have x = 3600 tanA.
.10/25 = . We use the suggestion that we need consider only those situations in which the destroyer heads straight for the battleship. Hence h = 25(sin A)A. Then cos e = . x. 21. Integrate and use sin2x ::= 2 sinx cos x to get the text's answer. Then Y"= u2 cos 3x. Let us determine 1> by the condition that C'D is to be a minimum.% cos 2x = 1/2 . We find dx/de and set it equal to O. Answer. y = x/2 +(sinx cosx)/2 + C. y == _1/4 cos 4x + C.8t.8.. (c) Use sinx = .A dA/dt. Hence x = 14.AC' = . Use (38). (e) Let u:= sinx. Then tan cp:= BD/ AB = % and sin 1> = 0. the destroyer does not know at what angle 1> to head.2 esc. But OB = 8/sin e. along the shore from the foot of the perpendicular from the beacon to the shore and taking A as the angle between this perpendicular and the beam. This gives sin e = %. then x = 2 cot A. (b) Write y'= Y4 sin (4x)4 and let u = 4x.400 1T/cos2A.OB)cos e.. When A = 30 and dA/dt = -15 = -1T/12 rad/sec.66
17./(1 cos 2x)/2. Then h = 25 (ill/5)1T . 20. we have that at thtsvalue of t.2/5. Then y = t/g sins 3x + C. However... When h = 40.(8/sin e)l_pose = 27 cos e -'8 cot e. 18.1/4 cos (2x)2. Hence write y'= l/S u2 cos 3x· 3 == Ysu2du/dx.. (d) The method is the same as in (c) except cos = .. Hence dx/dt = x = 3600 sec" A· A.
Then A = 5vm/k.
use
CHAPTER
10. Then A2 = tan x.A! = 8 cosx. 4. The error is replacing v(l-cos x)/2 by +sin(x/2) in thde entire t interval (O./3.(D/v'v~k7m + 02) cos v'k/m t].. where n is any integer. Then y . Hence ¢ = 1T/2 and C = .. This velocity y is greatest when the sine function is 1 or v'k/mt = rr/2 + 2n1T. We must find the point of intersection of y = sec-x and y = 8 cosx. Now we introduce an angle cp whose cosine JS /Qv'Vgm/k + 02 and whose sine is D/lv~m/k + D2. Hence B = O. FIRST SET
1. A2 = -13. sin(x/2) is negative an one mus -sin(x/2) .1»..68
9. Then A = Al .
°
. In (2n. (55) reduces to y = 5v'm/k sin v'k/m t and the amplitude is 5v'm/k. Then Al = 4v'3. From (60) we have y = Dv'k/ill sin v'k/m t. = sec" X. Solving simultaneously gives x = 1T/3. The area under y = 8 cos x from x = 0 to x = 71/3 is given by integrating . y must be O.) we must have at t = 0. y.JV~m/k+ D2sin ( k m t .
SECTION 6. But C = 0 will not satisfy the first condition. Then Al = 8 sinx. 0 = Cv'k/m cos cpo The right side can be 0 if C is 0 or ¢ = 1T/2. is greatest when the + cosine function is 1 or v'k/m t = + 2n71. Since tan (1T/3) =. At t = 0 we must have -D = C sin¢.D satisfies both initial conditions.Then y = .4n). From (55) we have 0 = B cosO.~ = 3v'3.
Y
7r/3
10. 3. To write (61) in the form (55) we rewrite (61) as y = v'vgm/k + D2[(vo/v'v~m/k+ D2) sin v'k/m i.:: Av'k/m cos v'k/m t.or when t == 2n71v'm/k. given by y. Then t = v'm/k (7T/2) 2n1Tvm!k.4n). Again substitute 11'/3and 0 and subtract the second from the first. Then so far y = A sin v'k/mt. The area under y::::sec+x from x = 0 to x:::: 1T/3 is obtained from if. 2.= 5. With these values of A and B. Now substitute 71/3 and 0 and subtract the second result from the first. At t = 0.. The amplitude is the radical. Since y = CVk/m cos (Vk/mt + 4. The acceleration. At t = 0.
If we compare this equation with (51) we see that it is of the same form with 2k replacing k.
B=O. Then y = (1/"'.P.768/4t.k(a .1!.
e
. The period of the motion is 21T/16 or about 0. it will stretch I1/2 in. m:X =-2kx.x). Similarly.When t = 2. 9. k = 768.£ and by pulling the particle a distance x more to the right the stretch is a . ~ = 27T2 rad/sec and jj = O.192cosv'192 t. y = 1. Then as on p.192)cos. Hence by (50). Then y = A"'.393 sec. Hence by (50).x and the force pulling the object to the right is k(a . or % ft when t = O. Hence by arguing as in Exercise 6 or by using (60) with . Hence A = 1/"'. By Newton's second law. When m = 1. Since (51) remains the same the solution is still (52).£ . y = A Sin"'. 8. Suppose.192. Hence in place of (60) we get y = 1/6 cos v'k/m t. When the mass of 3 pounds is attached to the spring.£ + x). 6. When the particle is at 0 the stretch (extension over the normal length £) is already a . say. The net force (in the positive x-direction) is k(a . Then. . If the "spring" constant is k. then by Hooke's law the force pulling the particle to the left is k(a . Then the pull to the left is the amount of stretch in the left hand portion of the string. we have y = D when t = 0 and y = 0 when t = O. The only difference from what was done on p. Then (50) is irrelevant and (51) remains the same except that y now means displacement from the: end of the unextended spring. If gravity is ignored the spring is not extended by the addition of the mass m to the lower end.D replaced by %4' with k = 768 and m = 3 we have y = %4 cos"'. The mass will fall until it reaches the lowest point on the cosine curve.x) . Then ~ = 21T2 cos 41Tt and =-87T3 sin41Tt. We are also told that m = 1/4 and d = % ft.5. We may start from (52) and then use the initial conditions. + x) = . the particle is a distance x. as Fig 10'-16 shows. the stretch of the right hand portion is a .ff92t. d = 1~4 ft. + x.2k/mt. Hence we may use (52) with 2k replacing k and y = A sin"'. Hence the final formula is y = % cosv'96t.£ . k = 768. As in Exercise 7. and then released.256 t = Va cos 16t.2kx:. 253 in applying the initial conditions there is that in the present exercise y = 2 in. When t = 0.P. We may first of all determine k by the use of (50). Then 8 = (1T/2) Sin 41Tt. The initial conditions here are y = 0 when t = and y = 1 when t = O. Hence the equilibrium position is %4 ft. 253 we can determine that A = D and
° °
10. We are told that a = 1T/2 and 21T /b = 1/2 so that b = 41T. to the right of O. Now if the particle is pulled initially a distance D to the right. If we let (} be the angular displacement then e is of the form e = a sin bt. Since the amplitude is % it will fall % ft or 3/1. so far. 7. or %4 ft. below the lower end of the unextended spring. If we use (52) and apply the condition y = 0 when t = we get B = O.2k/mt + B cos"'. By placing the mass on the spring and then releasing it suddenly we are fixing the initial condition that y = %4 when t = O.k (Vs) = 32(V4) or k =: 24.
s == £ A sin f327i_ t + £ B cos '.1fl732. 293 ) we have. We have as our initial conditions that e == 0. (Compare Exercise 3.!k.0.1. Hence the angular velocity varies sinusoidally with an amplitude of -0. by (66).1-v'32/i and the period is 21T-vV32. At such values is 311'2/64. Now follow the method of Exercise 2.05 when t == O. We saw in the derivation in the text that the component of the force of gravity which causes the motion (p.05. When the bob is at its highest point. its linear and angular velocities are O. 9.. Since B == .32/£ t. We use (66) with the initial conditions e == 0 when t = 0 and e = 0. To make T twice as large we must increase 1!. SECOND SET
1.
CHAPTER 10.32/£ t . Since s == £ (J (p. though the velocity is 0 at t == 0.
SECTION 6. The amplitude is 1T/16 and the period is % seconds.0. Sin v'327'£ t. Hence x == 100 cos vk t. 2.70
CHAPTER
10. as t increases the velocity depends on xo. As painted out in the text..100. 3. SECTION 6.05 when t = 0 A = -0.. If we use (51) we see that k in the present case replaces kim there. by a factor of 4. 4.f327l. The first condition substituted in (66) gives B ==' 0.f£732. In a longer tunnel x varies over a longer range of values and the motion starts out with a larger acceleration. The maximum velocity is the amplitude. 5.1 when t == 0 and == -0. Then x == . THIRD SET 1.1 cos -I327it. The period does not depend on the mass of the bob. Hence the object acquires more velocity.l) sin~t.1. 100.f327l. The maximum value occurs at t == 0 or any multiple of 21T.1 when t = O. We find that B = 0 and A == 0. 2. 6. Then from (70) S =
e
e
e
t::
-JI.) Then (a) 8 == (1T /16) sin (3rt/4) (b) The angular velocity is == (31T2/64) cos (31Tt/4). Moreover its velocity is small near that point of its path. namely. The path is longer but the greater velocity acquired compensates. Then 8 = -0. To meet the second condition we first obtain from (66) that B ==' /32lff_ Acos '.
. The motion is given by (76) with Xo== 100 and k is the constant GM/R3. s ttl.Jf732 == 4/31T so that f327I == 31T/4..05. This equation is of the form (51) or (65).f£732· sin -I327it + 0. From (67) we see that .Ji{ sin v'k t. This shows that. We use (66). If we now assume that the pendulum started from the equilibrium position with some initial velocity then we may use the form 8 =: A sin. (O.. Mathematically we see from (76) that x == -xo. From (70) we have e == . 7.J'k sinv'kt.J327£t.298 ) is PQ == kmx. 8. Hence the bob is momentarily stationary. The period is given by (67). 10.J"2/ft. B sin.
letting the letters M and R still stand for values of the earth's mass and radtus . The physical phenomenon discussed here is best compared with the phenomenon of the bob on the spring. And the quantity (c) is the quantity +ky below (50). If we start with (75) and apply the initial conditions x== 0 at t :=0 and x := 0 at t:= 0 we get in place of (76) just x :=O.}GM!R3 211"/.61TY. there is a downward force which is its weight in air minus the buoyant force on the portion of the cylinder still in the water.3.lkt. R by 3R/11 and then 32R2 by GM one obtains an answer in terms of the k of the text where M and R refer to
..}4.. the period is 21T/1k. 4. we can use (52) with k/ m replaced by 4.9921Tt.9921Tt. when the y-value is positive but less than d.}4. If we follow the procedure of fixing A and B that was used in the text. The period is still 211" This last fact is interesting be/.. Before we applied any initial conditions we derived (75). For x = 0 the period is 0 (as in Exercise 4). Then T = 21T/v'k 21T/. We start with the formula T:= 21T /-Jk where k = GM/R3. T = 21T/v'(GM!Sl)/(3R!11)3=-(54v'31T/1lv'li)(1/v'k).Ik. 7.3R2. Equation (a) is the analogue of (50). The initial conditions are that y =-2 when t = 0 and y = 0 when t = O. but with no point on the line segment at x:= O.}4. Hence the graph of period versus x is a line segment parallel to the x axis and extending from -xo to +x. This is the analogue of the downward force exerted in the case of the spring. When the cylinder rises above the equilibrium position. in which case GM = 5.}5. One may choose to let M and R stand for the mass and radius of the moon. Since the differential equation (my = net force) is y = -4. Equation (b) is the analogue of (49). Here when the cylinder is depressed below its equilibrium position the buoyant force of the water replaces the upward pull of the spring. we obtain x := (xo/3) cos.. the force is the weight of the bob minus the upward pull of the spring. cause the period is the same as when the object traverses the entire path UV. Since the theory of the text does not restrict M and R to be the mass and radius of the earth we may let them be the mass and radius of the moon. 5. 6..71
3.3R2/R3 ::::= == := 27rV'R/v'5. If in the next to the last expression one replaces 5..99211Y. and k still stands for GM/R3 but M and R refer to the moon.. If we start with (75) and apply the initial conditions x = x when t = 0 and and x = 0 when t = 0 we get x = x cos -Ii{ t in place of (76). Now our initial conditions are x e= xo/3 when t :=0 and x :=0 when t := O.99211. Then y =-2 cos.This is also a correct answer.3 by 32/6. Then.. The resulting differential equation is the analogue of (51) and is 100y::::= -32' 15. Thus for every value of x except 0.99211 t + B cps.. Then y = A sin .
if we use the more convenient form (80). Then 2 = 21Ta/Ysand a = 1/51T.
. From Exercise 1 we see that the maximum velocity is the amplitude a times 21T/r. The maximum value of y occurs when the cosine is 1 and then y is 21Ta/T.
SECTION 6. FOURTH SET
L The function.J21l/v'l1k.3 by 32/6 is an approximation. Here k has the value GM/R3 where M and R stand for the earth's values. 2. CHAPTER
10. This latter answer can be put in the form T = 6.72
values for the earth. Then y = a(21T/r) cos [(2'll/T)t +cp). where T is the period. Of course the replacement of 5. is y == a sin [(21T/r)t + CP]..
Then yl is in the form (30) with u = 3x/4. (c) Write yl = (1/a){1/[1 + (x/a)2]).h Y Y' (dy/dx). Then du/dx = 2.
.1)](du/dx).for 31T/2 ~ x ~ 21T. Since dy/dx =l/(dx/dy).' 5. Hence iJ == (vfa) eos" B.(4x2/5). Then yl is in the form (28) with u = 3x/4. except for the factor %. for 11"/2 x -s 311"/2. To find () use (24) with u = gt2/2a.[(reg:~/dd ~)//(ddxy/ads )2J' a function of y with y as a function of x.x +C. Then we have the form (28). (d) Write s'» (1/a2){1/[1 + (bx/a)2J}.0). . Now to differentiate with resp~ct/to2 x.J5. (g) Write v= 1~6{1/[1+ (3x/4)2]). 9. Then y = % sec-12x + C. The idea here is the same as in (a) except that we have the derrvative of tan " x at x = 1. (i ) yl is in the form (32) [or (33)J with u = 2x. to (b): x = O. From (24) we see that the answer is %. Y = 11/2~ x.
(k)
(1/3) sin. Hence y = % secJx + C. From (24) we see that the answer is 1. except for the factor %. In each case find d2y/dx2 and set it equal to 0 to fwd the abscrssa of the point of inflection.(3x/4)2).1 . 11"/2. This constant factor can be introduced and the resnlt is in the text.
e
e
CHAPTER 11. (11) Write v= 1/4(1/'. Then d y dx = -d-. Hence write yI= %[1/(u'. we have the text result. The answer is in the text. Then du/dx = b/a. (e) Except for the factor ~. Then we have the form (30) with u = x/a. 11-12 we see that
e = tan ' (y/a). Y = 51T/2 ~ x. From (24) we find that dB/dy = a/(a2 + y2) = cos" B/a. Use
Now use
f (x)
(28). Also dy/dt = v. Then du/dx = 2/.y = ::. (a) tarr-x/x = (tarr+x ~ tan= d(tan-1x)/dx
(b)
x~
at x == O. Then the text answer follows at once and to find jj merely differentiate with respect to t. f (x ) = fxdx/ (x4+3) = (1/2) f2xdx/
=
(30). Now = dB/dt = (dB/dy)(dy/dt).(2v2/a2) cos" B sin B.sin IJ)e = . . Now lJ = dB/dt = (v/a)2 cos 8(. dy/dx = 1/(dx/dy). Let u == bx/a.u2 . (0 Write v= (1/v'5)(1/v'l .
=
(13/6)tan-l (x2/13)+c.
B. Ans. 6. [ (x2) 2+3J. B = tarr ' (gt2/2a). Hence y = (1/ab) tarr" (bx/a) + C. .74
4. the s' is in the form (32) (or (33}J. 10)/(x . For 0 ~ x -s 11/2. Then y' is in the form (28) with u = 2x/v'5. (b) Let u = xS. The answer is in the text. Here B is a function of t. (a) Let u = x2• Then we have the form (28).
(j) Let
f (x)
u
=
x3
and write
1 2
f(x)
=
(1/3)f3x2dx/f~1--~(x-3-)-2. Hence write y/= (1/ab){1/[1 + (bx/a)2]}{b/a). Then li!8-tan-lx/x 7. From Fig. SECTION 4 1.
a2. SECTION 5 1. (n) Write the given integral as dx/I(x+5) 2+5] = (1/5)Jdx/I«x+5)/i5)2 +1]. By our formula for area. The second integration uses (28). 3. 2. Our method of finding areas is to start with dA/dx = y. dA/dx = y or A =
(30). We can use the table to find that tan-l (13/3) = n/6 or happen to remember this. 2. (m) Break up the given integral into !xdx/.x2• U we now substitute a for x and then o for x and subtract the second result from the first we get nb/4. Hence A = (1/3)2 tan-1 (13/3). Then y = (~/5)tan-l [(y+5)/~]+C.
-13
J
13
dx/(9+x2).75
(t) Write the given integral as (1/2)!2sec xtan xdx/132+(2sec x)z. Hence read off the answer from (52). (c) Let x = 4/3 tan e and follow the procedure of (a) or (b).9 .x2• The integral in view of (53) is (ab/2) sin"! (x/a) + (bx/2a)"'. Then A = n/9. The first integral is evaluated by letting u = l-x2 and using the inverse of the power rule. Hence y = 8/3 + C = Ys sm " (3x/4) + C.9 sin2 8) 3 cos 0 = 1. Whenever a radical occurs a change of variable which eliminates the radical is usually helpful. Thus A = 1/3 [tan-l (13/3)-tan-1(-13/3)]. A = 1/3 tan-l (x/3). Use (30). The steps parallel those of the text where the substitution x = a sin 8 was used. Here y = (b/a) la 2 _b2• Hence we must integrate (b/a)-Ja2 . (b) Let x = ¥a sin O.r=xr + 3 Jdx/lf=XT.
. Then y = 8 + C or y = strr" (x/3) + C. Then y = (1/6)tan-1 [2sec(x/3)]+C.
Then with
CHAPTER 11. NOw substitute 13 and -13 for x and subtract. But to use change of variable let x = 3 sin O. (a) This can be done by factoring the 9 out of the radical and thus reducing to (28). Now let u = (x+5)/1S and use (30). The two answers agree because sin-1 (x/a) and cos"! (x/a) differ by a constant. But tan-1 (-13/3) = -tan-l (/3/3) . Then dy/ d8 = (dy/ dx}(dx/ de) =: 1/4(1/ cos e) 7'8 cos e. (d) This is (48) with a = 3. Then y = -~+3 sin-1x+C. Then dy/cW = (dy/ dx)(dx/ dO) = (1/"'.
(i ) Let x = 5 sin e.J2GM and t = %r3/2/. Let us use (7 O-~ Then we must first use (65) to find the value of e when r == 4000 ·5280.4 sin2e. 2. Replace sin261 by 2 sin e cos e and transform back to x. If we let t = 0 when r == R. Then y "" !tan2. C ==-2R3/2/3. The text answer is approximate. Since an acceleration of 32 ft/sec2 is greater than the actual gravitational acceleration the object will acquire greater velocity and take less time to fall 2000 miles. as in Exercise 1.sirr ' (x/5) + C.000 miles == 60R and r == The arithmetic is again R. Again the accuracy will depend on the number of decimal places carried. Replace sin2() by (1 . then dyIde = [1/(a2 + a2 tan28)3f2]a sec20 = \1/a2) cos 8.
. 4. 4. 5280 + 100. The time of flight is the same as if the object were dropped from a height of 100. a == -32.xi/x) . We could also use (7 a) and calculate the e for which R == (3R/2)cos2 e and then substitute in (70) . The procedure is the same as in Exercise 2. The calculation is extensive and the accuracy of the answer will depend on how many decimal plac es are carried.j2GM. As in Exercise 3 the time required is the same as if an object falls from a height of 240.. we can use (69) or (70). To calculate the time we use the method of Chapter 3. O 2 + 3R/2.2R3/2/3. s = 3R/2 when t :::: .000 ft and traveled to the surface of the earth. S-8. 5.GM/R. Thus. 3.000 miles to the surface of the earth. Here Vo is the velocity at the surface of the earth and v is the velocity at the distance r from the center of the earth.j2GM and so t = 2r3/2/3. 1 Then y = se .
(e)
If
CHAPTER 11. :::: /a x (g) Let x = 4 sin O..J2GM+ C. Then since v == dr/dt.000 and r 4000·5280. If tan 8 = x/a then sin 8 = x/.j2GM/R...8dS = J(sec2S-1)d8 = tar. dt/dr = r1f2/. lengthy and the accuracy depends on the number of decimal places carried. Then dyIde ::::(1/a2) sec28. Following the suggestion we start with v2/2 = vU2 + GM/r . That is. Use of (70) gives 98 sec.jR!32 = 812 Hence s = -16t seconds. SECTION 6 L Use (69) with r1 = R + 2000·5280 == 3R/2 and r = R. This gives e = 86020' approx.cos 2e)/2.. A result between 1250 and 1300 seconds is good enough for present purposes. However instead of the end values a and 0 we have here 2 and O.. Then y = -cot 8-S Transforming back to x gives y:::: (--v'25.J2GM!r. In either case r1::::4000 . y:::: (1/a2) tan () a 2.J2GM. We let Vo be the escape velocity . approx. Then dy/d() :::: 6 sin" (). Then y = (1/a2) sin 8 + C.x2 + C.76
x == a tan B.ja2 + x2• (f) Let x :::: sin 8.ja2 . Now calculate t when s = R. Then v2/2 = GM/r and v > . v » -32t + C.. (h) Let x = 5 sine. We can use (69) or (70). Then t = . Then s == 2 + C -16t and if s is measured from the height of 2000 miles. C = 0 because v = 0 when t == O. In either case r1 = 240. Then y = J (c s c ' 8··1) d8. Now transfor m back to x.
(b) We see from the expression for t that when r is infinite (and positive) t is . following the suggestion. then t :::: when r :::: and C = O. 3/2/-v'2GM Then t ::::-%r + C. This would not happen physically because the body never reaches the center. The force of attraction acting say on the sphere to the right is. Then 0 0 r :::: . starting from r :::: for t:= 0. Hence in the present situation the formula has no physical meaning for positive t. (c) We see from the expression for v that v is infinite when r is O. r must decrease with Incr-easmg t. the object 0 would have to move away from the center in the direction of incr eastng r as t increases. Then rnX == . and the distance between the cenm ters of the two spheres is 2x. We must understand t to be negative as the ( formula for t shows. Since t is 0 when the object reaches the center. Integrating gives v2/2 :::: M/r + C. by taking the acceleration negative and then the velocity negative. . Then.3/2)2/!f(2GM)1I3 t2l8. For positive t values r is positive. according to the law of gravitation. Hence it takes an infinite time to reach the center.Gm2/4x2. v dv/dr == -GM/r2. but replace GM by Gm/4. (d) Yes.77
6. (e) Since r increases with t. 8. (a) We cannot use (69) because letting r1 be infinite does not give a clear value for t. sec. Now our initial condition is that v ::::: when r becomes G 0 2/2 :::: M/r.. 7. dt/dr ::::rl/2/-v'2GM. 7. is . The factor (-3/2)213 is positive.Gm2/4x2 or x == -Gm/4x2.
. Since v::::dr /dt.=.=. Then v::::--v'GM/r. The answer of two hours is approximate because the accuracy of the answer depends on the number of decimal places carried. wherein now M. We may then go directly to (69) and replace GM there by Gm/4. In our case r1 == and r == because 3 1 the spheres are in contact when each is 1 foot away from the origin. We start with a == 2r/dt2 ==-GM/r2 and write dv/dt :=-GM/r2• d Then as in Chap. That is. it is negative previously. Hence C :::: and v 0 G sign enters because v is negative. the minus infinite. starting with (62). If we agree to measure time from the instant the body reaches the center. we take over the theory of the text. | 677.169 | 1 |
Calculus Lab Spring 2017
A laboratory to investigate ways in which the
computer can help in understanding the calculus and
in dealing with problems whose solutions involve calculus.
No programming required; a variety of existing
programs will be used. Prerequisite or corequisite for
Mathematics 235: Mathematics 225 (Calc 3);
There is no textbook required for this course.
Week 4: Feb 7/9, 2017
Week 5: Feb 14/16, 2017
This week, we'll continue writing up the solutions to the biorhythms lab.
Week 6: Feb 21 (A) and Mar 2 (B)
This week, we will look at more Maple commands- In particular, we'll
work through discussions of (1) Curves and Surfaces, and (2) Limits
and Derivatives, and then there is a set of 6 exercises to solve using
Maple. You may, as usual, work in groups. Due: Before our next class
session.
Week 7: Feb 28 (A) and Mar 9 (B), (Midsemester)
For the next two weeks, we'll be looking at the theorem from
Calculus III known as ``Clairaut's Theorem''. This first week,
we'll get some background questions answered. Next week, we'll work
on our writing.
Week 10: Apr 4 (A) and Apr 13 (B)
Week 11: Apr 18/20, 2017
This week, both sections are back in sync. The first half of today, we'll discuss
something called Beamer. The second half, you may put the finishing touches
on the lab. The is due before our next class meeting. | 677.169 | 1 |
Geometry Books
Geometry Books
geometry books. Are you looking for books for geometry students to better understand geometry topics?
Your child may need some help with his or her geometry class. Or you may be looking for geometry math resources for your students. TuLyn is the right place. Learning geometry is now easier.
We have hundreds of books for geometry students to practice. This page lists books on geometry. You can navigate through these pages to locate our geometry books.
Glencoe Geometry is a key program in our vertically aligned high school mathematics series developed to help all students achieve a better understanding of mathematics and improve their mathematics scores on today's high-stakes assessments.
E-Z Geometry
Known for many years as Barron's Easy Way Series, the new editions of these popular self-teaching titles are now Barron's E-Z Series. Brand-new cover designs reflect all new page layouts, which feature extensive two-color treatment, a fresh, modern typeface, and more graphic material than ever— charts, graphs, diagrams, instructive line illustrations, and where appropriate, amusing cartoons. Meanwhile, the quality of the books' contents remains at least as high as ever. Barron's E-Z books are self-help manuals focused to improve students' grades in a wide variety of academic and practical subjects. For most subjects, the level of difficulty ranges between high school and college-101 standards. Although primarily designed as self-teaching manuals, these books are also preferred by many teachers as classroom supplements—and for some courses, as main textbooks. E-Z books review their subjects in detail, and feature both short quizzes and longer tests with answers to help students gauge their learning progress. Subject heads and key phrases are set in a second color as an easy reference aid. Barron's E-Z Geometry covers the "how" and "why" of geometry, with examples, exercises, and solutions throughout, plus hundreds of drawings, graphs, and tables.
Math Subjects: Geometry
List Price: $16.99 Low Price: $9.82
geometry book
Contemporary's Number Power 4: Geometry: a real world approach to math (The Number Power Series)
Number Power Series (Revised) The first choice for those who want to develop and improve their math skills. Every Number Power book targets a particular set of math skills with straightforward explanations, easy-to-follow, step-by-step instruction, real-life examples, and extensive reinforcement exercises. Use these texts across the full scope of the basic math curriculum, from whole numbers to pre-algebra and geometry. NUMBER 4: GEOMETRY Introduces lines, angles, triangles, and other plane figures, and solid figures.
Math Subjects: Geometry
List Price: $21.92 Low Price: $11.98
geometry book
Geometry Workbook For Dummies
Geometry is one of the oldest mathematical subjects in history. Unfortunately, few geometry study guides offer clear explanations, causing many people to get tripped up or lost when trying to solve a proof—even when they know the terms and concepts like the back of their hand. However, this problem can be fixed with practice and some strategies for slicing through all the mumbo-jumbo and getting right to the heart of the proof.
Geometry Workbook For Dummies ensures that practice makes perfect, especially when problems are presented without the stiff, formal style that you'd find in your math textbook. Written with a commonsense, street-smart approach, this guide gives you the step-by-step process to solve each proof, along with tips, shortcuts, and mnemonic devices to make sure the solutions stick. It also gives you plenty of room to work out your solutions, providing you with space to breathe and a clear head. This book provides you with the tools you need to solve all types of geometry problems, including:
Congruent triangles
Finding the area, angle, and size of quadrilaterals
Angle-arc theorems and formulas
Touching radii and tangents
Connecting radii and chords
Parallel, perpendicular, and intersecting lines and planes
Slope, distance, and midpoint formulas
Line and circle equations
Handling rotations, reflections, and other transformations
Packed with tons of strategies for solving proofs and a review of key concepts, Geometry Workbook For Dummies is the ultimate study aid for students, parents, and anyone with an interest in the field.
Mastering Essential Math Skills GEOMETRY
This is a new title by America's math teacher and author, Richard W. Fisher. This book will provide students with all the essential geometry skills. Vocabulary, points, lines, planes, perimeter, area, volume, and the Pythagorean theorem are just some of the topics that are covered. Each lesson contains built-in review and easy-to-understand instruction that introduces new material with lots of examples. There is plenty of real-life problem solving that shows students the importance of geometry in the real world. An excellent summer review prior to taking high school geometry. CHECK OUT THE REST OF OUR MASTERING ESSENTIAL MATH SKILLS BOOKS AND DVD'S!
Math Subjects: Geometry
List Price: $14.95 Low Price: $8.60
geometry book
Geometric Structures: An Inquiry-Based Approach for Prospective Elementary and Middle School Teachers
This text provides a creative, inquiry-based experience with geometry that is appropriate for prospective elementary and middle school teachers. The coherent series of text activities supports each student's growth toward being a confident, independent learner empowered with the help of peers to make sense of the geometric world. This curriculum is explicitly developed to provide future elementary and middle school teachers.
Math Subjects: Geometry
List Price: $88.00 Low Price: $52.84
geometry book
Geometry: Practice Workbook
ISBN: 0618736956 EAN: 9780618736959 Binding: Paperback
Description:
No additional information available.
Math Subjects: Geometry
List Price: $5.73 Low Price: $3.88
geometry book
Geometry Demystified
This book helps learn geometry from an all-new angle! Now anyone with an interest in basic, practical geometry can master it - without formal training, unlimited time, or a genius IQ.In "Geometry Demystified", best-selling author Stan Gibilisco provides a fun, effective, and totally painless way to learn the fundamentals and general concepts of geometry. With "Geometry Demystified", you master the subject one simple step at a time - at your own speed. This unique self-teaching guide offers multiple-choice questions at the end of each chapter and section to pinpoint weaknesses, and a 100-question final exam to reinforce the entire book. Simple enough for beginners but challenging enough for advanced students, "Geometry Demystified" is your direct route to learning or brushing up on this essential math subject. Get ready to: learn all about points, lines, and angles; figure out perimeters, areas, and volumes; improve your spatial perception; envision warped space and hyperspace; and much more!
Math Subjects: Geometry
List Price: $19.95 Low Price: $3.99
geometry book
Geometry Success in 20 Minutes a Day
A good knowledge of geometry is essential to success on many standardized tests and applicable to a wide range of careers. Geometry Success in 20 Minutes a Day provides a thorough course in geometry skills that can be fit into any busy schedule. | 677.169 | 1 |
First Day Review Activity for Precalculus and Algebra 2
Be sure that you have an application to open
this file type before downloading and/or purchasing.
15 MB|13 pages
Product Description
There are 20 cards in this activity that review solving equations (linear and quadratic), finding the distance and midpoint, evaluating a function, finding the slope of a line, and writing an equation given 2 points. You can use these as task cards or set them up in centers.
I use this activity on the first day of class. I give each student a card when they walk in the door along with an answer sheet. They first need to solve their own question and check the answer with me. Then have to go around the room and ask people for the other questions. They need to solve the question that the student has. Besides the work/answer on the sheet, they also have to write the other students name and ask them a question to get to know them. This is a great icebreaker activity and review at the same time. | 677.169 | 1 |
This book explores the life and scientific legacy of Manfred Schroeder through personal reflections, scientific essays and Schroeder's own memoirs. Reflecting the wide range of Schroeder's activities, the first part of the book contains thirteen articles written by his colleagues and former students. Topics discussed include his early, pioneering... more...
This book examines the relationship between mathematics and data in the modern world. Indeed, modern societies are awash with data which must be manipulated in many different ways: encrypted, compressed, shared between users in a prescribed manner, protected from an unauthorised access and transmitted over unreliable channels. All of these operations... more...
This volume develops the depth and breadth of the mathematics underlying the construction and analysis of Hadamard matrices, and their use in the construction of combinatorial designs. At the same time, it pursues current research in their numerous applications in security and cryptography, quantum information, and communications. Bridges among diverse... more...
Arithmetic for the Mature Student provides information pertinent to the basic operations of arithmetic, including addition, subtraction, multiplication, and division. This book covers a variety of topics, including percentages, fractions, and factors. Organized into five chapters, this book begins with an overview of the instructions on how to use... more...
Generic group algorithms solve computational problems defined over algebraic groups without exploiting properties of a particular representation of group elements. This is modeled by treating the group as a black-box. The fact that a computational problem cannot be solved by a reasonably restricted class of algorithms may be seen as support towards... more...
Since the publication of the first edition in 1976, there has been a notable increase of interest in the development of logic. This is evidenced by the several conferences on the history of logic, by a journal devoted to the subject, and by an accumulation of new results. This increased activity and the new results - the chief one being that Boole's... more...
This book contains the proceedings of the NATO-Russia Advanced Study Institute (ASI) 'Boolean Functions in Cryptology and Information Security', which was held at September 8-18, 2007 in Zvenigorod, Moscow region, Russia. These proceedings consist of three parts. The first part contains survey lectures on various areas of Boolean function theory that... more...
This self-contained monograph explores a new theory centered around boolean representations of simplicial complexes leading to a new class of complexes featuring matroids as central to the theory. The book illustrates these new tools to study the classical theory of matroids as well as their important geometric connections. Moreover, many geometric... more...
This book contains selected papers on the language, applications, and environments of CafeOBJ, which is a state-of -the-art algebraic specification language. The authors are speakers at a workshop held in 1998 to commemorate a large industrial/academic project dedicated to CafeOBJ. The project involved more than 40 people from more than 10 organisations,... more... | 677.169 | 1 |
Review info & popularity
Brief course description
Welcome to the easily learning basic math, a place where you can come and learn the basic topics of math from scratch. In this course we will be covering the most important fundamentals math concepts, that will help you in everyday life, and will help you move on to Pre Algebra.
You maybe someone who hasn't practices the core math fundamentals that are taught in grade school and need a good refresher, or you maybe a college student who needs to pass a college entrance exam, or you just might just be a parent who wants to make sure your kid becomes excellent at math and doesn't to fall victim to the "Math is difficult syndrome."
In reality Math isn't hard at all, all that's require is learning a few necessary steps to solve any given problem, and thats what you'll learn in this course. | 677.169 | 1 |
algebra 2 cp 1 ppt
2.
About Ms. LeBlanc<br />This is my second year at Triton.<br />I graduated from the University of Tampa.<br />Algebra 2 is my favorite subject.<br />I enjoy reading, skiing, and traveling.<br />
3.
Class Expectations<br />Everyone received a copy of the "Course Expectations/Syllabus". Please read over this document carefully and go over it with your parents/guardians tonight.<br />Attached to the Course Expectations/Syllabus is a sheet that must be filled out and signed by you and your parent/guardian. <br />
4.
Rules and Responsibilities <br />Class Rules<br />Be respectful and courteous to the teacher and your peers<br />Follow directions and classroom procedures<br />Be honest and take pride in your work<br />Obey all School Rules<br /> Student Responsibilities <br />Arrive on time and ready to work<br />Ask for help when you need it<br />Take responsibility for your education, behavior, grades, and consequences<br />Participate in class appropriately<br />Remain seated during instruction<br />
5.
Classroom Procedures and Policies<br />Daily Procedures<br />Arrival to class<br />Obtain the Problem of the Day<br />Take your assigned seat<br />Take out the previous night's homework to be checked off<br />Complete the problem of the day<br />During class<br /> Once class has begun, it is expected that you listen, take notes, participate appropriately, and remain seated during instruction.<br />Dismissal<br /> It is expected that you remain seated and working until I dismiss<br />
6.
Classroom Procedures and Policies<br />Tardiness<br />You are expected to be in you assigned seat and ready to work when the bell rings.<br />Students who are not in their assigned seat when the bell rings will be marked tardy.<br />
7.
Classroom Procedures and Policies<br />Attendance<br />Consistent attendance is critical to academic success and students should strive to maintain good attendance.<br />If you are absent:<br />It is your responsibility to obtain missed notes and assignments. Assignments will be posted on the class calendar. <br />If you missed a test or a quiz it is your responsibility to make arrangements to make it up.<br />You have twice the number of days to complete missed work as the number of days you are absent.<br />When you return from an absence you are required to complete assignments that we are working on that day, this included tests and quizzes. If you were absent the day of a test review, you are still required to take the test.<br />
8.
Cell Phones and Other Technology<br />Use of / visibility of electronic devices (including headphones) (beepers, cell phones (including text messaging), ipods, mp3 players, radios, CD players (all with or without headphones) and laser pens during school hours is prohibited. Failure to comply will result in the forfeiture of the device and will result in disciplinary action. Triton Regional High School will hold no responsibility or involvement should there be a theft or loss of any such device. Parents who anticipate a need for emergency communication should contact a school administrator to discuss the specific situation. <br />
11.
Extra Help<br />I will be available on Thursdays after school for extra help.<br />I am also available by appointment for Mondays – Wednesdays after school.<br />
12.
Fire Drill<br />Fire Drill Procedures <br />1. Students and staff are to leave the building by the nearest exit when the fire alarm sounds. Please follow all instructions given by any staff members. <br />2. Move rapidly, but DO NOT RUN. <br />3. Students and staff are to move at least 50 yards from the building before stopping to wait for the signal to return to the building is given. <br />4. Attendance will be taken outside at the assembly point and upon return to class. <br />Go left out of our classroom, down the stairs (on you right), and exit the doors on the right of the cafeteria. Meet in the parking lot.<br /> | 677.169 | 1 |
Core Maths
Core Maths qualifications
Core Maths qualifications are designed for students who have achieved a standard pass in GCSE Mathematics, but who do not intend to take AS/A level Mathematics. They enable learners to strengthen and develop the mathematical knowledge and skills they have learnt at GCSE so that they can apply them to the problems that they will encounter in their other level 3 courses, further study, life and employment. MEI recommends that students with GCSE Mathematics grade 4 or better consider taking Core Maths.
Free resources suitable for all Core Maths courses
The DfE has funded MEI to develop a curriculum and resources suitable for post-16 students who have achieved grade C or better in GCSE Mathematics but who need a different set of skills to that provided by A level Mathematics. The freely available resources developed in this Critical Maths project and are suitable for all Core Maths courses.
Free resources tailored to the OCR(MEI) Core Maths courses
OCR has sponsored free online resources which cover all the content of both the OCR(MEI) Core Maths qualifications: Quantitative Reasoning and Quantitative Problem Solving. Access to the resources is by free online registration. | 677.169 | 1 |
Paperback | October 13, 2010
Pricing and Purchase Info
about
When you need just the essentials of elementary algebra, this Easy Outlines book is there to help
If you are looking for a quick nuts-and-bolts overview of elementary algebra, it's got to be Schaum's Easy Outline. This book is a pared-down, simplified, and tightly focused version of its Schaum's Outline cousin, with an emphasis on clarity and conciseness.
Graphic elements such as sidebars, reader-alert icons, and boxed highlights stress selected points from the text, illuminate keys to learning, and give you quick pointers to the essentials.
Perfect if you have missed class or need extra review
Gives you expert help from teachers who are authorities in their fields
About The Author
Barnett Rich earned his Ph.D. from Columbia University and a J.D. from New York University. He was a founder of the New York City High School of Music and Art and was chairman of Mathematics at the Brooklyn Technical High School. He also taught at City University of New York and Columbia From Arithmetic to Algebra;2.Simple Equations and Their Solutions;3.Graphs of Linear Equations;4.Monomials and Polynomials;5.Problem Solving;6.Factoring;7.Fractions;8.Roots and Radicals;9.Quadratic Equations;10.Introduction to Geometry;Index | 677.169 | 1 |
A's study guides for state-required exams are teacher-recommended and written by experts who have mastered the test.
Synopsis
REA Real review, Real practice, Real results.
REA's New York Grade 8 Math Test Prep!
Fully aligned New Yorks Core Curriculum Standards
Are you prepared to excel on this state high-stakes assessment exam?
As study guides for state-required exams are teacher-recommended and written by experts who have mastered the test.
Synopsis
Graphs
Frequency Distributions
Frequency Histograms and Relative
Frequency Histograms
Cumulative Frequency Distributions
Types of Frequency Curves
Deductive and Inductive Reasoning
Sampling
Statistical Plots and Graphs
Diagnostic Tests For Class and Homework Assignments
Arithmetic Diagnostic Test
Algebra Diagnostic Test
Geometry Diagnostic Test
Word Problems Diagnostic Test
Reference Information
NYSTP Mathematics Practice Test
Answer Key
Detailed Explanations of Answers
Answer Sheet
Index
Synopsis
New York State Grade 8 Mathematics Test
Completely aligned with the Learning Standards adopted by the New York State Regents REA helps Grade 8 students get ready for the New York State Grade 8 Mathematics test!
Provides all theinstruction and practice students need to excel on the test
Students at all levels will benefit from the many lessons and examples in this book
Diagnostic tests help students determine which topics require additional study | 677.169 | 1 |
This"Sinopsis" puede pertenecer a otra edición de este libro.
Reseña del editor:
AReseña del editor:
Endorsed by Cambridge International Examinations, this new series offers all the resources you will need to completely cover the Cambridge IGCSE Mathematics (0580) syllabus. A Coursebook, two Practice Books and a Teacher's Resource provide thorough explanation and ample practice opportunity to help students to understand their IGCSE maths course and support teachers in guiding them. Suitable for all classrooms, whether core and extended students are learning together or separately, these resources will help students to achieve their potential in maths. 2012. 1 Paperback 1 CD-ROM. Estado de conservación: NEW. 9781107606272 This listing is a new book, a title currently in-print which we order directly and immediately from the publisher. Nº de ref. de la librería HTANDREE0487909 636 Pages ( with CD-ROM 561227
Descripción Paperback. Estado de conservación: New. New Softcover International Edition, Printed in Black and White, Only USPS Media mail Shipping ONLY, Different ISBN, Same Content As US edition, Book Cover may be Different, in English Language. Nº de ref. de la librería 35471
Descripción CAMBRIDGE UNIVERSITY PRESS, United Kingdom, 2012. Mixed media product. Estado de conservación: New. 276 x 218 mm. Language: English . This book usually ship within 10-15 business days and we will endeavor to dispatch orders quicker than this where possible. APE9781107606272 | 677.169 | 1 |
Common Core
The Short Answer:
Everyone is talking about Common Core like it's some crazy new way of teaching math, as if in the span of a few years it's going to "fix" math by "catching U.S. students up" to the "rest of the world". It's intimidating, like everyone is suddenly going to have to be some kind of math genius doing calculus proofs at the board in the 8th grade. But that's just the crazy story being pushed forward by the press.
In reality, Common Core is just the same old math that your school has always covered, except with fancy new names for stuff. Algebra 1 is still "Algebra", but Algebra 2 is now "numbers and quantities" and "functions". Your school may switch books to one that has chapters with the right names for everything, they may even change the name of the class, but the topics are still the same. Lines are lines, variables are variables, and the Fundamental Theorem of Calculus is still the Fundamental Theorem of Calculus (that's a math joke).
Each class is still going to cover the same types of topics and problems, so That Tutor Guy has you covered. It's your school's job to design the math curriculum to comply with the standards. Your job, with ThatTutorGuy's help, is to learn the topics and take the tests which won't be much different from how they were pre-CC.
If you'd like to search by Common Core reference number...
You can usually find the standard reference number in the margin of your textbook at the beginning of the chapter. It will be some crazy combination of letters and numbers like N.VM.6 or 7.EE.1. Just enter this number into the search box. Our videos are tagged with the standard numbers for easy reference. (in progress) | 677.169 | 1 |
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The best part is that the explanations to math questions are accurate and clear. This book covers angles, triangles, quadrilaterals, area, perimeter, Pi, circle parts, polyhedra, surface area, volume, nets, congruent transformations, and symmetry. Price Area of a Triangle In this activity, you try to make a parallelogram linear systems worksheet a copy of workshheet given triangle.
Or linear systems worksheet our curated resources for Starting a Business. Worksgeet linear systems worksheet Business or for Professional Development. Art Lesson Plans Home The art lessons section has experienced a massive overhaul. Lessons are now categorized for easier retrieval. The images should be large so people can see them harry potter epub full. If your images are too large to submit, contact me. In this section are hundreds of free art lessons from preschool through the college level.
They justify their conclusions, communicate them to zystems, and respond to the arguments of others. They reason inductively about data, making plausible arguments that take into account the context from which the data arose. Elementary students can construct arguments linear systems worksheet concrete referents such as objects, drawings, diagrams, and actions. Such third grade projects ideas can make sense and be correct, even though they are not generalized or made formal until later grades. | 677.169 | 1 |
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