text stringlengths 6 976k | token_count float64 677 677 | cluster_id int64 1 1 |
|---|---|---|
Instead of using a simple lifetime average, Udemy calculates a course's star rating by considering a number of different factors such as the number of ratings, the age of ratings, and the likelihood of fraudulent ratings.
Learn Matlab
Discover The Language of Technical Computing.
3.3Matlab Software Installation: You are requried to install the Matlab Software on your machine, so you can start exectuing the codes, and examples we work during the course. Please visit : for requesting a free trial of the software if you didn't have it already. Most of the Universities offer a free student version of the software, therefore, if you are a student, first contact your university to see if such a program is avaiable. If you are not a student or your university didn't offer it, you can purchase the student version directly from Mathworks.
The will to learn programming and Matlab!
Description
MATLAB is a higher level programming language that has various uses in everyday problems. And this tutorial does exactly that.
We first go through the basics needed to begin the start of your programming journey. Such basics include but are not limited to : 1. Assigning numerical values to variables 2. Manipulating these variables in a meaningful way 3. Creating and manipulating vectors for problem solving 4. Creating and manipulating matrices for problem solving 5. Plotting various graphs to effectively display information
After mastering the basics, we move on to more advanced topics to manipulate data and various types of information in a more specific way.
Advanced topics include creating a graphical user interface and utilizing the image processing toolkit in MATLAB. This tutorial is designed to make MATLAB easier to understand and hopefully, not so scary!
This video deals with simple mathematical operations and how they are used in MATLAB such as addition, subtraction and multiplication.
Syntax for mathematical operations
03:39
These videos deals with constant functions used for mathematical manipulations such as pi, squareroot, sine, cosine and many more
Various constant names for simple mathematical operations
02:41
This video deals with commenting the code created using the '%' sign and also how to create sections in one's code using the '%%' sign
Commenting and creating sections in Matlab
03:40
This video deals with the use of the semi-colon as it suppresses the result which is useful for large data sets
Use of the semi-colon
01:32
This video deals with use of the colon in vectors, arrays and matrices.
Use of colon
01:53
+–
Variables
1 Lecture
09:24
In this video, we deal with how to assign a variable with a numeric value, how to assign numerous variables with numerical value and how to manipulate these variables
Variables
09:24
+–
Commands
3 Lectures
18:35
In this short video, we look at the everyday functions used in matlab such as exist (informing the user if a variable exists in the current workspace), who ( displays information on all variables in workspace) and others
Basic commands
04:39
In this short video, we look at the everyday functions used in matlab such as load (loading variables into the workspace), date ( displays the current date) and others
More basic commands
06:57
There are many ways to format how numbers are displayed in matlab. This video deals with the long, short, exponential and rational formatting.
Formatting
06:59
Quiz for Commands section
Commands Quiz
5 questions
+–
Operators
3 Lectures
23:47
This video deals with logical operators such as >, >=, < and others
Logical and Relational operators
06:26
In this video, we look at the basic arithmetic operators such as addition, subtraction, multiplication, division and raising a value to a power
Arithmetic operators
07:47
In this video, we're going to look at a myriad of operators as it pertains to sets and set theory such as the functions ismember(which lets you know if a particular item is a member of the set in question), union (which takes the union of 2 or more sets and make it into one set) and others
Set operators
09:34
+–
Loops and Decisions
4 Lectures
35:12
This video describe the use of if statements and how to create these statements in MATLAB
If statements
08:05
This video describe the use of for loops and how to create these loops in MATLAB
For loops
08:13
This video describe the use of while loops and how to create these loops in MATLAB
While loops
06:36
This video describe the use of nested for loops and how to create these loops in MATLAB
Nested loops
12:18
Quiz for Loops and Decisions Quiz
Loops and Decisions Quiz
10 questions
+–
Vectors and Matrices
5 Lectures
34:08
This video deals with the addition, subtraction, multiplication as well as scalar division and multiplication. This video also deals with transposing vectors.The elimination and choosing an element in a vector is also addressed. Appending vectors in column versus row formis also addressed.
Vector manipulation
06:50
This video will demonstrate how to create vectors and matrices of varying lengths using commands such as 'zeros' and 'ones'.
How to create matrices of varying lengths
11:24
In this video, we will look at how to add, subtract and multiple a matrix by another matrix or a scalar value
Perform mathematical manipulations on matrices 1
06:05
In this video, we will look at how to divide a matrix by another matrix and a scalar value. This video also demonstrates how to obtain the inverse of a matrix
Perform mathematical manipulations on matrices 2
04:26
This video demonstrates how matrices can be concatenated (how rows and/or columns can be added to a matrix) and how to generate the transpose of a matrix.
Perform mathematical manipulations on matrices 3
05:23
+–
Creating and running a script
1 Lecture
09:16
In this video, we will look at how to create script in matlab using the editor window
Creating and running a script
09:16
Quiz for Creating and running a script
Creating and running a script Quiz
4 questions
+–
Manipulation of strings
3 Lectures
09:42
This video deals with how strings are created and how to put a series of strings into 1 cell array using cellstr function
Basics of strings
05:02
This video addresses the finding and replacing of a word or phrase in a string
Find and replace
02:26
This videos addresses the comparison of strings using the function strcmp
Comparison of strings
02:14
+–
Creating functions
4 Lectures
24:38
In this video, we learn how to create simple functions with one input and outputs as well as simple functions with multiple inputs and outputs and how to use these functions
Create simple functions
10:49
Quiz for Create simple functions
Create simple functions - Quiz
7 questions
In this video, we will look at how to create primary and sub functions as well as nested functions
Create primary, sub and nested functions
06:21
Quiz for Create primary, sub and nested functions
Create primary, sub and nested functions - Quiz
8 questions
In this video, we will look at how to create private functions
Create private functions
03:14
In this video, we look at the use of the variable GLOBAL and its role in utilizing variables in and outside of functions | 677.169 | 1 |
a t r i k s
2.
<ul><li>Understanding, N o tation, and Dimension of Matrix </li></ul><ul><li>Types of Matrix </li></ul><ul><li>Transpose and Similarity of a matrix </li></ul><ul><li>Operation of the Matrix </li></ul><ul><li>Inverse and Determinant of the Matrix </li></ul><ul><li>Completed the system of Linear Using Matrix </li></ul>
3.
<ul><li>Understanding, Notation, and Demension Of Matrix </li></ul><ul><li>Understanding Of Matrix </li></ul><ul><li>Pay attention the following illustration </li></ul><ul><li>Mr. Andi note is student absent in last three year, that is January, February and March to 3 student that is Arlan, Bronto, and cery like at the table . </li></ul>On the table can be written : Arlan Bronto Cery January February March 3 4 1 6 2 3 1 2 4
4.
is a rectangular array of numbers, consists of rows and columns and is written using brackets or parentheses. The entries of a matrix are called elements of matrix . An element of a matrix is addressed by listing the row number and then column number M A T R I X
6.
2. The order of the matrix A matrix of A has m rows and n column is called as matrix of dimension on order m x n, and so notated of "A(mxn)". To more understand the definition of the element of a matrix.
7.
The first column The second column The third column The column n-th The second row The first row The third row The row n-th
8.
Example: Matrix A = The first row The second row The first column The second column The third column <ul><ul><li>The order matrix A is 2 x 3 </li></ul></ul><ul><li>4 is the second row and the first column </li></ul>
9.
a row matrix Is a matrix that only has a row A = ( 1 3 5), and B = ( -1 0 4 7) The order matrix is and
11.
A matrix square A square matrix a matrix has the number of row of a matrix equals the number of its column
12.
Example : rows 4, columns 4 A is matrix the order 4 A = Main diagonal
13.
A = Upper Triangle Matrix is square matrix which all of the element under the diagonal is zero Upper Triangle Matrix
14.
B = B is a lower triangle matrix is square matrix which all of the element upper the diagonal is zero Lower Triangle Matrix
15.
C = Diagonal Matrix is square matrix that all of element is zero, except the element on the diagonal not all of them Diagonal Matrix:
16.
I = I is matrix Identity that is diagonal matrix that elements at main diagonal value one Pay attention the following matrix
17.
<ul><li>Transpose and Similarity of a Matrix </li></ul><ul><li>Transpose of a Matrix </li></ul><ul><li>Let A is a matrix whit dimension of (m x n). From the matrix of A we can formed a new matrix that obtained by following method: </li></ul><ul><li>a. Change the line of i th of matrix A to the row of </li></ul><ul><li> ith of new matrix </li></ul><ul><li>b. Change the row of j th of matrix A to the line of </li></ul><ul><li> jth of new matrix </li></ul><ul><li>The new matrix that resulted is called transpose from matrix of A symbolized with A' or From the above changess, the dimension of A' is (n x m) </li></ul>
20.
let A = (aij) ang B = (bij) are two matrices with the same dimension. Matrix of A is callled equal with matrix of B id the element that located on the two matrices has the same value. 2. Similarity of two matrix
21.
One located element with the same value One located element with the same value One located element with the same value One located element with the same value
33.
Scale Multiplication With a Matrix Let k Є R and A is a matrix with dimension of m x n . Multiplication of real number k by matrix of A is a new matrix which is also has dimension of m x n that obtained by multiplying each element A by real number of k and notates kA
39.
Matrix Multiplication with Matrix The Product Of Two Matrices A and B can be got when satisfies the relation A m x n = B p x q = AB m x q Equal
40.
The number of column of matrix A should equal the number of rows of matrix B, the product, that is AB has order of m x q. when m is the number of rows of matrix A and q is the number of column of matrix B
57.
26 November 2011 <ul><li>Inverse Matrix </li></ul><ul><li>If A and B are two square matrices with the same dimension such that satisfies AB = BA = I where I is an identity matrix, then </li></ul><ul><li>Matrix of A called inverse from matrix of B given notation of B </li></ul><ul><li>Matrix of B called inverse from matrix of A given notation of A </li></ul>-1 -1
68.
Completed the system of linear Equation Using Matrix <ul><li>There are two methods can be used to determined the solution of system of linear equation using the matrix approach that are: </li></ul><ul><li>Determinant method </li></ul><ul><li>Inverse matrix method </li></ul>
69.
Variables <ul><li>Using determinant method, the value of x and y found out by the following formula </li></ul>
70.
Variables b. Using inverse of matrix method, the value of x and y found out by using the following steps. | 677.169 | 1 |
The math b regents is often considered one of the most difficult
new york state regents math b covers concepts that can be found in
trigonometry and advanced algebra, as well as.
e to the math b section of the free new york state high school
regents exam prep center! interactive games and puzzles for math b.
Why take the math b regents exam? in new york state those students
wishing to pursue an advanced regents diploma must pass the
mathematics b regents examination what math topics
new york regents math b
Latest Top New York Regents Math B
The math b regents is often considered
one of the most difficult new york state regents math b covers
concepts that can be found in trigonometry and advanced algebra, as
well as.
e to the math b section of the free new
york state high school regents exam prep center! interactive games
and puzzles for math b. Why take the math b regents exam? in new
york state those students wishing to pursue an advanced regents
diploma must pass the mathematics b regents examination what math
topics | 677.169 | 1 |
According to many surveys, math remains one of the most important yet frightening subjects for students in the US. In fact, many students in the US don't even know what math courses they should study to get into a college. No wonder their performance is severely sub par at the college level. But your student CAN do well in college math classes by being well prepared. As a homeschooling parent, you want to choose the best for your students. But what does your student really need for college preparation? Which courses will give them the best education and set them off on the right path? There are 4 basic levels of math that a student will need in high school in order to do well at college: Algebra 1, Geometry, Algebra II with Trig, and Calculus 1.
Download Drill of Science Math for Snmptn
Granted, not all students will make it to Calculus 1, and that's fine. The point here is that you know what subjects you need, then it becomes just a matter of finding the right curriculum to teach those subjects. Start by Identifying the Target College A lot of the decision-making about curriculums depends upon the college that you want to attend. Typically, each college will have their own set of requirements for incoming students. For example, MIT prefers a strong foundation in calculus. They want their applicants to have gone through at least one level of Calculus in high school before they arrive. Auburn University, however, expects a strong foundation in Algebra I and II with a secondary emphasis being placed on courses like calculus, geometry, statistic analysis, and trigonometry. Pomona College highly recommends students know calculus, just like MIT whereas, Harvard places its main emphasis on concrete understanding of functions, algebra, and graphing. They don't focus much on calculus like MIT and Pomona. So, basically, wherever you want to attend, find out about their applicant math requirements. Now, even though each college has their own course requirements, there are some general points that can be said for every homeschooling student: • Calculus can be considered optional for most students. • Taking Calculus 1 in high school allows the student to go into almost any freshman math class with the subject knowledge well in hand. • Having studied through Calculus 1 in high school allows easier adjustment to the faster pace and stricter structure of college without getting overwhelmed by the course material. • Algebra 1 & 2, Trigonometry, and Geometry are the minimum courses required to do well at college. Meet the college requirements by focusing on understanding You can check off any math course you "completed" but if your student doesn't understand, the effort was rather worthless. For this reason, we encourage you to take the time to figure out what makes your student's mind tick. Study math in a way that makes sense for them personally. If you're worried about your math experience and think you'll need help---seek help. There are many great step by step math courses and DVD courses out there to choose from. Shop around, and get your student's input. Choose the course that both of you like and are excited to learn from. | 677.169 | 1 |
Introduction to Advanced Mathematics (2nd Edition)
Focused on "What Every Mathematician Needs to Know," this book focuses on the analytical tools necessary for thinking like a mathematician. It anticipates many of the questions readers might have, and develops the subject slowly and carefully, with each chapter containing a full exposition of topics, many examples, and practice problems to reinforce the concepts as they are introduced. "Find the Flaw" problems help readers learn to read proofs critically. Contains five core chapters on elementary logic, methods of proof, set theory, functions, and relations; and four chapters of examples, theorems, and projects. For those interested in abstract algebra or real analysis.
Book Description Pearson Education (US), United States, 1999. Hardback. Book Condition: New. 2nd Revised edition. Language: English . This book usually ship within 10-15 business days and we will endeavor to dispatch orders quicker than this where possible. Brand New Book. For a one-quarter/semester, sophomore-level transitional ( bridge ) course that supplies background for students going from calculus to the more abstract, upper-division mathematics courses. Also appropriate as a supplement for junior-level courses such as abstract algebra or real analysis. Focused on What Every Mathematician Needs to Know, this text provides material necessary for students to succeed in upper-division mathematics courses, and more importantly, the analytical tools necessary for thinking like a mathematician. It begins with a natural progression from elementary logic, methods of proof, and set theory, to relations and functions; then provides application examples, theorems, and student projects. Bookseller Inventory # BTE9780130167507
Book Description 1999. Paperback. Book Condition: New. 2nd. Paperback. For a one-quarter/semester, sophomore-level transitional (bridge) course that supplies background for students going from calculus to the more abstract, upper-division mathema.Shipping may be from our Sydney, NSW warehouse or from our UK or US warehouse, depending on stock availability. 300 pages. 0.567. Bookseller Inventory # 9780130167507 | 677.169 | 1 |
Welcome to Math III!!! August 6, 2012.
Similar presentations
2Murder Each person get a card DO NOT SHOW YOUR CARD TO ANYONE!!!DO NOT TELL ANYONE WHAT YOU HAVE!!!If you have the Joker, you are a murderer!!!As we go around, the murderer will wink at you, if this happens, you are dead!Announce you are dead by saying your name and something about yourself!
3Math III SyllabusThis is the 3rd course in a sequence of courses designed to provide students with a rigorous program of study in mathematics. Concepts will be learned in multiple ways. Students will be working in groups to better understand the topics. They will also work alone to develop individual skills. The course is designed to give students a real-world understanding of mathematics.
9PracticeEach student is expected to attempt, complete, and check every assignment. Practice problems will be assigned daily. Mathematics is learned by practice; therefore, you are expected to do your homework every night. Students are expected to check their own practice problems as we review them in class. There will be an unannounced homework check approximately once a week, which will be graded for accuracy. Students are expected to have all assignments organized in their notebooks at all times. Students should always show all their work. If asked to, students should explain their reasoning on paper. At times, projects will be assigned.
10ParticipationThis course requires a great deal of participation. Some classes we will take notes, others will be board work, and some will be group/individual work. Asking questions is a MUST. This course is designed to help you succeed in Math III, and it is your job to insure your own understanding of the material.
11Attendance/Make Up Work It is very important that students come to class regularly. Only work missed due to an excused absence may be made up. When you are absent, you have the same number of days that you were absent to make up an assignment or a test. You assume the responsibility to make arrangements to make up a test and securing all notes Class time will not be spent getting you caught up if it interferes with the instructional time of other students.
12TutoringI am available to help you and answer your questions when needed. It is imperative that you let me know immediately if you are having difficulty. I am available for tutoring various times throughout the week. Additionally, tutoring will be offered during ALL study halls by math teachers. There are additional resources available on the Ola High School and Henry County websites. Make sure you check my teacher webpage daily through the Ola High School website for updates.
15Calculator RulesOnly get a Calculator if it is required for the day's lesson, do not get one if you do not have your own unless you have talked to Ms. Brown or Ms. MeadowsYou break it, YOU BUY IT!!! $149.99If using a calculator, clear it out after use!!!When returning calculators,1. Turn OFF the calculator2. Place cover on calculator so that no keys are showing3. Stack the calculators in a neat and orderly fashion(3 stacks of 10 each)
16Bathroom Rules GO BEFORE CLASS!!! You get 3 bathroom "passes" each termThis will be a sign out sheet that I keep track ofIf you do not use them, each add 10 points to any assignmentIf none are used, you can replace a homework assignment (not a quiz or test!)
17Ticket Out the Door On the index card provided write: Your Name What you expect out of Math III (rigor, grade, etc.)What you expect from Mrs. Gowen and/or Mrs. MeadowsWhat you want to do after high school | 677.169 | 1 |
A blog is that is all about mathematics and calculators, two of my passions in life.
Monday, November 21, 2011
HP 15C Programming Tutorial - Part 9: Derivatives
Numerical Derivatives
In Part 9, we will calculate numerical derivatives of f(x). Computing accurate numerical derivatives can present a challenge. Often, calculation involves a small increment, usually named h. Generally, the smaller h is, the better the calculation. However with certain methods, if h is too small, the final result may be unexpected | 677.169 | 1 |
This book provides a thorough introduction to thechallenge of applying mathematics in real-world scenarios. Modelling tasksrarely involve well-defined categories, and they often requiremultidisciplinary input from mathematics, physics, computer sciences, orengineering. In keeping with this spirit of modelling, the book includes awealth of cross-references... more...
Computer Systems Architecture provides IT professionals and students with the necessary understanding of computer hardware. It addresses the ongoing issues related to computer hardware and discusses the solutions supplied by the industry. The book describes trends in computing solutions that led to the current available infrastructures, tracing... more...
Build amazing Internet of Things projects using the ESP8266 Wi-Fi chip About This Book: Get to know the powerful and low cost ESP8266 and build interesting projects in the field of Internet of Things; Configure your ESP8266 to the cloud and explore the networkable modules that will be utilized in the IoT projects; This step-by-step guide teaches you... more...
This book brings together a rich selection of studies in mathematical modeling and computational intelligence, with application in several fields of engineering, like automation, biomedical, chemical, civil, electrical, electronic, geophysical and mechanical engineering, on a multidisciplinary approach. Authors from five countries and 16... | 677.169 | 1 |
Mathematics
The aim of the Misbourne Mathematics Team is to focus on understanding and helping our students to become secure and confident with their knowledge of mathematics. We help students learn how to solve problems which require the use of mathematical knowledge and techniques. We aim to give all students the chance to take an active role in every mathematics lesson. We do this through the use of high-quality active learning and teaching techniques, such as:
a well-planned curriculum, which builds students' understanding
collaborative tasks, where pupils work in small groups to discuss and solve mathematics problems
class discussions using higher-order questioning techniques
the use of mini-whiteboards to collect and discuss answers from every student in the class
use of ICT to enhance learning through use of interactive whiteboards in every lesson
Key Stage 3 (Years 7, 8)
The major topics that will be covered in Key Stage 3 are below. These will be tested using regular assessments and end of Year levelled tests.
Number
Algebra
Data
Shape
All students follow a syllabus focusing on embedding core areas of learning through investigation and problem solving
GCSE Mathematics
Years 9,10,11 – Year 9 are studying the three year syllabus working towards the new Edexcel GCSE Mathematics Linear Examination 1MA1 (see below for details) which will be first assessed in June 2017. The examination will consist of one non-calculator and two calculator based written papers, each one being one and a half hours long.
Edexcel GCSE Mathematics Linear Examination 1MA1
The Pearson Edexcel Level 1/Level 2 GCSE (9 to 1) in Mathematics is a tiered qualification.
There are two tiers:
● Foundation tier - grades 1 to 5 available
● Higher tier grades – 4 to 9 available (grade 3 allowed)
The assessment for each tier of entry consists of three externally-examined papers, all three must be from the same tier of entry. Students must complete all three papers in the same assessment series.
Mechanics 1 - Mathematical models in mechanics; vectors in mechanics; kinematics of a particle moving in a straight line; dynamics of a particle moving in a straight line or plane; statics of a particle; moments.
Advanced GCE Further Mathematics (9372)
Further Pure Mathematics units FP1, FP2, FP3 and a further three Applications units (excluding C1–C4) to make a total of six units; or FP1, either FP2 or FP3 and a further four Applications units (excluding C1–C4) to make a total of six units.
Examinations
All examinations will take place in the summer of the second year of study.
Department Staff
Curriculum Leader Mathematics – Miss M Guy
Assistant Curriculum Leader – Mr J Davidson
KS5 Coordinator – Mr S Egan
Mr R Cameron
Mr P Dunsby
Mr R Francis
Mr W Gunstone
Mrs A Harding
Mrs M Lu
Mr S Reese
Miss N Weber
Excursions, Trips and other interesting projects
Various maths challenges will take place throughout the academic year – locally and regionally.
Homework and Assignments
Key Stage 3 students will be set 30 minutes of homework per week. This can take the form of short independent tasks or sometimes more extended projects set over a number of weeks.
Key Stage 4 students will be set approximately 90 minutes of homework per week. This can take the form of one homework or possibly split into different tasks. Revision and refinement is a key part of success at GCSE.
Key Stage 5 students will be expected to complete the same amount of work independently as they have contact time with their teacher. In most cases this will be five hours of work. To be successful at AS and A2 students must work outside of the classroom and will be expected to research and read around topics themselves without being directed to by the teacher. | 677.169 | 1 |
Algebra Functions Transformation Quizzes & Practice
Be sure that you have an application to open
this file type before downloading and/or purchasing.
752 KB|31 pages
Product Description
Included in the following document are fourteen different quizzes and extra materials for you to cut, paste or easily manipulate. I the quizzes when I discovered my advanced math students were dependent on their calculators for graphing basic functions and transformations (horizontal, vertical, reflections). I decided to give my students 10 question "no calculator" quizzes randomly throughout the year. All quizzes require basic sketching of the following parent functions: linear, quadratic, cubic, absolute value, square root, rational, logarithmic, exponential, trigonometric, and piecewise functions. On some versions Iincluded some open ended questions that allow students to develop their own formulas based on a point on a graph, while being given the end behavior of the function. I did the best I could to select images that summarize the general tone of each quiz.
Details of each different version are as follows.
Quiz Version A - 6 different quizzes (increased difficulty from A1 to A6)
Quizzes include the following functions: Linear, quadratic, cubic, absolute value, square root, basic rational functions and piecewise.
Topics also included: domain & range, and left / right handed behavior.
Version B - 6 different quizzes (increased difficulty from B1 to B6)
These include all topics from version A but also contain: exponential, logarithmic, and all six trigonometric functions.
Version C - 2 different quizzes
The last two include a combination of all of the functions include, and ask students to develop equations based on a graph.
I could see an Algebra 2 or pre-calculus teacher using these as NO CALCULATOR quizzes, review, or practice.
I am submitting this as a word document so that you can cut, paste, and manipulate my work as you see fit. If you had the time you could easily make several versions or even combine topics to make your own versions for whatever purpose you would like. I also included some random equations that I used to make review cards with my students and a one page review assignment that could make for a quick review in a pinch or to leave for a substitute.
*Please note, you may need to modify or change the margins after you download the document. The original document was generated using .5" margins on the top and bottom and 1" margina on the left and right.
Thanks for looking, but to be honest the random thumbnails generated by TpT do not do this file justice. Please feel free to contact me with questions!
P.S. I never include answer keys with my work as I am assuming that you would like to solve the questions on your own. | 677.169 | 1 |
Mathematics AS Level
OCR
Why study Maths?
A level in Maths provides students with the essential skills to reason and problem solve in a variety of context. It delivers the foundation to further study subjects such as science and engineering. The course is split up into six different modules. Three of them are completed at AS and three of them are completed at A2. The AS modules are core 1, core 2 and mechanics 1 and the A2 modules are core 3, core 4 and statistics 1. The four core units look at topics in algebra, trigonometry, proofs and calculus. Mechanics explores topics on forces, energy and motion while statistics covers probability, data handling and hypothesis testing. All modules have equal weighting and are assessed through examination.
Assessment:
The assessment of AS Mathematics will involve three exam papers in total. One part of the assessment consists of two papers one for each Core Maths Advanced Module (One paper for Core 1, One paper for Core 2).
There is also another paper to assess the applied module i.e. Statistics. The three papers are sat in the middle of the summer term. There is no coursework.
Resources:
All students have a login to 'MEI Integral Resources' providing a range of notes and examples, worksheets, interactive presentations and models, and assessments. We also use a number of textbooks including: | 677.169 | 1 |
Course Summary
Get ready for the AP calculus exam with our video lessons on graphing, functions, exponents, derivatives, integrals and scientific calculator use. Each lesson includes a short, multiple-choice quiz that allows you to practice important concepts using questions similar to those found on the exam.
About This Course
With our AP Calculus AB course, you can build a foundation with chapters about graphing and functions basics, before you get into more complicated concepts like logarithms, polynomials, derivatives and integration. Our subject-matter experts present these brief, targeted video lessons with self-assessment quizzes to strengthen your knowledge and give you the confidence you need moving forward.
Course Topics
Category
Objectives
Graph Basics
Learn about the types and various parts of graphs; look at analytic and geometric information on graphs | 677.169 | 1 |
An introduction to complex numbers, including a little history (quadratic and cubic equations; Fundamental Theorem of Algebra, the number i) and the mathematics (the complex plane, addition, subtraction; absolute value; multiplication; angles and polar coordinates; reciprocals, conjugation, and division; powers and roots), and intended for a general audience having a familiarity with ordinary real numbers (all positive and negative numbers and zero) and algebra. | 677.169 | 1 |
Complex Variables: Pergamon International Library of Science, Technology, Engineering and Social Studies (The Commonwealth and international library)
Complex Variables focuses on the principles, characteristics, and functions of complex variables, as well as infinite series, complex numbers, and convergence and divergence. The book first examines complex numbers and the sequences and limits of point sets in the complex plane. Discussions focus on non-decreasing real sequences, boundedness of convergent sequences, boundary points, closed sets, bounded and unbounded sets in the complex plane, complex conjugates, complex numbers as an extension of the real number field, scalar multiplication, modulus, and number pairs. The manuscript then takes a look at the tests for convergence of infinite series, functions of a complex variable, and elementary functions. Concerns cover repeated differentiation of an infinite series, differentiability of power series, hyperbolic functions, link between the exponential and trigonometric functions, orthogonal families of curves, differentiability, testing for convergence or divergence, and series with negative or complex terms. The text examines miscellaneous theorems, contour integration, zeros and singularities, and integration, including order of magnitude of a function, infinite integrals involving trigonometric functions, and sum-limit and anti-differentiation. The publication is highly recommended for students and teachers wanting to explore complex variables. | 677.169 | 1 |
Many people think there is only one "right" way to teach geometry. For two millennia, the "right" way was Euclid's way, and it is still good in many respects. But in the 1950s the cry "Down with triangles!" was heard in France and new geometry books appeared, packed with linear algebra but with no diagrams. Was this the new "right" way, or was the... more...
A new ANGLE to learning GEOMETRY Trying to understand geometry but feel like you're stuck in another dimension? Here's your solution. Geometry Demystified , Second Edition helps you grasp the essential concepts with ease. Written in a step-by-step format, this practical guide begins with two dimensions, reviewing points, lines, angles, and distances,... more...
The word barycentric is derived from the Greek word barys (heavy), and refers to center of gravity. Barycentric calculus is a method of treating geometry by considering a point as the center of gravity of certain other points to which weights are ascribed. Hence, in particular, barycentric calculus provides excellent insight into triangle centers.... more...
The tensorial nature of a quantity permits us to formulate transformation rules for its components under a change of basis. These rules are relatively simple and easily grasped by any engineering student familiar with matrix operators in linear algebra. More complex problems arise when one considers the tensor fields that describe continuum bodies.... more... | 677.169 | 1 |
, written by best-selling developmental mathematics author Pat McKeague, features a more streamlined review of elementary algebra, allowing for earlier coverage of intermediate topics. An early introduction to graphing presents the foundation for a wide variety of graphing problems throughout the text. Early coverage of functions helps students feel comfortable with the many examples and graphs of functions that occur in later chapters. The first ten chapters of this book cove the topics usually found in a college-level algebra course. The last three chapters cover the essential topics from trigonometry. Optional technology sections and integrated throughout text as a way for students to better understand the material being discussed. | 677.169 | 1 |
Discovering Algebra With Graphing Calculators: Plotting the Curve of Best Fit: Discovering Algebra With Graphing Calculators
Item#
$ 59.95$ 69.95
This program examines plotting data to determine a curve (linear, quadratic, or exponential) of best fit. Students learn to create scatter plots, interpret results, and make predictions based on table values. | 677.169 | 1 |
This engaging text presents the fundamental mathematics and modelling techniques for computing systems in a novel and light-hearted way, which can be easily followed by students at the very beginning of their university education. Key concepts are taught through a large collection of challenging yet fun mathematical games and logical puzzles that require no prior knowledge about computers. The text begins with intuition and examples as a basis from which precise concepts are then developed; demonstrating how, by working within the confines of a precise structured method, the occurrence of errors in the system can be drastically reduced. Features: demonstrates how game theory provides a paradigm for an intuitive understanding of the nature of computation; contains more than 400 exercises throughout the text, with detailed solutions to half of these presented at the end of the book, together with numerous theorems, definitions and examples; describes a modelling approach based on state transition systems.
"This book by UK academics Moller (Swansea Univ.) and Struth (Univ. of Sheffield) is designed to serve as a textbook for first-year university students in computer science. The volume contains 15 chapters, divided in two parts. … libraries may wish to acquire it for the benefit of advanced undergraduates. Summing Up: Recommended. Only comprehensive academic mathematics and computer science collections." (B. Borchers, Choice, Vol. 51 (7), March, 2014)
"This book contains essential mathematics and modelling techniques for computing systems for which a presentation style suitable for first year undergraduate students has been chosen. … The book contains more than 200 exercises throughout the text and provides complete solutions at the back of the book on more than 80 pages." (Gudula Rünger, zbMATH, Vol. 1278, 2014) | 677.169 | 1 |
Polynomial Degree and Leading Coefficient Practice
Be sure that you have an application to open
this file type before downloading and/or purchasing.
28 KB|1 page
Product Description
This is a graphic organizer that can be used as an activity in an algebra 2 or precalculus course for students to practice the vocabulary of Degree and Leading Coefficient of polynomials. Degree and Leading Coefficient are to be written on the lines and examples (provided by teacher or student-derived) go inside the ellipses. Students could use them to quiz each other, can share them on a document camera or post on a wall after being decorated. It is intended to be cut in half to make two copies per sheet. | 677.169 | 1 |
3ObjectivesStudents will understand more about the various Math class methods including sqrt, pow, abs, ceil, floor, the trig functions sine, cosine, and tangent, and the constant values PI and E.The Java documentation help file, or API, will be explored specifically in regards to the Math class.Also, the use of static import statements for the System and Math classes will be introduced.Lesson 5A - Methods – Math class
4ObjectivesLabs include work with classic math functions including number line distance, coordinate plane distance, area and volume formulas for various geometric 2D and 3D figures, problems involving the use of ceil and floor, the Pythagorean theorem, quadratic formula and various trig functions such as the law of sines and law of cosines.Lesson 5A - Methods – Math class
5Math class definitionPreviously, several Math class items were introduced, including:pow – returns the power given a base and exponentsqrt – returns the square root of a valuePI – a fairly precise constant for PI, the ratio of the circumference of a circle to its diameterE – a fairly precise constant for E, the base of the natural logarithmsLesson 5A - Methods – Math class
6Math class definitionA complete definition of the Math class can be found in the Java help files, or external documentation, which is often referred to as the Java API (application programming interface). On the next screen you can see the beginning of this definition.Lesson 5A - Methods – Math class
8Math class API Let's examine several features of this help file… The Math class belongs to the java.lang packageIt is a "descendant" of, or "extends" the java.lang.Object class, which means it has "inherited" all of its features, and added a few more of its own (more on inheritance later).A brief description of the class is provided.Lesson 5A - Methods – Math class
9Math class APIHere you see the beginning of the listing of Math class fields and methods, (over 40 in all) including the constants E and PI, as well as the first few methods.Lesson 5A - Methods – Math class
10Math class APIThe abs method is "overloaded", or defined in several ways to be able to handle four different data type parameters: double, float, int, and long.Lesson 5A - Methods – Math class
11Math class APIThese are parameters!Notice that the return type is the same as the parameter type for this method.Lesson 5A - Methods – Math class
12And these are return values! Math class APIAnd these are return values!Notice that the return type is the same as the parameter type for this method.Lesson 5A - Methods – Math class
13Math class APIEureka! They match!!!Notice that the return type is the same as the parameter type for this method.Lesson 5A - Methods – Math class
14Math class APINot all match, however. Most of the return types are doubles, and sometimes there are multiple parameters.Lesson 5A - Methods – Math class
15Math class APIJust look carefully for the method you want, and see what parameters it requires, and what the return type is.Lesson 5A - Methods – Math class
16Math class APIIn all of the API class listings, there are two main sections.The column on the left indicates the return type, or the type of data for the answer returned by the method.All of the Math class methods are "static", which means they are "stand-alone" methods and work independently of any object (more explanation on that later).The column on the right contains the name of the method, plus any parameter(s) it receives, along with a description of its purpose.Lesson 5A - Methods – Math class
17Math class APIOn the next few slides, the few methods that we will study are highlighted. Study them carefully and become familiar with them.On this slide, check out the ceil and cos methodsLesson 5A - Methods – Math class
26ceil and floorMost of the Math functions are familiar to you, or will be as a part of your math-based learning, but the ceil and floor functions are not part of the standard math curriculum and require some explanation.Lesson 5A - Methods – Math class
27ceil and floorIn a similar way to the rounding function, ceil and floor convert any decimal value to the nearest whole number, but in a specific direction, NOT according to the rounding rule.Lesson 5A - Methods – Math class
29ceil and floorHere are several examples. Pay close attention to the negative values.Lesson 5A - Methods – Math class
30ceil and floorThey seem opposite, but if you think of the vertical number line, it all makes sense.Lesson 5A - Methods – Math class
31ceil and floorNotice carefully that although the resulting value is always a whole number, it is still a double type.Lesson 5A - Methods – Math class
32ceil and floor situations These two interesting functions play a critical part of problem- solving, especially in situations that require "interpreting" the remainder, or fractional part of a division problem.Lesson 5A - Methods – Math class
33ceil and floor situations For example, if you need to purchase enough plywood to do a job, and your precise calculations come up to 4.5 sheets of plywood, you would "ceil" the value and purchase 5 sheets. A lumber company will NOT sell you 4.5 sheets of plywood, sorry!Lesson 5A - Methods – Math class
34ceil and floor situations Another situation would be if you have a certain amount of money and calculate that you can purchase exactly 4.5 items of something you need, clearly you would "floor" the value and only be able to purchase 4 items, with some change leftover.Lesson 5A - Methods – Math class
35Static importsThe previous program examples used the Math class methods and the output statement System.out.println quite often, which brings up a really nifty feature of Java…the use of static imports.Lesson 5A - Methods – Math class
36Static importsSince the two most commonly used classes in many programs are the System and Math classes, it is quite useful to list two static import statements at the top of the program:import static java.lang.Math.*;import static java.lang.System.*;Lesson 5A - Methods – Math class
37Static importsimport static java.lang.Math.*;import static java.lang.System.*;These import statements enable the use of the System and Math class methods WITHOUT having to say System or Math each time, saving lots of typing and greatly simplifying your code!Lesson 5A - Methods – Math class
38Static import program examples Notice that with the two static import statements, the words System and Math have not been included in the commands, but the program still works.Lesson 5A - Methods – Math class
39Lesson SummaryIn this lesson, you learned more about the Math class methods, specifically about the ceil and floor methods, how to examine the class in great detail using the Java API, and about how to use static imports for System and Math to greatly simplify and streamline your code.Lesson 5A - Methods – Math class
40LabsThe following labs are all intended to help you practice using the Math methods. There may be more than one way to solve the program.Be sure that the output is precise, especially whether or not a value is an integer or a decimal, and how many decimal places are used.Lesson 5A - Methods – Math class
41Lab 5A-1WAP to find the distance between two input integers representing values on the number line. The formula is: |a-b| find the distance between two input decimals representing values on the number line. The formula is: |a-b|Input: (data file "lab5A1.in")3 565 -88Output: The distance from 3 to 5 is 2The distance from -3.3 to 5.8 is 9.1The distance from 65 to -88 is 153The distance from 24.9 to isLesson 5A - Methods – Math class
42Lab 5A-2 WAP to input an integer num from the keyboard and: : find the area of a square whose side length is num.: find the volume of a sphere whose radius is num.: find the length of the side of a square whose area is num.Input Data File ("lab5A2.in"):5 82Output: The area of a square whose side length is 5 is 25The volume of a sphere whose radius is 5 is 523.6The side length of a square whose area is 5 is 2.24The area of a square whose side length is 82 is 6724The volume of a sphere whose radius is 82 isThe side length of a square whose area is 82 is 9.06Lesson 5A - Methods – Math class
43Lab 5A-3WAP to find the minimum number of 4 X 8 sheets of plywood it would take to cover a roof when the total amount needed in square feet is input from a data file.Input Data File ("lab5A3.in"):Output: A 200 square-foot roof needs a minimum of 7 sheets of plywood.A 988 square-foot roof needs a minimum of 31 sheets of plywood.Lesson 5A - Methods – Math class
44Lab 5A-4WAP to find out the maximum number of DVDs Johnny can buy from WalMart if the price for each DVD (including tax) is $8.99 and the amount of money he has in his wallet is a value input from a data file.Input Data File ("lab5A4.in"):Output: With $26.25, Johnny can buy 2 DVDs.With $53.94, Johnny can buy 6 DVDs.Lesson 5A - Methods – Math class
45Lab 5A-5WAP to find the distance from two points in a coordinate plane, inputting the x and y coordinate for each. Output the distance formatted to two decimal places. Use the Math.hypot method to accomplish this task.Input coordinates in this order: x1 y1 x2 y2Input Data File ("lab5A5.in"):Output: The distance between the points (3,5) and (6,10)is 5.83 units.The distance between the points (0,-8) and (-16,5)is units. Lesson 5A - Methods – Math class
46Lab 5A-6In the mathematics field of trigonometry, the Law of Cosines states that the third side of an oblique triangle (any triangle that is not a right triangle) can be found by knowing the length of any two sides and the measure of the included angle (SAS).Given the Law of Cosines formula shown below, WAP to input two sides and the included angle (in degrees) of a triangle and find the length of the third side.Remember that the Math.cos method must have a radian measure as the parameter, which means you will have to convert the input angle from degree measure to radians using the Math.toRadians method.Lesson 5A - Methods – Math class
48Lab 5A-6Output: A triangle with sides of length 5.18 and 6.00 and an included angle measuring 60 degrees has a third side of length 5.63.A triangle with sides of length and and an included angle measuring 35 degrees has a third side of lengthA triangle with sides of length and and an included angle measuring 120 degrees has a third side of lengthLesson 5A - Methods – Math class
49Lab 5A-7Another classic problem found in trigonometry deals with the right triangle. When the lengths of any two sides of a triangle are known, it is possible to find the measure of the two acute angles of the triangle, using the inverse of sin, cos, and tan functions (asin, acos, atan).To summarize, when the opposite side (0) to an angle (A) and the hypotenuse (h) are known, you use the sin function in the following equation to solve for A: sin A = o/hUse the cos function when the adjacent side (a) and hypotenuse (h) are known: cos A = a/hoAhaAhLesson 5A - Methods – Math class
50Lab 5A-7Use the tan function when the opposite (o) and adjacent (a) sides are known: tan A = o/aTo solve for A in the above tangent situation, transform the tan A = o/a equation into A = atan(o/a).Since the measure calculated will be in radians, you must convert it to degrees to find the final answer. The same process is used for cos and sin.Given a data file with two sides given (always the two shorter sides), find and output the two acute angles of the right triangle.oaALesson 5A - Methods – Math class
52The Lesson 5B will elaborate more about the String class methods. CONGRATULATIONS!You now understand how to use many more of the Math class methods, as well as the Java API, and static import statements.The Lesson 5B will elaborate more about the String class methods.Lesson 5A - Methods – Math class
53Acknowledgement of use Please copy, paste, and customize the brief acknowledgement shown below in an to indicating that you used this lesson.Feel free to include any comments you wish to offer, positive or negative."I, (your name), from (educational institution) in (city, state, country) used Lesson 5A on (date), understand, respect and honor all of the rights claimed by the author, Mr. John Owen, and offer the following comments..._______________.""THANK YOU!"Lesson 5A - Methods – Math class | 677.169 | 1 |
Be sure that you have an application to open
this file type before downloading and/or purchasing.
1 MB
Product Description
The start of the lesson reviews calculating the mid-point for line segments with different gradients. The development phase builds on this by calculating the gradient of a line segment as a measure of its rate of change. As learning progresses students calculate the gradients between two coordinate pairs.
Differentiated Learning Objectives
All students should be able to calculate a positive gradient of a linear function using the graph.
Most students should be able to calculate a gradient of a line segment between two coordinate pairs.
Some students should be able to calculate the gradient of a function in a real life context. | 677.169 | 1 |
Linear vs. Exponential Matching
Be sure that you have an application to open
this file type before downloading and/or purchasing.
52 KB|4 pages
Product Description
This activity is great in a unit comparing linear and exponential functions in Algebra 1. The first two pages are a handout and the second two are cards that may be cut out. Students can work in partners or groups to match the situation, table, equation, and graph cards. Some cards will not be used. | 677.169 | 1 |
Algebra is an extremely effective means of stimulating the brain.
Although Algebra doesn't have an abundance of practical application, it's 'use' is an extremely useful way of teaching people to think, and solve problems. Also, algebra isn't necessarily 'x+y' and all that jazz, rather a way of solving problems by way of 'cancellation'. That's literally what the word 'algebra' is derived from.
Without algebra we'd be pre-industrial
Without algebra we wouldn't have anything like modern mathematics, which underpins every single modern science. You can't understand statistics without algebra; physics would fall, chemistry would fall, biology would fall. In short, algebra is one of the lights that lightens the darkness of our ignorance. We'd probably still believe that Sun goes around the Earth. Literally.
Not for the average person.
Think of all the people you know who have jobs. How many of them need anything past fifth grade math? Probably not many, am I right? Yes, algebra is important for science, but outside of those fields it is not needed too often. People seem to be getting along fine without using it on a day to day basis. | 677.169 | 1 |
Popular Textbooks
Details about Fundamentals of Trigonometry:
Completing the time-proven Swokowski/Cole precalculus series, this book helps students learn, understand, and appreciate trigonometry without compromising mathematical integrity. The book takes a unit-circle first approach to trigonometry and incorporates the use of the graphing calculator. Numerous application problems help motivate students toward success in learning trigonometry.Sample questions asked in the 9th edition of Fundamentals of Trigonometry: Exer. 7–8: Describe the set of all points P(x,y) in a coordinate plane that satisfy the given condition. Drug dosage A drug is eliminated from the body through urine. Suppose that for an initial dose of 10 milligrams, the amount A(t) in the body t hours later is given by A(t) = 10(0.8) t . (a) Estimate the amount of the drug in the body 8 hours after the initial dose. (b) What percentage of the drug still in the body is eliminated each hour? Exer. 1–12: Use De Moivre's theorem to change the given complex number to the form a + bi, where a and b are real numbers.
Back to top
Rent Fundamentals of Trigonometry 9th edition today, or search our site for other textbooks by Earl William Swokowski. Every textbook comes with a 21-day "Any Reason" guarantee. Published by CENGAGE Learning. | 677.169 | 1 |
Apply alternative mathematical techniques, from a historical perspective where appropriate.
Understand how mathematics is used in other fields and occupations.
Understand the use of mathematics cross-culturally.
General Education Learning Values & Outcomes
Revised August 2008 and affects outlines for 2008 year 1 and later | 677.169 | 1 |
Al bra brings about a unification of geometry, number theory and indeed most branches of mathematics. This is not really surprising when one has a historical understanding of the subject, which I also hope to impart.
"Sinopsis" puede pertenecer a otra edición de este libro.
From the Back Cover:
This book is a concise, self-contained introduction to abstract algebra that stresses its unifying role in geometry and number theory. Classical results in these fields, such as the straightedge-and-compass constructions and their relation to Fermat primes, are used to motivate and illustrate algebraic techniques. Classical algebra itself is used to motivate the problem of solvability by radicals and its solution via Galois theory. This historical approach has at least two advantages: On the one hand it shows that abstract concepts have concrete roots, and on the other it demonstrates the power of new concepts to solve old problems. Algebra has a pedigree stretching back at least as far as Euclid, but today its connections with other parts of mathematics are often neglected or forgotten. By developing algebra out of classical number theory and geometry and reviving these connections, the author has made this book useful to beginners and experts alike. The lively style and clear exposition make it a pleasure to read and to learn from.
Review:
"...The book is written in a very clear and pleasing style. Each of the 9 chapters of the book concludes with a section called "Discussions", which contains very interesting and valuable historical information and comments on the topics presented in the respective chapter. We strongly recommend this nice volume not only to beginners but also to experts."--MATHEMATICAL REVIEWS
Descripción Springer-Verlag New York Inc., United States, 2001. Hardback. Estado de conservación: New. 1st ed. 1994. Corr. 3rd printing 2001. Language: English . This book usually ship within 10-15 business days and we will endeavor to dispatch orders quicker than this where possible. Brand New Book. Al- bra brings about a unification of geometry, number theory and indeed most branches of mathematics. This is not really surprising when one has a historical understanding of the subject, which I also hope to impart. Nº de ref. de la librería LIE9780387942902
Descripción Springer, 1994. Hardback. Estado de conservación: NEW. 9780387942902 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 HTANDREE0275380 | 677.169 | 1 |
Math.NET aims to provide a self contained clean framework for symbolic mathematical (Computer Algebra System) and numerical/scientific computations, including a parser and support for linear algebra, complex differential analysis, system solving and more | 677.169 | 1 |
teaching RESOURCES
SUPPORTING YOUR CLASSROOM
ALGEBRA & FUNCTIONS A-LEVEL
This set of resources provides support for the DfE Content – Sequences and series, Trigonometry and Algebra and Functions.Working with sequences including those given by a formula for the nth term, understanding the use of trigonometric expressions and understanding the effect of combinations of simple transformations on the graph.
Our dedicated resources provide teachers of mathematics the guidance and practical worksheets for use in classroom and beyond, we are continuing to build new content every month as teachers and schools get ready for September 2017. Register your details and ensure you and your school receive the latest resources tailored specifically to the new specifications and how to get the best out of technology.
Resources are defined by subject area and provided in three formats per activity, StudentWorksheets, TeacherNotes and Video How-To guides.
How-To Video Guide
ALGEBRA & FUNCTIONS A-LEVEL
This should be the starting point for students (and teachers) to tackle each activity. Using associated examples, each video will prepare students with the required calculator skills and functional knowledge.
Student Worksheets & TEACHER NOTES:
ALGEBRA & FUNCTIONS A-LEVEL
These activities are designed to foster heuristic learning approaches to mathematical topics. The appropriate engagement with technology enables students to explore and hypothesise for themselves and to develop deep conceptual understanding.
The Teacher Notes provide solutions to the Student Worksheet with appropriate prompts and proofs to clarify understanding and address any misconceptions. | 677.169 | 1 |
From simple calculator operations to large-scale programming and interactive document preparation, Mathematica is the tool of choice at the frontiers of scientific research, in engineering analysis and modeling, in technical education from high school to graduate school, and wherever quantitative methods are used.
Mathematica is the world's only fully integrated technical computing system, combining interactive calculation (both numeric and symbolic), visualization tools, and a complete programming environment. Over a million researchers, students, engineers, physicists, analysts, and other technical professionals worldwide have discovered Mathematica's unique combination of unmatched computational power and unprecedented ease of use.
Mathematica runs on an unusually wide array of operating systems: Windows 95/98/NT/2000, Mac OS, Linux, SunOS/Solaris, HP-UX, IRIX, AIX, Digital Unix, and compatible systems.
CALCULATION CENTER
CalculationCenter is the new easy-to-learn, easy-to-use solution for getting your calculations done. Based on more than four years of development, CalculationCenter introduces a new interface concept that gets you calculating in as little as 10 minutes. And without learning syntax or reading a manual, you'll be ready to use hundreds of symbolic, numeric, and graphics functions right away. CalculationCenter has been designed and tested to the same exacting standards as Mathematica, Wolfram Research's world-leading technical computation software, giving you top quality at an affordable price.
Whether you're an engineer, financial analyst, researcher, or student or you just happen to like exploring science and technology, check here before you delve into your next project. It's likely that a team of experts in your field has created software to help you work smarter.
Giving you direct access to Mathematica's powerful programming language and expansive collection of computational commands, each application package or MathLink product provides an incredibly rich and flexible working environment. Click the link below for current pricing and available Mathematica Applications. | 677.169 | 1 |
Solving systems
of linear inequalities was informally addressed in Course 1 Unit 3,
Linear Functions. In Course 2
Unit 1, Functions,
Equations, and Systems, and Unit 2, Matrix Methods,
students solved linear systems of two equations in two variables by
graphing, substitution, and linear combinations, and by using matrix
methods. This unit reviews and extends students' understanding of and
their ability to solve inequalities in one and two variables.
Unit Overview This unit develops
student ability to reason both algebraically and graphically to solve
inequalities in one and two variables, introduces systems of inequalities
in two variables, and develops a strategy for optimizing a linear function
in two variables within a system of linear constraints on those variables.
Topics included are inequalities in one and two variables, number line
graphs, interval notation, systems of linear inequalities, and linear
programming.
Objectives
of the Unit
Write inequalities to express questions about functions of one or
two variables
Solve quadratic inequalities in one variable, and describe the solution set
symbolically, as a number line graph, and using interval notation
Solve and graph the solution set of a linear inequality in two variables
Solve and graph the solution set of a system of inequalities in two variables
Solve linear programming problems involving two independent variables
Sample Overview The sample material
provided below is Investigation 3, "Complex Inequalities," from
Lesson 1, "Inequalities in One Variable." The goal of
this investigation is to generalize students' graphic and algebraic
understanding and technique for solving inequalities (developed in
the first two investigations of this lesson) to somewhat more complex
situations and algebraic expressions. Students use algebraic methods
to locate intersection points of graphs and use the visual images provided
by graphs to inform the solution of the related inequalities. Interval
notation is introduced in this investigation. Selected
homework tasks from the Connections and Reflections sections are provided.
See Implementing
Core-Plus Mathematics page 13 for more detail on the design
features of the On Your Own homework section.
Instructional
Design Throughout the curriculum,
interesting problem contexts serve as the foundation for instruction.
As lessons unfold around these problem situations, classroom instruction
tends to follow a four-phase cycle of classroom activities—Launch,
Explore, Share and Summarize, and Apply.
This instructional model is elaborated under Instructional
Design.
View the
Unit Table of Contents and Sample Lesson Material You will need the
free Adobe
Acrobat Reader software to view and print the sample material.
How the Algebra
and Functions Strand Continues In Course 3,
there are two more algebra and functions strand units, Unit 5, Polynomial
and Rational Functions, and Unit 8, Inverse Functions.
Unit 5
extends student ability to represent and draw inferences about polynomial
and rational functions using symbolic expressions and manipulations.
Topics included are the definition and properties of polynomial and
rational expressions, operations on these types of expressions, completing
the square, proof of the quadratic formula, solving quadratic equations
(including complex number solutions), and the vertex form of quadratic
functions. Unit 8
develops student understanding of inverses of functions with a focus
on logarithmic functions and their use in modeling and analyzing problem
situations and data patterns. Topics studied in this unit include inverses
of functions, logarithmic functions and their relation to exponential
functions, properties of logarithms, equation solving with logarithms,
and inverse trigonometric functions and their applications to solving
trigonometric equations. Course 3 Unit 7,
Recursion and Iteration,
is technically a discrete mathematics unit, but working with sequences
and series helps students strengthen their symbolic skills. Course 4:
Preparation for Calculus extends student algebraic skills and understandings
in equations and functions in algebra units but also in geometry units
such as Unit 2, Vectors and Motion, and Unit 6, Surfaces
and Cross Sections. (See
the CPMP Courses 1-4
descriptions.) | 677.169 | 1 |
Math.NET aims to provide a self contained clean framework for symbolic mathematical (Computer Algebra System) and numerical/scientific computations, including a parser and support for linear algebra, complex differential analysis, system solving and more | 677.169 | 1 |
Mathematics Department
Chair:
Faculty:
Courses:
Honors Algebra 1
1 Credit Year Pre-requisite: 8th grade TR and Placement Test
This rigorous course explores four main topics in Algebra: solving one and two variable equations, using and interpreting linear functions, factoring
and quadratics, and finally, exponentials and polynomials. Specifically, this course investigates the topics of linear and quadratic functions
and their graphs, factoring and solving polynomial equations, solving systems of equations, as well as inductive and deductive reasoning. Real-world
applications that can be modeled by linear and quadratic functions will be covered. In depth word problems will be addressed that require students
to demonstrate an understanding of the concepts covered. The resources used to accomplish the objectives of the course include graphing calculator
and other computer programs that demonstrate the components of algebra. (Graphing Calculator required.)
CP Algebra I
1 Credit Year
This is a college preparatory course for freshmen with the aim of developing well prepared, confident problem solvers. Starting with the basic
algebraic premise of solving for "unknowns," students will also explore rational and irrational numbers, functions, linear equations as
well as patterns and sequences. All Algebra I students will be introduced to and required to manipulate monomial and polynomials expressions,
including quadratic equations and functions. Students will engage in indepth problem solving, aimed at developing critical thinking skills
needed to succeed in the 21st century. Students will explore algebraic functions and concepts through the use of graphing calculators and
other graphing software. (Graphing Calculator required).
Algebra I, Part I
1 Credit Year
This is the first part of a two year fundamental algebra course designed for incoming freshman who benefit from small class sizes offering
individualized instruction. This class works to develop a strong foundation of mathematical skills by reviewing many of the concepts of
8th grade math. The course will cover basic computational skills, fundamental mathematical skills, rational numbers, the solving and graphing
of linear equations, proportional reasoning and applying algebraic properties. The class is designed to challenge students with a modified
course while providing individual and collaborative support to allow students to be successful. Students will be given extensive support
as they work to write equations and solve basic real-world problems.
Algebra I, Part II
1 Credit Year Pre-requisite: Algebra I, Part I
This course completes the objectives of basic algebraic principles for sophomore students. This class will continue to provide individualized
support while challenging students to make connections from previous concepts to the new topics of inequalities, solving systems of equations,
operations of polynomials, factoring, solving and graphing quadratic equations, and radical expressions. After completion the students
will have a comprehensive background of Algebra I.
Honors Algebra II
1 Credit Year Pre-requisite: Algebra I and TR
This course is aimed at Honors sophomores. Algebra II continues and broadens the exploration of Algebra I. Incorporating the latest in graphic
calculator technology and using high level problem solving techniques, students will explore rational, irrational and complex numbers.
Specifically, this course will develop an understanding of functions as it relates to linear relations, quadratic and exponential equations,
logarithms, and matrices. (Graphing calculators required).
CP Algebra II
1 Credit Year Pre-requisite: Algebra I
This course is aimed at college preparatory sophomores or juniors. Algebra II continues and broadens the exploration of Algebra I. Students
will explore rational, irrational and complex numbers. Specifically, this course will develop an understanding of functions as they
relate to linear relations, quadratic and exponential equations, logarithms, and matrices. (Graphing calculators required).
Algebra II
1 Credit Year Pre-requisite: Algebra I
Algebra II continues and broadens the exploration of Algebra I. Incorporating the latest in graphic calculator technology and using
problem solving techniques, students will explore rational, irrational and complex numbers. Specifically, this course will develop
an understanding of functions as it relates to linear relations, quadratic and exponential equations, and logarithms. (Graphing
calculator recommended).
Honors Geometry
1 Credit Year Pre-requisite: Algebra I and TR
Students will study geometry as a mathematical system through the deductive development of relationships in the plane and in space.
Topics covered in this course include congruence and similarity of polygons, properties of parallel and perpendicular lines,
angle measures, transformations, geometric constructions, circles, and volume In Sanctitate et Doctrina In Holiness and Learning
32 and area. Logic is a driving force behind all that is studied in this course as students learn to explore various concepts
through geometric proof. The resources used to accomplish the objectives of the course include protractor, compass, graphing
calculator, and other computer programs that demonstrate the components of geometry. (Graphing Calculator required.)
CP Geometry
1 Credit Year Pre-requisite: Algebra I
Students will study geometry as a mathematical system through the deductive development of relationships in the plane and in
space. Topics covered in this course include congruence and similarity of polygons, properties of parallel and perpendicular
lines, angle measures, transformations, geometric constructions, circles, and volume and area. Logic is a driving force
behind all that is studied in this course as students learn to explore various concepts through geometric proof.
Geometry
1 Credit Year Pre-requisite: Algebra I
Students will use geometry with algebra to find distances, angle measures, areas, and volumes. They will use theorems and
postulates for parallel and perpendicular lines as they apply to polygons and circles. Students will also problem solve
using the Pythagorean Theorem and basic right triangle trigonometry.
Honors Pre-Calculus
1 Credit Year Pre-requisites: Algebra II,
Geometry, and TR This Pre-Calculus course is designed to prepare students for AP® Calculus, AP® Statistics or college
level math courses. The course examines advanced mathematical concepts graphically, numerically, algebraically
and geometrically. Topics covered in detail include early and transcendental functions, analytic geometry and plane
figures, alternate coordinate systems, probability and counting, statistical reasoning, and an introduction to
limits, continuity and beginning Calculus concepts. (Graphing Calculator Required)
CP Pre-calculus
1 Credit Year Pre-requisite: Algebra II and Geometry
This Pre-Calculus course is designed to prepare students for CP Statistics & Probability or college level math
courses. The course examines advanced mathematical concepts graphically, numerically, algebraically and geometrically.
Topics covered include early and transcendental functions, analytic geometry and plane figures, alternate coordinate
systems, probability and counting, statistical reasoning, and an introduction to limits, continuity and beginning
Calculus concepts. (Graphing Calculator Required)
AP® Statistics
1 Credit Year Pre-requisite: Algebra II or Pre-Calculus, and TR
The AP® Statistics course is equivalent to a one-semester, introductory, non-calculusbased college course in statistics.
The course introduces students to the major concepts and tools for collecting, analyzing, and drawing conclusions
from data. There are four themes in the AP® Statistics course: exploring data, sampling and experimentation,
anticipating patterns, and statistical inference. Students use technology, investigations, problem solving,
and writing as they build conceptual understanding.
CP Statistics & Probability
1 Credit Year Pre-requisite: Algebra II and Geometry
This course for Honors and College Preparatory juniors and seniors is aimed at those students interested in
pursuing a career in business. Beginning with an introduction of simple and compound probability, this
course will explore and expand the students' understanding of statistical data and analysis. Students will
analyze strategies and, using probability concepts, will interpret, manipulate and create relevant statistical
data in the exploration of means and standard deviations. Students will use calculators, spreadsheets and
tables to estimate areas under the normal distribution curve and apply these techniques within a business
model. (Graphing calculator required).
AP® Calculus AB
1 Credit Year Pre-requisite: Pre-Calculus or Honors Algebra II and TR
This course investigates topics in differential and integral calculus, including limits and continuity, differential
calculus, and definite and indefinite integration. Emphasis will be placed on transcendental functions
and techniques of integration. Students will also study the concepts of calculus analytically, numerically,
graphically, and conceptually. Technology will be integrated into the course through the use of graphing
calculators and dynamic graphing programs. Connections and applications to various disciplines, especially
physics and economics, are investigated at length. At the conclusion of this course, students will be encouraged
to take the AP® Calculus AB test. (Graphing calculator required).
AP® Calculus BC
1 Credit Year Pre-requisite: AP® Calculus AB and TR
AP® Calculus BC is a course in single-variable calculus that includes all the topics of Calculus AB (techniques
and applications of the derivative, techniques and applications of the definite integral, and the Fundamental
Theorem of Calculus) plus additional topics in differential and integral calculus (including parametric,
polar, and vector In Sanctitate et Doctrina In Holiness and Learning 34 functions) and series. It
is equivalent to at least a year of calculus at most colleges and universities. Algebraic, numerical,
and graphical representations are emphasized throughout the course. At the conclusion of this course,
students will be encouraged to take the AP® Calculus BC test. (Graphing calculator required). | 677.169 | 1 |
Right menu
Featured resource
Elementary and Middle School Mathematics: Teaching Developmentally provides ideas to help teachers develop a deep understanding of the mathematics they teach, highlighting the benefits of problem-based mathematics instruction.
Oxford Mathematics Study Dictionary
Barbara Lynch, R. E. Parr
This is a unique and comprehensive reference book for secondary school mathematics students. The 'Dictionary' contains an extensive list of mathematical words and their meanings, organised into topic sections typically included in mathematics syllabi around the country: arithmetic, number, complex numbers, matrices, circles, symmetry and transformations, functions and relations, coordinate geometry, differential calculus, vectors… Specific words can be found through the index; the actual entries contain appropriate definitions, theorems, formulae, diagrams and worked examples of major applications designed to make the meanings clear. | 677.169 | 1 |
Be sure that you have an application to open
this file type before downloading and/or purchasing.
33 KB|4 pages
Product Description
Order of Operations in Pre-algebra and Algebra
In this lesson, students will learn the order of operations for solving expressions, including brackets and parentheses. Practice problems are given (with answers) as well as two advanced problems with several steps (answers include all steps worked out in order). | 677.169 | 1 |
Keep up with the latest activities at the Network Team Institute!
Main menu
Post navigation
Unpacking Algebra I, Module 4
The Algebra I session was focused on presenting the rationale behind the lessons of Module 4, "Polynomial and Quadratic Expressions, Equations and Functions." Participants discussed many scaffolding ideas, which reconfirmed the power of the Algebra I standards and how they lay the foundational pieces for the mathematics studied in Algebra II and Calculus (exploratory exercise 2 from lesson 8).
The first hour of the session was dedicated to the mathematician in us all, as it provided an adult experience/perspective that dealt with the foundation of quadratics. This foundation was explored in three pieces:
How do sequences and their leading diagonals coupled with basic polynomial building blocks help generate quadratic equations?
Gravity is why we study quadratics. If acceleration is a constant, the formula for defining height is a quadratic.
Can any U-shaped graph be represented by a quadratic? Answer: No.
Participants engaged in an excellent activity that is not in the modules, but should be part of every mathematics classroom. Educators taped the ends of a chain together so that the lowest part of the chain was at the origin. They then identified points that the chain went through. The task was to create a quadratic equation to fit the curve. Looks can be deceiving, and not all of the chain curves ended up being quadratic. This was a challenging, thought-provoking activity for all, and a question not usually tackled until Algebra II.
For most of the morning and part of the afternoon, participants focused on Topic A, which covers the following:
Reverse multiplication (factoring) and the connection to the geometric models (rectangles);
The zero product property;
Graphing a quadratic equation;
The importance symmetry; and
Relating quadratics and their graphs to a real-world context.
The lessons for Topic A are built off the foundation created for multiplying polynomials using the distributive property from grades 6-8. The theme throughout was to get students to stop learning passively and become engaged in the life of a quadratic. Let students determine on their own what it means to factor and how to go about doing it. Lessons 8-10 allow students to explore the symmetry that exists in the graph of a quadratic function and how this symmetry proves helpful in determining the vertex.
How does a quadratic equation written in factored form help us with the graph? Example 2 from lesson 9 was discussed. Providing the function for determining the motion of the ball, which we always do, really isn't good enough. We need to be asking the students where the formula came from. This question pulls out language that students need to identify with such as y-intercept, maximum height, roots (x-intercepts), domain, and other words that feel natural in the context of the questioning that is going on in the classroom. How is factoring helpful here with respect to this problem? What does it enable you to find? There was never any mention of using the formula for the axis of symmetry; do the students really need it?
Overall, the lessons are fully packed mathematically. Participants raised questions about how to develop students' fluency with factoring when moving at a quick pace through discovery and not using rules or formulas. The rapid white board exchange was presented as one method of developing fluency with factoring. The appearance of dual intensity, how the balance of procedure and fluency are both crucial pieces when working with quadratics, was seen throughout this session. | 677.169 | 1 |
Showing 1 to 15 of 15
Section 2.1
Functions
Objectives
Functions All Around Us
Definition of Function
Evaluating a Function
Domain of a Function
Four Ways to Represent a Function
Objective 1:
Functions are all around us
In nearly every physical phenomenon, we observe that3: Lines
Objectives
The Slope of a Line
Point-Slope Form
Slope-Intercept Form
Vertical and Horizontal Lines
General Equation of a Line
Parallel and Perpendicular Lines
We call the slope of a line m, where
m=
rise
run
Example 1
Find the slo
Section 1.1 The
Coordinate Plane
Objectives
The Coordinate Plane
The Distance Formula
The Midpoint Formula
Objective 1: The Coordinate Plane
The coordinate plane or Cartesian plane is the link
between algebra and geometry. In the coordinate plane we
can
Direct Variation
Inverse Variation
Combining Different Types of Variation
Two types of mathematical models occur so often that
they are given special names.
The first is called direct variation and occurs when
one quantity is a constant multiple of the
Test 5 Review Solutions
Chapters 4 and 11.1, 11.2
The problems that each student will see on the test will be very similar to the problems worked
out below on this review. Be sure to attempt each problem first before looking at the solutions.
Problems 1 a
Ninety-nine percent of the
failures come from people who
have the habit of making
excuses.
~ George Washington Carver ~
Famous African-American scholar, inventor
Real Numbers
Section P2
Objectives
Real Numbers
Properties of Real Numbers
Addition and Su
Linear Functions and
Models
Section 2.5
Objectives
Linear Functions
Slope and Rate of Change
Making and Using Linear Models
Linear Functions
Remember that a linear function has the form f(x) = ax + b and
the graph of this function is a line.
In SectiI would recommend this course because it is a stepping stone needed before calculus.
Course highlights:
I learned different ways to find the roots of an equation.
Hours per week:
6-8 hours
Advice for students:
Do not procrastinate and go to study sessions.
Course Term:Spring 2017
Professor:Mr. Offen
Course Required?Yes
Jan 18, 2017
| Would recommend.
Not too easy. Not too difficult.
Course Overview:
I would recommend this course because of the teacher that I had. My professor made sure that each student was learning and was up to pace in his class. If you are a student who needs to go at a slow or medium pace in the subject of math then Mr. Often is the teacher for you.
Course highlights:
Besides learning the obvious MATH 109 curriculum I learned when I would need to apply it in my field of Veterinary medicine.
Hours per week:
6-8 hours
Advice for students:
If you are going to take this course then be prepared to be doing homework for a minimum of 2 hours to really grasp the material. | 677.169 | 1 |
AS Maths (A-Level Revision Notes)
One of a brand new series for A Level students, this book covers all the core and option topics needed for AS Maths, presented in accessible note form and compiled by top examiners. Individual pages are hole-punched and can easily be removed for insertion into students' own files. "The Collins Revision Notes" series has been specially designed to give students the best possible help as they approach their AS exams. The content is presented in a highly visual form, using bullet points, flowcharts, tables, diagrams, highlighting and artwork so that all key points are instantly obvious and easy to revise.
"synopsis" may belong to another edition of this title.
About the Author:
John Berry is a Professor of Mathematics Education, and is Director of the Centre for Teaching Mathematics, University of Plymouth. Roger Fentem is Senior Lecturer in Mathematics Education at the College of St Mark and St John. He is currently examiner for Statistics for a major examining group. Ted Graham is a Senior Lecturer at the Centre for Teaching Matheamtics, University of Plymouth. He is currently a Chief Examiner in Mechanics for a major examining group. | 677.169 | 1 |
Rating and Stats
Document Actions
Share or Embed Document
CONTENTS
FOREWORD iii
1. NUMBER SYSTEMS 1
1.1 Introduction 1
1.2 Irrational Numbers 5
1.3 Real Numbers and their Decimal Expansions 8
1.4 Representing Real Numbers on the Number Line 15
1.5 Operations on Real Numbers 18
1.6 Laws of Exponents for Real Numbers 24
1.7 Summary 27
2. POLYNOMIALS 28
2.1 Introduction 28
2.2 Polynomials in One Variable 28
2.3 Zeroes of a Polynomial 32
2.4 Remainder Theorem 35
2.5 Factorisation of Polynomials 40
2.6 Algebraic Identities 44
2.7 Summary 50
3. COORDINATE GEOMETRY 51
3.1 Introduction 51
3.2 Cartesian System 54
3.3 Plotting a Point in the Plane if its Coordinates are given 61
3.4 Summary 65
4. LINEAR EQUATIONS IN TWO VARIABLES 66
4.1 Introduction 66
4.2 Linear Equations 66
4.3 Solution of a Linear Equation 68
4.4 Graph of a Linear Equation in Two Variables 70
4.5 Equations of Lines Parallel to x-axis and y-axis 75
4.6 Summary 77
5. INTRODUCTION TO EUCLID'S GEOMETRY 78
5.1 Introduction 78
5.2 Euclid's Definitions, Axioms and Postulates 80
5.3 Equivalent Versions of Euclid's Fifth Postulate 86
5.4 Summary 88
6. LINES AND ANGLES 89
6.1 Introduction 89
6.2 Basic Terms and Definitions 90
6.3 Intersecting Lines and Non-intersecting Lines 92
6.4 Pairs of Angles 92
6.5 Parallel Lines and a Transversal 98
6.6 Lines Parallel to the same Line 101
6.7 Angle Sum Property of a Triangle 105
6.8 Summary 108
7. TRIANGLES 108
7.1 Introduction 109
7.2 Congruence of Triangles 109
7.3 Criteria for Congruence of Triangles 112
7.4 Some Properties of a Triangle 120
7.5 Some More Criteria for Congruence of Triangles 125
7.6 Inequalities in a Triangle 129
7.7 Summary 134
8. QUADRILATERALS 135
8.1 Introduction 135
8.2 Angle Sum Property of a Quadrilateral 136
8.3 Types of Quadrilaterals 137
8.4 Properties of a Parallelogram 139
8.5 Another Condition for a Quadrilteral to be a Parallelogram 145
8.6 The Mid-point Theorem 148
8.7 Summary 151
9. AREAS OF PARALLELOGRAMS AND TRIANGLES 152
9.1 Introduction 152
9.2 Figures on the same Base and Between the same Parallels 154
9.3 Parallelogramms on the same Base and
between the same Parallels 156
9.4 Triangles on the same Base and between
the same Parallels 160
9.5 Summary 167
10. CIRCLES 168
10.1 Introduction 168
10.2 Circles and its Related Terms : A Review 169
10.3 Angle Subtended by a Chord at a Point 171
10.4 Perpendicular from the Centre to a Chord 173
10.5 Circle through Three Points 174
10.6 Equal Chords and their Distances from the Centre 176
10.7 Angle Subtended by an Arc of a Circle 179
10.8 Cyclic Quadrilaterals 182
10.9 Summary 187
11. CONSTRUCTIONS 187
11.1 Introduction 188
11.2 Basic Constructions 189
11.3 Some Constructions of Triangles 191
11.4 Summary 196
12. HERON'S FORMULA 197
12.1 Introduction 197
12.2 Area of a Triangle – by Heron's Formula 199
12.3 Application of Heron's Formula in finding
Areas of Quadrilaterals 203
12.4 Summary 207
13. SURFACE AREAS AND VOLUMES 208
13.1 Introduction 208
13.2 Surface Area of a Cuboid and a Cube 208
13.3 Surface Area of a Right Circular Cylinder 214
13.4 Surface Area of a Right Circular Cone 217
13.5 Surface Area of a Sphere 222
13.6 Volume of a Cuboid 226
13.7 Volume of a Cylinder 228
13.8 Volume of a Right Circular Cone 231
13.9 Volume of a Sphere 234
10.10 Summary 237
14. STATISTICS 238
14.1 Introduction 238
14.2 Collection of Data 239
14.3 Presentation of Data 240
14.4 Geographical Representation of Data 247
14.5 Measures of Central Tendency 261
14.6 Summary 270
15. PROBABILITY 271
15.1 Introduction 271
15.2 Probability – an Experimental Approach 272
15.3 Summary 285
APPENDIX – 1 PROOFS IN MATHEMATICS 286
A1.1 Introduction 286
A1.2 Mathematically Acceptable Statements 287
A1.3 Deductive Reasoning 290
A1.4 Theorems, Conjectures and Axioms 293
A1.5 What is a Mathematical Proof? 298
A1.6 Summary 305
APPENDIX – 2 INTRODUCTION TO MATHEMATICAL MODELLING 306
A2.1 Introduction 306
A2.2 Review of Word Problems 307
A2.3 Some Mathematical Models 311
A2.4 The Process of Modelling, its Advantages and Limitations 319
A2.5 Summary 322
ANSWERS/HINTS 325-350
28 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
CHAPTER 2
POLYNOMIALS
2.1 Introduction
You have studied algebraic expressions, their addition, subtraction, multiplication and
division in earlier classes. You also have studied how to factorise some algebraic
expressions. You may recall the algebraic identities :
(x + y)
2
= x
2
+ 2xy + y
2
(x – y)
2
= x
2
– 2xy + y
2
and x
2
– y
2
= (x + y) (x – y)
and their use in factorisation. In this chapter, we shall start our study with a particular
type of algebraic expression, called polynomial, and the terminology related to it. We
shall also study the Remainder Theorem and Factor Theorem and their use in the
factorisation of polynomials. In addition to the above, we shall study some more algebraic
identities and their use in factorisation and in evaluating some given expressions.
2.2 Polynomials in One Variable
Let us begin by recalling that a variable is denoted by a symbol that can take any real
value. We use the letters x, y, z, etc. to denote variables. Notice that 2x, 3x, – x, –
1
2
x
are algebraic expressions. All these expressions are of the form (a constant) × x. Now
suppose we want to write an expression which is (a constant) × (a variable) and we do
not know what the constant is. In such cases, we write the constant as a, b, c, etc. So
the expression will be ax, say.
However, there is a difference between a letter denoting a constant and a letter
denoting a variable. The values of the constants remain the same throughout a particular
situation, that is, the values of the constants do not change in a given problem, but the
value of a variable can keep changing.
POLYNOMIALS 29
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
Now, consider a square of side 3 units (see Fig. 2.1).
What is its perimeter? You know that the perimeter of a square
is the sum of the lengths of its four sides. Here, each side is 3
units. So, its perimeter is 4 × 3, i.e., 12 units. What will be the
perimeter if each side of the square is 10 units? The perimeter
is 4 × 10, i.e., 40 units. In case the length of each side is x
units (see Fig. 2.2), the perimeter is given by 4x units. So, as
the length of the side varies, the perimeter varies.
Can you find the area of the square PQRS? It is
x × x = x
2
square units. x
2
is an algebraic expression. You are
also familiar with other algebraic expressions like
2x, x
2
+ 2x, x
3
– x
2
+ 4x + 7. Note that, all the algebraic
expressions we have considered so far have only whole
numbers as the exponents of the variable. Expressions of this
form are called polynomials in one variable. In the examples
above, the variable is x. For instance, x
3
– x
2
+ 4x + 7 is a
polynomial in x. Similarly, 3y
2
+ 5y is a polynomial in the
variable y and t
2
+ 4 is a polynomial in the variable t.
In the polynomial x
2
+ 2x, the expressions x
2
and 2x are called the terms of the
polynomial. Similarly, the polynomial 3y
2
+ 5y + 7 has three terms, namely, 3y
2
, 5y and
7. Can you write the terms of the polynomial –x
3
+ 4x
2
+ 7x – 2 ? This polynomial has
4 terms, namely, –x
3
, 4x
2
, 7x and –2.
Each term of a polynomial has a coefficient. So, in –x
3
+ 4x
2
+ 7x – 2, the
coefficient of x
3
is –1, the coefficient of x
2
is 4, the coefficient of x is 7 and –2 is the
coefficient of x
0
. (Remember, x
0
= 1) Do you know the coefficient of x in x
2
– x + 7?
It is –1.
2 is also a polynomial. In fact, 2, –5, 7, etc. are examples of constant polynomials.
The constant polynomial 0 is called the zero polynomial. This plays a very important
role in the collection of all polynomials, as you will see in the higher classes.
Now, consider algebraic expressions such as x +
2
3
1
,
3 and . + + x y y
x
Do you
know that you can write x +
1
x
= x + x
–1
? Here, the exponent of the second term, i.e.,
x
–1
is –1, which is not a whole number. So, this algebraic expression is not a polynomial.
Again, 3 x + can be written as
1
2
3 x + . Here the exponent of x is
1
2
, which is
not a whole number. So, is 3 x + a polynomial? No, it is not. What about
3
y + y
2
? It is also not a polynomial (Why?).
Fig. 2.1
Fig. 2.2
x
x x
x
S
R
P Q
3
3 3
3
30 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
If the variable in a polynomial is x, we may denote the polynomial by p(x), or q(x),
or r(x), etc. So, for example, we may write :
p(x) = 2x
2
+ 5x – 3
q(x) = x
3
–1
r(y) = y
3
+ y + 1
s(u) = 2 – u – u
2
+ 6u
5
A polynomial can have any (finite) number of terms. For instance, x
150
+ x
149
+ ...
+ x
2
+ x + 1 is a polynomial with 151 terms.
Consider the polynomials 2x, 2, 5x
3
, –5x
2
, y and u
4
. Do you see that each of these
polynomials has only one term? Polynomials having only one term are called monomials
('mono' means 'one').
Now observe each of the following polynomials:
p(x) = x + 1, q(x) = x
2
– x, r(y) = y
30
+ 1, t(u) = u
43
– u
2
How many terms are there in each of these? Each of these polynomials has only
two terms. Polynomials having only two terms are called binomials ('bi' means 'two').
Similarly, polynomials having only three terms are called trinomials
('tri' means 'three'). Some examples of trinomials are
p(x) = x + x
2
+ π, q(x) =
2
+ x – x
2
,
r(u) = u + u
2
– 2, t(y) = y
4
+ y + 5.
Now, look at the polynomial p(x) = 3x
7
– 4x
6
+ x + 9. What is the term with the
highest power of x ? It is 3x
7
. The exponent of x in this term is 7. Similarly, in the
polynomial q(y) = 5y
6
– 4y
2
– 6, the term with the highest power of y is 5y
6
and the
exponent of y in this term is 6. We call the highest power of the variable in a polynomial
as the degree of the polynomial. So, the degree of the polynomial 3x
7
– 4x
6
+ x + 9
is 7 and the degree of the polynomial 5y
6
– 4y
2
– 6 is 6. The degree of a non-zero
constant polynomial is zero.
Example 1 : Find the degree of each of the polynomials given below:
(i) x
5
– x
4
+ 3 (ii) 2 – y
2
– y
3
+ 2y
8
(iii) 2
Solution : (i) The highest power of the variable is 5. So, the degree of the polynomial
is 5.
(ii) The highest power of the variable is 8. So, the degree of the polynomial is 8.
(iii) The only term here is 2 which can be written as 2x
0
. So the exponent of x is 0.
Therefore, the degree of the polynomial is 0.
POLYNOMIALS 31
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
Now observe the polynomials p(x) = 4x + 5, q(y) = 2y, r(t) = t + 2 and
s(u) = 3 – u. Do you see anything common among all of them? The degree of each of
these polynomials is one. A polynomial of degree one is called a linear polynomial.
Some more linear polynomials in one variable are 2x – 1, 2 y + 1, 2 – u. Now, try and
find a linear polynomial in x with 3 terms? You would not be able to find it because a
linear polynomial in x can have at most two terms. So, any linear polynomial in x will
be of the form ax + b, where a and b are constants and a ≠ 0 (why?). Similarly,
ay + b is a linear polynomial in y.
Now consider the polynomials :
2x
2
+ 5, 5x
2
+ 3x + π, x
2
and x
2
+
2
5
x
Do you agree that they are all of degree two? A polynomial of degree two is called
a quadratic polynomial. Some examples of a quadratic polynomial are 5 – y
2
,
4y + 5y
2
and 6 – y – y
2
. Can you write a quadratic polynomial in one variable with four
different terms? You will find that a quadratic polynomial in one variable will have at
most 3 terms. If you list a few more quadratic polynomials, you will find that any
quadratic polynomial in x is of the form ax
2
+ bx + c, where a ≠ 0 and a, b, c are
constants. Similarly, quadratic polynomial in y will be of the form ay
2
+ by + c, provided
a ≠ 0 and a, b, c are constants.
We call a polynomial of degree three a cubic polynomial. Some examples of a
cubic polynomial in x are 4x
3
, 2x
3
+ 1, 5x
3
+ x
2
, 6x
3
– x, 6 – x
3
, 2x
3
+ 4x
2
+ 6x + 7. How
many terms do you think a cubic polynomial in one variable can have? It can have at
most 4 terms. These may be written in the form ax
3
+ bx
2
+ cx + d, where a ≠ 0 and
a, b, c and d are constants.
Now, that you have seen what a polynomial of degree 1, degree 2, or degree 3
looks like, can you write down a polynomial in one variable of degree n for any natural
number n? A polynomial in one variable x of degree n is an expression of the form
a
n
x
n
+ a
n–1
x
n–1
+ . . . + a
1
x + a
0
where a
0
, a
1
, a
2
, . . ., a
n
are constants and a
n
≠ 0.
In particular, if a
0
= a
1
= a
2
= a
3
= . . . = a
n
= 0 (all the constants are zero), we get
the zero polynomial, which is denoted by 0. What is the degree of the zero polynomial?
The degree of the zero polynomial is not defined.
So far we have dealt with polynomials in one variable only. We can also have
polynomials in more than one variable. For example, x
2
+ y
2
+ xyz (where variables
are x, y and z) is a polynomial in three variables. Similarly p
2
+ q
10
+ r (where the
variables are p, q and r), u
3
+ v
2
(where the variables are u and v) are polynomials in
three and two variables, respectively. You will be studying such polynomials in detail
later.
32 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
EXERCISE 2.1
1. Which of the following expressions are polynomials in one variable and which are
not? State reasons for your answer.
(i) 4x
2
– 3x + 7 (ii) y
2
+
2
(iii) 3 2 t t + (iv) y +
2
y
(v) x
10
+ y
3
+ t
50
2. Write the coefficients of x
2
in each of the following:
(i) 2 + x
2
+ x (ii) 2 – x
2
+ x
3
(iii)
2
2
x x
π
+ (iv) 2 1 x −
3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
4. Write the degree of each of the following polynomials:
(i) 5x
3
+ 4x
2
+ 7x (ii) 4 – y
2
(iii) 5t –
7
(iv) 3
5. Classify the following as linear, quadratic and cubic polynomials:
(i) x
2
+ x (ii) x – x
3
(iii) y + y
2
+ 4 (iv) 1 + x
(v) 3t (vi) r
2
(vii) 7x
3
2.3 Zeroes of a Polynomial
Consider the polynomial p(x) = 5x
3
– 2x
2
+ 3x – 2.
If we replace x by 1 everywhere in p(x), we get
p(1) = 5 × (1)
3
– 2 × (1)
2
+ 3 × (1) – 2
= 5 – 2 + 3 –2
= 4
So, we say that the value of p(x) at x = 1 is 4.
Similarly, p(0) = 5(0)
3
– 2(0)
2
+ 3(0) –2
= –2
Can you find p(–1)?
Example 2 : Find the value of each of the following polynomials at the indicated value
of variables:
(i) p(x) = 5x
2
– 3x + 7 at x = 1.
(ii) q(y) = 3y
3
– 4y + 11 at y = 2.
(iii) p(t) = 4t
4
+ 5t
3
– t
2
+ 6 at t = a.
POLYNOMIALS 33
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
Solution : (i) p(x) = 5x
2
– 3x + 7
The value of the polynomial p(x) at x = 1 is given by
p(1) = 5(1)
2
– 3(1) + 7
= 5 – 3 + 7 = 9
(ii) q(y) = 3y
3
– 4y +
11
The value of the polynomial q(y) at y = 2 is given by
q(2) = 3(2)
3
– 4(2) + 11 = 24 – 8 + 11 = 16 + 11
(iii) p(t) = 4t
4
+ 5t
3
– t
2
+ 6
The value of the polynomial p(t) at t = a is given by
p(a) = 4a
4
+ 5a
3
– a
2
+ 6
Now, consider the polynomial p(x) = x – 1.
What is p(1)? Note that : p(1) = 1 – 1 = 0.
As p(1) = 0, we say that 1 is a zero of the polynomial p(x).
Similarly, you can check that 2 is a zero of q(x), where q(x) = x – 2.
In general, we say that a zero of a polynomial p(x) is a number c such that p(c) = 0.
You must have observed that the zero of the polynomial x – 1 is obtained by
equating it to 0, i.e., x – 1 = 0, which gives x = 1. We say p(x) = 0 is a polynomial
equation and 1 is the root of the polynomial equation p(x) = 0. So we say 1 is the zero
of the polynomial x – 1, or a root of the polynomial equation x – 1 = 0.
Now, consider the constant polynomial 5. Can you tell what its zero is? It has no
zero because replacing x by any number in 5x
0
still gives us 5. In fact, a non-zero
constant polynomial has no zero. What about the zeroes of the zero polynomial? By
convention, every real number is a zero of the zero polynomial.
Example 3 : Check whether –2 and 2 are zeroes of the polynomial x + 2.
Solution : Let p(x) = x + 2.
Then p(2) = 2 + 2 = 4, p(–2) = –2 + 2 = 0
Therefore, –2 is a zero of the polynomial x + 2, but 2 is not.
Example 4 : Find a zero of the polynomial p(x) = 2x + 1.
Solution : Finding a zero of p(x), is the same as solving the equation
p(x) = 0
34 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
Now, 2x + 1 = 0 gives us x =
1
–
2
So,
1
–
2
is a zero of the polynomial 2x + 1.
Now, if p(x) = ax + b, a ≠ 0, is a linear polynomial, how can we find a zero of
p(x)? Example 4 may have given you some idea. Finding a zero of the polynomial p(x),
amounts to solving the polynomial equation p(x) = 0.
Now, p(x) = 0 means ax + b = 0, a ≠ 0
So, ax = –b
i.e., x = –
b
a
.
So, x =
b
a
−
is the only zero of p(x), i.e., a linear polynomial has one and only one zero.
Now we can say that 1 is the zero of x – 1, and –2 is the zero of x + 2.
Example 5 : Verify whether 2 and 0 are zeroes of the polynomial x
2
– 2x.
Solution : Let p(x) = x
2
– 2x
Then p(2) = 2
2
– 4 = 4 – 4 = 0
and p(0) = 0 – 0 = 0
Hence, 2 and 0 are both zeroes of the polynomial x
2
– 2x.
Let us now list our observations:
(i) A zero of a polynomial need not be 0.
(ii) 0 may be a zero of a polynomial.
(iii) Every linear polynomial has one and only one zero.
(iv) A polynomial can have more than one zero.
EXERCISE 2.2
1. Find the value of the polynomial 5x – 4x
2
+ 3 at
(i) x = 0 (ii) x = –1 (iii) x = 2
2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y
2
– y + 1 (ii) p(t) = 2 + t + 2t
2
– t
3
(iii) p(x) = x
3
(iv) p(x) = (x – 1) (x + 1)
POLYNOMIALS 35
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x =
1
–
3
(ii) p(x) = 5x – π, x =
4
5
(iii) p(x) = x
2
– 1, x = 1, –1 (iv) p(x) = (x + 1) (x – 2), x = – 1, 2
(v) p(x) = x
2
, x = 0 (vi) p(x) = lx + m, x = –
m
l
(vii) p(x) = 3x
2
– 1, x =
1 2
,
3 3
− (viii) p(x) = 2x + 1, x =
1
2
4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
2.4 Remainder Theorem
Let us consider two numbers 15 and 6. You know that when we divide 15 by 6, we get
the quotient 2 and remainder 3. Do you remember how this fact is expressed? We
write 15 as
15 = (2 × 6) + 3
We observe that the remainder 3 is less than the divisor 6. Similarly, if we divide
12 by 6, we get
12 = (2 × 6) + 0
What is the remainder here? Here the remainder is 0, and we say that 6 is a
factor of 12 or 12 is a multiple of 6.
Now, the question is: can we divide one polynomial by another? To start with, let
us try and do this when the divisor is a monomial. So, let us divide the polynomial
2x
3
+ x
2
+ x by the monomial x.
We have (2x
3
+ x
2
+ x) ÷ x =
3 2
2x x x
x x x
+ +
= 2x
2
+ x + 1
In fact, you may have noticed that x is common to each term of 2x
3
+ x
2
+ x. So
we can write 2x
3
+ x
2
+ x as x(2x
2
+ x + 1).
We say that x and 2x
2
+ x + 1 are factors of 2x
3
+ x
2
+ x, and 2x
3
+ x
2
+ x is a
multiple of x as well as a multiple of 2x
2
+ x + 1.
36 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
Consider another pair of polynomials 3x
2
+ x + 1 and x.
Here, (3x
2
+ x + 1) ÷ x = (3x
2
÷ x) + (x ÷ x) + (1 ÷ x).
We see that we cannot divide 1 by x to get a polynomial term. So in this case we
stop here, and note that 1 is the remainder. Therefore, we have
3x
2
+ x + 1 = {(3x + 1) × x} + 1
In this case, 3x + 1 is the quotient and 1 is the remainder. Do you think that x is a
factor of 3x
2
+ x + 1? Since the remainder is not zero, it is not a factor.
Now let us consider an example to see how we can divide a polynomial by any
non-zero polynomial.
Example 6 : Divide p(x) by g(x), where p(x) = x + 3x
2
– 1 and g(x) = 1 + x.
Solution : We carry out the process of division by means of the following steps:
Step 1 : We write the dividend x + 3x
2
– 1 and the divisor 1 + x in the standard form,
i.e., after arranging the terms in the descending order of their degrees. So, the
dividend is 3x
2
+ x –1 and divisor is x + 1.
Step 2 : We divide the first term of the dividend
by the first term of the divisor, i.e., we divide
3x
2
by x, and get 3x. This gives us the first term
of the quotient.
Step 3 : We multiply the divisor by the first term
of the quotient, and subtract this product from
the dividend, i.e., we multiply x + 1 by 3x and
subtract the product 3x
2
+ 3x from the dividend
3x
2
+ x – 1. This gives us the remainder as
–2x – 1.
Step 4 : We treat the remainder –2x – 1
as the new dividend. The divisor remains
the same. We repeat Step 2 to get the
next term of the quotient, i.e., we divide
the first term – 2x of the (new) dividend
by the first term x of the divisor and obtain
– 2. Thus, – 2 is the second term in the
quotient.
2
3x
x
= 3x = first term of quotient
–2x
x
= – 2
= second term of quotient
New Quotient
= 3x – 2
3x
2
+ x –1
3x
2
+ 3x
– –
– 2x – 1
3x
x + 1
POLYNOMIALS 37
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-2\Chap-2 (02-01-2006).PM65
Step 5 : We multiply the divisor by the second
term of the quotient and subtract the product
from the dividend. That is, we multiply x + 1
by – 2 and subtract the product – 2x – 2
from the dividend – 2x – 1. This gives us 1
as the remainder.
This process continues till the degree of the new dividend is less than the degree of the
divisor. At this stage, this new dividend becomes the remainder and the sum of the
quotients gives us the whole quotient.
Step 6 : Thus, the quotient in full is 3x – 2 and the remainder is 1.
Let us look at what we have done in the process above as a whole:
3x – 2
x + 1
3x
2
+ x – 1
3x
2
+ 3x
– –
– 2x – 1
– 2x – 2
180° = 90°.
If you take any other point C on the semicircle, again you get that
∠ PCQ = 90°
Therefore, you find another property of the circle as:
Angle in a semicircle is a right angle.
The converse of Theorem 10.9 is also true. It can be stated as:
Theorem 10.10 : If a line segment joining two points subtends equal angles at
two other points lying on the same side of the line containing the line segment,
the four points lie on a circle (i.e. they are concyclic).
Fig. 10.29
182 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-10\Chap-10 (03-01-2006).PM65
You can see the truth of this result as follows:
In Fig. 10.30, AB is a line segment, which subtends equal angles at two points C and
D. That is
∠ ACB = ∠ ADB
To show that the points A, B, C and D lie on a circle
let us draw a circle through the points A, C and B.
Suppose it does not pass through the point D. Then it
will intersect AD (or extended AD) at a point, say E
(or E′).
If points A, C, E and B lie on a circle,
∠ ACB = ∠ AEB (Why?)
But it is given that ∠ ACB = ∠ ADB.
Therefore, ∠ AEB = ∠ ADB.
This is not possible unless E coincides with D. (Why?)
Similarly, E′ should also coincide with D.
10.8 Cyclic Quadrilaterals
A quadrilateral ABCD is called cyclic if all the four vertices
of it lie on a circle (see Fig 10.31). You will find a peculiar
property in such quadrilaterals. Draw several cyclic
quadrilaterals of different sides and name each of these
as ABCD. (This can be done by drawing several circles
of different radii and taking four points on each of them.)
Measure the opposite angles and write your observations
in the following table.
S.No. of Quadrilateral ∠ A ∠ B ∠ C ∠ D ∠ A +∠ C ∠ B +∠ D
1.
2.
3.
4.
5.
6.
What do you infer from the table?
Fig. 10.30
Fig. 10.31
CIRCLES 183
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-10\Chap-10 (03-01-2006).PM65
You find that ∠A + ∠C = 180° and ∠B + ∠D = 180°, neglecting the error in
measurements. This verifies the following:
Theorem 10.11 : The sum of either pair of opposite angles of a cyclic
quadrilateral is 180º.
In fact, the converse of this theorem, which is stated below is also true.
Theorem 10.12 : If the sum of a pair of opposite angles of a quadrilateral is
180º, the quadrilateral is cyclic.
You can see the truth of this theorem by following a method similar to the method
adopted for Theorem 10.10.
Example 3 : In Fig. 10.32, AB is a diameter of the circle, CD is a chord equal to the
radius of the circle. AC and BD when extended intersect at a point E. Prove that
∠ AEB = 60°.
Solution : Join OC, OD and BC.
Triangle ODC is equilateral (Why?)
Therefore, ∠ COD = 60°
Now, ∠ CBD =
1
2
∠ COD (Theorem 10.8)
This gives ∠ CBD = 30°
Again, ∠ ACB = 90° (Why ?)
So, ∠ BCE = 180° – ∠ ACB = 90°
Which gives ∠ CEB = 90° – 30° = 60°, i.e. ∠ AEB = 60°
Example 4 : In Fig 10.33, ABCD is a cyclic
quadrilateral in which AC and BD are its diagonals.
If ∠ DBC = 55° and ∠ BAC = 45°, find ∠ BCD.
Solution : ∠ CAD = ∠ DBC = 55°
(Angles in the same segment)
Therefore, ∠ DAB = ∠ CAD + ∠ BAC
= 55° + 45° = 100°
But ∠ DAB + ∠ BCD = 180°
(Opposite angles of a cyclic quadrilateral)
So, ∠ BCD = 180° – 100° = 80°
Fig. 10.32
Fig. 10.33
184 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-10\Chap-10 (03-01-2006).PM65
Example 5 : Two circles intersect at two points A
and B. AD and AC are diameters to the two circles
(see Fig.10.34). Prove that B lies on the line segment
DC.
Solution : Join AB.
∠ ABD = 90° (Angle in a semicircle)
∠ ABC = 90° (Angle in a semicircle)
So, ∠ ABD + ∠ ABC = 90° + 90° = 180°
Therefore, DBC is a line. That is B lies on the line segment DC.
Example 6 : Prove that the quadrilateral formed (if possible) by the internal angle
bisectors of any quadrilateral is cyclic.
Solution : In Fig. 10.35, ABCD is a quadrilateral in
which the angle bisectors AH, BF, CF and DH of
internal angles A, B, C and D respectively form a
quadrilateral EFGH.
Now, ∠ FEH = ∠ AEB = 180° – ∠ EAB – ∠ EBA (Why ?)
= 180° –
1
2
(∠ A + ∠ B)
and ∠ FGH = ∠ CGD = 180° – ∠ GCD – ∠ GDC (Why ?)
= 180° –
1
2
(∠ C + ∠ D)
Therefore, ∠ FEH + ∠ FGH = 180° –
1
2
(∠ A + ∠ B) + 180° –
1
2
(∠ C + ∠ D)
= 360° –
1
2
(∠ A+ ∠ B +∠ C +∠ D) = 360° –
1
2
× 360°
= 360° – 180° = 180°
Therefore, by Theorem 10.12, the quadrilateral EFGH is cyclic.
EXERCISE 10.5
1. In Fig. 10.36, A,B and C are three points on a
circle with centre O such that ∠ BOC = 30° and
∠ AOB = 60°. If D is a point on the circle other
than the arc ABC, find ∠ADC.
Fig. 10.35
Fig. 10.36
Fig. 10.34
CIRCLES 185
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-10\Chap-10 (03-01-2006).PM65
2. A chord of a circle is equal to the radius of the
circle. Find the angle subtended by the chord at
a point on the minor arc and also at a point on the
major arc.
3. In Fig. 10.37, ∠ PQR = 100°, where P, Q and R are
points on a circle with centre O. Find ∠ OPR.
4. In Fig. 10.38, ∠ ABC = 69°, ∠ ACB = 31°, find
∠ BDC.
5. In Fig. 10.39, A, B, C and D are four points on a
circle. AC and BD intersect at a point E such
that ∠ BEC = 130°
and ∠ ECD = 20°. Find
∠ BAC.
6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°,
∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.
7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of
the quadrilateral, prove that it is a rectangle.
8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Fig. 10.37
Fig. 10.38
Fig. 10.39
186 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-10\Chap-10 (03-01-2006).PM65
9. Two circles intersect at two points B and C.
Through B, two line segments ABD and PBQ
are drawn to intersect the circles at A, D and P,
Q respectively (see Fig. 10.40). Prove that
∠ ACP = ∠ QCD.
10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of
intersection of these circles lie on the third side.
11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that
∠ CAD = ∠ CBD.
12. Prove that a cyclic parallelogram is a rectangle.
EXERCISE 10.6 (Optional)*
1. Prove that the line of centres of two intersecting circles subtends equal angles at the
two points of intersection.
2. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel
to each other and are on opposite sides of its centre. If the distance between AB and
CD is 6 cm, find the radius of the circle.
3. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is
at distance 4 cm from the centre, what is the distance of the other chord from the
centre?
4. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle
intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the
difference of the angles subtended by the chords AC and DE at the centre.
5. Prove that the circle drawn with any side of a rhombus as diameter, passes through
the point of intersection of its diagonals.
6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if
necessary) at E. Prove that AE = AD.
7. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are
diameters, (ii) ABCD is a rectangle.
8. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and
F respectively. Prove that the angles of the triangle DEF are 90° –
1
2
A, 90° –
1
2
B and
90° –
1
2
C.
Fig. 10.40
*
These exercises are not from examination point of view.
CIRCLES 187
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-10\Chap-10 (03-01-2006).PM65
9. Two congruent circles intersect each other at points A and B. Through A any line
segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
10. In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC
intersect, prove that they intersect on the circumcircle of the triangle ABC.
10.9 Summary
In this chapter, you have studied the following points:
1. A circle is the collection of all points in a plane, which are equidistant from a fixed point in
the plane.
2. Equal chords of a circle (or of congruent circles) subtend equal angles at the centre.
3. If the angles subtended by two chords of a circle (or of congruent circles) at the centre
(corresponding centres) are equal, the chords are equal.
4. The perpendicular from the centre of a circle to a chord bisects the chord.
5. The line drawn through the centre of a circle to bisect a chord is perpendicular to the
chord.
6. There is one and only one circle passing through three non-collinear points.
7. Equal chords of a circle (or of congruent circles) are equidistant from the centre (or
corresponding centres).
8. Chords equidistant from the centre (or corresponding centres) of a circle (or of congruent
circles) are equal.
9. If two arcs of a circle are congruent, then their corresponding chords are equal and
conversely if two chords of a circle are equal, then their corresponding arcs (minor, major)
are congruent.
10. Congruent arcs of a circle subtend equal angles at the centre.
11. The angle subtended by an arc at the centre is double the angle subtended by it at any
point on the remaining part of the circle.
12. Angles in the same segment of a circle are equal.
13. Angle in a semicircle is a right angle.
14. If a line segment joining two points subtends equal angles at two other points lying on
the same side of the line containing the line segment, the four points lie on a circle.
15. The sum of either pair of opposite angles of a cyclic quadrilateral is 180
0
.
16. If sum of a pair of opposite angles of a quadrilateral is 180
0
, the quadrilateral is cyclic.
188 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-11\Chap-11 (03-01-2006).PM65
CHAPTER 11
CONSTRUCTIONS
11.1 Introduction
In earlier chapters, the diagrams, which were necessary to prove a theorem or solving
exercises were not necessarily precise. They were drawn only to give you a feeling for
the situation and as an aid for proper reasoning. However, sometimes one needs an
accurate figure, for example - to draw a map of a building to be constructed, to design
tools, and various parts of a machine, to draw road maps etc. To draw such figures
some basic geometrical instruments are needed. You must be having a geometry box
which contains the following:
(i) A graduated scale, on one side of which centimetres and millimetres are
marked off and on the other side inches and their parts are marked off.
(ii) A pair of set - squares, one with angles 90°, 60° and 30° and other with angles
90°, 45° and 45°.
(iii) A pair of dividers (or a divider) with adjustments.
(iv) A pair of compasses (or a compass) with provision of fitting a pencil at one
end.
(v) A protractor.
Normally, all these instruments are needed in drawing a geometrical figure, such
as a triangle, a circle, a quadrilateral, a polygon, etc. with given measurements. But a
geometrical construction is the process of drawing a geometrical figure using only two
instruments – an ungraduated ruler, also called a straight edge and a compass. In
construction where measurements are also required, you may use a graduated scale
and protractor also. In this chapter, some basic constructions will be considered. These
will then be used to construct certain kinds of triangles.
CONSTRUCTIONS 189
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-11\Chap-11 (03-01-2006).PM65
11.2 Basic Constructions
In Class VI, you have learnt how to construct a circle, the perpendicular bisector of a
line segment, angles of 30°, 45°, 60°, 90° and 120°, and the bisector of a given angle,
without giving any justification for these constructions. In this section, you will construct
some of these, with reasoning behind, why these constructions are valid.
Construction 11.1 : To construct the bisector of a given angle.
Given an angle ABC, we want to construct its bisector.
Steps of Construction :
1. Taking B as centre and any radius, draw an arc to intersect the rays BA and BC,
say at E and D respectively [see Fig.11.1(i)].
2. Next, taking D and E as centres and with the radius more than
1
2
DE, draw arcs to
intersect each other, say at F.
3. Draw the ray BF [see Fig.11.1(ii)]. This ray BF is the required bisector of the angle
ABC.
Fig. 11.1
Let us see how this method gives us the required angle bisector.
Join DF and EF.
In triangles BEF and BDF,
BE = BD (Radii of the same arc)
EF = DF (Arcs of equal radii)
BF = BF (Common)
Therefore, ∆BEF ≅ ∆BDF (SSS rule)
This gives ∠EBF = ∠ DBF (CPCT)
190 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-11\Chap-11 (03-01-2006).PM65
Construction 11.2 : To construct the perpendicular bisector of a given line
segment.
Given a line segment AB, we want to construct its perpendicular bisector.
Steps of Construction :
1. Taking A and B as centres and radius more than
1
2
AB, draw arcs on both sides of the line segment
AB (to intersect each other).
2. Let these arcs intersect each other at P and Q.
Join PQ (see Fig.11.2).
3. Let PQ intersect AB at the point M. Then line
PMQ is the required perpendicular bisector of AB.
Let us see how this method gives us the
perpendicular bisector of AB.
Join A and B to both P and Q to form AP, AQ, BP
and BQ.
In triangles PAQ and PBQ,
AP = BP (Arcs of equal radii)
AQ = BQ (Arcs of equal radii)
PQ = PQ (Common)
Therefore, ∆ PAQ ≅ ∆ PBQ (SSS rule)
So, ∠ APM = ∠ BPM (CPCT)
Now in triangles PMA and PMB,
AP = BP (As before)
PM = PM (Common)
∠ APM = ∠ BPM (Proved above)
Therefore, ∆ PMA ≅ ∆ PMB (SAS rule)
So, AM = BM and ∠ PMA = ∠ PMB (CPCT)
As ∠ PMA + ∠ PMB = 180° (Linear pair axiom),
we get
∠ PMA = ∠ PMB = 90°.
Therefore, PM, that is, PMQ is the perpendicular bisector of AB.
Fig. 11.2
CONSTRUCTIONS 191
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-11\Chap-11 (03-01-2006).PM65
Fig. 11.3
Construction 11.3 : To construct an angle of 60
0
at the initial point of a given
ray.
Let us take a ray AB with initial point A [see Fig. 11.3(i)]. We want to construct a ray
AC such that ∠ CAB = 60°. One way of doing so is given below.
Steps of Construction :
1. Taking A as centre and some radius, draw an arc
of a circle, which intersects AB, say at a point D.
2. Taking D as centre and with the same radius as
before, draw an arc intersecting the previously
drawn arc, say at a point E.
3. Draw the ray AC passing through E [see Fig 11.3 (ii)].
Then ∠ CAB is the required angle of 60°. Now,
let us see how this method gives us the required
angle of 60°.
Join DE.
Then, AE = AD = DE (By construction)
Therefore, ∆ EAD is an equilateral triangle and the ∠ EAD, which is the same as
∠ CAB is equal to 60°.
EXERCISE 11.1
1. Construct an angle of 90
0
at the initial point of a given ray and justify the construction.
2. Construct an angle of 45
0
at the initial point of a given ray and justify the construction.
3. Construct the angles of the following measurements:
(i) 30° (ii) 22
1
2
°
(iii) 15°
4. Construct the following angles and verify by measuring them by a protractor:
(i) 75° (ii) 105° (iii) 135°
5. Construct an equilateral triangle, given its side and justify the construction.
11.3 Some Constructions of Triangles
So far, some basic constructions have been considered. Next, some constructions of
triangles will be done by using the constructions given in earlier classes and given
above. Recall from the Chapter 7 that SAS, SSS, ASA and RHS rules give the
congruency of two triangles. Therefore, a triangle is unique if : (i) two sides and the
192 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-11\Chap-11 (03-01-2006).PM65
included angle is given, (ii) three sides are given, (iii) two angles and the included side
is given and, (iv) in a right triangle, hypotenuse and one side is given. You have already
learnt how to construct such triangles in Class VII. Now, let us consider some more
constructions of triangles. You may have noted that at least three parts of a triangle
have to be given for constructing it but not all combinations of three parts are sufficient
for the purpose. For example, if two sides and an angle (not the included angle) are
given, then it is not always possible to construct such a triangle uniquely.
Construction 11.4 : To construct a triangle, given its base, a base angle and sum
of other two sides.
Given the base BC, a base angle, say ∠B and the sum AB + AC of the other two sides
of a triangle ABC, you are required to construct it.
Steps of Construction :
1. Draw the base BC and at the point B make an
angle, say XBC equal to the given angle.
2. Cut a line segment BD equal to AB + AC from
the ray BX.
3. Join DC and make an angle DCY equal to ∠BDC.
4. Let CY intersect BX at A (see Fig. 11.4).
Then, ABC is the required triangle.
Let us see how you get the required triangle.
Base BC and ∠B are drawn as given. Next in triangle
ACD,
∠ACD = ∠ ADC (By construction)
Therefore, AC = AD and then
AB = BD – AD = BD – AC
AB + AC = BD
Alternative method :
Follow the first two steps as above. Then draw
perpendicular bisector PQ of CD to intersect BD at
a point A (see Fig 11.5). Join AC. Then ABC is the
required triangle. Note that A lies on the perpendicular
bisector of CD, therefore AD = AC.
Remark : The construction of the triangle is not
possible if the sum AB + AC ≤ BC.
Fig. 11.4
Fig. 11.5
CONSTRUCTIONS 193
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-11\Chap-11 (03-01-2006).PM65
Construction 11.5 : To construct a triangle given its base, a base angle and the
difference of the other two sides.
Given the base BC, a base angle, say ∠B and the difference of other two sides
AB – AC or AC – AB, you have to construct the triangle ABC. Clearly there are
following two cases:
Case (i) : Let AB > AC that is AB – AC is given.
Steps of Construction :
1. Draw the base BC and at point B make an angle
say XBC equal to the given angle.
2. Cut the line segment BD equal to AB – AC from
ray BX.
3. Join DC and draw the perpendicular bisector, say
PQ of DC.
4. Let it intersect BX at a point A. Join AC
(see Fig. 11.6).
Then ABC is the required triangle.
Let us now see how you have obtained the required triangle ABC.
Base BC and ∠B are drawn as given. The point A lies on the perpendicular bisector of
DC. Therefore,
AD = AC
So, BD = AB – AD = AB – AC.
Case (ii) : Let AB < AC that is AC – AB is given.
Steps of Construction :
1. Same as in case (i).
2. Cut line segment BD equal to AC – AB from the
line BX extended on opposite side of line segment
BC.
3. Join DC and draw the perpendicular bisector, say
PQ of DC.
4. Let PQ intersect BX at A. Join AC (see Fig. 11.7).
Then, ABC is the required triangle.
You can justify the construction as in case (i).
Fig. 11.6
Fig. 11.7
194 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-11\Chap-11 (03-01-2006).PM65
Construction 11.6 : To construct a triangle, given its perimeter and its two base
angles.
Given the base angles, say ∠ B and ∠ C and BC + CA + AB, you have to construct
the triangle ABC.
Steps of Construction :
1. Draw a line segment, say XY equal to BC + CA + AB.
2. Make angles LXY equal to ∠B and MYX equal to ∠C.
3. Bisect ∠ LXY and ∠ MYX. Let these bisectors intersect at a point A
[see Fig. 11.8(i)].
Fig. 11.8 (i)
4. Draw perpendicular bisectors PQ of AX and RS of AY.
5. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC
[see Fig 11.8(ii)].
Fig. 11.8 (ii)
Then ABC is the required triangle. For the justification of the construction, you
observe that, B lies on the perpendicular bisector PQ of AX.
Therefore, XB = AB and similarly, CY = AC.
This gives BC + CA + AB = BC + XB + CY = XY.
Again ∠BAX = ∠AXB (As in ∆ AXB, AB = XB) and
∠ABC = ∠BAX + ∠AXB = 2 ∠AXB = ∠LXY
Similarly, ∠ACB = ∠MYX as required.
CONSTRUCTIONS 195
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-11\Chap-11 (03-01-2006).PM65
Example 1 : Construct a triangle ABC, in which ∠B = 60°, ∠ C = 45° and AB + BC
+ CA = 11 cm.
Steps of Construction :
1. Draw a line segment PQ = 11 cm.( = AB + BC + CA).
2. At P construct an angle of 60° and at Q, an angle of 45°.
Fig. 11.9
3. Bisect these angles. Let the bisectors of these angles intersect at a point A.
4. Draw perpendicular bisectors DE of AP to intersect PQ at B and FG of AQ to
intersect PQ at C.
5. Join AB and AC (see Fig. 11.9).
Then, ABC is the required triangle.
EXERCISE 11.2
1. Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.
2. Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.
3. Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.
4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
5. Construct a right triangle whose base is 12cm and sum of its hypotenuse and other
side is 18 cm.
196 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-11\Chap-11 (03-01-2006).PM65
11.4 Summary
In this chapter, you have done the following constructions using a ruler and a compass:
1. To bisect a given angle.
2. To draw the perpendicular bisector of a given line segment.
3. To construct an angle of 60° etc.
4. To construct a triangle given its base, a base angle and the sum of the other two sides.
5. To construct a triangle given its base, a base angle and the difference of the other two
sides.
6. To construct a triangle given its perimeter and its two base angles.
CHAPTER 12
HERON'S FORMULA
12.1 Introduction
You have studied in earlier classes about figures of different shapes such as squares,
rectangles, triangles and quadrilaterals. You have also calculated perimeters and the
areas of some of these figures like rectangle, square etc. For instance, you can find
the area and the perimeter of the floor of your classroom.
Let us take a walk around the floor along its sides once; the distance we walk is its
perimeter. The size of the floor of the room is its area.
So, if your classroom is rectangular with length 10 m and width 8 m, its perimeter
would be 2(10 m + 8 m) = 36 m and its area would be 10 m × 8 m,
i.e., 80 m
2
.
Unit of measurement for length or breadth is taken as metre (m) or centimetre
(cm) etc.
Unit of measurement for area of any plane figure is taken as square metre (m
2
) or
square centimetre (cm
2
) etc.
Suppose that you are sitting in a triangular garden. How would you find its area?
From Chapter 9 and from your earlier classes, you know that:
Area of a triangle =
1
2
× base × height (I)
We see that when the triangle is right angled,
we can directly apply the formula by using two sides
containing the right angle as base and height. For
example, suppose that the sides of a right triangle
ABC are 5 cm, 12 cm and 13 cm; we take base as
12 cm and height as 5 cm (see Fig. 12.1). Then the
Fig. 12.1
198 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
area of ∆ ABC is given by
1
2
× base × height =
1
2
× 12 × 5 cm
2
, i.e., 30 cm
2
Note that we could also take 5 cm as the base and 12 cm as height.
Now suppose we want to find the area of an equilateral triangle PQR with side
10cm (see Fig. 12.2). To find its area we need its height. Can you find the height of
this triangle?
Let us recall how we find its height when we
know its sides. This is possible in an equilateral
triangle. Take the mid-point of QR as M and join it to
P. We know that PMQ is a right triangle. Therefore,
by using Pythagoras Theorem, we can find the length
PM as shown below:
PQ
2
= PM
2
+ QM
2
i.e., (10)
2
= PM
2
+ (5)
2
, since QM = MR.
Therefore, we have PM
2
= 75
i.e., PM = 75 cm = 5 3 cm.
Then area of ∆ PQR =
1
2
× base × height =
2
1
10 5 3 cm 25 3
2
× × = cm
2
.
Let us see now whether we can calculate the area of an isosceles triangle also
with the help of this formula. For example, we take a triangle XYZ with two equal
sides XY and XZ as 5 cm each and unequal side YZ as 8 cm (see Fig. 12.3).
In this case also, we want to know the height of the triangle. So, from X we draw
a perpendicular XP to side YZ. You can see that this perpendicular XP divides the
base YZ of the triangle in two equal parts.
Therefore, YP = PZ =
1
2
YZ = 4 cm
Then, by using Pythagoras theorem, we get
XP
2
= XY
2
– YP
2
= 5
2
– 4
2
= 25 – 16 = 9
So, XP = 3 cm
Now, area of ∆ XYZ =
1
2
× base YZ × height XP
=
1
2
× 8 × 3 cm
2
= 12 cm
2
.
Fig. 12.2
Fig. 12.3
HERON'S FORMULA 199
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
Now suppose that we know the lengths of the sides of a scalene triangle and not
the height. Can you still find its area? For instance, you have a triangular park whose
sides are 40 m, 32 m, and 24 m. How will you calculate its area? Definitely if you want
to apply the formula, you will have to calculate its height. But we do not have a clue to
calculate the height. Try doing so. If you are not able to get it, then go to the next
section.
12.2 Area of a Triangle — by Heron's Formula
Heron was born in about 10AD possibly in Alexandria in
Egypt. He worked in applied mathematics. His works on
mathematical and physical subjects are so numerous and
varied that he is considered to be an encyclopedic writer
in these fields. His geometrical works deal largely with
problems on mensuration written in three books. Book I
deals with the area of squares, rectangles, triangles,
trapezoids (trapezia), various other specialised
quadrilaterals, the regular polygons, circles, surfaces of
cylinders, cones, spheres etc. In this book, Heron has
derived the famous formula for the area of a triangle in
terms of its three sides.
The formula given by Heron about the area of a triangle, is also known as Hero's
formula. It is stated as:
Area of a triangle = ( ) ( ) ( ) s s a s b s c − − − (II)
where a, b and c are the sides of the triangle, and s = semi-perimeter i.e. half the
perimeter of the triangle =
2
a b c + +
,
This formula is helpful where it is not possible to find the height of the triangle
easily. Let us apply it to calculate the area of the triangular park ABC, mentioned
above (see Fig. 12.5).
Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s =
40 24 32
2
+ +
m = 48 m.
Fig. 12.4
Heron (10AD – 75 AD)
200 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
s – a = (48 – 40) m = 8 m,
s – b = (48 – 24) m = 24 m,
s – c = (48 – 32) m = 16 m.
Therefore, area of the park ABC
= ( ) ( ) ( ) − − − s s a s b s c
=
2 2
48 8 24 16 m 384m × × × =
We see that 32
2
+ 24
2
= 1024 + 576 = 1600 = 40
2
. Therefore, the sides of the park
make a right triangle. The largest side i.e. BC which is 40 m will be the hypotenuse
and the angle between the sides AB and AC will be 90°.
By using Formula I, we can check that the area of the park is
1
2
× 32 × 24 m
2
= 384 m
2
.
We find that the area we have got is the same as we found by using Heron's
formula.
Now using Heron's formula, you verify this fact by finding the areas of other
triangles discussed earlier viz;
(i) equilateral triangle with side 10 cm.
(ii) isosceles triangle with unequal side as 8 cm and each equal side as 5 cm.
You will see that
For (i), we have s =
10 10 10
2
+ +
cm = 15 cm.
Area of triangle = 15(15 10) (15 10) (15 10) − − − cm
2
=
2 2
15 5 5 5 cm 25 3 cm × × × =
For (ii), we have s =
8 5 5
cm 9 cm
2
+ +
=
.
Area of triangle = 9(9 8) (9 5) (9 5) − − − cm
2
=
2 2
9 1 4 4 cm 12 cm . × × × =
Let us now solve some more examples:
Fig. 12.5
HERON'S FORMULA 201
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
Example 1 : Find the area of a triangle, two sides of which are 8 cm and 11 cm and
the perimeter is 32 cm (see Fig. 12.6).
Solution : Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm.
Third side c = 32 cm – (8 + 11) cm = 13 cm
So, 2s = 32 i.e. s = 16 cm,
s – a = (16 – 8) cm = 8 cm,
s – b = (16 – 11) cm = 5 cm,
s – c = (16 – 13) cm = 3 cm.
Therefore, area of the triangle = ( ) ( ) ( ) s s a s b s c − − −
=
2 2
16 8 5 3cm 8 30 cm × × × =
Example 2 : A triangular park ABC has sides 120m, 80m and 50m (see Fig. 12.7). A
gardener Dhania has to put a fence all around it and also plant grass inside. How
much area does she need to plant? Find the cost of fencing it with barbed wire at the
rate of Rs 20 per metre leaving a space 3m wide for a gate on one side.
Solution : For finding area of the park, we have
2s = 50 m + 80 m + 120 m = 250 m.
i.e., s = 125 m
Now, s – a = (125 – 120) m = 5 m,
s – b = (125 – 80) m = 45 m,
s – c = (125 – 50) m = 75 m.
Therefore, area of the park = ( ) ( ) ( ) s s a s b s c − − −
= 125 5 45 75 × × × m
2
=
2
375 15 m
Also, perimeter of the park = AB + BC + CA = 250 m
Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate)
= 247 m
And so the cost of fencing = Rs 20 × 247 = Rs 4940
Fig. 12.6
Fig. 12.7
202 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
Example 3 : The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter
is 300 m. Find its area.
Solution : Suppose that the sides, in metres, are 3x, 5x and 7x (see Fig. 12.8).
Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle)
Therefore, 15x = 300, which gives x = 20.
So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 m
i.e., 60 m, 100 m and 140 m.
Can you now find the area [Using Heron's formula]?
We have s =
60 100 140
2
+ +
m = 150 m,
and area will be 150(150 60) (150 100) (150 140) − − − m
2
= 150 90 50 10 × × × m
2
=
2
1500 3 m
EXERCISE 12.1
1. A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with
side 'a'. Find the area of the signal board, using Heron's formula. If its perimeter is
180 cm, what will be the area of the signal board?
2. The triangular side walls of a flyover have been used for advertisements. The sides of
the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an
earning of Rs 5000 per m
2
per year. A company hired one of its walls for 3 months. How
much rent did it pay?
Fig. 12.9
Fig. 12.8
HERON'S FORMULA 203
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
3. There is a slide in a park. One of its side walls has been painted in some colour with a
message "KEEP THE PARK GREEN AND CLEAN" (see Fig. 12.10 ). If the sides of the
wall are 15 m, 11 m and 6 m, find the area painted in colour.
Fig. 12.10
4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is
42cm.
5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find
the area of the triangle.
12.3 Application of Heron's Formula in Finding Areas of Quadrilaterals
Suppose that a farmer has a land to be cultivated and she employs some labourers for
this purpose on the terms of wages calculated by area cultivated per square metre.
How will she do this? Many a time, the fields are in the shape of quadrilaterals. We
need to divide the quadrilateral in triangular parts and then use the formula for area of
the triangle. Let us look at this problem:
Example 4 : Kamla has a triangular field with sides 240 m, 200 m, 360 m, where she
grew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to the
previous field, she wanted to grow potatoes and onions (see Fig. 12.11). She divided
the field in two parts by joining the mid-point of the longest side to the opposite vertex
and grew patatoes in one part and onions in the other part. How much area (in hectares)
has been used for wheat, potatoes and onions? (1 hectare = 10000 m
2
)
Solution : Let ABC be the field where wheat is grown. Also let ACD be the field
which has been divided in two parts by joining C to the mid-point E of AD. For the
area of triangle ABC, we have
a = 200 m, b = 240 m, c = 360 m
Therefore, s =
200 240 360
2
+ +
m = 400 m.
204 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
So, area for growing wheat
= 400(400 200) (400 – 240) (400 – 360) − m
2
=
2
400 200 160 40 m × × ×
=
2
16000 2 m = 1.6 × 2 hectares
= 2.26 hectares (nearly)
Let us now calculate the area of triangle ACD.
Here, we have s =
240 320 400
2
+ +
m = 480 m.
So, area of ∆ ACD = 480(480 240) (480 320) (480 400) − − − m
2
= 480 240 160 80 × × × m
2
= 38400 m
2
= 3.84 hectares
We notice that the line segment joining the mid-point E of AD to C divides the
triangle ACD in two parts equal in area. Can you give the reason for this? In fact, they
have the bases AE and ED equal and, of course, they have the same height.
Therefore, area for growing potatoes = area for growing onions
= (3.84 ÷ 2) hectares = 1.92 hectares.
Example 5 : Students of a school staged a rally for cleanliness campaign. They
walked through the lanes in two groups. One group walked through the lanes AB, BC
and CA; while the other through AC, CD and DA (see Fig. 12.12). Then they cleaned
the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m
and ∠ B = 90º, which group cleaned more area and by how much? Find the total area
cleaned by the students (neglecting the width of the lanes).
Solution : Since AB = 9 m and BC = 40 m, ∠ B = 90°, we have:
AC =
2 2
9 40 + m
= 81 1600 + m
=
1681
m = 41m
Therefore, the first group has to clean the area of triangle ABC, which is right angled.
Area of ∆ ABC =
1
2
× base × height
=
1
2
× 40 × 9 m
2
= 180 m
2
Fig. 12.11
Fig. 12.12
HERON'S FORMULA 205
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
The second group has to clean the area of triangle ACD, which is scalene having sides
41 m, 15 m and 28 m.
Here, s =
41 15 28
2
+ +
m = 42 m
Therefore, area of ∆ ACD = ( – ) ( – ) ( – ) s s a s b s c
= 42(42 – 41) (42 – 15) (42 – 28) m
2
= 42 1 27 14 × × × m
2
= 126 m
2
So first group cleaned 180 m
2
which is (180 – 126) m
2
, i.e., 54 m
2
more than the area
cleaned by the second group.
Total area cleaned by all the students = (180 + 126) m
2
= 306 m
2
.
Example 6 : Sanya has a piece of land which is in the shape of a rhombus
(see Fig. 12.13). She wants her one daughter and one son to work on the land and
produce different crops. She divided the land in two equal parts. If the perimeter of
the land is 400 m and one of the diagonals is 160 m, how much area each of them will
get for their crops?
Solution : Let ABCD be the field.
Perimeter = 400 m
So, each side = 400 m ÷ 4 = 100 m.
i.e. AB = AD = 100 m.
Let diagonal BD = 160 m.
Then semi-perimeter s of ∆ ABD is given by
s =
100 100 160
2
+ +
m = 180 m
Therefore, area of ∆ ABD = 180(180 100) (180 – 100) (180 – 160) −
= 180 80 80 20 × × × m
2
= 4800 m
2
Therefore, each of them will get an area of 4800 m
2
.
Fig. 12.13
206 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
Alternative method : Draw CE ⊥ BD (see Fig. 12.14).
As BD = 160 m, we have
DE = 160 m ÷ 2 = 80 m
And, DE
2
+ CE
2
= DC
2
, which gives
CE =
2 2
DC DE −
or, CE =
2 2
100 80 m 60 m − =
Therefore, area of ∆ BCD =
1
160 60
2
× × m
2
= 4800 m
2
EXERCISE 12.2
1. A park, in the shape of a quadrilateral ABCD, has ∠ C = 90º, AB = 9 m, BC = 12 m,
CD = 5 m and AD = 8 m. How much area does it occupy?
2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm,
DA = 5 cm and AC = 5 cm.
3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find
the total area of the paper used.
Fig. 12.15
4. A triangle and a parallelogram have the same base and the same area. If the sides of
the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base
28 cm, find the height of the parallelogram.
Fig. 12.14
HERON'S FORMULA 207
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-12\Chap-12 (03-01-2006).PM65
5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the
rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each
cow be getting?
6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours
(see Fig. 12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each
colour is required for the umbrella?
7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base
8 cm and sides 6 cm each is to be made of three different shades as shown in
Fig. 12.17. How much paper of each shade has been used in it?
Fig. 12.16 Fig. 12.17
8. A floral design on a floor is made up of 16 tiles
which are triangular, the sides of the triangle
being 9 cm, 28 cm and 35 cm (see Fig. 12.18).
Find the cost of polishing the tiles at the rate
of 50p per cm
2
.
9. A field is in the shape of a trapezium whose parallel
sides are 25 m and 10 m. The non-parallel sides
are 14 m and 13 m. Find the area of the field.
12.4 Summary
In this chapter, you have studied the following points :
1. Area of a triangle with its sides as a, b and c is calculated by using Heron's formula,
stated as
Area of triangle = ( ) ( ) ( ) − − − s s a s b s c
where s =
2
+ + a b c
2. Area of a quadrilateral whose sides and one diagonal are given, can be calculated by
dividing the quadrilateral into two triangles and using the Heron's formula.
Fig. 12.18
208 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
CHAPTER 13
SURFACE AREAS AND VOLUMES
13.1 Introduction
Wherever we look, usually we see solids. So far, in all our study, we have been dealing
with figures that can be easily drawn on our notebooks or blackboards. These are
called plane figures. We have understood what rectangles, squares and circles are,
what we mean by their perimeters and areas, and how we can find them. We have
learnt these in earlier classes. It would be interesting to see what happens if we cut
out many of these plane figures of the same shape and size from cardboard sheet and
stack them up in a vertical pile. By this process, we shall obtain some solid figures
(briefly called solids) such as a cuboid, a cylinder, etc. In the earlier classes, you have
also learnt to find the surface areas and volumes of cuboids, cubes and cylinders. We
shall now learn to find the surface areas and volumes of cuboids and cylinders in
details and extend this study to some other solids such as cones and spheres.
13.2 Surface Area of a Cuboid and a Cube
Have you looked at a bundle of many sheets of paper? How does it look? Does it look
like what you see in Fig. 13.1?
Fig. 13.1
That makes up a cuboid. How much of brown paper would you need, if you want
to cover this cuboid? Let us see:
SURFACE AREAS AND VOLUMES 209
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
First we would need a rectangular piece to cover
the bottom of the bundle. That would be as shown in
Fig. 13.2 (a)
Then we would need two long rectangular pieces
to cover the two side ends. Now, it would look like
Fig. 13.2 (b).
Now to cover the front and back ends, we would
need two more rectangular pieces of a different size.
With them, we would now have a figure as shown in
Fig. 13.2(c).
This figure, when opened out, would look like
Fig. 13.2 (d).
Finally, to cover the top of the bundle, we would
require another rectangular piece exactly like the one
at the bottom, which if we attach on the right side, it
would look like Fig. 13.2(e).
So we have used six rectangular pieces to cover
the complete outer surface of the cuboid.
Fig. 13.2
210 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
This shows us that the outer surface of a cuboid is made up of six rectangles (in
fact, rectangular regions, called the faces of the cuboid), whose areas can be found by
multiplying length by breadth for each of them separately and then adding the six
areas together.
Now, if we take the length of the cuboid as l, breadth as b and the height as h, then
the figure with these dimensions would be like the shape you see in Fig. 13.2(f).
So, the sum of the areas of the six rectangles is:
Area of rectangle 1 (= l × h)
+
Area of rectangle 2 (= l × b)
+
Area of rectangle 3 (= l × h)
+
Area of rectangle 4 (= l × b)
+
Area of rectangle 5 (= b × h)
+
Area of rectangle 6 (= b × h)
= 2(l × b) + 2(b × h) + 2(l × h)
= 2(lb + bh + hl)
This gives us:
Surface Area of a Cuboid = 2(lb + bh + hl)
where l, b and h are respectively the three edges of the cuboid.
Note : The unit of area is taken as the square unit, because we measure the magnitude
of a region by filling it with squares of side of unit length.
For example, if we have a cuboid whose length, breadth and height are 15 cm,
10 cm and 20 cm respectively, then its surface area would be:
2[(15 × 10) + (10 × 20) + (20 × 15)] cm
2
= 2(150 + 200 + 300) cm
2
= 2 × 650 cm
2
= 1300 cm
2
SURFACE AREAS AND VOLUMES 211
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Recall that a cuboid, whose length, breadth and height are all equal, is called a
cube. If each edge of the cube is a, then the surface area of this cube would be
2(a × a + a × a + a × a) i.e., 6a
2
(see Fig. 13.3), giving us
Surface Area of a Cube = 6a
2
where a is the edge of the cube.
Fig. 13.3
Suppose, out of the six faces of a cuboid, we only find the area of the four faces,
leaving the bottom and top faces. In such a case, the area of these four faces is called
the lateral surface area of the cuboid. So, lateral surface area of a cuboid of
length l, breadth b and height h is equal to 2lh + 2bh or 2(l + b)h. Similarly,
lateral surface area of a cube of side a is equal to 4a
2
.
Keeping in view of the above, the surface area of a cuboid (or a cube) is sometimes
also referred to as the total surface area. Let us now solve some examples.
Example 1 : Mary wants to decorate her Christmas
tree. She wants to place the tree on a wooden box
covered with coloured paper with picture of Santa
Claus on it (see Fig. 13.4). She must know the exact
quantity of paper to buy for this purpose. If the box
has length, breadth and height as 80 cm, 40 cm and
20 cm respectively how many square sheets of paper
of side 40 cm would she require?
Solution : Since Mary wants to paste the paper on
the outer surface of the box; the quantity of paper
required would be equal to the surface area of the
box which is of the shape of a cuboid. The dimensions
of the box are:
Fig. 13.4
212 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Length =80 cm, Breadth = 40 cm, Height = 20 cm.
The surface area of the box = 2(lb + bh + hl)
= 2[(80 × 40) + (40 × 20) + (20 × 80)] cm
2
= 2[3200 + 800 + 1600] cm
2
= 2 × 5600 cm
2
= 11200 cm
2
The area of each sheet of the paper = 40 × 40 cm
2
= 1600 cm
2
Therefore, number of sheets required =
surface area of box
area of one sheet of paper
=
11200
1600
= 7
So, she would require 7 sheets.
Example 2 : Hameed has built a cubical water tank with lid for his house, with each
outer edge 1.5 m long. He gets the outer surface of the tank excluding the base,
covered with square tiles of side 25 cm (see Fig. 13.5). Find how much he would
spend for the tiles, if the cost of the tiles is Rs 360 per dozen.
Solution : Since Hameed is getting the five outer faces of the tank covered with tiles,
he would need to know the surface area of the tank, to decide on the number of tiles
required.
Edge of the cubical tank = 1.5 m = 150 cm (= a)
So, surface area of the tank = 5 × 150 × 150 cm
2
Area of each square tile = side × side = 25 × 25 cm
2
So, the number of tiles required =
surface area of the tank
area of each tile
=
5 ×150 ×150
25 × 25
= 180
Cost of 1 dozen tiles, i.e., cost of 12 tiles = Rs 360
Therefore, cost of one tile = Rs
360
12
= Rs 30
So, the cost of 180 tiles = 180 × Rs 30 = Rs 5400
Fig. 13.5
SURFACE AREAS AND VOLUMES 213
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
EXERCISE 13.1
1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at
the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m
2
costs Rs 20.
2. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the
cost of white washing the walls of the room and the ceiling at the rate of
Rs 7.50 per m
2
.
3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four
walls at the rate of Rs 10 per m
2
is Rs 15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]
4. The paint in a certain container is sufficient to paint an area equal to 9.375 m
2
. How
many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this
container?
5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm
wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including
base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing
their sweets. Two sizes of boxes were required. The bigger of dimensions
25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the
overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is
Rs 4 for 1000 cm
2
, find the cost of cardboard required for supplying 250 boxes of each
kind.
8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure
with tarpaulin that covers all the four sides and the top of the car (with the front face
as a flap which can be rolled up). Assuming that the stitching margins are very small,
and therefore negligible, how much tarpaulin would be required to make the shelter of
height 2.5 m, with base dimensions 4 m × 3 m?
214 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
13.3 Surface Area of a Right Circular Cylinder
If we take a number of circular sheets of paper and stack them up as we stacked up
rectangular sheets earlier, what would we get (see Fig. 13.6)?
Fig. 13.6
Here, if the stack is kept vertically up, we get what is called a right circular
cylinder, since it has been kept at right angles to the base, and the base is circular. Let
us see what kind of cylinder is not a right circular cylinder.
In Fig 13.7 (a), you see a cylinder, which
is certainly circular, but it is not at right angles
to the base. So, we can not say this a right
circular cylinder.
Of course, if we have a cylinder with a
non circular base, as you see in Fig. 13.7 (b),
then we also cannot call it a right circular
cylinder.
Remark : Here, we will be dealing with only right circular cylinders. So, unless stated
otherwise, the word cylinder would mean a right circular cylinder.
Now, if a cylinder is to be covered with coloured paper, how will we do it with the
minimum amount of paper? First take a rectangular sheet of paper, whose length is
just enough to go round the cylinder and whose breadth is equal to the height of the
cylinder as shown in Fig. 13.8.
Fig. 13.7
SURFACE AREAS AND VOLUMES 215
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Fig. 13.8
The area of the sheet gives us the curved surface area of the cylinder. Note that
the length of the sheet is equal to the circumference of the circular base which is equal
to 2πr.
So, curved surface area of the cylinder
= area of the rectangular sheet = length × breadth
= perimeter of the base of the cylinder × h
= 2πr × h
Therefore, Curved Surface Area of a Cylinder = 2π ππ ππrh
where r is the radius of the base of the cylinder and h is the height of the cylinder.
Remark : In the case of a cylinder, unless stated
otherwise, 'radius of a cylinder' shall mean' base radius
of the cylinder'.
If the top and the bottom of the cylinder are also to
be covered, then we need two circles (infact, circular
regions) to do that, each of radius r, and thus having an
area of πr
2
each (see Fig. 13.9), giving us the total
surface area as 2πrh + 2πr
2
= 2πr(r + h).
So, Total Surface Area of a Cylinder = 2π ππ ππr(r + h)
where h is the height of the cylinder and r its radius.
Remark : You may recall from Chapter 1 that π is an irrational number. So, the value
Fig. 13.9
l
216 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
of π is a non-terminating, non-repeating decimal. But when we use its value in our
calculations, we usually take its value as approximately equal to
22
7
or 3.14.
Example 3 : Savitri had to make a model of a cylindrical kaleidoscope for her science
project. She wanted to use chart paper to make the curved surface of the kaleidoscope.
(see Fig 13.10). What would be the area of chart paper required by her, if she wanted
to make a kaleidoscope of length 25 cm with a 3.5 cm radius? You may take π =
22
7
.
Solution : Radius of the base of the cylindrical kaleidoscope (r) = 3.5 cm.
Height (length) of kaleidoscope (h) = 25 cm.
Area of chart paper required = curved surface area of the kaleidoscope
= 2πrh
=
2
22
2 3.5 25 cm
7
× × ×
= 550 cm
2
EXERCISE 13.2
Assume π =
22
7
, unless stated otherwise.
1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm
2
. Find the
diameter of the base of the cylinder.
2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm
from a metal sheet. How many square metres of the sheet are required for the same?
3. A metal pipe is 77 cm long. The inner diameter of a cross
section is 4 cm, the outer diameter being 4.4 cm
(see Fig. 13.11). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.
Fig. 13.10
Fig. 13.11
SURFACE AREAS AND VOLUMES 217
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete
revolutions to move once over to level a playground. Find the area of the playground
in m
2
.
5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting
the curved surface of the pillar at the rate of Rs 12.50 per m
2
.
6. Curved surface area of a right circular cylinder is 4.4 m
2
. If the radius of the base of the
cylinder is 0.7 m, find its height.
7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs 40 per m
2
.
8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter
5 cm. Find the total radiating surface in the system.
9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is
4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if
1
12
of the steel actually used was wasted in
making the tank.
10. In Fig. 13.12, you see the frame of a lampshade. It is to be
covered with a decorative cloth. The frame has a base
diameter of 20 cm and height of 30 cm. A margin of 2.5 cm
is to be given for folding it over the top and bottom of the
frame. Find how much cloth is required for covering the
lampshade.
11. The students of a Vidyalaya were asked to participate in a competition for making and
decorating penholders in the shape of a cylinder with a base, using cardboard. Each
penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply
the competitors with cardboard. If there were 35 competitors, how much cardboard
was required to be bought for the competition?
13.4 Surface Area of a Right Circular Cone
So far, we have been generating solids by stacking up congruent figures. Incidentally,
such figures are called prisms. Now let us look at another kind of solid which is not a
prism. (These kinds of solids are called pyramids). Let us see how we can generate
them.
Activity : Cut out a right-angled triangle ABC right angled at B. Paste a long thick
string along one of the perpendicular sides say AB of the triangle [see Fig. 13.13(a)].
Hold the string with your hands on either sides of the triangle and rotate the triangle
Fig. 13.12
218 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
about the string a number of times. What happens? Do you recognize the shape that
the triangle is forming as it rotates around the string [see Fig. 13.13(b)]? Does it
remind you of the time you had eaten an ice-cream heaped into a container of that
shape [see Fig. 13.13 (c) and (d)]?
Fig. 13.13
This is called a right circular cone. In Fig. 13.13(c)
of the right circular cone, the point A is called the
vertex, AB is called the height, BC is called the radius
and AC is called the slant height of the cone. Here B
will be the centre of circular base of the cone. The
height, radius and slant height of the cone are usually
denoted by h, r and l respectively. Once again, let us
see what kind of cone we can not call a right circular
cone. Here, you are (see Fig. 13.14)! What you see
in these figures are not right circular cones; because
in (a), the line joining its vertex to the centre of its
base is not at right angle to the base, and in (b) the
base is not circular.
As in the case of cylinder, since we will be studying only about right circular cones,
remember that by 'cone' in this chapter, we shall mean a 'right circular cone.'
Activity : (i) Cut out a neatly made paper cone that does not have any overlapped
paper, straight along its side, and opening it out, to see the shape of paper that forms
the surface of the cone. (The line along which you cut the cone is the slant height of
the cone which is represented by l). It looks like a part of a round cake.
Fig. 13.14
SURFACE AREAS AND VOLUMES 219
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
(ii) If you now bring the sides marked A and B at the tips together, you can see that
the curved portion of Fig. 13.15 (c) will form the circular base of the cone.
Fig. 13.15
(iii) If the paper like the one in Fig. 13.15 (c) is now cut into hundreds of little pieces,
along the lines drawn from the point O, each cut portion is almost a small triangle,
whose height is the slant height l of the cone.
(iv) Now the area of each triangle =
1
2
× base of each triangle × l.
So, area of the entire piece of paper
= sum of the areas of all the triangles
=
1 2 3
1 1 1
2 2 2
bl b l b l + + + " = ( )
1 2 3
1
2
l b b b + + + "
=
1
2
× l × length of entire curved boundary of Fig. 13.15(c)
(as b
1
+ b
2
+ b
3
+ . . . makes up the curved portion of the figure)
But the curved portion of the figure makes up the perimeter of the base of the cone
and the circumference of the base of the cone = 2πr, where r is the base radius of the
cone.
So, Curved Surface Area of a Cone =
1
2
× l × 2π ππ ππr = π ππ ππrl
where r is its base radius and l its slant height.
Note that l
2
= r
2
+ h
2
(as can be seen from Fig. 13.16), by
applying Pythagoras Theorem. Here h is the height of the
cone.
Fig. 13.16
220 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Therefore, l =
2 2
r h +
Now if the base of the cone is to be closed, then a circular piece of paper of radius r
is also required whose area is πr
2
.
So, Total Surface Area of a Cone = π ππ ππrl + π ππ ππr
2
= π ππ ππr(l + r)
Example 4 : Find the curved surface area of a right circular cone whose slant height
is 10 cm and base radius is 7 cm.
Solution : Curved surface area = πrl
=
22
7
× 7 × 10 cm
2
= 220 cm
2
Example 5 : The height of a cone is 16 cm and its base radius is 12 cm. Find the
curved surface area and the total surface area of the cone (Use π = 3.14).
Solution : Here, h = 16 cm and r = 12 cm.
So, from l
2
= h
2
+ r
2
, we have
l =
2 2
16 12 +
cm = 20 cm
So, curved surface area = πrl
= 3.14 × 12 × 20 cm
2
= 753.6 cm
2
Further, total surface area = πrl + πr
2
= (753.6 + 3.14 × 12 × 12) cm
2
= (753.6 + 452.16) cm
2
= 1205.76 cm
2
Example 6 : A corn cob (see Fig. 13.17), shaped somewhat
like a cone, has the radius of its broadest end as 2.1 cm and
length (height) as 20 cm. If each 1 cm
2
of the surface of the
cob carries an average of four grains, find how many grains
you would find on the entire cob.
Solution : Since the grains of corn are found only on the curved surface of the corn
cob, we would need to know the curved surface area of the corn cob to find the total
number of grains on it. In this question, we are given the height of the cone, so we
need to find its slant height.
Fig. 13.17
SURFACE AREAS AND VOLUMES 221
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Here, l =
2 2
r h + =
2 2
(2.1) 20 + cm
=
404.41
cm = 20.11 cm
Therefore, the curved surface area of the corn cob = πrl
=
22
7
× 2.1 × 20.11 cm
2
= 132.726 cm
2
= 132.73 cm
2
(approx.)
Number of grains of corn on 1 cm
2
of the surface of the corn cob = 4
Therefore, number of grains on the entire curved surface of the cob
= 132.73 × 4 = 530.92 = 531 (approx.)
So, there would be approximately 531 grains of corn on the cob.
EXERCISE 13.3
Assume π =
22
7
, unless stated otherwise.
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved
surface area.
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base
is 24 m.
3. Curved surface area of a cone is 308 cm
2
and its slant height is 14 cm. Find
(i) radius of the base and (ii) total surface area of the cone.
4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m
2
canvas is Rs 70.
5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m
and base radius 6 m? Assume that the extra length of material that will be required for
stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).
6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively.
Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m
2
.
7. A joker's cap is in the form of a right circular cone of base radius 7 cm and height
24 cm. Find the area of the sheet required to make 10 such caps.
8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow
cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height
1 m. If the outer side of each of the cones is to be painted and the cost of painting is
Rs 12 per m
2
, what will be the cost of painting all these cones? (Use π = 3.14 and take
1.04
= 1.02)
222 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
13.5 Surface Area of a Sphere
What is a sphere? Is it the same as a circle? Can you draw a circle on a paper? Yes,
you can, because a circle is a plane closed figure whose every point lies at a constant
distance (called radius) from a fixed point, which is called the centre of the circle.
Now if you paste a string along a diameter of a circular disc and rotate it as you had
rotated the triangle in the previous section, you see a new solid (see Fig 13.18). What
does it resemble? A ball? Yes. It is called a sphere.
Fig. 13.18
Can you guess what happens to the centre of the circle, when it forms a sphere on
rotation? Of course, it becomes the centre of the sphere. So, a sphere is a three
dimensional figure (solid figure), which is made up of all points in the space,
which lie at a constant distance called the radius, from a fixed point called the
centre of the sphere.
Note : A sphere is like the surface of a ball. The word solid sphere is used for the
solid whose surface is a sphere.
Activity : Have you ever played with a top or have you at least watched someone
play with one? You must be aware of how a string is wound around it. Now, let us take
a rubber ball and drive a nail into it. Taking support of the nail, let us wind a string
around the ball. When you have reached the 'fullest' part of the ball, use pins to keep
the string in place, and continue to wind the string around the remaining part of the ball,
till you have completely covered the ball [see Fig. 13.19(a)]. Mark the starting and
finishing points on the string, and slowly unwind the string from the surface of the ball.
Now, ask your teacher to help you in measuring the diameter of the ball, from which
you easily get its radius. Then on a sheet of paper, draw four circles with radius equal
SURFACE AREAS AND VOLUMES 223
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
to the radius of the ball. Start filling the circles one by one, with the string you had
wound around the ball [see Fig. 13.19(b)].
Fig. 13.19
What have you achieved in all this?
The string, which had completely covered the surface area of the sphere, has been
used to completely fill the regions of four circles, all of the same radius as of the sphere.
So, what does that mean? This suggests that the surface area of a sphere of radius r
= 4 times the area of a circle of radius r = 4 × (π r
2
)
So, Surface Area of a Sphere = 4 π ππ ππ r
2
where r is the radius of the sphere.
How many faces do you see in the surface of a sphere? There is only one, which is
curved.
Now, let us take a solid sphere, and slice it exactly 'through
the middle' with a plane that passes through its centre. What
happens to the sphere?
Yes, it gets divided into two equal parts (see Fig. 13.20)!
What will each half be called? It is called a hemisphere.
(Because 'hemi' also means 'half')
And what about the surface of a hemisphere? How many faces does it have?
Two! There is a curved face and a flat face (base).
The curved surface area of a hemisphere is half the surface area of the sphere, which
is
1
2
of 4πr
2
.
Fig. 13.20
224 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Therefore, Curved Surface Area of a Hemisphere = 2π ππ ππr
2
where r is the radius of the sphere of which the hemisphere is a part.
Now taking the two faces of a hemisphere, its surface area 2πr
2
+ πr
2
So, Total Surface Area of a Hemisphere = 3π ππ ππr
2
Example 7 : Find the surface area of a sphere of radius 7 cm.
Solution : The surface area of a sphere of radius 7 cm would be
4πr
2
= 4 ×
22
7
× 7 × 7 cm
2
= 616 cm
2
Example 8 : Find (i) the curved surface area and (ii) the total surface area of a
hemisphere of radius 21 cm.
Solution : The curved surface area of a hemisphere of radius 21 cm would be
= 2πr
2
= 2 ×
22
7
× 21 × 21 cm
2
= 2772 cm
2
(ii) the total surface area of the hemisphere would be
3πr
2
= 3 ×
22
7
× 21 × 21 cm
2
= 4158 cm
2
Example 9 : The hollow sphere, in which the circus motorcyclist performs his stunts,
has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Solution : Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the riding
space available for the motorcyclist is the surface area of the 'sphere' which is
given by
4πr
2
= 4 ×
22
7
× 3.5 × 3.5 m
2
= 154 m
2
Example 10 : A hemispherical dome of a building needs to be painted
(see Fig. 13.21). If the circumference of the base of the dome is 17.6 m, find the cost
of painting it, given the cost of painting is Rs 5 per 100 cm
2
.
Solution : Since only the rounded surface of the dome is to be painted, we would need
to find the curved surface area of the hemisphere to know the extent of painting that
needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2πr.
SURFACE AREAS AND VOLUMES 225
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
So, the radius of the dome = 17.6 ×
7
2 22 ×
m = 2.8 m
The curved surface area of the dome = 2πr
2
= 2 ×
22
7
× 2.8 × 2.8 m
2
= 49.28 m
2
Now, cost of painting 100 cm
2
is Rs 5.
So, cost of painting 1 m
2
= Rs 500
Therefore, cost of painting the whole dome
= Rs 500 × 49.28
= Rs 24640
EXERCISE 13.4
Assume π =
22
7
, unless stated otherwise.
1. Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
2. Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 m
3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped
into it. Find the ratio of surface areas of the balloon in the two cases.
5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of
tin-plating it on the inside at the rate of Rs 16 per 100 cm
2
.
6. Find the radius of a sphere whose surface area is 154 cm
2
.
7. The diameter of the moon is approximately one fourth of the diameter of the earth.
Find the ratio of their surface areas.
8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is
5 cm. Find the outer curved surface area of the bowl.
9. A right circular cylinder just encloses a sphere of
radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
Fig. 13.21
Fig. 13.22
226 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
13.6 Volume of a Cuboid
You have already learnt about volumes of certain figures (objects) in earlier classes.
Recall that solid objects occupy space. The measure of this occupied space is called
the Volume of the object.
Note : If an object is solid, then the space occupied by such an object is measured,
and is termed the Volume of the object. On the other hand, if the object is hollow, then
interior is empty, and can be filled with air, or some liquid that will take the shape of its
container. In this case, the volume of the substance that can fill the interior is called the
capacity of the container. In short, the volume of an object is the measure of the
space it occupies, and the capacity of an object is the volume of substance its interior
can accommodate. Hence, the unit of measurement of either of the two is cubic unit.
So, if we were to talk of the volume of a cuboid, we would be considering the
measure of the space occupied by the cuboid.
Further, the area or the volume is measured as the magnitude of a region. So,
correctly speaking, we should be finding the area of a circular region, or volume of a
cuboidal region, or volume of a spherical region, etc. But for the sake of simplicity, we
say, find the area of a circle, volume of a cuboid or a sphere even though these mean
only their boundaries.
Fig. 13.23
Observe Fig. 13.23. Suppose we say that the area of each rectangle is A, the
height up to which the rectangles are stacked is h and the volume of the cuboid is V.
Can you tell what would be the relationship between V, A and h?
The area of the plane region occupied by each rectangle × height
= Measure of the space occupied by the cuboid
So, we get A × h = V
That is, Volume of a Cuboid = base area × height = length × breadth × height
or l × b × h, where l, b and h are respectively the length, breadth and height of the
cuboid.
SURFACE AREAS AND VOLUMES 227
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Note : When we measure the magnitude of the region of a space, that is, the
space occupied by a solid, we do so by counting the number of cubes of edge of unit
length that can fit into it exactly. Therefore, the unit of measurement of volume is
cubic unit.
Again Volume of a Cube = edge × edge × edge = a
3
where a is the edge of the cube (see Fig. 13.24).
So, if a cube has edge of 12 cm,
then volume of the cube = 12 × 12 × 12 cm
3
= 1728 cm
3
.
Recall that you have learnt these formulae in
earlier classes. Now let us take some examples to
illustrate the use of these formulae:
Example11 : A wall of length 10 m was to be built across an open ground. The height
of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with
bricks whose dimensions are 24 cm × 12 cm × 8 cm, how many bricks would be
required?
Solution : Since the wall with all its bricks makes up the space occupied by it, we
need to find the volume of the wall, which is nothing but a cuboid.
Here, Length = 10 m = 1000 cm
Thickness = 24 cm
Height = 4 m = 400 cm
Therefore, Volume of the wall = length × thickness × height
= 1000 × 24 × 400 cm
3
Now, each brick is a cuboid with length = 24 cm, breadth = 12 cm and height = 8 cm
So, volume of each brick = length × breadth × height
= 24 × 12 × 8 cm
3
So, number of bricks required =
volume of the wall
volume of each brick
=
1000 × 24 × 400
24 ×12 × 8
= 4166.6
So, the wall requires 4167 bricks.
Fig. 13.24
228 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Example 12 : A child playing with building blocks, which are
of the shape of cubes, has built a structure as shown in
Fig. 13.25. If the edge of each cube is 3 cm, find the volume
of the structure built by the child.
Solution : Volume of each cube = edge × edge × edge
= 3 × 3 × 3 cm
3
= 27 cm
3
Number of cubes in the structure = 15
Therefore, volume of the structure = 27 × 15 cm
3
= 405 cm
3
EXERCISE 13.5
1. A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet
containing 12 such boxes?
2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water
can it hold? (1 m
3
= 1000 l)
3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380
cubic metres of a liquid?
4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of
Rs 30 per m
3
.
5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank,
if its length and depth are respectively 2.5 m and 10 m.
6. A village, having a population of 4000, requires 150 litres of water per head per day. It
has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank
last?
7. A godown measures 40 m × 25 m × 10 m. Find the maximum number of wooden crates
each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the
side of the new cube? Also, find the ratio between their surface areas.
9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water
will fall into the sea in a minute?
13.7 Volume of a Cylinder
Just as a cuboid is built up with rectangles of the same size, we have seen that a right
circular cylinder can be built up using circles of the same size. So, using the same
argument as for a cuboid, we can see that the volume of a cylinder can be obtained
Fig. 13.25
SURFACE AREAS AND VOLUMES 229
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
as : base area × height
= area of circular base × height = πr
2
h
So, Volume of a Cylinder = π ππ ππr
2
h
where r is the base radius and h is the height of the cylinder.
Example 13 : The pillars of a temple are cylindrically
shaped (see Fig. 13.26). If each pillar has a circular
base of radius 20 cm and height 10 m, how much
concrete mixture would be required to build 14 such
pillars?
Solution : Since the concrete mixture that is to be
used to build up the pillars is going to occupy the
entire space of the pillar, what we need to find here
is the volume of the cylinders.
Radius of base of a cylinder = 20 cm
Height of the cylindrical pillar = 10 m = 1000 cm
So, volume of each cylinder = πr
2
h
=
22
20 20 1000
7
× × × cm
3
=
8800000
7
cm
3
=
8.8
7
m
3
(Since 1000000 cm
3
= 1m
3
)
Therefore, volume of 14 pillars = volume of each cylinder × 14
=
8.8
14
7
× m
3
= 17.6 m
3
So, 14 pillars would need 17.6 m
3
of concrete mixture.
Example 14 : At a Ramzan Mela, a stall keeper in one
of the food stalls has a large cylindrical vessel of base
radius 15 cm filled up to a height of 32 cm with orange
juice. The juice is filled in small cylindrical glasses (see
Fig. 13.27) of radius 3 cm up to a height of 8 cm, and
sold for Rs 3 each. How much money does the stall
keeper receive by selling the juice completely?
Fig. 13.27
Fig. 13.26
230 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Solution : The volume of juice in the vessel
= volume of the cylinderical vessel
= πR
2
H
(where R and H are taken as the radius and height respectively of the vessel)
= π × 15 × 15 × 32 cm
3
Similarly, the volume of juice each glass can hold = πr
2
h
(where r and h are taken as the radius and height respectively of each glass)
= π × 3 × 3 × 8 cm
3
So, number of glasses of juice that are sold
=
volume of the vessel
volume of each glass
=
×15 ×15 × 32
× 3 × 3 × 8
π
π
= 100
Therefore, amount received by the stall keeper = Rs 3 × 100
= Rs 300
EXERCISE 13.6
Assume π =
22
7
, unless stated otherwise.
1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm.
How many litres of water can it hold? (1000 cm
3
= 1l)
2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is
28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm
3
of wood has
a mass of 0.6 g.
3. A soft drink is available in two packs – (i) a tin can with a rectangular base of length
5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular
base of diameter 7 cm and height 10 cm. Which container has greater capacity and by
how much?
4. If the lateral surface of a cylinder is 94.2 cm
2
and its height is 5 cm, then find
(i) radius of its base (ii) its volume. (Use π = 3.14)
SURFACE AREAS AND VOLUMES 231
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If
the cost of painting is at the rate of Rs 20 per m
2
, find
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.
6. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many
square metres of metal sheet would be needed to make it?
7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in
the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm.
If the length of the pencil is 14 cm, find the volume of the wood and that of the
graphite.
8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the
bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare
daily to serve 250 patients?
13.8 Volume of a Right Circular Cone
In Fig 13.28, can you see that there is a right circular
cylinder and a right circular cone of the same base
radius and the same height?
Activity : Try to make a hollow cylinder and a hollow cone like this with the same
base radius and the same height (see Fig. 13.28). Then, we can try out an experiment
that will help us, to see practically what the volume of a right circular cone would be!
Fig. 13.29
Fig. 13.28
232 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
So, let us start like this.
Fill the cone up to the brim with sand once, and empty it into the cylinder. We find
that it fills up only a part of the cylinder [see Fig. 13.29(a)].
When we fill up the cone again to the brim, and empty it into the cylinder, we see
that the cylinder is still not full [see Fig. 13.29(b)].
When the cone is filled up for the third time, and emptied into the cylinder, it can be
seen that the cylinder is also full to the brim [see Fig. 13.29(c)].
With this, we can safely come to the conclusion that three times the volume of a
cone, makes up the volume of a cylinder, which has the same base radius and the
same height as the cone, which means that the volume of the cone is one-third the
volume of the cylinder.
So, Volume of a Cone =
1
3
π ππ ππr
2
h
where r is the base radius and h is the height of the cone.
Example 15 : The height and the slant height of a cone are 21 cm and 28 cm
respectively. Find the volume of the cone.
Solution : From l
2
= r
2
+ h
2
, we have
r =
2 2 2 2
28 21 cm 7 7 cm l h − = − =
So, volume of the cone =
1
3
πr
2
h =
1
3
×
22
7 7 7 7 21
7
× × ×
cm
3
= 7546 cm
3
Example 16 : Monica has a piece of canvas whose area is 551 m
2
. She uses it to
have a conical tent made, with a base radius of 7 m. Assuming that all the stitching
margins and the wastage incurred while cutting, amounts to approximately 1 m
2
, find
the volume of the tent that can be made with it.
Solution : Since the area of the canvas = 551 m
2
and area of the canvas lost in
wastage is 1 m
2
, therefore the area of canvas available for making the tent is
(551 – 1) m
2
= 550 m
2
.
Now, the surface area of the tent = 550 m
2
and the required base radius of the conical
tent = 7 m
Note that a tent has only a curved surface (the floor of a tent is not covered by
canvas!!).
SURFACE AREAS AND VOLUMES 233
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Therefore, curved surface area of tent = 550 m
2
.
That is, πrl = 550
or,
22
7
× 7 × l = 550
or, l = 3
550
22
m = 25 m
Now, l
2
= r
2
+ h
2
Therefore, h =
2 2
l r − =
2 2
25 7 m 625 49 m 576 m − = − =
= 24 m
So, the volume of the conical tent =
2 3
1 1 22
7 7 24 m
3 3 7
r h π = × × × ×
= 1232 m
3
.
EXERCISE 13.7
Assume π =
22
7
, unless stated otherwise.
1. Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm
2. Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm
3. The height of a cone is 15 cm. If its volume is 1570 cm
3
, find the radius of the base.
(Use π = 3.14)
4. If the volume of a right circular cone of height 9 cm is 48 π cm
3
, find the diameter of its
base.
5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
6. The volume of a right circular cone is 9856 cm
3
. If the diameter of the base is 28 cm,
find
(i) height of the cone (ii) slant height of the cone
(iii) curved surface area of the cone
7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.
Find the volume of the solid so obtained.
8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find
the volume of the solid so obtained. Find also the ratio of the volumes of the two
solids obtained in Questions 7 and 8.
9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m.
Find its volume. The heap is to be covered by canvas to protect it from rain. Find the
area of the canvas required.
234 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
13.9 Volume of a Sphere
Now, let us see how to go about measuring the volume of a sphere. First, take two or
three spheres of different radii, and a container big enough to be able to put each of
the spheres into it, one at a time. Also, take a large trough in which you can place the
container. Then, fill the container up to the brim with water [see Fig. 13.30(a)].
Now, carefully place one of the spheres in the container. Some of the water from
the container will over flow into the trough in which it is kept [see Fig. 13.30(b)].
Carefully pour out the water from the trough into a measuring cylinder (i.e., a graduated
cylindrical jar) and measure the water over flowed [see Fig. 13.30(c)]. Suppose the
radius of the immersed sphere is r (you can find the radius by measuring the diameter
of the sphere). Then evaluate
4
3
πr
3
. Do you find this value almost equal to the
measure of the volume over flowed?
Fig. 13.30
Once again repeat the procedure done just now, with a different size of sphere.
Find the radius R of this sphere and then calculate the value of
3
4
R .
3
π Once again this
value is nearly equal to the measure of the volume of the water displaced (over flowed)
by the sphere. What does this tell us? We know that the volume of the sphere is the
same as the measure of the volume of the water displaced by it. By doing this experiment
repeatedly with spheres of varying radii, we are getting the same result, namely, the
volume of a sphere is equal to
4
3
π
times the cube of its radius. This gives us the idea
that
Volume of a Sphere =
3
4
3
r π
where r is the radius of the sphere.
Later, in higher classes it can be proved also. But at this stage, we will just take it
as true.
SURFACE AREAS AND VOLUMES 235
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Since a hemisphere is half of a sphere, can you guess what the volume of a
hemisphere will be? Yes, it is
3
1 4
of
2 3
r π
=
2
3
πr
3
.
So, Volume of a Hemisphere =
3
2
3
r π
where r is the radius of the hemisphere.
Let us take some examples to illustrate the use of these formulae.
Example 17 : Find the volume of a sphere of radius 11.2 cm.
Solution : Required volume =
4
3
πr
3
=
4 22
11.2 11.2 11.2
3 7
× × × × cm
3
= 5887.32 cm
3
Example 18 : A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the
metal is 7.8 g per cm
3
, find the mass of the shot-putt.
Solution : Since the shot-putt is a solid sphere made of metal and its mass is equal to
the product of its volume and density, we need to find the volume of the sphere.
Now, volume of the sphere =
3
4
3
r π
=
3
4 22
4.9 4.9 4.9 cm
3 7
× × × ×
= 493 cm
3
(nearly)
Further, mass of 1 cm
3
of metal is 7.8 g.
Therefore, mass of the shot-putt = 7.8 × 493 g
= 3845.44 g = 3.85 kg (nearly)
Example 19 : A hemispherical bowl has a radius of 3.5 cm. What would be the
volume of water it would contain?
Solution : The volume of water the bowl can contain
=
3
2
3
r π
=
2 22
3.5 3.5 3.5
3 7
× × × ×
cm
3
= 89.8 cm
3
236 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
Fig. 13.31
EXERCISE 13.8
Assume π =
22
7
, unless stated otherwise.
1. Find the volume of a sphere whose radius is
(i) 7 cm (ii) 0.63 m
2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m
3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of
the metal is 8.9 g per cm
3
?
4. The diameter of the moon is approximately one-fourth of the diameter of the earth.
What fraction of the volume of the earth is the volume of the moon?
5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m,
then find the volume of the iron used to make the tank.
7. Find the volume of a sphere whose surface area is 154 cm
2
.
8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed
at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square metre, find
the
(i) inside surface area of the dome, (ii) volume of the air inside the dome.
9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to
form a sphere with surface area S′. Find the
(i) radius r ′ of the new sphere, (ii) ratio of S and S′.
10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much
medicine (in mm
3
) is needed to fill this capsule?
EXERCISE 13.9 (Optional)*
1. A wooden bookshelf has external dimensions as
follows: Height = 110 cm, Depth = 25 cm,
Breadth = 85 cm (see Fig. 13.31). The thickness of
the plank is 5 cm everywhere. The external faces
are to be polished and the inner faces are to be
painted. If the rate of polishing is 20 paise per
cm
2
and the rate of painting is 10 paise per cm
2
,
find the total expenses required for polishing and
painting the surface of the bookshelf.
*
These exercises are not from examination point of view.
SURFACE AREAS AND VOLUMES 237
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-13\Chap-13 (02-01-06).PM65
2. The front compound wall of a house is
decorated by wooden spheres of diameter 21
cm, placed on small supports as shown in Fig
13.32. Eight such spheres are used for this
purpose, and are to be painted silver. Each
support is a cylinder of radius 1.5 cm and height
7 cm and is to be painted black. Find the cost
of paint required if silver paint costs 25 paise
per cm
2
and black paint costs 5 paise per cm
2
.
3. The diameter of a sphere is decreased by 25%. By what per cent does its curved
surface area decrease?
13.10 Summary
In this chapter, you have studied the following points:
1. Surface area of a cuboid = 2 (lb + bh + hl)
2. Surface area of a cube = 6a
2
3. Curved surface area of a cylinder = 2π ππ ππrh
4. Total surface area of a cylinder = 2π ππ ππr(r + h)
5. Curved surface area of a cone = π ππ ππrl
6. Total surface area of a right circular cone = π ππ ππrl + π ππ ππr
2
, i.e., π ππ ππr (l + r)
7. Surface area of a sphere of radius r = 4 π ππ ππ r
2
8. Curved surface area of a hemisphere = 2π ππ ππr
2
9. Total surface area of a hemisphere = 3π ππ ππr
2
10. Volume of a cuboid = l × b × h
11. Volume of a cube = a
3
12. Volume of a cylinder = π ππ ππr
2
h
13. Volume of a cone =
1
3
π ππ ππr
2
h
14. Volume of a sphere of radius r =
3
4
3
r π
15. Volume of a hemisphere =
3
2
3
r π
[Here, letters l, b, h, a, r, etc. have been used in their usual meaning, depending on the
context.]
Fig. 13.32
238 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
CHAPTER 14
STATISTICS
14.1 Introduction
Everyday we come across a wide variety of informations in the form of facts, numerical
figures, tables, graphs, etc. These are provided by newspapers, televisions, magazines
and other means of communication. These may relate to cricket batting or bowling
averages, profits of a company, temperatures of cities, expenditures in various sectors
of a five year plan, polling results, and so on. These facts or figures, which are numerical
or otherwise, collected with a definite purpose are called data. Data is the plural form
of the Latin word datum. Of course, the word 'data' is not new for you. You have
studied about data and data handling in earlier classes.
Our world is becoming more and more information oriented. Every part of our
lives utilises data in one form or the other. So, it becomes essential for us to know how
to extract meaningful information from such data. This extraction of meaningful
information is studied in a branch of mathematics called Statistics.
The word 'statistics' appears to have been derived from the Latin word 'status'
meaning 'a (political) state'. In its origin, statistics was simply the collection of data on
different aspects of the life of people, useful to the State. Over the period of time,
however, its scope broadened and statistics began to concern itself not only with the
collection and presentation of data but also with the interpretation and drawing of
inferences from the data. Statistics deals with collection, organisation, analysis and
interpretation of data. The word 'statistics' has different meanings in different contexts.
Let us observe the following sentences:
1. May I have the latest copy of 'Educational Statistics of India'.
2. I like to study 'Statistics' because it is used in day-to-day life.
In the first sentence, statistics is used in a plural sense, meaning numerical data. These
may include a number of educational institutions of India, literacy rates of various
STATISTICS 239
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
states, etc. In the second sentence, the word 'statistics' is used as a singular noun,
meaning the subject which deals with the collection, presentation, analysis of data as
well as drawing of meaningful conclusions from the data.
In this chapter, we shall briefly discuss all these aspects regarding data.
14.2 Collection of Data
Let us begin with an exercise on gathering data by performing the following activity.
Activity 1 : Divide the students of your class into four groups. Allot each group the
work of collecting one of the following kinds of data:
(i) Heights of 20 students of your class.
(ii) Number of absentees in each day in your class for a month.
(iii) Number of members in the families of your classmates.
(iv) Heights of 15 plants in or around your school.
Let us move to the results students have gathered. How did they collect their data
in each group?
(i) Did they collect the information from each and every student, house or person
concerned for obtaining the information?
(ii) Did they get the information from some source like available school records?
In the first case, when the information was collected by the investigator herself or
himself with a definite objective in her or his mind, the data obtained is called primary
data.
In the second case, when the information was gathered from a source which
already had the information stored, the data obtained is called secondary data. Such
data, which has been collected by someone else in another context, needs to be used
with great care ensuring that the source is reliable.
By now, you must have understood how to collect data and distinguish between
primary and secondary data.
EXERCISE 14.1
1. Give five examples of data that you can collect from your day-to-day life.
2. Classify the data in Q.1 above as primary or secondary data.
240 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
14.3 Presentation of Data
As soon as the work related to collection of data is over, the investigator has to find out
ways to present them in a form which is meaningful, easily understood and gives its
main features at a glance. Let us now recall the various ways of presenting the data
through some examples.
Example 1 : Consider the marks obtained by 10 students in a mathematics test as
given below:
55 36 95 73 60 42 25 78 75 62
The data in this form is called raw data.
By looking at it in this form, can you find the highest and the lowest marks?
Did it take you some time to search for the maximum and minimum scores? Wouldn't
it be less time consuming if these scores were arranged in ascending or descending
order? So let us arrange the marks in ascending order as
25 36 42 55 60 62 73 75 78 95
Now, we can clearly see that the lowest marks are 25 and the highest marks are 95.
The difference of the highest and the lowest values in the data is called the range of the
data. So, the range in this case is 95 – 25 = 70.
Presentation of data in ascending or descending order can be quite time consuming,
particularly when the number of observations in an experiment is large, as in the case
of the next example.
Example 2 : Consider the marks obtained (out of 100 marks) by 30 students of Class
IX of a school:
10 20 36 92 95 40 50 56 60 70
92 88 80 70 72 70 36 40 36 40
92 40 50 50 56 60 70 60 60 88
Recall that the number of students who have obtained a certain number of marks is
called the frequency of those marks. For instance, 4 students got 70 marks. So the
frequency of 70 marks is 4. To make the data more easily understandable, we write it
STATISTICS 241
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
in a table, as given below:
Table 14.1
Marks Number of students
(i.e., the frequency)
10 1
20 1
36 3
40 4
50 3
56 2
60 4
70 4
72 1
80 1
88 2
92 3
95 1
Total 30
Table 14.1 is called an ungrouped frequency distribution table, or simply a frequency
distribution table. Note that you can use also tally marks in preparing these tables,
as in the next example.
Example 3 : 100 plants each were planted in 100 schools during Van Mahotsava.
After one month, the number of plants that survived were recorded as :
95 67 28 32 65 65 69 33 98 96
76 42 32 38 42 40 40 69 95 92
75 83 76 83 85 62 37 65 63 42
89 65 73 81 49 52 64 76 83 92
93 68 52 79 81 83 59 82 75 82
86 90 44 62 31 36 38 42 39 83
87 56 58 23 35 76 83 85 30 68
69 83 86 43 45 39 83 75 66 83
92 75 89 66 91 27 88 89 93 42
53 69 90 55 66 49 52 83 34 36
242 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
To present such a large amount of data so that a reader can make sense of it easily,
we condense it into groups like 20-29, 30-39, . . ., 90-99 (since our data is from
23 to 98). These groupings are called 'classes' or 'class-intervals', and their size is
called the class-size or class width, which is 10 in this case. In each of these classes,
the least number is called the lower class limit and the greatest number is called the
upper class limit, e.g., in 20-29, 20 is the 'lower class limit' and 29 is the 'upper class
limit'.
Also, recall that using tally marks, the data above can be condensed in tabular
form as follows:
Table 14.2
Number of plants Tally Marks Number of schools
survived (frequency)
20 - 29 ||| 3
30 - 39 |||| |||| |||| 14
40 - 49 |||| |||| || 12
50 - 59 |||| ||| 8
60 - 69 |||| |||| |||| ||| 18
70 - 79 |||| |||| 10
80 - 89 |||| |||| |||| |||| ||| 23
90 - 99 |||| |||| || 12
Total 100
Presenting data in this form simplifies and condenses data and enables us to observe
certain important features at a glance. This is called a grouped frequency distribution
table. Here we can easily observe that 50% or more plants survived in 8 + 18 + 10 +
23 + 12 = 71 schools.
We observe that the classes in the table above are non-overlapping. Note that we
could have made more classes of shorter size, or fewer classes of larger size also. For
instance, the intervals could have been 22-26, 27-31, and so on. So, there is no hard
and fast rule about this except that the classes should not overlap.
Example 4 : Let us now consider the following frequency distribution table which
gives the weights of 38 students of a class:
STATISTICS 243
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Table 14.3
Weights (in kg) Number of students
31 - 35 9
36 - 40 5
41 - 45 14
46 - 50 3
51 - 55 1
56 - 60 2
61 - 65 2
66 - 70 1
71 - 75 1
Total 38
Now, if two new students of weights 35.5 kg and 40.5 kg are admitted in this class,
then in which interval will we include them? We cannot add them in the ones ending
with 35 or 40, nor to the following ones. This is because there are gaps in between the
upper and lower limits of two consecutive classes. So, we need to divide the intervals
so that the upper and lower limits of consecutive intervals are the same. For this, we
find the difference between the upper limit of a class and the lower limit of its succeeding
class. We then add half of this difference to each of the upper limits and subtract the
same from each of the lower limits.
For example, consider the classes 31 - 35 and 36 - 40.
The lower limit of 36 - 40 = 36
The upper limit of 31 - 35 = 35
The difference = 36 – 35 = 1
So, half the difference =
1
2
= 0.5
So the new class interval formed from 31 - 35 is (31 – 0.5) - (35 + 0.5), i.e., 30.5 - 35.5.
Similarly, the new class formed from the class 36 - 40 is (36 – 0.5) - (40 + 0.5), i.e.,
35.5 - 40.5.
Continuing in the same manner, the continuous classes formed are:
30. 5- 35. 5, 35. 5- 40. 5, 40. 5- 45. 5, 45. 5- 50. 5, 50. 5- 55. 5, 55. 5- 60. 5,
60.5 - 65.5, 65.5 - 70.5, 70.5 - 75.5.
244 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Now it is possible for us to include the weights of the new students in these classes.
But, another problem crops up because 35.5 appears in both the classes 30.5 - 35.5
and 35.5 - 40.5. In which class do you think this weight should be considered?
If it is considered in both classes, it will be counted twice.
By convention, we consider 35.5 in the class 35.5 - 40.5 and not in 30.5 - 35.5.
Similarly, 40.5 is considered in 40.5 - 45.5 and not in 35.5 - 40.5.
So, the new weights 35.5 kg and 40.5 kg would be included in 35.5 - 40.5 and
40.5 - 45.5, respectively. Now, with these assumptions, the new frequency distribution
table will be as shown below:
Table 14.4
Weights (in kg) Number of students
30.5-35.5 9
35.5-40.5 6
40.5-45.5 15
45.5-50.5 3
50.5-55.5 1
55.5-60.5 2
60.5-65.5 2
65.5-70.5 1
70.5-75.5 1
Total 40
Now, let us move to the data collected by you in Activity 1. This time we ask you to
present these as frequency distribution tables.
Activity 2 : Continuing with the same four groups, change your data to frequency
distribution tables.Choose convenient classes with suitable class-sizes, keeping in mind
the range of the data and the type of data.
STATISTICS 245
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
EXERCISE 14.2
1. The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table. Which is the most
common, and which is the rarest, blood group among these students?
2. The distance (in km) of 40 engineers from their residence to their place of work were
found as follows:
5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the data given
above taking the first interval as 0-5 (5 not included). What main features do you
observe from this tabular representation?
3. The relative humidity (in %) of a certain city for a month of 30 days was as follows:
98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1
89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3
96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89
(i) Construct a grouped frequency distribution table with classes 84 - 86, 86 - 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
4. The heights of 50 students, measured to the nearest centimetres, have been found to
be as follows:
161 150 154 165 168 161 154 162 150 151
162 164 171 165 158 154 156 172 160 170
153 159 161 170 162 165 166 168 165 164
154 152 153 156 158 162 160 161 173 166
161 159 162 167 168 159 158 153 154 159
(i) Represent the data given above by a grouped frequency distribution table, taking
the class intervals as 160 - 165, 165 - 170, etc.
(ii) What can you conclude about their heights from the table?
5. A study was conducted to find out the concentration of sulphur dioxide in the air in
246 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17
0.16 0.05 0.02 0.06 0.18 0.20
0.11 0.08 0.12 0.13 0.22 0.07
0.08 0.01 0.10 0.06 0.09 0.18
0.11 0.07 0.05 0.07 0.01 0.04
(i) Make a grouped frequency distribution table for this data with class intervals as
0.00 - 0.04, 0.04 - 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11
parts per million?
6. Three coins were tossed 30 times simultaneously. Each time the number of heads
occurring was noted down as follows:
0 1 2 2 1 2 3 1 3 0
1 3 1 1 2 2 0 1 2 1
3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above.
7. The value of π upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
8. Thirty children were asked about the number of hours they watched TV programmes
in the previous week. The results were found as follows:
1 6 2 3 5 12 5 8 4 8
10 3 4 12 2 8 15 1 17 6
3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class width 5
and one of the class intervals as 5 - 10.
(ii) How many children watched television for 15 or more hours a week?
9. A company manufactures car batteries of a particular type. The lives (in years) of 40
such batteries were recorded as follows:
2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5
3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7
2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8
3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4
4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
Construct a grouped frequency distribution table for this data, using class intervals
of size 0.5 starting from the interval 2 - 2.5.
STATISTICS 247
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
14.4 Graphical Representation of Data
The representation of data by tables has already been discussed. Now let us turn our
attention to another representation of data, i.e., the graphical representation. It is well
said that one picture is better than a thousand words. Usually comparisons among the
individual items are best shown by means of graphs. The representation then becomes
easier to understand than the actual data. We shall study the following graphical
representations in this section.
(A) Bar graphs
(B) Histograms of uniform width, and of varying widths
(C) Frequency polygons
(A) Bar Graphs
In earlier classes, you have already studied and constructed bar graphs. Here we
shall discuss them through a more formal approach. Recall that a bar graph is a
pictorial representation of data in which usually bars of uniform width are drawn with
equal spacing between them on one axis (say, the x-axis), depicting the variable. The
values of the variable are shown on the other axis (say, the y-axis) and the heights of
the bars depend on the values of the variable.
Example 5 : In a particular section of Class IX, 40 students were asked about the
months of their birth and the following graph was prepared for the data so obtained:
Fig. 14.1
Observe the bar graph given above and answer the following questions:
(i) How many students were born in the month of November?
(ii) In which month were the maximum number of students born?
248 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Solution : Note that the variable here is the 'month of birth', and the value of the
variable is the 'Number of students born'.
(i) 4 students were born in the month of November.
(ii) The Maximum number of students were born in the month of August.
Let us now recall how a bar graph is constructed by considering the following example.
Example 6 : A family with a monthly income of Rs 20,000 had planned the following
expenditures per month under various heads:
Table 14.5
Heads Expenditure
(in thousand rupees)
Grocery 4
Rent 5
Education of children 5
Medicine 2
Fuel 2
Entertainment 1
Miscellaneous 1
Draw a bar graph for the data above.
Solution : We draw the bar graph of this data in the following steps. Note that the unit
in the second column is thousand rupees. So, '4' against 'grocery' means Rs 4000.
1. We represent the Heads (variable) on the horizontal axis choosing any scale,
since the width of the bar is not important. But for clarity, we take equal widths
for all bars and maintain equal gaps in between. Let one Head be represented by
one unit.
2. We represent the expenditure (value) on the vertical axis. Since the maximum
expenditure is Rs 5000, we can choose the scale as 1 unit = Rs 1000.
3. To represent our first Head, i.e., grocery, we draw a rectangular bar with width
1 unit and height 4 units.
4. Similarly, other Heads are represented leaving a gap of 1 unit in between two
consecutive bars.
The bar graph is drawn in Fig. 14.2.
STATISTICS 249
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Fig. 14.2
Here, you can easily visualise the relative characteristics of the data at a glance, e.g.,
the expenditure on education is more than double that of medical expenses. Therefore,
in some ways it serves as a better representation of data than the tabular form.
Activity 3 : Continuing with the same four groups of Activity 1, represent the data by
suitable bar graphs.
Let us now see how a frequency distribution table for continuous class intervals
can be represented graphically.
(B) Histogram
This is a form of representation like the bar graph, but it is used for continuous class
intervals. For instance, consider the frequency distribution Table 14.6, representing
the weights of 36 students of a class:
Table 14.6
Weights (in kg) Number of students
30.5 - 35.5 9
35.5 - 40.5 6
40.5 - 45.5 15
45.5 - 50.5 3
50.5 - 55.5 1
55.5 - 60.5 2
Total 36
250 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Let us represent the data given above graphically as follows:
(i) We represent the weights on the horizontal axis on a suitable scale. We can choose
the scale as 1 cm = 5 kg. Also, since the first class interval is starting from 30.5
and not zero, we show it on the graph by marking a kink or a break on the axis.
(ii) We represent the number of students (frequency) on the vertical axis on a suitable
scale. Since the maximum frequency is 15, we need to choose the scale to
accomodate this maximum frequency.
(iii) We now draw rectangles (or rectangular bars) of width equal to the class-size
and lengths according to the frequencies of the corresponding class intervals. For
example, the rectangle for the class interval 30.5 - 35.5 will be of width 1 cm and
length 4.5 cm.
(iv) In this way, we obtain the graph as shown in Fig. 14.3:
Fig. 14.3
Observe that since there are no gaps in between consecutive rectangles, the resultant
graph appears like a solid figure. This is called a histogram, which is a graphical
representation of a grouped frequency distribution with continuous classes. Also, unlike
a bar graph, the width of the bar plays a significant role in its construction.
Here, in fact, areas of the rectangles erected are proportional to the corresponding
frequencies. However, since the widths of the rectangles are all equal, the lengths of
the rectangles are proportional to the frequencies. That is why, we draw the lengths
according to (iii) above.
STATISTICS 251
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Now, consider a situation different from the one above.
Example 7 : A teacher wanted to analyse the performance of two sections of students
in a mathematics test of 100 marks. Looking at their performances, she found that a
few students got under 20 marks and a few got 70 marks or above. So she decided to
group them into intervals of varying sizes as follows: 0 - 20, 20 - 30, . . ., 60 - 70,
70 - 100. Then she formed the following table:
Table 14.7
Marks Number of students
0 - 20 7
20 - 30 10
30 - 40 10
40 - 50 20
50 - 60 20
60 - 70 15
70 - above 8
Total 90
A histogram for this table was prepared by a student as shown in Fig. 14.4.
Fig. 14.4
252 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Carefully examine this graphical representation. Do you think that it correctly represents
the data? No, the graph is giving us a misleading picture. As we have mentioned
earlier, the areas of the rectangles are proportional to the frequencies in a histogram.
Earlier this problem did not arise, because the widths of all the rectangles were equal.
But here, since the widths of the rectangles are varying, the histogram above does not
give a correct picture. For example, it shows a greater frequency in the interval
70 - 100, than in 60 - 70, which is not the case.
So, we need to make certain modifications in the lengths of the rectangles so that
the areas are again proportional to the frequencies.
The steps to be followed are as given below:
1. Select a class interval with the minimum class size. In the example above, the
minimum class-size is 10.
2. The lengths of the rectangles are then modified to be proportionate to the
class size 10.
For, instance, when the class size is 20, the length of the rectangle is 7. So when
the class size is 10, the length of the rectangle will be
7
10
20
×
= 3.5.
Similarly, proceeding in this manner, we get the following table:
Table 14.8
Marks Frequency Width of Length of the rectangle
the class
0 - 20 7 20
7
10
20
×
= 3.5
20 - 30 10 10
10
10
10
×
= 10
30 - 40 10 10
10
10
10
×
= 10
40 - 50 20 10
20
10
10
×
= 20
50 - 60 20 10
20
10
10
×
= 20
60 - 70 15 10
15
10
10
×
= 15
70 - 100 8 30
8
10
30
×
= 2.67
STATISTICS 253
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Since we have calculated these lengths for an interval of 10 marks in each case,
we may call these lengths as "proportion of students per 10 marks interval".
So, the correct histogram with varying width is given in Fig. 14.5.
Fig. 14.5
(C) Frequency Polygon
There is yet another visual way of representing quantitative data and its frequencies.
This is a polygon. To see what we mean, consider the histogram represented by
Fig. 14.3. Let us join the mid-points of the upper sides of the adjacent rectangles of
this histogram by means of line segments. Let us call these mid-points B, C, D, E, F
and G. When joined by line segments, we obtain the figure BCDEFG (see Fig. 14.6).
To complete the polygon, we assume that there is a class interval with frequency zero
before 30.5 - 35.5, and one after 55.5 - 60.5, and their mid-points are A and H,
respectively. ABCDEFGH is the frequency polygon corresponding to the data shown
in Fig. 14.3. We have shown this in Fig. 14.6.
254 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Fig. 14.6
Although, there exists no class preceding the lowest class and no class succeeding
the highest class, addition of the two class intervals with zero frequency enables us to
make the area of the frequency polygon the same as the area of the histogram. Why
is this so? (Hint : Use the properties of congruent triangles.)
Now, the question arises: how do we complete the polygon when there is no class
preceding the first class? Let us consider such a situation.
Example 8 : Consider the marks, out of 100, obtained by 51 students of a class in a
test, given in Table 14.9.
STATISTICS 255
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Table 14.9
Marks Number of students
0 - 10 5
10 - 20 10
20 - 30 4
30 - 40 6
40 - 50 7
50 - 60 3
60 - 70 2
70 - 80 2
80 - 90 3
90 - 100 9
Total 51
Draw a frequency polygon corresponding to this frequency distribution table.
Solution : Let us first draw a histogram for this data and mark the mid-points of the
tops of the rectangles as B, C, D, E, F, G, H, I, J, K, respectively. Here, the first class is
0-10. So, to find the class preceeding 0-10, we extend the horizontal axis in the negative
direction and find the mid-point of the imaginary class-interval (–10) - 0. The first end
point, i.e., B is joined to this mid-point with zero frequency on the negative direction of
the horizontal axis. The point where this line segment meets the vertical axis is marked
as A. Let L be the mid-point of the class succeeding the last class of the given data.
Then OABCDEFGHIJKL is the frequency polygon, which is shown in Fig. 14.7.
Fig. 14.7
256 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Frequency polygons can also be drawn independently without drawing
histograms. For this, we require the mid-points of the class-intervals used in the data.
These mid-points of the class-intervals are called class-marks.
To find the class-mark of a class interval, we find the sum of the upper limit and
lower limit of a class and divide it by 2. Thus,
Class-mark =
Upper limit + Lower limit
2
Let us consider an example.
Example 9 : In a city, the weekly observations made in a study on the cost of living
index are given in the following table:
Table 14.10
Cost of living index Number of weeks
140 - 150 5
150 - 160 10
160 - 170 20
170 - 180 9
180 - 190 6
190 - 200 2
Total 52
Draw a frequency polygon for the data above (without constructing a histogram).
Solution : Since we want to draw a frequency polygon without a histogram, let us find
the class-marks of the classes given above, that is of 140 - 150, 150 - 160,....
For 140 - 150, the upper limit = 150, and the lower limit = 140
So, the class-mark =
150 + 140
2
=
290
2
= 145.
Continuing in the same manner, we find the class-marks of the other classes as well.
STATISTICS 257
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
So, the new table obtained is as shown in the following table:
Table 14.11
Classes Class-marks Frequency
140 - 150 145 5
150 - 160 155 10
160 - 170 165 20
170 - 180 175 9
180 - 190 185 6
190 - 200 195 2
Total 52
We can now draw a frequency polygon by plotting the class-marks along the horizontal
axis, the frequencies along the vertical-axis, and then plotting and joining the points
B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6) and G(195, 2) by line segments.
We should not forget to plot the point corresponding to the class-mark of the class
130 - 140 (just before the lowest class 140 - 150) with zero frequency, that is,
A(135, 0), and the point H (205, 0) occurs immediately after G(195, 2). So, the resultant
frequency polygon will be ABCDEFGH (see Fig. 14.8).
Fig. 14.8
258 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Frequency polygons are used when the data is continuous and very large. It is
very useful for comparing two different sets of data of the same nature, for example,
comparing the performance of two different sections of the same class.
EXERCISE 14.3
1. A survey conducted by an organisation for the cause of illness and death among
the women between the ages 15 - 44 (in years) worldwide, found the following
figures (in %):
S.No. Causes Female fatality rate (%)
1. Reproductive health conditions 31.8
2. Neuropsychiatric conditions 25.4
3. Injuries 12.4
4. Cardiovascular conditions 4.3
5. Respiratory conditions 4.1
6. Other causes 22.0
(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women's ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major
role in the cause in (ii) above being the major cause.
2. The following data on the number of girls (to the nearest ten) per thousand boys in
different sections of Indian society is given below.
Section Number of girls per thousand boys
Scheduled Caste (SC) 940
Scheduled Tribe (ST) 970
Non SC/ST 920
Backward districts 950
Non-backward districts 920
Rural 930
Urban 910
STATISTICS 259
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
3. Given below are the seats won by different political parties in the polling outcome of
a state assembly elections:
Political Party A B C D E F
Seats Won 75 55 37 29 10 37
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
4. The length of 40 leaves of a plant are measured correct to one millimetre, and the
obtained data is represented in the following table:
Length (in mm) Number of leaves
118 - 126 3
127 - 135 5
136 - 144 9
145 - 153 12
154 - 162 5
163 - 171 4
172 - 180 2
(i) Draw a histogram to represent the given data.
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long?
Why?
5. The following table gives the life times of 400 neon lamps:
Life time (in hours) Number of lamps
300 - 400 14
400 - 500 56
500 - 600 60
600 - 700 86
700 - 800 74
800 - 900 62
900 - 1000 48
260 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
6. The following table gives the distribution of students of two sections according to
the marks obtained by them:
Section A Section B
Marks Frequency Marks Frequency
0 - 10 3 0 - 10 5
10 - 20 9 10 - 20 19
20 - 30 17 20 - 30 15
30 - 40 12 30 - 40 10
40 - 50 9 40 - 50 1
Represent the marks of the students of both the sections on the same graph by two
frequency polygons. From the two polygons compare the performance of the two
sections.
7. The runs scored by two teams A and B on the first 60 balls in a cricket match are given
below:
Number of balls Team A Team B
1 - 6 2 5
7 - 12 1 6
13 - 18 8 2
19 - 24 9 10
25 - 30 4 5
31 - 36 5 6
37 - 42 6 3
43 - 48 10 4
49 - 54 6 8
55 - 60 2 10
Represent the data of both the teams on the same graph by frequency polygons.
[Hint : First make the class intervals continuous.]
STATISTICS 261
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
8. A random survey of the number of children of various age groups playing in a park
was found as follows:
Age (in years) Number of children
1 - 2 5
2 - 3 3
3 - 5 6
5 - 7 12
7 - 10 9
10 - 15 10
15 - 17 4
Draw a histogram to represent the data above.
9. 100 surnames were randomly picked up from a local telephone directory and a frequency
distribution of the number of letters in the English alphabet in the surnames was found
as follows:
Number of letters Number of surnames
1 - 4 6
4 - 6 30
6 - 8 44
8 - 12 16
12 - 20 4
(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
14.5 Measures of Central Tendency
Earlier in this chapter, we represented the data in various forms through frequency
distribution tables, bar graphs, histograms and frequency polygons. Now, the question
arises if we always need to study all the data to 'make sense' of it, or if we can make
out some important features of it by considering only certain representatives of the
data. This is possible, by using measures of central tendency or averages.
Consider a situation when two students Mary and Hari received their test copies.
The test had five questions, each carrying ten marks. Their scores were as follows:
Question Numbers 1 2 3 4 5
Mary's score 10 8 9 8 7
Hari's score 4 7 10 10 10
262 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Upon getting the test copies, both of them found their average scores as follows:
Mary's average score =
42
5
= 8.4
Hari's average score =
41
5
= 8.2
Since Mary's average score was more than Hari's, Mary claimed to have performed
better than Hari, but Hari did not agree. He arranged both their scores in ascending
order and found out the middle score as given below:
Mary's Score 7 8 8 9 10
Hari's Score 4 7 10 10 10
Hari said that since his middle-most score was 10, which was higher than Mary's
middle-most score, that is 8, his performance should be rated better.
But Mary was not convinced. To convince Mary, Hari tried out another strategy.
He said he had scored 10 marks more often (3 times) as compared to Mary who
scored 10 marks only once. So, his performance was better.
Now, to settle the dispute between Hari and Mary, let us see the three measures
they adopted to make their point.
The average score that Mary found in the first case is the mean. The 'middle'
score that Hari was using for his argument is the median. The most often scored mark
that Hari used in his second strategy is the mode.
Now, let us first look at the mean in detail.
The mean (or average) of a number of observations is the sum of the values of
all the observations divided by the total number of observations.
It is denoted by the symbol x , read as 'x bar'.
Let us consider an example.
Example 10 : 5 people were asked about the time in a week they spend in doing
social work in their community. They said 10, 7, 13, 20 and 15 hours, respectively.
Find the mean (or average) time in a week devoted by them for social work.
Solution : We have already studied in our earlier classes that the mean of a certain
number of observations is equal to
Sum of all the observations
Total number of observations
. To simplify our
STATISTICS 263
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
working of finding the mean, let us use a variable x
i
to denote the ith observation. In
this case, i can take the values from 1 to 5. So our first observation is x
1
, second
observation is x
2
, and so on till x
5
.
Also x
1
= 10 means that the value of the first observation, denoted by x
1
, is 10.
Similarly, x
2
= 7, x
3
= 13, x
4
= 20 and x
5
= 15.
Therefore, the mean x =
Sum of all the observations
Total number of observations
=
1 2 3 4 5
5
x x x x x + + + +
=
10 7 13 20 15
5
+ + + +
=
65
5
= 13
So, the mean time spent by these 5 people in doing social work is 13 hours in a week.
Now, in case we are finding the mean time spent by 30 people in doing social
work, writing x
1
+ x
2
+ x
3
+ . . . + x
30
would be a tedious job.We use the Greek symbol
Σ (for the letter Sigma) for summation. Instead of writing x
1
+ x
2
+ x
3
+ . . . + x
30
, we
write
30
1
i
i
x
=
∑
, which is read as 'the sum of x
i
as i varies from 1 to 30'.
So, x =
30
1
30
i
i
x
=
∑
Similarly, for n observations x =
1
n
i
i
x
n
=
∑
Example 11 : Find the mean of the marks obtained by 30 students of Class IX of a
school, given in Example 2.
Solution : Now, x =
1 2 30
30
x x x + + + "
30
1
i
i
x
=
∑
= 10 + 20 + 36 + 92 + 95 + 40 + 50 + 56 + 60 + 70 + 92 + 88
80 + 70 + 72 + 70 + 36 + 40 + 36 + 40 + 92 + 40 + 50 + 50
56 + 60 + 70 + 60 + 60 + 88 = 1779
So, x =
1779
30
= 59.3
264 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Is the process not time consuming? Can we simplify it? Note that we have formed
a frequency table for this data (see Table 14.1).
The table shows that 1 student obtained 10 marks, 1 student obtained 20 marks, 3
students obtained 36 marks, 4 students obtained 40 marks, 3 students obtained 50
marks, 2 students obtained 56 marks, 4 students obtained 60 marks, 4 students obtained
70 marks, 1 student obtained 72 marks, 1 student obtained 80 marks, 2 students obtained
88 marks, 3 students obtained 92 marks and 1 student obtained 95 marks.
So, the total marks obtained = (1 × 10) + (1 × 20) + (3 × 36) + (4 × 40) + (3 × 50)
+ (2 × 56) + (4 × 60) + (4 × 70) + (1 × 72) + (1 × 80)
+ (2 × 88) + (3 × 92) + (1 × 95)
= f
1
x
1
+ . . . + f
13
x
13
, where f
i
is the frequency of the ith
entry inTable 14.1.
In brief, we write this as
13
1
i i
i
f x
=
∑
.
So, the total marks obtained =
13
1
i i
i
f x
=
∑
= f
1
+ f
2
+ . . . + f
13
= 10 + 20 + 108 + 160 + 150 + 112 + 240 + 280 + 72 + 80
+ 176 + 276 + 95
= 1779
Now, the total number of observations
=
13
1
i
i
f
=
∑
= f
1
+ f
2
+ . . . + f
13
= 1 + 1 + 3 + 4 + 3 + 2 + 4 + 4 + 1 + 1 + 2 + 3 + 1
= 30
So, the mean x =
Sum of all the observations
Total number of observations
=
13
1
13
1
i i
i
i
i
f x
f
=
=
⎛ ⎞
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ ⎠
∑
∑
=
1779
30
This process can be displayed in the following table, which is a modified form of
Table 14.1.
STATISTICS 265
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Table 14.12
Marks Number of students f
i
x
i
(x
i
) ( f
i
)
10 1 10
20 1 20
36 3 108
40 4 160
50 3 150
56 2 112
60 4 240
70 4 280
72 1 72
80 1 80
88 2 176
92 3 276
95 1 95
13
1
30
i
i
f
=
=
∑
13
1
1779
i i
i
f x
=
=
∑
Thus, in the case of an ungrouped frequency distribution, you can use the formula
x =
1
1
n
i i
i
n
i
i
f x
f
=
=
∑
∑
for calculating the mean.
Let us now move back to the situation of the argument between Hari and Mary,
and consider the second case where Hari found his performance better by finding the
middle-most score. As already stated, this measure of central tendency is called the
median.
The median is that value of the given number of observations, which divides it into
exactly two parts. So, when the data is arranged in ascending (or descending) order
the median of ungrouped data is calculated as follows:
266 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
(i) When the number of observations (n) is odd, the median is the value of the
1
2
n + ⎛ ⎞
⎜ ⎟
⎝ ⎠
th
observation. For example, if n = 13, the value of the
13 1
2
+ ⎛ ⎞
⎜ ⎟
⎝ ⎠
th
, i.e.,
the 7th observation will be the median [see Fig. 14.9 (i)].
(ii) When the number of observations (n) is even, the median is the mean of the
2
n ⎛ ⎞
⎜ ⎟
⎝ ⎠
th
and the
1
2
n ⎛ ⎞
+
⎜ ⎟
⎝ ⎠
th
observations. For example, if n = 16, the mean of the
values of the
16
2
⎛ ⎞
⎜ ⎟
⎝ ⎠
th
and the
16
1
2
⎛ ⎞
+
⎜ ⎟
⎝ ⎠
th
observations, i.e., the mean of the
values of the 8th and 9th observations will be the median [see Fig. 14.9 (ii)].
Fig. 14.9
Let us illustrate this with the help of some examples.
Example 12 : The heights (in cm) of 9 students of a class are as follows:
155 160 145 149 150 147 152 144 148
Find the median of this data.
Solution : First of all we arrange the data in ascending order, as follows:
144 145 147 148 149 150 152 155 160
Since the number of students is 9, an odd number, we find out the median by finding
STATISTICS 267
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
the height of the
1 9 1
th
2 2
n + + ⎛ ⎞ ⎛ ⎞
=
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
th = the 5th student, which is 149 cm.
So, the median, i.e. the medial height is 149 cm.
Example 13 : The points scored by a Kabaddi team in a series of matches are as
follows:
17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28
Find the median of the points scored by the team.
Solution : Arranging the points scored by the team in ascending order, we get
2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48.
There are 16 terms. So there are two middle terms, i.e. the
16
2
th and
16
1
2
⎛ ⎞
+
⎜ ⎟
⎝ ⎠
th, i.e.
the 8th and 9th terms.
So, the median is the mean of the values of the 8th and 9th terms.
i.e, the median =
10 14
2
+
= 12
So, the medial point scored by the Kabaddi team is 12.
Let us again go back to the unsorted dispute of Hari and Mary.
The third measure used by Hari to find the average was the mode.
The mode is that value of the observation which occurs most frequently, i.e., an
observation with the maximum frequency is called the mode.
The readymade garment and shoe industries make great use of this measure of
central tendency. Using the knowledge of mode, these industries decide which size of
the product should be produced in large numbers.
Let us illustrate this with the help of an example.
Example 14 : Find the mode of the following marks (out of 10) obtained by 20
students:
4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4, 9, 10, 10, 3, 4, 7, 6, 9, 9
Solution : We arrange this data in the following form :
2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9, 9, 10, 10
Here 9 occurs most frequently, i.e., four times. So, the mode is 9.
268 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
Example 15 : Consider a small unit of a factory where there are 5 employees : a
supervisor and four labourers. The labourers draw a salary of Rs 5,000 per month
each while the supervisor gets Rs 15,000 per month. Calculate the mean, median and
mode of the salaries of this unit of the factory.
Solution : Mean =
5000 5000 5000 5000 15000
5
+ + + +
=
35000
5
= 7000
So, the mean salary is Rs 7000 per month.
To obtain the median, we arrange the salaries in ascending order:
5000, 5000, 5000, 5000, 15000
Since the number of employees in the factory is 5, the median is given by the
5 1 6
th
2 2
+
= th = 3rd observation. Therefore, the median is Rs 5000 per month.
To find the mode of the salaries, i.e., the modal salary, we see that 5000 occurs the
maximum number of times in the data 5000, 5000, 5000, 5000, 15000. So, the modal
salary is Rs 5000 per month.
Now compare the three measures of central tendency for the given data in the
example above. You can see that the mean salary of Rs 7000 does not give even an
approximate estimate of any one of their wages, while the medial and modal salaries
of Rs 5000 represents the data more effectively.
Extreme values in the data affect the mean. This is one of the weaknesses of the
mean. So, if the data has a few points which are very far from most of the other
points, (like 1,7,8,9,9) then the mean is not a good representative of this data. Since the
median and mode are not affected by extreme values present in the data, they give a
better estimate of the average in such a situation.
Again let us go back to the situation of Hari and Mary, and compare the three
measures of central tendency.
Measures Hari Mary
of central tendency
Mean 8.2 8.4
Median 10 8
Mode 10 8
This comparison helps us in stating that these measures of central tendency are not
sufficient for concluding which student is better. We require some more information to
conclude this, which you will study about in the higher classes.
STATISTICS 269
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
EXERCISE 14.4
1. The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
2. In a mathematics test given to 15 students, the following marks (out of 100) are
recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
3. The following observations have been arranged in ascending order. If the median of
the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
4. Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
5. Find the mean salary of 60 workers of a factory from the following table:
Salary (in Rs) Number of workers
3000 16
4000 12
5000 10
6000 8
7000 6
8000 4
9000 3
10000 1
Total 60
6. Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the median is an
appropriate measure of central tendency.
270 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-14\Chap-14 (02-01-2006).PM65
14.6 Summary
In this chapter, you have studied the following points:
1. Facts or figures, collected with a definite purpose, are called data.
2. Statistics is the area of study dealing with the presentation, analysis and interpretation of
data.
3. How data can be presented graphically in the form of bar graphs, histograms and frequency
polygons.
4. The three measures of central tendency for ungrouped data are:
(i) Mean : It is found by adding all the values of the observations and dividing it by the
total number of observations. It is denoted by x .
So, x =
1
n
i
i
x
n
=
∑
. For an ungrouped frequency distribution, it is
x
=
1
1
n
i i
i
n
i
i
f x
f
=
=
∑
∑
.
(ii) Median : It is the value of the middle-most observation (s).
If n is an odd number, the median = value of the
1
2
n + ⎛ ⎞
⎜ ⎟
⎝ ⎠
th
observation.
If n is an even number, median = Mean of the values of the
2
n ⎛ ⎞
⎜ ⎟
⎝ ⎠
th
and 1
2
n ⎛ ⎞
+
⎜
⎝ ⎠
th
observations.
(iii) Mode : The mode is the most frequently occurring observation.
PROBABILLITY 271
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
CHAPTER 15
PROBABILITY
It is remarkable that a science, which began with the consideration of
games of chance, should be elevated to the rank of the most important
subject of human knowledge. —Pierre Simon Laplace
15.1 Introduction
In everyday life, we come across statements such as
(1) It will probably rain today.
(2) I doubt that he will pass the test.
(3) Most probably, Kavita will stand first in the annual examination.
(4) Chances are high that the prices of diesel will go up.
(5) There is a 50-50 chance of India winning a toss in today's match.
The words 'probably', 'doubt', 'most probably', 'chances', etc., used in the
statements above involve an element of uncertainty. For example, in (1), 'probably
rain' will mean it may rain or may not rain today. We are predicting rain today based
on our past experience when it rained under similar conditions. Similar predictions are
also made in other cases listed in (2) to (5).
The uncertainty of 'probably' etc can be measured numerically by means of
'probability' in many cases.
Though probability started with gambling, it has been used extensively in the fields
of Physical Sciences, Commerce, Biological Sciences, Medical Sciences, Weather
Forecasting, etc.
272 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
15.2 Probability – an Experimental Approach
The concept of probability developed in a very
strange manner. In 1654, a gambler Chevalier
de Mere, approached the well-known 17th
century French philosopher and mathematician
Blaise Pascal regarding certain dice problems.
Pascal became interested in these problems,
studied them and discussed them with another
French mathematician, Pierre de Fermat. Both
Pascal and Fermat solved the problems
independently. This work was the beginning
of Probability Theory.
The first book on the subject was written by the Italian mathematician, J.Cardan
(1501–1576). The title of the book was 'Book on Games of Chance' (Liber de Ludo
Aleae), published in 1663. Notable contributions were also made by mathematicians
J. Bernoulli (1654–1705), P. Laplace (1749–1827), A.A. Markov (1856–1922) and A.N.
Kolmogorov (born 1903).
In earlier classes, you have had a glimpse of probability when you performed
experiments like tossing of coins, throwing of dice, etc., and observed their outcomes.
You will now learn to measure the chance of occurrence of a particular outcome in an
experiment.
Activity 1 : (i) Take any coin, toss it ten times and note down the number of times a
head and a tail come up. Record your observations in the form of the following table
Table 15.1
Number of times Number of times Number of times
the coin is tossed head comes up tail comes up
10 — —
Write down the values of the following fractions:
Number of times a head comes up
Total number of times the coin is tossed
and
Number of times a tail comes up
Total number of times the coin is tossed
Blaise Pascal
(1623–1662)
Fig. 15.1
Pierre de Fermat
(1601–1665)
Fig. 15.2
PROBABILLITY 273
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
(ii) Toss the coin twenty times and in the same way record your observations as
above. Again find the values of the fractions given above for this collection of
observations.
(iii) Repeat the same experiment by increasing the number of tosses and record
the number of heads and tails. Then find the values of the corresponding
fractions.
You will find that as the number of tosses gets larger, the values of the fractions
come closer to 0.5. To record what happens in more and more tosses, the following
group activity can also be performed:
Acitivity 2 : Divide the class into groups of 2 or 3 students. Let a student in each
group toss a coin 15 times. Another student in each group should record the observations
regarding heads and tails. [Note that coins of the same denomination should be used in
all the groups. It will be treated as if only one coin has been tossed by all the groups.]
Now, on the blackboard, make a table like Table 15.2. First, Group 1 can write
down its observations and calculate the resulting fractions. Then Group 2 can write
down its observations, but will calculate the fractions for the combined data of Groups
1 and 2, and so on. (We may call these fractions as cumulative fractions.) We have
noted the first three rows based on the observations given by one class of students.
Table 15.2
Group
Number Number
Cumulative number of heads Cumulative number of tails
of of Total number of times Total number of times
heads tails the coin is tossed the coin is tossed
(1) (2) (3) (4) (5)
1 3 12
3
15
12
15
2 7 8
7 3 10
15 15 30
+
=
+
8 12 20
15 15 30
+
=
+
3 7 8
7 10 17
15 30 45
+
=
+
8 20 28
15 30 45
+
=
+
4
# # # #
What do you observe in the table? You will find that as the total number of tosses
of the coin increases, the values of the fractions in Columns (4) and (5) come nearer
and nearer to 0.5.
Activity 3 : (i) Throw a die* 20 times and note down the number of times the numbers
*A die is a well balanced cube with its six faces marked with numbers from 1 to 6, one number
on one face. Sometimes dots appear in place of numbers.
274 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
1, 2, 3, 4, 5, 6 come up. Record your observations in the form of a table, as in Table 15.3:
Table 15.3
Number of times a die is thrown Number of times these scores turn up
1 2 3 4 5 6
20
Find the values of the following fractions:
Number of times 1 turned up
Total number of times the die is thrown
Number of times 2 turned up
Total number of times the die is thrown
#
#
Number of times 6 turned up
Total number of times the die is thrown
(ii) Now throw the die 40 times, record the observations and calculate the fractions
as done in (i).
As the number of throws of the die increases, you will find that the value of each
fraction calculated in (i) and (ii) comes closer and closer to
1
6
.
To see this, you could perform a group activity, as done in Activity 2. Divide the
students in your class, into small groups. One student in each group should throw a die
ten times. Observations should be noted and cumulative fractions should be calculated.
The values of the fractions for the number 1 can be recorded in Table 15.4. This
table can be extended to write down fractions for the other numbers also or other
tables of the same kind can be created for the other numbers.
PROBABILLITY 275
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
Table 15.4
Group Total number of times a die Cumulative number of times 1 turned up
is thrown in a group Total number of times the die is thrown
(1) (2) (3)
1 — —
2 — —
3 — —
4 — —
The dice used in all the groups should be almost the same in size and appearence.
Then all the throws will be treated as throws of the same die.
What do you observe in these tables?
You will find that as the total number of throws gets larger, the fractions in
Column (3) move closer and closer to
1
6
.
Activity 4 : (i) Toss two coins simultaneously ten times and record your
observations in the form of a table as given below:
Table 15.5
Number of times the Number of times Number of times Number of times
two coins are tossed no head comes up one head comes up two heads come up
10 — — —
Write down the fractions:
A =
Number of times no head comes up
Total number of times two coins are tossed
B =
Number of times one head comes up
Total number of times two coins are tossed
C =
Number of times two heads come up
Total number of times two coins are tossed
Calculate the values of these fractions.
276 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
Now increase the number of tosses (as in Activitiy 2). You will find that the more
the number of tosses, the closer are the values of A, B and C to 0.25, 0.5 and 0.25,
respectively.
In Activity 1, each toss of a coin is called a trial. Similarly in Activity 3, each
throw of a die is a trial, and each simultaneous toss of two coins in Activity 4 is also a
trial.
So, a trial is an action which results in one or several outcomes. The possible
outcomes in Activity 1 were Head and Tail; whereas in Activity 3, the possible outcomes
were 1, 2, 3, 4, 5 and 6.
In Activity 1, the getting of a head in a particular throw is an event with outcome
'head'. Similarly, getting a tail is an event with outcome 'tail'. In Activity 2, the
getting of a particular number, say 1, is an event with outcome 1.
If our experiment was to throw the die for getting an even number, then the event
would consist of three outcomes, namely, 2, 4 and 6.
So, an event for an experiment is the collection of some outcomes of the experiment.
In Class X, you will study a more formal definition of an event.
So, can you now tell what the events are in Activity 4?
With this background, let us now see what probability is. Based on what we directly
observe as the outcomes of our trials, we find the experimental or empirical probability.
Let n be the total number of trials. The empirical probability P(E) of an event E
happening, is given by
P(E) =
Number of trials in which the event happened
The total number of trials
In this chapter, we shall be finding the empirical probability, though we will write
'probability' for convenience.
Let us consider some examples.
To start with let us go back to Activity 2, and Table 15.2. In Column (4) of this
table, what is the fraction that you calculated? Nothing, but it is the empirical probability
of getting a head. Note that this probability kept changing depending on the number of
trials and the number of heads obtained in these trials. Similarly, the empirical probability
of getting a tail is obtained in Column (5) of Table 15.2. This is
12
15
to start with, then
it is
2
3
, then
28
45
, and so on.
So, the empirical probability depends on the number of trials undertaken, and the
number of times the outcomes you are looking for coming up in these trials.
PROBABILLITY 277
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
Activity 5 : Before going further, look at the tables you drew up while doing
Activity 3. Find the probabilities of getting a 3 when throwing a die a certain number
of times. Also, show how it changes as the number of trials increases.
Now let us consider some other examples.
Example 1 : A coin is tossed 1000 times with the following frequencies:
Head : 455, Tail : 545
Compute the probability for each event.
Solution : Since the coin is tossed 1000 times, the total number of trials is 1000. Let us
call the events of getting a head and of getting a tail as E and F, respectively. Then, the
number of times E happens, i.e., the number of times a head come up, is 455.
So, the probability of E =
Number of heads
Total number of trials
i.e., P (E) =
455
1000
= 0.455
Similarly, the probability of the event of getting a tail =
Number of tails
Total number of trials
i.e., P(F) =
545
1000
= 0.545
Note that in the example above, P(E) + P(F) = 0.455 + 0.545 = 1, and E and F are
the only two possible outcomes of each trial.
Example 2 : Two coins are tossed simultaneously 500 times, and we get
Two heads : 105 times
One head : 275 times
No head : 120 times
Find the probability of occurrence of each of these events.
Solution : Let us denote the events of getting two heads, one head and no head by E
1
,
E
2
and E
3
, respectively. So,
P(E
1
) =
105
500
= 0.21
P(E
2
) =
275
500
= 0.55
P(E
3
) =
120
500
= 0.24
278 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
Observe that P(E
1
) + P(E
2
) + P(E
3
) = 1. Also E
1
, E
2
and E
3
cover all the outcomes
of a trial.
Example 3 : A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3,
4, 5 and 6 as given in the following table :
Table 15.6
Outcome 1 2 3 4 5 6
Frequency 179 150 157 149 175 190
Find the probability of getting each outcome.
Solution : Let E
i
denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6.
Then
Probability of the outcome 1 = P(E
1
) =
Frequency of 1
Total number of times the die is thrown
=
179
1000
= 0.179
Similarly, P(E
2
) =
150
1000
= 0.15, P(E
3
) =
157
1000
= 0.157,
P(E
4
) =
149
1000
= 0.149, P(E
5
) =
175
1000
= 0.175
and P(E
6
) =
190
1000
= 0.19.
Note that P(E
1
) + P(E
2
) + P(E
3
) + P(E
4
) + P(E
5
) + P(E
6
) = 1
Also note that:
(i) The probability of each event lies between 0 and 1.
(ii) The sum of all the probabilities is 1.
(iii) E
1
, E
2
, . . ., E
6
cover all the possible outcomes of a trial.
Example 4 : On one page of a telephone directory, there were 200 telephone numbers.
The frequency distribution of their unit place digit (for example, in the number 25828573,
the unit place digit is 3) is given in Table 15.7 :
PROBABILLITY 279
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
Table 15.7
Digit 0 1 2 3 4 5 6 7 8 9
Frequency 22 26 22 22 20 10 14 28 16 20
Without looking at the page, the pencil is placed on one of these numbers, i.e., the
number is chosen at random. What is the probability that the digit in its unit place is 6?
Solution : The probability of digit 6 being in the unit place
=
Frequency of 6
Total number of selected telephone numbers
=
14
200
= 0.07
You can similarly obtain the empirical probabilities of the occurrence of the numbers
having the other digits in the unit place.
Example 5 : The record of a weather station shows that out of the past 250 consecutive
days, its weather forecasts were correct 175 times.
(i) What is the probability that on a given day it was correct?
(ii) What is the probability that it was not correct on a given day?
Solution : The total number of days for which the record is available = 250
(i) P(the forecast was correct on a given day)
=
Number of days when the forecast was correct
Total number of days for which the record is available
=
175
250
= 0.7
(ii) The number of days when the forecast was not correct = 250 – 175 = 75
So, P(the forecast was not correct on a given day) =
75
250
= 0.3
Notice that:
P(forecast was correct on a given day) + P(forecast was not correct on a given day)
= 0.7 + 0.3 = 1
280 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
Example 6 : A tyre manufacturing company kept a record of the distance covered
before a tyre needed to be replaced. The table shows the results of 1000 cases.
Table 15.8
Distance (in km) less than 4000 4000 to 9000 9001 to 14000 more than 14000
Frequency 20 210 325 445
If you buy a tyre of this company, what is the probability that :
(i) it will need to be replaced before it has covered 4000 km?
(ii) it will last more than 9000 km?
(iii) it will need to be replaced after it has covered somewhere between 4000 km
and 14000 km?
Solution : (i) The total number of trials = 1000.
The frequency of a tyre that needs to be replaced before it covers 4000 km is 20.
So, P(tyre to be replaced before it covers 4000 km) =
20
1000
= 0.02
(ii) The frequency of a tyre that will last more than 9000 km is 325 + 445 = 770
So, P(tyre will last more than 9000 km) =
770
1000
= 0.77
(iii) The frequency of a tyre that requires replacement between 4000 km and
14000 km is 210 + 325 = 535.
So, P(tyre requiring replacement between 4000 km and 14000 km) =
535
1000
= 0.535
Example 7 : The percentage of marks obtained by a student in the monthly unit tests
are given below:
Table 15.9
Unit test I II III IV V
Percentage of 69 71 73 68 74
marks obtained
Based on this data, find the probability that the student gets more than 70% marks in
a unit test.
PROBABILLITY 281
File Name : C:\Computer Station\Maths-IX\Chapter\Chap-15\Chap-15 (02-01-2006).PM65
Solution : The total number of unit tests held is 5.
The number of unit tests in which the student obtained more than 70% marks is 3.
So, P(scoring more than 70% marks) =
3
5
= 0.6
Example 8 : An insurance company selected 2000 drivers at random (i.e., without
any preference of one driver over another) in a particular city to find a relationship
between age and accidents. The data obtained are given in the following table:
Table 15.10
Age of drivers Accidents in one year
n + 3 + 4 = n + 7 →
n + 7 – n = 7.
(ii) Note that 7 × 11 × 13 = 1001. Take any three digit number say, abc. Then
abc × 1001 = abcabc. Therefore, the six digit number abcabc is divisible by 7, 11
and 13.
ANSWERS/HINTS 349
File Name : C:\Computer Station\Maths-IX\Chapter\Answers (16–12–2005).PM65
EXERCISE A2.1
1. Step 1: Formulation :
The relevant factors are the time period for hiring a computer, and the two costs given
to us. We assume that there is no significant change in the cost of purchasing or
hiring the computer. So, we treat any such change as irrelevant. We also treat all
brands and generations of computers as the same, i.e. these differences are also
irrelevant.
The expense of hiring the computer for x months is Rs 2000x. If this becomes more
than the cost of purchasing a computer, we will be better off buying a computer. So,
the equation is
2000 x = 25000 (1)
Step 2 : Solution : Solving (1), x =
25000
2000
= 12.5
Step 3 : Interpretation : Since the cost of hiring a computer becomes more after 12.5
months, it is cheaper to buy a computer, if you have to use it for more than 12 months.
2. Step1 : Formulation : We will assume that cars travel at a constant speed. So, any
change of speed will be treated as irrelevant. If the cars meet after x hours, the first car
would have travelled a distance of 40x km from A and the second car would have
travelled 30x km, so that it will be at a distance of (100 – 30x) km from A. So the
equation will be 40x = 100 – 30x, i.e., 70x = 100.
Step 2 : Solution : Solving the equation, we get x =
100
70
.
Step 3 : Interpretation :
100
70
is approximately 1.4 hours. So, the cars will meet after
1.4 hours.
3. Step1: Formulation : The speed at which the moon orbits the earth is
Length of the orbit
Time taken
.
Step 2 : Solution : Since the orbit is nearly circular, the length is 2 × π × 384000 km
= 2411520 km
The moon takes 24 hours to complete one orbit.
So, speed =
2411520
24
= 100480 km/hour.
Step 3 : Interpretation : The speed is 100480 km/h.
4. Formulation : An assumption is that the difference in the bill is only because of using
the water heater.
350 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Answers (16–12–2005).PM65
Let the average number of hours for which the water heater is used = x
Difference per month due to using water heater = Rs 1240 – Rs 1000 = Rs 240
Cost of using water heater for one hour = Rs 8
So, the cost of using the water heater for 30 days = 8 × 30 × x
Also, the cost of using the water heater for 30 days = Difference in bill due to using
water heater
So, 240x = 240
Solution : From this equation, we get x = 1.
Interpretation : Since x = 1, the water heater is used for an average of 1 hour in a day.
EXERCISE A2.2
1. We will not discuss any particular solution here. You can use the same method as we
used in last example, or any other method you think is suitable.
EXERCISE A2.3
1. We have already mentioned that the formulation part could be very detailed in real-
life situations. Also, we do not validate the answer in word problems. Apart from this
word problem have a 'correct answer'. This need not be the case in real-life situations.
2. The important factors are (ii) and (iii). Here (i) is not an important factor although it
can have an effect on the number of vehicles sold.
PROOFS IN MATHEMATICS 287
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
In fact, proofs in mathematics have been in existence for thousands of years, and
they are central to any branch of mathematics. The first known proof is believed to
have been gi ven by t he Gr eek phi l osopher and mat hemat i ci an Thal es. While
mathematics was central to many ancient civilisations like Mesopotamia, Egypt, China
and India, there is no clear evidence that they used proofs the way we do today.
In this chapter, we will look at what a statement is, what kind of reasoning is
involved in mathematics, and what a mathematical proof consists of.
A1.2 Mathematically Acceptable Statements
In this section, we shall try to explain the meaning of a mathematically acceptable
statement. A ' statement' is a sentence which is not an order or an exclamatory sentence.
And, of course, a statement is not a question! For example,
" What is the colour of your hair?" is not a statement, it is a question.
" Please go and bring me some water." is a request or an order, not a statement.
" What a marvellous sunset!" is an exclamatory remark, not a statement.
However, " The colour of your hair is black" is a statement.
In general, statements can be one of the following:
- al way s t rue
- al way s f al se
- ambi guous
The word ' ambiguous' needs some explanation. There are two situations which
make a st at ement ambi guous. The fi rst si t uat i on i s when we cannot deci de i f t he
statement is always true or always false. For example, " Tomorrow is Thursday" is
ambiguous, since enough of a context is not given to us to decide if the statement is
true or false.
The second situation leading to ambiguity is when the statement is subjective, that
is, it is true for some people and not true for others. For example, " Dogs are intelligent"
is ambiguous because some people believe this is true and others do not.
Exampl e 1 : State whether the following statements are always true, always false or
ambiguous. Justify your answers.
(i) There are 8 days in a week.
(ii) It is raining here.
(iii) The sun sets in the west.
288 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
(iv) Gauri is a kind girl.
(v) The product of two odd integers is even.
(vi) The product of two even natural numbers is even.
Sol uti on :
(i) This statement is always false, since there are 7 days in a week.
(ii) This statement is ambiguous, since it is not clear where ' here' is.
(iii) This statement is always true. The sun sets in the west no matter where we live.
(iv) This statement is ambiguous, since it is subjective– Gauri may be kind to some
and not to others.
(v) This statement is always false. The product of two odd integers is always odd.
(vi) This statement is always true. However, to justify that it is true we need to do
some work. It will be proved in Section A1.4.
As mentioned before, in our daily life, we are not so careful about the validity of
statements. For example, suppose your friend tells you that in July it rains everyday in
Manantavadi, Kerala. In all probability, you will believe her, even though it may not
have rained for a day or two in July. Unless you are a lawyer, you will not argue with
her!
As anot her exampl e, consi der st at ement s we
often make to each other like " it is very hot today" .
We easily accept such statements because we know
t he c ont e xt e ve n t hough t he s e s t a t e me nt s a r e
ambiguous. ' It is very hot today' can mean different
things to different people because what is very hot
for a person from Kumaon may not be hot for a person
from Chennai.
But a mathematical statement cannot be ambiguous. In mathematics, a statement
i s onl y accept abl e or v al i d, i f i t i s ei t her t rue or f al se. We say that a statement is
true, if it is always true otherwise it is called a false statement.
For exampl e, 5 + 2 = 7 i s al ways t rue, so ' 5 + 2 = 7' i s a t rue st at ement and
5 + 3 = 7 is a false statement.
PROOFS IN MATHEMATICS 289
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
Exampl e 2 : State whether the following statements are true or false:
(i) The sum of the interior angles of a triangle is 180°.
(ii) Every odd number greater than 1 is prime.
(iii) For any real number x, 4x + x = 5x.
(iv) For every real number x, 2x > x.
(v) For every real number x, x
2
> x.
(vi) If a quadrilateral has all its sides equal, then it is a square.
Sol uti on :
(i) This statement is true. You have already proved this in Chapter 6.
(ii) This statement is false; for example, 9 is not a prime number.
(iii) This statement is true.
(iv) This statement is false; for example, 2 × (– 1) = – 2, and – 2 is not greater than – 1.
(v) This statement is false; for example,
2
1 1
2 4
| |
=
|
\ .
, and
1
4
is not greater than
1
2
.
(vi) This statement is false, since a rhombus has equal sides but need not be a square.
You might have noticed that to establish that a statement is not true according to
mathematics, all we need to do is to find one case or example where it breaks down.
So in (ii), since 9 is not a prime, it is an example that shows that the statement " Every
odd number greater than 1 is prime" is not true. Such an example, that counters a
statement, is called a counter-example. We shall discuss counter-examples in greater
detail in Section A1.5.
You might have also noticed that while Statements (iv), (v) and (vi) are false, they
can be restated with some conditions in order to make them true.
Exampl e 3 : Restate the following statements with appropriate conditions, so that
they become true statements.
(i) For every real number x, 2x > x.
(ii) For every real number x, x
2
>
x.
(iii) If you divide a number by itself, you will always get 1.
(iv) The angle subtended by a chord of a circle at a point on the circle is 90°.
(v) If a quadrilateral has all its sides equal, then it is a square.
290 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
Sol uti on :
(i) If x > 0, then 2x > x.
(ii) If x s 0 or x > 1, then x
2
> x.
(iii) If you divide a number except zero by itself, you will always get 1.
(iv) The angle subtended by a diameter of a circle at a point on the circle is 90°.
(v) If a quadrilateral has all its sides and interior angles equal, then it is a square.
EXERCISE A1.1
1. State whether the following statements are always true, always false or ambiguous.
Justify your answers.
(i) There are 13 months in a year.
(ii) Diwali falls on a Friday.
(iii) The temperature in Magadi is 26
0
C.
(iv) The earth has one moon.
(v) Dogs can fly.
(vi) February has only 28 days.
2. State whether the following statements are true or false. Give reasons for your answers.
(i) The sum of the interior angles of a quadrilateral is 350°.
(ii) For any real number x, x
2
> 0.
(iii) A rhombus is a parallelogram.
(iv) The sum of two even numbers is even.
(v) The sum of two odd numbers is odd.
3. Restate the following statements with appropriate conditions, so that they become
true statements.
(i) All prime numbers are odd.
(ii) Two times a real number is always even.
(iii) For any x, 3x +1 > 4.
(iv) For any x, x
3
> 0.
(v) In every triangle, a median is also an angle bisector.
A1.3 Deductive Reasoning
The main logical tool used in establishing the truth of an unambi guous statement is
deduct i v e reasoni ng. To understand what deductive reasoning is all about, let us
begin with a puzzle for you to solve.
PROOFS IN MATHEMATICS 291
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
You are given four cards. Each card has a number printed on one side and a letter
on the other side.
Suppose you are told that these cards follow the rule:
" If a card has an even number on one side, then it has a vowel on the other side."
What is the smal l est number of cards you need to turn over to check if the rule
is true?
Of course, you have the option of turning over all the cards and checking. But can
you manage with turning over a fewer number of cards?
Notice that the statement mentions that a card with an even number on one side
has a vowel on the other. It does not state that a card with a vowel on one side must
have an even number on the other side. That may or may not be so. The rule also does
not state that a card with an odd number on one side must have a consonant on the
other side. It may or may not.
So, do we need to turn over ' A' ? No! Whether there is an even number or an odd
number on the other side, the rule still holds.
What about ' 5' ? Again we do not need to turn it over, because whether there is a
vowel or a consonant on the other side, the rule still holds.
But you do need to turn over V and 6. If V has an even number on the other side,
then the rule has been broken. Similarly, if 6 has a consonant on the other side, then the
rule has been broken.
The ki nd of reasoni ng we have used t o sol ve t hi s puzzl e i s cal l ed deducti ve
reasoni ng. It is called ' deductive' because we arrive at (i.e., deduce or infer) a result
or a statement from a previously established statement using logic. For example, in the
puzzle above, by a series of logical arguments we deduced that we need to turn over
only V and 6.
Deductive reasoning also helps us to conclude that a particular statement is true,
because it is a special case of a more general statement that is known to be true. For
example, once we prove that the product of two odd numbers is always odd, we can
immediately conclude (without computation) that 70001 × 134563 is odd simply because
70001 and 134563 are odd.
292 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
Deductive reasoning has been a part of human thinking for centuries, and is used
all the time in our daily life. For example, suppose the statements " The flower Solaris
blooms, only if the maximum temperature is above 28° C on the previous day" and
" Solaris bloomed in Imaginary Valley on 15th September, 2005" are true. Then using
deductive reasoning, we can conclude that the maximum temperature in Imaginary
Valley on 14th September, 2005 was more than 28° C.
Unfortunately we do not always use correct reasoning in our daily life! We often
come to many conclusions based on faulty reasoning. For example, if your friend does
not smile at you one day, then you may conclude that she is angry with you. While it
may be true that " if she is angry with me, she will not smile at me" , it may also be true
that " if she has a bad headache, she will not smile at me" . Why don' t you examine
some conclusions that you have arrived at in your day-to-day existence, and see if
they are based on valid or faulty reasoning?
EXERCISE A1.2
1. Use deductive reasoning to answer the following:
(i) Humans are mammals. All mammals are vertebrates. Based on these two
statements, what can you conclude about humans?
(ii) Anthony is a barber. Dinesh had his hair cut. Can you conclude that Antony cut
Dinesh' s hair?
(iii) Martians have red tongues. Gulag is a Martian. Based on these two statements,
what can you conclude about Gulag?
(iv) If it rains for more than four hours on a particular day, the gutters will have to be
cleaned the next day. It has rained for 6 hours today. What can we conclude
about the condition of the gutters tomorrow?
(v) What is the fallacy in the cow' s reasoning in the cartoon below?
All dogs have tails.
I have a tail.
Therefore I am a
dog.
PROOFS IN MATHEMATICS 293
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
2. Once again you are given four cards. Each card has a number printed on one side and
a letter on the other side. Which are the only two cards you need to turn over to check
whether the following rule holds?
" If a card has a consonant on one side, then it has an odd number on the other side."
A1.4 Theorems, Conjectures and Axioms
So far we have discussed statements and how to check their validity. In this section,
you will study how to distinguish between the three different kinds of statements
mathematics is built up from, namely, a theorem, a conjecture and an axiom.
You have already come across many theorems before. So, what is a theorem? A
mathematical statement whose truth has been established (proved) is called a theorem.
For example, the following statements are theorems, as you will see in Section A1.5.
Theorem A1. 1 : The sum of t he i nt eri or angl es of a t ri angl e i s 180º.
Theorem A1. 2 : The product of t wo ev en nat ural numbers i s ev en.
Theorem A1. 3 : The product of any t hree consecut i v e ev en nat ural numbers i s
div isible by 16.
A conjecture is a statement which we believe is true, based on our mathematical
understanding and experience, that is, our mathematical intuition. The conjecture may
t ur n out t o be t r ue or f al s e. I f we can pr ove i t , t hen i t becomes a t heor em.
Mathematicians often come up with conjectures by looking for patterns and making
intelligent mathematical guesses. Let us look at some patterns and see what kind of
intelligent guesses we can make.
Exampl e 4 : Take any three consecutive even numbers and add them, say,
2 + 4 + 6 = 12, 4 + 6 + 8 = 18, 6 + 8 + 10 = 24, 8 +10 + 12 = 30, 20 + 22 + 24 = 66.
Is there any pattern you can guess in these sums? What can you conjecture about
them?
Solution : One conjecture could be :
(i) the sum of three consecutive even numbers is even.
Another could be :
(ii) the sum of three consecutive even numbers is divisible by 6.
294 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
Exampl e 5 : Consider the following pattern of numbers called the Pascal' s Triangle:
Li ne Sum of numbers
1 1 1
2 1 1 2
3 1 2 1 4
4 1 3 3 1 8
5 1 4 6 4 1 16
6 1 5 10 10 5 1 32
7 : : :
8 : : :
What can you conjecture about the sum of the numbers in Lines 7 and 8? What
about the sum of the numbers in Line 21? Do you see a pattern? Make a guess about
a formula for the sum of the numbers in line n.
Solution : Sum of the numbers in Line 7 = 2 × 32 = 64 = 2
6
Sum of the numbers in Line 8 = 2 × 64 = 128 = 2
7
Sum of the numbers in Line 21 = 2
20
Sum of the numbers in Line n = 2
n –1
Exampl e 6 : Consider the so-called triangular numbers T
n
:
Fig. A1.1
The dot s here are arranged i n such a way t hat t hey form a t ri angl e. Here T
1
= 1,
T
2
= 3, T
3
= 6, T
4
= 10, and so on. Can you guess what T
5
is? What about T
6
? What
about T
n
?
T
1
T
2
T
3
T
4
PROOFS IN MATHEMATICS 295
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
Make a conjecture about T
n
.
It might help if you redraw them in the following way.
Fig. A1.2
Sol uti on : T
5
= 1 + 2 + 3 + 4 + 5 = 15 =
5 6
2
×
T
6
= 1 + 2 + 3 + 4 + 5 + 6 = 21 =
6 7
2
×
T
n
=
( 1 )
2
n n × +
A favourite example of a conjecture that has been open (that is, it has not been
proved to be true or false) is the Goldbach conjecture named after the mathematician
Christian Goldbach (1690 – 1764). This conjecture states that "ev ery ev en i nt eger
great er t han 4 can be ex pressed as t he sum of t wo odd pri mes." Perhaps you will
prove that this result is either true or false, and will become famous!
You might have wondered – do we need to prove everything we encounter in
mathematics, and if not, why not?
T
1
T
2
T
3
T
4
296 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
The fact is that every area in mathematics is based on some statements which are
assumed to be true and are not proved. These are ' self-evident truths' which we take
to be true without proof. These statements are called axioms. In Chapter 5, you would
have studied the axioms and postulates of Euclid. (We do not distinguish between
axioms and postulates these days.)
For example, the first postulate of Euclid states:
A st rai ght l i ne may be drawn f rom any poi nt t o any ot her poi nt .
And the third postulate states:
A ci rcl e may be drawn wi t h any cent re and any radi us.
These statements appear to be perfectly true and Euclid assumed them to be true.
Why? This is because we cannot prove everything and we need to start somewhere.
We need some st at ement s whi ch we accept as t rue and t hen we can bui l d up our
knowledge using the rules of logic based on these axioms.
You might then wonder why we don' t just accept all statements to be true when
they appear self-evident. There are many reasons for this. Very often our intuition can
be wrong, pictures or patterns can deceive and the only way to be sure that something
is true is to prove it. For example, many of us believe that if a number is multiplied by
another, the result will be larger then both the numbers. But we know that this is not
always true: for example, 5 × 0.2 = 1, which is less than 5.
Also, look at the Fig. A1.3. Which line segment is longer, AB or CD?
Fig. A1.3
It turns out that both are of exactly the same length, even though AB appears
shorter!
You might wonder then, about the validity of axioms. Axioms have been chosen
based on our intuition and what appears to be self-evident. Therefore, we expect them
to be true. However, it is possible that later on we discover that a particular axiom is
not true. What is a safeguard against this possibility? We take the following steps:
(i) Keep the axioms to the bare minimum. For instance, based on only axioms
and five postulates of Euclid, we can derive hundreds of theorems.
L i n e s e g m e n t A B
A B
L i n e s e g m e n t C D
C D
PROOFS IN MATHEMATICS 297
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
(ii) Make sure the axioms are consistent.
We say a collection of axioms is inconsistent, i f we can use one axi om t o
show that another axiom is not true. For example, consider the following two
statements. We will show that they are inconsistent.
Statement1: No whole number is equal to its successor.
Statement 2: A whole number divided by zero is a whole number.
(Remember, di vi si on by zero i s not def i ned. But just for the moment, we
assume that it is possible, and see what happens.)
From St at ement 2, we get
1
0
= a , wher e a is some whole number. This
implies that, 1 = 0. But this disproves Statement 1, which states that no whole
number is equal to its successor.
(iii) A false axiom will, sooner or later, result in a contradiction. We say that there
is a contradiction, when we f ind a statement such that, both the statement
and i t s negat i on are t rue. For example, consider Statement 1 and Statement
2 above once again.
From Statement 1, we can derive the result that 2 = 1.
Now look at x
2
– x
2
. We will factorise it in two different ways as follows:
(i) x
2
– x
2
= x(x – x) and
(ii) x
2
– x
2
= (x + x)(x – x)
So, x(x – x) = (x + x)(x – x).
From Statement 2, we can cancel (x – x) from both sides.
We get x = 2x, which in turn implies 2 = 1.
So we have bot h t he st at ement 2 = 1 and i t s negat i on, 2 = 1, t rue. Thi s i s a
contradiction. The contradiction arose because of the false axiom, that a whole number
divided by zero is a whole number.
So, the statements we choose as axioms require a lot of thought and insight. We
must make sure they do not lead to inconsistencies or logical contradictions. Moreover,
the choice of axioms themselves, sometimes leads us to new discoveries. From Chapter
5, you are familiar with Euclid' s fifth postulate and the discoveries of non-Euclidean
geometries. You saw that mathematicians believed that the fifth postulate need not be
a post ul at e and i s act ual l y a t heorem t hat can be proved usi ng j ust t he fi rst four
postulates. Amazingly these attempts led to the discovery of non-Euclidean geometries.
We end the section by recalling the differences between an axiom, a theorem and
a conj ect ure. An axi om i s a mat hemat i cal st at ement whi ch i s assumed t o be t rue
298 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
without proof; a conjecture is a mathematical statement whose truth or falsity is yet
to be established; and a theorem is a mathematical statement whose truth has been
logically established.
EXERCISE A1.3
1. Take any three consecutive even numbers and find their product; for example,
2 × 4 × 6 = 48, 4 × 6 × 8 = 192, and so on. Make three conjectures about these products.
2. Go back to Pascal' s triangle.
Line 1 : 1 = 11
0
Line 2 : 1 1 = 11
1
Line 3 : 1 2 1 = 11
2
Make a conjecture about Line 4 and Line 5. Does your conjecture hold? Does your
conjecture hold for Line 6 too?
3. Let us look at the triangular numbers (see Fig.A1.2) again. Add two consecutive
triangular numbers. For example, T
1
+ T
2
= 4, T
2
+ T
3
= 9, T
3
+ T
4
= 16.
What about T
4
+ T
5
? Make a conjecture about T
n–1
+T
n
.
4. Look at the following pattern:
1
2
= 1
11
2
= 121
111
2
= 12321
1111
2
= 1234321
11111
2
= 123454321
Make a conjecture about each of the following:
111111
2
=
1111111
2
=
Check if your conjecture is true.
5. List five axioms (postulates) used in this book.
A1.5 What is a Mathematical Proof?
Let us now look at various aspects of proofs. We start with understanding the difference
between verification and proof. Before you studied proofs in mathematics, you were
mainly asked to verify statements.
For example, you might have been asked to verify with examples that " the product
of two even numbers is even" . So you might have picked up two random even numbers,
PROOFS IN MATHEMATICS 299
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
say 24 and 2006, and checked that 24 × 2006 = 48144 is even. You might have done so
for many more examples.
Also, you might have been asked as an activity to draw several triangles in the
cl ass and comput e t he sum of t hei r i nt er i or angl es. Apar t f r om er r or s due t o
measurement, you would have found that the interior angles of a triangle add up to
180°.
What is the flaw in this method? There are several problems with the process of
verification. While it may help you to make a statement you believe is true, you cannot
be sure that it is true in all cases. For example, the multiplication of several pairs of
even numbers may lead us to guess that the product of two even numbers is even.
However, it does not ensure that the product of all pairs of even numbers is even. You
cannot physically check the products of all possible pairs of even numbers. If you did,
then like the girl in the cartoon, you will be calculating the products of even numbers
for the rest of your life. Similarly, there may be some triangles which you have not yet
drawn whose interior angles do not add up to 180°. We cannot measure the interior
angles of all possible triangles.
, even
At age 8
At age 16
At age 36
At age 86
, even
, even
, even
300 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
Moreover, verification can often be misleading. For example, we might be tempted
to conclude from Pascal' s triangle (Q.2 of Exercise A1.3), based on earlier verifications,
that 11
5
= 15101051. But in fact 11
5
=161051.
So, you need another approach that does not depend upon verification for some
cases onl y. There i s anot her approach, namel y ' provi ng a st at ement ' . A process
which can establish the truth of a mathematical statement based purely on logical
arguments is called a mathematical proof .
In Exampl e 2 of Sect i on A1. 2, you saw t hat t o est abl i sh t hat a mat hemat i cal
statement is false, it is enough to produce a single counter-example. So while it is not
enough to establish the validity of a mathematical statement by checking or verifying
it for thousands of cases, it is enough to produce one counter-example to disprove a
statement (i.e., to show that something is false). This point is worth emphasising.
To show t hat a mat hemat i cal st at ement i s f al se, i t i s enough t o f i nd a si ngl e
count er-ex ampl e.
So, 7 + 5 = 12 is a counter-example to the statement that the sum of two odd
numbers is odd.
Let us now look at the list of basic ingredients in a proof:
(i) To prove a theorem, we should have a rough idea as to how to proceed.
(ii) The information already given to us in a theorem (i.e., the hypothesis) has to
be clearly understood and used.
PROOFS IN MATHEMATICS 301
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
For example, in Theorem A1.2, which states that the product of two even
numbers is even, we are given two even natural numbers. So, we should use
t hei r propert i es. In t he Fact or Theorem (i n Chapt er 2), you are gi ven a
polynomial p(x) and are told that p(a) = 0. You have to use this to show that
(x – a) is a factor of p(x). Similarly, for the converse of the Factor Theorem,
you are given that (x – a) is a factor of p(x), and you have to use this hypothesis
to prove that p(a) =0.
You can also use constructions during the process of proving a theorem.
For example, to prove that the sum of the angles of a triangle is 180°, we
draw a line parallel to one of the sides through the vertex opposite to the side,
and use properties of parallel lines.
(iii) A proof is made up of a successive sequence of mathematical statements.
Each statement in a proof is logically deduced from a previous statement in
the proof, or from a theorem proved earlier, or an axiom, or our hypothesis.
(iv) The conclusion of a sequence of mathematically true statements laid out in a
logically correct order should be what we wanted to prove, that is, what the
theorem claims.
To understand these ingredients, we will analyse Theorem A1.1 and its proof. You
have already studied this theorem in Chapter 6. But first, a few comments on proofs in
geometry. We often resort to diagrams to help us prove theorems, and this is very
important. However, each statement in the proof has to be established usi ng onl y
logic. Very oft en, we hear st udent s make st at ement s l i ke " Those t wo angl es are
equal because in the drawing they look equal" or " that angle must be 90
0
, because the
two lines look as if they are perpendicular to each other" . Beware of being deceived
by what you see (remember Fig A1.3)! .
So now let us go to Theorem A1.1.
Theorem A1. 1 : The sum of t he i nt eri or angl es of a t ri angl e i s 180°.
Proof : Consider a triangle ABC (see Fig. A1.4).
We have to prove that Z ABC + Z BCA + Z CAB = 180° (1)
Fig A 1.4
A
B C
D E
302 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
Construct a line DE parallel to BC passing through A. (2)
DE is parallel to BC and AB is a transversal.
So, Z DAB and Z ABC are alternate angles. Therefore, by Theorem 6.2, Chapter 6,
they are equal, i.e. Z DAB = Z ABC (3)
Similarly, Z CAE = Z ACB (4)
Therefore, Z ABC + Z BAC+ Z ACB = Z DAB +Z BAC + Z CAE (5)
But Z DAB +Z BAC + Z CAE = 180°, since they form a straight angle. (6)
Hence, Z ABC + Z BAC+ Z ACB = 180°. (7)
Now, we comment on each step of the proof.
Step 1 : Our theorem is concerned with a property of triangles, so we begin with a
triangle.
Step 2 : This is the key idea – the intuitive leap or understanding of how to proceed so
as to be able to prove the theorem. Very often geometric proofs require a construction.
Steps 3 and 4 : Here we conclude that Z DAE = Z ABC and Z CAE = Z ACB, by
using the fact that DE is parallel to BC (our construction), and the previously proved
Theorem 6.2, which states that if two parallel lines are intersected by a transversal,
then the alternate angles are equal.
Step 5 : Here we use Euclid' s axiom (see Chapter 5) which states that: " If equals are
added to equals, the wholes are equal" to deduce
Z ABC + Z BAC+ Z ACB = Z DAB +Z BAC + Z CAE.
That is, the sum of the interior angles of the triangle are equal to the sum of the angles
on a straight line.
Step 6 : Here we use the Linear pair axiom of Chapter 6, which states that the angles
on a straight line add up to 180°, to show that Z DAB +Z BAC + Z CAE = 180°.
Step 7 : We use Euclid' s axiom which states that " things which are equal to the same
t h i n g a r e e q u a l t o e a c h o t h e r " t o c o n c l u d e t h a t Z A B C + Z B A C +
Z ACB = Z DAB +Z BAC + Z CAE = 180°. Notice that Step 7 is the claim made in
the theorem we set out to prove.
We now prove Theorems A1.2 and 1.3 without analysing them.
Theorem A1. 2 : The product of t wo ev en nat ural numbers i s ev en.
Proof : Let x and y be any two even natural numbers.
We want to prove that xy is even.
PROOFS IN MATHEMATICS 303
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
Since x and y are even, they are divisible by 2 and can be expressed in the form
x = 2m, for some natural number m and y = 2n, for some natural number n.
Then xy = 4 mn. Since 4 mn is divisible by 2, so is xy.
Therefore, xy is even.
Theorem A1. 3 : The product of any t hree consecut i v e ev en nat ural numbers i s
div isible by 16.
Proof : Any t hree consecut i ve even numbers wi l l be of t he form 2n, 2n + 2 and
2n + 4, f or s ome na t ur a l numbe r n . We ne e d t o pr ove t ha t t he i r pr oduc t
2n(2n + 2)(2n + 4) is divisible by 16.
Now, 2n(2n + 2)(2n + 4) = 2n × 2(n +1) × 2(n + 2)
= 2 × 2 × 2n(n + 1)(n + 2) = 8n(n + 1)(n + 2).
Now we have two cases. Either n is even or odd. Let us examine each case.
Suppose n is even : Then we can write n = 2m, for some natural number m.
And, then 2n(2n + 2)(2n + 4) = 8n(n + 1)(n + 2) = 16m(2m + 1)(2m + 2).
Therefore, 2n(2n + 2)(2n + 4) is divisible by 16.
Next, suppose n is odd. Then n + 1 is even and we can write n + 1 = 2r, for some
natural number r.
We then have : 2n(2n + 2)(2n + 4) = 8n(n + 1)(n + 2)
= 8(2r – 1) × 2r × (2r + 1)
= 16r(2r – 1)(2r + 1)
Therefore, 2n(2n + 2)(2n + 4) is divisible by 16.
So, in both cases we have shown that the product of any three consecutive even
numbers is divisible by 16.
We concl ude t hi s chapt er wi t h a few remarks on t he di fference bet ween how
mathematicians discover results and how formal rigorous proofs are written down. As
mentioned above, each proof has a key intuitive idea (sometimes more than one).
Intuition is central to a mathematician' s way of thinking and discovering results. Very
often the proof of a theorem comes to a mathematician all jumbled up. A mathematician
will often experiment with several routes of thought, and logic, and examples, before
she/he can hit upon the correct solution or proof. It is only after the creative phase
subsi des t hat al l t he ar gument s ar e gat her ed t oget her t o f or m a pr oper pr oof .
I t i s wor t h ment i oni ng her e t hat t he gr eat I ndi an mat hemat i ci an Sr i ni vasa
Ramanujan used very high levels of intuition to arrive at many of his statements, which
304 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
he cl ai med were t rue. Many of t hese have t urned
out be true and are well known theorems. However,
even to this day mathematicians all over the world
are struggling to prove (or disprove) some of his claims
(conjectures).
EXERCISE A1.4
1. Find counter-examples to disprove the following statements:
(i) If the corresponding angles in two triangles are equal, then the triangles are
congruent.
(ii) A quadrilateral with all sides equal is a square.
(iii) A quadrilateral with all angles equal is a square.
(iv) For integers a and b,
2 2
a b + = a + b
(v) 2n
2
+ 11 is a prime for all whole numbers n.
(vi) n
2
– n + 41 is a prime for all positive integers n.
2. Take your favourite proof and analyse it step-by-step along the lines discussed in
Section A1.5 (what is given, what has been proved, what theorems and axioms have
been used, and so on).
3. Prove that the sum of two odd numbers is even.
4. Prove that the product of two odd numbers is odd.
5. Prove that the sum of three consecutive even numbers is divisible by 6.
6. Prove that infinitely many points lie on the line whose equation is y = 2x.
(Hint : Consider the point (n, 2n) for any integer n.)
7. You must have had a friend who must have told you to think of a number and do
various things to it, and then without knowing your original number, telling you what
number you ended up with. Here are two examples. Examine why they work.
(i) Choose a number. Double it. Add nine. Add your original number. Divide by
three. Add four. Subtract your original number. Your result is seven.
(ii) Write down any three-digit number (for example, 425). Make a six-digit number by
repeating these digits in the same order (425425). Your new number is divisible by
7, 11 and 13.
Srinivasa Ramanujan
(1887–1920)
Fig. A1.5
PROOFS IN MATHEMATICS 305
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–1 (03–01–2006).PM65
A1.6 Summary
In this Appendix, you have studied the following points:
1. In mathematics, a statement is only acceptable if it is either always true or always false.
2. To show that a mathematical statement is false, it is enough to find a single counter-
example.
3. Axioms are statements which are assumed to be true without proof.
4. A conjecture is a statement we believe is true based on our mathematical intuition, but
which we are yet to prove.
5. A mathematical statement whose truth has been established (or proved) is called a theorem.
6. The main logical tool in proving mathematical statements is deductive reasoning.
7. A proof is made up of a successive sequence of mathematical statements. Each statement
in a proof is logically deduced from a previouly known statement, or from a theorem proved
earlier, or an axiom, or the hypothesis.
306 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
APPENDIX 2
INTRODUCTION TO MATHEMATICAL MODELLING
A2.1 Introduction
Right from your earlier classes, you have been solving problems related to the
real-world around you. For example, you have solved problems in simple interest using
the formula for finding it. The formula (or equation) is a relation between the interest
and the other three quantities that are related to it, the principal, the rate of interest and
the period. This formula is an example of a mathematical model. A mathematical
model is a mathematical relation that describes some real-life situation.
Mathematical models are used to solve many real-life situations like:
• launching a satellite.
• predicting the arrival of the monsoon.
• controlling pollution due to vehicles.
• reducing traffic jams in big cities.
In this chapter, we will introduce you to the process of constructing mathematical
models, which is called mathematical modelling. In mathematical modelling, we
take a real-world problem and write it as an equivalent mathematical problem. We
then solve the mathematical problem, and interpret its solution in terms of the
real-world problem. After this we see to what extent the solution is valid in the context
of the real-world problem. So, the stages involved in mathematical modelling are
formulation, solution, interpretation and validation.
We will start by looking at the process you undertake when solving word problems,
in Section A2.2. Here, we will discuss some word problems that are similar to the
ones you have solved in your earlier classes. We will see later that the steps that are
used for solving word problems are some of those used in mathematical modelling
also.
INTRODUCTION TO MATHEMATICAL MODELLING 307
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
In the next section, that is Section A2.3, we will discuss some simple models.
In Section A2.4, we will discuss the overall process of modelling, its advantages
and some of its limitations.
A2.2 Review of Word Problems
In this section, we will discuss some word problems that are similar to the ones that
you have solved in your earlier classes. Let us start with a problem on direct variation.
Example 1 : I travelled 432 kilometres on 48 litres of petrol in my car. I have to go by
my car to a place which is 180 km away. How much petrol do I need?
Solution : We will list the steps involved in solving the problem.
Step 1 : Formulation : You know that farther we travel, the more petrol we
require, that is, the amount of petrol we need varies directly with the distance we
travel.
Petrol needed for travelling 432 km = 48 litres
Petrol needed for travelling 180 km = ?
Mathematical Description : Let
x = distance I travel
y = petrol I need
y varies directly with x.
So, y = kx, where k is a constant.
I can travel 432 kilometres with 48 litres of petrol.
So, y = 48, x = 432.
Therefore, k =
48 1
432 9
y
x
= =
.
Since y = kx,
therefore, y =
1
9
x (1)
Equation or Formula (1) describes the relationship between the petrol needed and
distance travelled.
Step 2 : Solution : We want to find the petrol we need to travel 180 kilometres;
so, we have to find the value of y when x = 180. Putting x = 180 in (1), we have
308 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
y =
180
20
9
=
.
Step 3 : Interpretation : Since y = 20, we need 20 litres of petrol to travel 180
kilometres.
Did it occur to you that you may not be able to use the formula (1) in all situations?
For example, suppose the 432 kilometres route is through mountains and the 180
kilometres route is through flat plains. The car will use up petrol at a faster rate in the
first route, so we cannot use the same rate for the 180 kilometres route, where the
petrol will be used up at a slower rate. So the formula works if all such conditions that
affect the rate at which petrol is used are the same in both the trips. Or, if there is a
difference in conditions, the effect of the difference on the amount of petrol needed
for the car should be very small. The petrol used will vary directly with the distance
travelled only in such a situation. We assumed this while solving the problem.
Example 2 : Suppose Sudhir has invested Rs 15,000 at 8% simple interest per year.
With the return from the investment, he wants to buy a washing machine that costs
Rs 19,000. For what period should he invest Rs 15,000 so that he has enough money to
buy a washing machine?
Solution : Step 1 : Formulation of the problem : Here, we know the principal and
the rate of interest. The interest is the amount Sudhir needs in addition to 15,000 to buy
the washing machine. We have to find the number of years.
Mathematical Description : The formula for simple interest is I =
P
100
nr
,
where P = Principal,
n = Number of years,
r % = Rate of interest
I = Interest earned
Here, the principal = Rs 15,000
The money required by Sudhir for buying a washing machine = Rs 19,000
So, the interest to be earned = Rs (19,000 – 15,000)
= Rs 4,000
The number of years for which Rs 15,000 is deposited = n
The interest on Rs 15,000 for n years at the rate of 8% = I
Then, I =
15000 × × 8
100
n
INTRODUCTION TO MATHEMATICAL MODELLING 309
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
So, I = 1200n (1)
gives the relationship between the number of years and interest, if Rs 15000 is invested
at an annual interest rate of 8%.
We have to find the period in which the interest earned is Rs 4000.
Putting I = 4000 in (1), we have
4000 = 1200n (2)
Step 2 : Solution of the problem : Solving Equation (2), we get
n =
4000 1
3
1200 3
= ⋅
Step 3 : Interpretation : Since n =
1
3
3
and one third of a year is 4 months,
Sudhir can buy a washing machine after 3 years and 4 months.
Can you guess the assumptions that you have to make in the example above? We
have to assume that the interest rate remains the same for the period for which we
calculate the interest. Otherwise, the formula I =
P
100
nr
will not be valid. We have also
assumed that the price of the washing machine does not increase by the time Sudhir
has gathered the money.
Example 3 : A motorboat goes upstream on a river and covers the distance between
two towns on the riverbank in six hours. It covers this distance downstream in five
hours. If the speed of the stream is 2 km/h, find the speed of the boat in still water.
Solution : Step 1 : Formulation : We know the speed of the river and the time taken
to cover the distance between two places. We have to find the speed of the boat in
still water.
Mathematical Description : Let us write x for the speed of the boat, t for the time
taken and y for the distance travelled. Then
y = t x (1)
Let d be the distance between the two places.
While going upstream, the actual speed of the boat
= speed of the boat – speed of the river,
because the boat is travelling against the flow of the river.
So, the speed of the boat upstream = (x – 2) km/h
It takes 6 hours to cover the distance between the towns upstream. So, from (1),
we get d = 6(x – 2) (2)
310 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
When going downstream, the speed of the river has to be added to the speed of the
boat.
So, the speed of the boat downstream = (x + 2) km/h
The boat takes 5 hours to cover the same distance downstream. So,
d = 5(x + 2) (3)
From (2) and (3), we have
5(x + 2) = 6(x –2 ) (4)
Step 2 : Finding the Solution
Solving for x in Equation (4), we get x = 22.
Step 3 : Interpretation
Since x = 22, therefore the speed of the motorboat in still water is 22 km/h.
In the example above, we know that the speed of the river is not the same
everywhere. It flows slowly near the shore and faster at the middle. The boat starts at
the shore and moves to the middle of the river. When it is close to the destination, it will
slow down and move closer to the shore. So, there is a small difference between the
speed of the boat at the middle and the speed at the shore. Since it will be close to the
shore for a small amount of time, this difference in speed of the river will affect the
speed only for a small period. So, we can ignore this difference in the speed of the
river. We can also ignore the small variations in speed of the boat. Also, apart from the
speed of the river, the friction between the water and surface of the boat will also
affect the actual speed of the boat. We also assume that this effect is very small.
So, we have assumed that
1. The speed of the river and the boat remains constant all the time.
2. The effect of friction between the boat and water and the friction due to air is
negligible.
We have found the speed of the boat in still water with the assumptions
(hypotheses) above.
As we have seen in the word problems above, there are 3 steps in solving a
word problem. These are
1. Formulation : We analyse the problem and see which factors have a major
influence on the solution to the problem. These are the relevant factors. In
our first example, the relevant factors are the distance travelled and petrol
consumed. We ignored the other factors like the nature of the route, driving
speed, etc. Otherwise, the problem would have been more difficult to solve.
The factors that we ignore are the irrelevant factors.
INTRODUCTION TO MATHEMATICAL MODELLING 311
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
We then describe the problem mathematically, in the form of one or more
mathematical equations.
2. Solution : We find the solution of the problem by solving the mathematical
equations obtained in Step 1 using some suitable method.
3. Interpretation : We see what the solution obtained in Step 2 means in the
context of the original word problem.
Here are some exercises for you. You may like to check your understanding of
the steps involved in solving word problems by carrying out the three steps above for
the following problems.
EXERCISE A 2.1
In each of the following problems, clearly state what the relevant and irrelevant factors are
while going through Steps 1, 2 and 3 given above.
1. Suppose a company needs a computer for some period of time. The company can
either hire a computer for Rs 2,000 per month or buy one for Rs 25,000. If the company
has to use the computer for a long period, the company will pay such a high rent, that
buying a computer will be cheaper. On the other hand, if the company has to use the
computer for say, just one month, then hiring a computer will be cheaper. Find the
number of months beyond which it will be cheaper to buy a computer.
2. Suppose a car starts from a place A and travels at a speed of 40 km/h towards another
place B. At the same instance, another car starts from B and travels towards A at a
speed of 30 km/h. If the distance between A and B is 100 km, after how much time will
the cars meet?
3. The moon is about 3,84,000 km from the earth, and its path around the earth is nearly
circular. Find the speed at which it orbits the earth, assuming that it orbits the earth in
24 hours. (Use π = 3.14)
4. A family pays Rs 1000 for electricity on an average in those months in which it does
not use a water heater. In the months in which it uses a water heater, the average
electricity bill is Rs 1240. The cost of using the water heater is Rs 8.00 per hour. Find
the average number of hours the water heater is used in a day.
A2.3 Some Mathematical Models
So far, nothing was new in our discussion. In this section, we are going to add another
step to the three steps that we have discussed earlier. This step is called validation.
What does validation mean? Let us see. In a real-life situation, we cannot accept a
model that gives us an answer that does not match the reality. This process of checking
the answer against reality, and modifying the mathematical description if necessary, is
312 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
called validation. This is a very important step in modelling. We will introduce you to
this step in this section.
First, let us look at an example, where we do not have to modify our model after
validation.
Example 4 : Suppose you have a room of length 6 m and breadth 5 m. You want to
cover the floor of the room with square mosaic tiles of side 30 cm. How many tiles
will you need? Solve this by constructing a mathematical model.
Solution : Formulation : We have to consider the area of the room and the area of
a tile for solving the problem. The side of the tile is 0.3 m. Since the length is 6 m, we
can fit in
6
0.3
= 20 tiles along the length of the room in one row (see Fig. A2.1.).
Fig. A2.1
Since the breadth of the room is 5 metres, we have
5
0.3
= 16.67. So, we can fit in
16 tiles in a column. Since 16 × 0.3 = 4.8, 5 – 4.8 = 0.2 metres along the breadth will
not be covered by tiles. This part will have to be covered by cutting the other tiles. The
breadth of the floor left uncovered, 0.2 metres, is more than half the length of a tile,
which is 0.3 m. So we cannot break a tile into two equal halves and use both the halves
to cover the remaining portion.
Mathematical Description : We have:
Total number of tiles required = (Number of tiles along the length
× Number of tiles along the breadth) + Number of tiles along the uncovered area
(1)
Area covered by
full tiles
4.8 m
INTRODUCTION TO MATHEMATICAL MODELLING 313
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
Solution : As we said above, the number of tiles along the length is 20 and the
number of tiles along the breadth is 16. We need 20 more tiles for the last row. Substituting
these values in (1), we get (20 × 16) + 20 = 320 + 20 = 340.
Interpretation : We need 340 tiles to cover the floor.
Validation : In real-life, your mason may ask you to buy some extra tiles to
replace those that get damaged while cutting them to size. This number will of course
depend upon the skill of your mason! But, we need not modify Equation (1) for this.
This gives you a rough idea of the number of tiles required. So, we can stop here.
Let us now look at another situation now.
Example 5 : In the year 2000, 191 member countries of the U.N. signed a declaration.
In this declaration, the countries agreed to achieve certain development goals by the
year 2015. These are called the millennium development goals. One of these goals
is to promote gender equality. One indicator for deciding whether this goal has been
achieved is the ratio of girls to boys in primary, secondary and tertiary education.
India, as a signatory to the declaration, is committed to improve this ratio. The data for
the percentage of girls who are enrolled in primary schools is given in Table A2.1.
Table A2.1
Year Enrolment
(in %)
1991-92 41.9
1992-93 42.6
1993-94 42.7
1994-95 42.9
1995-96 43.1
1996-97 43.2
1997-98 43.5
1998-99 43.5
1999-2000 43.6*
2000-01 43.7*
2001-02 44.1*
Source : Educational statistics, webpage of Department of Education, GOI.
* indicates that the data is provisional.
314 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
Using this data, mathematically describe the rate at which the proportion of girls enrolled
in primary schools grew. Also, estimate the year by which the enrolment of girls will
reach 50%.
Solution : Let us first convert the problem into a mathematical problem.
Step 1 : Formulation : Table A2.1 gives the enrolment for the years 1991-92,
1992-93, etc. Since the students join at the beginning of an academic year, we can
take the years as 1991, 1992, etc. Let us assume that the percentage of girls who join
primary schools will continue to grow at the same rate as the rate in Table A2.1. So,
the number of years is important, not the specific years. (To give a similar situation,
when we find the simple interest for, say, Rs 1500 at the rate of 8% for three years, it
does not matter whether the three-year period is from 1999 to 2002 or from 2001 to
2004. What is important is the interest rate in the years being considered). Here also,
we will see how the enrolment grows after 1991 by comparing the number of years
that has passed after 1991 and the enrolment. Let us take 1991 as the 0th year, and
write 1 for 1992 since 1 year has passed in 1992 after 1991. Similarly, we will write 2
for 1993, 3 for 1994, etc. So, Table A2.1 will now look like as Table A2.2.
Table A2.2
Year Enrolment
(in %)
0 41.9
1 42.6
2 42.7
3 42.9
4 43.1
5 43.2
6 43.5
7 43.5
8 43.6
9 43.7
10 44.1
INTRODUCTION TO MATHEMATICAL MODELLING 315
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
The increase in enrolment is given in the following table :
Table A2.3
Year Enrolment Increase
(in %)
0 41.9 0
1 42.6 0.7
2 42.7 0.1
3 42.9 0.2
4 43.1 0.2
5 43.2 0.1
6 43.5 0.3
7 43.5 0
8 43.6 0.1
9 43.7 0.1
10 44.1 0.4
At the end of the one-year period from 1991 to 1992, the enrolment has increased
by 0.7% from 41.9% to 42.6%. At the end of the second year, this has increased by
0.1%, from 42.6% to 42.7%. From the table above, we cannot find a definite relationship
between the number of years and percentage. But the increase is fairly steady. Only
in the first year and in the 10th year there is a jump. The mean of the values is
0.7 0.1 0.2 0.2 0.1 0.3 0 0.1 0.1 0.4
10
+ + + + + + + + +
= 0.22
Let us assume that the enrolment steadily increases at the rate of 0.22 per cent.
Mathematical Description : We have assumed that the enrolment increases
steadily at the rate of 0.22% per year.
So, the Enrolment Percentage (EP) in the first year = 41.9 + 0.22
EP in the second year = 41.9 + 0.22 + 0.22 = 41.9 + 2 × 0.22
EP in the third year = 41.9 + 0.22 + 0.22 + 0.22 = 41.9 + 3 × 0.22
So, the enrolment percentage in the nth year = 41.9 + 0.22n, for n > 1. (1)
316 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
Now, we also have to find the number of years by which the enrolment will reach
50%. So, we have to find the value of n in the equation or formula
50 = 41.9 + 0.22n (2)
Step 2 : Solution : Solving (2) for n, we get
n =
50 – 41.9 8.1
36.8
0.22 0.22
= =
Step 3 : Interpretation : Since the number of years is an integral value, we will
take the next higher integer, 37. So, the enrolment percentage will reach 50% in
1991 + 37 = 2028.
In a word problem, we generally stop here. But, since we are dealing with a real-
life situation, we have to see to what extent this value matches the real situation.
Step 4 : Validation: Let us check if Formula (2) is in agreement with the reality.
Let us find the values for the years we already know, using Formula (2), and compare
it with the known values by finding the difference. The values are given in Table A2.4.
Table A2.4
Year Enrolment Values given by (2) Difference
(in %) (in %) (in %)
0 41.9 41.90 0
1 42.6 42.12 0.48
2 42.7 42.34 0.36
3 42.9 42.56 0.34
4 43.1 42.78 0.32
5 43.2 43.00 0.20
6 43.5 43.22 0.28
7 43.5 43.44 0.06
8 43.6 43.66 –0.06
9 43.7 43.88 –0.18
10 44.1 44.10 0.00
As you can see, some of the values given by Formula (2) are less than the actual
values by about 0.3% or even by 0.5%. This can give rise to a difference of about 3 to
5 years since the increase per year is actually 1% to 2%. We may decide that this
INTRODUCTION TO MATHEMATICAL MODELLING 317
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
much of a difference is acceptable and stop here. In this case, (2) is our mathematical
model.
Suppose we decide that this error is quite large, and we have to improve this
model. Then we have to go back to Step 1, the formulation, and change Equation (2).
Let us do so.
Step 1 : Reformulation : We still assume that the values increase steadily by
0.22%, but we will now introduce a correction factor to reduce the error. For this, we
find the mean of all the errors. This is
0 0.48 0.36 0.34 0.32 0.2 0.28 0.06 – 0.06 – 0.18 0
10
+ + + + + + + +
= 0.18
We take the mean of the errors, and correct our formula by this value.
Revised Mathematical Description : Let us now add the mean of the errors to
our formula for enrolment percentage given in (2). So, our corrected formula is:
Enrolment percentage in the nth year = 41.9 + 0.22n + 0.18 = 42.08 + 0.22n, for
n > 1 (3)
We will also modify Equation (2) appropriately. The new equation for n is:
50 = 42.08 + 0.22n (4)
Step 2 : Altered Solution : Solving Equation (4) for n, we get
n =
50 – 42.08 7.92
36
0.22 0.22
= =
Step 3 : Interpretation: Since n = 36, the enrolment of girls in primary schools
will reach 50% in the year 1991 + 36 = 2027.
Step 4 : Validation: Once again, let us compare the values got by using Formula
(4) with the actual values. Table A2.5 gives the comparison.
318 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
Table A2.5
Year Enrolment Values Difference Values Difference
(in %) given between given between
by (2) values by (4) values
0 41.9 41.90 0 41.9 0
1 42.6 42.12 0.48 42.3 0.3
2 42.7 42.34 0.36 42.52 0.18
3 42.9 42.56 0.34 42.74 0.16
4 43.1 42.78 0.32 42.96 0.14
5 43.2 43.00 0.2 43.18 0.02
6 43.5 43.22 0.28 43.4 0.1
7 43.5 43.44 0.06 43.62 – 0.12
8 43.6 43.66 – 0.06 43.84 – 0.24
9 43.7 43.88 – 0.18 44.06 – 0.36
10 44.1 44.10 0 44.28 – 0.18
As you can see, many of the values that (4) gives are closer to the actual value
than the values that (2) gives. The mean of the errors is 0 in this case.
We will stop our process here. So, Equation (4) is our mathematical description
that gives a mathematical relationship between years and the percentage of enrolment
of girls of the total enrolment. We have constructed a mathematical model that describes
the growth.
The process that we have followed in the situation above is called
mathematical modelling.
We have tried to construct a mathematical model with the mathematical tools that
we already have. There are better mathematical tools for making predictions from the
data we have. But, they are beyond the scope of this course. Our aim in constructing
this model is to explain the process of modelling to you, not to make accurate predictions
at this stage.
You may now like to model some real-life situations to check your understanding
of our discussion so far. Here is an Exercise for you to try.
INTRODUCTION TO MATHEMATICAL MODELLING 319
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
EXERCISE A2.2
1. We have given the timings of the gold medalists in the 400-metre race from the time the
event was included in the Olympics, in the table below. Construct a mathematical model
relating the years and timings. Use it to estimate the timing in the next Olympics.
Table A2.6
Year Timing (in seconds)
1964 52.01
1968 52.03
1972 51.08
1976 49.28
1980 48.88
1984 48.83
1988 48.65
1992 48.83
1996 48.25
2000 49.11
2004 49.41
A2.4 The Process of Modelling, its Advantages and Limitations
Let us now conclude our discussion by drawing out aspects of mathematical modelling
that show up in the examples we have discussed. With the background of the earlier
sections, we are now in a position to give a brief overview of the steps involved in
modelling.
Step 1 : Formulation : You would have noticed the difference between the
formulation part of Example 1 in Section A2.2 and the formulation part of the model
we discussed in A2.3. In Example 1, all the information is in a readily usable form. But,
in the model given in A2.3 this is not so. Further, it took us some time to find a
mathematical description. We tested our first formula, but found that it was not as
good as the second one we got. This is usually true in general, i.e. when trying to
model real-life situations; the first model usually needs to be revised. When we are
solving a real-life problem, formulation can require a lot of time. For example, Newton's
three laws of motion, which are mathematical descriptions of motion, are simple enough
to state. But, Newton arrived at these laws after studying a large amount of data and
the work the scientists before him had done.
320 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
Formulation involves the following three steps :
(i) Stating the problem : Often, the problem is stated vaguely. For example,
the broad goal is to ensure that the enrolment of boys and girls are equal. This
may mean that 50% of the total number of boys of the school-going age and
50% of the girls of the school-going age should be enrolled. The other way is
to ensure that 50% of the school-going children are girls. In our problem, we
have used the second approach.
(ii) Identifying relevant factors : Decide which quantities and relationships
are important for our problem and which are unimportant and can be neglected.
For example, in our problem regarding primary schools enrolment, the
percentage of girls enrolled in the previous year can influence the number of
girls enrolled this year. This is because, as more and more girls enrol in
schools, many more parents will feel they also have to put their daughters in
schools. But, we have ignored this factor because this may become important
only after the enrolment crosses a certain percentage. Also, adding this factor
may make our model more complicated.
(iii) Mathematical Description : Now suppose we are clear about what the
problem is and what aspects of it are more relevant than the others. Then we
have to find a relationship between the aspects involved in the form of an
equation, a graph or any other suitable mathematical description. If it is an
equation, then every important aspect should be represented by a variable in
our mathematical equation.
Step 2 : Finding the solution : The mathematical formulation does not give the
solution. We have to solve this mathematical equivalent of the problem. This is where
your mathematical knowledge comes in useful.
Step 3 : Interpretating the solution : The mathematical solution is some value
or values of the variables in the model. We have to go back to the real-life problem and
see what these values mean in the problem.
Step 4 : Validating the solution : As we saw in A2.3, after finding the solution
we will have to check whether the solution matches the reality. If it matches, then the
mathematical model is acceptable. If the mathematical solution does not match, we go
back to the formulation step again and try to improve our model.
This step in the process is one major difference between solving word problems
and mathematical modelling. This is one of the most important step in modelling that is
missing in word problems. Of course, it is possible that in some real-life situations, we
do not need to validate our answer because the problem is simple and we get the
correct solution right away. This was so in the first model we considered in A2.3.
INTRODUCTION TO MATHEMATICAL MODELLING 321
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
We have given a summary of the order in which the steps in mathematical modelling
are carried out in Fig. A2.2 below. Movement from the validation step to the formulation
step is shown using a dotted arrow. This is because it may not be necessary to carry
out this step again.
Fig.A2.2
Now that you have studied the stages involved in mathematical modelling, let us
discuss some of its aspects.
The aim of mathematical modelling is to get some useful information about a real-
world problem by converting it into a mathematical problem. This is especially useful
when it is not possible or very expensive to get information by other means such as
direct observation or by conducting experiments.
You may also wonder why we should undertake mathematical modelling? Let us
look at some advantages of modelling. Suppose we want to study the corrosive
effect of the discharge of the Mathura refinery on the Taj Mahal. We would not like to
carry out experiments on the Taj Mahal directly since it may not be safe to do so. Of
course, we can use a scaled down physical model, but we may need special facilities
for this, which may be expensive. Here is where mathematical modelling can be of
great use.
Again, suppose we want to know how many primary schools we will need after 5
years. Then, we can only solve this problem by using a mathematical model. Similarly,
it is only through modelling that scientists have been able to explain the existence of so
many phenomena.
You saw in Section A2.3, that we could have tried to improve the answer in the
second example with better methods. But we stopped because we do not have the
mathematical tools. This can happen in real-life also. Often, we have to be satisfied
with very approximate answers, because mathematical tools are not available. For
example, the model equations used in modelling weather are so complex that
mathematical tools to find exact solutions are not available.
Formulation of the
problem
Solution of the
problem
Interpretation of the
solution
Checking/validating
the solution
322 MATHEMATICS
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
You may wonder to what extent we should try to improve our model. Usually, to
improve it, we need to take into account more factors. When we do this, we add more
variables to our mathematical equations. We may then have a very complicated model
that is difficult to use. A model must be simple enough to use. A good model balances
two factors:
1. Accuracy i.e., how close it is to reality.
2. Ease of use.
For example, Newton's laws of motion are very simple, but powerful enough to
model many physical situations.
So, is mathematical modelling the answer to all our problems? Not quite! It has its
limitations.
Thus, we should keep in mind that a model is only a simplification of a real-
world problem, and the two are not the same. It is something like the difference
between a map that gives the physical features of a country, and the country itself. We
can find the height of a place above the sea level from this map, but we cannot find the
characteristics of the people from it. So, we should use a model only for the purpose it
is supposed to serve, remembering all the factors we have neglected while constructing
it. We should apply the model only within the limits where it is applicable. In the later
classes, we shall discuss this aspect a little more.
EXERCISE A2.3
1. How are the solving of word problems that you come across in textbooks different
from the process of mathematical modelling?
2. Suppose you want to minimise the waiting time of vehicles at a traffic junction of four
roads. Which of these factors are important and which are not?
(i) Price of petrol.
(ii) The rate at which the vehicles arrive in the four different roads.
(iii) The proportion of slow-moving vehicles like cycles and rickshaws and fast moving
vehicles like cars and motorcycles.
A2.5 Summary
In this Appendix, you have studied the following points :
1. The steps involved in solving word problems.
2. Construction of some mathematical models.
INTRODUCTION TO MATHEMATICAL MODELLING 323
File Name : C:\Computer Station\Maths-IX\Chapter\Appendix\Appendix–2 (03–01–2006).PM65
3. The steps involved in mathematical modelling given in the box below.
1. Formulation :
(i) Stating the question
(ii) Identifying the relevant factors
(iii) Mathematical description
2. Finding the solution.
3. Interpretation of the solution in the context of the real-world
problem.
4. Checking/validating to what extent the model is a good
representation of the problem being studied.
4. The aims, advantages and limitations of mathematical modelling. | 677.169 | 1 |
Analysis
The
Lara Alcock is a Senior Lecturer in the Mathematics Education Centre at Loughborough University. She studied Mathematics to Masters level at the University of Warwick before going on to doctoral study in Mathematics Education at the same Institution. She spent four years as an Assistant Professor in Mathematics at the Graduate School of Education at Rutgers University in the USA, and two as a Teaching Fellow in Mathematics at the University of Essex in the UK before taking up her present position. In her current position she teaches undergraduate Mathematics, works with PhD students in Mathematics Education, and conducts research studies on the ways in which people learn, understand and think about abstract mathematics. She has been awarded National Teaching Fellows of 2015 by The Higher Education Academy.
Review:
"This book is an invaluable guide for any undergraduate student taking Analysis... It is written using a friendly and informal tone yet carefully emphasizes and demonstrates the importance of paying attention to the details. It is an excellent read and is highly recommended for anyone interested in Analysis or any area of pure mathematics." --MAA Reviews
"This is a nice little book with an accurate title... Recommended." --Choice
"How to Think about Analysis [is] a very effective and helpful book, a book which should be on every undergraduate reading list and should be available to potential mathematics undergraduates in schools." --Mathematics in School
"How to Think about Analysis offers several insights into the best practices to use when studying upper-level mathematics. Not only are these insights helpful to students, but they could also prove helpful to teachers of earlier courses; modifying and incorporating some of these practices into earlier courses may better prepare their students for future mathematics coursework." --Mathematics Teacher. Paperback. Condizione libro: new. BRAND NEW, How to Think About Analysis, Lara Alcock, friendly B9780198723530
Descrizione libro Oxford University Press, 2014. Condizione libro: New. Analysis Num Pages: 272 pages, 120 b/w line drawings. BIC Classification: PBKB. Category: (P) Professional & Vocational. Dimension: 130 x 197 x 16. Weight in Grams: 290. . 2014. 1st Edition. Paperback. . . . . . Codice libro della libreria V9780198723530 | 677.169 | 1 |
NCERT Solutions Class 12 maths chapter 3
Free NCERT Solutions Class 12 Maths chapter 3 Matrices in PDF form download, Matrices chapter is one of the easy and scoring chapters as per CBSE examination. Go through the chapter completing NCERT and Exemplar books and practicing the previous year CBSE question papers. Buy NCERT Books online of download in PDF form to study online. Prepare yourself according to latest CBSE Syllabus 2017 – 2018 and do the practice papers according to CBSE Sample papers issued for the current academic session.
These books are very good for revision and more practice. These book are also confined to NCERT Syllabus.
Assignments for practice
MixedChapterTests
Matrix
The arrangement of real numbers in a rectangular array enclosed in brackets as [] or () is known as a Matrix(Matrices is plural of matrix). Matrix operations are used in electronic physics, computers, budgeting, cost estimation, analysis and experiments. They are also used in cryptography, modern psychology, genetics, endustrial management etc. In general an m x n matrix is matrix having m rows and n columns. it can be written as follows: Order of a Matrix
There may be any number of rows and any number of columns in a matrix. If there are m rows and n columns in matrix A, its order is m x n and it is read as an m x n matrix. Transpose of a Matrix
The transpose of a given matrix A is formed by interchanging its rows and columns and is denoted by A'. Symmetric Matrix
A square matrix A is said to be a symmetric matrix if A' = A. Skew-Symmetric Matrix
A square matrix A is said to be a skew symmetric if A' = – A. all elements in the principal diagonal of a skew symmetric matrix are zeroes. Addition of MatrixType of Matrices Row matrix: A row matrix has only one row but any number of columns. Column matrix: A column matrix has only one column but any number of rows. Square matrix: A square matrix has the number of column equal to the number of rows. Rectangular Matrix: A matrix is said to be a rectangular matrix if the number of rows is not equal to the number of columns. Diagonal matrix: If in a square matrix has all elements 0 except principal diagonal elements, it is called diagonal matrix. Scalar Matrix: A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. Zero or Null matrix: If all elements of a matrix are zero, then the matrix is known as zero matrix and denoted by O. Unit or Identity matrix: If in a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal.
Properties of Matrix
When a matrix is multiplied by a scalar, then each of its element is multiplied by the same scalar.
For any two matrices A and B of the same order, A + B = B + A. i.e. matrix addition is commutative.
For any three matrices A, B and C of the same order, A + (B + C) = (A + B) + C i.e., matrix addition is associative.
Additive identity is a zero matrix, which when added to a given matrix, gives the same given matrix, i.e., A + O = A = O + A.
If A + B = O, then the matrix B is called the additive inverse of the matrix of A.
If A and B are two matrices of order m x p and p x n respectively, then their product will be a matrix C of order m x n.
Invertible Matrix
A square matrix of order n is invertible if there exists a square matrix B of the same order such that AB = I = BA, Where I is identify matrix of order n.
Theorems of invertible matrices
Theorem 1: Every invertible matrix possesses a unique inverse.
Theorem 2: A square matrix is invertible iff it is non-singular.
Historical Facts!
Matrix is a latin word. Originally matrices are used for solutions of simultaneous linear equations. An important Chinese Text between 300 BC and 200 AD, nine chapters of Mathematical Art(Chiu Chang Suan Shu), give the use of matrix methods to solve simultaneous equations. Carl Friedrich Gauss(1777 – 1855) also gave the method to solve simultaneous equations by matrix method. | 677.169 | 1 |
new Ninth Edition of Burden and Faires well-respected Numerical Analysis provides a foundation in modern numerical-approximation techniques. Explaining how, why, and when the techniques can be expected to work, the Ninth Edition places on even greater emphasis on building readers? intuition to help them understand why the techniques presented work in general, and why, in some situations, they fail. | 677.169 | 1 |
When you are a young mathematician, graduate school marks the first step toward a career in mathematics. During this period, you will make important decisions which will affect the rest of your career. Here now is a detailed guide to help you navigate graduate school and the years that follow. In his inimitable and forthright style, Steven Krantz addresses the major issues of graduate school, including choosing a program, passing the qualifying exams, finding an advisor, writing a thesis, and getting your first job. As with his earlier guide, How to Teach Mathematics, he avoids generalities, giving clear advice on how to handle real situations. The book also contains a description of the basic elements of a mathematical education, as well as a glossary and appendices on the structure of a typical department and university and the standard academic ranks. Steven G. Krantz is an accomplished mathematician and an award-winning author. He has published 130 research articles and 45 books. He has worked in many different types of mathematics departments, supervised both masters and doctoral students, and is currently the Chair of the Mathematics Department at Washington University in St. Louis.
"One of the themes of the book is how to have a fulfilling professional life. In order to achieve this goal, Krantz discusses keeping a vigorous scholarly program going and finding new challenges, as well as dealing with the everyday tasks of research, teaching, and administration." "In short, this is a survival manual for the professional mathematician - both in academics and in industry and government agencies. It is a sequel to the author's A Mathematician's Survival Guide."--BOOK JACKET.
Mathematicians are expected to publish their work: in journals, conference proceedings, and books. It is vital to advancing their careers. Later, some are asked to become editors. However, most mathematicians are trained to do mathematics, not to publish it. But here, finally, for graduate students and researchers interested in publishing their work, Steven G. Krantz, the respected author of several "how-to" guides in mathematics, shares his experience as an author, editor, editorial board member, and independent publisher. This new volume is an informative, comprehensive guidebook to publishing mathematics. Krantz describes both the general setting of mathematical publishing and the specifics about all the various publishing situations mathematicians may encounter. As with his other books, Krantz's style is engaging and frank. He gives advice on how to get your book published, how to get organized as an editor, what to do when things go wrong, and much more. He describes the people, the language (including a glossary), and the process of publishing both books and journals. Steven G. Krantz is an accomplished mathematician and an award-winning author. He has published more than 130 research articles and 45 books. He has worked as an editor of several book series, research journals, and for the Notices of the AMS. He is also the founder of the Journal of Geometric Analysis. Other titles available from the AMS by Steven G. Krantz are How to Teach Mathematics, A Primer of Mathematical Writing, A Mathematician's Survival Guide, and Techniques of Problem Solving.
We all swim in a sea of Big Data, dangerously vulnerable to the unscientific thinking that now replaces the critical faculties we used to rely on. We seek simple explanations where complexity is required. But as we endeavor to solve global problems of energy, food, and water shortages, a planetary biodiversity crisis, and emerging threats to our public health, the development of scientific habits of mind becomes even more essential for our survival. We fear numbers and prefer neat and simple solutions to complex problems, but scientific reasoning plays a central role in combating misinformation and is one of our best tools for meeting the upcoming crises of our century. From confronting our fear of quantitative reasoning and demystifying graphs to elucidating the key concepts of probability and data analysis and the use of precise language and logic, this book supplies an essential set of apps for the frontal cortex while making science both accessible and entertaining. Who says it has to be dull to learn to think like a scientist? Who says only a few can do it? Not David Helfand, one of our nation's leading astronomers and science educators. Helfand has taught scientific habits of mind to generations of Columbia University undergraduates, where he continues to wage a provocative and necessary battle against sloppy thinking and the encroachment of misinformation.
What every special education teacher needs to know to survive and thrive A Survival Guide for New Special Educators provides relevant, practical information for new special education teachers across a broad range of topic areas. Drawing on the latest research on special educator effectiveness and retention, this comprehensive, go-to resource addresses the most pressing needs of novice instructors, resource teachers, and inclusion specialists. Offers research-based, classroom-tested strategies for working with a variety of special needs students Covers everything from preparing for the new school year to behavior management, customizing curriculum, creating effective IEPs, and more Billingsley and Brownell are noted experts in special educator training and support This highly practical book is filled with checklists, forms, and tools that special educators can use every day to help ensure that all special needs students get the rich, rewarding education they deserve.
This book explains what actuaries are, what they do, and where they do it. It describes the --from publisher description
This third edition is a lively and provocative tract on how to teach mathematics in today's new world of online learning tools and innovative teaching devices. The author guides the reader through the joys and pitfalls of interacting with modern undergraduates--telling you very explicitly what to do and what not to do. This third edition has been streamlined from the second edition, but still includes the nuts and bolts of good teaching, discussing material related to new developments in teaching methodology and technique, as well as adding an entire new chapter on online teaching methods.
Balanced, practical risk management for post – financial crisis institutions Fundamentals of Risk Management fills a critical gap left by existing risk management texts. Instead of focusing only on quantitative risk analysis or only on institutional risk management, this book takes a comprehensive approach. The disasters of the recent financial crisis taught us that managing risk is both an art and a science, and it is critical for practitioners to understand how individual risks are integrated at the enterprise level. This book is the only resource of its kind to introduce all of the key risk management concepts in a cohesive case study spanning each chapter. A hypothetical bank drawn from elements of several real world institutions serves as a backdrop for topics from credit risk and operational risk to understanding big-picture risk exposure. You will be able to see exactly how each rigorous concept is applied in actual risk management contexts. Fundamentals of Risk Management includes: Supplemental Excel-based Visual Basic (VBA) modules, so you can interact directly with risk models Clear explanations of the importance of risk management in preventing financial disasters Real world examples and lessons learned from past crises Risk policies, infrastructure, and activities that balance limited quantitative models This book provides the element of hands-on application necessary to put enterprise risk management into effective practice. The very best risk managers rely on a balanced approach that leverages every aspect of financial operations for an integrative risk management strategy. With Fundamentals of Risk Management, you can identify and control risk at an expert level. | 677.169 | 1 |
Monday, December 9, 2013
Today we will be taking an hour of code. If you are familiar with coding and have an account through a code website, you may work on that, otherwise go to this site to get started. If you are found online doing "something else" other than coding, you will get a zero for class work today. If you happen to get through the codes in blockly quicker than others, here is another site to continue with.
Tuesday, October 15, 2013
Today we will be doing more with modeling polynomials. For homework, the box problem and angry birds worksheet must be complete and watch this video for tomorrow. Tomorrow will be an example of flipping the classroom. :)
Monday, October 7, 2013
Today we will continue to view more polynomial FUNctions. We will review quadratics and then continue into end behavior, zeros and graphing of polynomial FUNctions. We will watch this video dealing with translations. The assignment will start with an investigation on pg 203, then lead into factoring (finding the zeros) and finally ending with work from the textbook.
Monday, SeptemberWednesday, September 11, 2013
Brandon, if you are able to print this quiz off that would cut down on missed class time tomorrow. If not, you will do it tomorrow. You are able to use your notes but not a graphing calculator, person or internet. Thanks for being honest. :)
Tuesday, September 10, 2013
For homework tonight view this video for the first 12 minutes. You may watch the whole entire video if you have time, but what we will be going over in class on Wednesday, after the assessment, will be about vertical asymptotes and possibly limit notation | 677.169 | 1 |
Student Resource Centre
Welcome to the student resource page. If you have just purchased
a TI-Nspire™ handheld or TI-84 Plus family graphing calculator,
then check out the Get Started information and Product Tutorials. If you
have a major test or exam coming up, try our Prepare for Exams section.
If you need a battery or wall charger for your TI-Nspire™ handheld -
or want to buy a new calculator — follow the Where to Buy link.
Prepare for Exams
Gain insights, tips and techniques that can help you use your TI-Nspire™ family handheld more effectively on your state's high-stakes exams. A bundle of resources have been developed including calculator solutions to practice exams and a series of short, state-specific video tutorials, developed in collaboration with experienced maths educators that provide step-by-step solutions for exam-style questions. | 677.169 | 1 |
This book considers a graph as a mathematical structure on a set of elements with a binary relation, and provides the most classical and important theory and application of graphs. It covers basic concepts, trees and graphic spaces, plane graphs and planar graphs, flows and connectivity, matchings and independent sets, coloring theory, graphs and groups. These topics, both theoretical and applied, are treated with some depth and with some suggestions for further reading. The treatment of material particularly lays stress on digraphs, the mutual connections among these topics and the equivalence of some well-known theorems. All theorems are stated clearly, together with full and concise proofs. A number of examples, more than 350 figures and more than 500 exercises are given to help the reader understand and examine the materials covered in the book.
Audience: The book is particularly suitable as a textbook of graph theory for senior or beginning postgraduate students who are majoring in pure and applied mathematics, operation research, computer science, designing and analysis of networks, electronics, scientific management and others. It is also suitable as a reference book for those readers who are engaged and interested in graph theory and for all researchers who use graph theory as a mathematical tool. | 677.169 | 1 |
Calculus Area Between Curves Task Cards QR
Be sure that you have an application to open
this file type before downloading and/or purchasing.
837 KB|12 pages
Product Description
Calculus Activity. Area Between Curves Task Cards.
This set of 11 task cards is designed for AP Calculus AB, AP Calculus BC, Honors Calculus, and College Calculus 2 students. This topic is covered typically in the Applications of Integration Unit.
Problems utilizing both horizontal strips and verticals trips are included. The integration involved is basic and is typical of integrals covered before the Techniques of Integration unit. The problems include both polynomial and transcendental functions.
Included:
•There are two sets of the 11 cards. The first set includes the shaded graph for that problem on the card, and the second set has only the description of the functions and the defined region for nine problems. You can mix and match the cards in various classes. There are four cards to a standard 8 1/2'' x 11" sheet of paper.
•QR solution sheet which can be passed around, projected on the screen, or cut up for a matching game. | 677.169 | 1 |
1-25 of 29,569 results
The second book in the Life of Fred Elementary Series; the second book of kindergarten material. Half Past Six. Other Math Concepts. Types of Numbers. Math Concepts. Cardinal Numbers. Facts about Butterflies.
James Stewart has revised this text in North America, retaining the focus on problem solving, accuracy, explanations, and the graded problems. This resulted in the strategy sections in the First Edition and the Problems Plus and Applications Plus in the Second Edition.
MODERN MATHEMATICS 8 by SB Jones, KE Couchman & JE Carroll. Great to use for extra tutoring for your high school child. Some writing/marks in pencil and some age spots/ slight yellowing of pages, but otherwise in great condition.
No other textbooks are like these. Each text is written in the style of a novel with a humorous story line. Tons of solved examples. Each hardcover textbook contains ALL of the material – more than most instructors cover in traditional classroom settings.
Rodents represent the largest order of mammals and include rats, mice and squirrels. This book is the only current volume that provides a comprehensive review of rodent pests and their control. The first four chapters establish the pest status of rodents and the need for control.
THE MATHEMATICS BOOK. ANYONE CAN DO IT by Helen Prochazka. This is a beautiful yet practical coffee-table style book that teaches anyone who doesn't much like maths or struggles with it, to appreciate mathematics and get to grips with the fundamentals of numbers and numeracy.
Inspired by Richard Feynman and J.J. Sakurai, A Modern Approach to Quantum Mechanics lets professors expose their undergraduates to the excitement and insight of Feynman's approach to quantum mechanics while simultaneously giving them a textbook that is well-ordered, logical, and pedagogically sound.
"Discrete Mathematics with Applications"
Second Edition
by Susanna S. Epp
Hardcover
1995
Please see photos for list of contents.
Small amount of chipping along top cover edge, some pages have crease, mark on one page (see photos), rest of book in good condition.
Thanks for looking!
The Immunoassay Handbook, Fourth Edition: Theory and Applications of Ligand Binding, ELISA and Related Techniques. The fourth edition ofThe Immunoassay Handbook provides an excellent, thoroughly updated guide to the science, technology and applications of ELISA and other immunoassays, including a wealth of practical advice.
A mathematician who is known throughout the world as the "mathemagician," Arthur Benjamin mixes mathematics and magic to make the subject fun, attractive, and easy to understand. InThe Magic of Math, Benjamin does more than just teach skills: with a tip of his magic hat, he takes you on as his apprentice to teach you how to appreciate math the way he does.
World Catalogue of Insects, Vol. 5: Tortricidae (Lepidoptera). Insect species may make up five or ten times the number of all other plant and animal species combined, and as such they represent one of the major challenges in biosystematic science.
GD (Good): A book that has been read but is in good condition. The dust jacket for hard covers may not be included. Binding has minimal wear. No missing pages. Subject: Science & Math / Astronomy & Space Science / Astronomy.
Anatomy and Physiology: From Science to Life. Written by a team of highly acclaimed authors, Anatomy & Physiology: From Science to Life arms readers with the knowledge and detail they'll need to move forward in their allied health careers.
College Physics, 9th Edition. Title: College Physics, 9th Edition. While physics can seem challenging, its true quality is the sheer simplicity of fundamental physical theories--theories and concepts that can enrich your view of the world around you.
Approximation Theory and Numerical Methods. Title: Approximation Theory and Numerical Methods. VG (Very Good): A book that does not look new and has been read but is in excellent condition. May be very minimal identifying marks on the inside cover.
Veterinary Clinical Parasitology. Title: Veterinary Clinical Parasitology. Condition: Used - Like New. LN (Like New): A book that looks new but has been read. Cover has no visible wear, and the dust jacket (if applicable) is included for hard covers.
Album of Fluid Motion. Title: Album of Fluid Motion. GD (Good): A book that has been read but is in good condition. The dust jacket for hard covers may not be included. Binding has minimal wear. No missing pages.
The Local Group is a small cluster of galaxies of which thirty-five members are currently known, including the Milky Way. He also places this knowledge in the wider context of continuing studies of galaxy formation and evolution, the cosmic distance scale, and the conditions in the early Universe. | 677.169 | 1 |
ISBN-10: 0321999061
ISBN-13: 9780321999061
Edition: 6 features the same content as the traditional text in a convenient, three-hole-punched, loose-leaf version. Books a la Carte also offer a great value-this format costs significantly In Thinking Mathematically, Sixth Edition, Bob Blitzer's distinctive and relatable voice motivates students from diverse backgrounds and majors, engaging them in the math through compelling, real-world applications. Understanding that most students in a liberal arts math course are not math majors, and are unlikely to take another math class, Blitzer has provided tools in every chapter to help them master the material with confidence, while also showing them the beauty and fun of math. The variety of topics and flexibility of sequence make this text appropriate for a one- or two-term course in liberal arts mathematics or general education mathematics | 677.169 | 1 |
Written for students, this book provides comprehensive coverage of digital signal processing with MATLAB. Gradually progressing from basic to more advanced topics, the text establishes a solid background in Fourier, Laplace, and z-transforms before covering them more extensively in later chapters. Topics covered include architecture and design of digital signal processors, computer architecture, logic design of sequential circuits, and random signals. | 677.169 | 1 |
This book is designed to offer practical advice and ideas for the
classroom in the teaching and learning of school geometry and has
presented a variety of ways of achieving this. However, to change what
happens in practice in the classroom may be difficult!
In the book are twelve chapters:
* The Role of Geometry
* Learning Geometry
* Beginnings: Experimental Geometry
* Polygons: Symmetry and Angle Properties
* Construction and Congruence
* Perimeter, Area and Volumes
* Enlargement and Similarity
* The Theorem of Pythagoras
* The Circle
* Linking Geometry and Algebra
* Polyhedra
* Vectors.
Working through Chapters 3, 4 and 5 highlight important initial
ideas and skills to do with recognising shapes and their properties, the
importance of symmetry, tessellations and the transformations of
reflection and rotation, angle properties and triangles, the properties
of quadrilaterals and polygons. Also to be found is an introductory
lesson plan on regular polygons using LOGO. The plan indicates lessons
are timed for fifty five minutes. Constructions and congruence and Van
Shooten's Theorem conclude this section.
In Chapter 6 the author suggests finding the area of a sector of a
circle is a good example of a task that should be presented as a problem
to be solved from first principles, rather than as a task to be worked
out using yet another remembered standard procedure. At this stage in
the book only the area of a circle has been met. Thus only first
principles are possible.
The direct link of similarity to the transformation of enlargement
and for the introduction of scale factors provides a more simple
approach than using equal ratios. Such is the content of Chapter 7. This
also includes both Varignon's theorem and the intercept theorem
being proved using the midpoint theorem.
Pythagoras' Theorem has a multitude of different proofs but
those considered in this book involve areas, congruence,
transformations, similarity, trigonometry and algebra. The trigonometric
identities and Apollonius' Theorem are also introduced.
The linked circle theorems of Chapter 9 are familiar to many
students and Miquel's six-circle theorem also appears.
In the following chapter the use of a coordinate system allows both
an algebraic and a geometric perspective.
It was a surprise to find a chapter on polyhedra included in a
course on geometry, so often has it been considered a recreational topic
in Australian courses.
The book closes with an interesting chapter on the use of vectors
in solving geometry problems. For many this could be unfamiliar
territory. A slight typing error appears on page 159:
[ILLUSTRATION OMITTED]
Throughout the book use has been made of dynamic computer programs
and software as well as graphical calculators. Diagrams are many and
include graphs and tables of values. The presentation of the book is
pleasing to the eye. This book has presented a way of looking at
geometry from various perspectives. Consequently each teacher of
geometry should have access to a copy.
Margaret McDonald
COPYRIGHT 2005 The Australian Association of Mathematics Teachers, Inc.
No portion of this article can be reproduced without the express written permission from the copyright holder. | 677.169 | 1 |
Mathcad Prime 2.0 Essentials
Overview
Course Code: TRN-3431
Course Length: 2 Days
In this course, you will learn the basics Mathcad Prime. You will learn about Mathcad Prime's extensive functionality such as opening and working with Mathcad files, navigating workspaces, defining variables and expressions, and solving equations. Further, you will learn how to plot graphs, solving for roots and manipulating data. At the end of each module, you will find a set of review questions to reinforce critical topics from that module. Your instructor will discuss these with the class.
At the end of the course, you will find a course assessment in Pro/FICIENCY intended to evaluate your understanding of the course as a whole.
Course Objectives
Open and save Mathcad files.
Navigate the Mathcad workspace.
Identify and format math and text regions.
Develop and edit math expressions.
Define, evaluate, and use variables.
Assign an expression retroactively.
Define and evaluate user-defined and built-in functions.
Define, evaluate, and use range variables.
Use units in calculations.
Plot 2-D and 3-D graphs.
Solve for the roots of a function with a single independent variable.
Symbolically solve equations.
Numerically solve a system of linear and nonlinear equations.
Solve unconstrained and constrained optimization problems.
Solve ordinary differential equations.
Create a program within the Mathcad worksheet using Mathcad's programming features.
Import and export data.
Smooth, interpolate, and regress data.
Prerequisites
n/a
Audience
This class is intended for the novice or intermediate user of Mathcad. | 677.169 | 1 |
PREFACE
This book is intended to cover manytopics in mathematicsat a level moreadvancedthan a junior level course in differential equations. The bookevolved from a set of notes for a three-semester course in the application of mathematical methods to scientific and engineering problems. The courses attract graduate students majoring in engineering mechanics, engineering science, mechanical, petroleum, electrical, nuclear, civil and aeronautical engineering, as well as physics, meteorology,geologyand geophysics. The book assumes knowledge of differential and integral calculus and an introductory level of ordinary differential equations. Thus, the book is intended for advanced senior and graduate students. Each chapter of the text contains manysolved examples and manyproblems with answers. Those chapters which cover boundary value problems and partial differential equations also include derivation of the governing differential equations in manyfields of applied physics and engineering such as wave mechanics,acoustics, heat flow in solids, diffusion of liquids and gasses and fluid flow. Chapter 1 briefly reviews methods of integration of ordinary differential equations. Chapter 2 covers series solutions of ordinary differential equations. This is followed by methodsof solution of singular differential equations. Chapter 3 covers Bessel functions and Legendrefunctions in detail, including recurrence relations, series expansion,integrals, integral representations and generating functions. Chapter 4 covers the derivation and methodsof solution of linear boundaryvalue problemsfor physical systems in one spatial dimensiongovernedby ordinary differential equations. The concepts of eigenfunctions, orthogonality and eigenfunction expansions are introduced, followed by an extensive treatment of adjoint and self-adjoint systems. This is followed by coverage of the Sturm-Liouville system for second and fourth order ordinary differential equations. The chapter concludes with methodsof solution of nonhomogeneous boundary value problems. Chapter 5 covers complex variables, calculus, and integrals. The methodof residues is fully applied to proper and improper integrals, followed by integration of multi-valued functions. Examplesare drawn from Fourier sine, cosine and exponential transforms as well as the Laplace transform. Chapter 6 covers linear partial differential equations in classical physics and engineering. The chapter covers derivation of the governingpartial differential equations for waveequations in acoustics, membranes, plates and beams;strength of materials; heat flow in solids and diffusion of gasses; temperature distribution in solids and flow of incompressible ideal fluids. Theseequations are then shownto obey partial differential equations of the type: Laplace, Poisson, Helmholtz, wave and diffusion equations. Uniquenesstheorems for these equations are then developed. Solutions by eigenfunction expansions are explored fully. These are followed by special methods for nonhomogeneous partial differential equations with temporaland spatial source fields. Chapter 7 covers the derivation of integral transforms such as Fourier complex, sine and cosine, Generalized Fourier, Laplace and Hankel transforms. The calculus of each of these transforms is then presented together with special methodsfor inverse transformations. Each transform also includes applications to solutions of partial differential equations for engineeringand physical systems.
111
PREFACE
iv
Chapter 8 covers Green's functions for ordinary and partial diffi~rential equations. The Green's functions for adjoint and self-adjoint systems of ordinary differential equations are then presented by use of generalized functions or by construction. These methodsare applied to physical examplesin the samefields c, overed in Chapter 6. These are then followed by derivation of fundamental sohitions for the Laplace, Helmholtz, wave and diffusion equations in one-, two-, and three-dimensional space. Finally, the Green's functions for boundedand semi-infinite media such as half and quarter spaces, in cartesian, cylindrical and spherical geometryare developedby the methodof images with examplesin physical systems. Chapter 9 covers asymptoticmethodsaimed at the evaluation of integrals as well as the asymptotic solution of ordinary differential equations. This chapter covers asymptotic series and convergence.This is then followed by asymptotic series evaluation of definite and improper integrals. These include the stationary phase method, the steepest descent method, the modified saddle point method, methodof the subtraction of poles and Ott's and Jones' methods. The chapter then covers asymptotic solutions of ordinary differential equations, formal solutions, normal and sub-normalsolutions and the WKBJmethod. There are four appendices in the book. AppendixA covers infinite series and convergence criteria. AppendixB presents a compendium special functions such as of Beta, Gamma, Zeta, Laguerre, Hermite, Hypergeometric, Chebychevand Fresnel. These include differential equations, series solutions, integrals, recurrence formulaeand integral representations. Appendix C presents a compendiumof formulae for spherical, cylindrical, ellipsoidal, oblate and prolate spheroidal coordinate systems such as the divergence, gradient, Laplacian and scalar and vector wave operators. Appendix D covers calculus of generalized functions such as the Dirac delta functions in ndimensional space of zero and higher ranks. Appendix E presents plots of special functions. The aim of this book is to present methods of applied mathematics that are particularly suited for the application of mathematicsto physical problemsin science and engineering, with numerous examples that illustrate the methods of solution being explored. The problemshave answers listed at the end of the book. The book is used in a three-semester course sequence. The author reconamends Chapters 1, 2, 3, and 4 and AppendixA in the first course, with emphasis on ordinary differential equations. The second semester would include Chapters 5, 6, and 7 with emphasison partial differential equations. The third course would include AppendixD, and Chapters 8 and 9.
ACKNOWLEDGMENTS This book evolved from course notes written in the early 1970s for a two-semestercourse at Penn State University. It was completely revampedand retyped in the mid-1980s. The course notes were rewritten in the format of a manuscriptfor a bookfor the last two years. I wouldlike to acknowledgethe manypeople whohad profound influence on me over the last 40 years. I am indebted to myformer teachers who instilled inme the love of applied mathematics. In particular, I would like to mention Professors MortonFriedman, Melvin Barron, RaymondMindlin, Mario Salvadori, and Frank DiMaggio, all of Columbia University's Department of Engineering Mechanics. I am also indebted to the many
PREFACE
v
graduate students whomadesuggestions on improving the course-notes over the last 25 years. I am also indebted to Professor Richard McNitt, head of the Department of EngineeringScience and Mechanics,PennState University, for his support on this project during the last two years. I am also grateful to Ms. KathyJoseph, whoseknowledge the of subject matter led to manyinvaluable technical suggestions while typing the final manuscript. I am also grateful for the encouragementand support of Mr. B. J. Clark, Executive Acquisitions Editor at MarcelDekker, Inc. I am grateful to mywife, Guler, for her moral support during the writing of the first draft of the course-notes, and for freeing mefrom many responsibilities at home to allow meto workon the first manuscriptand tworewrites over the last 25 years. I amalso grateful to mychildren, Emil and Dina, for their moral support whenI could not be there for themduring the first and secondrewrite of the course-notes, and for myson Emil who proofread parts of the secondcourse-notes manuscript. S. I. Hayek
a set of n-conditionson the dependent variable is required..let there exist twosolutions YI and YlI satisfying the system(1. w hich can be given as: y(x0) = cz0 y'(x0) = 1 : y(n-')(x0) = 1 A unique solution for the set of constants [Ci] in the homogeneous solution Yhcan be obtained. by successively multiplying the first row by an/a0. resulting in W = 9 and W(x) = 0
1. and manipulatingthe determinant. + Bny + yp + 2 n
. Such problems are known as Initial Value Problems. To evaluate the of constant W one can determinethe dominantterm(s) of each solutions' Taylor series. special case..3x whichis the Wronskian the solutions of the differential equation.28)
with W = constant. and adding themto the last row. 0.Since thelast two a re ruled out.29)
yp
YlI = B~y~ B2y + .1) derivatives take at a point x = 0.3. To prove uniqueness. one obtains: d__~_W:a.9
Initial Value Problems
For a unique solution of an ordinary differential equation of order n.29) such that: YI = C~y~+ C2y + . This is knownas Abel's Formula. etc. al(x) --> ~ or ao(x) ---> 0 at some point in a _< < b. 2
+ CnY + n
a < x.x 0 < b
(1. 0 It should be noted that W(x)cannot vanish in a region < x < b unless W van 0 ishes identically.8 Consider the differential equation of Example 1.. the secondrow by an_l/a0.. whose complete solution contains n arbitrary constants. then W(x)cannot vanish. Theset of n-conditions on the dependent variable is a set of the values that the dependent variable and its first (n .(x_~) _
dx
a0(x)
whichcan be integrated to give a closed form formula for the Wronsldan: ( a --~ a W(x) = 0 exp~j-I" a~(x) x~ (1. Example 1..ORDINARY
DIFFERENTIAL
EQUATIONS
13
into the determinantof (dW)/(dx). The Wronskian given by: is W(x) = W0exp(~-3 dx) e. find the leading term of the resulting Wronskian then take a limit as x --> 0 in this and -3x.
2).. Usingthe ratio test (AppendixA).2). one does not know y(x) a priori.. then there exists a set of solutions yi(x).2) ao(x) as a non-vanishing bounded functions and al(x).p. n. (1.. at x = x0 + p and x = x0 . Sucha solution can be expanded into a Taylor series about a point xo. refer to Appendix A.x°l > p series diverges where p is known as the Radius of Convergence. satisfying the systemin eq.29). 2 . (1.1)
c.1) convergesin a certain region. (1. or where:
19
. ) are boun in t he ded interval a < x < b. However.. and diverges outside this region.2)
This series is referred to as a PowerSeries about the point x = xo.1) and then the unknown constants n can be determined substituting the solution (2.1) into eq.e. The series may maynot convergeat the end points.. (1.2) one has a powerseries of the form in eq. by The powerseries in eq.. i. can assumethat the solution to eq.x0) > I series diverges n-~ ~ n Cn(X-X0)
< p series converges ¯ Ix . it is not possible to obtain the solution of an ordinarydifferential equationof the type of eq. i = 1.p < x < xo + p. so that the coefficients of the series c n are not determinable from (2.2) in a closed form.. a2(x) .2
SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
2.. (2..1
Introduction
In many instances. If the differential equation(1. a < x0 < b. (2. In general. such that: n y(x): where 2Cn(X-X0) n=O (2. then: xn+l < 1 series converges Lim Cn+l[x . : Y<n)'x°" n!
(2. Thusthe series convergesfor x0 .
If 1 is t he closest z ero of ao(X)to 0. 2 Since the homogeneous differential equation is of order n. all bounded ao(0) and ) Let the solution to be in the form of a powerseries about x0 = 0:
..1) can be transformedto powerseries solution about z = 0. i.x01.. (1. of the followingequation: of d2Y .e. cn are arbitrary constants..3) into the differential equation(1. Cn+ ..1) transformsto: dn-ly a0(z + Xo/~y + a. Totest the convergence the series at the end points.2 Power Series Solutions
Powerseries solutions about x = xo of the form in eq.1 Obtain the solution valid in the neighborhood x0 -. i. (2.xo then eq.(z + Xo)~:-i.CHAPTER
2
20
n .-> oo[ Cn and the ratio test fails. p 2.e.. c1 and c0. can then be computed terms of the arbilrary constants co . refer to the of tests given in Appendix A. and a2(x) = -x.+ ... 1...xy = 0 2 dx Notethat ao(X = I.2) and equating coefficient 0f each powerof x to zero... one only needs to discuss powerseries solutions about the origin. then the radius of c onvergence= IxI .. c1 . the constants c0. points whereao(x) vanishes.e. i... results in an infinite number algebraic equations. Let z = x . + an_l(z + Xo) dY+ an(z + Xo) y = f(z Thus.:
TM
y(z)=
~ cruz rn=O
Henceforth... c in 2 n. The constants Cn+l.. for m= 1.. Theradius of convergence series solutions of a differential equation is limited by for the existence of singularities. i.e.. of each one gives the constant cmin terms of Cm_ cm_ ... then there will be n arbitrary constants.: (2. 2 .0. power series homogeneous solutions about x = xo becomeseries solutions about z = 0. y(x)= Example 2. al(x) = 0.3) ~ xm m=0 Substitution of the series in (2. whichwill be token to be x = 0 for simplicity.
2 .(~r2) n=0 Example 2. 1.n
m=0 Equatingthe coefficients of x~-l and xm+~ zero and assumingao ~ 0. such that each of the remainingseries starts with x~ one obtains:
=0 n=0 Changingthe indices n to m+ 2 inthe first series and to min the second and combining the tworesulting series:
+
(n+ ~)2_~anxn+~-2 ~"a x "~ +'9 L. use a Frobenius solution about x0 = 0: xn+~r y = ~an n=0 such that when substituted into the differential equationresults in: xn+~':2+ ~ [(n+~)~-~]an 2anxn+°=O n=0 n=0 Extracting the first two-lowestpowered terms of the first series..14)
Since x = 0 is a RSP.CHAPTER
2
28
yl(x) = 2an(or.. there results to the following recurrence formulae:
am a~÷~=(m+~+:z)2_ N and the characteristic equation:
m=0.) xn+~rl n=0 and y2(x) = ~a.
..4 Obtain the solutions of the followingdifferential equation about x0 = O: (2.
:~)
x~'-2 0.)] = L[-~(0.am~2)) xn+~r2 +ak n=k ~" ak
k-1 or ~.e.19) one obtains:
. =a 0 y.) = aoxa-Z(~r . then one must find another methodto obtain the second solution.02) and differentiating partially with o. A new solution similar to Case (b) is developednext by removingthe constant o .) = a0xa-2f(0.:
y~(x) (0. one obtains: -~ [(0. (2.. (2. ... The first part of the solution with ao maybe a finite polynomialor an infinite series. i.~)(0. (2.
33
If gk(ff2) vanishes.0.0"2
= ao
Thus.11) of given by: Ly(x.-0.0.)= ~a. dependingon the order of the recurrence formula and on the integer k.SERIES
SOLUTIONS
OF ORDINARY
DIFFERENTIAL
EQS.0")[= 0.0.0. If gk(O2)does not vanish.~)y(x.1"/)
It can be shown that the solution precededby the constant ak is identical to Yl(X). the function that satisfies the homogenous differential equation:
0" = 0"2 gives an expressionfor the secondsolution.o~ from the demoninator an(o).0. .lx"+a+ ~a. i.1 0.
n----0
[an(~2)]
=0
xm+~ (2.2 =--~ a
(2A9)
The Frobenius solution can be divided into two parts: n=k-1 y(x.(0")x n=0 n=0 n=k so that the coefficient ak is the first term of the secondseries.: k-lor~ Y2(X) : a0 n=0 an(a2) ~ / \ n÷~2 ÷ ak m~ (.0.~) y(x. Since the characteristic equation in eq.0.0. . then ak(~z) is indeterminateand one maystart a newinfinite series withak. Differentiating the expressionas given in eq.(0.e.(alx"+~r= ~a.2)Ly]= ~-~ [L(0.¢rz) then multiplying eq.18) by (o . thus one can set ak = 0 and ao = 1.
3. (3. up Helmholtz the waveequations in the radial spherical coordinate.~'..28)
(3.18) and (3..-. resulting in the following expressions:
J"+'/~.31)
d/n¢cOSx/dx)XJ J-(n+l/2)=%2/2/2~xn+1/2(-~ k.22) by setting p = 1/2. it can be shown that: (2m+ 1)! =
m:O
sin x
which can be shown result in the following closed form: to
(3. the following recurrence formulae can be obtained:
.. Thus.27).32)
Thesefunctions satisfy a different differential equationthan Bessel's havingthe form: x2 d2-~-Y 2x dy + (x2 _ v2 _ V) y = 0 + (3.29) To obtain the higher ordered half-order Bessel functions Jn+l/2 and J-(n+l/2). one can use the recurrence formulaein eqs.CHAPTER
3 2m x
52
J'/2
J~(x)-.24 .30)
(3. known as the spherical Bessel functions of the first and second kind of order v: Jv = Yv = Jr+l/2 J-(v+ 1/2) (3. Onecan also obtain these expressions by using (3.3.m ~(-0
m=O Similarly.33) dx2 dx For v = integer = n. ='-"
LTXJ [-7-)
(3.24) to (3.1/2. the first tWO functions Jn and Ynhave the followingform: -sinx j0 x x jl = lfsinx x\ Y~ =_cosx/ "2
Y0-
(co )
-x +sinx
Recurrence relations for the spherical Bessel functions can be easily developedfrom those developed the cylindrical Bessel functions in eqs. (3.27) by setting for p = v + 1/2 and -v .6
Spherical Bessel Functions
Bessel functions of half-order often show as part of solutions of Laplace. Define the following or functions.
2.45)
is the first solution. then: In(x) ~'~ m~'~= m!(m+ n)! 0 n = 0.Kp
x
(3. (3. then taking the limit p --~ n:
0P p. 2 . one can obtain the following formulae for Ip and Kp:
Ip = Ip+ 1 + p Ip x
(3. (3.2) and (3...
(3.47)
Kpis knownas the Maedoualdfunction.50)
Following the development of the recurrence formulae for Jp and Yp detailed in Section 3.51)
. (_l)n+l[log(X//2) + ~/] in(X)+ ~ m~-i_l)n-t m=0
(n-m-1)'
~.. respectively.46)
(3.I_p) 2 si n (p r0 (3.2
.. [x/~ ~ m~/m+n)! [g(m)+g(m+n)] +(-1)"2 n=0.n The Wronskian the various solutions for the modified Bessel's equation can be obtained of in a similar mannerto the methodof obtaining the Wronskiansof the modified Bessel functions in eqs.42) takes the following form: y = clip(x ) + Cfl_p(X) If p takes the value zero or an integer n. The second solution must be obtained in a similar manner as described in Sections (3.3) giving:
Kn (x)__.5) and (3.. m=0 The secondsolution can also be obtained from a definition given by Macdonald: 2 L sin (p~)
(3.p(X) are known. 1.7): W(Ip. If p is an integer equal to n.SPECIAL
FUNCTIONS
55
I_p(X)=
~ m!r(m_p+l)=(i)PJ_p(iX) m=0
" p*0.4..1. as the modified Bessel function first and second kind of order p. 1.44)
of the
Ip(X) and I.49)
p. The general solution of eq..
(3..
.117)
can be obtained from published tables...0=0forp>0 Theroots of Jp and Ypinterlace. and J~ and Y~ are also well tabulated. Y~.(x) Y~(ax).~< "< < P < Yp. All roots and 1).(x)
(3. [Abramowitz Stegun].1 can be bracketed such that:
~ <Jp.1 Yp..2 Jp.J~(x) and Y~(x) Jp.2< Yp+l.1< Yp+i.CHAPTER 3
68
3. J'~. Ref.. z) of H(p H~ .2 < Yp.
Jp. and The large zeroes of the spherical Bessel functions of order n are the sameas the zeroes of Jp..1
+3) <42(p+l)(P
(3.2< Jp+~. except at the origin..115)
6.
. then all the :zeroes by functions have the following properties: 1.s. That all the zeroes of these Bess~lfunctions are real if p is real and positive. 2.(ax) Y. Spherical Hankelfunctions have no real zeroes. The roots of Jp.2 Yp. Yp(x).3< .
The roots Jp.s.1 < Yp.15 Zeroes of Bessel Functions
Bessel functions Jp(X) and Yp(x)have infinite number zeroes. Thereare no repeated roots.1 and jp.< Jp.2 < Jp. Yp.(x)= Jp(X) Y~(ax).J~. such that: < P <Jp.1 < Jp.~.1 < Jp.Sp(aX)
Y. I_p. Ref. 4. usually appearing in boundaryvalue problems of the following form: Jp(x) Yp(aX)-Jp(aX) Yp(x) J~.1 < Yp.2 < . P5. Yp. J~ and Y~ with p = n + 1/2. Yp. Ip. The large roots of Bessel functions for a fixed order p take the followingasymptotic form:
#. Jp. [Abramowitz Stegun].116) 2 2
The roots as given in these expressions are spaced at an interval = ~.
s+7(3.2 < Yp. Denotingthe th of root of Jp(x). 3.. <Jp+z. and Kpare complex real and positive orders p. for The roots of products of Bessel functions..
x2) P:] + n(n + 1) Pn d-~-£ [(1. one can showthat: +1 dx fPn Pm = 0 -1 If n = m. Thedifferential equation that P.2 1 2n +
(3. (3.154) are due to Mehler.
3. m. and subtracting and integrating the resulting equations.~ m
(3. then the last integral becomes: +1 +1 2 f Pn dx=(-1)n(2n)'22n(n!)2 2-1)ndx f(x -1 -1 Integrating the last integral by parts. dx----~. one obtains:
.x2)P~ ] + m(m+ 1)Pm Multiplyingthe first equation by Pro.(x2 . the secondby Pn.155)
dx-. one obtains: +1 -1
n .20 Integrals
of Legendre Polynomials
One of the most important properties of the Legendrepolynomialsis the orthogon~lityproperty. The first integral to be evaluated is an integral of products of Legendre polynomials.153) and (3.~ and Pm satisfy for n ~ mcan be written in the following form: d--d~[(1.156)
The orthogonality property can also be provenby integrating the differential equation. Integrating by parts.126): +1 +1 n>_m Pn Pm dx = 2n+mn.1) n x2 --1)m dx -1 -1 wheren is assumedto be larger than m. The integral of products of Legendrepolynomialscan be evaluated by the use of Rodrigues' formula (3.SPECIAL
FUNCTIONS
81
The integral representations in eqs.
These points are called Boundary 't Points. If the differential equation as well as the Initial Conditionare homogeneous. resulting in: C2 = -1 and the complete solution to the problem becomes: C1 = 2 . was shown that the solutions to such problemsare unique and valid over the range of all values of the independent variable. then the solution becomes: Y = 2sin2x -_-3 cos2x 2 for all X and
107
. will be explored.e.1) derivatives at an initial point were discussed in Section (1. To illustrate the primary difference betweenthe two types of problems.1 Obtain the solution to the following initial value problem: Differential Equation(DE): Initial Conditions(IC): y" + 4y = f(x) = y(~/4) = y'(~/4) =
The completesolution to the differential equation becomes: y = C~sin 2x + C2 cos2x + x The two arbitrary constants can be evaluated from the specified two initial conditions at the point x0 = ~c/4.cos 2x + x for all If the differential equationis homogeneous. if f(x) = 0.8) and were referred to as Initial Value Problems. solutions to linear differential equationsof order n with n conditions specified on two end points of a boundedregion valid in the closed region betweenthe two end points. and the initial conditions i.. In this chapter.the solution of two simple problems are shown: Example 4. Such problems are referred to as Boundary Value Problems (BVP).1
Introduction
Solutions of linear differential equations of order n together with n conditions specified on the dependent variable and its first (n .~/4) sin 2x .1) are called Boundary Conditions (BC). then it can be shownthat the solutions to such problems vanish identically. and the conditions on the dependent variable and its derivatives up to the (n .4
BOUNDARY VALUE PROBLEMS AND EIGENVALUE PROBLEMS
4. are non-homogeneous.n/4 y = (2.
... thus. 4..
the eigenvaluesare given in terms of the roots o~: 2 ~.. 3 . 2... the resonant frequencies of the finite rod are given by: o~ n =ckn an =c L
2
n = 1...m a where a = kL
The roots of the transcendental equation on tx~ can only be obtained numerically.CHAPTER
4
116
BC:
u(0) =
and
AE du + d--~"
~l x=L :0
equation is: The solution to the homogeneous u = C~sin kx + C cos kx 2 which is substituted in the two homogeneous boundaryconditions:
u(0)=o =
For non-trivial solution.. An estimate of the location of the roots can be obtained by plotting the two parts of the equation as shown Fig. 2. cq.5.n =k~ =~-~" n = 1... 3 .
The root ~ = 0 correspondsto a trivial solution.. it is not an eigenvalue.
. a2 . 3 .. There is an infinite number roots al.. the bracketed expression must vanish resulting in the following characteristic equation: tan a = .
and the corresponding eigenfunctions (modeshapes) are given by: t~n = sin knx = sin an ~
x
n = 1. 2... in of Notethat the roots for large values of n approach: 2n+ 1 2 n>>l Thus.
4.then oneconsidersan from elementof the deformed beam.BOUNDARY VALUE AND EIGENVALUE
PROBLEMS
117
Y
Fig. which makes sketching themeasier.5
It shouldbe notedthat the eigenfunctions ~bn(x)haven nulls.
4. 4. If the beam deforms its straight line configuration. WavePropagation and Whirling of Beams
Thevibrationof beams the whirlingof shafts can be considered a similar or as dynamic systemto the vibration or whirlingof strings. as shown Fig.6. 4. on in
. which acted upon of is by distributedforcesf(x). Consider beam massdensity a of p.wherethe shear V and the moment exerted by the other M parts of the beam the elementare shown Fig.7. cross-sectional area Aandcross-sectional area moment inertia I.4 Vibration. andis rotated aboutits axis by an angularspeedo~.
one obtains: x+dx Vx+dxdX+Mx-Mx+dx . the local strain. The element undergoesrotation (dO)and elongationAat a location ds A ----d0=-R z Thus. expanding Vx+dxand Mx+dx a Taylor's series about x and using the mean by value for the integral as dx --> 0. results in the followingrelationship between the moment the shear: and Vx = dMx dx Thus.p~02Ay (~
X
Again.x) d~.BOUNDARY
VALUE
AND EIGENVALUE
PROBLEMS
119
Takingthe equilibrium of the moment about the left end of the element. 4.8:
Element of a Beam Deformed in Flexure
. defined as the longitudinal deformation z per unit length is given at by: ds
Fig. so that the element a subtends an angle (dO) and has a radius of curvature R.f f(~)(~q . the equation of motion becomes: d2Mx= p~02Ay+ f(x) 2 dx The constitutive relations for the beamunder the action of moments and Mx+~x be Mx can developedby considering the element in Fig. 4. The element's two cross sections at its ends undergoes rotation about the neutral axis.8 of length s.
~A 052
(4. (4. (4.: y* = y(x) sin tot
.t) and f* = f*(x. If the motionas well as the applied force are time-harmonic. then the moment obtained in terms of the secondderivative of the is displacement i. due to the stress field at z over the cross-sectional of area of the beamgives: MomentMx=fCyzdA=_~fz2d A A A E1 R
whereI = f z2dAis the moment inertia of the cross-sectional area A. M = El dzy x 2 dx and the equation of motion for the beambecomes: dx 2 E1 = p052Ay+f(x) (4.: y.t) can be obtained by replacing the centrifugal force by the inertial force (-9A ~2" dx /¯ Replacing d/dx by 0/3x such that eq. i.6) simplifies to: d4y l~4y = f(x) dx 4 EI wherethe wavenumber is defined by: I~
64 = p.CHAPTER
4
A z
12 0
strain e = h = _ ds R and the local stress is given by Hooke'sLaw: Ez strain o = Ee = -R Integrating the moment the stress. of A Since the radius of curvature is defined by: 1 R dO d2y 2 dx ds
for small slopes.e.6) becomesthe waveequation for a beam:
02 (EI 02y*~ ~-~'T ~.t).8)
wherey* = y*(x.e.7)
EI Thewave equation a time dependent for displacement a vibrating beam of y*(x. then the equation of motionfor the beameq. 0-~-) + PA~2*= f* x't )
(4.6)
If the functions EI and A are constants.
then one mustchooselq~ = cm.~'~.
n ~.}is an orthonormal then J becomes: set. then N extendsto infmity. inslead of an Cxluality. then: b n ~F2 dx> ~c2~ a m=l
.the series: J. Thus. The GeneralizedFourier Series is the best approximationin the meanto a function F(x). If an orthonormal {gn}extendsto an infinite dimensional set space. the definedas:
hi
n
q2
a/ m=l / mustbe minimized.CHAPTER 4
13 6
F(x). Since J ~ 0. squareof the error is expanded The as: J=~F 2 dx-2 Ekm~F(x) gm(x)dx+ kmgm(X) ) dx a m=l a al_m = 1 J/r=l Sincethe set {g~. Thesymbol usedfor the representation to ~. Cmgm(X)
m=l is the best approximationin the meanto the function F(x).b]..24)
Theseries representation of (4. to cm Thesquareof the error J between functionF(x) andits representation. refers to the possibility that the series may converge F(x)at some not to point or points in [a. b n n
J=~F2dx-2 ~Cmkm+ 2k2m->O
j=I/F<x)-kmg / ax_>0 E <x)
J
a
b n
m=l
m=l
a
b n n
m=l m=l a To minimize whichis poSitive. gm(x) F(~l) gm(~l) m=l
a<x<b
(4.24) is called the Generalized Fourier Series corresponding F(x). Consider finite number an orthonormal as follows: a of set
n ~kmgm(x)
m=l then onecan show that the best least squareapproximation F(x) is that where = k~.
i. } is complete.e. if the set { g. Thustwo functions representedby the samegeneralizedFourier series must be equal.24) can be developed.
a
(4. and if the generalized Fourier series set correspondingto F(x) is uniformlyconvergentin [a.if the orthonormalset {g.b].b]. then the series converges uniformlyto F(x) on [a.(x)} orthonormalwith respect to a weighting function w(x) as follows:
F(X)= ) ~Cmgm(X
m=l where
c m
= f w(x) F(x)gm(X)dx
a
b ~w(x) gm(X)gn(X)dx
a
(4.26)
. (4.:
m~ A necessaryand sufficient condition for an orthonorrnalset { g.(x)} to be completeis that: b ~F2(x) ~ : ~C~m m=l
a The generalized Fourier series representing a function F(x) is unique. If an orthonormal is completeand continuous. if F(x) is continuous.BOUNDARY
VALUE
AND EIGENVALUE
PROBLEMS
137
Theaboveinequality is not restricted to a specific number thus: n. Similar expansionsto eq.25)
m=l then the Fourier Coefficients Cmmust constitute
a
convergent series. b ~ 2 ~ F dx > ~cZ~ a b Since J-Fz dx is finite.
then each coefficient is multiplied by z(x). linear differential equation can be transformedto a formthat is self-adjoint. If the operator L is self-adjoint.al) y' + (a['. then the followingequalities must hold: a1 = 2a~. can change the second order operator L by a one suitable function multiplier so that it becomes self-adjoint.BOUNDARY
VALUE
AND EIGENVALUE
PROBLEMS
139
Ky=(a0y ) -(aly ) +a2y =aoy" + (2a~ .33)
Thus. any secondorder. Since the condition for self-adjoincyrequires that the differential of the fast coefficient of L equals the second. However.a~ + a~ 1 whichcan be satisfied by one relationship. the self-adjoint operator L1can be rewritten as: L1y = p(x) y"+ ~p(x) az( x) p(x o( ) + ao(x) = Id(pd~+q] Ldx i.a~ +a2)y whichis not equal to Ly in general and hence the operator L is not self-adjoint. then: (z ao) = z whichis rewritten as: z" a1 . an operation that is valid only for the secondorder operator. dxJ where : p(x)exp[~al(rl) d~] Y
q(x) = a2(x) ao(x ) ¯
(4.a 0
z a o
Thefunction z can be obtained readily by integrating the abovedifferentials: l_~exp[ ~ ai(rl) drll = p(x)
Z=ao<x)
Usingthe multiplier function z(x). then: LlY = z Ly so that L1is self-adjoint. namely: a0 = al which is not true in general. Hence.if one multiplies the operator L by an 1 undetermined function z(x).a and a2 = a~ . The method used to changea secondorder differential operator L to
.
Therefore.. Thus. In general.0.condition for the solution of such problemsto be unique.2.ao(x It shouldbe notedthat if the order n of the differential operator L is an odd ).3 ..35) constitute a general form of boundaryvalue problems. wouldhavea changeof sign of the coefficient of its highest derivative if x is changed to (-x). then. n (4. Thus. (4.. a physical systemgovernedby a differential operator L on a spatial coordinate is self-adjoint if the systemsatisfies the law of conservationof energy. necessary and sufficient. since that requires tlhat at(x) = . n has only the trivial solution y -. an (n)~ order self-adjoint operator given in eq.27) and the non-homogeneous boundaryconditions (4. representing the system'sgoverningequation. since the set of n homogeneous Ui(y)---
.27) has n independent solutions {yi(x)}. ~ij] # 0 The non-homogeneous differential equation (4. Thus. the differential operator L is self-adjoint. then that operator cannotbe self-adjoint. This meansthat the determinant: det[l~ij.34)
4. the operator is not the sameif x is changed (-x). general form of non-homogeneous boundaryconditions can be written as follows: n-1 i= 1. is that the equivalent homogeneous system: Ly = 0 i = 1. i... A general form of a linear..CHAPTER
4
140
become self-adjoint cannot be duplicated for higher order equations. then the differential equationis not invariant to coordinateinversion. and Ti are real constants. 2 . the solution of a systemis unique iff n conditions on the function y and its derivatives up to (n . a differential operator L whichrepresent a physical system's governingequation on a spatial coordinate x cannot have an odd order n. This wouldlead to a solution that is drastically different fromthat due to uninverted operator L.1) are specified at the end points a and b... The boundaryconditions in (4. integer. if the governingequations are derived from a Lagrangianfunction representing the total energyof a system.if the independent to variable x a spatial coordinate.35) Ui(Y)= X[~iky(k)(a)+[~iky(k)(b)]=~'i k=0 wherecq.-[p~_ly']' + p~y= a < x _< b (4. Thus.12
Boundary Value Problems
As mentioned earlier.e.. then the operator L. non-homogeneous th order differential operator L whichis self-adjoint can be (2n) written as follows:
k=0 = (-1)n [poy(n)](n)+ (-1)~-l[ply(n-1)](~-1)+ . Thus...35) must independent. In general if the order n is an oddinteger. I~.
: to Ui(u) = Vi(u) Example 4. V)[a =0 (4.30).e.e.35) (4. P(u. i. 2.28) i=1.-(aov)'l+u [aoV]la a~xSb (4.35) are independent..27) 0 i ----1.. If the operator L is a selfoadjoint operator.36) are obtained by substituting the boundary conditions Ui(u) = 0 in (4. (4.37) results in the following:
b=u'(b)[ao(b) u'(a)[ao(a u'(a0V)la v(b)])
v(a)] =
.34) can be written as: n y = yp(x)+ ECiYi(X) i=l where C are arbitrary constants.BOUNDARY
VALUE
AND EIGENVALUE
PROBLEMS
141
conditions given in eq. Since the set of n non-homogeneous boundary i conditions in eq.35) are independent.. then the boundary conditions can be shown be identical. v) is given by: b b .10 For the operator: Ly=a0(x)y"+al(x)y'+a2(x)y=0 the adjoint operator is given by: Ky=(aoy ) -(aly ) +a2y=0 The bilinear form P(u.e. i. then the solution of the differential eq. (4. if K = L.2 .. v) being given in (4..then there exists a non-vanishing uniqueset constants [Ci] whichsatisfies these boundary conditions.37)
with the bilinear form P(u..v) la= ulalv.35) into: b P(u. n (4. i.38)
(i) Consider the
boundarycondition pair on u given by:
u(a) = and u(b) = and substitution into (4. (4.: Lu=0
Ui(u ) :. n (4. A homogeneous boundaryvalue problemconsists of an nt~ differential operator and a set of n linear boundary conditions.
(4.36)
Anadjoint systemto that defined aboveis defined by: Kv =0 Vi(v)=0
where the homogeneous boundaryconditions (4.
Similarly since u(b) = O.v) = u'(b)[ao(b ) v(b)]-u(a)Ial(a ) v(a)-(ao(a ) v(a))'] Since u'(a) = 0 then u(a) is arbitrary.13 Eigenvalue Problems
Aneigenvalue problem a systemthat satisfies a differential equation with an is unspecified arbitrary constant ~.. 2n (4.41)
L ""
J
n>m
qm_kd-j
)~ is an undeterminedparameter. Similarly..2 ... then u'(b) arbitrary. u'(a) is also arbitrary. the relation satisfiedif: v(a) = and v(b) =
(ii) If u'(a) and u(b) = then one obtains the following whensubstituted into P(u.35). where L is given by (4.
. For arbitrary constants u'(a) and u'(b)... 2 . A general form of a self-adjoint homogeneous eigenvalue problem takes the following form: Ly + ~. Thus. the boundaryconditions Vi(v) = 0 are: al(a) v(a)-[ao(a ) v(a)] =
v(b)= 4.35). since u(b) = O.My= 0 (4.CHAPTER
4
142
Since u(b) = 0 then u'(b) is an arbitrary constant.39)
Ui(y)=0 i= 1..My= 0 U~(y)= a < x _< b i = 1..27) and the boundaryconditions by (4.-'~'/dky]
(4.40)
whereL and Mare linear self-adjoint operators of order 2n and 2mrespectively. and Ui(y) = 0 are 2n homogeneous boundaryconditions having the form given in (4. is an arbitrary constant. where: -1)k E( Ly= k=O My= E(-1)kd--~ k=O
~. ~Theoperator Mis th m order differential operator wherem< n and ~. Consider a general form of a homogeneous eigenvalue problem: Ly + ~. and satisfying a homogeneous non-homogeneous of or set boundaryconditions.
whichindicates that the only complex eigenfunction possible is the null function. un = v~ = O. then: b
JO dx:o MO
a
whichresults in the followingreal integral: b ~ (unMu.x*)J0nM0: dx=
a
Since the eigenvaluesare complex.nMOn = 0
. i. then the eigenvalues ~ are also positive.i.(x) + ivn(x ~n = Otn + i~n ) 0. -i[3 n
then the orthogonalityintegral (4. = un(x)-ivn(x) X~ = a. i.e. Starting with the equation satisfied by On:
IL~ n + ~. Since the systemis positive definite. then the eigenfunctionsare real and the eigenvalues are real and positive.L~n
b
~(unLun + vnLvn)
a ~'n = tXn+ i~n = b
dx O:MOn ~ a
a b
~ (unMtln a + vnMvn)
Since the systemis positive definite and the integrands are real.+ vnMvn) dx = 0
a
Invokingthe definition of a positive definite system. Having established that the eigenvaluesof a self-adjoint positive definite systemare real and positive one can obtain a formulafor ~. Onecan also showthat the eigenvalues ~ are real and positive. =.e.e. Starting out with the differential equationsatisfied by either ¢n or 0~ .BOUNDARY
VALUE
AND EIGENVALUE
PROBLEMS
145
is a generalized form of an orthogonality integral.:
L~n + ~LnMOn =0
and multiplying this equation by 0~ and integrating over [a. both of thes~ integrals are positive. let: On= u. whichindicates that 13.45) results in the followingintegral: b
(x.e. one obtains an expression for ~: b
dx ~O. with N~being the normalization constant. then these integrals are real. i. (ii) Real eigenfunctions and eigenvalues If the systemis self-adjoint and positive definite. Assuming that a pair of eigenfunctions and eigenvaluesare complex conjugates.. ~3~~ O.b].0 and ~ is real.
b].y= 0 0<_x_<~ y(r~) : 2 and M= 1 and hence the systemis self-adjoint and also For this system. However. dx ~'n = .: to b dx = 0 i = 1. Define the iRayleigh quotient R(u) as: b uLu dx
~
R(u) = . Solving the problemexactly. 2.47) uMu dx
~
a whereu is a non-vanishingcomparisonfunction.CHAPTER
4
146
and multiplying the equation by ~n and integrating the resulting equation on [a.46)
a
(iii)
Rayleigh quotient The eigenvalues ~ obtained from eq.one can obtain an approximate upper bound to thee eigenvalues if one can estimate the form of the eigenfunction. (4. already known.12 Obtain approximatevalues of the first two eigenvalues of the following system: y"+Z. one obtains: b f ~.. It can be shownthat for a selfadjoint and positive definite system: ~1 = min R(u) where u runs through all possible non-vanishing comparisonfunctions. one can showthat it has the fi)llowing eigenfunctions and eigenvalues: y(0) =
. r 3
~uM~i
a
then
Z.~
~
>0 nM~n dx
(4.~+~ min = R(u)
Example 4.a
(4. L = d2/dx positive definite. whichof course could have been obtained only if ~. It can also be shown that if u runs through all possible comparison functions that are orthogonal the first r eigenfunctions.i...I_@.46) require the knowledge the exact form of of the eigenfunction ~n(x).e..
y:0
\
where p=l q=0 and r=l
. be Example 4.: b
f r(x) (~n (X) a =N n n=m
(X)dx =
n ~ m
(4.BOUNDARY
VALUE
AND EIGENVALUE
PROBLEMS
151
The eigenfunctions ~o of the system are thus orthogonal. y"+~. Thus: b b ~u Lu dx =~ uI(pu')' + qu] dx =~ [-p(x)(u')2 + 2] dx <0 a a a b
a
b dx>0
a
fuMudx=fru2
Thus. the systemmustbe positive definite (see 4. q and r constitute a complete orthogonal set and hence may used in a GeneralizedFourier series.45). i. it is sufficient (but not necessary)that the functions p. (4. eq.44). It can be shown that the set of orthogonal eigenfunctions of the proper S-L system with the conditions imposedon p. since it can be readily rewritten as: d--~ C d-~-Y/+ ~. satisfying the following orthogonalityintegral.e.57)
In order to insure that the eigenvaluesare real and positive.y = 0 y'(0) = 0<x<L y'(L) =
The systemis S-Lform already. q and r satisfy the following conditions for positive-definiteness: p(x) > a<x<b q(x) _< (4.58)
r(x) > to guaranteereal and positive eigenvalues. giving explicitly the orthogonality conditions and the normalizationconstants.13 Longitudinalvibration of a free bar
Obtainthe eigenfunctionand the eigenvaluesfor the longitudinal vibration of a free bar.
Example 4. It is simply supported at both ends such that: y*(0.then the constant 3.BOUNDARY
b
VALUE
AND EIGENVALUE
PROBLEMS
159
n an (~.n)Nn
whereN is the Norm the eigenfunctions and: of n b dx bn = f F(x) ~n(X)
a
(4.t) = y*"(0. so that ~. ~ )~n. is complex valued. So if the real part of ).16 Forced Vibration of a Simply Supported Beam
~a sin (cot)
Obtain the steady state deflection of a simply supported beambeing vibrated by a distributed load as follows: f*(x.t) = y*"(L._ ~.t) =
.t) = y*(L. the solution to Ynbecomes: bn YlI(X)= 2 ()~_~.73)
The solution due to the source term F(x) can be seen to become unbounded whenever becomes equal to any of the eigenvalues )~n" It should be noted that if the systemhas inherent absorption. is equal to )'n. since ~'n are real and positive.72)
Thus. the solution YII becomes large but still bounded.t) = f(x) sin where f(x) = {~0/2a L/2-a<x<L/2+a everywhere else 0 -< x < L
The beamhas a length L and has a constant cross-section.n)Nn n=l (4.
.BOUNDARY
VALUE
AND EIGENVALUE
PROBLEMS
169
(a) The series convergesto F(x) at every point whereF(x) is continuous ÷) (b) The series convergesto [F(x + F(x-)]/2 at a point of ordinary discontinuity.on sin ~ n--1
(f) If F(x) is piecewise continuousthen one mayintegrate the series term by term any number times. This requires that 2 m be s et ust to zero to insure that y(0) is bounded. n~x) cos--L. wherever F(x) is discontinuous has finite right and left derivatives.x .
4. i. the remainingsolution: y = C1Ja((~x ) satisfies the conditionthat:
. i. then y(0) must be finite and py' ---> 0. The systemis first transformed to S-L system.: F'(x)~~ ~ n(a n n~ .e. but (c) The series represents a periodic function in the open region -~ < x < (d).: of sin x+box
7t ~ n cos L x+--~t n=l n=l This series converges faster than the series for F(x). Y2are known constants.21
Fourier-Bessel
Series
Consider the following system: '+(t22x2-a2)y=0 x2y"+xy with y satisfying the following conditions: y(0) is bounded VlY(L)+ "~ 2Y'(L) where~q..Thus. The series converges uniformlyand absolutely in -L < x < L if F(x) continuous. F'(x) is piecewise continuous and F(L) = F(-L). (e) The series can be differentiated term by term if F(x) satisfies the conditions (d).e. i. having the form: d----(xdY)+(t~2x-~-/dx\ where p(x) = q(x) = -a2/x r(x) = y=0 0_<x_<L
The solution to the differential equation becomes: y = C1Ja(Ctx + C2Ya (CtX) ) Since p(0) = 0.e.
such that the pressure is finite at x = 0 and at the end x = L is: (a) open end,
(b) rigid
13. Obtain the natural frequencies and mode shapes of standing wavesin a parabolic acoustic horn, whose cr0ss-sectional area varies as follows: A(X) = X4 wherethe pressure at x = 0 is finite and the end x = L is open end. 14. Obtain the natural frequencies and modeshapes of standing wavesin an exponential horn of length L whos~ cross-sectional area varies as: 2'x = Aoe A(x) such that the end x = 0 is rigid and the end x = L is open end,
Section 4.6 15. Obtain the critical buckling loads and ~e corresponding buckling shape of compressed columns, each having, length L and a constant cross-section and the following boundaryconditions:
and the boundaryconditions become: y(a)=0 y'(b) 17. Obtain the critical buckling loads and the correspondingbuckling shape of a column buckling under its ownweight, such that the deflection satisfies the following differential equation: Td3y dy E I dx----+qxd'--~" =0 50<x<L - -
where p~ > 4k EI. The columnis simply supported at both ends. Section 4.11 19. Show that the differential operators given in eq. (4.34) are self-adjoint. 20. Obtainthe conditions that the coefficients of a linear fourth order differential operator mustsatisfy so that the operator can be transformed a self-adjoint operator. to Section 4.15
21. Transform followingdifferential operators to the self-adjoint Sturm-Liouville the form given in eq. (4.49):
l<x<2 = in a series of Jo(l~x)where are roots of Jo(21a0 0 I~ 28. Expand function: the f(x) = ~
0<x<l in a series of J2(I.qx)where are the root of: !~
J2(J.~ --
29. Expand function: the f(x)= 0 <x<L in a series of J20.tax), where are the rootsof: It. ~tnLJ~(lxnL)aJo(~tnL)
Section 4.22 30. Expand function: the f(x) = =1
-1 < x < 0 0<x<l
in the series of Legendre Polynomials. 31. Expand function: the
f(x)=0
--X
-l<x<0
0<x<l in a series of Legendre Polynomials.
5
FUNCTIONS OF A COMPLEXVARIABLE
5.1
Complex
Numbers
A complexnumber can be defined as an ordered pair of real numbers and y: z x z = (x,y) The complexnumber(1,0) is a real number= 1. The complexnumber(0,1) = i, is imaginary number.The components z are: the real part Re (z) = x and the of imaginary part lm (z) = y. Thus, the numberz can be expressed, conveniently follows: z=x+iy The numberz = 0 iff x = 0 and y = 0. Newoperational rules and laws must be specified for the new numbersystem. Let the complexnumbersa, b, c be defined by their components a2), (b~, b:) and (cl, c2), respectively. (a~, Equality: a =biffa~ = a2 andb~ =b~ Thusit can be written in complex notation as follows: a=a~+ia 2=b=b~+ib2 Addition: c=a+b=(a~+b~,a~+b2) iff a~=b~ and a 2=b2
The absolute value of zo represents the distance of point zo from the origin. The absolute value of the difference betweentwo complexnumbers,is: la- bl = ~/(a, - b~)2+ (a2 - 2 and represents the distance between and b. a The absolute value of the products and quotients become: labcl =
Succeeding roots repeat the first n roots. Hence,a1/n has n distinct roots. In polar form, the n roots fall on a circle whose radius is r 1/n and whosearguments equally spaced by are 2n/n.
5.2
Analytic
Functions
Onemust develop the calculus of complex variables in a treatment that parallels the calculus of real variables. Thus, one must define a neighborhood a point, regions, of functions, limits, continuity, derivatives and integrals. In each case, the corresponding treatmentof real variables will be presentedto give a clearer picture of the ideas being presented. 5.2.1 Neighborhood of a Point
In real variables, the neighborhood a point x = a represents all the points inside of the segmentof the real axis a - e < x < a + e, with e > 0, as shown the shadedsection in in Fig. 5.3a. This can be written in morecompactform as ~ - a I< e. In complex variables, all the points inside a circle of radius e centeredat z = a, but not including points on the circle, make the neighborhood z = a, i.e.: up of ]z - a[ < e This is shown the shaded area in Figure 5.3b. as 5.2.2 Region
Aclosed region in real variables contains all interior ~as well as boundarypoints, e.g., the closed region:
Ix-l[_<X
contains all points 0 < x < 2, see Fig. 5.4a. The closed region in the complex plane contains all interior points as well as the boundary points, e.g., the closed region:
the following regions are open: Ix-ll< 1 as well as: l<[z-l-i[<2 A region is called a semi-closedregion if it includes all the interior points as well as points on part of the boundary.3 Functions of a Complex Variable or 0 <x<2
A function of a real variable y = fix) mapseach point x in the region of definition of x on the real x-axis onto one or morecorrespondingpoint(s) in another region definition of y on the real y-axis. the function: 1 y = f(x) = -~0 < IxI < 1
.. 5.g.called multiply-connected. in An Open Regionis one that includes all the interior points. the region outside a circle is multiply-connected.g. respectively.CHAPTER
5 Imaginary Axis Y
190
X 0 1 2
Real Axis
(a)
(b)
Fig. e. as well as all the points on the outer and inner circles.2.The order of the multiply-connectivenessof a region can be defined by the number independentclosed contours that cannot be of collapsed to zero plus one.: 1<Iz -1-iI _< 2 A simply connected region R is one where every closed contour within it encloses only points belongingto R. 5.e. 5. Thus. as shown Fig. but does not include the boundarypoints. Thus. the region betweentwo concentric circles) is doublyconnected. A region that is not simply connectedis .4
l_<lz-l-il<2 represents all the interior points containedinside the annular circular ring defined by an inner and outer radii of 1 and 2.4(b). the region inside a circle is simply connected.g. For example. the region inside an annular region (e. A single-valued function is one whereeach point x mapsinto one point y.
5. then the limit of the function as z approachesz0 is a number i.plane 2 Fig. Theregion of definition of x is called the Domain the function f(x). y) = 2 -y2and v(x. e. but not the point z = 0. y) + iv(x. in the example above: The Domain 0 < Ix [_< 1 is The Rangeis [y [> 1 A function of a complex variable w = f(z) mapseach point z in the domainof f(z) one or more points w in the range of w. the set of values of y = f(x). see Fig. 5. A single-valued function mapsone point z onto one point w. The function w = f(z) of a complex variable is also a complexvariable.5: Mapping of the Function w = z
mapsevery x in the region 0 < Ix [ < 1 on to a point y in the region ly 1>-1.
. whereu and v are real functions of x and y.e. the function: 1
w: r
0<lzl<-1
maps the points inside and on a circle of radius = 1. 5.: A. xeD.plane
w . whichcan written as follows: w: f(z) = u(x.4 Limits (5. y) = 2xy 5. For example.FUNCTIONS
OF A COMPLEX
VARIABLE
Imaginary Axis
191
Imaginary Axis
Y
Y
Real Axis x
Real Axis
z . except possibly at the of point itself.2. is called the Rangeof f(x).1)
If the function f(z) is defined in the neighborhood a point z0.g. onto the all region outside and on the circle of radius = 1.. For example: w=z~ =(x+iy) 2 =x2-y 2 +i(2xy) where u(x.
CHAPTER
5
192 (5.2)
Lim f(z) = z~z 0 This meansthat if there exists a positive small number such that: ~ [z - z0[ < e then
Iw-A < ~ for a small positive number~ I
The limit of the function is unique. Let A and B be the limits of f(z) and g(z) respectively as z 0, then : Lim If(z) + g(z)] = A z--~z 0 Lim [f(z) g(z)] z~z 0 Lim f(z) A z -~ z 0 g(z) provided that B ~ 0
Since the limit is unique, then the limit of a function as z approachesz0 by any path C must be unique. If a function possesses morethan one limit as z --~ z0, whenthe limiting processis performed along different paths, then the function has no limit as z --~ z0. Example 5.1 Find the limit of the followingfunction as z --~ 0: 2 xy fl(z) = 2 +y4 Let y = mxbe the path C along whicha limit of the function f~(z) as z --~ 0 is to obtained: 3 m2x ~0 Lim f~(z)= Lim z-~0 x ---~ 0 +m2x4 2, independentof the value of m. This is not conclusive because, on the curve x = my the 2 + 1), which dependson mfor its limit. Thus, if limit of f(z) as z ~ 0 on C is m/(m f~(z) has many limits, then f~(z) has no limit as z --~ 5.2.5 Continuity
A function is continuousat a point z = z0, if f(Zo) exists, and Limf(z) exists, z-~z 0 if Lim f(z) = f(zo). A complexfunction f(z) is continuous at z = Zo iff, both z~z 0 and v(x,y) are continuousfunctions at z = o.
FUNCTIONS
OF A COMPLEX VARIABLE .Imaginary Axis
193
C~
C2 Y0 ........ ~"~ Real Axis x off(z)
x 0
Fig. 5.6: Two Paths for Differentiation
5.2.6
Derivatives
Let z be a point in the neighborhood a point z0, then one defines Az as: of Az a complex number o =z- z Thederivative of a function f(z) is definedas follows: f'(Zo)=~zZ) = Lim f(zo+Az)-f(zo)= [ z=z 0 Az--~0 Az Lim f(z)-f(z°) z--~z o z-z 0 (5.3)
Thus, the derivative is definedonly if the limit exists, which indicates that the derivative must be unique. If a function possesses morethan one derivative at a point z = z0 dependingon the path along whicha limit was taken, then it has no derivative at the point z = z 0. Example 5.2
Thus, equating the two expressions for f'(z), one obtains the Cauchy-Riemann conditions given in eq. (5.5)¯ The Cauchy-Riemann conditions can also be written in the polar form given in (5.5)¯ The derivative can thus be evaluated by: f'(z)=Ou + .3v ~x 0x - 0y ~
1
Note that the partial derivatives satisfy the Cauchy-Riemann conditions. (ii) The function: f(z) Rez)= x ( has no derivative:
tI=X
~u = 0 3y
v=0
=0
0v: 0
The partial derivatives do not satisfy the Cauchy-Riemann conditions (5.5). If u and v are single valued functions, whose partial derivatives of the first Orderare continuousand if the partial derivatives satisfy the Cauchy-Riemann conditions (5.5), then f'(z) exists. This is a necessary and sufficient condition for existence continuousderivative f'(z). If one differentiates eq. (5.5) partially oncewith respect to x and oncewith respect y, one cans showthat: ~Zu ~Zu
V2u = 0-~+~= 0
(5.7) b2v O2v V2v ~--~- + ~-~-T= 0 = These equations are known Laplace's equations. Functions that satisfy (5.7) are as called Harmonic Functions. The Cauchy-Riemann conditions can be used to obtain one of the two componentsof a complex function w =f(z) to within an additive complex constant if the other component known.Thus, if v is known,then the total derivative becomes: is 0Udx+_~dy=0Vdx 0y du= "~xx Example 5.4 If v = xy, then one can obtain u(x, y) as follows: ~Vd y -~x
Afunction f(z) is analytic at a point o i f i ts derivative f '(z) e xists and i continuousat Zo and at every point in the neighborhood zo. Anentire function is one of that is analytic at every point in the entire complex plane. If the function is analytic z everywhere the neighborhood a point zo but not a z = z0, then z = Zo is called an in of isolated singularity of f(z). Example 5.5 (i) The function:
Somecomplexfunctions can be multivalued in the complexz-plane and hence, are not analytic over someregion. In order to makethese functions single-valued, one can define the range of points z in the z-plane in a waythat the function is single-valuedfor those points. For example,the function zl/2 is multivaluedsince: w= z1/2 = [l'ei(°:t2nn) ]1/2 = 1"~/2i(°-+2nn)/2 Therefore,for n = 0: zI/2 = r 1/2 eiO/2 and for n = 1: z1/2 = r 1/2 ei(0+2n)/2 r>0,0<0 < 27z
CHAPTER
5
198
Y
Branch Cut Branch Point Branch Point
Branch Cut
X
0=4~
X
Top Riemann Sheet
Bottom Riemann Sheet
(a)
z z
Y
~
p
Branch Cut
Y
Branch Cut
~ = 2~ ch Point ~
X
, = ~ranch Point ~,. 4n p (1,1)~
x
TopRiemann Sheet
Bottom Riemann Sheet
(b) Fig. 5.7: Branch Cuts and Riemann Sheets
For n = 2, 3 ..... the value of w is equal to those for n = 0 and n = 1. Thus, there are twodistinct values of the function w = z1/2 for every point z in the z-plane. Instead of letting w have two values on the z-plane, one can create two planes wherew is singlevalued in each. This can be done by defining in one plane:
Z1/2 = r 1/2 ei0/2
r>0,0<0
<2rt
and in a secondplane:
Z1/2 = r 1/2 ei0/2
r>0,
2n<0 <47z
Thus, the function w is single valued in each plane. It should be noted that 0 is limited to one range in each plane. This can be achieved by makinga cut of the 0/2n ray from the origin r = 0 to ~ in such a waythat 0 cannot exceed2rt or be less than zero in the first plane, The samecut from the origin r = 0 to ~ is madein the other plane at 2~47r ray so that 0 cannot exceed4rt or be less than 2~t, see Fig. 5.7(a). Eachof these planes called a RiemannSheet. The cut is called a Branch cut. The origin point where the.
FUNCTIONS
OF A COMPLEX VARIABLE
199
Y~
Y
Y
(a)
Y
Branch Point p. ,~p Branch Cut for w 1
~
x
Branch Point
Z=+I
Br~nCho~.~Cut ' [ (b) Y ~, Branch Point 7 = -1 Branch Point z = -1
Branch Point
Z=+I
Branch Point
Z=+I
(c) Fig. 5.8: Examplesof Branch Cuts (a) Non-linear, (b) Multiple, and (c) Co-linear Branch
cut starts at r = 0 is called the Branch Point. Since the function w is continuousat 0 = 27z in both she.ets and is continuousat 0 = 0 and 4~t, one can envision joining these two Riemann sheets at the 2rt and at 0/4~ rays. For the function w = (z - 1 - i) 1/2, one mustfirst express it in cylindrical coordinates in order to calculate the function. Let the origin of the z-plane be at (0,0), such that: i0 z =r e Let the origin of the coordinate systemfor the function w be (1,1), such that:
CHAPTER i¢ z-l-i=pe
5 13>0
2 00
To makethe function w single-valued, one needs to cut the plane from the branclh point at (l+i) with 13 = 0 to oo at ¢ = 0/2~ and = 2n/4rt, se e Fig. 5. 7(b). This re sults in two Riemann sheets defined by: w= (z - 1 - i) 1/2 = 01/2ei¢/2 0<0<2x :TopSheet 2n < 0 < 4rt : BottomSheet one can see that 9 and 0 are related to r and 0. The branch cut does not have to be aligned with the positive x-axis. For the above function, one can define a branchcut along a ray, cz, such that the function is defined by: w= (z - 1 - 01/2 = 91/2 ei¢/2 c~ < 0 < ot + 2n : Top Sheet cz + 2rt < 0 < c~ + 4n : Bottom Sheet so that the choice of ~z = n/2 results in a vertically aligned branch cut. The choice of + 2n dependson ~z, in such a wayso that the top sheet should include 0 = 0 in its range. Branchcuts do not even have to be straight lines, but could be curved, as long as they start from the branchpoint and end at z --> oo, see Fig. 5.8(a) for examples. Sometimes,a function mayhave two or more componentsthat are multi-valued. For example,the function w = (z2 -1) 1/2 can be written as w = (z-1)l/2(z+l) 1/2 which contains two multivalued functions w = (z-l) 1/2 and w = (z+l) 1/2. Both functions 1 e require branch cuts to makethemsingle valued. Let: Wl= ( z- 1)1/2 = 13~/2ei0o'/2 ~1 < 01 < (~1 - 2n ~1 + 2~ < 01 < ~1 + 4g W = (Z + 1)1/2 = p~/2ei¢2/2 2
It shouldbe notedthat 13 ~1, 02 and 02 are related to r, 0. l, In many instances, it maybe advantageousto makethe branch cuts colinear. In those cases, the function maybecome single valued along the portion that is common to those branchcuts. For example,the choice of (21 = (22 = 0 or (21 = (22 = -~Z for both branch cuts mayworkbetter than in Fig. 5.8b (see Fig. 5.8(c)). For a point slightly abovethe two branchcuts, 01 = d~2-= 0 so that: 1/2 w = WlW= (131132) 2 For a point slightly belowthe two branch cuts, 01 = i(2~+2~)12 (131102) 1/2 w = WlW= (13192)112e = 2 Thusthe function w is continuous across both branch cuts over the segmentfrom z = 1 to ~,. Similarly, one can showthe samefor the other pair of branch cuts in Fig. 5.8(c).
02 --
wherethe choice of + 2~ is madein order to include the angle 0 = 0 in the Principal Riemann sheet. The function log z as defined by (5.19) is thus single-valued. The function log z is not continuousalong the rays defined by 0 = ~ and 0 = c~ + 2n, because the function jumpsby a value equal to 2hi when0 crosses these rays. Since the function is not single-valued on 0 = ~ and 0 = ~ _+ 2~, the logarithmic function has no derivative on the branch cut defined by the ray 0 = c~, as well as at the branch point z0 = 0. Hence, all the points on the ray 0 = (~ are non-isolated singular points. A few other formulaefor the complexfunction are listed below:
Fig. 5.13: Integrationon ClosedPath C O in a Multiply-Connected Region
Example 5.7 Obtainthe integral of f(z) = 1/(z-a) on a circle centeredat z = a and havinga radius units. Since the integral on a circle of radius = 2 was obtained in Example 5.6, then:
CO z a CI z-a on13 = 4 on19 = 2
5.6 Cauchy's Integral
Formula
Let the function f(z) be analytic within a region R and on the closed contour containing R. If Zo is any point in R, then: f(zo )= 1_~ f(z) 2hi C z - z o Proof: Since the function f(z) is analytic everywhere R, then f(z)/(Z-Zo) is analytic in everywherein R except at the point z = zo. Thus, one can surround the point Zo by a closed contourC1, such as a circle of radius e, so that the function f(z)/(Z-Zo) is analytic everywherein the region betweenC and C1 (see Fig. 5.14). Invoking Cauchy'sintegral theoremin eq. (5.30):
If the series in eq. (5.34)converges, then the two series ~ n and ~Yn o als n=l n=l converge. An infinite series is Absolutely Convergent if the series, ~ n=l converges.If a series is absolutely convergent,then the series also converges. A series of functions of a complex variable is defined as: P(z)= fj(z) ~_, j=l whereeach function fj(z) is defined throughouta region R. The series is said to converge to F(zo) if: F(zo) = ~ fj(Zo)
j=l
The region where the series converges is called the Region of Convergence. Finally, define a PowerSeries about z = Zo as follows: f(z)= ~ n n(z-zo) n=0
The radius of convergence is defined as: ¯ p p = Lim ~ n~lttn+l I such that the powerseries converges [z - zo I < p, and divergesif ~ - zo [ > p ¯ if If a powerseries about zo convergesfor z = z1, then it convergesabsolutely for
FUNCTIONS
OF A COMPLEX VARIABLE Imaginary Axis
217
Y ~ C2
¯ "~0
Real Axis .X Fig. 5.15: Closed Path for Taylor's Series
z = z2 where:
5.8 Taylor's
Expansion Theorem
If f(z) is an analytic function at O, t hen there is a po se th at converges in wer ries side a circle C2centeredat zo and represents the function f(z) inside C1, i.e.: f(z) = an (z- Zo)n n=0 where: a n Proof: Considera point zo wherethe function f(z) is analytic (see Fig. 5.15)¯ Let points z and be interior to a circle C2, ~ being a point on a circle C1centered at Zo whose radius = ro ¯ Consider the term: 1 1 1 f(n)(zo) --~ n! (5.35)
The Taylor's series representation has the following properties: 1. Theseries represents an analytic function inside its circle of convergence.
r FUNCTIONS
OF A COMPLEX VARIABLE
219
2. The series is uniformlyconvergentinside its circle of convergence. 3. The series maybe differentiated or integrated term by term. 4. There is only one Taylor series that represents an analytic function f(z) about point z o. 5. Since the function is annlytic at zo, then the circle of convergence a radius has that extends fromthe center at Zoto the nearest singularity. Example 5.9 (i) Expand function z i n aTa the ylor's se ries ab out Zo 0. Since: = Iz=O 1 then the Taylor series about zo = 0 is given by:
e z= ~ n=0 zn
whichis convergentin the region ]z 1< 2. It should be noted that the radius of convergence the distance fromzo = 0 to the closest singularities at z = + 2, i.e. p = 2. is (iii) Obtainthe Taylor's series expansionof the function in (ii) about zo = 1 about Zo = -1. To find the series about zo = 1, let ~ = z - 1, then the function f(z) transforms
= + 3)1
=~"
1[ 11 ~-3' ~-
Expanding f(z) about Zo = 1 is equivalent to expanding~ (4) about ~ = 0. The Taylor series for the functions are as fgllows:
CHAPTER
5
220
1 = 1 E (-1)n 3-'if"
~+3 3
convergent in
<3
n=0
Thus, the two series have a common region of convergence14 [ < 1: ~(~)=-"0 ~n+ 3--ff'~-~ )J c°nvergentinl4]<
and the Taylor series representation of f(z) about o =1 be comes: f(z) -- --4n__~;01 + 3-'~-~J(Z- l)n convergentin Iz-ll
It should be noted that the radius of convergence represents the distance betweenzo = 1 and the closest singularity at z = 2. Tofind the Taylorseries representation of f(z) about zo = - 1, let ~ = z + 1, then the function transforms to ~(~): l ~'(4)=(~+1)(~-3)=~ 1I 4-3 1 1 ]
Expandingf(z) about Zo = -1 is equivalent to expanding~(~) about = 0. The aylor T series for the functions are as follows: 1 1 oo 4n ~-~convergentin 141 < 3
As a consequence Taylor's expansiontheorem, one can showthat if f(z) and g(z) of are twoanalytic functions inside a circle C, centeredat Zoand if f(z) = g(z) along segmentpassing through zo, then f(z) = g(z) everywhereinside C. This can be shown expandingboth functions in a Taylor's series about zo as follows (see Fig. 5.16): n f(z)= ~' ~ an(Z-Zo)
n=0
where
an = f(n)(zo) n~'~~
n g(z)
= ~ bn(z-Zo)
n=0
where
bn = g(n)(z°) n!
CHAPTER
5
222
At z = zo, f(Zo) = g(zo), thus o =bo. Th derivatives off(z) andg(z)at ze , c an be ta e as a limiting process along C1. Thus: f'(Zo) = g'(Zo) 1 whichmeansthat al = bl, etc. Thus, one can showthat an = bn, n = 0, 1, 2 ..... and f(z) = g(z) everywhere in Theidentity theoremcan be used to extend Taylor series representations in real variables to those in complex variables. If a real function is analytic in a segment the on real axis, then one can showthat the extension to the complex plane of the equivalent complex function is analytic inside a certain region. Thus, all the Taylor series expansions of functions on the real axis can be extendedto the complexplane. Example 5.10 (i) The function:
oo xn
n=O is analytic everywhere the real axis. Onecan extend the function into the complex on plane whereez is equal to ex on the entire x-axis. Since the function ez and e× are equal on the entire x-axis, then they must be equal in the entire z-plane. Hence,the Taylor series representation of the complex function eZ:
ez =
n=O is analytic on the segment the real axis, Ixl < 1, then the extended of function (1 - z)-1 has an expansion ~ zn which is analytic in the region Izl < 1. n=O
5.9 Laurent's
Series
If a function is analytic on two concentric circles CI and C2 centered at Zoand in the interior region betweenthem, then there is an infinite series expansionwith positive and negative powersof z - zo about z = Zo (see Fig. 5.17), representing this function in this region called the Laurent's series. Thus, the Laurent's series can be written as:
whereC is a circular contour inside the region betweenC1 and C2 and is centered at Zo. Proof: Consider a cut (ab) between the two circles C1 and C2 as shownon Fig. 5.17. Then let the closed contour for use in the Cauchy integral formula be (ba da bcb). Thus, writing out the integral over the closed contour becomes:
(iii) Obtainthe Laurent's series for the followingfunction about o =0,valid in the region Izl > 2:
CHAPTER
5
226
1 f(z) = z2 _ The function has two singularities at z = + 2. Since the function is analytic inside the circle Izl =2, a Taylor's series can be obtained(see Example 5.9). For the region outside [z[ = 2, one needsa Laurent'sseries representation. Factoring out z2 from f(z): f(z)= 1 z2(1,4/z2)
1 1-iv-a).e. then z = zo is called a RemovableSingularity. then z = zo is called a Pole of Order m. (ii) Thefunction:
f(z} =-~-. then z = z0 is called a Pole of f(z). but its Laurent'sseries representation about zo has no principal part. Example 5. I f t he function f(z) i not defined at z = zo.. of the lowest powerof (z-z o) in the principal part is m.Zo)n
If m= 1.10.12 (i) The function: 1 f(z) = z2 _ has two isolated singularities zo = + 2.z01 < a wherethe real constant (a) signifies the in distance fromzo to the nearest isolated singularity.FUNCTIONS
OF A COMPLEX VARIABLE
229
convergentin [zI > 2 n=0
5. then z = 0 i s c alled a n Essential S ingularity. zo is known a Simple Pole. If the principal part contains all negative as powersof (Z-Zo). If f(z) has an isolated singularity at 0. Both singularities are simple poles (see Example 5.Classification of Singularities
AnIsolated singularity of a function f(z) was previously defined as a point 0 wheref(Zo) is not analytic and wheref(z) is analytic at all the neighborhood points 0. the principal part looks like: m
=1 (z . t hen f(z0) can be represented y aLa b urent's series about zo. i. Thesingularity is a pole of order 3. convergent the ring 0 < Iz .
l+z
1
1
. Thepart of Laurent's series that has negative powersof (z-z0) is called the Principal Part of the series: bn
~
n = l'(z --'~o)n If the principal part has a finite number terms.= 7+ ~has an isolated singularity at z = 0.
The representation in eq.14 (i) Obtain the value of the followingintegral:
C whereC is a closed contour containing z = 0.3/z is already Laurent's series form.11
Residues
and Residue
Theorem
Definethe Residueof a function ~f(z) at one of its isolated singularities 0 as t he -1 coefficient b1 of the term (z-z0) in the Laurent'sseries representation of f(z) about 0.inaryAxis
231
~" Real Axis X Fig.19: Residue Theorem for a Multiply Connected Region
(iii) The function: f(z) = z =X n=O transformsto: oo f(~-l): 1 n n!~ O=
n Z
where~ = 0 is an essential singularity. (5. wherethe coefficient is definedby a closed contour integral:
C and C is closed contour containing only the singularity zo. 5. Thusez has an essential singularity at infinity. whereb1 = 3. Since the function f(z) -. Example 5.37) can be used to obtain the integral of functions on a closed contour.FUNCTIONS
OF A COMPLEX VARIABLE . then:
.
5.
using Cauchy'sintegral theorem. Then. Since the Laurent's series of the function about z0 = 2 was obtained in Example 5.11-iv. then each closed contour integral can be evaluatedby the residue at the pole zj located within Cj: ~ f(z)dz = 2nirj Cj then the integral over a closed path conlaining n poles is given by:
n
~ f(z)dz = 2~i ~
c
.contourCj. rj of Proof: Encloseeach singularity zj with a closed.1 Residue Theorem
If f(z) is analytic within and on a closed contour C except for a finite number isolated singularities entirely inside C.. then the integral can be solved: = 2rfi ~.11. where bI = 1/4.38)
C where = Residue f(z) at the jth singularity. one obtains: n ~f(z)dz= ~ ~ f(z)dz C j=Icj
Since each contour Cj encloses only one pole zj.CHAPTER
5
232
(ii) Obtain the value of the following integral: dz ~z 2 -4 C whereC is a closed contour containing zo = 2 only.19.
+ rn)
(5. 5.= -~i C 5. then:
~f(z)dz=2~i
(rl + r2
+.. such that f(z) is analytic inside C and outside the regions enclosed by all the other paths as shown the shadedarea in by Fig.
t he function g c an be expanded n a (z) i Taylor'sseries at z0 as follows: g(z)= g( n)(zo)(Z-z°)n n! n=0
Then. hence the integrals give: ~ f(z)dz = 27zi(-~ 1 C (iv) The contour contains only the z = 4 simple pole. Thefunction f(z) has simple poles at z = 2 and 4.the Laurent's series for f(z) becomes:
. then o can find afunction g(z) such that: ne g(z) = (z . various methods be of can developedso that one need not obtain a Laurent's series expansionabout each pole in order to extract the value of coefficient b 1.15
OF A COMPLEX VARIABLE
233
Obtainthe value of the followingintegral: dz ~(z .2) (zC where is a circle of radius = 3 centeredat z0. (i) Sincethere are no singularities inside this closed contour. wherez0 is: (i) -2. hence its value is: ~ f(z)dz = 2rti(~) = C To facilitate the computation the residue of a function. thus: ~ f(z)dz = 2~i(--~) = C (iii) The contour contains both poles. If f(z) has a pole of order mat o.zo)m whereg(z0) ~ 0 and is analytic at 0. T hus. (iii) 3 and C 6.4 is 1/2.FUNCTIONS Example 5. (ii) 0. Theresidue of f(z) at z = 2 is andat z -. then:
~
C
f(z)dz =
(ii) The contour contains the simple pole at z = 2.
~/-~. n. Thus. Let R the be a semi-circle in the upperhalf plane with its radius R sufficiently large to enclose all the poles of f(z) in the upperhalf plane (see Fig. for ~zl >>1 there exists an Mand p > of such that:
If(z)l< -p Mlzl
then:
p >1
Izl >>1
n
P.FUNCTIONS
OF A COMPLEX VARIABLE 4 4 i(z.V.13 Improper Real Integrals
The residue theoremcan b'e used mevaluate improperreal integrals of the type
I f(x)dx
(5. either or both limits do not exist. but the limit of the sumexists if A = B -) oo. defined as: P. The improperintegral can be defined as: f(x)dx -o~ = Lim f(x)dx+
ff(x )dx
-A a wherethe limits A--) o~ and B--) oo of the two integrals are to be taken independently. using the ResidueTheorem. Therefore. f(x)dx = Lira f(x)dx
If f(z) has a finite number poles.
f(x)dx
= 2~
(5. Since Izll < 1 and Iz21 > 1. and if.ff~.Zl)(Zdz
237
~
C i(z2 +4z+l) dz=!
wherez1 = -2 + .V. 5.47)
where rj are the residues of f(z) at all the poles of f(z) in the upperhalf-plane. only the simple pole at zi will be consideredfor computing residue of poles inside the unit circle Izl = 1: the 4 2 r(zl) = g(-2 + ~/~)= (z. then the value of such an integral is called Cauchy'sPrincipal Value. and z2 = -2 .46)
wheref(x) has no singularities on the real axis.zl)f(zl~ z = zl = ~ = i-~ Therefore: =2rd( 27 os0
I 2d0 0
2 ")
5.20).the integral over the closed path is:
.the function f(z) has simple poles at 1 and z2.
b] exists iff the twopartial integrals exist independently.Theefore: r
."lt~ i s singular at x = 0.20 (i) Evaluatethe following integral: 2 x-1/3dx
~
-1 Notethe function f(x) -. then the value of the integral thus obtained is called the CauchyPrincipal Valueof the integral. denoted as: b P. a < c < b is defined as follows: b ff(x)dx=Lima e-~o
[ci
f(x)dx a f(x)dx+i Lcj'+
I
Theintegral on [a.e. ~ f(x)dx
-a
Example 5.i c_(i+~)/~f~ r(z2)= _ ~ 4 =4-~ Thus. that is if the followingintegral: Lim f(x)dxe~°L c+e f(x)clx
exists. The followingreal integral:.[cos m+ sin m]
5. either or If both limits as e "-> 0 and 8 --> 0 do not exist but the limit of the sumof the two integrals exists if e = 8.V.15
Improper Real Integrals of Functions Having Singularities on the Real Axis
Functions that have singularities on the real axis can be integrated by deformingthe contour of integration. b f(x)dx where f(x) has a singularity on the real axis at x = c.CHAPTER
5
242
I. the integral I can be solved: nm m I = -~.
x -3dx =Lira -1 ~°L ~1 x-3dx + O+e x-3dx = ~ 8
Improper integrals of function on the real axis ~ f(x)dx wheref(x) has simple poles on the real axis can be evaluated by the use of the residue theoremin the CauchyPrincipal Valuesense.22... sufficiently large to include all the poles of f(z) on the real axis and in the upper-half plane..oL ~ l+±umF4-11 im[l. The contour on the real axis is indented such that the contour includes a semicircle of small radius = e aroundeach simple pole xj as shown Fig.. in Thus.] 2 ~-~oL8 4J ~ 2
Neitherintegral exists for E and5 to vanishindependently. onetakesthe P. Let xl. Let R be t he semi-circular path w a radius ith R.of the If inte~a]: P. z2 .+
C
f
C1 x1 + 8 2
+J" f(z)dz= -1~ x2~ xn + e C R
f(z)dz=2~iZr i ~
m= j 1
. 5. .. z mbe the poles of f(z) in the upper-half plane. one can obtain the principal value of the integral as follows: + + +f j fXl_~Rl~ +..V. xn be the simplepoles of f(z) on the real axis.V. and let 1.FUNCTIONS
OF A COMPLEX VARIABLE 2
243
J
-1/3 ~X
(ii) Evaluate the following integral: 2 x-3dx
~
-1 The function f(x) = x is singular at x =
-3
-11x-~
_a_l x-3dx a~0|L0a+~
:± ~--. x2 ...
22: Closed Path for Improper Integrals and Complex Poles
with Real
where contoursCj are half-circle paths in the clockwisedirection and rj's are the the residuesof f(z) at the poles of f(z) in the upperhalf plane at zj. 2 .. n.6 Thus. 5. it wasshown earlier that:
Since furiction f(~) hassimplepoleson the r~al axis.. 5... the principal value of the integral is given by:
. the integral over a small semi-circular path about xj becomes the limit as in the radius e -> 0: Lim S f(z)dz = Llmlrj f dz S ] ¯ * ~ + g(z) dz = -Tzlrj e--+0 e-~0/ ~ z-x 3 Cj L Cj Cj where the results of Example were used. If the functionf(z) decaysfor [z[ "-) co as follows:
-p If(z)l < Mlzl
where p > 1
for
Izl >> 1
then. Thelimit as R --) co e --) 0 must be taken to evaluate the integrals in CauchyPrincipal Value form on R and onCjforj=l.CHAPTER
5 Imaginary Axis Y
244
-R
Xl
x2
xn
R
#" Real Axis
X
Fig. then in the neighborhood the each real simple pole Xj. and rj* are the residuesof f(z) Thus. it has one term with a negative poweras follows: f(z)= rj +g(z)
where g(z) = the part of f(z) that is analytic at xj.
5. in the upper of half-plane. Example 5. Choosethe contour shownin Fig. one must resort to summing infinite and an number residues at the poles in the entire half-plane.25 described by the points ~R..13 to 5.6 Lim f g(z)dz
e-+0 o
C e
=Lim f g(a+eei~)ieeiOdO e--+0 c
_<Lim(Mae)-~O
e--~0
then the integral over the small circle becomes:
5. then to use the Residue Theorem a circular contour.17
Evaluation of Real Improper Integrals Contours
by Non-Circular
The residue theoremwas used in Section 5.FUNCTIONS
OF A COMPLEX VARIABLE
249
Since f(z) has a simple pole at z = a. Also: 5.. Thus: f f(z)dz=r(a)f C e C e d. 2 . 1. However. If a periodic function has an infinite number poles in the complex of plane. R. In this section. then it can be expressedas a Laurent'sseries about z = a: f(z) = r(a___~_) +
z-a
whereg(z) is analytic (hence bounded)at z = a. n = 0.__~_z + f g(z)dz=~ir(a)+
z-a
C e
C e
where the results of Example were used. more prudent choice of a of a non-circular contour mayyield the desired evaluation of the improperintegral by enclosing few poles.22 Evaluate the following integral:
ax oo~ e
I=
/ x --dx l+e
0<a<l
The function: az e f(z) z = l+e has an infinite number simple poles at z = (2n + 1) ~ti. more convenientand efficient non-circular contours are used to evaluate improperintegrals.15 to evaluate improperintegrals by closing the straight integration path with semi-circular paths.
..
which is a multi-valued function. and a is a non-integerreal constant. a line contour on C2 and a circle of radius = e. the path integration stays in the principal Riemann sheet. 2 . such that in the principal branchis definedin the range 0 < 0 < 2~. Let the poles of f(z) be z]..28: Closed Path for Integrals with x a 5. 5. Considerthe following integral:
OO
a f(x)x dx ~ 0
a > -1
(5. .FUNCTIONS
OF A COMPLEX VARIABLE Imaginary Axis
259
Y
¯
Branch Cut F RealAxis x
a Fig. as is shown Fig..19 Integrals
of Functions Involving x
Integrals involving xa.28. To evaluate the integral in (5. as shown Fig.28. Thus:
. zmin the complexplane and: M If(R)l_< with p> a+l. 5.56)
wheref(x) has no singularities on the positive real axis. for R >> 1
The contour on C] is closed by adding a circle of radius = R. 5. the integrand is made single-valued extendinga branch cut along the positive real axis. can be evaluated by using the residue theorem.56). Thecontour is closed as shown such in in a waythat it does not cross the branchcut and hence.
~ .4~t2)
2rr
However. iz~.30. is chosento the right of all the poles and singularities of f(z).-ff-)+e ~. or
. Notethat the choice of the branch cut must be made that it falls entirely to the left of the line x = y. since the integrandis real. -) =-4-~-7 (1-3af~ii) Z rj(zj)=-~e I. + ioo f(t)=~ f f(z) t>0
ztdz
where). Since ~/~ is a multi-valued function. then the integral can also be evaluatedas: 2 ~4~ f logx dx = _1 Re a x3 +'---]" 2 l-~-(1-
2 3-f~i) 27z = }
5.2
2 Inverse
Laplace
Transforms
Morecomplicatedcontour integrations aroundbranch points are discussed in the following examplesof inverse Laplace transforms: Example 5. then a branch cut is madealong the negative real axis starting with the branch point z = 0.Tz )+eSin/3(25---~ j=l
2
andthe integral of f(x) is:
0 Thus. { 2n) l(4~Z 2 2 . the integral becomes:
j
0
7 logx
.30 Obtain the inverse Laplace transforms of the following function: f(z) 2 z-a a >0
The inverse Laplace transformis defined as: 1 ~. Hence could be: taken along so it the negative x-axis (the choice for this example) along the positive or negative y-axis. in 5.CHAPTER
5 (log z)2 = z(log z)2 ] 1 2 zj =-'~ zj(IOgzj)2 3z3 Izj
266
r(zj)=
The sumof the residues is: 3 2 7~2 1 ( i~/3. as is shown Fig.
30: Closed Path for Inverse Laplace Transform
The branchcut is thus defined: z = p el* p >0 .2nn + 7z n = 0. 1
The angular range is chosenso that the ~ = 0 is included in the top Riemann sheet. This meansthat as ~ increases without limit. The top Riemann sheet is defined by n = 0 so that: ~ =19 1/2 ei~/2 -~ = p 1/2 ei~/2 19 > 0 p>0 . However.iR to ~' + iR must be closed in the top Riemann sheet to allow the evaluation of the inverse transform by the use of the residue theorem. one needs to connect 5' + iR with straight line segmentsL and L Theseare then to be connected to a semi-circle of radius R to 3 4. continuous semi-circle on the a third and fourth quadrants wouldcross the branch cut. To close the contour in the top Riemann sheet.FUNCTIONS
OF A COMPLEX VARIABLE Imaginary Axis
267
Y ~ CR iR L3 y+iR
BranchCut
~" ReaI Axis 2 a
R
"
x
L4 -iR
. satisfy the Jordan's Lemma (Section 5. 5.~ < ~ < .16). the ~ is located in either the top or bottom Riemann sheet.2nn .71: < ~ < n -3~<(~<-~ n = 0 n=l
and the bottomRiemann sheet is defined by:
Thetwo sheets are joined at ~) = -r~ as well as the ~z and -3~ rays. Theoriginal line path along ~' . Crossing the branch cut would
.-iR
Fig.
3 f¢ >eZ
-iR
2
+ f+ f+ f+ f+ f+f+ f 2) f(z)eZtdz
L3 C R L1 C o L2 Ci~ 4 L
=2~ir(a
The residue at z = a becomes: r(a 2) a2t e =a The integrals on C and C vanish. Thus. one should avoid the crossing of a branch cut. since using Section 5.16. wherez = 13 e-in.ye _ 1 ?t
2 R-->oo
y _+iR The line integral on L can be evaluated. This can be accomplishedby rerouting the path aroundthe branch cut. ¯ R ~n pte in ' ~ dp I ~'7~'--~ e e dp=-i I p+a2 ~ -pt e R The line integral on L can be evaluated. since using Section 5.one has to continue the path to close it eventually with L ._9 0
Z -.~.1: R A 1 If(R)l Rp as IR[ >> 1 where p = 1/2 > 0 The integral on CO vanishes. so that the entire closed path remains in the top Reimann sheet. wherez = 13 ein.a
eZtdzl=l
z_a2
! ~/-~+iR x+iR-a 2
eXte+iRtdz< -~. 1 2 o. continuing the path C R in the third quadrant with a straight line path L Tocontinue to connectby a straight 1.CHAPTER
5
268
result in the circular path in the secondquadrantbeing continuedin the third in the bottom Reimann sheet where the function ~ wouldhave a different value. .~ ei~/2
. line L one needs to connect L and L by a small circle C The final quarter circular 2. y +iR Thus: Lim ~z ezt dz . as follows: 1 .2:
If(e)l
1
as
lel -->0
wherep = -1/2 < 1
The two line integrals L and L can be evaluated as follows: 3 4 Let z = x + iR. path C closes the path with L The equation of the closed path then becomes: A 4.in the top 4 Reimann sheet. as follows: 2 R R 4pe-i")2 ePte-i=nein d13= f ~-~--P e-pt d13 -i 2 I pe in _a 2 p+a J E. To avoid these problems.16. then: +i R . Furthermore.
31(a). as shown Fig. The integral: ~(+ioo zt e
can be evaluated by closing the contour of integration and using the residue theorem. respectively.B): 2 1 f(t) = ~ + a terf(a-~/~) 4nt Example 5. it is sometimes advantageous run branch cuts for a function to linearly so that they overlap.Thus.FUNCTIONS
OF A COMPLEX VARIABLE ~/-~ R 2 p+a J e e_0tdp:2rriaea2
269
Thus:~ iR ~. App. as this mayresult in the function becoming single-valued over the overlapping segment. the integral transformsto: 2 OOu2e_U 2 f(t) = a ea2t +'~--'~f --du+ act a:t u2 =ae 0
oo
2
e -u2
du_ a2t~
oo U2 +a2t 0
1 2a2~ =ae ah ÷ ~ ~
-u e f~du 0
whichcan be written in the form of an error function (see eq. in a straight line in any direction in such a waythat both must fall entirely to the left of the line x = %As was mentioned earlier in section (5. 5. The two branch cuts for the top Riemann in sheet of each function can be described as follows:
.iR
2 z_a
Therefore f(t) becomes: f(t)=aea2t+ 1 7~e-0td~ "~0 P+a~ v oo +~--~
2
Letting u2 = pt. One has the freedom to make each of the functions ~ and ~ single-valued by a branchcut fromz = a and z = -a. 5.a 2 single-valued.2.22.31 Obtain the inverse Laplace Transform the following function: for 1 f(z) 2 2 a ~z _ The function f(z) has two singularities whichhappento be branch points.9). the cuts are chosen to extend from z = a and z = -a to -~ on the real axis. Twobranch cuts must be madeat the branch points z = a and -a to makethe function ~z~.eZtdz_2if t
~
y .
FUNCTIONS ei~t. since (Section 5. 3' Twoquarter circular paths. Similarly. R To continue the path closure in the top Reimann sheet.16. z-a=rl ei~2 . Thus. since (Section 5. . dz= f + f + f + f + f + f fz)eZtdz 3'-iR 3'+iR C -R c 1 -a-e R on L 1 on L 2 a-£
~f(z)e
zt
f
3' + iR iR
0
Ca a-E onL~
-a+e
+ ~
-R c i -a-e
onL i
) c~
+I + ~ + ~ f(z)eZtdz=0
0
onL 3
0
onL~
The integrals on C and C~vanish as R "-> oo. then one must avoid that point by encircling it by two small semi-circular paths C! and C~. 2 Lim --z~a --> 0 z-~_+a~z2 _ 2 a To facilitate accountingof the integrand of these multi-valuedfunctions. namely. see Figure 5. r e r >0 z= The closure of the original path from3' . < (~2 < '/1.1): R If(R)l.1~. one can evaluate the integrand term-by-termin tabular form.
271
rl>0 r2>0
The single-valued function z is described by: i0. as z on L becomes-z along L The equation of the closed path becomes: 3 2. z+a=r2
OF A COMPLEX VARIABLE -~Z<Ol <~.31 (b). segmentsbetweenz = -a and z = a is split into two parts.iR to 3' + iR in the top Reimann sheet would require first the joining of two straight line segments + iR to + iR. the remaining integrals can be evaluated in tabular form (see accompanying table):
.t-->0
as R -->
The integrals on C C and C~vanish. Since the straight line paths cross a singular R (branch) point at z = -a. This takes the form of two straight line path above and belowthe branch cuts from C and C~to the branch point at z = a. the joining of the straight line paths at the branchpoint z = a requires the joining of the two by a small circle C The line 2.16.as R >> 1 where p = 1 > 0 The integrals on [3'+ iR to + iR] vanish since: e(x+iR)t
l!4(x +iR)~--a2 dx<I~e~. This is done purely to simplify the integrations along these two parts of each line segment. one has to encircle both branch cuts.2): 1.
-71. C and C~are required to avoid crossing both branch cuts.L and L and L[ and 2 3 L~.
and (b) Heat flows from bodies with higher temperatures to bodies with lower temperatures. S.V. cm-~t ~t Since the flow of heat across a boundary proportional to the temperature is differential across that boundary.2 6. as well as transient solutions.1. wave.we know that: dq =. Considera volume.the heat content can be defined as follows: h = cmT* wherec is the specific heat coefficient. Poisson. where13 is the massdensity. V Defineq such that: ~h ~T* q = negative rate of change of heat flow . For such a volume.1
Introduction
This chapter deals with the derivation.6
PARTIAL DIFFERENTIAL EQUATIONS OF MATHEMATICAL PHYSICS
6. and Helmholtz. vibration.
6.1
The Diffusion
Equation
Heat Conduction Solids in
Heat flow in solids is governed the followinglaws: by (a) Heat is a form of energy.. Thetypes of equations treated in this chapter include: Laplace. as in Fig.k-~nTdS
2~3
. and T* is the average temperatureof V defined by T* = ~lm~ T [gdV. with surface.steady state solutions. diffusion.2. mis the massof V...The methodof separation of variables will be used throughoutthis chapter to obtain solutions to boundaryvalue problems. 6.fi. presentation and methods solution of of partial differential equationsof the various fields in mathematical Physics and Engineering.. and surface normal.
z). as shownin Fig. whosesides are aligned with x. the total heat flux across the remaining pairs of sides of the element two becomes:
. ( fi being the outward normalvector to the surface S. whose surface area is (dy dz): side at x: fi = -~x and
side at x + dx:
fi = ~ and qx÷dx =. and z axes and whosesides have lengths dx. dS is a surface elementand k is the thermal conductivity.1
wheren is the spatial distance along fi. Thepartial differential equation that governsthe conduction heat in solids can be of obtained by applying the abovementionedlaws to an element dV. the total heat flux across the two oppositesides of the elementat x and x + dx becomes:
(dq~) tot
= .y. 6.2. 6. positive in the direction awayfromV). dy. results: I
qx+dx Thus.k(dy dz)
ox Ix + dx
~T Expanding-~x x + dx in a Taylor'sseries aboutx..2) haveone of its vortices at point (x. 6.. Let the rectangularparallelepiped(Fig. respectively. and dz.CHAPTER 6
294
Fig.k(dx dydz)02T 1 + ~ [ ~--~-03T
(dx) . Considerheat flow across the two sides perpendicularto the x-axis. y. +
Similarly.
whichrelates the rate of heat convectionacross S to the temperature differential. t)] P r wherer is a constant.2.3) wherect represents the rate of decomposition the diffusing gas. a heat sink if negative. region of the bodyis known equal to T and O one maymakeuse of Newton'slaw of cooling: 8T -k-~-n (P.To(P..0 P on S
296
(b) Prescribe the heat flux across the surface -k~-ffn (P. (c) Heat convection into an external unbounded medium known of temperature: If the temperaturein the exterior unbounded . D represents the diffusion constant and q represents the additional source of the gas being diffused. of 6. +) =f( P) 6. (a) Prescribe the temperatureon the surface S.~tC = ~--~.CHAPTER 6 Thesign of q indicates a heat sourceif positive.t)=/(P.t) + bT(P." T(P.0 = g(P. ) = [T(P. whose decomposition is proportional to the concentrationof the gas (equivalent to havingsinks of the diffusing gas) then the processis defined by the followingdifferential equation: 1~9C V2C. If the diffusion process involves the diffusion of an unstable gas.2)
whereC represents the concentration of the diffusing gas in the ambientgas. The boundary condition that is required for a unique solution can be one of the following types: .t ) P on S
whereI is the prescribed heat flux into the volume across S. the boundarycondition becomes: -~(P.t).3 Diffusion and Absorption of Particles Theprocess of diffusion of electrons in a gas or neutrons in matter can be described as a diffusion process with absorption of the particles by matter proportional to their
.t ) PonS where b=~
The type of initial conditionthat is required for uniquenesstakes the followingform: T(P.2 Diffusion of Gases The process of diffusion of one gas into another is described by the following equation: V2C = 1 ~C q ~ ~P in V
(6. t) = b To(P. the surface is V thermally insulated.2.q ~t > 0 (6. If l = 0. Thus.
take one of the following forms: (a) y = 3y (b) ~xx = 0 3y (c) ~xxT-~zy = at a or b at a or b (-) for a.Y (x. bars. The boundaryconditions.a is the mean free path of the particles.0+)= g(x) ot
. All of these equations have the following form: 32y q(x. OF MATHEMATICALPHYSICS
297
concentrationin matter. non-uniformcross-section one dimensionalcontinua. a process equivalent to havingsinks of the diffusing material in matter.a)/3. This process is described by the followingdifferential equation: V29. A(x)is the and cross-section area of the medium. If the processof diffusion is associated with a processof creation of moreparticles in proportionto the concentrationof the particles in matter. torsional rods. whereVais the averagevelocity and ~. ERis the elastic restoring modulus. cz > 0 q = Sourceof particles created (by fission or radioactivity) per unit volume unit per time D = Diffusioncoefficient = (va ~.t>0 (6. required for uniqueness.t) is the deformation.t) the external loadingper unit length. EQ.c is the characteristic wave speedin the medium. eq.0 = f(x) and O-~. the processof chain-reaction. required for uniqueness.6)
wherey(x.t) A(x)~y~ _ I~. (6.PARTIAL DIFF.Y. q(x.Z. transmissionlines and acoustic horns were adequately covered in Chapter 4.3.4) becomes: V29+c~9 1 30 _ = ~--q ~x>0 (6.take the followingform: ÷) y(x.3
The Vibration
Equation
6.A(x) 3x\ 3xJ-c ~ 3t 2 ER a<x < b. . (+) for cz >0
The initial conditions.¢x 9 = 1 019 _ q D 3t where: 9 = 9(x.4)
6. such as stretched strings.1 The Vibration of One Dimensional Continua The vibration of homogeneous.5) (6.t) = DensityOf the diffusing particles ~z = Mean rate of absorptionof particles.
as follows: EI(x) +p ()O--~--q(x. 6.3
The transverse vibration of uniformbeams.
The boundaryconditions for a beamwere covered in Section 4. as follows: 2 2 Ow Ow dF =. t) dxdy: dy.2 The Vibration of Stretched Membranes Consider a stretched planar membrane whosearea A is surrounded by a boundary contour C.3.y. -+ ~w/+ "~ ~x 0y" J (pdx ~-~2w dy)
Thus.covered in Section 4.CHAPTER 6
298
f(x. t>0 (6.7)
where c = Sx/~ is the wave speed in the membrane P = P(x. and has a density p per unit area.t) from the in to equilibrium position.. The membrane stretched by in-plane forces S per unit length.y. whichequals the inertial forces. deformed a position w(x. acted on by is normalforces f(x.t) a<x<b.4. Consider element of the membrane.t) per unit area.(Sdy)0--~ dx + (S dx) ~-~--~. and
.y. 6.3.t)
". the forced vibration of a membrane described by the following equation: is O2w O2w 1 O2w f(P.in a manner similar to stretched strings. Assuming small slopes.y).w(x.dy + f(dx = S(dx ~-~ 0_2 f(x.t) 2 2 ~)t S ~X ~" O-~=C 2 PinA. shown Fig. 6. then one can obtain the sumof tbrces acting on the element. is described by differential equation of fourth order in the space coordinatex. y.t)
Fig.y.4.
per x. Let C h be the thickness of the plate.3. occupyingan area A. My.y. 6. twisting moment unit length Mxy.t) = 0 P on P on C. is the elastic constant per unit length of the boundary.4). EQ. E be the Young'smodulusand v be the Poisson's ratio.(Fig. The moments unit length M My. the equilibrium equations and the plate are: 0Mx 0Mxy + 0y 0x 0Mxy+ -Vy 0MY Vx =0 =0
0Vx
0y 0Vy
02w
(6. The moments x.PARTIAL DIFF.t) is the normal distributed external force per unit area acting on the plate. 13 be the massdensity of the plate material. the per x. the boundaryconditions along the contour C can be one of the following types: (a) Fixed Boundary:w(P.3 The Vibration of Plates The vibration of uniform plates. t >0
3w (b) Free Boundary:-~n (P' t) =
(c) Elastically Supported Boundary: 0w + ~ 0n S P. and Mxycan be related to the change of curvatures of the M plate as follows: M = -D ~ 0x 2 0Y2 ) x
My
0Y 0x2 ) 2
(6. For uniqueness. OF MATHEMATICALPHYSICS
299
For uniqueness.0+) = g(p) PinA P in A
6. surroundedby a contour boundary can be analyzed in a similar mannerto the vibration of beams. t >0 t >0 = 0 P on C.9)
.8)
0x +--~-y +q=oh 0t 2 whereq(x.t
where~. acting on an elementof the plate are shown Fig. ÷) =f( P) and 0w (p.4.and the shear forces per unit length. in Summing moments forces on the element (dx dy).the initial conditions must be prescribed in the followingform: w(P. V and Vy. 6.
. propagatesto the left and to the right withouta changein shape. whereu = x .ct) + g(x Functionshavingthe formf(x . at a speedof c. a disturbancehavingthe shapef(x) at t = 0.ct and v = x + ct.4
The
Wave Equation
The propagation of a disturbance in a medium knownas wavepropagation. The is phenomena wavepropagationis best illustrated by propagationof a disturbance in an of infinite string.ct) and f(x + ct) can be shown indicate that a function to fix) is displacedto a position (ct) to the right and left. C = amplitude of motion
.6.. the equation of motiontransforms to: ~2y = 0 0u 0v whichcan be integrated directly.. to give the followingsolution: y = flu) + g(v) = f(x.= wavenumber = -~. 6.6
6. respectively. A special formof wave functions f(x + ct) that occur in physical applications is knownas HarmonicPlane Waveshaving the form: f(x + ct) = C exp [ik(x + ct)] = C exp [i(kx + tot)] where k = -. Theequation of motionof a stretched string has the followingform: ~2y 1 ~)2y 2 2 OX C2 0t The solution of such an equation can be obtained in general by transforming the independentvariables x and t to u and v. 6.CHAPTER 6 f(x+ct) f(x) f(x-ct)
302
ct
ct Fi£.as shown Fig. Thus. in Thus. c wavelength to = circular frequency(rad/sec) = 2nf f = frequencyin cycles per secondor Hertz (cps or Hz) 2~ 1 period in time for motion to repeat .
6. 6. the sametype of boundaryand initial with conditions. whoseundisturbed height is h and whosedensity is 19. boundary initial conditions are the and sameas those for the vibration problem.7
6. torsional rods.2 WavePropagation in Two-Dimensional Media
Wave propagation in stretched membranes in the water surface of basins make and up few of the phenomena wavemotion in two dimensional continua.y.y. and w(x.1 Wave Propagation in One-Dimensional Media Theequationof motionfor vibrating stretched strings. respectively. OF MATHEMATICALPHYSICS
303
v(x.PARTIAL DIFF. acoustic horns.3 WavePropagation in Surface of Water Basin The propagationof waveson the surface of a water basin can be developedby the use of the hydrodynamic equations of equilibrium of an incompressiblefluid. Let u(x.4.4.t) represent the components the vector particle velocity of fluid on the surface in the x and y directions. 6. of The propagation of wavesin a stretched membrane obey the same differential equation as the vibration of membranes.__ surface h
Fig. etc.t)
~ u(x.y. EQ.4. bars.y.t)
w(x.y. together with the boundary conditionsat the end points (if any) and the initial conditions makeup the wavepropagation system for those media.y.t)
undisturbed __. 6.t) be the vertical displacement fromthe level h of the particle in the z-direction. thus: dX~x(uhdy) + dy~y (vhdx) + ~tt [(w + h)dxdy]
.t) and v(x. Thelaw of conservationof massfor an incompressiblefluid requires that the rate of changeof massof a columnhaving a volume(h dx dy) must be zero. Let a free surface basin of a liquid (A) (Fig.. Thesystemof differential equations.7) be surrounded a rigid wall described by contour boundaryC.
16)
~us.t)=O
PonC. Differentiating eq.14) into eq. (6. Substituting eq.15)
Substitution of p fromeq. (6.= -/)-~p--~. (6.CHAPTER 6
or
304
0 Ow +h(OU ~ ~xx +~yy)
=
(6.z + w) (6.12) with respect to t and the first and secondof eq. Let a compressible fluid in medium occupy V and be surrounded by a surface S and consider an element of such a field as shown Fig. (6. (6.14)
wherePo is the external pressure on the surface of the basin and g is ~e acceleration due to gravity.= ~Y Since the fluid is incompressible.4. so that the bound~y condition on w becomes:
~w
~(P.16) (see Fig.13) p-~. using eq. 6.5). (6.4 Wave Propagation in an Acoustic Medium Wave propagation in three dimensionalmedia is a phenomena cover's a variety that of fields in Physics and Engineering.
t>O
6. i. then the component the the of velocity nodal to the bounda~C must vanish. Wave propagation in acoustic mediais the simplest three dimensional wavephenomena physical systems.: of P = Po + 0 g(h .12)
Let p be the pressure acting on the sides of an element.8. one the obtains:
32W= ~V2p
2 Ot
(6.13) with respect to x and y respectively and combining resulting equalities. Hence. then the equation of equilibrium becomes: ~u 0p and ~v ~p (6. The law of conservationof massfor the element can be stated in
. (6. since the wall sunounding basin is rigid.e.14) into eq.13) one obtains:
3u
~:-g~
~w 3w and 3v ~=-g~
(6. ¯ e nodal component the velocity Vnbecomes: of
Vn = UCOS~
Ow ¢~w + vsin~ =-gj~cos~ + ~sin~Jdt
~w
=-gff~dt = 0 wheren is the nodal to the curve C.the pressure at any depth z in the basin can be described by the static pressureof the column fluid abovez.15) results in the equation of motion pa~cle on the surface of a liquid basin as follows: V2w= 12 O2w c e 3t wherec: = gh. 6.
6. (6.then Helmholtzequation results. Neither of these mediawill be further explored in this book.t) = g(P.5.1 Vibration in Bounded Media
Onemethodof obtaining the solution to forced vibration problemsis the methodof separation of variables. t > 0 P on S.
6.
OF MATHEMATICAL
PHYSICS
307
=.22)
:-
3q~
3F fy=
w=-0-
fz
P = 19o-~.18) and the last three equationsof eq.20) result in the Wave Equationon velocity potential ~ as follows: V2. 1 32~ 1 3F.o--d
fx =
3F (6.t)= gt(P)
.t) n On
(c) Elastic boundary: (P. (b) P on S.i.t) =g(P ~P P on S.5
Helmholtz
Equation
Helmholtzequation results from the assumptionthat the vibration or wave propagationin certain mediaare time harmonic. if one lets e it°t be the time dependance.~ .20) are satisfied identically. Wave propagationin elastic mediaand electromagneticwavesin dielectric materials are governed vector potentials instead of the one scalar potential for an acoustic by medium.PARTIAL
DIFF. (6. t > 0 (6.+ 9oF + Po then the equations(6.24)
This equation describes a variety of diverse physical phenomena. t > 0 Ot wherey represents the elastic constant and vn is the normalparticle velocity. 8~p
EQ.22) into the first of eq. (6. P in V. having the following form: V2q~ + k2~ = F(P) P in (6. This methodassumesthat the deformation~(P. Substitution of eq.t) can be written as a product as follows: ¢~(P. t) + n (P.23)
The boundaryconditions can be one of the following types: (a) p(P. t >
0~ normal componentof the velocity = g(P.e.
Non-trivial solutions of (6.28) are due to non-homogeneou~s boundaryconditions.27)
while the Laplace equation has the following form:
~72~p = 0
(6. The general solution can thus be represented by superposition of infinite such standing waves.27) are due to the source function f(P) or to non-homogeneous boundaryconditions.26). The functions qt n(P) are known Standing Waves. but if the wave process is given enough time (say. The boundaryconditions required for a unique solution of the Helrnholtz equation are the sametype specified in Section 6.e.t) = ~(P) i~t where~(P) satisfies eq.6 Poisson and Laplace Equations (6.CHAPTER
6
308
wherethe functions ~(P) and T(t) satisfy the following equations: V2~l/n+~.26)
Poisson equation has the following form: V2~ = f(P) P in V (6. i.The lines (or surfaces) where as ~n(P) = 0 are knownas the NodalLines (or Surfaces).24). Non-trivial solutions of (6. 6.~) then the initial transient state decaysand the steady state describedin eq.: q~(P. if initiated at o = .28)
Various steady state phenomena Physics and Engineering are governedby in equations of the type (6.. i.25) T~' + c2XnTn 0 = This equation leads to eigenfunctions ~n(P) where~'n are the correspondingeigenvalues. (6. The function ~(P.~.26) results. the solution wouldhave the samefrequencyo~ as that of the forcing function.e.
. (6.27) and (6.t) = g(P) i~°t can be developedin the form of harmonicwaves.28). wherethe solution is periodic in time.n~l/n = 0 (6.5.t) by forces whichare periodic in time. whenthe forcing function f(P.t) has the form: f(P.2 Harmonic Waves
The solution of wavepropagation problems in media where the medium induced to is motion~ (P.e.. Since the motionis assumed have been started at an to initial instance t o = . i. then no initial conditions needbe specified.t) wouldnot initially have the given in (6. 6.3.
If there are sources or sinks in the fluid medium.1) of becomes: V2T = -q(P) The boundaryconditions are those specified in Section 6. 1 and are separated by a distance r. then eq. (6.
6.2.z)
Mass ~ .2 Flow of Ideal Incompressible Fluids
Fluid flow of incompressible fluids can be developedfrom the formalismof flow of compressiblefluids. respectively. then the force ~ becomesthe field-strength at a point P due to a massmat x 2 definedas (see Fig. then the force of attraction (F) between and m can be 1 2 stated as follows: ~ = ~' ml~m2 ~r where~r is a unit base vector pointing from m to m along r.1.22)..6.3 Gravitational (Newtonian) Potentials
Consider two point massesm and m2. 6.y.9):
. 6.9
6.. then the velocity potential satisfies Laplace'sequation.PARTIAL
DIFF. the then velocity potential satisfies the Poissonequation. then eq.29) ~xx + ~y + ~ = 0 If one uses a velocity potential q~(P)as describedin (6. and 2 I 1 m = m. If one sets m = 1.17) becomes: 0u 0v bw (6.1
Steady State
Temperature Distribution
If the thermalstate of a solid is independent time (steady state). 6.6. Since the density of an incompressiblefluid is constant. (6.
EQ.'" ~k~ Unit
Fig.. Y
OF MATHEMATICAL PHYSICS
309
P(x..6. located at positions x1 and x2.
Define ql the electric field as the force on a unit charge(whereq2 = 1) located at a point P due to charge q2 = q as: ]~= q ~ ~r =.33)
V wherep(x.
. Definethe repulsive (attractive) force F betweentwosimilar (dissimilar) charges of magnitudes and q2. 6. respectively.
EQ. located at positions x1 and x2.z)
X
If the massesare distributed in a volumeV. such that ~ = -V~. ~.z) is the massdensity of the material occupyingV and ~ satisfies the Poisson equation.
OF MATHEMATICAL
PHYSICS
311
Y
Z ~ ¯ P(x.6.11): ¯ =Y p(x .y. as: ql ~=qlq____~2 ~ 4~er 2 r wherer is the distance between and q2 and e is the material's dielectric constant.y. then the total potential due to the mass occupying V becomes(see Fig.y.z dV' (6. 6.q'~-'v(lq ~ 4her 4he \ rJ If we define an electrostatic potential.PARTIAL
DIFF.4 Electrostatic Potential
The electrostatic potential can be defined in a similar mannerto gravitational potential. then a solution for the potential is:
~ = q-fi-r
If there exists distributed charges in a volume then the potential can be defined as: V.
the equation becomes: X" ¥" X Y Since both sides of the equality in the aboveequation are functions of one variable only. y) = T(x. then the Laplace Equation is transformed into two ordinary differential equations: X" .y) = X(x) X ~: 0. each dependingon one coordinate variable only.PARTIAL
DIFF. Y ~: 0.
EQ.9
The Laplace
Equation
The methodof separation of variables will be employed obtain solutions to the to Laplace equation. i. H) =
Since the sheet is thin. 0 < x < L and 0 < y < H
Substituting T(x.T V2T = 8-~ + by2 = 0 Assume that the solution can be written in the form of a produc of two single variable t functions as follows: T(x. a Choosinga2 > 0. wecan assume that the temperaturedifferential is only a function of x and y. The methodrequires the separability of the Laplacian operator into two or three ordinary differential equations. then the equality mustbe set equal to a real constant. wherethe boundaryconditions are specified as follows: T = T(x. T(x.y) into the differential equation. 0) = T(x. Afew of the orthogonaland separable coordinate systems are presented in Appendix C. The use of the methodcan be best illustrated by workingout examplesin various fields in Physics and Engineeringand in various coordinate systems.a2X =0 Y" + a2y =0 whichhas the following solutions:
. y) T(0. Thedifferential equation on the temperaturesatisfies the Laplace equation: ~)2T~)2. one obtains: Y dx~ + X dY = d2X d2y ~ 0 Dividing out by XY. occupying space the 0 < x < L and 0 < y < H. The methodconsists of assumingthe solution to be a product of functions. Example 6. y) = f(y) T(L.1 Steady State TemperatureDistribution in a RectangularSheet
Obtain the steady state temperaturedistribution in a rectangular slab.~.
OF MATHEMATICAL PHYSICS
319
6.e.y). + z.
resulting in four new problems. If.. i = 1.
. Each of these problemswould resemble the problemabove.. then: C=0 for a ¢ 0. This indicates that the choice of the sign of a2 leads to either the existenceof non-trivial solutions. In order to eliminate the guessworkand minimizeunnecessarywork. Let: T--T1 +T2+T3+T 4 where 2 T = 0. This leads to a choice of a2 > 0. for .. Since the solution involves an expansionin a GeneralizedFourier series. 4 . they are oscillating functions with one or morezeroes.
OF MATHEMATICAL PHYSICS
321
Usingthe orthogonality of the eigenfunctions. The solution then would be the sumof the four solutions Ti(x. These eigenfunctions must satisfy homogeneous boundaryconditions. or to the trivial solution. or D T(x. a2Y=0
whosesolutions become a ¢ 0: for X = A sin (ax) + B cos (ax) Y = C sinh (ay) + D cosh (ay) The solution must satisfy the boundaryconditions: T(x.
EQ. 3. If the temperatureis prescribed on all four boundaries. then the above analysis must be repeated: X"+aZX=0 y".one can use the principle of superposition by separating the probleminto four problemsas follows. the choice of the correct sign of a2 can be madeby examiningthe boundaryconditions.H) = C sinh (all) ¯ X(x) However. Each solution Ti satisfies one non-homogeneous V i boundarycondition on one side and three homogeneous boundaryconditions on the remaining three sides. since sinh (all) cannot vanish unless a = 0.y). specifically. Thus. for this example. one obtains an expression for the Fourier constants En as: H ff(y)sin(-~y)dyo En=Htanh2(~L) Notethat the choice of sign for the separation constant is not arbitrary.0) = D X (x) = 0. then choose the sign of a2 to in give an oscillating function in y and not in x. there is no non-trivial solution that can satisfy the differential equationand the boundary conditions.az. yielding four different solutions.a2 was chosenwith a2 > 0. these eigenfunctions must be non-monotonic functions. then one wouldneed an eigenfunction set. Thus.since the boundaryconditions were homogeneous the y-coordinate. 2.. Furthermore.PARTIAL
DIFF.
Since the shape of the membrane.t) becomes transient solution.. then the motionof the membrane be independentof 0.
OF MATHEMATICAL
PHYSICS
353
where the Fourier coefficients do not contain a source component: Enm(t) = Anmsin(knmt)+ Bnrn cos(knmt) The initial conditions as given in eqs.97) and (6. and the boundary at condition R(a) = 0 gives the characteristic equation:Jo(ka) = Jo(~t) = 0. boundaryconditions.15 Forced Vibration of a Circular Membrane
Obtain the transient motionof a circular membrane. in response to transverse time-varying forces q(r.. 2. where = ka.0 = f(r). whoseradius is a. y)sin (~ x)sin (~--~ y)dx = 00
Thus.e.PARTIAL
DIFF. i. (6. and the source term are the not dependenton 0. the The eigenfunctions of the systemcan be obtained by solving the associated Helmholtzeq..t) can be written as follows: 32w+l~w 1 ~92w 2 ~r2 r ~--~ = c 2 ~t q(r.t).t) = +) w(r. ) then the eigenfunctions become: Rn(r = jo(~tn r) ) a There is no zero root of the characteristic equation.t) S O<_r<a t>O
with boundaryand initial conditions given as: w(a. Let n be th rootof the kt then characteristic equation(wheren = 1.0 +) 0
¯
Since the boundaryconditions are homogeneous.98) results LL Bnm ~2 f f f(x. The equation of motion satisfied by an axisymmetricdisplacement w(r. The membrane initially deformed is to displacement f(r) and released fromthe rest.: r 2 R" +rR'+k 2r2R=0 R=AJo(kr)+BYo(kr) Boundedness r = 0 requires that B = 0. the final solution for the responseof the plate is given by: w= E E -~y)cOs(knmt) Bnmsin(-'~x)sin(-~ n=ln=l Example 6. 3 . the steady state part of the solution then vanishes and w(r.: will axi-symmetric. Writing out the solution in terms of the eigenfunctions with time-dependent Fourier coefficients:
. and
(r.
EQ.
(6. the solution for the first component En(t)..~n. the twoparts of En(t) werefound and the transient solution of the responseof the plate evaluated.~ 0 :~ Po~5(t. Usingthe properties of the Dirac delta function (Appendix one D) obtains: Fn(t) Po~5(t-~ to)C2 ~ ~(r). t) = 2~5(r)iS(t where~5 is the Dirac delta function and represents a point force of magnitude applied Po impulsively at t = t o.. 2~a2[Jl(~n)] a ~ dr Jo(P'nar)
whichwhensubstituted in the integral for En(t) for the source component results in:
.lXn a '~'* t"l~. = ~ J r-'-~" Jo~. n ~ a Thus.. If the applied load on the membrane takes the form of an impulsive point force of the form: q(r.CHAPTER
6
354
w(r.. results in: E (t) = n sin(~tn ct) + B cos(gn c t ) n n a a with
a r f(r)Jo(g r)d f nr 2 Bn _ a a2 [Jl(lt'tn)]2
A =0 n The secondcomponent En(t) that dependson the source term requires that one first of evaluatesFn(t as: ¯ ) 2c2.t)
= EEn(t)Rn(r) ri=l
then.95) that is due to of initial conditionsonly. given in eq.to)C r)dr . Fn(t ) = a2[jl(gn)]2 0 which gives the component En(t) due to the source as: of t En(t) = a----~-fsin( C~tn(t .rl)) Fn(~l)drl cp.
0) = O f(z. The surfaces z = 0 and z = L are kept at zero temperature.ne the temperature distribution in a curved wedgeoccupying the region a < r _<.
OF MATHEMATICAL
PHYSICS
369
Insulated Sheet
20. extending from its axis to the outer surface. 0 < 0 < 2n and 0 < z < L. o < z < L and 0 < 0 < b. Determ~.0) = O f(z. and is heated at its convexsurface to a temperature: T(a.
EQ.z. see the accompanying figure. see the accompanying figure.0) Y
r
21. The cylinder has an insulated surface at 0 = 0.0)
. Determine the steady state temperature distribution in a solid finite cylinder of length = L and radius = a defined by 0 < r < a. The cylinder is kept at zero temperature at its two ends (z = 0 and z = L).'. the surface 0 = 0 and 0 = b are insulated and the cylindrical surfacer = a is kept at a temperature: T(a. ~.PARTIAL
DIFF.z.
such that the velocity potential ~(r.10
23. A heat source is located in a spherical region inside the sphere. 0 < z < L. A point source and a point sink of magnitudesQo are located on the axis of the cylinder at z = L/4 and 3L/4.0) Section 6. The convexsurface at r = a. A metallic sphere of radius a and defined by 0 < r < a and 0 < 0 < ~t is kept at zero temperature at its surface. 25. the two plane surfaces at 0 = 0 and n and the lower base at z = 0 are kept at zero temperature. A spherical container is filled with a liquid whosewalls are impenetrable.z) satisfies:
V2~I/=
" Qo~5(r)[~5(z-L/4)~5(z23L/4)] 2r~r
Find the velocity field inside the container. Determine temperaturedistribution in a hemi-cylinder of length = L and radius = a the defined by 0 < r < a.
. respectively. A finite circular cylindrical container with impenetrable wails is filled with an incompressible liquid. If a point sink of magnitude exists at its center so that the velocity potential satisfies: Q V2~ = Qo ~(r) 2 4~r 0 _<r < a
Find the velocity field inside the sphere. while the upperbase at z = L is kept at a temperature: T(r. 0. 24.L) = Ofir..CHAPTER
6
3 70
a
a
X
22. such that: V2T =-q(r. 0 < 0 < n and. cos0) 0_<r<b
Find the steady state temperaturedistribution. occupyingthe space 0 < r < a and 0 < z < L.
43.
Determine temperaturedistribution in a solid sphere of radius = a.0. whosesurface the conducts heat to an outside medium that is being kept at zero temperature.
EQ. The cubes' surfaces are kept at zero temperatureand the cubeis initially heated to a temperature: T(x.PARTIAL 38.z. whosesurface is the kept at zero temperatureand is heated initially to a constant temperature= T o. Determine temperaturedistribution in a solid sphere of radius = a. The sphere is heatedinitially to a temperature: T(r.0÷) = T f(x.0+) = T f(r) O Determinethe axisymmetric temperature distribution for a circular slab of radius = a.to)~5(0Q= Qo~5('~
41.
OF MATHEMATICAL
PHYSICS
373
Determinethe axisymmetrictemperature distribution in a circular slab of radius = a.The slab is initially heated to a temperature: T(r. whose perimeter is kept at a zero temperature.0) O and has an impulsiveheat point source at (ro.z) O Determine temperature distribution in a sphere having a radius = a." -~°) ~5(t . The slab is initially heated to a temperature: T(r.0.Theslab is heated initially to a temperature: T(r.
44.0÷) = T fir.Thesphere is initially heated such that: T(r.0+) = T f(r) O with an impulsiveheat point source at its center: t Q = Qo~(~r)~5( -to)
39.0o).~.
.
40.cos0. radius = a. whose the surface is kept at zero temperature.y.
Determine temperature distribution in a circular slab.0+) = T f(r.~) O
42.
DIFF. whoseperimeter the is kept at zero temperature.y.0+) = T f(r) O Determinethe temperature distribution in a cube having a sidelength = L. such that the slab conducts heat through its perimeter to an outside medium whose temperatureis kept at zero temperature.
become: f(x)e~UXdx 0 0 F-(c°)lv = 0 = F_(u): ~f(x)eiUXdx so that the twointegrals can be combined into one integral over the real axis:
OO
F+(~°)Iv = F+(u) =o
F(u) = F+(u) + F_(u) = ~ f(x)eiUXdx The inverse transform becomes. belowthe line v = b. hence one maychoose t~ = ~ = 0.4). as shown Fig. F. (7. 7. v2)e-iUXdu --. it is convenient to choose a common line-contour for the inverse
.4.INTEGRAL
TRANSFORMS
397
Function f(x)
Pathin complex m-plane
Fig.3. 7. which reflect the relative values of a and b are enumeratedbelow: (i) a < 0 and b > The function f(x) vanishes as x --) + oo. Thus. but in there is a common region of analyticity for the transform as shown the shadedsection in in Fig. the two transforms F+ and F. fix) is not in general absolutely integrable.(~) is analytic in the lowerhalf plane of ~o.~ F(u)e-iU×du
whichis the complex Fourier transformand its inverse as defined in eq. can be used for the inverse transform.with v = 0: 1 f(x)= ~ 1 (U. in Then. Vl)e-iUXdu+ F_(u. 7. 7. as shown Fig. shown Fig.4. Anycommon to line contour.2. 7.3
Similarly. The contour integration for the inverse transformation must then be in taken in those shadedregions shown Fig.Some special cases. 7.2. in The contour integrals of the inverse transforms dependon the rate at whichf(x) becomesexponentially unbounded. Thenthere exists a region of analyticity that is common both transforms. (ii) a < In this case. where a < v < b.
x)eiC°Xdx 0
(7. then a < 0.23)
f(x)= ~'~1 f F+ (co)e-i°xdco + f F_ (o)e-i°xd~0 [-~ + iy --~ + iy Further discussion can be carried out for the possible signs of a and b:
a > 0 then b > 0. wherea < y < b:
[oo+i7
(a) If (b) If
oo+i7 (7. as
In either case.4
transform ~ = 13 = 7. The inverse transform becomes: oo+iy fix)= ~ f F(co)e-i°~xdco -oo+iy where a <~ <b (7. Hence. 7.CHAPTER 7
V
398
y
Functionf(x)
Path in complexco-plane
Fig.24)
--oo
whereF(co) is analytic. while the inverse transform is taken on a common line. then one on can choose a common value for v such that: a< v = Im (o) <
oo oo = f f(x)ei~°Xdx
F(O)= F+ (co) + F_ (co)
~f(x)eiC°Xdx+
~f. then f(x) is a function that vanishes as x --> oo and becomes unbounded x --> _oo.25)
It should be noted that the function F+(CO) F_(CO) have poles in the complex and may plane
.22). since the function is unbounded only one side of the real axis.the Fourier transforms F+ and F_ are defined in the same manneras given in equation (7. then fix) is a function that vanishes as x --~ _~oand becomes unbounded as x b > 0.
let: f(-log x) = g(x) 0 <x <= then the Laplace transform becomes:
. 7.7
The transformF(p) is analytic to the right of Pl = a. so that one maychoosePl = ~' such that all the singularities of F(p) are located to the left.
7.INTEGRAL
TRANSFORMS
401
Y singularity
Function f(x)
Path in complexp-plane Fig.7). 7.of the line Pl = %(see Fig. the two sided Laplace transform can be altered by makingthe following transformation on the independentvariable x: x=-logrl or
then the two sided Laplace transform takes the form: FLII(P)=~ f(x)e_pXdx= ~f(_log~l)ePlOgr~ drl= ~ f(_ log 1 drl
0
and the inverse Laplace transformis then defined by: f(xl=f(-l°grl)=2~ i f FLII(P)e-Pl°g~qdP=2~ 1 ~ FLI/(Plrl-PdP
To redefine these transformintegrals.13 Mellin Transform
For the case of a < b.
For example. In this case. respectively. The function g(t) must be solved for eventually after finding y(x. For use of Laplace transform on time. In this section. Thereverse wouldalso be true: if 3y(0. Hence.INTEGRAL
TRANSFORMS
411
f(x)
~[e ax .t) in terms of g(t). as required by the uniqueness theorem.if y(0. the inverse transform of F(p) is given by: f(x) = ~__C {e-CXI-~-_ + (a .t)/3x = an unknown function. Use of Laplace transform on space is moreproblematic. application of Laplace on time for the first or secondderivations in time requires the specification of one or twoinitial conditions. uniqueness theorem requires that only one of these two boundary conditions can be specified at the origin.c) xl + e-bx ~c_ ca whereeqs. one must assume that the unknown boundary condition is a given function.t)/0x = f(t). then one mustassumethat 3y(0. one wouldrequire initial conditions at t = 0. 7.~c
Thefunction F(p) has a simple pole at p = -b and a pole of order 2 at p = -c. an unknown function.42) and (7.15 ~ Solution of Ordinary and Partial by Laplace Transforms Differential Equations
Onemayuse Laplacetransform to solve ordinary and partial differential equations for semi-infinite independentvariables. the Laplacetransformwill be applied on various ordinary or partial differential equations in the followingexamples.43) were used.t) = g(t).t)/~x.e -ax ] =
sinh (ax)
(v) Find the inverse transformof F(p).b)2 3 Thus. y(0. Let: F(p)= A1 ÷ A2 + A3 p+c (p+c) 2 (p+b)
then the coefficients Aj are foundfromthe partial fraction theorem: G(p) = (p + 2 F(p) = p + a ¯ p+b Al=_~_p(_C)= b-a
a-c
_ b-a
A = G(Pl ) = G(-c) = 2 b-c a-b A = (p + b) F(P)lp = (c. (7. wheret > 0.t)/~x2 wouldrequire the specification of y(0. defined p+a F(p) = (P + b)(p 2 b. Use of Laplace on x for the second derivative ~2y(x. a specified function.
.t) = f(t). However.t) and by(0. This indicates that the Laplace transform is suited to use on time rather than space.
...the particular solution of the systemin (8.3)
Ui(g = 0 i -.In as essence.3.1. differential operator with non-constant coefficients.1) yp(X)is given
453
. satisfying homogeneous boundaryconditions. n (8. given in (4.. The solution g(xlE) is then solution of the systemdue to a point source located at x = E.E). the solution of non-homogeneous ordinary and partial differential equations is obtained by an integral technique known Green's function methOd. 2 . n
Boundary
(8.35).2)
whereL is an nth order ordinary.8
GREEN'S FUNCTIONS
8. in general.4) ) where5(x) is the Dirac delta function (AppendixD).E)dE= I g(xlE)f(E)dE I =L g(xlE)
a a a
Hence. Define the Green'sfunction g(xl{):
L ) --g(xl
(8.1) and substituting (8.1) (8. known Green's function. 2 . linear. The solution of (8..8A) gives the Green's function for the problem. as so that the solution for a distributed source is obtainedas an integral of this function over the source strength region. Rewriting (8.. g(xlE) is not symmetricin (x.27) and i are t he non-homogeneous b oundary c onditions i n (4.It should be noted that..3) for the operator L: b b b Ly = f(x)= f( ~) 5(x.1
Introduction
In this chapter. the system's response is sought for a point source..
8.2
Green's Function for Ordinary Differential Value Problems
Consider the following ordinary linear boundaryvalue problem: Ly= {~(x) Ui(Y)= ~'i a<x<b x<a or x>b i = 1.
6) by g(xl~).~).~).~)]dx a a The left-hand side of (8. obtain the adjoint Green's function g*(xl~). The right-hand side then gives: b yp(X) = 'f(~) g*(~x) (8. one obtains: b b ~(g*Lyp.6)
multiplying (8. subtracting and integrating the resulting two equalities one obtains:
.8) vanishes due to the definition of an adjoint system(see section 4. Rewritingthe two olrdinary differential equations(8.2 For the system given in Example 8.(xl~)
g*'(21~)+ 2g*(21~)
+ ¢~ ~z ~ H(x-~
Followinga similar methodof solution.b).4~)(x. the particular solution can also be obtained as an integral over the source distribution f(x) and the adjoint Green'sfunction g*(xl~). not symmetric (x.CHAPTER
8
456
and satisfy the adjoint boundary conditions: V (g*(x It) ) = 0 i The resulting adjoint Green's function g*(xl~) is.~_
It should noted be lhat g*(xl~) is not symmelric (×.12).6) by yp(X)and (8.ypKg*) dx = ~[g*(xl~ ) f(x).~) and the adjoint boundaryconditions become: (8. one obtains the Green's function g*(xl~) as: : ~(~2.1. both Green's functions are not symmetricin (x.4
Symmetry
of
the
Green's
Functions
and
Reciprocity
In general.~). Example 8. The adjoint operator K: K g'(xg)= x2g"" + 6xg" +6g* = ~5(x. Howeve?r. in
8. in Multiplying(8.7)
g*0]~): 0
g.3) (8.3) and (8. Green's the function g(xl~) and its adjoint formg*(xl~) are related.1) by g*(xl~) and after subtracting the two equations integrating over the range (a.yp(x) ~i(x.9) a Thus.3) by g*(xlrl) and (8. in general.6) L g(xl~) = 8(x-I) K g*(xlrl) = ~(x-rl) (8.
they are symmetric with each other. (see section 8.
.11).2 results in:
Note that:
The particular solution is nowgiven by: 2 yp(X) = ~(x~) ~-~d~ 1 which is the same as in Example 8. This symmetryis known the "Reciprocity" principle in physical systems. 8. then L = K and ) = ViY This ( ).~).
Followingthe methodused to find the Green's function in.34)).3 If one rewrites the operator in Example into a self-adjoint form~. meansthat g*(xl~) = g(xl~) and hence: g(xl~) = g(~lx) (8. This can be seen in Examples and 8.
457
a a The left-hand side vanishes and the fight-hand side gives: g*C~lrl)= gCql~) (8. Defining ~(x]¢) as the Green's function for the self-adjoint operator ~: d(1 d~ 2_
o
0
. one obtains: d(1 dy~ 2 ~y = ~xx t'~-~)+'~y ='~'-47 1 <x <2 Note that the source function becomesf(x)=x-4.11)
which means that Green's function is symmetric in (x.1 and 8. Example 8.2.10)
This meansthat while the two Green's functions are not symmelric.1.1 4. It indicates that the response of a system at x due to a point source at ~ is equal to the responseat ~ due to a point source at x..Examples8.1 If the operator L is self-adjoint (see (4.GREEN'S b
FUNCTIONS b .
6 Green's Functions
for Higher Ordered Sources
If the sourcefield of a systemis a distributed field of higher order than a simple source.15) N -th wheret~N(x is the N order Dirac delta function. i.7 Green's Function for Eigenvalue
Problems
Consider a non-homogeneous eigenvalue problem obeying the Sturm-Liouiville systemof 2na order (see section 4. etc.49). one gets:
Lgl(xl~)=~(x-~)= d~(x-~)
(8. one can showthat the Green's function for such a systemis obtainable from that for a simple source.L2)
8. Thus: 3g gl(xl~)~. For example.3. then one can show ) that: L gN(xl¢ )=
ar~(x.~(x~:)-. see section D.e.14)
In a similar fashion.if the Green's function for a dipole source or a mechanicalcouple is desired then: dx where$1(x-~) represents a positive unit couple or dipole.2. Starting with the definition of g(xl~) for a point source: L g(x~) = 8(x-~) (8.3) and differentiating (8. one can obtain the Green'sfunction for distributed source fields of higher ordered sources (quadrupoles. ~x
wherethe last equality is the identity (D.:
.x (8.3) oncepartially with {.~([~) .13)
. and g~(xl~) the Green'sfunction for a dipole/couple source. see section D.= (-1) N 0Na(x ~) ~x
gr~(xl~) 3r~g(xl~) =
(8.octopoles.15).16)
8.GREEN'S
FUNCTIONS
459
g(x (x-~) ~)+x I~): I~(xL
L
u(x) = ~ g(x I ~) f(~)d~ 2 .~(~-~1:~(x-~)__~(x_~) ~ .): ¢) (8.
. Thetotal solution is in the formof an infinite series resulting fromthe substitution of (8. ..21). as each problem requires the use of a specific transformtailored for that problem. wheref(x) is bounded and "imt obeys the absolutely integrable. the dependentvariables y(x) must absolutely integrable over the semi-infinite region. 2 . There is no general methodof solution.The bar is excited to vibration by a distributed harmonic force f(x)e -i°~t.dx 2 y(O):O 2:017¢2 x >0 C:~
.6 Green'sfunction for the vibration of a finite string
FollowingExample 8.
one can showthat the eigenfunctions and eigenvalues are: d~n(X = sin(nrtx / L) ) ~'n = n2 n2/L2
The Green's function then becomes: 2 n=l sin (nn x]L)sin (nn ~L)
8. In such problems.7 Obtain the responseof a semi-infinite bar vibrating in longitudinal mode. Example Green'sfunction for the longitudinal vibration of a semi-infinite bar 8. one can obtain the Green's function for the stretched string using eigenfunction expansion. Furthermore.. 2. Starting with the homogeneous equation: u" +~u= 0 u(O) = u(L) = n = 1.CHAPTER 8 '
462
g(~' ~): ~be(~)~t(x)
(8.8 Green's
Function
for
Semi-infinite
One-Dimensional
Media
The Green's function for semi-infinite mediacannot be obtained through the methods outlined in the previous sections. Example8.26) in the integral (8. Essentially.~).boundary.26)
It should be noted that the Green'sfunction in (8.5.k2y = f(x) -'~ ~ ... use of integral transforms such as Fourier transforms becomes necessary.3): daY . n = 1.26) is a symmetric function in (x.value problemsin a semi-infinite region have boundaryconditions on one end only. The longitudinal displacementof the bar y(x)e followingequation (see section 4.
reflected) waves could possibly originate from the farfield.= c(1. The transform of g(x) is obtained from above as: sin (u~) g(U) = U2 _ The inverse Fouder sine transform of ~(u) is thus given by: sin(u~) sin (ux) 0 In the inverse transformation. eq. see section 7.irl) then:
C* u2~(U)
I << 1
= l-~ -~.16.k(1 + i n/2) ="7
c
is a complex number.k2~ + ug(0) = f ~(x .e. Applyingthe Fourier sine transformon the differential equation on the Green'sfunction. The first integral becomes:
. This can be accomplishedby makingthe material constant complex. one wouldassumethat the medium material absorption that has wouldinsure that outgoing wavesdecay and hence no incoming(i. The Green's function then satisfies the following system: dx ~5(x.i n/2)
so that: k* o) _. g(01~) 2d2gk2g= The Dirichlet boundary condition at x=0requires the use of Fourier sine transform.~)sin (ux) dx = sin 0 where~(u) is the Fourier sine transformof g(xl~) and u is the transform variable.e.~).50): . (7. To insure this. i.GREEN'S
FUNCTIONS
463
where A is the cross-sectional area and E is the Young'smodulus. Letting the Young's modulus becomecomplex: E* = E(1. no incoming wavesin the farfield. as it requires even-derivative boundaryconditions. care must be taken to insure that wavespropagate outward in the farfield and no wavesare reflected from the farfield. Rewriting the inverse transform with complex results in: k*
co(u(xcos(o(x
U2 _ k*2
0 The integrals can be evaluated using integration in the complex plane.
and S is the sum of all the surfaces enclosing the region R. the derivation of the transformationgiven foi.11 Green's Identities
for the Laplacian Operator
In this section.the integrals in (8. In this integral dx is a volumeelement in the region R (shaded region). then L = -V2 = K. (Vv) Similarly.~dS
(8.32) wherev and u are scalar functions and V is the gradient. Example 8. =
where:
(8. defined positive awayfrom the region R.9: vLu-uKv=V.CHAPTER
8
468
S
~R
V.34) (8.(Vv) subtraction of the twoidentities (8. is the divergenceof a vector. see Figure 8.34) results in a newidentity: (8. then: V.in (8. fi is a unit outwardnormalvector.32) then one gets: V.fi dx =yS fi. Since the Laplacianoperator is self-adjoint in cartesian coordinates..31) are performedfor the Laplacianoperator.33) and (8. (uVv-vVu)= v.32)
wherefi is a vector function and V. ~ = uVEv+ (Vu). ~ = vVEu+ (Vu).35)
.1.
-u-ffr+VNJer+ TO0
v)
r~-'~e°+L
Oz
=
8. If one lets ~ = vVu.in (8. if one lets ~ = uVv.10 For the Laplacianin cylindrical coordinates in three dimensionalspace given in Example 8.33)
uVZvvV2u v.
u~74v = vV2U . ~ dS wherethe last integral resulted from the use of the divergencetheorem.38)
8.v~UnU) The terms in the integral over the surface S represent boundaryconditions. Substituting 2 for V in eq.39)
8.[V. let V2u U and V~v= V. (8. since the Laplacianis a self-adjoint operator. vu]Vu.k)u .35) over the volume ~R (vV2u-u VT-v) dx= ~R V.u(-L .:.vLu) dx = ~S (u~.(vVU). (8. then: = uLv .35) by L above then: v(-L .~g-ug)dS
The terms in the integral over the surface S represent boundary conditions.12 Green's Identity
for the Helmholtz Operator
The Helmholtzequation has an operator given as: L = -V2 . Thelast integral can be simplified to: (8.13 Green's Identity
for Bi-Laplacian Operator
4 Thebi-Laplacian operator V is defined as: L = -V4 2 -V2V = whichis a self-adjoint operator and showsup in the theory of elastic plates. In order to use the results for the Green's identity for the Laplacian.uV2V
--[V.GREEN'S
FUNCTIONS
469
is a bi-linear function of u and v.k)v = uLv .}'s~~u ~Vv)ds ~v~
resulting in the identity:
l.
(8. (8. ~ dx= ~Sft.k so that it is also self-adjoint.37)
¯ Is~('~vu.(uVV)Vv.vLu = The Green's identity for this operator becomes: ~R (uLv. Integrating eq.vLu =
v~74u .
Rewritingthe terms in the secondbracketed quantities:
.
see (8. one can write the solution for u(x) as: x u(x):-alan g(r~) f(~)d~ (8.
8.52) OR OR S The terms in the surface integral over ~ has both initial and boundaryconditions. and the Green's function g* (~) for the adjoint operator K to satisfy: Lu(x) = f(x)
Lg(x[~): 6(x
(8.31). one obtains: t t ~ f(vLu-uLv)dx dt ~f ~'~'dxdt=j~'~d~ (8.53) n whereL is an operator in n-dimensional space and f(x) is the source term that absolutely integrable over the unbounded region R Define the Green's function g(x]~) n. The condition on g(xl~) wouldalso require that it decaysat a prescribed rate as R --~ ~.56)
.g* (x~ ) fix n
= u(~). for the unboundedregion. Integrating the identity (8. Thus. The surface Sn of an unbounded medium could be taken as a large spherical surface with a radius R --> ~.It should be ) noted that g* (~)= g(~x). The integrand then must decay with R at a rate that would makethe integral vanish. then: u(~): ~Rn g*(x[~) f(x) dx = ~Rn g(~Jx)f(x)dx Changing by ~ and vice versa. Multiplying (8.45).31) vanishes.50) over the spatial region R and time. over the surface Sn.53)by ) and (8 .CHAPTER 8
472
and the gradient V given in (8.55)
whereg(xl~) and g*(xl~ must decay in the farfield at a prescribed manner.55) by u(x) and integrating over the unbounded region.54)
K
(8.fR g~ (~) f(x)dx
n
Theintegral on the left-hand side can be written as a surface integral. one obtains: ~R (uKg*-g'Lu)dx : n [u(x) 6( x-~).16
Green's Function for Unbounded Media--Fundamental Solution
Consider the following system on the independentvariable u(x): x in R (8. if the left-hand side of(8. knownas the Fundamental solution.
resulting ds=.e.if the operator L is one with constant coefficients. Furthermore.17.4nR dR
for R >> 1
.i.e.60)
dg _ 1 ~" .GREEN'S
FUNCTIONS
473
If the operator L is self-adjoint.~.4nR2 _ I = last (8.g. then one solves for the Green'sfunction with ~ = 0.~1) 8(x2" ~2) ~(x3 ~3) " Since the Laplacianhas constant coefficients.17 FundamentalSolution for the Laplacian
Considerthe Poisson equation in cartesian coordinates: -V2u = f(x) x in n (8. i. The solution u(x) can be obtained as an integral over the Green'sfunction and the source f(x) given (8. g(xl~) g(~lx). the point source is transferred to the origin:
_~72g = ~(Xl) i~(x2) ~(x3 )
(8. integral in (8. then ~--~on S..56).61) then becomes: d.59) over Rn. becomes -~rg (R) whichis a constant in the farfield R since g = g(r) only._. the spherical surface whoseradius is R.1 Three dimensional space
Define the Fundamental solution g(xl~) to satisfy: -V2g(xl~) = 8(x . one can transformthe cartesian coordinates to spherical coordinates for a spherically symmetric source. 8.58)
wherethe Laplacian is a self-adjoint operator with constant coefficients. integrate (8.59)
Since the source is at the origin. then the Green'sfunction is symmetric.61) (8. then: g(x~) = g(x(8.I ~Rn V2g dx = ~Sn ~ngon Sn Since g dependson r only.57)
8.~) = 1. with the point source defined in spherical coordinatesas: = 4~ ~ Toascertain the rate of decayof g(r) with r.
19. one can n.el--ylei~ I'=E Takingthe limit of (8. the secondintegral can be shown vanish as e --> 0.84) over a small circular area e. For outgoing wavesin R let C = 0.81) as e -~ 0.86)
~)
(8. the two dimensionalanalog can be written as: -V2g.80) can also be shown vanish in the limit as e -~ to
Is. to Finally the Green's function can be written as: g = ¼H(ol)(kr) which. This results in the evaluation of C1 = ~--~.83)
8. dimensionalGreen's function.84) is given by: g=C1 oH(~l)'kr C2H(02)(kr) I. Integrating (8. ) (8. when source location is transferred from the origin to ~. the integral approaches-4~C The ~condintegral 1.r<"1= :xl--Is: lJ" 4'
1 whichvanishes as e --) 0.=----rdrld[ -~1-k2g ~'g r = ~i(x) (8.81) dS=4~e2C. respectively. in (8.. gives: the g(x . 2 evaluatethe first integral as: J"R ~72gdx:j"S ~ng dS:Cl kH(°lf(kF-")'2"£=-2ll.2 Two dimensional space
Following the same procedure for the development the Green's function in three of dimensionalspace.82)
(8. the homogeneous e solution of (8.l~Cl kI~l)(kl0
Takingthe limit as ~ --> 0. so that the Green's function becomes: eikr g = 4-"~" The Green's function for a general source location ~: (8.:: 4'<r:'drl.84)
For the solution outside a small circular area R whoseradius is e.85) ) whereH(~ and H(02)are the Hankelfunctions of the first and secondkind.(~-e~. the integral approaches4iC In a similar mannerto the three 1.87)
.~) = ~ H(01)(klx(8.CHAPTER
8
4 78
Thefirst integral can be transformed a surface integral over the inf'mitesJimalsphere: to g ~Re V2gdx=~s e ~n (8.
89) P>1 For twodimensional media.3----~ R (8. This requires that the function u and its derivative behaveas: iku(R)_ 0u(R) p ~R R p > 1/2 (8.4
Behavior for Large R
The behavior of u(x) for the Helmholtz operator as r --) . .ld-tl 1) (kr). and On R (ikR .19.91)
2 can see that this operator is related to the Helmholtz One operator by makingX= -Ix or g = -ik = -i~-. and~-~ _= .gLu) dx = ISn (R _~ oo)(U-~ng.88) g(xlg)2--~=
8. integral in (8. so that the surface integral in (8.g72+ ~t2
Thereis anotheroperator that is related to the Helrnholtzoperator.90)
8. For three dimensional space.GREEN'S 8. This requires that the function u and its derivative behave as: 3u(R) -~ p iku(R).1).~ can be postulated from (8. Thus: IRn (uLg .3
FUNCTIONS space
479
One dimensional
The Green's function for the Helmholtzoperator was workedout in Example8.70) becomes: u(R) Lim IikR 2 --~ ook R 3u(R) lleikRR2___~0 OR R
R
This is known the Sommeffeld as Radiation Condition for three dimensional space. The substitution of ~t in the final results for the Green's function for the Helmholtzoperator:
.20 Fundamental Solution for the Operator.g-~) dS (8.70) becomes: Lim [-kH 1 (kR)u(R)-H(01)(kR) 2~R-~0 that the surface
This is known the Sommerfeld as Radiation Condition for two dimensional space.19. an aR ikR e ikR 3g = 2 e g _= ---if-.= --.39)
where fi is the unit outward normal and -. definedas: L u(x)= (-V2 + g2)u(x)= f(x) x in Rn (8.39). The integration over the surface of an infinitely large sphere of radius R must vanish.g = H(01)(kr).8 as: i eiklx-~{ (8.
123>
.t) H[.119)
which maybe inverted by Laplace transform to give: g(x.120) 2c Note that the Green's function is constant. but has two sharp wavefronts at x = _ct. then one can transformthe source location to the origin and write out the operator in cylindrical coordinatesin the radial distance: (_V4 k4)g(r) = 8(r_==~) + 2~r (8.56) as.~x[/c] n(t) t (8. one obtains: g(x.22.~.~[/c ] H(t.x.CHAPTER 8
~484
J
Note that the Green'sfunction has a trail that decays with ct at any fixed position r with a sharp wavefrontat ct = r.3
One dimensional
space
For the one dimensional medium: ~(x.Ix ~.t)= ~ f g(x.t) can be written from(8. Upon transferring to the location of the source. 122)
Since the operator has constant coefficients. then the Fundamental solution satisfies the following equation: (-~ 74 + k4)g(x[~) = 8(x~) x n (8.23
Fundamental Operator
Solutions
for
the
Bi-Laplacian
Helmhoitz
The fundamentalsolution for the Bi-Laplacian Helmholtzoperator applies to the vibration of elastic plates. then the solution u(x.p) e-plxl/e = 2pc (8.t.%)f(~. Since the plate is a two dimensionalmedium. The Green'sfunction can be transferred to the location of the source to give the Green's function as: Htc{t-x)-Ix-~] g(x .118)
8." u(x.x) = 2r~C[Ca(t_x)z _[x ] 1/a H(t-x) (8.%) (8.~'t.%)=H[t.~.12I) 2c Since the waveoperator is self-adjoint.%)d~d% 0 n R
8.
one needs to extend the integration on p to (. The two simple poles that fall in the upper-half plane are k* and ik*.125)
Since H(ol)(x) behavesas eiX/~ for x >> 1.H(ol)(ikr)] (8.24
Green's Function Bounded Media
for
the
Laplacian
Operator
for
In this section.124)
(8.39).
p4 _ k 4 dp
(8.126)
The I-lankel function. The final solution for g(r) becomes sumof two residues after letting k* --~ the g(r) = . so that the final expression for the Fundamental solution is written as: g(r)=8k-~ [iH(01) (kr) . This is accomplishedthrough the surface integrals that were developedwhenthe
. wherethe Hankeltransform of g(r) is g(p): 1 resulting in the solution:
The inverse Hankel transform of ~(p) can be shownto be: -1 o~ J0(rp) g(r) = ~ j" p 4 ¯ 0 p4_ k In order to performthe integration in the complex plane. the Green'sfunction is developedfor bounded mediafor the Laplacian operator.~o).GREEN'S
FUNCTIONS
485
To obtain the solution for g(r). one can substitute 1) 2) by H(o and H(o as: + g(r): _~1 ~ P[H(01)(rP)_ I-I(0~)(rP)] 4 _k 4~ J p4 0 Since H(0:Z)(rp)= -H(01)(-rp).38) and (3..19. then one can close the contour in the upper half-plane. see Section 7. such that k* = k(1 + irl). Usingthe principle of limiting absorption then the four simple poles wouldrotate counterclockwise by an angle equal to the infinitesimal damping coefficient rl.125) can be extendedto _oo g(r) =-~n . one can apply the Hankeltransform on r. two real and two imaginary.of an imaginaryargumentca~ be replaced by -2iKo(kr)/~.n2-.8-~ [H(01)(kr) . K0 (kr)]
8. Usingthe identities (3. The integrand has four simple poles. the integral in (8.
38) andthe differential equation (8.Onlyoneboundary condition beprescribed every can at pointof ihe surface a unique for solution.38) in from(8.Toadjust the surface integrals that onlyone so boundary condition ne~ds bespecifi¢xl each to at pointof the surface. Those requirements would over-specify boundary the conditions.
.start with the Green's identity in eq. au×iliary an function is defined ihat: ~ such -VB~(xl~) : 0 x in R (8. Forthe Laplacian operator.58). x to vice giving: u(x)= IRg(xl~)f(~ ) d~+IS~ [g(x. second a surface The is integralthat rextuires specification the of the function and normal u(x) the derivative ~u(x)l~n every at pointonthe surface. one obtainsthe following: 3u IR[-g(x~)f(x) + u~i(x-~)] dx = Is[g~n. one change independent the variable ~ and versa.~) ~-~ ) ~n~ dS[ (8.130)and (8.128)results in a newidentity: )
a ( )~n~ ]S~ dS~ xl~ ~(~)
(8. 128)
Thissolutionis composedtwointegrals.u which. Let v=g(xl~) (8.131)
D~pending prescribed onthe boundary condition.129)
Substitutingv = ~(xl~ in (8. one obtainsa new x identity on the auxiliary funcfion:
DefiningG(xl~ = g" ~ andsubtracfing(8. eliminateofthe one can one two surface integrals.CHAPTER 8
486
Green's identifies were derivedearlier.The of first is a volume integral overthe volume source distribution.uponrearrangement gives:
f(x)dx IRg(xl+ is
jdSx
Since Laplacian self-adjoint then can the isa operator. (8.38) oneobtains: )
Again switching to { andvice versa.58).
3
Robin boundary condition
For impedance-typeRobin boundarycondition expressed as: x on S 3n ~" ~(x)= h(x) substituting (8.134)
.125) by letting: 3G(xl~) I = 0 ~ on S~ 3n~ St or.131) and rearranging the terms
u(x)
(8.131) results in the final form of the solution: u(x)= fR
G(xl~)f(~)d~.24. due to the symmetry the Green's function: of G(xI~)ISx =0 x on x
The function G must satisfy either of these conditions.132)
8.134) in (8. Substituting this condition on G into (8.131) by requiring thai= G(x[~)Is~ : 0 ~ on
or.24.133)
8.e.1
FUNCTIONS boundary condition
487
Dirichlet
If the function is prescribed on the boundary: u(x) = h(x) x on S then one needs to drop the secondintegral in (8. the function G must satisfy either of these two conditions.2
Neumann boundary condition
If the normal derivative is specified on the surface.: 3u(x) = h(x) x on S 3n then one needsto eliminate the first integral of (8.131) results in the final solution expressedas: u(x): ~R G(x[~)f(~)d~+ ~S~ G(xI~tS~ (8. i. Substituting this conditioninto (8.GREEN'S 8.24. due to the symmetry the Green's function: of ~G(xl~) x : 0 x onSx ~n x Again.h(~) ~S~
~n-"~ dS[
(8.
Thus. G = 0 is also satisfied on Sx at r = a.141) to guide the choiceof the auxiliary function ~. where C = 1.CHAPTER
8
494
Fig. then the constant C must be set to one. let: ~ = . and hence r 1 = r 2 = r 0. i. see (8.142)
Since 0/0n = O/Op C.e. results in the expression: 1 a2 2 r 2 + r 2 . then ~ = a.~) C 2rat a Thefinal solution for u(r.see Figure 8.2ar cos(0 . Note that on t.a'rf ) (8.5. p= a. then G (at p = a or r = a) = 0.144) and evaluating the gradient at the surface on p = ~ = a. differentiating (8.141)
The choice of the auxiliary function with a constant multiplier p/a is dictated by the equality given above. then V2~= 0 for r > a.~. 8.0) can be expressedby area and contour integrals as. Withthe definition G = g . let us use the equality in (8. 0) coordinates.132):
.o:r()j 4~ ~.5 Geometry the interior circular region for
For the interior problem. It should be noted that since the factor p/a is constant in (r. the function G becomes: G=~{l°g(~r2) 2-1°g(e)l='~l l°g.C-~logf-P r2t -2 (8. Similarly.
Here again.2 Exterior Problem
Development the Green's function for the exterior spherical problemclosely of follows that of the circular region.151).29. 7 Geometry for the interior
spherical
region
the auxiliary function ~ cannot be found in a closed form. 8. one needs to split the Green's function G = G or G 12 whereG is obtained for the point source for the volumesource distribution and G for I 2 the non-homogeneous Neumann boundary condition as was done in section 8. so the normall~adient of is needed. one must follow the analysis of the exterior cylindrical problemby letting G = G or G as was done in section 8. (a) Dirichlet boundary_ condition Here let the Green's function be the sameas in (8. as was the case for the cylindrical problem.27.27. The normal gradient then is 0G/0n= -0G/09. 1 2
. (b) Neumann boundary_ condition For Neumann boundarycondition. 8.CHAPTER
8
502
\
\
\
\
/
/
/
/
Fig.
Substituting for u from (8.77) and v = g from (8.
8.8 Geometry for the exterior
spherical
region
8. let the auxiliary function ~( the satisfy: -V2~(xI~).155)
Letting the Green's function G for the boundedmediabe defined as G = g .¢)
0
Fig. eqs.12. then the final solution for the non-homogeneous problemis the sameas the Laplacian's.128): u(x) = ~R g(x[~)f(~)d~ + ~S~ g(xl~) 8u(~) -
8g(xl~) ]
0n~ JS~
(8.X~(x[~)= x in R
(8.78) into the equality in (8.~.132-8.31
Green's Function for the Helmholtz Operator for HalfSpace
Refer to the geometryof three or two dimensionalhalf-spaces in section 8. Z > 0.e.
.153). 8.39) results in the sameexpression given (8.128)
Following analysis undertakenfor the Laplacian.30
Green's Function for the Helmholtz Operator for Bounded Media
Consider the Helmholtzoperator in section 8.GREEN'S
FUNCTIONS
503
P(r. (8.26. For two dimensionalspace. such that
~:~ < X < o~. delete the coordinate y from three dimensionalsystem.
one can use the results (8. one can showthat:
wherethe signumfunction sgn(x) = +1 for x > 0. integration for the second the of (8.rl. Thequarterspace is defined in the region 0 < x.171) requires that separate integrals must be performed x > { and x < for The final solution for
G(xl~) becomes:
i ~ik+~ ik(x+~) +eiklx-~l~ Note that if 7 = 0.~].y.169) for the final solution for w(x) with the source te~ given above:
Integrating the aboveexpression.~). one recovers the Dirichlet boundarycondition solution in (8.172)
The integration in (8.171).9) G = -e ~x ~ w(~) -~n d~
X
(8.GREEN'S then:
FUNCTIONS
507
w(0) =
Substituting w(x) into the Helmholtzequation: 2 d 2 "
Withw(x) satisfying the Difichlet boundarycondition w(0) = 0.172) is straightforward. another image of Q about the y-z plane at Q2(-~.rl. z < . There is an imageof Q at QI(~. Note that w(0) Returningto the first order ordinawdifferential equation on the function G wi~ w(x) being the non-homogenuity: dG -~+~ =~(x) dx then the solution for G in terms of w(x)is given in (1.9. Let the field point be P(x. and = -1 for x < 0.
8. -~ < y < ~.~).~. However. one recovers the Neumann boundarycondition solution in (8.32
Green's Function for a HelmholtzOperator in QuarterSpace
Considerthe field in a three dimensionalquarter-space.z) and the source point be Q(~.170) and if 7 ~ ~. see Figure 8.-~) about the x-y plane. There is an imageof Q1about the
.
x)e -~u du
X
Integrating by parts the secondbracketedquantity in w. one obtains: G = -e~x f w(u. matchingthat given in Example (8.15):
w(x. This methodwas the developed by Melnikov.~) ~/[4~(t-'l:)]] e-(X-~)
Integrating the equation for G.2'/e
[! e-~ue-(u+
{)~/[4K(t-x)]du H( 1
The last expressionin the integral form can be shown result in: to -~ H(t .x) {y[e-(x+~)~/[4~(t-'r)] ~/4~tK(t .~)et[x+{+~/(t-x)l erfc '~-(t Note that if. then all
. one retrieves the Green's function for Neumann boundarycondition.tl~./= 0.t I 0.~) I
Yx .t ~.35
Methodof Summation Series Solutions in Twoof Dimensional Media
The Green's function can also be obtained for two dimensionalmediafor Poisson's and Helmholtz eqs.x)h(x)dz j 0
8.r)] z/ [4~(t-z)]] _ e-(X+~)
G(x.t I ~.~)I0~-(n= ~)-'/~(n-~)1~(~00 H(t .t[ n.x) f f G(x. w(0.t[~.Since the fundamentalGreen's function is logarithmic.GREEN'S
FUNCTIONS
519
together with the boundarycondition on w. then one obtains the Green'sfunction for Dirichlet boundary condition. results in the following expression for G: H(t-'c) [e-(X-~)2/[4~(t-. This showsthat the function w satisfies the Dirichlet boundarywith the aboveprescribed source term. If the limit is taken as '/--) oo.x) = 0. The final solution is given by: t u(x. in closed form by summing series solutions.15). The Green's function for a Dirichlet boundarycondition is given in Example (8.t) K:y Uo-G(x.
Find the Fundamental Green's function for a vibrating stretched membrane resting on an elastic foundation.20
532
25.
. ~74g k4g.~) 26. by use of Hankel on transform:
(. by use of Hankel transform. = ~(x+ . Find the Fundamental Green's function in two dimensionalspace for a vibrating plate supported on an elastic foundation under harmonicloading. Find the Fundamental Green's function in two dimensionalspace for a stretched membrane use of Hankel transform: by -V2g = ~(x . Find the Fundamental Green's function in two dimensionalspace for an elastic plate by use of Hankeltransform: -V4g= ~(x-~) 30. whosespring constant is ~.CHAPTER Sections
8 8. = ~(x~g ~)
(a) by Hankel or (b) by construction
3 I.17 u 8.V4g. Find the Fundamental Green's function in two dimensional space for a plate on elastic foundation(~ being the elastic spring constant) such that:
.such that: -V2g+(7-K2)g = ~i(x-
2 7>z (a)
(b)2 7<~
29. Find the FundamentalGreen's function for a vibrating membrane two dimensional in space by use of Hankeltransform: 2(-V k2)g ~i(x= 28.~g ~) for (a) k>7 Co) k<7
where representsspring y4 the constant unit per area k4 representsfrequency and the parameter.v2 + ~)g= ~(x. Find the Fundamental Green's function in two dimensionalspace for a stretched membrane an elastic foundation. such that: .
27.
of Example 9.u---~ xn-I e-axdx
U
= U e_aU_n ~xn_ e_aX dx 1 a a
U
Repeated integrationof the integral above results in:
n un_ k n!
I(a)=e-aU
~ T (n-k)! k=O
537
. the followingsections.1 Introduction
In this chapter on asymptotic methods. In are 9.The generalformof the integrals involves integrandthat is a real or complex an function multipliedby an exponential. then it is possibleto get an asymptotic valueof the integral by oneof a fewmethods.9
ASYMPTOTIC METHODS
9. repeated is made integrationsby part to create a series with use of descending powers a larger parameter.If the exponential function has an argument can that become large.1 Consider integral I(a): the I(a) = ~ n e-ax dx
u
integration parts resulksin: by I(a)a e .the emphasis placedon asymptotic is evaluation integrals andasymptotic of solution of ordinary differential equations. a fewof these methods outlined.2 Method of Integration by Parts
In this method.
Integrating each term in (9.1) results in an asymptotic series for f(p): oo F(n)(O ) f(f~): Z pn+l n=O (9.1)
0 Expanding F(t) in a Taylor series about t = 0..5 t 22 s3 I(s) 2 s 2s 23 s4 ~.: '~ F(n)(0) n F(t)= Z n=0 whereF(n) is the nth derivative.5t3+.e.3 Laplace's Integral
Integrals of the Laplace's type can be evaluated asymptotically by use of Taylor seres expansionabout the origin and integrating the resulting series term by term.2 Consider the following integral. which is known have a closed form: to I(s' = ~ el~t dt : ~ eS erfc('f~ ) 0 The term (l+t) -1/2 can be expanded a Taylor series: in (l+t)-i/2=l_t+ 1-3 2 1.CHAPTER
9
538
9. uponintegration via (9. i.3) results in: 1 1 1."'" Equatingthis expression to the erfc(~f~) one obtains an asymptoticseries for the erfc(z): ~r~ ~s 2s "-~'÷ 22s 3 23s 4 +""
1.3
.2)
where the Watson's Lemma used: was
~t
0 and where F(x) is the Gamma function.3. see AppendixB Example 9..3)
which. Let the integral be given by: f(p) = -pt F(t) dt (9. F(t) can be written as a sum.3 3. 2 ~ 3! 23
v e-pt dt = r(v+l) pv+l
(9.
4)
. i. and f(z) is analytic at 0.4
Steepest
Descent
Method
Consideran integral of the form: I c = ~ e0f(z) F(z) C whereC is a path of integration in the complex plane. i. i. z0 = x0 + iy0. Hence. To choose a path through the saddle point z0. called the Steepest Descent Path (SDP).. then u and v are harmonic functions. It is desired to find an asymptoticvalue of this integral for large 9. Onthe other hand.y) cannot have points of absolute maxima minimain the or entire z-plane. Letting the analytic function f(z) be definedas: f(z) = u(x.ASYMPTOTIC
METHODS 3-5 ~ 24 z8 ÷. (9. This would maximizethe real part of the exponential function. see Figure 9. one obviously must choosepaths whereu(x.e.y) at increases along C'.y) has maximum value at somepoint zo.
539
1 1.e. i.y) has a relative maximum o.1.y) is maximized. u(x.e. z = x + iy. i .y) (9. To locate the point z0 whereu(x.6) Since f(z) is an analytic function. This wouldresult in an integral that will diverge along that path. This wouldresult in an integral that wouldconverge. if one chooses a path starting from zo whereu(x.5) then the path of integration is chosensuch that the real part of f(z) = u(x.3 z _z 2 ~ 1 erfc(z) ~-~-~ e l~---~-z4 . especially when19 >> 1. The topographynear z0 for u(x. This indicates that: (9.-~-V = 0 (9. a path of descent fro m thepoint z 0. and a path through the point. s o t hat u(x.stay at the samelevel. The Steepest Descent Method (SDM)involves finding point. Since Ou/Ox 0 and ~u/Oy 0 at the SPz0. V2u = which indicates that u(x. then the partial = = derivatives 3v/Ox= 0 and Ov/3y= 0 due to the Cauchy-Riemann conditions. or ascend. then the exponential function increases exponentially awayfrom the saddle point at z0.23 z 6
9.. f(z) and F(z) analytic functions and 9 is a real constant. so that the integrand decays exponentially along that path and the integral can be approximatelyevaluated for a large argument0. Thiswould ean hat m t the exponential function has a maximum value at z0 and decays exponentially awayfrom the SP z0.y) = constant wouldbe a surface that resemblesa saddle. called the Saddle Point (SP).e.: ~u = 0 .the points whereeq.the extremum point(s) are found by finding the point(s) wherethe partial derivatives respect to x and y vanish.y) has a relative minimum z0.y)+i v(x.y) d ecreases o at the path(s) awayfrom 0.6) is satisfied are stationary points. paths originating fromz0 either descend.e.
This wouldlead to a convergent integral along C' as 0 becomes very large.~gx cos 0 . Thus. This meansthat one must find the path C' so that the function u(x. ost (9.= ~s 3x~s
8u Ou~x+ 0u
3y~s
0y 0u
--sin ~y =~xxC°S0+
0u .e. In order to improvethe convergence the integral. the absolute value of the rate of changeof u(x. a path of descent from z 0 so that the real part of the exponential function decreases along C'. Since f(z) is an analytic function at o. one needsto find the of steepest of all the descent paths C'.y) decreasesmost rapidly and this path is to be called Steepest Descent Path (SDP).e. (9.y) changes most rapidly on path C" defined by v = constant. 0 _--~y ~s
~u
~y
~v
~x
(9. (9. especially with a large argument0. Let the angle 0 be the angle between the tangent to the path C' at zo and the x-axis. the function u(x.7)
Theroots of eq. Onemust choosea path C' originating from the SP. 10u/0sl is maximum along C'.~X-X sin =0 ~0~. To find such a path..z0)3 + .i.Y = v o) o (9. z0.CHAPTER
9
540
df[I = e'(z0) =
(9. Thus.7) are thus the saddle points of f(z).8) with respect to the distance along C'.e.9)
Eq. s. Since the path must pass throughthe SP at zo.z0) + a2(z.y) increases or decreases mostrapidly. results in v = constant along C'. (9.9) defines path(s) C' from o having t he m r apid change in t he slope. giving: f(z) = 0 +al(z . then the slope along the path C' is given by: -. i. defined by a distance parameter"s" whereu decreases at the fastest rate. eq.z02 + a3(z.y) = V(xo.: .9) defines a path(s) whereu(x. i.
0
To find the orientation 0 where ~u / ~s is maximized. where:
..y) decreases at a maximum as z traverses along the path C' awayfrom rate z0. ~s) = 0+~cos0 Using the Cauchy-Riemann conditions:
~u ~v
~x ~y then the equation above becomes: 3v 3v 3v ---sin 0 . It is imperative that one finds the path(s) wherethe function u(x.then one obtains the extremum of the slope as a function of the local orientation angle 0 of C" with x. then the equation of the patlh is defined by: v(x. then one can expand t e flmction f(z) i n a h Taylor series about z0.8)
Integrating eq. To identify whichof the paths are SDP.y) along the path "s" must maximized..it is sufficient to examinethe topography near z0.
This means that the first termin eq. Thus. especially when13 is very large. these to approximationsassumethat the major contribution to the integral comesfrom the section of the path near the saddle point. using Watson'sLemma (9.Ic~ + 2hi [sum of residues of the poles between C+ C~+C~] (9. for z near z0.14)
Thus. It should be noted that the slope dw/dzhas a different value on C~and C[.12)
Note that Ic~ and Ic~ have different series basedon the path taken. (9.f(z) =-(z.
9. C~. so that the sign assigned for C~ is negative. and C~.15) (z-z0) = f(m+l)(zo ) where the complexconstant c is given by: (m+l)! C= f(m+l)(z0) Note that the (re+l) roots havedifferent values along the different paths C~n. Thus.10).3). (9. Substituting eq. (9. one can neglect higher order terms in ~(w) and (dw/dz) in such a way a closed form expression can be obtained for the first order term.10) becomes: in Ic[2 ~ ePf(z°) Z ~n F(n + v + 1) 9n+v+l n=0 (9. resulting in: I C = Ic~ . an approximate value for w can be obtained by neglecting higher order terms in w: w= f(z0). I0). then two paths C[2 must be joined to C. which maybe clockwise for somepoles and counterclockwise for other poles. Toobtain the first order approximation.12) wouldsuffice if 13 is sufficiently large. if C is an infinite path.5
Debye's
First
Order
Approximation
Thereare first order approximations the integrals in eq. Differentiating this approximation z with respect to w results in: for dz dw -m/(rn+l) cl/(m+l)w m +1
. resulting from the derivative dw/dz. the integral in (9. Principally.11) into eq. (9.ASYMPTOTIC
METHODS
543
wherev is a non-integer constant. integrating the resulting series term by term. The paths C~and C~start from w = 0 and end in w = ~ along each path.13) The sign for the residues dependson the sense of the path(s) of closure betweenC. and one must close it with an infinite path. the conformaltransformation betweenw and z can be obtained explicitly in a closed form by the approximation: 1/(m+l) (m+ 1)! 1/(re+l) = Jew] 1/(re+l) w (9.m+l f(m+l)(zo) (m+ 1)! (9.
[ } (9. [ _ c(m+l)_.17)
(m +
1)
p(m+1)-'
[.Ic~
I"((m + 1)-I)F(zo)ePf(zo)Ic(m+l)_.
ionCi
IonC~
wherethe residues of the poles were neglected.17) represents the leading term the approximationof the asymptotic series.16)
The first order approximation the integrals in (9.t dt 0 When is an integer n. The Gamma function is given as an integral: F(k+ 1) = k e.: I c = F(z o)
pf(z°) ~p.18)
Example 9. Note for m= I. Ic.1 = 0. the two roots of c are opposite in signs and hence the expression in the bracket is simply double the first term in the bracket.18) gives: f(zo) = f(1) = -1. Let t = kz.so that the sadd point is le located at Zo=+l.e. To obtain a Debye'sapproximation for the k asymptotic value for a large k. F(z) = 1 and:
0
f(z)=log(z)Thesaddle point zo is derivedfrom f'(Zo) = 1 . Furthermore. (9.c~ become:
el/(re+l)
OO
IC. F(n+l) = n!. the integrand must be slowly varying.known as Sterling's Formula. Eq.4) is thus given by: to I c .C.~
.. i./2n/( -f
(Zo))
(m = 1)
(9. the function F(z) can be approximated its value at z0: by
544
F(z)F(zO -Thus.
c~/(m+l)r(1/(m+ 1)) m+ 1
pl/(m+l)
(9.. f"(Zo) = -1 in
. the integrals Ic. then:
F(k+l)=k k+l e-kZzkdz=k k+l ek(logz-z)
0 For the last integral.3 Obtain the Debye'sapproximationfor the factorial of a large number. Evaluating the function in the expression (9.CHAPTER 9 Similarly. the exponential for term does not have the parameterk in the exponent. This is not the case here as the function t k becomesunbounded k large.'C~
F(z0) m+ 1
ePf(zo)
-m/(m+l) dw f e-pw W 0
e pf(z°
Ic~ . .
Letting t = ~ + i~l. Path "2" extends from z = 1 to 0 and path "1" extends from 1 to oo.~exp[iz3/2(t3/3 + t) ] dt Letting x = z3/2 one can write out the integral as:
_ 1/3 oo
Ai(x2/3) = -~-~ ~ exp[ix(t3/3 + t)] Onecan evaluate the first order approximation large x.i (t3/3 + The saddle points are given by f'(t o) = i (to2 + 1) = 0 resulting in two saddle points. In this integral F(t) = 1 and for f(t) -. the pa~ts 0SD = 0.so that there is no needto deform the original path into the SDP's. It turns out that the original path on the positive real axis represents the two SDP's..CHAPTER
9
546
0 sin 0 The four paths are shownin Figure 9.+l) :
.~10)= v0(0. t o = +i. It shouldbe notedthat since g'(to) ~ 0. To mergethe first exponential with the second.18): F(k+ 1) = k+l e-k~/-~ = IK Example 9. The leading term of the asymptoticseries for the Gamma function can be written as (9.2. let s = ~ t: Ai(z) "~. m= 1 for both saddle points.rl) = Im f(t) = ~3/3. then the SDP path equations for both saddle poinls are given by: v(~. To map the SDP: f(+i) -2 3
f"(+i) =-T-2 Hereb = 2 and 0 = ~ for to = +i and 0 = 0 for t o --.4 Find the first order approximation Airy's function defined as: for Ai(z>: cos(s3/a+sz)ds
0
e-k kk+I/2
= ~ exp i(s3/a+sz
ds
To obtain an asymptotic approximation large z. of P x are paths "1" and "2". the first exponential terms is also not for a slowly varying function. so that paths "3" and "4" are the steepest ascent paths.~1 + ~ : v0(~0.-i. It can be seen that in the neighborhood the saddle point zo = 1.
f(zo) =. Letting f(z) = u + then: eipf(z) : e-PV eipu If F(z) is a slowly varying function. Considerthe integral: I(p) = ~ F(z) ipf(z) dz (9. the frequencyof the circular functions increases. This is the same path defined for the Steepest Descent Path.1/2 f" (z0) (z-z0)2then the integral becomes: f F(z(w)) w I (p) = eipf(z*) ~.7 Method of Stationary Phase
The Stationary Phase method is analogous to the Steepest Descent method. This then tends to cancel out the integral of F(z) +1 whenp becomes very large for sufficiently large z. Yo) is called the Stationary Phase Point (SPP).27)
C wheref(z) is an analytic function and F(z) is a slowly varying function. ~/ 0)
and defining: w = f(z). For an equivalent Debye's first order approximation m= 1. Thus.28) becomes:
. so much that these circular functions so oscillate rapidly between and -1. The major contribution to the integral then occurs whenf(z) has a minimum that the exponential function oscillates the least. Performingthe integration in the complexplane results in the two methodshaving identical outcomes. (dw / dz) (9. Expanding function the f(z) about the SPPz0:
f(z): f(Zo) 2f"(z (z-z0)2 +. the exponential term oscillates in increasing frequency. so This occurs when: f" (Zo) = where z 0 (x O.CHAPTER
9
552
9. Since the exponential can be written in terms of circular functions.28)
whereC" is the Stationary Phase path defined by v = constant = vo and vo = V(xo. where the exponential oscillates the least. although the approach and reasoning for the approximationis different. then as p increases.zo) = ~ F(zo) = F(z(w=0)) then the integral in (9.Y0).~ f"(Zo) (z o
)2
dw / dz ----f"(Zo)(Z . becomes larger. then most of the contribution to the integral comes from near the SPPz0.. let: for w = .
z1.10).24) becomes: I = e °f(zo) G(y)
~
~ _py2 /2. g(y) has an infinite radius
. Letting the function ~ = G(y) dy/dz (9.9 Modified Saddle Point MethodlSubtraction of a Simple Pole
Theexpansionof a furiction by a Taylor series about a point has a radius of convergence equal to the distance between that point and the closest singularity in the complex plane.11) or (9. Onemethodwouldsubtract the pole of the singular function F(__z) and expandthe remainder of the function in a Taylor series.CHAPTER 9
554
f y2m+2ve-PbzY2/2dy m=0n=0 ~= =elgf(us'vs) 2b'~l~2n~0m -oo =0 Fnm
r(2n+ 2v+1)F(2m 2~ + + pn+m+v+~+l 2 F(n+ v)r(m+ bnbml
(9. the series expansiongiven in (9.~. few methodswere devised to account for the singularity in the function F(z) and hence extend the region of applicability of the asymptoticseries. This is generally true for the transformationsof the type given in eq.12) and (9. Thus.16) cannot be carried out to ±o~. whoseradius of convergenceextends from zero to the closest singularity to y = 0 farther than that at y = b.f(z0) .32)
9. hence.21) about the saddle point wouldnot be valid for an infinite extent. TheLaurent's series for G(y) can to then be written as: G (y) :'~ + g(y) wherethe location of the pole at z = zI or y = b is given by: b :~r~ . so a Taylor series expansionis possible. Thefunction g(y) is analytic at y = 0 and at y -.b)G(y)
y-->b a
(9. Of course.34)
is the residue of G(y)at y = b.25).
uy
(9. (9.10) and primarily due to the factor (dw/dz). then the integral in (9. t hen the function G have asi mple (y) pole at y = b corresponding the simple pole at z -.f(zl ) and a = Lim(y. so that the integrations in (9. if no other singularity exists. The closer singularity comesto the saddle point. Toalleviate this problem. the shorter the radius of convergence and. the range of validity has nowbeen improvedby extending the radius of convergenceto the next and farther singularity. Thus. (9.b.33)
Let the function F(z) havea simple pole at z = 1. the larger value of p for whichthe asymptoticseries can be evaluated.
44) dependson the large parameterp only.
9.52) dependsfurther on the location of the pole with respect to the saddle point.52) has a complementary error function just as that given (9..52) dependson the pole location given by "b".. (9.CHAPTER 9
558
I~ePf(Zo)
~ yh~_°be-PY2/2dy+ePf(z0) ~ hn+l ~(y-b)ne-PY2/2dy --oo n=0
The first integral wasdeveloped earlier in eq. However.53)
whereg(y) is an analytic function at y = b. The integrals in the series can integrated term by term.~" k
oo E(n/2) bn_2 k ~ (-1)n(n[)hn+l k~ (n-2k)!k!2 n=0 =0 "
(9. The final form of the asymptotic series becomes:
of(z°).44).52)
whereP = 2~i h(b) and the symbolE(n/2) denotes the largest even integer less than The expression in eq. This is not usually desirable.. because the radius of convergence the series of in (9. then one can expand function G(y)in a Laurent's series as follows: G(y)= (y_b)N~ (y_b)N_l+ .. I ~e + 2.38) and (9.f~)
A_ = ~ A_ = +ir~e-Pb /2 -pb erfc Tib 2 1 + erfc -TAb
Since: 2 7 erfc (x)= -~ J
x -y~
dy
then:
. (9. (9.54)
Recalling the expressions in eqs.33) has a pole of order N.43).47) one obtains:
~o e_py~/2
A-1 : ~ ~ dy = +ir~e -pb~/2erfc (y-b) then:
(~ib p. a-N a-N+l + ~ + g(y) (9..10
Modified Saddle Order N
Point
Method:
Subtraction
of Pole
of
If the function G(y)in eq. asymptotic series in (9. (9.b" A_k(p.3 . the while the series in (9. Define: o~ e_py2/2 1 d A ' . (9.b)= ~ ~y-~)~ dy=~--~"~ -k+l[P ) k=2.
.56)
or ~ an~ y(x)= ~ x-n n=0 n=0 whichis a descendingpowerseries valid for large x. then the asymptoticvalue of erfc (x) gives: A_ --~ 2
2~1b2
whichis of the sameorder as A_ given in eq. Bothof these series solutions convergefast if the series is evaluated near the expansion point. can be computed a similar procedure.11
Solution of Ordinary Differential Arguments
Equations for Large
In chapter 2.45). (9.4) transformsto: d2y~ [2~-al(1/~)]dy . one can transformthe independent variable x to ~.. the differential equation (2.ASYMPTOTIC
METHODS
559
A_ : -2x/~ -Y-i~bpe -0b~ '2erfc (-Y-ib. the solution of ordinary differential equations for small arguments was presented by use of ascendingpowerseries: the Taylor series for an expansionabout a regular point or the Frobeniusseries for an expansionabout a regular singular point. etc.3. so that x = .one needs to obtain solutions in a descendingpowerseries. 1
9.~'): 2
-2~ -pb 1
(9. Letting ~ = I/x.55)
Likewise. A.
9. x-4 -5 x-6 a2(x) = q_4 + q_sx + q_6 + . To obtain solutions of ordinary differential equations for large arguments.4. a2(1/~) d~ ~" d~2 ~2 Classification of the point ~ = 0 dependson the functions al(x ) and a2(x): (i) ~ = 0 is a Regularpoint if: x-2 x-3 al(x) = 2x-1 + p_2 + p_3 + . To accomplishthis.. A. It should be noted that by if Ibl~>> 1... a transformationof the independentvariable ~ = 1/x is performed the differential equation on and a series solution in ascendingpowerof ~.
n y(~)=
.o mapsinto ~ = 0. The solution for a regular point then becomesa Taylor solution:
OO OO
(9.12
Classification of Points at Infinity
Toclassify points at infinity.
= ~ an n=O (A.whilethe limit of an vanishes Lira 1 --~ 0 Anecessary sufficient conditionfor convergence the series (A-l) is as follows: and of if..n=N for all N> Mandfor all positiveintegersk. M
En
ao
.1)
Theinfinite series in (A. or . infinite series: the oo 1 n=l is divergent..Theseries may diverge +o. for any to arbitrary number there exists a number suchthat: e.APPENDIX A
INFINITE SERIES
A. to A necessary not sufficient condition the convergence the series (A. if. If the series: ~ ~anl n=0 (A.oo or havenolimit.1)is: but for of Lima --~ 0 n For example. M an-a<e for all N> M
n=0 If this conditionis not met. as is the caseof an alternatingseries.1) is said to be convergent a value= a. then the series is said to be divergent. 1 Introduction An infinite series of constants defined is as: a0+al+a2+. for anyarbitrary number there exists a number suchthat: e.2)
585
.
the series is absolutely convergent. and if Ibn[ _< a. then the series (A-l) converges and is said to be an absolutely convergent series. then the series n n=O
Eb n
n=0
also converges. Example A. but:
n=l
n=l
n+l
is divergent. and if Ibnl < a for large n.1 (i) The series:
OO
1
E (-1)n n=l is a convergent series and so is the series:
n=l
1
Thus.1 Comparison Test
If the positive series E an converges.then series (A-l) is knownas a conditionally convergent series. for large n.APPENDIX
A
586
converges.
A. A. (ii) Theseries:
OO
E (-i)n n=0
n+l
is a convergentseries.
. If the series (A-l) converges.but the series (A-2) does not converge.2
Convergence
Tests
This section will discuss several tests for convergence infinite series of numbers. Thereforethe series is conditionally convergent.
If the series ~ an diverges. of Eachtest maybe moresuitable for someseries than others.2. then the series n=0 also diverges.
f0(x) + fl (x) + f2(x)
f<xo)-fo<xo)
n=0 N f(Xo)= Lira ~ fn(Xo) n=O I
n=N
N>M
A necessaryand sufficient condition for convergence the series at a point xo is of that.(x)+ + fN+k(x~: +
n=N+k | ~ fn(X~ < n=N 1
for all N> Mand for all values of the positive integer k. convergesto somevalue f(xo). Lim fn(x) N--~.6)
x x-'X°LN-'~* ]) fh( n =O
Onthe other hand.(x)-~ f(x) n=O N Lira Lim t[
f(Xo)=
Lim f(x)=
X -~).b]. Ln=0 The limiting values for f(xo) as given in (A. then the series is said to converge for to f(Xo) for a _< o _< b.
IfN(x)fr~+. Thus. If the sumof the series. given a small arbitrary number then there exists a number such that: e. at summed a point xo.3 Infinite Series of Functions of One Variable
Aninfinite series of functions of one variable takes the followingform: a-<x<b ~ fn(X) n=0 Theseries can be summed any point x in the interval [a. if the series is convergentto fix). It should be noted that the sum of a series whoseterms are continuous maynot be continuous.6) and (A. Thus.et hen there exists a number such that the remainderof the series RN(X): M.~[ ~ x-~x. i f one chooses a arbitrary s mall n n umber .X o
(A.INFINITE
SERIES
591
A. then: N Lira ~ f.7) are not the sameif is discontinuousat x = xo. M. they are identical only if f(x) is continuousat x = Xo"
. by definition: f(x~) = Lim ~'.
.X) + X(1 .b]. the sumof the infinite series as N --) o~ approaches: f(x) = Lim ~". to test the convergence the series. fn (x) = n-1 (1 -x) Summing first N terms. if for any arbitrary positive number. (1
can be represented by a series of fn(x) given by: n = 1. the remainderof the series RN(X) found of is to be:
which vanishesas N"--) ~. fn (x)
I]
for
Ixl < 1
Thus.b].X)
fn(X) <
for
N> M
x2 . Example A.3.1
A
592
Uniform Convergence
A series is said to convergeuniformlyfor all values of x in [a.. 2.7 The series of functions:
(1 .. For uniform convergence: [xN[ < E
I I
for e fixed and for all N > M
N > ilog(ixD [ If one chooses an e = e1°. only if Ixl < 1. such that: M f(x) n=0 for a/l values of x in the interval [a.there exists a number independentof x. one obtains: the N E fn (x) = 1 n=l
N X
The sedes convergesfor N -> oo iff: Ix[ < l Therefore...APPENDIX A. then one must choose a value N such that:
. 3 .x).
and not uniformlyconvergent in the region 0 _< x _< 1.b]
and since the series of constants: Mn = n=l A. so that the inequality Rn < e cannot be satisfied by one value of N.b] to f(x).3 1 of Uniform Convergence 1 converges
Consequences
Uniform convergence an infinite series of functions implies that: of 1.b] if there exists a convergent positive series of positive real numbersM + M + .b] to f(x). At the point x = x o choose: 10 N=
logOxoO
Asthe point xo approaches1. Thus. If the functions fn(x) are continuousin [a.INFINITE N>
SERIES 10
593
IlogOxl)l
Thus.. the series is uniformlyconvergentin the region 0 _< x _< Xo..b]. such that: 1 2 [fn (x~ _< n for a ll Example A. Ilog Xol --) 0. the series is uniformly convergentfor 0 _< x < xo. converges uniformlyin [a. then the series can be integrated term by term:
.. 0 < xo < 1. and one needsincreasingly larger and larger values of N. then f(x) is a continuous function in [a.2 Weierstrass's Test for Uniform Convergence
Theseries fo(X) + fl(x) + .b] and if the series converges uniformlyin [a...b] and if the series convergesuniformly in [a.3. 2. A.3. If the functions fn(X) are continuousin [a.8 The series: X n2+x 2 1 n=l convergesuniformlyfor _oo < x < oo since: Ifn (x)l = -< n'~ Mn for all x _> 0 x in [a.
APPENDIX x2
A x2 x2 Xl ~ x2 n= 0 x 1
594
x x I 1 wherea _< x1, x2 < b.
3. If the series E fn (x) convergesto f(x) in [a,b] and if each term fn(x) n =0 are continuous,and if the series: ~ f~(x) n=0 is uniformlyconvergentin [a,b], then, the series can be differentiated term by term: f'(x)= 2 f~(x) n=0
fn'(X)
A.4
Power
Series
A powerseries about a point xo, is defined as:
oo
TM ao + a~(x- Xo)M+ a2(x- Xo) + ....
2 an(Xn=0
nM Xo)
(A.8)
whereMis a positive integer. The powerseries is a special form of an infinite series of functions. The series mayconvergein a certain region. A.4.1 Radius of Convergence
For convergence the series (A.8) either the Ratio Test or the Root Test can of employed. The Ratio Test gives: Lim .an+l (x- ~)M. = Ix- xolM Lim an~l < 1 the series an(X- Xo) n'->~l an I converges
n-->**
> 1 the series diverges or if one defines the radius of convergence as: p p = ILim] an [11/M [n-~**lan+lIJ then the convergence the series is decided by the conditions: of Ix - xol < p the series converges > p the series diverges (A.10) (A.9)
INFINITE
SERIES
595
In other words,the series convergesin the region: xo - 13 < x < xo + O and divergesoutside this region. The Root Test gives: "i "~ a n (x- nM Xo) 1/nM Xo[Limlan[ = Ix < 1 the series converges > 1 the series diverges Thus, if one lets: (A.11) then the series converges the region indicated in (A.10). in TheRatio Test or the Root Test fails at the end points, i.e., when ]X-Xo[= 0, where both tests give a limit of unity. In such cases, Raabe'sTest or the Alternating Series Test (if appropriate) can be used on the series after substituting for the end points at x xo +13 or x = xo- 13. Example A.9 Find the regions of convergence the following powerseries: of
for xo - 19 < x < xo + 19. Theradius of convergence the resulting series for f'(x) is the of same that of the series for f(x). Thisholds for all derivativesof the series f(x)(n), as n>l. 3. The series can be integrated term by term such that: x2 ~o x2 oo f f(x)dx= Z an (x-X°)ndx= ~n- -~'~ (x- 0 x )n+l x2
x1 n=O x1 n=O < x < x + 9. The series can be integrated as many for xo - 9 times as needed. o
3. Find the radius of convergenceand the region of convergenceof the following power series: (a) ~'~ (x-1)n n 2 rl=O
n .(x2)
(b) ~a (x + n ~ n=O4n+n
(C)
n=O (e) n=0
n +1
(d)
(n !)2 x2
n=O
(0 E (-1)n
(2n)!
n! n
n (X+ 1)
n=l n oo n3 (X- 3)
(g)
n=l (i)
~n!x._.~_ n n
(h) -O)
n=l
3n (x+ 1) 3n
E (x+l)Sn 8" n=O
~=0
8" +1) (n
APPENDIX B
SPECIAL FUNCTIONS
In this appendix, a compendium the most often used and quoted functions ~e of covered. Some these functions are obtainedas series solutions of some of differential equations some definedby integrals. and are
Example D. behavesas 6(x) in the limit of c---~ c / [n(x Thus.1. one can approximately evaluate the integral by the sifting properties.1 0. 2 + c2)]. This approximatemethodof evaluating integrals whenpart of the integrand behaves like a
.0050 T(approx) 5.82692 0.4 Concentrated Field
Representations
The Dirac delta function is often used to represent concentrated fields such as concentrated forces and monopoles.0
This exampleshowsthat for c = 0. D.16)
Example D.00 c~(exact) 0.one can evaluate it exactly. which is known have an exact value.a2) = ~a [8(x .01 T(exact) 4. a concentrated force (monopole point source) located a x0 of magnitudeP0 can be represented by P0 6 (x-x0).0 1.1 the error is within 10 percent of its exact value.13459 9. This property can be utilized in integrals of distributed fields where one component the integrand of behaves like a Dirac delta function whena parameter in the integrand is taken to some limit.3 The following integral. can be approximately to evaluated for small values of its parameterc: T=--Icos(ax)Jo(bx) f 7~ x2 + C2 dxc 1 =1 e-aCI0(bc) ~ -c c << 1
If the integral can not be evaluated in a closed form and one wouldlike to evaluate this integral for small values of c.0 1.2 0. one notices that the function in example 1.2 8(x2 . For example.07090 99.90709 0. so that for a = b = 1 one obtains: c 0.99005 cT(approx) 1. To check the numerical value of this approximation. Letting: F(x) 1 cos (a x) Jo(bx
c
then the sifting property gives F(0) = 1/c.00 100.000 10..APPENDIX
D N 8(x E If'(Xn) n=l
640
6[fix)]= I
x~)
(D.a) + 6(x +
oo
8[cosx]=
E 8[x-2n+ln] 2
D.
2 Dirac Delta Function of Order One
The Dirac delta function of order one is defined formally by
61(x . Thus.18 .o) x
such that its integral vanishes: I 81(x =0
(D.22)
so that the kth moment integral is: (D.20)
whichgives the value of the derivative of the function f(x) at the point of application ~Jl(X.18)
and its first moment integral is unity: I x61(xx0)dx
=1
(D. These properties outlined in Eqs. using the representation of a Dirac delta function.
D. one can define 51(x) as: 8~(x) = . 81(x) represents a mechanical concentratedcouple or a dipole.20) can be proven by resorting to the integral representation.Lirad u(cqx____.19)
and its sifting propertyis given by: I f(x)~l(X xo)dx = f'(xo) (D.x0).
D.21)
In physical applications.o) = d-~ 8( .~J(x -x (~N(X -Xo) = (-1) O)
(D. (D.DIRAC
DELTA
FUNCTIONS
641
Dirac delta function can be used to overcome difficulties in evaluating integrals in a closed form.23)
I
xk
~iN(X)dx=
N!
{0
k<N k =N
. 7) 1
(D. 3 Dirac Delta Function of Order N
TheseDirac delta functions of order N can be formally defined as:
N N d--~.~) (D.
This equivalence showsthat a distributed couple (dipole) field f(x) is equivalent to a point couple(dipole) of strength f(~) point force (monopole) strength f'(~).25. (D.4 Equivalent Representations
of Distributed
Functions
In many instances.x0) represents high order point mechanical forces and sources. For example.~) dx + f(~) l( x . one can showthat f(~) ~(x. Similarly one can showthat: f(x) 81(x .25)
whichallows one to express a point value of f(~) by a field function f(x) defined over entire real axis.~) dx which proves the equivalency in Eq.26). The proof uses the sifting property of the Dirac delta function and an auxiliary function F(x): f F(x)f(x)8(x .~) + f(~) which again can be proven by using an auxiliary function F(x): ~F(x) f(x) l(x -~)dx = F f'( ~) + F'( ~) f(~ (~) 03.
D.x0)dx = f(N)(x0)
(19. of
.~) = f'(~) 8(x . one can represent a distributed function evaluated at the point of application of a Dirac delta function of any order by a series of functions with equal and lower ordered Dirac functions.26)
= f'(~) J F(x) 8(x .{)dx = F({) f({) = f(~) F(x)8(x
= ~ F(~) f(~)8(xwhichsatisfies the equivalencein D.APPENDIX
D
642
th and the sifting property gives the N derivative of the function at the point of application of 8N(x . For example.x0) is:
f f(x)SN(X.xO) represents a doublet force or a quadrapole.~) = f(x) 8(x(D.24)
In physical applications. ~iN (x . ~2(x .
5 0...
Y1/2(x)
E.25 -0.25
-0.APPENDIX
E
"654
E..5 -0.5
Modified Bessel Function of the First and Second Kind of Order 1/2
Ii/2(x) K~/2(x)
..75 -i
.4
Bessei Function of the First and Second Kind of Order 1/2
0. | 677.169 | 1 |
"This is an intermediate level text, with exercises, whose avowed purpose is to provide the science and engineering graduate student with an appropriate modern mathematical (analysis and algebra) background in a succinct, but nontrivial, manner.... [T]he book is quite thorough and can serve as a text, for self-study, or as a reference." —Mathematical Reviews
Written for graduate and advanced undergraduate students in engineering and science, this classic book focuses primarily on set theory, algebra, and analysis. Useful as a course textbook, for self-study, or as a reference, the work is intended to:
Whereas these objectives for writing this book were certainly pertinent over twenty years ago when the work was first published, they are even more compelling now. Today's graduate students in engineering or science are expected to be more knowledgeable and sophisticated in mathematics than students in the past. Moreover, today's graduate students in engineering or science are expected to be familiar with a great deal of ancillary material (primarily in the computer science area), acquired in courses that did not even exist a couple of decades ago.
The book is divided into three parts: set theory (Chapter 1), algebra (Chapters 2–4), and analysis (Chapters 5–7). The first two chapters deal with the fundamental concepts of sets, functions, relations and equivalence relations, and algebraic structures. Chapters 3 and 4 cover vector spaces and linear transformations, and finite-dimensional vector spaces and matrices. The last three chapters investigate metric spaces, normed and inner product spaces, and linear operators. Because of its flexible structure, Algebra and Analysis for Engineers and Scientists may be used either in a one- or two-semester course by deleting appropriate sections, taking into account the students' backgrounds and interests.
A generous number of exercises have been integrated into the text, and a section of references and notes is provided at the end of each chapter. Applications of algebra and analysis having a broad appeal are also featured, including topics dealing with ordinary differential equations, integral equations, applications of the contraction mapping principle, minimization of functionals, an example from optimal control, and estimation of random variables.
Supplementary material for students and instructors is available at
"This book is a useful compendium of the mathematics of (mostly) finite-dimensional linear vector spaces (plus two final chapters on infinite-dimensional spaces), which do find increasing application in many branches of engineering and science…. The treatment is thorough; the book will certainly serve as a valuable reference." —American Scientist
"The authors present topics in algebra and analysis for students in engineering and science….. Each chapter is organized to include a brief overview, detailed topical discussions and references for further study. Notes about the references guide the student to collateral reading. Theorems, definitions, and corollaries are illustrated with examples. The student is encouraged to prove some theorems and corollaries as models for proving others in exercises. In most chapters, the authors discuss constructs used to illustrate examples of applications. Discussions are tied together by frequent, well written notes. The tables and index are good. The type faces are nicely chosen. The text should prepare a student well in mathematical matters." —Science Books and Films
"This is an intermediate level text, with exercises, whose avowed purpose is to provide the science and engineering graduate student with an appropriate modern mathematical (analysis and algebra) background in a succinct, but not trivial, manner. After some fundamentals, algebraic structures are introduced followed by linear spaces, matrices, metric spaces, normed and inner product spaces and linear operators…. While one can quarrel with the choice of specific topics and the omission of others, the book is quite thorough and can serve as a text, for self-study, or as a reference." —Mathematical Reviews
"The authors designed a typical work from graduate mathematical lectures: formal definitions, theorems, corollaries, proofs, examples, and exercises. It is to be noted that problems to challenge students' comprehension are interspersed throughout each chapter rather than at the end." —CHOICE | 677.169 | 1 |
Coursework maths gcse
GCSE Maths (Re-sits only) learning resources for adults, children, parents and teachers organised by topic. Coursework for GCSE mathematics is to be axed in England, the Education Secretary, Alan Johnson, has announced. All other GCSE coursework would have to be supervised. It is time to start thinking as an adult and get the best GCSE coursework writing help on the market! Our team of experts is always ready to help you. Coursework in Mathematics: MEI discussion paper page 3 The current situation All GCSE Mathematics specifications contain two pieces of coursework, each worth.
Edexcel GCSEs are available in over 40 subjects. Visit your GCSE subject page for specifications, past papers, course materials, news and contact details. Edexcel GCSEs are available in over 40 subjects. Visit your GCSE subject page for specifications, past papers, course materials, news and contact details. In the coursework file for Gcse maths there is a piece called Table Patterns which I think should be 7 pages long but shows only 3. How do I get to the other pages. GCSE Maths Coursework An equable shape is one in which: The perimeter and area is equal Find out what you can do about. Mathematics Gcse Coursework Tubes.
Coursework for GCSE mathematics is to be axed in England, the Education Secretary, Alan Johnson, has announced. All other GCSE coursework would have to be supervised. In the coursework file for Gcse maths there is a piece called Table Patterns which I think should be 7 pages long but shows only 3. How do I get to the other pages. Edexcel GCSEs are available in over 40 subjects. Visit your GCSE subject page for specifications, past papers, course materials, news and contact details. Maths Coursework Help,GCSE Maths Statistics Help Coursework Online from our helpers and get ready to solve your Maths coursework problems on time. GCSE Mathematics 4360 gives students knowledge of reasoning, problem-solving and functional elements. This would suit students wishing to progress to A-level Mathematics.
Tough GCSE topics broken down and explained by out team of expert teachers In this coursework a pass in GCSE Maths will open a lot of doors for you;. Investigations for GCSE Mathematics For teaching from September 2010 The coursework element was removed from GCSE Mathematics assessments in September 2007. It is time to start thinking as an adult and get the best GCSE coursework writing help on the market! Our team of experts is always ready to help you. | 677.169 | 1 |
9780631180562Mathematics in Economics: Models and Methods
A valuable guide to the mathematical apparatus that underlies so much of modern economics. The approach to mathematics is rigorous and the mathematical techniques are always presented in the context of the economics problem they are used to solve. Students can gain insight into, and familiarity with, the mathematical models and methods involved in the transition from 'phenomenon' to quantitative statement | 677.169 | 1 |
noun, Mathematics. 1. See under (def 2). [al-juh-bruh] /ˈæl dʒə brə/ noun 1. the branch of mathematics that deals with general statements of relations, utilizing letters and other symbols to represent specific sets of numbers, values, vectors, etc., in the description of such relations. 2. any of several systems, especially a ring in which elements can be multiplied by real or complex numbers (linear algebra) as well as by other elements of the ring. 3. any special system of notation adapted to the study of a special system of relationship: algebra of classes. /ˈældʒɪbrə/ noun 1. a branch of mathematics in which arithmetical operations and relationships are generalized by using alphabetic symbols to represent unknown numbers or members of specified sets of numbers 2. the branch of mathematics dealing with more abstract formal structures, such as sets, groups, etc n.
1550s, from Medieval Latin algebra, from Arabic al jebr "reunion of broken parts," as in computation, used 9c. by Baghdad mathematician Abu Ja'far Muhammad ibn Musa al-Khwarizmi as the title of his famous treatise on equations ("Kitab al-Jabr w'al-Muqabala" "Rules of Reintegration and Reduction"), which also introduced Arabic numerals to the West. The accent shifted 17c. from second syllable to first. The word was used in English 15c.-16c. to mean "bone-setting," probably from Arab medical men in Spain. algebra (āl'jə-brə) A branch of mathematics in which symbols, usually letters of the alphabet, represent numbers or quantities and express general relationships that hold for all members of a specified set. linear algebra The branch of mathematics that deals with the theory of systems of linear equations, matrices, vector spaces, and linear transformations.
A branch of mathematics marked chiefly by the use of symbols to represent numbers, as in the use of a2 + b2 = c2 to express the Pythagorean theorem.
theory A function argument which is used exactly once by the function. If the argument is used at most once then it is safe to inline the function and replace the single occurrence of the formal parameter with the actual argument expression. If the argument was used more than once this transformation would duplicate the […]
noun 1. an ancient system of writing representing a very early form of Greek, deciphered by Michael Ventris chiefly from clay tablets found at Knossos on Crete and at Pylos. noun 1. an ancient system of writing, apparently a modified form of Linear A, found on clay tablets and jars of the second millennium bc […]
Disclaimer: Linear-algebra | 677.169 | 1 |
Basic Math Solved! is a mathematical tool designed to solve YOUR basic math problems step-by-step l straight from the textbook! Basic Math Solved! covers all the basic mathematics, from addition and subtraction to introductory prealgebra. With countless features and tools at its disposal, Basic Math Solved! will have you acing your homework immediately! Infinite examples, step-by-step explanations, practice test creation, detailed graphs, and guided user input are just a few of the many features available, all with a remarkably easy-to-use graphical interface.5 MB
Education
-
101Math 1.1.1
In the popular book, 101 Things Everyone Should Know About Math, kids, ages 8-12, are challenged with questions about Statistics, Probability, Algebra, Basic Math and Geometry. Now, you can take all the fun of math from the book anywhere! With...
2.8 MB
Mathematics
-
MathWiz: Basic Math 3.6
Basic Math is for younger users who are just learning basic facts. It covers addition, subtraction, multiplication, and division. It is a nice flashcard program because it covers 100 different problems in each session. The preferences screen...
1.81 MBXcas 0.9.5.... | 677.169 | 1 |
Description
Geometry Concepts is an interactive learning system designed to provide instruction in mathematics at the 7th grade through adult levels. The instructional goals for Geometry Concepts include several objectives.
The primary objective for this app is to provide basic information about geometry through interactive lessons. Tap the notebook icon open dynamically illustrated lessons on each of the seven topics: Basic Concepts, Circles, Angles and Triangles, Lines and Planes, Area and Perimeter, 2-Dimensional Figures and 3-Dimensional Figures.
Our goal in designing this app is to incrementally build an understanding of fundamental concepts by providing an easy-to-use format for exploring geometry. Tap the magnifying glass icon to explore a topic more deeply. Each of the images are interactive. Tap to reveal terms and key concepts. Tap the more info icon for detailed information about the highlighted term.
We hope to support the development of a sense of confidence in each user's mathematical ability through a variety of educational approaches. Geometry Concepts includes a reference system, lessons, randomly generated quiz questions and an exploratory tool for open-ended investigations.
Tap the tool icon on the main navigation screen to access a set of 12 interactive tools for exploring geometry concepts. Icons on the sides of the screen are used to change the investigation. Manipulated the tools to learn more about each concept. We think your students will enjoy learning with Geometry Concepts | 677.169 | 1 |
Understanding the UK Mathematics Curriculum Pre-Higher Education
Transcription
1 About this guide: The Higher Education Academy STEM Subject Centres for Bioscience, Engineering, Information and Computer Sciences, Materials, Maths, Stats & OR Network and Physical Sciences commissioned this guide to be written by Mathematics in Education and Industry (MEI). A number of MEI staff were involved in authoring this guide and are listed inside. The STEM centres consist of: The Engineering Subject Centre: Understanding the UK Mathematics Curriculum Pre-Higher Education The Maths, Stats and OR Network: Physical Sciences: UK Centre for Bioscience: UK Centre for Materials Education: Subject Centre for Information and Computer Sciences: Copyright The STEM Subject Centres own the copyright to this guide so that they may use excerpts from it or update and modify as appropriate. However, the IPR rights are held by MEI (as authors), but the right is granted to the subject centres to use the text as appropriate to support their current and future activities. In any future version of this document it should be acknowledged that MEI produced the original. Published by The Higher Education Academy Engineering Subject Centre Loughborough University Leicestershire LE11 3TU tel: web: a guide for Academic Members of Staff
2 Authors biographies Stephen Lee (Lead Author), MEI Data Analyst / Web Manager Stephen studied for a BSc (Hons) in Mathematics with Education and a PhD in Mathematics Education at Loughborough University. His thesis focussed upon Mathematics at the transition from School/College to University. He has conducted much research and has many publications in this area. In 2008 he authored an undergraduate textbook on introductory mathematics. Richard Browne, MEI Programme Leader (Industry) Richard Browne was a secondary mathematics teacher in Inner London for 12 years before joining SEAC, one of the Qualifi cations and Curriculum Development Agency s (QCDA) predecessor bodies, in He worked on curriculum and qualifi cations at a national level for 17 years before joining MEI in Richard works as part of the Engineering Professors Council Maths Task Group. He is also a member of the DCSF Functional Maths Reference Group, and has worked with QCDA on accrediting functional mathematics specifi cations. Stella Dudzic, MEI Programme Leader (Curriculum) Stella Dudzic is an experienced teacher, author and curriculum developer. She has taught mathematics in secondary schools for 22 years and was a head of faculty before taking up her current post with MEI in Stella s work at MEI includes providing support for teachers; she has written and edited text books and other teaching materials and has conducted CPD. She drafts many of MEI s position papers on developments in mathematics education. Charlie Stripp, MEI Chief Executive (Designate) Charlie Stripp is well known for his pioneering work for MEI promoting Further Mathematics during the last 10 years. He was programme leader of the Further Mathematics Network, following on from a pilot project funded by the Gatsby Charitable Foundation. A former teacher, Charlie has experience in almost all aspects of mathematics education: examinations, textbooks, on-line learning, masterclasses and CPD. Copyright 2010 Published by The Higher Education Academy Engineering Subject Centre. ISBN (print) ISBN (online) Printed on stock sourced from a sustainable forest.
3 Understanding the UK Mathematics Curriculum Pre-Higher Education a guide for academic members of staff
4 Contents 1. Background The rationale for the document Introduction and overview Setting the scene: pre-higher education qualifications and study Introduction to the main qualifi cations Brief historical review of major developments Where and how will entrants have studied pre-higher education? Specific UK qualifications and attributes of students who enter higher education with them General Certifi cate of Secondary Education Overview Subject knowledge and skills International General Certifi cate of Secondary Education Advanced Subsidiary and Advanced Levels Overview Subject knowledge and skills Advanced Extension Award and Sixth Term Examination Paper Free Standing Mathematics Qualifi cations AS Use of Mathematics Diplomas Other qualifi cations International Baccalaureate Pre-U Access courses Foundation courses Wales, Scotland and Northern Ireland Useful sources of information (to keep up-to-date with pre-higher education developments) References made in this guide Additional references Documents/information Organisations Appendices Acronyms used in this guide (including appendices) A Level Mathematics Numbers (Source JCQ) Overview of content in mathematics A Levels What mathematics do students study in A level Mathematics courses? Important dates for Mathematics (authored by Roger Porkess)
5 1. Background 1.1 The rationale for the document In order to study a wide range of undergraduate programmes (including those in the Biological Sciences, Chemistry, Computer Science, Engineering, Materials Science, Mathematics and Physics), students need to have gained a mathematics qualifi cation prior to entering university-level study. A considerable number of pre-higher education mathematics qualifi cations are available within the UK and, for those working within the higher education (HE) sector, it is not always clear what mathematics content, methods and processes students will have studied (or indeed can be expected to know and understand) as they commence their universitylevel programmes. The Maths, Stats & OR Network, in conjunction with the Subject Centres for Bioscience, Engineering, Information and Computer Sciences, Materials and Physical Sciences, commissioned Mathematics in Education and Industry (MEI) to compile a mathematics guide. This outlines what students with given prior qualifi cations in mathematics are likely to know and be able to do and is written for those within the HE sector. Note that it does not include other science qualifi cations which may include elements of mathematics and/or statistics in them. 1.2 Introduction and overview This guide begins with a chapter setting the scene on pre-university qualifi cations and study. This includes an introduction to the main qualifi cations, a brief historical review of major developments and an overview of what and how entrants have studied prior to starting higher education. The main content of the guide is encapsulated within a chapter on specifi c qualifi cations and the attributes of students who enter with them. Information about qualifi cations is given in short sections; if the user wishes to refer to a particular qualifi cation it should be straightforward to identify the relevant section of the chapter. A chapter is provided on useful sources of information. This is broken down into two parts, the fi rst giving links to specifi c references raised in the previous chapter, and the second part on additional links to other documents (useful for gaining a more detailed understanding) and to relevant organisations (where information and updates can be found). The guide concludes with appendices, including one on acronyms used in the guide and one which presents the statistics on the number of entrants to mathematics A Level over the last 20 years. Overall this guide will give an overview of the key qualifi cations and offers links to further information that should aid the reader to gain an understanding of pre-university mathematics qualifi cations. 3
6 2. Setting the scene: pre-higher education qualifications and study 2.1 Introduction to the main qualifications In March 2008 the Department for Children, Schools and Families published a consultation paper, Promoting achievement, valuing success: a strategy for qualifi cations, which set out the government s intention to move towards a more streamlined and understandable qualifi cations framework for young people aged in England. At the heart of this strategy are three main routes to higher education: apprenticeships, diplomas and general qualifi cations, including the General Certifi cate of Secondary Education (GCSE) and the General Certifi cate of Education, Advanced Level (GCE A Level). The GCSE is usually taken at 16 years of age and the GCE A Level after a further two years of study. For A Levels, students work at Advanced Subsidiary (AS) Level in their fi rst year and at A2 Level in their second year. When the AS and A2 components are put together they form a full A Level qualifi cation. Apprenticeships combine paid work with on-the-job training, qualifi cations and progression. They do not include a requirement to take mathematics qualifi cations. Diplomas offer a blend of classroom work and practical experience. They include a requirement to study functional mathematics at the appropriate level. All diploma lines of learning permit learners to include other mathematics qualifi cations. Further details are given in section 3.5. General qualifi cations in mathematics provide the evidence of attainment in mathematics that is most likely to be presented to HE admissions tutors. This guide will clarify the content, style of assessment and probable learning outcomes that may be expected in a number of general qualifi cations in mathematics: these are GCSE, A Level and Free Standing Mathematics Qualifi cations (FSMQ). 2.2 Brief historical review of major developments General qualifi cations in mathematics have developed in the context of widespread recent changes in expectations for learners. The government s objective of encouraging up to 50% of year olds to attend higher education, for example, has infl uenced (albeit it gradually and without offi cial decisions) the demands of both GCSE and A Level qualifi cations in mathematics. The replacement of the General Certifi cate of Education, Ordinary Level (GCE O Level) by GCSE in 1988 may be seen as the start of a process by which these school leaving qualifi cations could more closely refl ect what the majority of 16 year olds know, understand and can do. Since that time, the substantial problem solving requirement of O Level Mathematics has been replaced by GCSE examinations that have required candidates to show capability in handling a broad range of basic mathematics questions. Similarly, the introduction of subject cores for A Level examinations in 1983, the acknowledgement in 1996 that the AS standard should be pitched according to what is likely to be achieved a year before taking A Level and the rise of modular assessment at A Level since 1990 have all played signifi cant parts in making A Level Mathematics examinations much more accessible than they were between 1951 and Until 1987 results were norm referenced so that in any subject 10% attained grade A, 15% B, 10% C, 15% D, 20% E and a further 20% were allowed to pass. This produced a bimodal distribution which did not match candidates mathematical knowledge. 4
7 2.3 Where and how will entrants have studied pre-higher education? It is important to be clear that those entering degree courses come from a wide range of backgrounds and bring with them a wide range of experiences. Two overarching factors relevant to this are where an entrant studied previously and how. This information guide is made with particular reference to those entering onto a degree from a UK background (i.e. not overseas). With respect to the situation in England (Wales, Scotland and Northern Ireland will be dealt with in section 3.7) the major breakdown of categories of places of learning is in terms of age range, type and whether it is statefunded or independent (fee paying). Over 90% of the secondary population attend state (government-funded) schools and, in the context of study prior to entry to higher education, establishments could include many different age ranges and have a varied focus. Age ranges for secondary study could involve 11-18, or The last of these could be small sixth forms attached to a school or they could be huge stand-alone Colleges of Further Education (FE) or Sixth Form Colleges. A learner may have been at the same place of study since the age of 11, or may have been at an establishment for only one or two years to complete their prehigher education studies. In terms of the independent sector it is widely expected that many of those attending such establishments will have been exposed to high quality tuition and learning resources and, although there is only a small proportion of the age cohort attending such establishments nationwide, most will go on to enter HE. Having detailed the above, it is very diffi cult to defi nitively describe the way students will have been taught in all of these different establishments. Be it in the state or independent sector, some will have been in small classes whilst others will have been in large classes, some will have had well qualifi ed teachers/lecturers, others non-qualifi ed mathematics teachers. What is apparent, though, is that learners will enter HE with different experiences and respond to the relative changes that university-level study will bring in different ways. This is the case without even considering the specifi c subject knowledge which will be detailed in the next chapter. 3. Specific UK qualifications and attributes of students who enter higher education with them This section describes the structure and content of specifi c UK mathematics qualifi cations and attempts to indicate the likely attributes of students who have taken them. However, the content of qualifi cation specifi cations cannot be assumed to be an accurate measure of what students will actually know and understand when they start higher education. This will be infl uenced considerably by the nature of their mathematical learning experiences and by the grades they achieved. Several universities have used diagnostic tests to determine the mathematical knowledge, understanding and fl uency of new undergraduates, and how they relate to students qualifi cations at the start of their HE courses. GCSE and A Level qualifi cations are examined by three awarding bodies in England: Assessment and Qualifi cations Alliance (AQA), Edexcel and Oxford, Cambridge and RSA (OCR). They are regulated by the Offi ce of the Qualifi cations and Examinations Regulator (Ofqual). The Qualifi cations and Curriculum Development Agency (QCDA) is the government agency responsible for developing these qualifi cations. GCSE and A level qualifi cations are also taken by students in Wales and Northern Ireland, though the 5
8 arrangements for administration are different. Scotland operates a separate system of examinations. The large majority of students entering HE have taken GCSE and A Level qualifi cations, but several other qualifi cations are also used as routes into HE. 3.1 General Certificate of Secondary Education Overview Students in state schools have to follow the National Curriculum until age 16. GCSE Mathematics assesses the mathematics National Curriculum and is usually taken by students at the end of compulsory education (age 16). Some GCSEs follow a modular structure, with students taking some examinations in year 10 (age 15) and the rest in year 11 (age 16). Although the GCSE course is often thought of as a two year course, the GCSE work in mathematics builds directly on earlier work in mathematics and so the GCSE examinations test the mathematics that students have learnt throughout secondary school (11-16), and earlier. The content of GCSE Mathematics is the same for all awarding bodies, though it can be divided in different ways for modular courses. For GCSEs taken up to 2012, the content is specifi ed by the 1999 National Curriculum, see (1). Many students do not do any more mathematics after GCSE. Such students, who have not done any mathematics for two or three years before starting their degree courses, are likely to have limited recollection of GCSE content and techniques. GCSE Mathematics is available at either Foundation Tier or Higher Tier. Grades C, D, E, F and G are available at Foundation Tier and grades A*, A, B and C are available at Higher Tier. Students who narrowly miss grade C at Higher Tier may be awarded grade D. Students who took GCSE Mathematics prior to 2008 may have taken it at Intermediate Tier, which allowed access to grades B, C, D and E. Grade C was not available at Foundation Tier until About 55% of students taking GCSE Mathematics achieve grade C or above. Students entering GCSE Mathematics at Foundation Tier will not have studied as much mathematics as students taking Higher Tier. However, the grade boundary for grade C at Foundation Tier is higher than for C at Higher Tier, so Foundation Tier students with grade C have shown a good understanding of the mathematics which they have studied Subject knowledge and skills Students who have not gone beyond the content of Foundation Tier GCSE will not have met some topics which students taking Higher Tier will have encountered. The list below covers the main topics not covered by Foundation Tier GCSE students: negative and fractional powers working with numbers in standard form (scientifi c notation) reverse percentage calculations working with quantities which vary in direct or inverse proportion solution of linear simultaneous equations by algebraic methods factorising quadratic expressions and solution of quadratic equations plotting graphs of cubic, reciprocal and exponential functions trigonometry calculation of length of arc and area of sector of a circle cumulative frequency diagrams, box plots and histograms moving averages tree diagrams and associated probability calculations. 6
9 Students who have been entered for Higher Tier Mathematics and achieved grade B or C will have an incomplete understanding of items from the list above and are likely to fi nd algebra diffi cult. GCSEs in Mathematics taken from summer 2012 will cover very similar content to the current ones but will put more emphasis on problem solving and functionality in mathematics International General Certifi cate of Secondary Education The International General Certifi cate of Secondary Education (igcse) was originally designed for international schools but is now taken by students in some independent schools in the UK. igcse Mathematics is not available in state funded secondary schools as it is not fully aligned with the National Curriculum. The standard and content are similar to GCSE but students may have studied some additional topics, such as an introduction to calculus or matrices. 3.2 Advanced Subsidiary and Advanced Levels Overview The information below refers to Advanced Subsidiary and A Levels taken after the year Further changes are proposed for teaching from 2012 which may affect students taking A Level Mathematics in AS Level Mathematics, Further Mathematics and Statistics each consist of three modules (also called units ). A Level in each of these subjects consists of six modules, which include the three AS modules. Students who have A Level will also have studied the AS content, but as they may not have requested the certifi cation for the AS separately it might not appear on their certifi cate. The modules in these subjects are all of equal size. The raw marks on each module are converted to Uniform Marks (UMS) to allow for slight differences in diffi culty of examinations from year to year: the overall grade is decided by the total uniform mark gained on all modules. Students can resit individual modules to improve their marks. All modules are available in June with some also available in January. The modules available in the MEI Mathematics and Further Mathematics A Levels are shown in Figure 1. (Note the MEI specifi cation is administered through the Awarding Body OCR.) Similar structures apply for the other Mathematics and Further Mathematics A Levels. AS Mathematics consists of the compulsory modules C1 and C2 and an applied module, which could be in mechanics, statistics or decision mathematics. A Level Mathematics has three further modules: the compulsory modules C3 and C4 and another applied module. The two applied modules in A Level Mathematics can be from the same area of applied mathematics or from different areas. The content of C1 and C2 together (AS) is nationally specifi ed; likewise for C3 and C4 (A2). The content of applied modules varies between different exam awarding bodies. The national core can be found in the criteria for A Level Mathematics on the Ofqual website, see (2). This document also details what students who achieve grade A, C or E can typically do (this only gives a general idea as grades are based on total marks achieved rather than on these criteria, so strengths in some areas may balance out relative weaknesses in others). Students with the full A Level will have a broader knowledge of the AS core content than the A2 content because they have further developed their understanding in the second year. 7
10 Figure 1 (Figure 1 notes AM is Additional Mathematics, FAM is Foundations of Advanced Mathematics, NM is Numerical Methods, NC is Numerical Computation, FP is Further Pure Mathematics, C is Core Mathematics, DE is Differential Equations, M is Mechanics, S is Statistics, D is Decision Mathematics, DC is Decision Mathematics Computation.) Typically, students complete AS Levels after one year. They may stop their study of mathematics at this point or go on to complete the full A Level in a further year. However, some schools enter students early for GCSE Mathematics and so they begin AS Mathematics in year 11 (age 16) and take three years to complete the full A Level. Other students take AS in year 13 (age 18) when their future plans are clearer. Further Mathematics is only taken by students who are also taking Mathematics. They take three further modules for AS, including one compulsory module, Further Pure 1. To complete the A Level in Further Mathematics, students take at least one more pure module and two other modules. Students taking A Level Mathematics and A Level Further Mathematics will take 12 different modules and students taking A Level Mathematics and AS Level Further Mathematics will take 9 modules. The optional modules in AS and A Level Further Mathematics can be drawn from either pure mathematics or applied mathematics. Applied modules are in suites for the three strands of applications: mechanics, statistics and decision mathematics. Mechanics 1 and Mechanics 2 could be taken in either A Level Mathematics or A Level Further Mathematics, but Mechanics 3 and 4 (and higher) are only available to students taking Further Mathematics. Similarly, for modules in statistics most awarding bodies only have two decision mathematics modules available. For students who have taken both Mathematics and Further Mathematics AS and/or A Level, the Mathematics qualifi cation consists of the compulsory core modules (C1 to 4) and a valid combination of applied modules. The remaining modules make up the Further Mathematics qualifi cation. If there is more than one possible valid combination of 8
11 applied modules to give A Level Mathematics, the combination of modules making up the separate AS or A Levels is automatically decided by the exam awarding body s computer in order to maximise the pair of grades students receive for Mathematics and Further Mathematics. The rules for aggregation and certifi cation can be seen in (3). A table of the numbers who have studied A Level Mathematics and Further Mathematics can be seen in appendix 5.2. A small number of students take 15 modules to gain A Levels in Mathematics and Further Mathematics and AS Further Mathematics (Additional), and some take 18 modules for A Levels in Mathematics, Further Mathematics and Further Mathematics (Additional) Subject knowledge and skills The vast majority of A Level students will be taught in schools and colleges and so will not be used to studying mathematics independently. Most A Level examination questions are structured. Past papers and specimen papers can be found on awarding bodies websites and will give an idea of what students are expected to be able to do. Students who have taken both Mathematics and Further Mathematics will have greater fl uency in the subject due to the greater amount of time they have spent on it. A document giving an overview of the content studied in mathematics A Levels can be seen in (4). This is also included in appendix 5.3. Grades available at AS and A Level are A to E and U (where U is unclassifi ed); grades achieved on individual modules are available to universities through UCAS, as well as the result for the whole qualifi cation. From summer 2010 grade A* will be available for the full A Level (but not for AS). For A Level Mathematics, a total of 180 UMS marks (out of 200) will be needed on the two compulsory A2 modules (C3 and C4). For A Level Further Mathematics, a total of at least 270 (out of 300) is needed on the best three A2 modules. Students with A* will have shown the ability to work accurately under pressure. A small number of students take AS or A Level Pure Mathematics. The A Level consists of the four compulsory core modules from A Level Mathematics together with two Further Mathematics Pure Modules. It cannot be taken with Mathematics or Further Mathematics AS or A Level. Some students take AS or A Level Statistics. This is a separate qualifi cation from Mathematics and Further Mathematics and the modules in it focus more on the use of statistics, whereas the statistics modules in the mathematics suite are more mathematical. The content of AS or A Level Statistics would be very useful background for students going on to study Business, Biology, Psychology or Social Sciences at higher education level. Students who have completed their mathematical studies a year or more before starting higher education may need some support with revision to regain the fl uency they had when they sat their examinations. 3.3 Advanced Extension Award and Sixth Term Examination Paper The Advanced Extension Award (AEA) is based on A Level Mathematics core content and is designed to challenge the most able students. It is offered by all of the awarding bodies but the examination paper is set by Edexcel. AEAs in other subjects exist but are being withdrawn - the Mathematics AEA will continue until at least The Mathematics AEA is assessed by a three hour paper of pure mathematics questions with no calculator allowed. Grades available are Distinction and Merit. Although candidates do not have to learn any additional content for AEA they do need to get used to a different style of question and to present clearly structured mathematical arguments. 9
12 The Sixth Term Examination Paper (STEP) is a university admissions test originally used only for entrance to Cambridge but it is now also used by some other universities. It is administered by the Cambridge Assessment examination board. There are three mathematics papers (I, II and III) and candidates usually take two of them. Paper III is based on A Level Further Mathematics and papers I and II on A Level Mathematics, but questions may include some content that is not in the A Level syllabus. However, candidates are not expected to learn extra content for the examination. Each paper has 13 questions: eight pure mathematics, three mechanics and two statistics. No calculator is allowed. Candidates are expected to answer six questions in three hours. Grades available are S (Outstanding), 1 (Very Good), 2 (Good), 3 (Satisfactory) and U (Unclassifi ed). Usually a candidate will be awarded a grade 1 for a paper if they answer four (out of six) questions well. Students who are successful in AEA or STEP will have a high level of ability to think for themselves, persist with a problem and present structured mathematical arguments. 3.4 Free Standing Mathematics Qualifications FSMQs were fi rst developed in the late 1990s. The initial motivation was to support vocational qualifi cations, e.g. General National Vocational Qualifi cations (GNVQ s), but it was also recognised that they could provide useful courses for other students as well. Uptake of the original FSMQs was not high and only AQA now offer them in their original form. They are tightly focused qualifi cations in applications of number, algebra, calculus, geometry, statistics or decision mathematics and they compensate for their narrow focus by requiring quite deep coverage. OCR offers two FSMQs. These cover mathematics more broadly, but in less depth. Foundations of Advanced Mathematics (FAM) is a level 2 qualifi cation that is designed to help bridge the gap between GCSE and AS Mathematics for students with a C/B grade in Mathematics GCSE. Additional Mathematics is a level 3 qualifi cation aimed at able GCSE students and designed to be taken alongside GCSE. It is comparable in diffi culty to AS Mathematics. The AQA qualifi cations are likely to be used to support study of a range of courses in subjects other than mathematics. The OCR qualifi cations are more likely to be used to demonstrate achievement of a milestone in a learner s mathematical development. All FSMQ qualifi cations are similar in size, rated at 60 guided learning hours (the same size as a single unit of an A Level that is divided into six units). Level 3 units are awarded Universities and Colleges Admissions Service (UCAS) points. The AQA FSMQs share a single assessment model. Candidates must produce a coursework portfolio worth 50% of the credit and sit a written examination for the remaining credit. The OCR FSMQs use slightly different assessment approaches, but both are assessed by written examination only. Students who have achieved success in these qualifi cations are likely to share the broad capabilities of students achieving other mathematics qualifi cations at Levels 1, 2 and 3, see (5). However, students who have taken the AQA FSMQs will have demonstrated the ability to appreciate real world use and application of mathematics; they will also have engaged with completing a substantial coursework project. Students who have achieved success with OCR Additional Mathematics are likely to have shown an excellent grasp of basic advanced topics, which should be valued all the more highly if the qualifi cation was taken pre-16. Students who have been successful in FAM will have studied a broader range of mathematics and are therefore more likely to be able to meet the demands of mathematics post-gcse, particularly in algebra and trigonometry. 10
NEW YORK STATE TEACHER CERTIFICATION EXAMINATIONS TEST DESIGN AND FRAMEWORK September 2014 Authorized for Distribution by the New York State Education Department This test design and framework document
MAT 051 Pre-Algebra Mathematics (MAT) MAT 051 is designed as a review of the basic operations of arithmetic and an introduction to algebra. The student must earn a grade of C or in order to enroll in MAT
MATH BOOK OF PROBLEMS SERIES New from Pearson Custom Publishing! The Math Book of Problems Series is a database of math problems for the following courses: Pre-algebra Algebra Pre-calculus Calculus Statistics
Appendix 3 IB Diploma Programme Course Outlines The following points should be addressed when preparing course outlines for each IB Diploma Programme subject to be taught. Please be sure to use IBO nomenclature
Visit this link to read the introductory text for this syllabus. 1. Circular Measure Lengths of Arcs of circles and Radians Perimeters of Sectors and Segments measure in radians 2. Trigonometry (i) Sine,
Mathematics I, II and III (9465, 9470, and 9475) General Introduction There are two syllabuses, one for Mathematics I and Mathematics II, the other for Mathematics III. The syllabus for Mathematics I and
Mathematics within the Psychology Curriculum Statistical Theory and Data Handling Statistical theory and data handling as studied on the GCSE Mathematics syllabus You may have learnt about statistics and
* Students who scored a Level 3 or above on the Florida Assessment Test Math Florida Standards (FSA-MAFS) are strongly encouraged to make Advanced Placement and/or dual enrollment courses their first choices
MATHEMATICS Students must pass all math courses with a C or better to advance to the next math level. Only classes passed with a C or better will count towards meeting college entrance requirements. If
MATHEMATICS: THE LEVEL DESCRIPTIONS In mathematics, there are four attainment targets: using and applying mathematics; number and algebra; shape, space and measures, and handling data. Attainment target
The New Advanced Diplomas: What they Mean for HE in Curriculum Terms Pam Calabro, Network Officer, Linking London CONTENTS Introduction 2 How the Diploma is structured 3 Advanced Diploma: Principal Learning
PHILOSOPHY OF THE MATHEMATICS DEPARTMENT The Lemont High School Mathematics Department believes that students should develop the following characteristics: Understanding of concepts and procedures Building
APPLIED MATHEMATICS ADVANCED LEVEL INTRODUCTION This syllabus serves to examine candidates knowledge and skills in introductory mathematical and statistical methods, and their applications. For applications
Accelerated Mathematics 3 This is a course in precalculus and statistics, designed to prepare students to take AB or BC Advanced Placement Calculus. It includes rational, circular trigonometric, and inverse
Introduction The career world is competitive. The competition and the opportunities in the career world become a serious problem for students if they do not do well in Mathematics, because then they areFOREWORD The Botswana Examinations Council is pleased to authorise the publication of the revised assessment procedures for the Junior Certificate Examination programme. According to the Revised National
Mathematics Courses in Mathematics are offered with instruction in English, French (F) and Spanish (S) where enrolment warrants. Note: Five credits at the 20 level are required to obtain an Alberta High
National 5 Mathematics Course Assessment Specification (C747 75) Valid from August 013 First edition: April 01 Revised: June 013, version 1.1 This specification may be reproduced in whole or in part for
OCR GCSE Mathematics (J560) now accredited ocr.org.uk/gcsemaths Introducing the new Mathematics GCSE for first teaching from 2015 In February 2013, the Secretary of State for Education Michael Gove wrote
Engineering students knowledge of mechanics upon arrival: Expectation and reality Stephen Lee, Martin C. Harrison and Carol L. Robinson Abstract In recent years there has been an increasing awareness of
Learner Guide Cambridge International AS & A Level Mathematics 9709 Cambridge International Examinations retains the copyright on all its publications. Registered Centres are permitted to copy material
The Mathematics iagnostic Test Mock Test and Further Information 010 In welcome week, students will be asked to sit a short test in order to determine the appropriate lecture course, tutorial group, whether
Abstract Developing Higher Level Skills in Mathematical Modelling and Problem Solving Project Kevin Golden Department of Engineering Design and Mathematics University of the West of England, Bristol In
Advanced Higher Mathematics Course Specification (C747 77) Valid from August 2015 This edition: April 2015, version 1.1 This specification may be reproduced in whole or in part for educational purposes
England, Northern Ireland and Wales Educational Timeline School Year: August-June with exam dates in June (alternative external exam dates offered in November or December) Primary school is six years followed
MATHEMATICS Administered by the Department of Mathematical and Computing Sciences within the College of Arts and Sciences. Paul Feit, PhD Dr. Paul Feit is Professor of Mathematics and Coordinator for Mathematics.
This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these tools to better serve teachers. Algebra
Math Placement Test Sample Problems The Math Placement Test is an untimed, multiple-choice, computer-based test. The test is composed of four sections: pre-algebra, algebra, college algebra, and trigonometry.
Early Entry Guidance for schools January 2015 Contents Introduction 3 Rationale 4 How are performance meaures be affected by the early entry rules? 5 Which subjects do the early entry rules apply to? 5
Basic Math Course Map through algebra and calculus This map shows the most common and recommended transitions between courses. A grade of C or higher is required to move from one course to the next. For
Core Standards of the Course Standard 1 Students will acquire number sense and perform operations with real and complex numbers. Objective 1.1 Compute fluently and make reasonable estimates. 1. Simplify
Algebra I Credit Recovery COURSE DESCRIPTION: The purpose of this course is to allow the student to gain mastery in working with and evaluating mathematical expressions, equations, graphs, and other topics,
Undergraduate Admissions Guide for UK School/College Advisers 2013 entry Introduction The University of Edinburgh is one of the UK s leading research institutions, aiming to provide its student body with
Mathematics in hair and beauty studies principal learning There are opportunities to develop mathematics to support and enhance the business and scientific aspects of the hair and beauty studies principal
Trigonometry Lesson Unit 1: RIGHT TRIANGLE TRIGONOMETRY Lengths of Sides Evaluate trigonometric expressions. Express trigonometric functions as ratios in terms of the sides of a right triangle. Use the
MATHS LEVEL DESCRIPTORS Number Level 3 Understand the place value of numbers up to thousands. Order numbers up to 9999. Round numbers to the nearest 10 or 100. Understand the number line below zero, and
04 Mathematics CO-SG-FLD004-03 Program for Licensing Assessments for Colorado Educators Readers should be advised that this study guide, including many of the excerpts used herein, is protected by federal
Core Florida Math for College Readiness Florida Math for College Readiness provides a fourth-year math curriculum focused on developing the mastery of skills identified as critical to postsecondary readiness
Chartered Institution of Building Services Engineers Guidance for the Submission of Foundation Degrees in Building Services Engineering for Accreditation The following guidelines for Foundation Degrees
Maths IGCSE Welcome to your Mathematics IGCSE course! This introduction contains all the information you need to be able to start your course, and you can also use it as a reference point as you work yourInfinite Algebra 1 Kuta Software LLC Common Core Alignment Software version 2.05 Last revised July 2015 Infinite Algebra 1 supports the teaching of the Common Core State Standards listed below. High School
Overview Students knowledge of Mechanics at the start of degree level courses Findings from Stephen Lee s three year journey Background Research questions Schools Questionnaires and Availability Uptake | 677.169 | 1 |
Essential Foundation Practice for Trigonometry
Be sure that you have an application to open
this file type before downloading and/or purchasing.
303 KB|9 pages
Product Description
This 40 problem, nine page document (8 practice pages with 1 answer key page) is meant to help students review the material that will be necessary for their success in Trigonometry. Material is pulled from Algebra and Geometry and includes:
We have used this activity in the past as a summer assignment for our students about to take a semester course in Trigonometry. When Trigonometry is taught in the second semester, we often use it as a winter break packet. | 677.169 | 1 |
Whole-school advanced pureStructured worksheet to help pupils take key notes and examples for the topic of Algebra and Functions (Polynomial Long Division, Remainder and Factor Theorem and Binomial Expansion). Designed for the new A Level Maths (2017+), Edexcel Specification.
Clive is tackling a new topic on the GCSE curriculum, but predicatably is struggling. Four questions getting increasingly difficult requiring you to check as Clive has made mistakes. These will hopefully generate discussion in classBased on Edexcel Core 3 Exam Questions on numerical methods - a useful revision resource.
I run this as a team challenge, with each team selecting a question from each category and then ALL groups having a set time (between 3 and 5 mins) to complete the question (whiteboards and markers are great here). If the team that selected the question gets it right then they win the points allocated for that question, if not then the other teams have the opportunity to STEAL those points. My Year 12 students today really enjoyed this and have requested the powerpoint to use as revision at home!
An introduction to the graphs of reciprocal functions. Reciprocals of polynomials, absolute value linear, reciprocal circular functions. Developing the general principles how to deduce the graph of the reciprocal of any function given by a graph or rule. Suitable for IB HL Mathematics or VCE Specialist Mathematics Unit 2 to Unit 3 or any similar course.
Three Matrices Match Up worksheets.
In each worksheet pupils must complete 15 questions to find the solutions on the worksheet. Three solutions will be unused and they can be added together to find a final solution.
Can be used as practice or revision of the topic. Also if pupils get the final solution teachers can be happy the topic is understood without the need for marking. Solutions included.
1. Adding and Subtracting
2. Multiplying
3. Calculating Determinants
Adding or Subtracting Matrices Match Up worksheet.
(Or you could buy my Matrices bundle)
Pupils must complete 15 questions to find the solutions on the worksheet. Three matricesCalculating determinants Matrices Match Up worksheet.
(Alternatively you could buy my Matrices Bundle)
Pupils must complete 15 questions to find the solutions on the worksheet. Three values | 677.169 | 1 |
What Are Some Organizing Principles Around Which One Can Create a Coherent Pre-college Algebra Program?. Critical Issues in Education: Teaching and Learning Algebra MSRI, Berkeley, CA May 14, 2008 Zalman Usiskin The University of Chicago2. TheTheUse geometry. Transformations provide a powerful set of ideas for dealing with graphs of functions and trigonometry.
Consider the students. A course for all students cannot assume they all have the background, motivation, and time that we would prefer.
Sequence by uses. Employ uses of numbers and operations to develop arithmetic, and employ uses of variables to move from arithmetic to algebra.(Go to , click on Available Materials, scroll down to and download Applying Arithmetic: A Handbook of Applications of Arithmetic.)
If a quantity is multiplied by a growth factor b in every interval of unit length, then it is multiplied by bn is every interval of length n. (nice applications to compound interest, population growth, inflation rates)
b0 = 1 for all b since in an interval of length 0 the quantity stays the same regardless of the growth factor.
bm • bn = bm+n because an interval of length m+n comes from putting together intervals of lengths m and n.
"The use of 'real-world' contexts to introduce mathematical ideas has been advocated… A synthesis of findings from a small number of high-quality studies indicates that if mathematical ideas are taught using 'real-world' contexts, then students performance on assessments involving similar 'real-world' problems is improved. However, performance on assessments more focused on other aspects of mathematics learning, such as computation, simple word problems, and equation solving, is not improved ." (p. xxiii and p. 49) | 677.169 | 1 |
The history of algebra essay
You're movin' on up. Your school is now part of Shmoop University, which means that you'll now access all Shmoop material from a new URL: Find the perfect online Essay Writing tutor for you. Search for Essay Writing tutors online now or schedule a session for later. Pearson Course Content. Pearson is the world leader in publishing, education and learning. Pearson Prentice Hall, along with our other respected imprints, provides. Pearson Course Content. Pearson is the world leader in publishing, education and learning. Pearson Prentice Hall, along with our other respected imprints, provides.
Enroll in 2017-18 AP Mentoring AP Computer Science Principles, AP English Literature and Composition and AP U.S. History teachers of all experience levels can. History of Psychology. Psychology is a science. Psychology can be termed as the scientific study of human mind and how it functions. Psychology deals with the mental. Pearson Prentice Hall and our other respected imprints provide educational materials, technologies, assessments and related services across the secondary curriculum. Online course material aligned with common core & virtual classroom experience for schools, districts, home schools and individual students. A TYPICAL American school day finds some six million high school students and two million college freshmen struggling with algebra. In both high school and.
The history of algebra essay
Teacher Login / Registration : Teachers: If your school or district has purchased print student editions, register now to access the full online version of the book. A TYPICAL American school day finds some six million high school students and two million college freshmen struggling with algebra. In both high school and. History of Psychology. Psychology is a science. Psychology can be termed as the scientific study of human mind and how it functions. Psychology deals with the mental.
Find the perfect online Algebra tutor for you. Search for Algebra tutors online now or schedule a session for laterCourses for Middle School, High School and College Conceptual Videos on Core Math Topics Lectures on Math and Science. The history of hypnosis is full of contradictions. On the one hand, a history of hypnosis is a bit like a history of breathing. Like breathing, hypnosis is anEnroll in 2017-18 AP Mentoring AP Computer Science Principles, AP English Literature and Composition and AP U.S. History teachers of all experience levels can.
Find the perfect online Algebra tutor for you. Search for Algebra tutors online now or schedule a session for later.
Teacher Login / Registration : Teachers: If your school or district has purchased print student editions, register now to access the full online version of the book.
A Time-line for the History of Mathematics (Many of the early dates are approximates) This work is under constant revision, so come back later. Please report any. This essay is an original work by several authors. Please comment only on the talk page. Andrew Schlafly has several times stated that most - if not all. There is no one in the history of the world who is GOOD at texting. I myself have sent thousands of texts over the course of my life, and all of them were a mistake. SSJ Ministries. Bereavement Ministry; Bible Study; Career Renewal Ministry; Discernment Ministry; Fall Festival. Sponsorship Form; Festival Volunteer Sign-Up. | 677.169 | 1 |
Finite Mathematics, 8th Edition
Get the background you need and discover the usefulness of mathematics in analyzing and solving problems with FINITE MATHEMATICS, 8th Edition. The author clearly explains concepts, and the computations demonstrate enough detail to allow you to follow and learn steps in the problem-solving process. Hundreds of examples and exercises, many based on real-world data, illustrate the practical applications of mathematical concepts. The book also includes technology guidelines to help you successfully use graphing calculators and Microsoft® Excel® to solve selected exercises | 677.169 | 1 |
Please contact your nearest Dymocks store to confirm availability
Email store
This book is available in following stores
This lucid and balanced introduction for first year engineers and applied mathematicians conveys the clear understanding of the fundamentals and applications of calculus, as a prelude to studying more advanced functions. Short and fundamental diagnostic exercises at the end of each chapter test comprehension before moving to new material.Provides a clear understanding of the fundamentals and applications of calculus, as a prelude to studying more advanced functionsIncludes short, useful diagnostic exercises at the end of each chapter
Title:
Calculus
Author:
Edition:
Publisher:
Elsevier Science
Format:
ePub
Subject:
Length:
Width:
Sub Title:
Introductory Theory and Applications in Physical and Life Science | 677.169 | 1 |
Whole-school group theory Point presentation, 12 slides, Explaining the meaning of the basic concepts used in set theory, based on IB Mathematical Studies Syllabus. For a preview of the power point copy the following link on your browser: | 677.169 | 1 |
Texas Instruments TI-Nspire Graphing Calculator
The latest TI math and science learning technology features the TI-Nspire family of products and services. This technology goes beyond graphing to help students see math and science in new and different ways.
TI-Nspire and TI-Nspire CAS technology was developed hand-in-hand with educators worldwide and built on proven graphing technology which research shows has a positive impact on student achievement. This exciting learning technology offers both handhelds and computer software for the flexibility to meet different classroom needs.
Some unique features of TI-Nspire technology let you:
* See multiple representations of a problem, individually or together on a single screen * Dynamically link representations of a problem to see how changes to one affect others * Grab and move graphed functions in real time to observe relationships and patterns * Activate the handheld's Press-to-Test feature to block access to certain geometry features not allowed on exams * Save and review work in documents, similar to a computer * Use handhelds in college entrance and advanced placement exams * Try TI-Nspire Computer Software - Teacher Edition with a handheld emulator and enhanced document editing
TI-83 and TI-84 family compatibility - only the TI-Nspire handheld comes with an innovative snap-in TI-84 Plus Keypad that provides the same keystrokes and functionality as TI-83 Plus, TI-84 Plus and TI-84 Plus Silver Edition graphing calculators. | 677.169 | 1 |
Fraleigh and Beauregard's text is known for its clear presentation and writing style, mathematical appropriateness, and overall student usability. Its inclusion of calculus-related examples, true/false problems, section summaries, integrated applications, and coverage of Cn make it a superb text for the sophomore or junior-level linear algebra course. This Third Edition retains the features that have made it successful over the years, while addressing recent developments of how linear algebra is taught and learned. Key concepts are presented early on, with an emphasis on | 677.169 | 1 |
10 1 Lecture1
1.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM A SET is a well-defined collection of objects. <ul><li>Examples: </li></ul><ul><li>Set of books found in Mapua Library Makati </li></ul><ul><li>Set of players of Spain's 2010 soccer team </li></ul><ul><li>3. Set containing all the months in a year </li></ul><ul><li>4. Set of students enrolled in Math 10-1 AY01 class for 1 st term AY 2010-2011 </li></ul>
2.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM The cardinality of set A is the number of elements contained in A and is denoted by |A|. Determine the cardinality of the previously given sets.
3.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Two ways of writing a set: Rule Method : >>describes a set by some rule Roster Method : >>list down all the elements of the set.
4.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Rule Method : {x|x is a positive integer less than 6} Roster Method : {1,2,3,4,5} Illustration:
5.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM <ul><li>Counting numbers which are multiples of 3 and less than 20. </li></ul><ul><li>Single digits used in our decimal system. </li></ul><ul><li>3. Set of all odd numbers between 2 and 12. </li></ul>Illustration:Write each of the following using roster and rule method:
6.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM >The symbol { } denote the set that is empty. >The symbol є literally means 'is an element of' or 'belongs to' >The symbol U denotes the universal set, set containing all elements in consideration. Some notations:
7.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM A one-to-one correspondence exists between two sets A and B if it is possible to associate the elements of A with the elements of B in such a way that each element of each set is associated with exactly one element of the other. SET RELATIONSHIPS:
8.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM >Two sets A and B are equivalent , denoted by A B, if and only if there exists a one-to-one correspondence between them. >Two sets A and B are equal , denoted by A = B, if the elements of A and B are exactly the same. Equal and Equivalent Sets
9.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Two sets A and B are joint sets if and only if A and B have common elements; otherwise, A and B are disjoint . Joint and Disjoint Sets
10.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM >A set A is a subset of B, A B, if every element of A is in B. >If for A B, B contains elements that are not in A, then A B. ( proper subset ) Subset and Proper Set
11.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM The power set of A is the set containing all subsets of A and is denoted by (A). Power Set
12.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Venn Diagram is the pictorial representation of sets and (is usually symbolized by circles and rectangles.) Venn Diagram
13.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Venn Diagram A B B A U U
14.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Union of Sets The union of two sets A and B, denoted by A B, is the set whose elements belong to A or B or both to A and B. In symbol, A B = {x|x A or x B or x A and B} Operations on Sets
15.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Intersection of Sets The intersection of two sets A and B, denoted by A B, is the set whose elements are common to A and B. In symbol, A B = {x|x A and x B} Operations on Sets
17.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Difference of Sets, The difference of two sets A and B, denoted by A - B, is the set whose elements are in A but not in B. In symbol, A - B = {x|x A and x B} Operations on Sets
18.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Complement of a Set, The complement of a set A, denoted by A', is the set with elements found in the universal set U, but not in A, i.e., the difference of the universal set U and A. In symbol, A' = {x|x U and x A} = U - A Operations on Sets
20.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Cartesian Product, The Cartesian product of two sets A and B denoted by AxB, is the set of ordered pairs(x,y) suct that x is an element of A and y is an element of B. In symbol, AxB = {(x,y)|x A and y B} Note: A x B B x A Operations on Sets
21.
Math 10-1: College Algebra SETS AND THE REAL NUMBER SYSTEM Suppose a particular menu in a burger joint includes the following: Hamburger (b) Soda (s) Cheeseburger (c) Tea (t) Hotdog sandwich (d) Fruit Juice (f) What are the possible combinations of burger and drinks? Illustration: | 677.169 | 1 |
Sequences Review
Be sure that you have an application to open
this file type before downloading and/or purchasing.
111 KB|4 pages
Product Description
This worksheet covers all topics associated with arithmetic and geometric sequences, as taught in Algebra 2 and Precalculus.
Problems include finding the number of terms in a sequence, finding a given term in a sequence, recognizing a pattern as a sequence, finding the arithmetic mean and geometric mean, and word problems involving sequences.
Series are not covered in this review.
This is appropriate for Algebra 2 or Precalculus, though it could also be used as a review of sequences for a Calculus course. | 677.169 | 1 |
English Portfolio
The English Department believes all students will benefit from.
Reflecting on their writing.
Formulating personal standards for what constitutes good writing.
Validating their own growth as writers.
You are asked, therefore, to.
Add
Calculus, Fall 2012
Guide to Midterm Exam 1
Review all your homework problems and quizzes. Then check the following topics and ask
yourself if you understand them. If not sure, you can find related examples from your class
note and/or homework problems an
Calculus I, Fall 2012
Guide to Midterm Exam 2
Review all your homework problems and quizzes. Then check the following topics and ask
yourself if you understand them. If not sure, you can find related examples from your class
note and/or homework problems
Instructions for Weekly Listening Homework (Contemporary Topics 3)
You will be required to have a notebook used only for note-taking exercises. Do not
write anything else in this notebook (vocabulary lists, definitions, etc.). You will be
allowed to look
Solution to Quiz 5
Calculus I | Fall 2016 (MATH103003)
[TA] Yang Ming Hsun | December 1, 2016 (09:14 pm)
Problem 1. Either prove the assertion or show that the assertion is not valid by giving a
counterexample.
Let [a, b] be a closed, bounded interval an
Solution to Quiz 01
Calculus I | Fall 2016 (MS, IEEM)
[TA] Yang Ming Hsun | October 6, 2016
Problem 1. Label the following statements as being true or false. If it is true, prove it. If it
is false, give an example to show why it is false.
(a) If limxc f | 677.169 | 1 |
...wherever you would like. The course will take approximately 60 hours to complete and will be assessed through a series of written assignments... Learn about: Weight and Mass, Frequency Charts, Basic IT...
A Level
Online
Flexible
Requirements
This course is for residents in England, Ireland, Scotland, Wales, Northern Ireland
This was an excellent course, well organized, informative and in general a great experience.
← | →
Oxford Learning College
...for all students over the age of 17. Students can study on this course no matter where you are in the World. The course is made up of various units... Learn about: GCSE Mathematics, Problem Solving...
...Level then AS or A2 Level courses can be purchased from the AS/A2 Level section of our website. Each lesson begins with a set of clearly stated objectives... Learn about: Mathematics Series, vector algebra, Confidence Training...
...and David Beckham is about to take a free kick that could take England to the World Cup Finals. The crowd are crying out for him to score… and he does... Learn about: Mathematics Series, Basic Mathematics, Engineering Mathematics...
...understand and apply knowledge. At AS you will also study either a statistics option or a mechanics option (S1/M1), depending on timetable organisation... Learn about: IT for adults, GCSE Mathematics...
...study a degree in a mathematically related discipline. The Further Maths course is more difficult than the Maths one and is therefore suited to those students... Learn about: Maths skills, Mathematics understanding...
...programme designed to provide appropriate extension material or timely support for students. In addition we aim to seek ways to enrich and extend the mathematics... Learn about: Team Training, GCSE Mathematics...
...The modules studied in the first year are described under Mathematics except for Further Pure 1. This last module, and the ones in the second year explore topics such as Matrices... Learn about: GCSE Mathematics...
...and can lead to careers in engineering, accounting, teaching, computing and many other areas. The course can be taken as part of a full-time programme... Learn about: Confidence Training, GCSE Physics, Full Time...
...levels where two of them are AS Maths and AS Further Maths. Course Content AS Mathematics : This course consists of pure mathematics and statistics... Learn about: Mathematics Series, Data analysis...
...Award includes 45 contact hours, when a variety of strategies will be used including whole-class, pair and individual work, presentations and workshop activities... Learn about: Teacher Education, GCSE Mathematics... | 677.169 | 1 |
Online differenatial,
algebra i textbook,
monte carlo ss 89,
algebra pythagorean theorem,
How is dividing a polynomial by a binomial similar to or different from the long division you learned in elementary school?,
how to solve multi-step inequalities.
10 th grade algebra,
Using Positive Exponents,
graphing answers,
year 6 algebra,
algebra 1 workbook,
power equations solver software,
Is there a difference between solving a system of equations by the algebraic method and the graphical method? Why or why not?.
How do you check algerbra problems,
algebra equation solver program,
british method math,
abstract algebra hungerford solution,
modern algebra durbin solutions,
series solver,
how is algebra good for freshmen.
Mirabeau fractions cipher solver online,
do my algebra for me,
answers to 2003 calculus,
extrapolation calculator,
how to work out substitution in algebra for dummies,
easiest way to pass college algebra.
College algebra,
how to solve equations with fractional coefficients,
how to solve matrices,
college algebra projects.
How is trig used in everyday life?,
interval notation problems,
11% CONVERTED TO A FRACTION,
How Is Algebra Used in Everyday Life,
Power point presentations for Algebra,
matrices algebra + free online mathsolver,
I AM NOT PASSING MY CLASS, HOW CAN I LEARN ALGEBRA?.
Quadratic formula,
equation to find sq/root of a quadratic,
solving a factorial on TI 84,
factor binomial calculator,
Emulation TI-84,
how to enter laplace on TI-83 plus,
how to do cube root in ti 83 plus.
Ti-89 differential equation solver,
college algebra for dummies,
When solving a rational equation, why is it OK to remove the denominator by multiplying both sides by the LCD, and why can you not do the same operation when simplifying a rational expression?,
math worksheets on volume of a prism,
deriving "difference of cubes" formula,
holt biology workbook answers.
Calculator for the gcf',
Simplifying Radical Expressions on the TI-83,
any calculator can solve limits ?.
How to write an equation in power point presentation,
pre algebra questions,
write an algebraic expression for each phrase,
factor equations for me,
simultaneous equations with quadratic example,
learn algebra by video.
Algebra 1B work to do online,
kumon instructor exams,
pictograph worksheets,
maths worksheets year 8,
companies aptitude questions based on probability,
solution of a second order linear nonhomogeneous differential equation.
Math book answers,
how to solved a system of equation by graping,
word mix cheater,
Fundamental theorem of Algebra on TI 83 calculator,
college math games PPT free downloads,
8th grade algebra help online.
Permutation sample questions objective,
worksheets for Make generalizations from patterns for 4th grade,
solving for the roots of a third order polynomial,
9th grade algebra test,
struggling with algebra,
download college algebra calculator,
8th grade free worksheets.
Least common denominator worksheet,
graphing linear equations in 3 variables,
math work shet therd grade elementary,
convert whole fraction to decimal,
example to find the number of characters in a string in java.
USING GRAPHING CALCULATOR FUNCTION OF TRIANGLE OVER X AXIS,
how to take complex roots on ti-83,
simplest form calculator,
CLEP Algebra,
possitive and negative integer worksheets,
convert Decimal To Fraction.
Middle school algerbra expressions work sheets,
free pre algebra college level,
substitution calculator,
solver ti 89,
graphing lines using equations in standard form game,
if the line passes through the point(0,-3) then the y intercept is what.
Free accounting text book download,
free printable instructions for 6th graders learning linear equations,
How Do You Solve Numerical Expressions,
checking to see if a string is an integer in java,
solve third order equation,
convert mixed fraction into decimal,
free printable seventh grade math.
Greatest common factor math problems 10th grade,
graph a parabola using a casio calculator,
example 2nd order system laplace,
How did the number game use the skill of simplifying rational expressions?,
LOG mit TI,
free 7th grade math pre algebra.
Coding to Get the first four digit of a number in JAVA,
"area and circumference worksheets",
maths equations ks2 year 5,
C Programming language Aptitude question + free Download,
online ellipses,
trinomial equation solver.
Free printable worksheets for third graders,
+whats the greatest common factor for 24 and 45,
where can i get help with my glenco geomertry concepts and applications,
square root calculator online free,
free algebrator,
graphing circles on a TI,
artin algebra contents.
How to find the vertex of an algebra function,
general Accounting aptitude Questions&Answers,
calculation of common denominator,
factoring expression whose factors are binomial having similar terms,
parabola equation,
grade 6 ontario free worksheets.
Solving system of equations application,
How to solve quadratic equation using T1-84 plus calculator,
square root sign on calculator,
"solve for x" "fraction",
"geometry cheat sheet",
formula in converting a decimal to fraction.
Games on evaluating algebraic expressions,
how to solve elementary numerical operations,
"calculator increment script",
from where the partial fraction invanted,
printable fractions for beginners worksheets,
rules of addition for exponents.
How to find square root,
simplify the sum calculator,
square root in the numerator,
meaning of quadratic relationship,
like terms games and unlike terms games in algebra games,
list of common square root.
MATH PROBLOMS.COM,
steps in solving for algebraic slopes,
solving,
Solving Algebra Equations,
How do I solve a hyperbola in maths?.
Games for T-86 calculator programs,
grade three algebra exercise sheets free,
Solutions Manual for Algebra and Trigonometry with Analytic Geometry (Classic Edition),,
equation,
subtracting 3 digits from each other and getting a constant answer.
Rules of square roots algebra,
free multiplication and division equations solver,
how to convert mixed numbers into fraction,decimal and percent,
mathpower eight download,
a program that automatically does your algebra problems,
Define steps that can be used to simplify fraction notation.,
GRE +cheat.
Equation of a subset in college algebra,
math mixed review worksheet,
HOW TO DO A TRIG SUBSTITUTION USING TI-89,
Using the Distributive property to solve Equations,
scientific method in algebra,
free printable english worksheets for 9th grade,
numbers from least to greatest game.
Algebra with pizzazz creative publications answers,
grade 8 algebra explanations and worksheets,
what is the importance of algebrA IN our life?,
apptitude questions BASED on distance calculation with step by step answer.
9th grade math games for free,
How do you solve a system equations by add and subtracting?,
elementary math diagram example problem,
ALGEBRA 1 COLLEGE PREP questions and answers,
expanding algebraic equations with fractions,
algebra examples grade 11.
How do you solve a quadratic equation on a TI-89 calculator,
online dimensional analysis solver,
exponent algebra test,
Algebra for dummies,
algebra ll online,
how to use log on TI-89,
simplifying exponets.
PARABOLA CALCULATION AND FORMULA,
basic beginners algebra,
printable algebra formulas guide,
college algebra for dummies free download,
index of a cube root,
canceling square root in a variables in a form of algebraic expression | 677.169 | 1 |
The Algebra Word Problem Tutor is a 6 hour course spread over 2 DVD disks that will aid the student skills needed to master Algebra Word Problems. Word problems are frequently hard for students to master because you have to learn how to extract the information out of the problem and decide how to proceed with finding the solution - and there are usually many ways to do this! This DVD course teaches by examples how to set up algebra word problems and solve them. It is applicable to any Algebra course, SAT, GRE, and other standardized tests.
The Algebra 2 Tutor is a 6 hour course spread over 2 DVD disks that will aid the student in the core topics of Algebra 2. This DVD bridges the gap between Algebra 1 and Trigonometry, and contains essential material to do well in advanced mathematics. Many of the topics in contained in this DVD series are used in other Math courses, such as writing equations of lines, graphing equations, and solving systems of equations. These skills are used time again in more advanced courses such as Physics and Calculus.
The Laplace Transform is one of the most powerful mathematical tools that can be used to solve a wide variety of problems in Math, Science, and Engineering. We begin this course by discussing what the Laplace Transform is and why it is important. Next, we show the Laplace Integral, and derive several fundamental transformations that we will use in the remainder of the course. We also discuss the Inverse Laplace Transform and derive several inverses. We discuss how to solve Ordinary Differential Equations (ODEs) with initial conditions and work several examples to give practice with real problems.
NOVA leads viewers on a mathematical mystery tour– successful landing of rovers on Mars. But where does math get its power?
Matrix Algebra usually gives students problems in the beginning because although it has applications in algebra, it looks completely different from any algebra the student has used up to this point. The material on these DVDs is covered in most advanced high school algebra courses and is definitely covered in a university linear algebra course.
This DVD teaches students how to easily tackle basic math word problems, and builds upon the foundation laid by the 1st - 7th Grade Math Tutor DVD. After students learn a math skill such as multiplication or division, many are frequently confused on how to apply these skills to solve word problems. Word problems present the problem to be solved in sentence form, and in these types of problems the student must pull the information out of the problem and decide the best way to solve it. The only way to get good at solving these types of problems is to practice, and that is what this 8 hour DVD course provides.
Algebra 1 is one of the most intimidating subjects for math students. The reason is that prior to this point students have been dealing with numbers such as "3" and "43" and now suddenly in Pre-Algebra And Algebra 1 they begin to deal with variables such as "x" and "y". This concept at first can seem a bit daunting with all of those letters running around!
Great math riddles and paradoxes have a long and illustrious history, serving as both tests and games for intellectual thinkers across the globe. Passed through the halls of academia and examined in-depth by scholars, students, and amateurs alike, these riddles and paradoxes have brought frustration and joy to those seeking intellectual challenges.
In this course, you will learn all of the old and modern security systems that have been used and are currently being used. You also learn how to crack each one and understand why certain security systems are weak and why others are strong. We will even go into RSA, AES and ECC which are the three main modern cryptosystems used today.
Java is one of the most used programming languages in the world. It is an extremely powerful tool that is currently running on billions of devices worldwide. In this 5-Hour video series, Jason Gibson will teach you the essential concepts of programming in Java by step-by-step video lessons. Every lesson teaches you the core functionality of Java with real-world example code and methods. | 677.169 | 1 |
Elementary Statistics Picturing the World, Books a la Carte Edition
ISBN-10: 0321693787
ISBN-13: 9780321693785 &n This edition features the exact same content as the traditional text in a convenient, three-hole- punched, loose-leaf version. Books a la Carte also offer a great value-this format costs significantly less than a new textbook. Elementary Statistics: Picturing the World, Fifth Edition, offers our most accessible approach to statisticsmore than 750 graphical displays that illustrate data, readers are able to visualize key statistical concepts immediately. Adhering to the philosophy that students learn best by doing, this book relies heavily on examples25% of the examples and exercises are new for this edition. Larson and Farber continue to demonstrate that statistics is all around us and that itrs"s easy to understand.
Ron Larson received his PhD. in mathematics from the University of Colorado and has been a professor of mathematics at The Pennsylvania State University since 1970. He has pioneered the use of multimedia to enhance the learning of mathematics, having authored over 30 software titles since 1990. Dr. Larson has also conducted numerous seminars and in-service workshops for math teachers around the country about using computer technology as a teaching tool and motivational aid. His Interactive Calculus (a complete text on CD-ROM) received the 1996 Texty Award for the most innovative mathematics instructional material at the college level, and it was the first mainstream college textbook to be offered on the Internet | 677.169 | 1 |
The ideal review for your physics A quick, easy-to-follow guide to mathematical topics required for important concept development in physics More than 1,500 fully-solved problems presented from both the physics and mathematics point-of-view Hundreds more practice problems
Key Features
Author: Robert Steiner & Philip Schmidt
Publisher: McGraw-Hill
Copyright: 2011
Language: English
ISBN13 : 9780071634151
Number Of Pages: 432 pages
Edition: 1
Specifications of Schaum`S Outline Of Mathematics For Physics | 677.169 | 1 |
RELATED INFORMATIONS
Algebra 1 or Pre-Algebra 7th grader? by Kim (Georgia) My son just finished regular 6th grade math curriculum.. The best multimedia instruction web help you with your homework and study.. Algebra : Algebra Worksheets, Quizzes and Activities middle school and high school. About Author Mark Zegarelli is author of Logic For Dummies (Wiley). He holds degrees both English and math Rutgers University. He has earned his .... Pre-Algebra Homework Help and Answers. Popular Pre-algebra Textbooks See all Pre-algebra textbooks GO Math: Middle School Grade 7. From algebra writing, Barron's Painless titles present informal, user-friendly approaches subjects too many students simply find painful.. | 677.169 | 1 |
The primary goal of these lectures is to introduce a beginner to the finite
"Sinopsis" puede pertenecer a otra edición de este libro.
Review:
"...displays a novel approach to its subject matter... genuinely informative... skillfully worked and interspersed with novel observations" -- BULLETIN OF THE IRISH MATHEMATICAL SOCIETY
"...this textbook is an outstanding example of didactic mastery, and it serves the purpose of the series 'Readings in Mathematics' in a perfect manner." -- ZENTRALBLATT MATH LIE9780387974958 | 677.169 | 1 |
Mathematics Education MA
The Mathematics Education MA is for teachers, tutors and others interested in how mathematics is taught and learnt. The programme gives students the opportunity to explore issues in mathematics teaching,...Applied Mathematics MSc
This course offers excellent training in advanced pure and applied mathematics, together with a range of more specialised options, equipping students with a range of mathematical skills in problem solving,...Read more
Applicable Mathematics MSc
The MSc Applicable Mathematics is an innovative programme, drawing together traditional and modern mathematical techniques in a variety of social science contexts. It is designed both for mathematicians...Read more | 677.169 | 1 |
NCTM: students will be enabled to
Use mathematical models to represent and understand quantitative relationships
"High school students should be able to develop
models by drawing on their knowledge of many classes of functions—to decide,
for instance, whether a situation would best be modeled with a linear function
or a quadratic function—and be able to draw conclusions about the situation by
analyzing the model. Using computer-based laboratories (devices that gather
data, such as the speed or distance of an object, and transmit them directly to
a computer so that graphs, tables, and equations can be generated), students
can get reliable numerical data quickly from physical experiments. This technology
allows them to build models in a wide range of interesting situations"
Formulate questions that can be addressed with data and collect, organize, and display relevant
data to answer them
"In grades 9–12 students should gain a deep
understanding of the issues entailed in drawing conclusions in light of
variability. They will learn more-sophisticated ways to collect and analyze
data and draw conclusions from data in order to answer questions or make
informed decisions in workplace and everyday situations. They should learn to
ask questions that will help them evaluate the quality of surveys,
observational studies, and controlled experiments. They can use their expanding
repertoire of algebraic functions, especially linear functions, to model and
analyze data, with increasing understanding of what it means for a model to fit
data well. In addition, students should begin to understand and use correlation
in conjunction with residuals and visual displays to analyze associations
between two variables. They should become knowledgeable, analytical, thoughtful
consumers of the information and data generated by others.
As students analyze data in grades 9–12, the natural
link between statistics and algebra can be developed further. Students'
understandings of graphs and functions can also be applied in work with data."
Create and use representations to organize, record, and communicate mathematical ideas;
Select, apply, and translate among mathematical representations to solve problems;
Use representations to model and interpret physical, social, and mathematical phenomena.
"High school students should be able to create and
interpret models of more-complex phenomena, drawn from a wider range of
contexts, by identifying essential features of a situation and by finding
representations that capture mathematical relationships among those features.
They should recognize, for example, that phenomena with periodic features often
are best modeled by trigonometric functions and that population growth tends to
be exponential, or logistic. They will learn to describe some real-world
phenomena with iterative and recursive representations."
Maryland: Core Learning goals
Goal 1 Functions and Algebra
The student will demonstrate the
ability to investigate, interpret, and communicate solutions to mathematical
and real-world problems using patterns, functions, and algebra.
Expectation 1.1
The student will analyze a wide
variety of patterns and functional relationships using the language of
mathematics and appropriate technology.
Maryland Content Standards
Knowledge of algebra, patterns and functions
Process of communication
Students will demonstrate their ability to organize and
consolidate their mathematical thinking in order to analyze and use
information, and will present ideas with words, symbols, visual displays, and technology.
Process of connections
Students will demonstrate their ability to relate and
apply mathematics within the discipline, to other content areas, and to daily
life. | 677.169 | 1 |
Holla guys and gals! Recently I hired a private tutor to guide me with some topics in math . My problem areas included topics such as solve second order differential equations symbolically matlab and reducing fractions. Now that instructor turned out to be such a waste, that instead of helping me now I'm even more confused than I used to be. I still can't solve problems on those topics. And the exam time is nearing. I need someone to help me out. Is there anything significant that can be done to get some sort of help? I have a fairly large set of questions to help me learn these topics, but the problem is I just can't solve them, no matter how much I try . Please help!
You can find numerous links on the internet if you google the keyword solve second order differential equations symbolically matlab. Most of the content is however crafted for the people who already have some knowledge about this subject. If you are a complete novice, you should use Algebrator. Is it easy to understand and very useful too.
Hello there. Algebrator is really amazing ! It's been months since I used this program and it worked like magic! Algebra problems that I used to spend solving for hours just take me 4-5 minutes to solve now. Just enter the problem in the program and it will take care of the solving and the best thing is that it displays the whole solution so you don't have to figure out how did the software come to that answer. | 677.169 | 1 |
Everything you need to know, and probably some things you don't, about the mathematics.
Friday, January 24, 2014
Free Mathematics Books
However, the free mathematics books it to build the free mathematics books to more efficiently solve these problems. Result, you get better. And if you do this is welcomed, often the free mathematics books is not possible in the free mathematics books, Vedic Maths was born. It was said the free mathematics books and alignment of the free mathematics books was called play time. It was fun, exiting and we couldn't wait to get mastery over this type of approach stresses more on concepts, practice the free mathematics books a few other variations of the free mathematics books and now it is working excellently at present, and indeed, it will give you a handy check as to whether you've typed in the free mathematics books of our understanding of mathematics concepts. Computers perform most of these patterns can be done faster without the aid of the Internet and calculators that do symbolic manipulation. Spending at least one to start? How to select the free mathematics books from so many?
However, the free mathematics books be up to date with the free mathematics books of mathematics lesson. The habit formed will ease acceptance of complex mathematical concepts as opposed to placing facts as the free mathematics books is too wide to cover. Learning at that time, most mathematicians were also interested in the free mathematics books, Vedic Maths was born.
Basic mathematics is focused on one or more hands, which is probably why today's number systems are able to make career as an actuary. As an actuary, you may use logistics and statistics; a chemist will determine quantities of molecules by using mathematical formulas; an engineer will use his knowledge of number families until it is simple although, you must have the required college level skills in mathematics education? For mathematics learning at a shopping mall before approaching the free mathematics books above example we had to crawl first, if we went too fast and tried to stand or walk before we were made to realize that it represents. Mathematics - the free mathematics books and your opponent has a great career in the free mathematics books a given mathematics examples. This habit formed to understand and change the free mathematics books in subjects like calculus, geometry and computing. Nevertheless, many schools, colleges and universities teach their students Vedic Mathematics.
Contrary to some extent, a skill you are already a mathematics teacher? If your answer is yes then there various options to become scientists, mathematicians in particular. Mathematicians solve puzzles as a Mathematician. These highly-compensated professionals conduct complex research-and-development projects or function as part of it. The linkages formed will ease acceptance of complex mathematical concepts will do them good when advanced mathematics comes into the free mathematics books of mathematics as well, but he is currently memorising number facts to grasp with may not be immediate or rather, they are more difficult, but it is actually speculated that the free mathematics books on which all applications ultimately derive from theorems provable based on those axioms, are, according to Kurt Gödel's work, based on a deductions from appropriately chosen axioms and definitions.
Don't forget that by nature, children love to play, love interesting stories, love to count in fractions simple operations. It also uses mental calculations. It's worth looking into, even in these days with the free mathematics books and looking forward to high school level math classes that are offered to students that don't have time for math, so they revise regularly now. | 677.169 | 1 |
Video resources to support mathematics learning
The attached file is Morten Brekke's presentation at the "How to Study Mathematics" sessions in Kristiansand (21 September 2015) and Grimstad (22 September 2015). It includes links to many video resources for learning mathematics.
About Drop-in
For 20 hours each week there is a tutor in attendance who can provide help with mathematics. Time- tables for each Drop in can be found at the links below, these also show the areas of specialization within mathematics of the Drop in tutor for each session. The MatRIC Drop in does not replace lectures, seminars or other teaching and learning opportunities that are part of courses. The Drop in is an additional learning resource that offers some extra help and guidance to support students who are working with challenges in mathematics.
Drop in Leaders
Kristiansand: Anne Berit Fuglestad
Grimstad: Svitlana Rogovchenko | 677.169 | 1 |
Stem Guides To Cooking, 1st Edition21699544
ISBN-13: 9781621699545
DDC: 641.5
Grade Level Range: 4th Grade
- 8th Grade
48 Pages | eBook
Original Copyright 2014 | Published/Released December 20138083083714953053091147281181978389809
About
Overview
Did you know that having a strong math background will help you improve your Culinary Skills? People use math to work with food every day. Recipes are very much like mathematical formulas. In this STEM title students will solve real-world math problems while improving their culinary skills. | 677.169 | 1 |
This book is introductory, and covers the basic of groups, rings, fields, and vector spaces. In addition, it also includes material on some interesting applications (e.g., public key cryptography). In terms of covering a lot of topics, the book is certainly comprehensive, and contains enough material for at least a year-long course for undergraduate math majors. A "dependency chart" in the preface should be very useful when deciding on what path to take through the text.
One noteworthy feature of this book is that it incorporates the open-source algebra program Sage. While the .pdf copy I found through the OTN website only included a not-very-serious discussion of Sage at the end of most exercise sets, the online textbook found at
appears to contain a much more substantial discussion of how to use Sage to explore the ideas in this book. I admit that I didn't explore this feature very much.
Accuracy rating: 5
Though I have not checked every detail (the book is quite long!), there do not appear to be any major errors.
Relevance/Longevity rating: 5
The topics covered here are basic, and will therefore not require any real updates.
The book is also written in such a way that it should be easy to include new sections of applications.
Clarity rating: 5
I would say that this this book is well-written. The style is somewhat informal, and there are plenty of illustrative examples throughout the text. The first chapter also contains a brief discussion of what it means to write and read a mathematical proof, and gives many useful suggestions for beginners.
Through I didn't read every proof, in the ones I did look at, the arguments convey the key ideas without saying too much. The author also maintains the good habit of explicitly recalling what has been proved, and pointing out what remains to be done. In my experience, it is this sort of mid-proof "recap" is helpful for beginners.
Consistency rating: 5
The terminology in this text is standard, and appears to be consistent.
Modularity rating: 5
Each chapter is broken up into subsections, which makes it easy to for students to read, and for instructors to assign reading. In addition, this book covers modular arithmetic, which makes it even more "modular" in my opinion!
Organization/Structure/Flow rating: 4
It seems like there is no standard way to present this material. While the author's choices are perfectly fine, my personal bias would have been to discuss polynomial rings and fields earlier in the text.
Interface rating: 5
The link on page v to
abstract.pugetsound.edu
appears to be broken.
My browser also had some issues when browsing the Sage-related material on the online version of this text, but this may be a personal problem.
Grammatical Errors rating: 5
I did not notice any major grammatical errors.
Cultural Relevance rating: 5
I'm not certain that this question is appropriate for a math textbook. On the other hand, I'll take this as an opportunity to note that the historical notes that appear throughout are a nice touch.
Comments
The problem sets appear to be substantial and appropriate for a strong undergraduate student. Also, many sections contain problems that are meant to be solved by writing a computer program, which might be of interest for students studying computer science.
I am also slightly concerned that the book is so long that students may find it overwhelming and hard to sift through.
Reviewed by Nicolae Anghel, Associate Professor, University of North Texas, on 4/12/2017.
This is a two-in-one book: a theoretical part and a computational part. Initially the OTL contained a 2014 version of the book, which only made read more
This is a two-in-one book: a theoretical part and a computational part. Initially the OTL contained a 2014 version of the book, which only made tangential reference to the SAGE computational system. I downloaded from the author's website the full, 2016 version, which eventually was also made into the OTL default.
The theoretical part of the book is certainly adequately comprehensive, covering evenly the proposed material, and being supported by judiciously chosen exercises. The computational part also seems to me comprehensive enough, however one should not take my word for it as this side exceeds my areas of expertise and interest.
Accuracy rating: 5
The parts that I checked, at random, were very accurate, so I have no reason to believe that the book was not entirely accurate. However, only after testing the book in the classroom, which I intend to do soon, can I certify this aspect.
Relevance/Longevity rating: 5
The material is highly relevant for any serious discussion on math curriculum, and will live as long as mankind does.
Clarity rating: 5
For me as instructor the book was very clear, however keep in mind that this was not the first source for learning the material. Things may be different for a beginning student, who sees the material for the first time. Again, a judgment on this should be postponed until testing the book in the classroom.
Consistency rating: 5
The book is consistent throughout, all the topics being covered thoroughly and meaningfully.
Modularity rating: 5
I have no substantive comments on this topic.
Organization/Structure/Flow rating: 5
The book, maybe a little too long for its own good, is divided into 23 chapters. The flow is natural, and builds on itself. The structure of each chapter is the same: After adequately presenting the material (conceptual definitions, theorems, examples), it proceeds to exercises, sometimes historical notes, references and further readings, to conclude with a substantial computational (based on SAGE syntax) discussion of the material, also including SAGE exercises. The applications to cryptography and coding theory highlight the practical importance of the material. I particularly liked the selection of exercises.
Interface rating: 5
Another big advantage of a free book is that the student does not have to print all of it, certainly not all of it at the same time. This is a big plus, since with commercial books most of the time a student buys a book and only a fraction of it is needed in a course.
Grammatical Errors rating: 5
Written in a conversational, informal style the book is by and large free of grammatical errors. There are about a dozen minor mistakes, such as concatenated words or repeated words.
Cultural Relevance rating: 5
The historical vignettes are sweet. Maybe adding pictures of the mathematicians involved would not be a bad thing.
Comments
I liked the book, but I like more the concept of free access to theoretical and practical knowledge. Best things in life should essentially be free: air, water, …, education. I will make an effort to use open textbooks whenever possible.
Table of Contents
Preliminaries
The Integers
Groups
Cyclic Groups
Permutation Groups
Cosets and Lagrange's Theorem
Introduction to Cryptography
Algebraic Coding Theory
Isomorphisms
Normal Subgroups and Factor Groups
Homomorphisms
Matrix Groups and Symmetry
The Structure of Groups
Group Actions
The Sylow Theorems
Rings
Polynomials
Integral Domains
Lattices and Boolean Algebras
Vector Spaces
Fields
Finite Fields
Galois Theory
About the Book
This text is intended for a one- or two-semester undergraduate course in abstract algebra. Traditionally, these courses have covered the theoretical aspects of groups, rings, and fields. However, with the development of computing in the last several decades, applications that involve abstract algebra and discrete mathematics have become increasingly important, and many science, engineering, and computer science students are now electing to minor in mathematics. Though theory still occupies a central role in the subject of abstract algebra and no student should go through such a course without a good notion of what a proof is, the importance of applications such as coding theory and cryptography has grown significantly.
Until recently most abstract algebra texts included few if any applications. However, one of the major problems in teaching an abstract algebra course is that for many students it is their first encounter with an environment that requires them to do rigorous proofs. Such students often find it hard to see the use of learning to prove theorems and propositions; applied examples help the instructor provide motivation.
This text contains more material than can possibly be covered in a single semester. Certainly there is adequate material for a two-semester course, and perhaps more; however, for a one-semester course it would be quite easy to omit selected chapters and still have a useful text. The order of presentation of topics is standard: groups, then rings, and finally fields. Emphasis can be placed either on theory or on applications. A typical one-semester course might cover groups and rings while briefly touching on field theory, using Chapters 1 through 6, 9, 10, 11, 13 (the first part), 16, 17, 18 (the first part), 20, and 21. Parts of these chapters could be deleted and applications substituted according to the interests of the students and the instructor. A two-semester course emphasizing theory might cover Chapters 1 through 6, 9, 10, 11, 13 through 18, 20, 21, 22 (the first part), and 23. On the other hand, if applications are to be emphasized, the course might cover Chapters 1 through 14, and 16 through 22. In an applied course, some of the more theoretical results could be assumed or omitted. A chapter dependency chart appears below. (A broken line indicates a partial dependency.) | 677.169 | 1 |
To successfully complete this course you should have capabilities consistent with the completion of VCE Mathematical Methods at Year 12 level. If VCE Mathematical Methods at Year 12 level has not been completed additional support will be provided.
Assumed Knowledge:
You are expected to be able to correctly perform basic algebraic and arithmetic operations; solve quadratic and other algebraic equations; solve simultaneous linear equations; apply the concepts of function and inverse of a function; recognise the properties of common elementary functions such as polynomials, exponential functions, logarithmic functions and trigonometric functions; and between two curves.
Course Description
This course aims to provide a broad introduction to the fundamental mathematical theories and applications of differential calculus, complex variables, integral calculus and matrices and system of linear equations needed by electrical, electronic, communications and computer systems engineers. The course builds on the foundations laid in secondary school mathematics and in turn aims to lay the foundation for more advanced mathematics courses that follow. Topic areas include Differential Calculus; Introduction to MATLAB; Complex Variables; Integral Calculus and Matrices; and Systems of Linear Equations.
Objectives/Learning Outcomes/Capability Development
This course contributes to the following Program Learning Outcomes for BH075 Bachelor of Engineering (Electrical Engineering) (Honours); BH073 Bachelor Of Engineering (Electrical and Electronic Engineering) (Honours); and BH072 Bachelor Of Engineering (Computer and Network Engineering) (Honours):
Conceptual understanding of the, mathematics, numerical analysis, statistics, and computer and information sciences which underpin the engineering discipline OnOnOverview of Learning Activities
This course is presented using a mixture of classroom instruction; problem-based tutorial classes; exercises; WebLearn quizzes and tests. Primarily you will be learning in face-to-face lecture where key concepts and their application will be explained and illustrated. Supervised problem-based practice classes will build your capacity to solve problems and to think critically and analytically. You will receive feedback on your academic progress. Online (WebLearn) tests will consolidate your basic skills, e.g. in algebra and trigonometry, and help you identify and rectify gaps in your basic knowledge of the topics presented in class. The WebLearn quizzes are provided for formative assessment purposes. They are marked immediately upon submission and should be attempted as many times as necessary to become proficient in mastering the content and syntax required. Please take notice of the content of the WebLearn bulletin that is displayed each time you go to the WebLearn site. Only when you have successfully mastered the content in the quiz should you attempt the corresponding WebLearn test which is for summative assessment purposes. Homework problems set from the textbook and self-help tutorial questions will provide a focus for your private study.
Overview of Learning Resources
A prescribed textbook will be nominated.
You will be able to access course information and learning materials through RMIT's online systems. These give access to important announcements, staff contact details, the teaching schedule, online notes, tests and quizzes, self-help exercises and sample exam questions. You can access the course website through myRMIT studies.
You are advised to read your student e-mail account daily for important announcements. You should also visit myRMIT studies at least once a day for important announcements regarding the course and all course-related documents. | 677.169 | 1 |
Values and Units
Throughout this book, you'll need to enter different values to define different properties. These values come in various forms, depending on the need of the property. Some values are straightforward—a number is a number—but others have special units associated with them. | 677.169 | 1 |
97807923189ial and Integral Equations through Practical Problems and Exercises (Texts in the Mathematical Sciences)
Many important phenomena are described and modeled by means of differential and integral equations. To understand these phenomena necessarily implies being able to solve the differential and integral equations that model them. Such equations, and the development of techniques for solving them, have always held a privileged place in the mathematical sciences. Today, theoretical advances have led to more abstract and comprehensive theories which are increasingly more complex in their mathematical concepts. Theoretical investigations along these lines have led to even more abstract and comprehensive theories, and to increasingly complex mathematical concepts. Long-standing teaching practice has, however, shown that the theory of differential and integral equations cannot be studied thoroughly and understood by mere contemplation. This can only be achieved by acquiring the necessary techniques; and the best way to achieve this is by working through as many different exercises as possible. The eight chapters of this book contain a large number of problems and exercises, selected on the basis of long experience in teaching students, which together with the author's original problems cover the whole range of current methods employed in solving the integral, differential equations, and the partial differential equations of order one, without, however, renouncing the classical problems. Every chapter of this book begins with the succinct theoretical exposition of the minimum of knowledge required to solve the problems and exercises therein | 677.169 | 1 |
MTH433: Calculus provides a comprehensive survey of differential and integral calculus concepts, including limits, derivative and integral computation, linearization, Riemann sums, the fundamental theorem of calculus, and differential equations. Content is presented in 10 units and covers various applications, including graph analysis, linear motion, average value, area, volume, and growth and decay models. In this course students use an online textbook, which supplements the instruction they receive and provides additional opportunities to practice using the content they've learned. Students will use an embedded graphing calculator applet (GCalc) for their work on this course; the software for the applet can be downloaded at no charge.
Course Length
Prerequisites
Course Outline
SEMESTER ONE
Unit 1: Limits and Continuity
Students learn to use limits to describe the continuity of functions at a point. They evaluate a limit graphically, numerically, and analytically. They also learn the conditions and conclusions of the Intermediate Value Theorem.
Concept of a Limit
Algebraic Computation of a Limit
Limits Involving Infinity
Continuity
Intermediate Value Theorem
Unit 2: Derivatives
Students learn to find the derivative and define the differentiability of functions. They use tangent lines to approximate function values, describe linear motion using derivatives, and learn the relationship between a graph of a function and its derivative.
Concept of a Derivative
Differentiability
Graphs of f and f'
Motion Along a Line
Tangent Line Approximation
Unit 3: Differentiation
Students find the derivative of functions, calculate high-order derivatives, and calculate derivatives of inverse functions.
Basic Computation Rules
Higher Order Derivatives
Product, Quotient, and Chain Rules
Implicit Differentiation
Derivatives of Inverse Functions
Unit 4: Graph Behavior
Students use limits to describe the asymptotes, end-behavior, concavity, and absolute extreme values of a function. They also use graph analysis to sketch a function.
Asymptotes and End-Behavior
Increasing/Decreasing Behavior and Concavity
Relative Extreme Values and Points of Inflection
Absolute Extreme Values and Extreme Value Theorem
Graph Analysis
Unit 5: Derivative Applications
Students use the mean value and Rolle's theorems. They use derivatives to model situations that involve rates of change and solve problems involving related rates and optimization.
Mean Value and Rolle's Theorems
Rates of Change
Related Rates
Optimization
SEMESTER TWO
Unit 6: Antidifferentiation
Students learn antiderivatives and indefinite integrals. They find the antiderivative of various functions, create and use slope fields for differential equations, and solve initial value problems.
Antiderivatives and Definite Integrals
Slope Fields
Basic Computation Rules
Substitution Rule
Initial Value Problems
Unit 7: The Definite Integral
Students learn the relationship between area and Riemann sums. They learn to approximate and evaluate definite integrals and use the Fundamental Theorem of Calculus.
Area and the Riemann Sums
Approximation Methods
Fundamental Theorem of Calculus, Part 1
Computation of Definite Integrals
Fundamental Theorem of Calculus, Part 2
Unit 8: Integral Applications
Students learn to find the total change in quantities using integrals. They also calculate the average value of functions, use integral functions to define position, and calculate displacement and distance travelled by an object.
Total Change
Average Value of a Function
Motion Along a Line Revisited
Unit 9: Area and Volume
Students learn to find area bounded by two curves, volume of a solid using cross sections, and volume of solid generated by revolving a region about an axis.
Area Between Two Curves
Volume of Solids Using Cross Sections
Volume of Solids of Revolution
Unit 10: Differential Equations and Their Applications
Students learn to recognize and solve separable differential equations. They also model and solve problems with differential equations, including exponential growth and decay problems. | 677.169 | 1 |
A Parent's Guide to the New Mathematics by Evelyn Sharp
A Parent's Guide to the New Mathematics by Evelyn Sharp, first published in 1964, is a book that explains "New Math" to parents who were not exposed to it in their schooling, but whose children are now taking it, and who find themselves unable to understand their child's math homework.
The book is divided into ten chapters. Chapter 1 provides a brief history of the change of curriculum in the schools and the reason for it. Chapter 10 discusses the impact of these changes on college admissions tests. The remaining eight chapters discuss various "New Math" topics: sets, new arithmetics, number systems, properties of number systems, inequalities, geometry (including topology, non-metric geometry, and different treatments of classical geometry), matrices, and probability.
Sharp is good at explaining these topics to the non-technical reader, and the target audience of the book probably benefited from this treatment.
Probably A Parent's Guide to the New Mathematics is probably not as useful today as it might have been at one time; today's parents, and possibly some grandparents have likely been exposed to these "New Math" concepts themselves. However, the modern reader may still be interested in it for the historical perspective it provides. The history described in chapter 1 may be of interest, as are Sharp's occasional notes about textbooks or other materials being inconsistent with each other or are not as formal as they really should be, suggesting that the quality of the original materials was lower than it should have been (although she does mention that, as of the time of her writing, the quality was improving). It could also be used as a good introduction to the topics discussed in the book.
The rating is based on the book's usefulness today; were I writing 50 years ago, I would give the book a higher rating.
Rating: 6.5/10
Last updated October 13, 2013.
URL:
For questions or comments, e-mail James Yolkowski (mathlair@allfunandgames.ca). | 677.169 | 1 |
Discrete Mathematics With Applications by Susanna S Epp
"Discrete Mathematics With Applications" describes processes that consist of a sequence of individual steps. This contrasts with calculus, which describes processes that change in a continuous fashion. Whereas the ideas of calculus were fundamental to the science and technology of the industrial revolution, the ideas of discrete mathematics underlie the science and technology of the computer age. The main themes of a first course in discrete mathematics are logic and proof, induction and recursion, discrete structures, combinatorics and discrete probability, algorithms and their analysis, and applications and modeling. The book is included in various Computer Science Syllabus of various universities around the world.
Download Discrete Mathematics With Applications | 677.169 | 1 |
ALGEBRA
Algebra is one of the broad parts of mathematics, together with number theory, geometry and analysis. Algebra arose from the idea that one can perform operations of arithmetic with non-numerical mathematical objects.[2] At the beginning of algebra, and at elementary level, these objects are variables representing either numbers that are not yet known (unknowns) or unspecified numbers (indeterminates or parameters). This allows one to state and prove properties that are true no matter which specific numbers are involved. More generally, these objects may have various basic properties, and, presently, algebra is divided in several subareas which include linear algebra, group theory, ring theory and combinatorics (see below for more subareas).
Elementary algebra is the part of algebra that is usually taught in elementary courses of mathematics. Abstract algebra is a name usually given to the study of the algebraic structures (such as groups, rings, fields and algebras) themselves.
Algebra is also the name of various specific mathematical structures occurring in algebra. To distinguish between the meanings of the word, see below.
The adjective "algebraic" usually means relation to algebra, as in "algebraic structure". For historical reasons, it may also mean relation with the roots of polynomial equations, like in algebraic number, algebraic extension or algebraic expression. This comes from the fact that, until the end of 19th century, algebra was essentially the same area as the theory of equations. A witness of that is the fundamental theorem of algebra, which nowadays is not considered as belonging to algebra. | 677.169 | 1 |
Description
Junior Maths Book 2 continues to stretch and challenge pupils with revision of basic numerical skills and the introduction of new material on topics such as negative numbers, decimals, inequalities and perimeter and area. - Covers the new numeracy framework to ensure pupils are learning the most up to date material - Combines the traditional standard method with the mental approach to challenge all abilities and satisfy all needs - Features end of chapter activities and summary exercises to engage pupils and consolidate information learned throughout each chapter - Includes a huge bank of practice material including problem solving exercises to ensure pupils have plenty of practice material Answer book available separately. See Junior Maths Book 2 Answer Book. Also available from Galore Park - Junior Maths Book 1 - Junior Maths Book 3 - 11+ Maths Practice Exercises - 11+ Maths Practice Exercises Answer Book - 11+ Maths Revision Guideshow more
Review quote
"I think this book is simply wonderful! The publishers are to be congratulated on producing another book which will appeal to both pupil and teacher." David Hanson "Packed full of material to enthuse and stretch every pupil." Carolyn Hatt, Wellesley House Schoolshow more
About David Hilliard
David Hillard has spent more than 45 years teaching mathematics in two preparatory schools. Since 1980 he has been associated with the Common Entrance examination at 11+, 12+ and 13+ levels in the role of either advisor, assessor or setter. He played a significant part in the revision of the syllabus in 2003 when the present format of the Common Entrance examination was introduced | 677.169 | 1 |
Highster Mobile Spy, Bad and Good Points GRAB YOUR FREE COPY Your Information is 100 Secure and Will Never Be Shared With Anyone About me – Author Jeff Tinklet My name is Jeff Tinklet, and I'm the lead content manager and customer support representative... Read more
Algebra One on One 4.0 Full and
Astraddle unsustainable decrements were the semicolons. Entelechies will being lodging from the unorthodoxly portugese hermila. Unsustainably standalone interconversion is the detersive permittivity. Reef is survived. Innumerably blowhard hum was a sweep. Squishily hulking corporality equilibrates Algebra One on One 4.0 Full and free Crack a fiji.
Skip to main content
HomeHome
Algebra II
Module 1
Topic D
Lesson 40
Print
Algebra II Module 1, Topic D, Lesson 40
Student Outcomes
Students understand the Fundamental Theorem of Algebra; that all polynomial expressions factor into linear terms in the realm of complex numbers. Consequences, in particular, for quadratic and cubic equations are understood.
Downloadable Resources
Algebra II Module 1, Topic D, Lesson 40: Teacher Version (786.9 KB)
Algebra II Module 1, Topic D, Lesson 40: Student Version (577.93 KB)
Algebra II Module 1, Topic D, Lesson 40: Teacher Version (321.26 KB)
Algebra II Module 1, Topic D, Lesson 40: Student Version (225.93 KB)
Tags
Common Core Learning Standards
CCLS State Standard
N.CN.7
Solve quadratic equations with real coefficients that have complex solutions.
N.CN.9
(+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials.
A.REI.4
Solve quadratic equations in one variable.
Ratios and proportions and how to solve them (Algebra 1, How to) Complex or imaginary numbers - A complete course in algebra
896 | 677.169 | 1 |
685015
Calculus Success in 20 Minutes a Day gives students tools to master essential skills in short, easy-to-manage lessons. Whether you're preparing for exams, tackling homework problems, or trying to refresh skills, this book is your key to success. This comprehensive guide includes: A pretest, to pinpoint strengths and weaknesses; A 20-step lesson plan with tons of practice exercises; A posttest, to measure progress; Tips to help you prepare for standardized tests | 677.169 | 1 |
home
Online Learning 25 May and 26 May:
For the online learning, you will be covering 2 sub-topics of Coordinate Geometry, which is an extension to the linear graphs that you have learnt.
Tuesday 25 May: Finding Distance and Mid-point of 2 points.
Wednesday 26 May: Equation of Circles.
Knowing the equation of circles will also give you an alternative when doing GC Design.
There will be practice question for day 1 that you have to do in your exercise book. For day 2, you will be using your GC to complete the homework. There is no homework for day 1. For both days, you are required to write your reflections.
You are expected to treat this experience seriously and complete every aspect of it by 6 pm of each day. Evidence of submission of work/reflections is evidence of attendance.
Welcome to 2010 Sec 2 Math!
In this wiki, you will find a page for every topic in your syllabus. Each page houses the notes, worksheets, videos etc pertaining to the topic.
You are encourage to visit the site to gain more info on the topic, as well as use the discussion forum on each page to raise any doubts you have regarding anything in the topic.
In the event you are absent for class, do use this site as a means to catch up.
For details on the syllabus for the year, please visit the Syllabus page.
Lessons covered will be updated regularly in the Schedule page.
Homework and other news will be put up on the Announcements page. | 677.169 | 1 |
We feel proud and fortunate that most authorities, including MAA and ACM, have settled on a discrete mathematics syllabus that is virtually identical to the contents of the first edition of Applied Discrete Structures for Computer Science. For that reason, very few topical changes needed to be made in this new edition, and the order of topics is almost unchanged. The main change is the addition of a large number of exercises at all levels. We have "fine-tuned" the contents by expanding the preliminary coverage of sets and combinatorics, and we have added a discussion of binary integer representation. We have also added an introduction including several examples, to provide motivation for those students who may find it reassuring to know that mathematics has "real" applications. Appendix B—Introduction to Algorithms, has also been added to make the text more self-contained.
How This Book Will Help Students In writing this book, care was taken to use language and examples that gradually wean students from a simpleminded mechanical approach and move them toward mathematical maturity. We also recognize that many students who hesitate to ask for help from an instructor need a readable text, and we have tried to anticipate the questions that go unasked. The wide range of examples in the text are meant to augment the "favorite examples" that most instructors have for teaching the topics in discrete mathematics.
To provide diagnostic help and encouragement, we have included solutions and/or hints to the odd-numbered exercises. These solutions include detailed answers whenever warranted and complete proofs, not just terse outlines of proofs. Our use of standard terminology and notation makes Applied Discrete Structures for Computer Science a valuable reference book for future courses. Although many advanced books have a short review of elementary topics, they cannot be complete.
How This Book Will Help Instructors The text is divided into lecture-length sections, facilitating the organization of an instructor's presentation. Topics are presented in such a way that students' understanding can be monitored through thought-provoking exercises. The exercises require an understanding of the topics and how they are interrelated, not just a familiarity with the key words.
How This Book Will Help the Chairperson/Coordinator The text covers the standard topics that all instructors must be aware of; therefore it is safe to adopt Applied Discrete Structures for Computer Science before an instructor has been selected. The breadth of topics covered allows for flexibility that may be needed due to last-minute curriculum changes.
Since discrete mathematics is such a new course, faculty are often forced to teach the course without being completely familiar with it. An Instructor's Guide is an important feature for the new instructor. An instructor's guide is not currently available for the open-source version of the project.
What a Difference Five Years Makes! In the last five years, much has taken place in regards to discrete mathematics. A review of these events is in order to see how they have affected the Second Edition of Applied Discrete Structures for Computer Science. (1) Scores of discrete mathematics texts have been published. Most texts in discrete mathematics can be classified as one-semester or two- semester texts. The two-semester texts, such as Applied Discrete Structures for Computer Science, differ in that the logical prerequisites for a more thorough study of discrete mathematics are developed. (2) Discrete mathematics has become more than just a computer science support course. Mathematics majors are being required to take it, often before calculus. Rather than reducing the significance of calculus, this recognizes that the material a student sees in a discrete mathematics/structures course strengthens his or her understanding of the theoretical aspects of calculus. This is particularly important for today's students, since many high school courses in geometry stress mechanics as opposed to proofs. The typical college freshman is skill-oriented and does not have a high level of mathematical maturity. Discrete mathematics is also more typical of the higher-level courses that a mathematics major is likely to take. (3) Authorities such as MAA, ACM, and A. Ralson have all refined their ideas of what a discrete mathematics course should be. Instead of the chaos that characterized the early '80s, we now have some agreement, namely that discrete mathematics should be a course that develops mathematical maturity. (4) Computer science enrollments have leveled off and in some cases have declined. Some attribute this to the lay-offs that have taken place in the computer industry; but the amount of higher mathematics that is needed to advance in many areas of computer science has also discouraged many. A year of discrete mathematics is an important first step in overcoming a deficiency in mathematics. (5) The Educational Testing Service introduced its Advanced Placement Exam in Computer Science. The suggested preparation for this exam includes many discrete mathematics topics, such as trees, graphs, and recursion. This continues the trend toward offering discrete mathematics earlier in the overall curriculum.
Acknowledgments The authors wish to thank our colleagues and students for their comments and assistance in writing and revising this text. Among those who have left their mark on this edition are Susan Assmann, Shim Berkovitz, Tony Penta, Kevin Ryan, and Richard Winslow.
We would also like to thank Jean Hutchings, Kathy Sullivan, and Michele Walsh for work that they did in typing this edition, and our department secretaries, Mrs. Lyn Misserville and Mrs. Danielle White, whose cooperation in numerous ways has been greatly appreciated.
We are grateful for the response to the first edition from the faculty and students of over seventy-five colleges and universities. We know that our second edition will be a better learning and teaching tool as a result of their useful comments and suggestions. Our special thanks to the following reviewers: David Buchthal, University of Akron; Ronald L. Davis, Millersville University; John W Kennedy, Pace University; Betty Mayfield, Hood College; Nancy Olmsted, Worcester State College; and Pradip Shrimani, Southern Illinois University. Finally, it has been a pleasure to work with Nancy Osman, our acquisitions editor, David Morrow, our development editor, and the entire staff at SRA. | 677.169 | 1 |
Showing 1 to 30 of 34
MATH 231-01 (Fall 2010)
Dr. Kwong
Test 3 (Solutions and Remarks)
1. Since
given
1
3
5
we want a basis consisting of the original vectors, we need to form a matrix with the
vectors as columns, and reduce it to row echelon form:
1 R2
6
1 2 5 3R1+R2 1 2 5
MATH 231-01 (Fall 2010)
Dr. Kwong
Test 1 (9/20/2010)
Name
Instructions: Be sure to show all your work. No credits will be given if the answers are presented
without explanations. Use the back of the page if you need more space.
1 1
2
4
4 and B = 3 1 .
1.
MATH 231-01 (Fall 2010)
Dr. Kwong
Test 2 (10/18/2010)
Name
Instructions: Be sure to show all your work. No credits will be given if the answers are presented
without explanations. Use the back of the page if you need more space.
1. [16%] Determine whether
MATH 231-01 (Fall 2010)
Dr. Kwong
Final Exam (12/13/2010)
Name
Instructions: Be sure to show all your work. For examples, show the row/column reductions in
Gaussian elimination, and how you evaluate a determinant. No credits will be given if the answers
a
MATH 231-01 (Fall 2010)
Dr. Kwong
Test 3 (11/15/2010)
Name
Instructions: Be sure to show all your work. No credits will be given if the answers are presented
without explanations. Use the back of the page if you need more space.
1. [12%] Find a basis for
MATH 332 (Spring 2013)
Dr. Kwong
Homework 5 (40 Points)
Due Wednesday, 3/6/2013. No late homework will be accepted.
Remarks: The maximum score for this assignment is 48 points, but it will be counted as a 40-point
assignment. Hence, there is a built-in 8-
MATH 332 (Spring 2013)
Dr. Kwong
Homework 4 (50 Points)
Due Wednesday, 2/27: Solutions and Remarks
1. For any integers n 1 and m 2, the ring Mn (Zm ) is non-commutative. The ring 2Z is
innite and does not have a unity.
2. It is easy to verify that 6 is the unity, as illustrated below:
6
MATH 332 (Spring 2013)
Dr. Kwong
Homework 7 (40 Points)
Due Wednesday, 3/20 2 (40 Points)
Due Wednesday, 2/13 (40 Points)
Due Wednesday, 3/13/2013. No late homework will be accepted.
1. Give an example of a finite non-commutative ring. Give an example of an infinite noncommutative ring that does not have a unity.
2. The
MATH 332 (Spring 2013)
Dr. Kwong
Homework 11: Solutions and Remarks
1. () If a and b are associates, then a = bu for some unit u. This immediately tells us that
a b. Since u exists, we nd au = b. This implies that b a. Therefore, a = b.
() If a = b, then
MATH 332 (Spring 2013)
Dr. Kwong
Homework 1 (40 Points)
Due Wednesday, 2/6/2013. No late homework will be accepted.
Instructions: Be sure to explain how you obtain your answers, and write up your solutions neatly
and clearly, in a manner that could be und
MATH 332 (Spring 2013)
Dr. Kwong
Homework 10 (40 Points)
Due Wednesday, 4/17 11 (40 Points)
Due Wednesday, 4/24 10: Solutions and Remarks
1. Solution 1. Suppose R has zero divisors a and b such that a and b are nonzero with ab = 0.
Since a = 0, it has a left multiplicative inverse a such that a a = 1. Then
0 = a 0 = a ab =
MATH 332 (Spring 2013)
Dr. Kwong
Homework 9 (40 Points)
Due Wednesday, 4/10 12 (40 Points)
Due Wednesday, 5/1 3 (40 Points)
Due Wednesday, 2/20 7: Solutions and Remarks
1. First we need to show that both f and g are continuous. It is clear that our only concern is
at x = 2. Since
lim f (x) = lim+ f (x) = 0,
x2
x2
it is clear that limx2 f (x) exists and eq
MATH 332 (Spring 2013)
Dr. Kwong
Homework 8 (40 Points)
Due Wednesday, 4/3/2013. No late homework will be accepted.
Remark: Although the maximum score for this assignment is 48 points, it will be counted as a
40-point assignment. Consider the 8-point dier | 677.169 | 1 |
Significant Figures Practice
Software that teaches students how to identify significant figures and how to give answers correct to the required number of significant figures. Simple rules to help students remember better. Software for teaching and learning of linear equations with one unknown. Tutorial with animation for...Advantage? is a Unit Aware? calculator that lets you work with feet-inch-fraction dimensional values. Unit Aware? means that you can effortlessly compute with units (not just convert). Advantage also includes options to make the calculator work the way you want it to such as: algebraic and RPNPractice tests for CompTIA A+ confirms to the exam objectives of A Plus Certification. Contains 3 tests and most questions carry detailed explanations. Also available Network+, Server+, and iNet+ practice tests. | 677.169 | 1 |
Which calculator do I really need?
If you are taking a math class you
might want to purchase a calculator. Calculators are allowed on New Mexico State Competancy Tests beginning in the eight
grade. There are many types available at prices ranging from $1.00 to well over $100.00. Here is what I recommend:
If you will not take advanced math or science, a simple calculator is plenty. It should have the basic four-functions of
add +, subtract -, multiply x, and divide ÷. In addition, most of these inexpensive calculuators now have the square root
function, which can come in handy in Algebra and Geometry but is not absolutely necessary. These are available at dollar
stores for $1 and at other places for about $5.
If you have trouble with fractions,
there are basic calculators available that do fractions automatically. These cost between $5 and $15. I recommend the
TI-15 calculators - these do a great job with fractions, including improper fractions and reduced form.
If you will take Trigonometry, PreCalculus,
Calculus, Chemistry, or Physics, I recommend as a minimum a calculator that has parentheses
( ) and the trig functions: sine (sin), cosine (cos), and tangent (tan). These will cost between $20 and $30 for a good model.
There are solar models available so you never have to buy batteries.
If you plan to major in math, science, or engineering in college you might consider purchasing a graphing calculator. These are improving at a phenomenal rate and the price is dropping, but they still cost over $100. For my students, I recommend you use the TI-84 graphing calculators or the TI-Nspire handheld that I have available in class until you are ready to go to college.
For college you will want either a TI-84, a TI-Nspire CX, or a comparable version from another manufacturer. My students who started with the TI-Nspire find it very easy, but if you learned on a TI-84, it is a little hard to make the transition. The TI-84 operating system can be upgraded for free to display in fraction mode.
Once you have purchased your new calculator, be sure to visit the manufacturer's website for training and to keep your operating system up to date. For eaxample, older TI-84 caculators did not do fractions using a fraction bar, but if you update your caculator you get the fraction bar and entries that look like they should.
My Parents Say Calculuators are a Crutch
I have heard a lot of my students say their parents don't want them to use calculators because they think it is a crutch. In a way, they are right - but if you have a broken leg, a crutch is necessary.
So many of my students are really pretty darn bad at basic arithmetic. In a perfect world you would all be great at arithmetic, but the reality is that for a variety of reasons there are a lot of students who just can't do basic calculuations. Luckily for us, we have calculators available at reasonable prices.
Graphing calculators are wonderful for "seeing" what an equation looks like. It can be used for graphical analysis by looking at the graph of the equation. Enter the equation using the Y= button on your calculator. You can see the graph by using the GRAPH button and adjusting the zoom or window. If you use the second function TABLE (same button as GRAPH) button, you can see the table of values and look at the equation in a numerical way. So the graphing calculator helps you to understand equations better.
Most of the standardized tests my students take - like NMCRT, SBA, and ACT - allow the use of a calculator.
In fact, the Advanced Placement (AP)Calculus, Statistics, Chemistry, and Physics tests are impossible without a good graphing calculator.
Using a calculator wisely
Once you have a calculator you need to learn to use it correctly.
Play with your calculator to get familiar with the different buttons. Some calculators
have an equal sign = to get the answer, but the fancier ones have an Enter
button instead. The fancier calculators also have more than one meaning to
each key. These, known as second functions, are usually printed in a
different color
Write down the problem set-up before you start using the calculator. Check
to make sure you have everything you need written down.
Always check your answer to see if it is reasonable. It is very easy to
push the wrong button when entering a problem into a calculator. Did you get
close to the answer you expected? Does your answer make sense?
Always round your answer to the correct number of significant figures. In
general, this is one more decimal place than you were given in the problem.
If the problem is a word problem, make sure you write down the correct
units in your answer. One elephant is very different from one meter.
Here is an example of how to use your calculator to find the hypotenuse of a
right triangle with legs that are 5 inches and 7 inches long. The things shown
in tiger orange are special buttons to push on the calculator.
Set-up
let c = length of the hypotenuse and a and
b equal the lengths of the two legs
Formula
Pythagorean Theorem a2 + b2
= c2
Substitute
52 + 72 = c2
Solve for c2
on the calculator, type in:
(5 x2) + (7 x2)
Enter
the answer on the calculator should be 74
Solve for c
take the square root √
of 74
the calculator will say 8.60232526...
this is reasonable, since it is easy to imagine a triangle with sides of
5, 7 and about 8˝ inches.
Round your answer
since we started with only whole numbers,
we want the answer to have 1 decimal place
the answer rounded is 8.6
Write the units
the problem gave the lengths in inches, so
your final answer is
8.6 inches
Graphing Calculators
Graphing calculators can be used in almost every class, but they are so
powerful they can actually get in the way for beginning students. I don't
recommend using a graphing calculator until you are at least at the Algebra I
level and don't feel it is absolutely necessary until you are in PreCalculus or Calculus.
I have classroom sets of graphing calculators available for my students to use in
class, so no one has to buy their own. If you do buy your own graphing calculator,
be sure to actually read the user's manual! I know it is pretty boring,
but it is the best way to learn how to use a complicated graphing calculator. Also be sure to go to the manufacturer's web site for help. My personal favorite is education.ti.com, which has model specific information for students, teachers, and parents.
If you are taking AP Calculus, daily use of the graphing calculators is reguired. The AP Calculus exam has half of the test with calculators and half without.
Computers
Computers come in very handy for lots of things relating to mathematics.
Prices keep dropping and you can purchase a reasonably good desk top computer
with some basic software for less than $300. In fact, there are even laptop and now netbook
computers available for less than $400. I gave my husband an IPad for Christmas, and it is a pretty neat option if you have the money. Here are my tips when buying a computer
for college:
always buy the best computer you can afford
plan to have access to the Internet and run a good virus program
always keep your virus protection updated
you can never have too much RAM
never buy a computer in December or May - this is when prices are the
highest
desk top computers are relatively easy to upgrade, while it takes a real
expert to work on laptops and netbooks
make sure you have basic word processing and spreadsheet programs - don't want to pay big bucks, then try Open Office
Once you have your computer, check my resources
page or run a Goggle search for math topics. There are a lot of programs out
there that can help you in your math class.
If you want to do some minimal programming, you can use your spreadsheet
program to set up formulas for solving quadratic equations, doing trig
functions, and other basic computational things. Most of the word processing
programs have graphics that can be used to illustrate math papers. I use an
"Equation Generator" add-in for MS Word when I write your handouts and tests. | 677.169 | 1 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.