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Algebra for College Students 9780495105107 ISBN: 0495105104 Edition: 8 Pub Date: 2006 Publisher: Thomson Learning Summary: Kaufmann and Schwitters have built this text's reputation on clear and concise exposition, numerous examples, and plentiful problem sets. This traditional text consistently reinforces the following common thread: learn a skill; use the skill to help solve equations; and then apply what you have learned to solve application problems. This simple, straightforward approach has helped many students grasp and apply fundam...ental problem solving skills necessary for future mathematics courses in an easy-to-read format. The new Eighth Edition of ALGEBRA FOR COLLEGE STUDENTS includes new and updated problems, revised content based on reviewer feedback and a new function in iLrn. This enhanced iLrn homework functionality was designed specifically for Kaufmann/Schwitters' users. Textbook-specific practice problems have been added to iLrn to provide additional, algorithmically-generated practice problems, along with useful support and assistance to solve the problems for students. Kaufmann Schwitters Staff is the author of Algebra for College Students, published 2006 under ISBN 9780495105107 and 0495105104. Eighty eight Algebra for College Students textbooks are available for sale on ValoreBooks.com, twenty eight used from the cheapest price of $0.20
Mathematics : Practical Odyssey (012734) - 6th edition Summary: Well known for its clear writing and unique variety of topics, MATHEMATICS: A PRACTICAL ODYSSEY demonstrates how mathematics is usable and relevant to students. Throughout the book, the authors emphasize problem solving skills, practical applications, and the history of mathematics. Students encounter topics that will be useful in their daily lives, such as calculating interest and understanding voting systems. They are encouraged to recognize the relevance of mathemati...show morecs and appreciate its human aspect. To offer flexibility in content, the book contains more information than could be covered in a one-term course. The chapters are independent of each other so instructors can select the ideal topics for their courses. ...show less Place Systems. Arithmetic in Different Bases. Primes and Perfect Numbers. The Fibonacci Sequence and the Golden Ratio. 8. GEOMETRY. Perimeter and Area. Volume and Surface Area. Egyptian Geometry. The Greeks. Right Triangle Trigonometry. Conic Sections and Analytic Geometry. Non-Euclidean Geometry. Fractal Geometry. The Perimeter and Area of a Fractal. Review of Linear Inequalities. The Geometry of Linear Programming. Appendices: Using a Scientific Calculator. Using a Graphing Calculator. Graphing with a Graphing Calculator. Finding Points of Intersection with a Graphing Calculator. Dimensional Analysis. Body Table for the Standard Normal Distribution. Answers to Selected Exercises. 0495012734booksfromca Simi Valley, CA No comments from the seller10.49 +$3.99 s/h Good Highfield Books PA Pittsburgh, PA 2006-09-22 Hardcover Good 6th edition. Has underlining. Binding is secure. Cover is in good shape. Unless noted, used texts do NOT include supplements such as CDs & codes. Orders packed carefully a...show morend shipped daily with tracking # emailed to you. Canadian and international orders welcomed!182237.32
Mathematics Information The THEA (Texas Higher Education Assessment) is a state test that covers reading, writing, and mathematics skills (this was previously call the TASP test). If a student has not taken THEA, they will be given Accuplacer, which is an alternate test. If a student fails one or more sections of either test, they will be required to developmental coursework in the area of deficiency. Students will be given an individualized "success plan" showing which courses need to be completed. In mathematics, once a student is initially placed in the developmental sequence, they work their way through the sequence. After completion of the developmental sequence, students can continue into college level mathematics. The Kilgore College Mathematics Department uses a variety of test scores to verify student placement. In addition to THEA or Accuplacer, the department also uses ACT, SAT, or a additional tests required by the department for placement. Contact the Testing Office, (903) 983-8215 for further information or to set up and appointment. See individual course listings for placement information. Summer Bridge Program The mathematics department offers a three-week Summer Bridge Program designed for students who need a refresher of mathematics skills to prepare for college-level mathematics. Please contact the department chair for more details. Sequences The Mathematics Department offers several courses in sequence to satisfy particular degree plans. See the individual course listings for more information. 1.Calculus and Differential Equations. The calculus sequence has three 4-hour courses (MATH 2413,MATH 2414, and MATH 2415). Differential equations is a 3-hour course that is only offered in the summer. These courses are for mathematics, engineering, and physics majors who will take the entire sequence, as well as other science majors such as chemistry, biological sciences such as pre-med., pre-dent, etc. who will take Calculus I and maybe Calculus II. 2.Pre calculus. These courses are College Algebra (MATH 1314), Trigonometry (MATH 1316), and Pre calculus (MATH 2412) and are designed to prepare students for the study of calculus. MATH 1314 and MATH 1316 are 3-hour courses that also meets the needs of students that just need 3-6 hours of mathematics to satisfy the general education requirement for most degree plans. After completion of College Algebra and Trigonometry, the student may take Calculus I. MATH 2412 is a 4-hour accelerated course that is essentially College Algebra and Trigonometry together in one course. This course is specifically designed for the strong mathematics student coming out of high school who needs to review topics from algebra and trigonometry before beginning the calculus sequence. 3.Business. The Finite Mathematics and Business Calculus sequence (MATH 1324 and MATH 1325) is provided for business, finance, and accounting majors. 4.General. Mathematical Topics, MATH 1333, is a course that is provided for liberal or fine arts majors or for students who are working toward and Associate of Applied Science degree. 5.Education. The Mathematics for Elementary Teachers I and II sequence (MATH 1350 and MATH 1351) is designed specifically for students planning to teach at the elementary level. These courses do not have to be taken sequentially and do require College Algebra as a prerequisite. 6.Statistics. Introduction to Probability and Statistics (MATH 1342) is a course for many students majoring nursing or social science fields. This course does not require College Algebra as a prerequisite and will count as a 3-hour mathematics requirement for the associate degree. 7. Developmental. The developmental mathematics courses are designed to prepare students for college level mathematics. The sequence consists of Pre-Algebra (MATH 0304), Beginning Algebra (MATH 0306), and Intermediate Algebra (MATH 0308). All students are placed into the appropriate course using a test score from either THEA or our local placement test, Accuplacer. Students are not permitted to retest once they have started the sequence. After completing the sequence, students will have the necessary mathematics skills to be successful in whichever college level mathematics course is required in their degree plan.
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Course Details MATH 231 CALCULUS I Functions, limits, continuity, and rates of change are studied numerically, symbolically, and graphically. Definition and rules of differentiation; applications of the derivative to analyzing functions, solving equations, and computing extrema; antiderivatives.
Precalculus with Unit-Circle Trigonometry - 3rd edition Summary: This book introduces trigonometry through the unit circle. Cohen emphasizes graphing to explain complex concepts in an uncomplicated style, and provides supplementary graphing-calculator exercises at the end of most sections for additional perspective and reinforcement. Trigonometric Functions of Real Numbers. Graphs of the Sine and the Cosine Functions. Graphs of y = A sin(Bx-C) and y = A cos(Bx - C). Simple Harmonic Motion. Graphs of the Tangent and the Reciprocal Functions. Right-Triangle Applications. The Law of Sines and the Law of Cosines. Vectors in the Plane, a Geometric Approach. Vectors in the Plane, an Algebraic Approach. Parametric Equations. Introduction to Polar Coordinates. Curves in Polar Coordinates. PART X. SYSTEMS OF EQUATIONS. Systems of Two Linear Equations in Two Unknowns. Gaussian Elimination. Matrices. The Inverse of a Square Matrix. Determinants and Cramer's Rule. Nonlinear Systems of Equations. Systems of Inequalities. PART XI. ANALYTIC GEOMETRY. The Basic Equations. The Parabola. Tangents to Parabolas (Optional). The Ellipse. The Hyperbola. The Focus-Directrix Property of Conics. The Conics in Polar Coordinates. Rotation of Axes. PART XII. ROOTS OF POLYNOMIAL EQUATIONS. The Complex Number System. Division of Polynomials. Roots of Polynomial Equations : The Remainder Theorem and the Factor Theorem. The Fundamental Theorem of Algebra. Rational and Irrational Roots. Conjugate Roots and Descartes' Rule of Signs. Introduction to Partial Fractions. More About Partial Fractions49 +$3.99 s/h Acceptable AlphaBookWorks Alpharetta, GA 053435275815.98 +$3.99 s/h New Directtext4u El Monte, CA Hardcover New 0534352758 New book may have school stamps or class set numbers on the side but was not issued to a student. 100% guaranteed fast shipping! ! $19.95 +$3.99 s/h LikeNew arcfoundationthriftstore Ventura, CA 0534352758 Your purchase benefits those with developmental disabilities to live a better quality of life. $26.01 +$3.99 s/h New AZBOOKA Oceanside, CA 0534352758 New book in flawless condition
Precalculus This course reviews the concepts from Algebra II that are central to calculus and explores several discrete math topics. Calculus topics focus on the study of functions: polynomial, trigonometric, logarithmic, and exponential. Discrete topics include polar coordinates, sequences and series, permutations and combinations, the Binomial Theorem, and conic sections. Throughout the course, students are expected to use the graphing calculator to solve problems in each topic area.
Preface This book contains a collection of general mathematical results, formulas, and integrals that occur throughout applications of mathematics. Many of the entries are based on the updated fifth edition of Gradshteyn and Ryzhik's "Tables of Integrals, Series, and Products," though during the preparation of the book, results were also taken from various other reference works. The material has been arranged in a straightforward manner, and for the convenience of the user a quick reference list of the simplest and most frequently used results is to be found in Chapter 0 at the front of the book. Tab marks have been added to pages to identify the twelve main subject areas into which the entries have been divided and also to indicate the main interconnections that exist between them. Keys to the tab marks are to be found inside the front and back covers. The Table of Contents at the front of the book is sufficiently detailed to enable rapid location of the section in which a specific entry is to be found, and this information is supplemented by a detailed index at the end of the book. In the chapters listing integrals, instead of displaying them in their canonical form, as is customary in reference works, in order to make the tables more convenient to use, the integrands are presented in the more general form in which they are likely to arise. It is hoped that this will save the user the necessity of reducing a result to a canonical form before consulting the tables. Wherever it might be helpful, material has been added explaining the idea underlying a section or describing simple techniques that are often useful in the application of its results. Standard notations have been used for functions, and a list of these together with their names and a reference to the section in which they occur or are defined is to be found at the front of the book. As is customary with tables of indefinite integrals, the additive arbitrary constant of integration has always been omitted. The result of an integration may take more than one form, often depending on the method used for its evaluation, so only the most common forms are listed. A user requiring more extensive tables, or results involving the less familiar special functions, is referred to the short classified reference list at the end of the book. The list contains works the author found to be most useful and which a user is likely to find readily accessible in a library, but it is in no sense a comprehensive bibliography. Further specialist references are to be found in the bibliographies contained in these reference works. Every effort has been made to ensure the accuracy of these tables and, whenever possible, results have been checked by means of computer symbolic algebra and integration programs, but the final responsibility for errors must rest with the author. xix Preface to the Fourth Edition The preparation of the fourth edition of this handbook provided the opportunity to enlarge the sections on special functions and orthogonal polynomials, as suggested by many users of the third edition. A number of substantial additions have also been made elsewhere, like the enhancement of the description of spherical harmonics, but a major change is the inclusion of a completely new chapter on conformal mapping. Some minor changes that have been made are correcting of a few typographical errors and rearranging the last four chapters of the third edition into a more convenient form. A significant development that occurred during the later stages of preparation of this fourth edition was that my friend and colleague Dr. Hui-Hui Dai joined me as a co-editor. Chapter 30 on conformal mapping has been included because of its relevance to the solu- tion of the Laplace equation in the plane. To demonstrate the connection with the Laplace equation, the chapter is preceded by a brief introduction that demonstrates the relevance of conformal mapping to the solution of boundary value problems for real harmonic functions in the plane. Chapter 30 contains an extensive atlas of useful mappings that display, in the usual diagrammatic way, how given analytic functions w = f(z) map regions of interest in the complex z-plane onto corresponding regions in the complex w-plane, and conversely. By form- ing composite mappings, the basic atlas of mappings can be extended to more complicated regions than those that have been listed. The development of a typical composite mapping is illustrated by using mappings from the atlas to construct a mapping with the property that a region of complicated shape in the z-plane is mapped onto the much simpler region compris- ing the upper half of the w-plane. By combining this result with the Poisson integral formula, described in another section of the handbook, a boundary value problem for the original, more complicated region can be solved in terms of a corresponding boundary value problem in the simpler region comprising the upper half of the w-plane. The chapter on ordinary differential equations has been enhanced by the inclusion of mate- rial describing the construction and use of the Green's function when solving initial and boundary value problems for linear second order ordinary differential equations. More has been added about the properties of the Laplace transform and the Laplace and Fourier con- volution theorems, and the list of Laplace transform pairs has been enlarged. Furthermore, because of their use with special techniques in numerical analysis when solving differential equations, a new section has been included describing the Jacobi orthogonal polynomials. The section on the Poisson integral formulas has also been enlarged, and its use is illustrated by an example. A brief description of the Riemann method for the solution of hyperbolic equations has been included because of the important theoretical role it plays when examining general properties of wave-type equations, such as their domains of dependence. For the convenience of users, a new feature of the handbook is a CD-ROM that contains the classified lists of integrals found in the book. These lists can be searched manually, and when results of interest have been located, they can be either printed out or used in papers or xxi xxii Preface worksheets as required. This electronic material is introduced by a set of notes (also included in the following pages) intended to help users of the handbook by drawing attention to different notations and conventions that are in current use. If these are not properly understood, they can cause confusion when results from some other sources are combined with results from this handbook. Typically, confusion can occur when dealing with Laplace's equation and other second order linear partial differential equations using spherical polar coordinates because of the occurrence of differing notations for the angles involved and also when working with Fourier transforms for which definitions and normalizations differ. Some explanatory notes and examples have also been provided to interpret the meaning and use of the inversion integrals for Laplace and Fourier transforms. Alan Jeffrey alan.jeffrey@newcastle.ac.uk Hui-Hui Dai mahhdai@math.cityu.edu.hk Notes for Handbook Users The material contained in the fourth edition of the Handbook of Mathematical Formulas and Integrals was selected because it covers the main areas of mathematics that find frequent use in applied mathematics, physics, engineering, and other subjects that use mathematics. The material contained in the handbook includes, among other topics, algebra, calculus, indefinite and definite integrals, differential equations, integral transforms, and special functions. For the convenience of the user, the most frequently consulted chapters of the book are to be found on the accompanying CD that allows individual results of interest to be printed out, included in a work sheet, or in a manuscript. A major part of the handbook concerns integrals, so it is appropriate that mention of these should be made first. As is customary, when listing indefinite integrals, the arbitrary additive constant of integration has always been omitted. The results concerning integrals that are available in the mathematical literature are so numerous that a strict selection process had to be adopted when compiling this work. The criterion used amounted to choosing those results that experience suggested were likely to be the most useful in everyday applications of mathematics. To economize on space, when a simple transformation can convert an integral containing several parameters into one or more integrals with fewer parameters, only these simpler integrals have been listed. For example, instead of listing indefinite integrals like eax sin(bx + c)dx and eax cos(bx + c)dx, each containing the three parameters a, b, and c, the simpler indefinite inte- grals eax sin bxdx and eax cos bxdx contained in entries 5.1.3.1(1) and 5.1.3.1(4) have been listed. The results containing the parameter c then follow after using additive prop- erty of integrals with these tabulated entries, together with the trigonometric identities sin(bx + c) = sin bx cos c + cos bx sin c and cos(bx + c) = cos bx cos c−sin bx sin c. The order in which integrals are listed can be seen from the various section headings. If a required integral is not found in the appropriate section, it is possible that it can be transformed into an entry contained in the book by using one of the following elementary methods: 1. Representing the integrand in terms of partial fractions. 2. Completing the square in denominators containing quadratic factors. 3. Integration using a substitution. 4. Integration by parts. 5. Integration using a recurrence relation (recursion formula), xxiii xxiv Notes for Handbook Users or by a combination of these. It must, however, always be remembered that not all integrals can be evaluated in terms of elementary functions. Consequently, many simple looking integrals cannot be evaluated analytically, as is the case with sin x a + bex dx. A Comment on the Use of Substitutions When using substitutions, it is important to ensure the substitution is both continuous and one-to-one, and to remember to incorporate the substitution into the dx term in the integrand. When a definite integral is involved the substitution must also be incorporated into the limits of the integral. When an integrand involves an expression of the form √ a2 −x2, it is usual to use the substitution x = \|a sin θ\| which is equivalent to θ = arcsin(x/ \|a\|), though the substitution x = \|a\| cos θ would serve equally well. The occurrence of an expression of the form √ a2 + x2 in an integrand can be treated by making the substitution x = \|a\| tan θ, when θ = arctan(x/ \|a\|) (see also Section 9.1.1). If an expression of the form √ x2 −a2 occurs in an integrand, the substitution x = \|a\| sec θ can be used. Notice that whenever the square root occurs the positive square root is always implied, to ensure that the function is single valued. If a substitution involving either sin θ or cos θ is used, it is necessary to restrict θ to a suitable interval to ensure the substitution remains one-to-one. For example, by restricting θ to the interval −1 2 π ≤ θ ≤ 1 2 π, the function sin θ becomes one-to-one, whereas by restricting θ to the interval 0 ≤ θ ≤ π, the function cos θ becomes one-to-one. Similarly, when the inverse trigonometric function y = arcsin x is involved, equivalent to x = sin y, the function becomes one-to-one in its principal branch − 1 2 π ≤ y ≤ 1 2 π, so arcsin(sin x) = x for −1 2 π ≤ x ≤ 1 2 π and sin(arcsin x) = x for −1 ≤ x ≤ 1. Correspondingly, the inverse trigonometric function y = arccos x, equivalently x = cos y, becomes one-to-one in its principal branch 0 ≤ y ≤ π, so arccos(cos x) = x for 0 ≤ x ≤ π and sin(arccos x) = x for −1 ≤ x ≤ 1. It is important to recognize that a given integral may have more than one representation, because the form of the result is often determined by the method used to evaluate the integral. Some representations are more convenient to use than others so, where appropriate, integrals of this type are listed using their simplest representation. A typical example of this type is dx √ a2 + x2 = arcsinh(x/a) ln x + √ a2 + x2 where the result involving the logarithmic function is usually the more convenient of the two forms. In this handbook, both the inverse trigonometric and inverse hyperbolic functions all carry the prefix "arc." So, for example, the inverse sine function is written arcsin x and the inverse hyperbolic sine function is written arcsinh x, with corresponding notational conventions for the other inverse trigonometric and hyperbolic functions. However, many other works denote the inverse of these functions by adding the superscript −1 to the name of the function, in which case arcsin x becomes sin−1 x and arcsinh x becomes sinh−1 x. Elsewhere yet another notation is in use where, instead of using the prefix "arc" to denote an inverse hyperbolic Notes for Handbook Users xxv function, the prefix "arg" is used, so that arcsinh x becomes argsinh x, with the corresponding use of the prefix "arg" to denote the other inverse hyperbolic functions. This notation is preferred by some authors because they consider that the prefix "arc" implies an angle is involved, whereas this is not the case with hyperbolic functions. So, instead, they use the prefix "arg" when working with inverse hyperbolic functions. Example: Find I = x5 √ a2−x2 dx. Of the two obvious substitutions x = \|a\| sin θ and x = \|a\| cos θ that can be used, we will make use of the first one, while remembering to restrict θ to the interval −1 2 π ≤ θ ≤ 1 2 π to ensure the transformation is one-to-one. We have dx = \|a\| cos θdθ, while √ a2 −x2 = a2 −a2 sin2 θ = \|a\| 1−sin2 θ = \|a cos θ\|. However cos θ is positive in the interval −1 2 π ≤ θ ≤ 1 2 π, so we may set √ a2 −x2 = \|a\| cos θ. Substituting these results into the integrand of I gives I = \|a\| 5 sin5 θ \|a\| cos θdθ \|a\| cos θ = a4 \|a\| sin5 θdθ, and this trigonometric integral can be found using entry 9.2.2.2, 5. This result can be expressed in terms of x by using the fact that θ = arcsin (x/ \|a\|), so that after some manipulation we find that I = − 1 5 x4 a2 −x2 − 4a2 15 a2 −x2 2a2 + x2 . A Comment on Integration by Parts Integration by parts can often be used to express an integral in a simpler form, but it also has another important property because it also leads to the derivation of a reduction formula, also called a recursion relation. A reduction formula expresses an integral involving one or more parameters in terms of a simpler integral of the same form, but with the parameters having smaller values. Let us consider two examples in some detail, the second of which given a brief mention in Section 1.15.3. Example: (a) Find a reduction formula for Im = cosm θdθ, and hence find an expression for I5. (b) Modify the result to find a recurrence relation for Jm = π/2 0 cosm θdθ, and use it to find expressions for Jm when m is even and when it is odd. Notes for Handbook Users xxvii Example: The following is an example of a recurrence formula that contains two param- eters. If Im,n = sinm θ cosn θdθ, an argument along the lines of the one used in the previous example, but writing Im,n = sinm−1 θ cosn θd(− cos θ), leads to the result (m + n)Im,n = − sinm−1 θ cosn+1 θ + (m−1)Im−2,n, in which n remains unchanged, but m decreases by 2. Had integration by parts been used differently with Im,n written as Im,n = sinm θ cosn−1 θd(sin θ) a different reduction formula would have been obtained in which m remains unchanged but n decreases by 2. Some Comments on Definite Integrals Definite integrals evaluated over the semi-infinite interval [0, ∞) or over the infinite interval (−∞, ∞) are improper integrals and when they are convergent they can often be evaluated by means of contour integration. However, when considering these improper integrals, it is desirable to know in advance if they are convergent, or if they only have a finite value in the sense of a Cauchy principal value. (see Section 1.15.4). A geometrical interpretation of a Cauchy principal value for an integral of a function f(x) over the interval (−∞, ∞) follows by regarding an area between the curve y = f(x) and the x-axis as positive if it lies above the x-axis and negative if it lies below it. Then, when finding a Cauchy principal value, the areas to the left and right of the y-axis are paired off symmetrically as the limits of integration approach ±∞. If the result is a finite number, this is the Cauchy principal value to be attributed to the definite integral ∞ −∞ f(x)dx, otherwise the integral is divergent. When an improper integral is convergent, its value and its Cauchy principal value coincide. There are various tests for the convergence of improper integrals, but the ones due to Abel and Dirichlet given in Section 1.15.4 are the main ones. Convergent integrals exist that do not satisfy all of the conditions of the theorems, showing that although these tests represent sufficient conditions for convergence, they are not necessary ones. Example: Let us establish the convergence of the improper integral ∞ a sin mx xp dx, given that a, p > 0. To use the Dirichlet test we set f(x) = sin x and g(x) = 1/xp . Then lim x→∞ g(x) = 0 and ∞ a \|g (x)\|dx = 1/ap is finite, so this integral involving g(x) converges. We also have F(b) = b a sin mxdx =(cos ma − cos mb)/m, from which it follows that \|F(b)\| ≤ 2 for all xxviii Notes for Handbook Users a ≤ x ≤ b < ∞. Thus the conditions of the Dirichlet test are satisfied showing that ∞ a sin x xp dx is convergent for a, p > 0. It is necessary to exercise caution when using the fundamental theorem of calculus to evaluate an improper integral in case the integrand has a singularity (becomes infinite) inside the interval of integration. If this occurs the use of the fundamental theorem of calculus is invalid. Example: The improper integral a −a dx x2 with a > 0 has a singularity at the origin and is, in fact, divergent. This follows because if ε, δ > 0, we have lim ε→0 −ε −a dx x2 + lim δ→0 b δ dx x2 = ∞. However, an incorrect application of the fundamental theorem of calculus gives a −a dx x2 = −1 x a x=−a = −2 a . Although this result is finite, it is obviously incorrect because the integrand is positive over the interval of integration, so the definite integral must also be positive, but this is not the case here because a > 0 so −2/a < 0. Two simple results that often save time concern the integration of even and odd functions f(x) over an interval −a ≤ x ≤ a that is symmetrical about the origin. We have the obvious result that when f(x) is odd, that is when f(−x) = −f(x), then a −a f(x)dx = 0, and when f(x) is even, that is when f(−x) = f(x), then a −a f(x)dx = 2 a 0 f(x)dx. These simple results have many uses as, for example, when working with Fourier series and elsewhere. Some Comments on Notations, the Choice of Symbols, and Normalization Unfortunately there is no universal agreement on the choice of symbols used to identify a point P in cylindrical and spherical polar coordinates. Nor is there universal agreement on the choice of symbols used to represent some special functions, or on the normalization of Fourier transforms. Accordingly, before using results derived from other sources with those given in this handbook, it is necessary to check the notations, symbols, and normalization used elsewhere prior to combining the results. Symbols Used with Curvilinear Coordinates To avoid confusion, the symbols used in this handbook relating to plane polar coordinates, cylindrical polar coordinates, and spherical polar coordinates are shown in the diagrams in Section 24.3. Notes for Handbook Users xxix The plane polar coordinates (r, θ) that identify a point P in the (x, y)-plane are shown in Figure 1(a). The angle θ is the azimuthal angle measured counterclockwise from the x-axis in the (x, y)-plane to the radius vector r drawn from the origin to the point P. The connection between the Cartesian and the plane polar coordinates of P is given by x = r cos θ, y = r sin θ, with 0 ≤ θ < 2π. 0 y r x ␪ P (r, ␪) Figure 1(a) We mention here that a different convention denotes the azimuthal angle in plane polar coordinates by θ, instead of by φ. The cylindrical polar coordinates (r, θ, z) that identify a point P in space are shown in Figure 1(b). The angle θ is again the azimuthal angle measured as in plane polar coordinates, r is the radial distance measured from the origin in the (x, y)-plane to the projection of P onto the (x, y)-plane, and z is the perpendicular distance of P above the (x, y)-plane. The connection between cartesian and cylindrical polar coordinates used in this handbook is given by x = r cos θ, y = r sin θ and z = z, with 0 ≤ θ < 2π. 0 z z y x ␪ P (r, ␪, z) r Figure 1(b) xxx Notes for Handbook Users Here also, in a different convention involving cylindrical polar coordinates, the azimuthal angle is denoted by φ instead of by θ. The spherical polar coordinates (r, θ, φ) that identify a point P in space are shown in Fig- ure 1(c). Here, differently from plane cylindrical coordinates, the azimuthal angle measured as in plane cylindrical coordinates is denoted by φ, the radius r is measured from the origin to point P, and the polar angle measured from the z-axis to the radius vector OP is denoted by θ, with 0 ≤ φ < 2π, and 0 ≤ θ ≤ π. The cartesian and spherical polar coordinates used in this handbook are connected by x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ. 0 z y x ␾ ␪ P (r, ␪, ␾) Figure 1(c) In a different convention the roles of θ and φ are interchanged, so the azimuthal angle is denoted by θ, and the polar angle is denoted by φ. Bessel Functions There is general agreement that the Bessel function of the first kind of order ν is denoted by Jν(x), though sometimes the symbol ν is reserved for orders that are not integral, in which case n is used to denote integral orders. However, notations differ about the representation of the Bessel function of the second kind of order ν. In this handbook, a definition of the Bessel function of the second kind is adopted that is true for all orders ν (both integral and fractional) and it is denoted by Yν(x). However, a widely used alternative notation for this same Bessel function of the second kind of order ν uses the notation Nν(x). This choice of notation, sometimes called the Neumann form of the Bessel function of the second kind of order ν, is used in recognition of the fact that it was defined and introduced by the German mathematician Carl Neumann. His definition, but with Yν(x) in place of Nν(x), is given in Section 17.2.2. The reason for the rather strange form of this definition is because when the second linearly independent solution of Bessel's equation is derived using the Frobenius Notes for Handbook Users xxxi method, the nature of the solution takes one form when ν is an integer and a different one when ν is not an integer. The form of definition of Yν(x) used here overcomes this difficulty because it is valid for all ν. The recurrence relations for all Bessel functions can be written as Zν−1(x) + Zν+1(x) = 2ν x Zν(x), Zν−1(x) − Zν+1(x) = 2Zν(x), Zν(x) = Zν−1(x) − ν x Zν(x) Zν(x) = −Zν+1(x) + ν x Zν(x), (1) where Zν(x) can be either Jν(x) or Yν(x). Thus any recurrence relation derived from these results will apply to all Bessel functions. Similar general results exist for the modified Bessel functions Iν(x) and Kν(x). Normalization of Fourier Transforms The convention adopted in this handbook is to define the Fourier transform of a function f(x) as the function F(ω) where F(ω) = 1 √ 2π ∞ −∞ f(x)eiωx dx, (2) when the inverse Fourier transform becomes f(x) = 1 √ 2π ∞ −∞ F(ω)e−iωx dω, (3) where the normalization factor multiplying each integral in this Fourier transform pair is 1/ √ 2π. However other conventions for the normalization are in common use, and they follow from the requirement that the product of the two normalization factors in the Fourier and inverse Fourier transforms must equal 1/(2π). Thus another convention that is used defines the Fourier transform of f(x) as F(ω) = ∞ −∞ f(x)eiωx dx (4) and the inverse Fourier transform as f(x) = 1 2π ∞ −∞ F(ω)e−iωx dω. (5) To complicate matters still further, in some conventions the factor eiωx in the integral defining F(ω) is replaced by e−iωx and to compensate the factor e−iωx in the integral defining f(x) is replaced by eiωx . xxxii Notes for Handbook Users If a Fourier transform is defined in terms of an angular frequency, the ambiguity concerning the choice of normalization factors disappears because the Fourier transform of f(x) becomes F(ω) = ∞ −∞ f(x)e2πixs dx (6) and the inverse Fourier transform becomes f(x) = ∞ −∞ F(ω)e−2πixω dω. (7) Nevertheless, the difference between definitions still continues because sometimes the expo- nential factor in F(s) is replaced by e−2πixs , in which case the corresponding factor in the inverse Fourier transform becomes e2πixs . These remarks should suffice to convince a reader of the necessity to check the convention used before combining a Fourier transform pair from another source with results from this handbook. Some Remarks Concerning Elementary Ways of Finding Inverse Laplace Transforms The Laplace transform F(s) of a suitably integrable function f(x) is defined by the improper integral F(s) = ∞ 0 f(x)e−xs dx. (8) Let a Laplace transform F(s) be the quotient F(s) = P(s)/Q(s) of two polynomials P(s) and Q(s). Finding the inverse transform L−1 {F(s)} = f(x) can be accomplished by simplifying F(s) using partial fractions, and then using the Laplace transform pairs in Table 19.1 together with the operational properties of the transform given in 19.1.2.1. Notice that the degree of P(s) must be less than the degree of Q(s) because from the limiting condition in 19.11.2.1(10), if F(s) is to be a Laplace transform of some function f(x), it is necessary that lim s→∞ F(s) = 0. The same approach is valid if exponential terms of the type e−as occur in the numerator P(s) because depending on the form of the partial fraction representation of F(s), such terms will simply introduce either a Heaviside step function H(x − a), or a Dirac delta function δ(x − a) into the resulting expression for f(x). On occasions, if a Laplace transform can be expressed as the product of two simpler Laplace transforms, the convolution theorem can be used to simplify the task of inverting the Laplace transform. However, when factoring the transform before using the convolution theorem, care must be taken to ensure that each factor is in fact a Laplace transform of a function of x. This is easily accomplished by appeal to the limiting condition in 19.11.2.1(10), because if F(s) is factored as F(s) = F1(s)F2(s), the functions F1(s) and F2(s) will only be the Laplace transforms of some functions f1(x) and f2(x) if lim s→∞ F1(s) = 0 and lim s→∞ F2(s) = 0. Notes for Handbook Users xxxiii Example: (a) Find L−1 {F(s)} if F(s) = s3 +3s2 +5s+15 (s2+1)(s2+4s+13) . (b) Find L−1 {F(s)} if F(s) = s2 (s2+a2)2 . To solve (a) using partial fractions we write F(s) as F(s) = 1 s2+1 + s+2 s2+4s+13 . Taking the inverse Laplace transform of F(s) and using entry 26 in Table 19.1 gives L−1 {F(s)} = sin x + L−1 s + 2 s2 + 4s + 13 . Completing the square in the denominator of the second term and writing, s+2 s2+4s+13 = s+2 (s+2)2+32 , we see from the first shift theorem in 19.1.2.1(4) and entry 27 in Table 19.1 that L−1 s+2 (s+2)2+32 = e−2x cos 3x. Finally, combining results, we have L−1 {F(s)} = sin x + e−2x cos 3x. To solve (b) by the convolution transform, F(s) must be expressed as the product of two factors. The transform F(s) can be factored in two obvious ways, the first being F(s) = s2 (s2+a2) 1 (s2+a2) and the second being F(s) = s (s2+a2) s (s2+a2) . Of these two expressions, only the second is the product of two Laplace transforms, namely the product of the Laplace transforms of cos ax. The first result cannot be used because the factor s2 /(s2 + a2 ) fails the limiting condition in 19.11.2.1(10), and so is not the Laplace transform of a function of x. The inverse of the convolution theorem asserts that if F(s) and G(s) are Laplace transforms of the functions f(x) and g(x), then L−1 {F(s)G(s)} = x 0 f(τ)g(x−τ)dτ. (9) So setting F(s) = G(s) = cos ax, it follows that f(x) = L−1 s2 (s2 + a2)2 = x 0 cos τ cos(x−τ)dτ = sin ax 2a + x cos ax 2 . When more complicated Laplace transforms occur, it is necessary to find the inverse Laplace transform by using contour integration to evaluate the inversion integral in 19.1.1.1(5). More will be said about this, and about the use of the Fourier inversion integral, after a brief review of some key results from complex analysis. xxxiv Notes for Handbook Users Using the Fourier and Laplace Inversion Integrals As a preliminary to discussing the Fourier and Laplace inversion integrals, it is necessary to record some key results from complex analysis that will be used. An analytic function A complex valued function f(z) of the complex variable z = x + iy is said to be analytic on an open domain G (an area in the z-plane without its boundary points) if it has a derivative at each point of G. Other names used in place of analytic are holomorphic and regular. A function f(z) = u(x, y) + ν(x, y) will be analytic in a domain G if at every point of G it satisfies the Cauchy-Riemann equations ∂u ∂x = ∂ν ∂y and ∂u ∂y = − ∂ν ∂x . (10) These conditions are sufficient to ensure that f(z) had a derivative at every point of G, in which case df dz = ∂u ∂x + i ∂ν ∂x = ∂ν ∂y − i ∂u ∂y . (11) A pole of f (z) An analytic function f(z) is said to have a pole of order p at z = z0 if in some neighborhood the point z0 of a domain G where f(z) is defined, f(z) = g(z) (z−z0)p , (12) where the function g(z) is analytic at z0. When p = 1, the function f(z) is said to have simple pole at z = z0. A meromorphic function A function f(z) is said to be meromorphic if it is analytic everywhere in a domain G except for isolated points where its only singularities are poles. For example, the function f(z) = 1/(z2 + a2 ) = 1/ [(z − ia)(z + ia)] is a meromorphic function with simple poles at z = ± ia. The residue of f (z) at a pole If a function has a pole of order p at z = z0, then its residue at z = z0 is given by Residue (f(z) : z = z0) = lim z→z0 1 (p−1)! dp−1 dzp−1 (z−z0) p f(z) . For example, the residues of f(z) = 1/(z2 + a2 ) at its poles located at z = ± ia are Residue (1/(z2 + a2 ) : z = ia) = −i/(2a) and Residue (1/(z2 + a2 ) : z = −ia) = i/(2a). Notes for Handbook Users xxxv The Cauchy residue theorem Let be a simple closed curve in the z-plane (a non- intersecting curve in the form of a simple loop). Denoting by f(z)dz the integral of f(z) around in the counter-clockwise (positive) sense, the Cauchy residue theorem asserts that f(z)dz = 2πi × (sum of residues of f(z) inside ). (13) So, for example, if is any simple closed curve that contains only the residue of f(z) = 1/(z2 + a2 ) located at z = ia, then 1/(z2 + a2 )dz = 2πi × (−i/(2a)) = π/a. Jordan's Lemma in Integral Form, and Its Consequences This lemma take various forms, the most useful of which are as follows: (i) Let C+ be a circular arc of radius R located in the first and/or second quadrants, with its center at the origin of the z-plane. Then if f(z) → 0 uniformly as R → ∞, lim R→∞ C+ f(z)eimz dz = 0, where m > 0. (ii) Let C− be a circular arc of radius R located in the third and/or fourth quadrant with its center at the origin of the z plane. Then if f(z) → 0 uniformly as R → ∞, lim R→∞ C− f(z)e−imz dz = 0, where m > 0. (iii) In a somewhat different form the lemma takes the form π/2 0 e−k sin θ dθ ≤ π 2k 1 − e−k . The first two forms of Jordan's lemma are useful in general contour integration when estab- lishing that the integral of an analytic function around a circular arc of radius R centered on the origin vanishes in the limit as R → ∞. The third form is often used when estimating the magnitude of a complex function that is integrated around a quadrant. The form of Jordan's lemma to be used depends on the nature of the integrand to which it is to be applied. Later, result (iii) will be used when determining an inverse Laplace transform by means of the Laplace inversion integral. The Fourier Transform and Its Inverse In this handbook, the Fourier transform F(ω) of a suitably integrable function f(x) is defined as F(ω) = 1 √ 2π ∞ −∞ f(x)eiωx dx, (14) xxxvi Notes for Handbook Users while the inverse Fourier transform becomes f(x) = 1 √ 2π ∞ −∞ F(ω)e−iωx dω, (15) it being understood that when f(x) is piecewise continuous with a piecewise continuous first derivative in any finite interval, that this last result is to be interpreted as f(x−) + f(x+) 2 = 1 √ 2π ∞ −∞ F(ω)e−iωx dω, (16) with f(x±) the values of f(x) on either side of a discontinuity in f(x). Notice first that although f(x) is real, its Fourier transform F(ω) may be complex. Although F(ω) may often be found by direct integration care is necessary, and it is often simpler to find it by converting the line integral defining F(ω) into a contour integral. The necessary steps involve (i) integrating f(x) along the real axis from −R to R, (ii) joining the two ends of this segment of the real axis by a semicircle of radius R with its center at the origin where the semicircle is either located in the upper half-plane, or in the lower half-plane, (iii) denoting this contour by R, and (iv) using the limiting form of the contour R as R → ∞ as the contour around which integration is to be performed. The choice of contour in the upper or lower half of the z-plane to be used will depend on the sign of the transform variable ω. This same procedure is usually necessary when finding the inverse Fourier transform, because when F(ω) is complex direct integration of the inversion integral is not possible. The example that follows will illustrate the fact that considerable care is necessary when working with Fourier transforms. This is because when finding a Fourier transform, the transform vari- able ω often occurs in the form \|ω\|, causing the transform to take one form when ω is positive, and another when it is negative. Example: Let us find the Fourier transform of f(x) = 1/(x2 + a2 ) where a > 0, the result of which is given in entry 1 of Table 20.1. Replacing x by the complex variable z, the function f(z) = eiωz /(z2 + a2 ), the integrand in the Fourier transform, is seen to have simple poles at z = ia and z = −ia, where the residues are, respectively, −ie−ωa /(2a) and ieωa /(2a). For the time being, allowing CR to be a semicircle in either the upper or the lower half of the z-plane with its center at the origin, we have F(ω) = lim R→∞ 1 √ 2π R −R eiωx (x2 + a2) dx + lim R→∞ 1 √ 2π CR eiωz (z2 + a2) dz. To use the residue theorem we need to show the second integral vanishes in the limit as R → ∞. On CR we can set z = Reiθ , so dz = iReiθ dθ, showing that 1 √ 2κ CR eiωz (z2 + a2) dz = 1 √ 2π CR eiωR(cos θ+i sin θ) iR eiθ (R2e2iθ + a2) e−ωR sin θ dθ. Notes for Handbook Users xxxvii We now estimate the magnitude of the integral on the right by the result 1 √ 2π CR eiωz (z2 + a2) dz ≤ 1 √ 2π R \|R2 − a2\| CR e−ωR sin θ dθ. The multiplicative factor involving R on the right will vanish as R → ∞, so the integral around CR will vanish if the integral on the right around CR remains finite or vanishes as R → ∞. There are two cases to consider, the first being when ω > 0, and the second when ω < 0. If ω = 0 the integral will certainly vanish as R → ∞, because then the integral around CR becomes CR dθ = π. The case ω > 0. The integral on the right around CR will vanish in the limit as R → ∞ provided sin θ ≥ 0 because its integrand vanishes. This happens when CR becomes the semicircle CR+ located in the upper half of the z-plane. The case ω < 0. The integral around CR will vanish in the limit as R → ∞, provided sin θ ≤ 0 because its integrand vanishes. This happens when CR becomes the semicircle CR− located in the lower half of the z-plane. We may now apply the residue theorem after proceeding to the limit as R → ∞. When ω > 0 we have CR = CR+, in which case only the pole at z = ia lies inside the contour at which the residue is −ie−ωa /(2a), so 1 √ 2π ∞ −∞ eiωx (x2 + a2) dx = 2πi × 1 √ 2π − ie−ωa 2a = π 2 e−ωa a , (ω > 0). Similarly, when ω < 0 we have CR = CR−, in which case only the pole at z = −ia lies inside the contour at which the residue is ieωa /(2a). However, when integrating around CR− in the positive (counterclockwise) sense, the integration along the x-axis occurs in the negative sense, that is from x = R to x = −R, leading to the result 1 √ 2π −∞ ∞ eiωx (x2 + a2) dx = 2πi × 1 √ 2π ieωa 2a = − π 2 eωa a , (ω < 0). Reversing the order of the limits in the integral, and compensating by reversing its sign, we arrive at the result 1 √ 2π ∞ −∞ eiωx (x2 + a2) dx = π 2 eωa a , (ω < 0). Combining the two results for positive and negative ω we have shown the Fourier transform F(ω) of f(x) = 1/(x2 + a2 ) is F(ω) = π 2 e−a\|ω\| a , (a > 0). xxxviii Notes for Handbook Users The function f(x) can be recovered from its Fourier transform F(ω) by means of the inversion integral, though this case is sufficiently simplest that direct integration can be used. f(x) = 1 √ 2π ∞ −∞ π 2 e−iωx e−a\|ω\| a dω = 1 2a ∞ −∞ e−a\|ω\| (cos(ωx) − i sin(ωx)) dω. The imaginary part of the integrand is an odd function, so its integral vanishes. The real part of the integrand is an even function, so the interval of integration can be halved and replaced by 0 ≤ ω < ∞, while the resulting integral is doubled, with the result that f(x) = 1 a ∞ 0 e−aω cos(ωx)dω = 1 x2 + a2 . The Inverse Laplace Transform Given an elementary function f(x) for which the Laplace transform F(s) exists, the determi- nation of the form of F(s) is usually a matter of routine integration. However, when finding f(x) from F(s) cannot be accomplished by use of a table of Laplace transform pairs and the properties of the transform, it becomes necessary to make use of the Laplace inversion formula f(x) = 1 2πi γ+i∞ γ−i∞ F(s)esx ds. (17) Here the real number γ must be chosen such that all the poles of the integrand lie to the left of the line s = γ in the complex s-plane. This integral is to be interpreted as the limit as R → ∞ of a contour integral around the contour shown in Figure 2. This is called the Bromwich contour after the Cambridge mathematician T.J.I'A. Bromwich who introduced it at the beginning of the last century. Example: To illustrate the application of the Laplace inversion integral it will suffice to consider finding f(x) = L−1 {1/ √ s}. The function 1 √ s has a branch point at the origin, so the Bromwich contour must be modified to make the function single valued inside the contour. We will use the contour shown in Figure 3, where the branch point is enclosed in a small circle about the origin while the complex s-plane is cut along the negative real axis to make the function single valued inside the contour. Let CR1 denote the large circular arc and CR2 denote the small circle around the origin. Then on CR1 s = γ + Reiθ for π 2 ≤ θ ≤ 3π 2 , and for subsequent use we now set θ = π 2 + φ, so s = γ + iReiφ with 0 ≤ φ ≤ π. Consequently, ds = −Reiφ dφ, with the result that \|ds\| = Rdφ. Thus, when R is sufficiently large \|s\| = γ + iReiφ ≥ Reiφ − \|γ\| = R − γ. Also for subsequent use, we need the result that \|esx \| = \|exp x [(γ − R sin φ) + iR cos φ] \| = eγx exp [−Rx sin φ] . Notes for Handbook Users xxxix R 0 Re{s} Pole Im{s} ␪ ␥ Figure 2. The Bromwich contour for the inversion of a Laplace transform. The integral around the modified Bromwich contour is the sum of the integrals along each of its separate parts, so we now estimate the magnitudes of the respective integrals. The magnitude of the integral around the large circular arc CR1 can be estimated as IR = ABEF esx √ s ds ≤ ABEF \|esx \| \|s\| 1/2 \|ds\| ≤ eγx R (R−γ)1/2 π 0 exp [−Rx sin φ]dφ. The symmetry of sin φ about φ = 1 2 π allows the inequality to be rewritten as IR ≤ 2eγx R (R−γ)1/2 π/2 0 exp [−Rx sin φ]dφ, so after use of the Jordan inequality in form (iii), this becomes IR ≤ πeγx (R−γ)1/2x 1 − e−Rx , when x > 0. This shows that when x > 0, lim R→∞ IR = 0, so that the integral around CR1 vanishes in the limit as R → ∞. xl Notes for Handbook Users R E D CB Re{s} F A 0 Im{s} ␥ ␧ Figure 3. The modified Bromwich contour with an indentation and a cut. On the small circle CR2 with radius ε we have s = εeiθ , so ds = iεeiθ dθ and s1/2 = eiθ/2 √ ε, so the integral around CR2 becomes π −π 1 eiθ/2 √ ε exp [εx (cos θ + i sin θ)] iεeiθ dθ, but this vanishes as ε → 0, so in the limit the integral around CR2 also vanishes. Along the top BC of the branch cut s = reπi = −r, so √ s = eπi/2 √ r = i √ r, so that ds = −dr. Along the bottom BC of the branch cut the situation is different, because there s = re−πi = −r, so √ s = e−πi/2 √ r = −i √ r, where again ds = −dr. The construction of the Bromwich contour has ensured that no poles lie inside it, so from the Cauchy residue theorem, in the limit as R → ∞ and ε → 0, the only contributions to the contour integral come from integration along opposite sides of the branch cut, so we arrive at the result 1 2πi γ+i∞ γ−i∞ esx √ s ds = 1 2πi − 0 ∞ ie−rx √ r dr + ∞ 0 ie−rx √ r dr = 1 π ∞ 0 e−rx √ r dr. Notes for Handbook Users xli Finally, the change of variable r = u2 , followed by setting ν = u √ x, changes this result to 1 2πi γ+i∞ γ−i∞ esx √ s ds = 2 π √ x ∞ 0 e−v2 dν. This last definite integral is a standard integral, and from entry 15.3.1(29) we have ∞ 0 e−ν2 dν = √ π/2, so we have shown that L−1 1 √ s = 1 √ πx , for Re{s} > 0. The inversion integral can generate an infinite series if an infinite number of isolated poles lie along a line parallel to the imaginary s-axis. This happens with L−1 1 s cosh s , where the poles are actually located on the imaginary axis. We omit the details, but straightforward reasoning using the standard Bromwich contour shows that f(x) = L−1 1 s cosh s = 1 + 4 π ∞ n=0 (−1)n+1 cos [(2n + 1)πx/2] 2n + 1 . To understand why this periodic representation of f(x) has occurred, notice that F(s) = 1/[s cosh s] is the Laplace transform of the piecewise continuous function f(x) =    0, 0 < x < 1 2, 1 < x < 3 0, 3 < x < 4, that is periodic with period 4 and defined for x ≥ 0. So f(x) is in fact the Fourier series representation of this function with period 4 when it is defined for all x. Here the term period is used in the usual sense that X is the period of f(x) if f(X + x) = f(x) is true for all x and X is the smallest value for which this result is true. 0.8 Geometry 13 Parallelogram Area A = ah = ab sin α. The centroid C is located at the point of intersection of the diagonals. Trapezium A quadrilateral with two sides parallel, where h is the perpendicular distance between the parallel sides. Area A = 1 2 (a + b) h. The centroid C is located on PQ, the line joining the midpoints of AB and CD, with QC = h 3 (a + 2b) (a + b) . Rhombus A parallelogram with all sides of equal length. Area A = a2 sin α. 14 Chapter 0 Quick Reference List of Frequently Used Data The centroid C is located at the point of intersection of the diagonals. Cube Area A = 6a2 . Volume V = a3 . Diagonal d = a √ 3. The centroid C is located at the midpoint of a diagonal. Rectangular parallelepiped Area A = 2 (ab + ac + bc) . Volume V = abc. Diagonal d = a2 + b2 + c2. 0.8 Geometry 15 The centroid C is located at the midpoint of a diagonal. Pyramid Rectangular base with sides of length a and b and four sides comprising pairs of identical isosceles triangles. Area of sides AS = a h2 + (b/2)2 + b h2 + (a/2)2. Area of base AB = ab. Total area A = AB + AS. Volume V = 1 3 abh. The centroid C is located on the axis of symmetry with OC = h/4. 0.8 Geometry 17 The centroid C is located on the line from the centroid O of the base triangle to the vertex, with OC = h/4. Oblique prism with plane end faces If AB is the area of a plane end face and h is the perpendicular distance between the parallel end faces, then Total area = Area of plane sides + 2AB. Volume V = ABh. The centroid C is located at the midpoint of the line C1C2 joining the centroid of the parallel end faces. Circle Area A = πr2 , Circumference L = 2πr, where r is the radius of the circle. The centroid is located at the center of the circle. Arc of circle Length of arc AB : s = rα (α radians). 0.8 Geometry 19 The centroid C is located on the axis of symmetry with OC = a3 12A . Annulus Area A = π(R2 − r2 ) [r < R] . The centroid C is located at the center. Right circular cylinder Area of curved surface AS = 2πrh. Area of plane ends AB = 2πr2 . Total Area A = AB + AS. Volume V = πr2 h. 20 Chapter 0 Quick Reference List of Frequently Used Data The centroid C is located on the axis of symmetry with OC = h/2. Right circular cylinder with an oblique plane face Here, h1 is the greatest height of a side of the cylinder, h2 is the shortest height of a side of the cylinder, and r is the radius of the cylinder. Area of curved surface AS = πr(h1 + h2). Area of plane end faces AB = πr2 + πr r2 + h1 − h2 2 2 . Total area A = πr  h1 + h2 + r + r2 + (h1 − h2) 2 2   . Volume V = πr2 2 (h1 + h2) . The centroid C is located on the axis of symmetry with OC = (h1 + h2) 4 + 1 16 (h1 − h2) 2 (h1 + h2) . 0.8 Geometry 21 Cylindrical wedge Here, r is radius of cylinder, h is height of wedge, 2a is base chord of wedge, b is the greatest perpendicular distance from the base chord to the wall of the cylinder measured perpendicular to the axis of the cylinder, and α is the angle subtended at the center O of the normal cross-section by the base chord. Area of curved surface AS = 2rh b (b − r) α 2 + a . Volume V = h 3b a 3r2 − a2 + 3r2 (b − r) α 2 . Right circular cone Area of curved surface AS = πrs. Area of plane end AB = πr2 . Total area A = AB + AS. Volume V = 1 3 πr2 h. 0.8 Geometry 23 General cone If A is a area of the base and h is the perpendicular height, then Volume V = 1 3 Ah. The centroid C is located on the line joining the centroid O of the base to the vertex P with OC = 1 4 OP. Sphere Area A = 4πr2 (r is radius of sphere). Volume V = 4 3 πr3 . 24 Chapter 0 Quick Reference List of Frequently Used Data The centroid is located at the center. Spherical sector Here, h is height of spherical segment cap, a is radius of plane face of spherical segment cap, and r is radius of sphere. For the area of the spherical cap and conical sides, A = πr(2h + a). Volume V = 2πr2 h 3 . The centroid C is located on the axis of symmetry with OC = 3 8 (2r − h). Spherical segment Here, h is height of spherical segment, a is radius of plane face of spherical segment, r is radius of sphere, and a = h(2r − h). Area of spherical surface AS = 2πrh. Area of plane face AB = πa2 . Total area A = AB + AS. Volume V = 1 3 πh2 (3r − h) = 1 6 πh(3a2 + h2 ). The centroid C is located on the axis of symmetry with OC = 3 4 (2r − h) 2 (3r − h) . 0.8 Geometry 25 Spherical segment with two parallel plane faces Here, a1 and a2 are the radii of the plane faces and h is height of segment. Area of spherical surface AS = 2πrh. Area of plane end faces AB = π(a2 1 + a2 2). Total area A = AB + AS. Volume V = 1 6 πh(3a2 1 + 3a2 2 + h2 ). Ellipsoids of revolution Let the ellipse have the equation x2 a2 + y2 b2 = 1. When rotated about the x-axis the volume is Vx = 4 3 πab2 . When rotated about the y-axis the volume is Vy = 4 3 πa2 b. 1.1 Algebraic Results Involving Real and Complex Numbers 31 1.1.2.8 Carleman's Inequality If a1, a2, . . . , an is any set of positive numbers, then the geometric and arithmetic means satisfy the inequality n k=1 Gk ≤ eAn or, equivalently, n k=1 (a1a2 · · · ak) 1/k ≤ e a1 + a2 + · · · + an n , where e is the best possible constant in this inequality. The next inequality to be listed is of a somewhat different nature than that of the previous ones in that it involves a function of the type known as convex. When interpreted geometrically, a function f(x) that is convex on an interval I = [a, b] is one for which all points on the graph of y = f(x) for a < x < b lie below the chord joining the points (a, f(a)) and (b, f(b)). Definition of convexity. A function f(x) defined on some interval I = [a, b] is said to be convex on I if, and only if, f[(1 − λ) a + λf(b)] ≤ (1 − λ)f(a) + λf(b), for a = b and 0 < λ < 1. The function is said to be strictly convex on I if the preceding inequality is strict, so that f[(1 − λ) a + λf(b)] < (1 − λ)f(a) + λf(b). 1.1.2.9 Jensen's Inequality Let f(x) be convex on the interval I = [a, b], let x1, x2, . . . , xn be points in I, and take λ1, λ2, . . . , λn to be nonnegative numbers such that λ1 + λ2 + · · · + λn = 1. 50 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus sets of increasingly accurate approximations are generated until at the Mth stage only one approximation remains in the set SM , and this is the required improved approximation to S. The approach is illustrated is the following tabulation where it is applied to the following alternating series with a known sum ∞ r=0 (−1) r (r + 1) 2 = π2 12 . Setting sn = n r=0 (−1)2 (r + 1)2 and applying the averaging method to s5, s6, s7, s8, and s9 gives the following results, where each entry in column Sm is obtained by averaging the entries that lie immediately above and below it in the previous column Sm−1. n sn S1 S2 S3 S4 5 0.810833 0.821037 6 0.831241 0.822233 0.823429 0.822421 7 0.815616 0.822609 0.822457 0.821789 0.822493 8 0.827962 0.822376 0.822962 9 0.817962 The final approximation 0.822457 in column S4 should be compared with the correct value that to six decimal places is 0.822467. To obtain this accuracy by direct summation would necessitate summing approximately 190 terms. The convergence of the approximation can be seen by examining the number of decimal places that remain unchanged in each column Sm. Improved accuracy can be achieved either by using a larger set of values sn, or by starting with a large value of n. 1.4 DETERMINANTS 1.4.1 Expansion of Second- and Third-Order Determinants 1.4.1.1 The determinant associated with the 2 × 2 matrix 1. A = a11 a12 a21 a22 , (see 1.5.1.1.1) with elements aij comprising real or complex numbers, or functions, denoted either by \|A\| or by det A, is defined by 2. \|A\| = a11a22 − a12a21. This is called a second-order determinant. 58 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus 2. 1 = a1, 2 = a1 1 a3 a2 , 3 = a1 1 0 a3 a2 a1 a5 a4 a3 , · · · , n = a1 1 0 0 . . . 0 a3 a2 a1 1 . . . 0 a5 a4 a3 a2 . . . 0 ... ... ... ... ... ... a2n−1 a2n−2 a2n−3 a2n−4 . . . an , and set ar = 0 for r > n. 3. Then the necessary and sufficient conditions for the zeros of P(λ) all to have negative real parts (the Routh−Hurwitz conditions) are i > 0 [i = 1, 2, . . . , n]. 1.5 MATRICES 1.5.1 Special Matrices 1.5.1.1 Basic Definitions 1. An m × n matrix is a rectangular array of elements (numbers or functions) with m rows and n columns. If a matrix is denoted by A, the element (entry) in its i'th row and j'th column is denoted by aij, and we write A = [aij]. A matrix with as many rows as columns is called a square matrix. 2. A square matrix A of the form A =        λ1 0 0 · · · 0 0 λ2 0 · · · 0 0 0 λ3 · · · 0 ... ... ... ... ... 0 0 0 · · · λn        in which all entries away from the leading diagonal (the diagonal from top left to bottom right) are zero is called a diagonal matrix. This diagonal matrix is often abbreviated as A = diag{λ1, λ2, . . . , λn}, where the order in which the elements λ1, λ2, . . . , λn appear in this notation is the order in which they appear on the leading diagonal of A. 3. The identity matrix, or unit matrix, is a diagonal matrix I in which all entries in the leading diagonal are unity. 4. A null matrix is a matrix of any shape in which every entry is zero. 1.5 Matrices 59 5. The n × n matrix A = [aij] is said to be reducible if the indices 1, 2, . . . , n can be divided into two disjoint nonempty sets i1, i2, . . . , iµ; j1, j2, . . . , jv (µ + ν = n), such that aiα jβ = 0 [α = 1, 2, . . . , µ; β = 1, 2, . . . , ν] . Otherwise A will be said to be irreducible. 6. An m × n matrix A is equivalent to an m × n matrix B if, and only if, B = PAQ for suitable nonsingular m × m and n × n matrices P and Q, respectively. A matrix D is said to be nonsingular if \|D\| = 0. 7. If A = [aij] is an m × n matrix with element aij in its i'th row and the j'th column, then the transpose AT of A is n × m matrix AT = [bij] with bij = aji; that is, the transpose AT of A is the matrix derived from A by interchanging rows and columns, so the i'th row of A becomes the i'th column of AT for i = 1, 2, . . . , m. 8. If A is an n × n matrix, its adjoint, denoted by adj A, is the transpose of the matrix of cofactors Cij of A, so that adj A = [Cij] T . (see 1.4.2.1.3) 9. If A = [aij] is an n × n matrix with a nonsingular determinant \|A\|, then its inverse A−1 , also called its multiplicative inverse, is given by A−1 = adj A \|A\| , A−1 A = AA−1 = I. 10. The trace of an n × n matrix A = [aij], written tr A, is defined to be the sum of the terms on the leading diagonal, so that tr A = a11 + a22 + · · · + ann. 11. The n × n matrix A = [aij] is symmetric if aij = aji for i, j = 1, 2, . . . , n. 12. The n × n matrix A = [aij] is skew-symmetric if aij = −aji for i, j = 1, 2, . . . , n; so in a skew-symmetric matrix each element on the leading diagonal is zero. 13. An n × n matrix A = [aij] is of upper triangular type if aij = 0 for i > j and of lower triangular type if aij = 0 for j > i. 14. A real n × n matrix A is orthogonal if, and only if, AAT = I. 15. If A = [aij] is an n × n matrix with complex elements, then its Hermitian transpose A† is defined to be A† = [aji], with the bar denoting the complex conjugate operation. The Hermitian transpose operation is also denoted by a superscript H, so that A† = AH = ¯A T . 60 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus A Hermitian matrix A is said to be normal if A and A† commute, so that AA† = A† A or, in the equivalent notation, AAH = AH A. 16. An n × n matrix A is Hermitian if A = A† , or equivalently, if A = A T , with the overbar denoting the complex conjugate operation. 17. An n × n matrix A is unitary if AA† = A† A = I. 18. If A is an n × n matrix, the eigenvectors X satisfy the equation AX = λX, while the eigenvalues λ satisfy the characteristic equation \|A − λI\| = 0. 19. An n × n matrix A is nilpotent with index k, if k is the smallest integer such that Ak−1 = 0 but Ak = 0. For example, if A =   0 0 0 3 0 0 1 2 0  , A2 =   0 0 0 0 0 0 6 0 0   and A3 = 0, so A is nilpotent with index 3. 20. An n × n matrix A is idempotent if A2 = A. 21. An n × n matrix A is positive definite if xT Ax > 0, for x = 0 an n element column vector. 22. An n × n matrix A is nonnegative definite if xT Ax ≥ 0, for x = 0 an n element column vector. 23. An n × n matrix A is diagonally dominant if \|aii\| > j=i \|aij\| for all i. 24. Two matrices A and B are equal if, and only if, they are both of the same shape and corresponding elements are equal. 25. Two matrices A and B can be added (or subtracted) if, and only if, they have the same shape. If A = [aij] , B = [bij], and C = A + B, with C = [cij] , then cij = aij + bij. Similarly, if D = A − B, with D = [dij] , then dij = aij − bij. 26. If k is a scalar and A = [aij] is a matrix, then kA = [kaij] . 27. If A is an m × n matrix and B is a p × q matrix, the matrix product C = AB, in this order, is only defined if n = p, and then C is an m × q matrix. When the matrix product 1.5 Matrices 63 3. Q (x) ≡ (x, Ax) . If the n × n matrix A is Hermitian, so that A T = A, where the bar denotes the complex conjugate operation, the quadratic form associated with the Hermitian matrix A and the vector x, which may have complex elements, is the real quadratic form 4. Q (x) = (x, Ax). It is always possible to express an arbitrary quadratic form 5. Q (x) = n i=1 n j=1 αijxixj in the form 6. Q (x) = (x, Ax), in which A = [aij] is a symmetric matrix, by defining 7. aii = αii for i = 1, 2, . . . , n and 8. aij = 1 2 (αij + αji) for i, j = 1, 2, . . . , n and i = j. 9. When a quadratic form Q in n variables is reduced by a nonsingular linear transformation to the form Q = y2 1 + y2 2 + · · · + y2 p − y2 p+1 − y2 p+2 − · · · − y2 r , the number p of positive squares appearing in the reduction is an invariant of the quadratic form Q, and does not depend on the method of reduction itself (Sylvester's law of inertia). 10. The rank of the quadratic form Q in the above canonical form is the total number r of squared terms (both positive and negative) appearing in its reduced form (r ≤ n). 11. The signature of the quadratic form Q above is the number s of positive squared terms appearing in its reduced form. It is sometimes also defined to be 2s − r. 12. The quadratic form Q (x) = (x, Ax) is said to be positive definite when Q (x) > 0 for x = 0. It is said to be positive semidefinite if Q (x) ≥ 0 for x = 0. 1.5.2.2 Basic Theorems on Quadratic Forms 1. Two real quadratic forms are equivalent under the group of linear transformations if, and only if, they have the same rank and the same signature. 2. A real quadratic form in n variables is positive definite if, and only if, its canonical form is Q = z2 1 + z2 2 + · · · + z2 n. 3. A real symmetric matrix A is positive definite if, and only if, there exists a real nonsingular matrix M such that A = MMT . 4. Any real quadratic form in n variables may be reduced to the diagonal form Q = λ1z2 1 + λ2z2 2 + · · · + λnz2 n, λ1 ≥ λ2 ≥ · · · ≥ λn 64 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus by a suitable orthogonal point-transformation. See 32 in Section 1.5.1 for the diago- nalization of a matrix. 5. The quadratic form Q = (x, Ax) is positive definite if, and only if, every eigenvalue of A is positive; it is positive semidefinite if, and only if, all the eigenvalues of A are nonnegative; and it is indefinite if the eigenvalues of A are of both signs. 6. The necessary conditions for a Hermitian matrix A to be positive definite are (i) aii > 0 for all i, (ii) aiiaij > \|aij\| 2 for i = j, (iii) the element of largest modulus must lie on the leading diagonal, (iv) \|A\| > 0. 7. The quadratic form Q = (x, Ax) with A Hermitian will be positive definite if all the principal minors in the top left-hand corner of A are positive, so that a11 > 0, a11 a12 a21 a22 > 0, a11 a12 a13 a21 a22 a23 a31 a32 a33 > 0, · · · . 8. Let λ1, λ2, . . . , λn be the eigenvalues (they are real) of the n × n real symmetric matrix A associated with a real quadratic form Q (x), and let x1,x2, . . . , xn be the corresponding normalized eigenvectors. Then if P = [x1,x2, . . . , xn] is the n × n orthogonal matrix with xi as its ith column, the matrix D = P−1 AP is a diagonal matrix with λ1, λ2, . . . , λn as the elements on its leading diagonal. The change of variable x = Py, with y = [y1, y2, . . . , yn] T transforms Q (x) into the standard form Q (x) = λ1y2 1 + λ2y2 2 + · · · + λny2 n. Setting λmin = min{˘1, ˘2, . . . , ˘n} and λmax = max{˘1, ˘2, . . . , ˘n} it follows directly that λminyT y ≤ Q (x) ≤ λmaxyT y. 1.5.3 Differentiation and Integration of Matrices 1.5.3.1 If the n × n matrices A(t) and B(t) have elements that are differentiable function of t, so that A (t) = [aij (t)] , B (t) = [bij (t)] , then 1. d dt A (t) = d dt aij (t) 2. d dt [A (t) ± B (t)] = d dt aij (t) ± d dt bij (t) = d dt A (t) ± d dt B (t) . 1.6 Permutations and Combinations 67 Taking the inverse Laplace transform of each element of the matrix to transform back from the Laplace transform variable s to the original variable z gives eAz = L−1 {[sI − A] −1 } = e3z ze3z 0 e3z . 1.5.5 The Gerschgorin Circle Theorem 1.5.5.1 The Gerschgorin Circle Theorem Each eigenvalue of an arbitrary n × n matrix A = [aij] lies in at least one of the circles C1,C2,. . . , Cn in the complex plane, where the circle Cr with radius ρr has its center at arr, where arr is the r'th element of the leading diagonal of A, and ρr = n j=1 j=r \|arj\| = \|ar1\| + \|ar2\| + · · · + \|ar,r−1\| + \|ar,r+1\| + · · · + \|arn\| . 1.6 PERMUTATIONS AND COMBINATIONS 1.6.1 Permutations 1.6.1.1 1. A permutation of n mutually distinguishable elements is an arrangement or sequence of occurrence of the elements in which their order of appearance counts. 2. The number of possible mutually distinguishable permutations of n distinct elements is denoted either by n Pn or by nPn, where n Pn = n (n − 1) (n − 2) · · · 3.2.1 = n! 3. The number of possible mutually distinguishable permutations of n distinct elements m at a time is denoted either by n Pm or by nPm, where n Pm = n! (n − m)! [0! ≡ 1]. 4. The number of possible identifiably different permutations of n elements of two different types, of which m are of type 1 and n − m are of type 2 is n! m! (n − m)! = n m . (binomial coefficient) This gives the relationship between binomial coefficients and the number of m element subsets of an n set. Expressed differently, this says that the coefficient of xm in (1 + x) n is the number of ways x's can be selected from precisely m of the n factors of the n-fold product being expanded. 68 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus 5. The number of possible identifiably different permutations of n elements of m different types, in which mr are of type r, with m1 + m2 + · · · + mr = n, is n! m1!m2! · · · mr! . (multinomial coefficient) 1.6.2 Combinations 1.6.2.1 1. A combination of n mutually distinguishable elements m at a time is a selection of m elements from the n without regard to their order of arrangement. The number of such combinations is denoted either by n Cm or by nCm, where n Cm = n m . 2. The number of combinations of n mutually distinguishable elements in which each element may occur 0, 1, 2, . . . , m times in any combination is n + m − 1 m = n + n − 1 n − 1 . 3. The number of combinations of n mutually distinguishable elements in which each element must occur at least once in each combination is m − 1 n − 1 . 4. The number of distinguishable samples of m elements taken from n different elements, when each element may occur at most once in a sample, is n (n − 1) (n − 2) · · · (n − m + 1). 5. The number of distinguishable samples of m elements taken from n different elements, when each element may occur 0, 1, 2, . . . , m times in a sample is nm . 1.7 PARTIAL FRACTION DECOMPOSITION 1.7.1 Rational Functions 1.7.1.1 A function R (x) of the from 1. R (x) = N (x) D (x) , 1.7 Partial Fraction Decomposition 69 where N (x) and D (x) are polynomials in x, is called a rational function of x. The replacement of R (x) by an equivalent expression involving a sum of simpler rational func- tions is called a decomposition of R (x) into partial fractions. This technique is of use in the integration of arbitrary rational functions. Thus in the identity 3x2 + 2x + 1 x3 + x2 + x = 1 x + 2x + 1 x2 + x + 1 , the expression on the right-hand side is a partial fraction expansion of the rational function 3x2 + 2x + 1 x3 + x2 + x . 1.7.2 Method of Undetermined Coefficients 1.7.2.1 1. The general form of the simplest possible partial fraction expansion of R (x) in 1.7.1.1 depends on the respective degrees of N (x) and D (x), and on the decomposition of D (x) into real factors. The form of the partial fraction decomposition to be adopted is determined as follows. 2. Case 1. (Degree of N (x) less than the degree of D (x)). (i) Let the degree of N (x) be less than the degree of D (x), and factor D (x) into the simplest possible set of real factors. There may be linear factors with multiplicity 1, such as (ax + b); linear factors with multiplicity r, such as (ax + b) r ; quadratic factors with multiplicity 1, such as (ax2 + bx + c); or quadratic factors with mul- tiplicity m such as ax2 + bx + c m , where a, b, . . . , are real numbers, and the quadratic factors cannot be expressed as the product of real linear factors. (ii) To each linear factor with multiplicity 1, such as (ax + b), include in the partial fraction decomposition a term such as A ax + b , where A is an undetermined constant. (iii) To each linear factor with multiplicity r, such as (ax + b) r , include in the partial fraction decomposition terms such as B1 ax + b + B2 (ax + b) 2 + · · · + Br (ax + b) r , where B1, B2, . . . , Br are undetermined constants. (iv) To each quadratic factor with multiplicity 1, such as (ax2 + bx + c), include in the partial fraction decomposition a term such as C1x + D1 ax2 + bx + c , where C1, D1 are undetermined constants. 72 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus 1.8 CONVERGENCE OF SERIES 1.8.1 Types of Convergence of Numerical Series 1.8.1.1 Let {uk} with k = 1, 2, . . . , be an infinite sequence of numbers, then the infinite numerical series 1. ∞ k=1 uk = u1 + u2 + u3 + · · · is said to converge to the sum S if the sequence {Sn} of partial sums 2. Sn = n k=1 uk = u1 + u2 + · · · + un has a finite limit S, so that 3. S = lim n→∞ Sn. If S is infinite, or the sequence {Sn} has no limit, the series 1.8.1.1.1 is said to diverge. 4. The series 1.8.1.1.1 is convergent if for each ε > 0 there is a number N (ε) such that \|Sm − Sn\| < ε for all m > n > N. (Cauchy criterion for convergence) 5. The series 1.8.1.1.1 is said to be absolutely convergent if the series of absolute values n k=1 \|uk\| = \|u1\| + \|u2\| + \|u3\| + · · · converges. Every absolutely convergent series is convergent. If series 1.8.1.1.1 is such that it is convergent, but it is not absolutely convergent, it is said to be conditionally convergent. 1.8.2 Convergence Tests 1.8.2.1 Let the series 1.8.1.1.1 be such that uk = 0 for any k and 1. lim k→∞ uk+1 uk = r. Then series 1.8.1.1.1 is absolutely convergent if r < 1 and it diverges if r > 1. The test fails to provide information about convergence or divergence if r = 1. (d'Alembert's ratio test) 1.8.2.2 Let the series 1.8.1.1.1 be such that uk = 0 for any k and 1. lim k→∞ \|uk\| 1/k = r. 1.8 Convergence of Series 73 The series 1.8.1.1.1 is absolutely convergent if r < 1 and it diverges if r > 1. The test fails to provide information about convergence or divergence if r = 1. (Cauchy's n'th root test) 1.8.2.3 Let the series ∞ k=1 uk be such that lim k→∞ k uk uk+1 − 1 = r. Then the series is absolutely convergent if r > 1 and it is divergent if r < 1. The test fails to provide information about convergence or divergence if r = 1. (Raabe's test: This test is a more delicate form of ratio test and it is often useful when the ratio test fails because r = 1.) 1.8.2.4 Let the series in 1.8.1.1.1 be such that uk ≥ 0 for k = 1, 2, . . . , and let ∞ k=1 ak be a convergent series of positive terms such that uk ≤ ak for all k. Then 1. ∞ k=1 uk is convergent and ∞ k=1 uk ≤ ∞ k=1 ak (comparison test for convergence) 2. If ∞ k=1 ak is a divergent series of nonnegative terms and uk ≥ ak for all k, then ∞ k=1 uk is divergent. (comparison test for divergence) 1.8.2.5 Let ∞ k=1 uk be a series of positive terms whose convergence is to be determined, and let ∞ k=1 ak be a comparison series of positive terms known to be either convergent or divergent. Let 1. lim k→∞ uk ak = L, where L is either a nonnegative number or infinity. 2. If ∞ k=1 ak converges and 0 ≤ L < ∞, ∞ k=1 uk converges. 3. If ∞ k=1 ak diverges and 0 < L ≤ ∞, ∞ k=1 uk diverges. (limit comparison test) 1.8.2.6 Let ∞ k=1 uk be a series of positive nonincreasing terms, and let f(x) be a nonincreasing function defined for k ≥ N such that 80 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus 1.10.1.2 Let D be a region in which the functional series 1.10.1.1.1 converges for each value of x. Then the series is said to converge uniformly in D if, for every ε > 0, there exists a number N(ε) such that, for n > N, it follows that 1. ∞ k=n+1 fk(x) < ε, for all x in D. The Cauchy criterion for the uniform convergence of series 1.10.1.1.1 requires that 2. \|fm(x) + fm+1(x) + · · · + fn(x)\| < ε, for every ε > 0, all x in D and all n > m > N. 1.10.1.3 Let {fk(x)}, k = 1, 2, . . . , be an infinite sequence of functions, and let {Mk}, k = 1, 2, . . . , be a sequence of positive numbers such that ∞ k=1 Mk is convergent. Then, if 1. \|fk(x)\| ≤ Mk, for all x in a region D and all k = 1, 2, . . . , the functional series in 1.10.1.1.1 converges uniformly for all x in D. (Weierstrass's M test) 1.10.1.4 Let the series 1.10.1.1.1 converge for all x in some region D, in which it defines a function 1. f(x) = ∞ k=1 fk(x) . Then the series is said to converge uniformly to f(x) in D if, for every ε > 0, there exists a number N(ε) such that, for n > N, it follows that 2. f(x) − n k=0 fk(x) < ε for all x in D. 1.10.1.5 Let the infinite sequence of functions {fk (x)}, k = 1, 2, . . . , be continuous for all x in some region D. Then if the functional series ∞ k=1 fk (x) is uniformly convergent to the function f(x) for all x in D, the function f(x) is continuous in D. 1.11 Functional Series 81 1.10.1.6 Suppose the series 1.10.1.1.1 converges uniformly in a region D, and that for each x in D the sequence of functions {gk (x)}, k = 1, 2, . . . is monotonic and uniformly bounded, so that for some number L > 0 1. \|gk (x)\| ≤ L for each k = 1, 2, . . . , and all x in D. Then the series 2. ∞ k=1 fk (x) gk (x) converges uniformly in D. (Abel's theorem) 1.10.1.7 Suppose the partial sums Sn (x) = n k=1 fk (x) of 1.10.1.1.1 are uniformly bounded, so that 1. n k=1 fk (x) < L for some L, all n = 1, 2, . . . , and all x in the region of convergence D. Then, if {gk (x)}, k = 1, 2, . . . , is a monotonic decreasing sequence of functions that approaches zero uniformly for all x in D, the series 2. ∞ k=1 fk (x) gk (x) converges uniformly in D. (Dirichlet's theorem) 1.10.1.8 If each function in the infinite sequence of functions {fk (x)}, k = 1, 2, . . . , is integrable on the interval [a, b], and if the series 1.10.1.1.1 converges uniformly on this interval, the series may be integrated term by term (termwise), so that 1. b a ∞ k=1 fk (x) dx = ∞ k=1 b a fk (x)dx [a ≤ x ≤ b] . 1.10.1.9 Let each function in the infinite sequence of functions {fk (x)}, k = 1, 2, . . . , have a continuous derivative fk (x) on the interval [a, b]. Then if series 1.10.1.1.1 converges on this interval, and if the series ∞ k=1 fk (x) converges uniformly on the same interval, the series 1.10.1.1.1 may be differentiated term by term (termwise), so that d dx ∞ k=1 fk (x) dx = ∞ k=1 fk (x) . 82 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus 1.11 POWER SERIES 1.11.1 Definition 1.11.1.1 A functional series of the form 1. ∞ k=0 ak (x − x0) k = a0 + a1 (x − x0) + a2 (x − x0) 2 + · · · is called a power series in x expanded about the point x0 with coefficients ak. The following is true of any power series: If it is not everywhere convergent, the region of convergence (in the complex plane) is a circle of radius R with its center at the point x0; at every interior point of this circle the power series 1.11.1.1.1 converges absolutely, and outside this circle it diverges. The circle is called the circle of convergence and its radius R is called the radius of convergence. A series that converges at all points of the complex plane is said to have an infinite radius of convergence (R = +∞). 1.11.1.2 The radius of convergence R of the power series in 1.11.1.1.1 may be determined by 1. R = lim k→∞ ak ak+1 , when the limit exists; by 2. R = 1 limk→∞ \|ak\| 1/k , when the limit exists; or by the Cauchy−Hadamard formula 3. R = 1 lim sup \|ak\| 1/k , which is always defined (though the result is difficult to apply). The circle of convergence of the power series in 1.11.1.1.1 is \|x − x0\| = R, so the series is absolutely convergent in the open disk \|x − x0\| < R and divergent outside it where x and x0 are points in the complex plane. 1.11.1.3 The power series 1.11.1.1.1 may be integrated and differentiated term by term inside its circle of convergence; thus 88 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus + 1 2! (x − x0) 2 ∂2 f ∂x2 (x0,y0) + 2 (x − x0) (y − y0) ∂2 f ∂x∂y (x0,y0) + (y − y0) 2 ∂2 f ∂y2 (x0,y0) + · · · . In its simplest form the remainder term Rn(x, y) satisfies a condition analogous to 1.12.1.3, so that 2. Rn (x, y) = 1 (n + 1)! (Dn+1f)(x0+θ1(x−x0),y0+θ2(y−y0)) [0 < θ1 < 1, 0 < θ2 < 1] where 3. Dn ≡ (x − x0) ∂ ∂x + (y − y0) ∂ ∂y n . 1.12.4 Order Notation (Big O and Little o) When working with a function f(x) it is useful to have a simple notation that indicates its order of magnitude, either for all x, or in a neighborhood of a point x0. This is provided by the so-called 'big oh' and 'little oh' notation. 1. We write f(x) = O [ϕ (x)] , and say f(x) is of order ϕ (x), or is 'big oh' ϕ (x), if there exists a real constant K such that \|f(x)\| ≤ K \|ϕ (x)\| for all x. The function f(x) is said to be of the order of ϕ (x), or to be 'big oh' ϕ (x) in a neighborhood of x0, written f(x) = O (ϕ (x)) as x → x0 if \|f(x)\| ≤ K \|ϕ (x)\| as x → x0. 2. If in a neighborhood of a point x0 two functions f(x) and g(x) are such that lim x→x0 f(x) g (x) = 0 the function f(x) is said to be of smaller order than g(x), or to be 'little oh' g(x), written f(x) = o (g (x)) as x → x0. In particular, these notations are useful when representing the error term in Taylor series expansions and when working with asymptotic expansions. 90 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus 2. h 0 \|f(x0 + t) + f(x0 − t) − f(x0 + 0) − f(x0 − 0)\| t dt exists, where it is assumed that f(x) is either continuous at x0 or it has a finite jump dis- continuity at x0 (a saltus) at which both the one-sided limits f(x0 − 0) and f(x0 + 0) exist. Thus, if f(x) is continuous at x0, the Fourier series of f(x) converges to the value f(x0) at the point x0, while if a finite jump discontinuity occurs at x0 the Fourier series converges to the average of the values f(x0 + 0) and f(x0 − 0) of f(x) to the immediate left and right of x0. (Dini's condition) 1.13.1.3 A function f(x) is said to satisfy Dirichlet conditions on the interval [a, b] if it is bounded on the interval, and the interval [a, b] can be partitioned into a finite number of subintervals inside each of which the function f(x) is continuous and monotonic (either increasing or decreasing). The Fourier series of a periodic function f(x) that satisfies Dirichlet conditions on the interval [a, b] converges at every point x0 of [a, b] to the value 1 2 {f(x0 + 0) + f(x0 − 0)} . (Dirichlet's result) 1.13.1.4 Let the function f(x) be defined on the interval [a, b], where a < b, and let the interval be partitioned into subintervals in an arbitrary manner with the ends of the intervals at 1. a = x0 < x1 < x2 < · · · < xn−1 < xn = b. Form the sum 2. n k=1 \|f(xk) − f(xk−1)\|. Then different partitions of the interval [a, b] that is, different choices of the points xk, will give rise to different sums of the form in 1.13.1.4.2. If the set of these sums is bounded above, the function f(x) is said to be of bounded variation on the interval [a, b]. The least upper bound of these sums is called the total variation of the function f(x) on the interval [a, b]. 1.13.1.5 Let the function f(x) be piecewise-continuous on the interval [a, b], and let it have a piecewise- continuous derivative within each such interval in which it is continuous. Then, at every point x0 of the interval [a, b], the Fourier series of the function f(x) converges to the value 1 2 {f(x0 + 0) + f(x0 − 0)}. 1.13.1.6 A function f(x) defined in the interval [0, l], can be expanded in a cosine series (half-range Fourier cosine series) of the form 1.14 Asymptotic Expansions 93 which f (x) exists, the Fourier series for f(x) may be differentiated term by term to yield a Fourier series that converges to f (x). Thus, if f(x) has the Fourier series given in 1.13.1.1.2, f (x) = ∞ k = 1 d dx ak cos kπx l + bk sin kπx l [−l ≤ x ≤ l] . (differentiation of Fourier series) 1.13.1.14 Let f(x) be a piecewise-continuous over [−l, l] with the Fourier series given in 1.13.1.1.2. Then 1 2 a2 0 + ∞ k=1 a2 k + b2 k ≤ 1 l l −l [f(x)] 2 dx. (Bessel's inequality) 1.13.1.15 Let f(x) be continuous over [−l, l], and periodic with period 2l, with the Fourier series given in 1.13.1.2. Then 1 2 a2 0 + ∞ k=1 a2 k + b2 k = 1 l l −l [f(x)] 2 dx. (Parseval's identity) 1.13.1.16 Let f(x) and g(x) be two functions defined over the interval [−l, l] with respective Fourier coefficients ak, bk and αk, βk, and such that the integrals l −l [f(x)] 2 dx and l −l [g (x)] 2 dx are both finite [f(x) and g(x) are square integrable]. Then the Parseval identity for the product function f(x)g(x) becomes a0α0 2 + ∞ k = 1 (akαk + bkβk) = 1 l l −l f(x) g (x) dx. (generalized Parseval identity) 1.14 ASYMPTOTIC EXPANSIONS 1.14.1 Introduction 1.14.1.1 Among the set of all divergent series is a special class known as the asymptotic series. These are series which, although divergent, have the property that the sum of a suitable number of terms provides a good approximation to the functions they represent. In the case of an alternating asymptotic series, the greatest accuracy is obtained by truncating the series at the term of smallest absolute value. When working with an alternating asymptotic series representing a function f(x), the magnitude of the error involved when f(x) is approximated by summing only the first n terms of the series does not exceed the magnitude of the (n + 1)'th term (the first term to be discarded). 94 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus An example of this type is provided by the asymptotic series for the function f(x) = ∞ x ex − t t dt. Integrating by parts n times gives f(x) = 1 x − 1 x2 + 2! x3 − · · · + (−1) n − 1 (n − 1)! xn + (−1) n n! ∞ x ex − t tn + 1 dt. It is easily established that the infinite series 1 x − 1 x2 + 2! x3 − · · · + (−1) n − 1 (n − 1)! xn + . . . is divergent for all x, so if f(x) is expanded as an infinite series, the series diverges for all x. The remainder after the n'th term can be estimated by using the fact that n! ∞ x ex −t tn + 1 dt < n! xn + 1 ∞ x ex − t dt = n! xn + 1 , from which it can be seen that when f(x) is approximated by the first n terms of this divergent series, the magnitude of the error involved is less than the magnitude of the (n + 1)'th term (see 1.8.2.7.2). For any fixed value of x, the terms in the divergent series decrease in magnitude until the N'th term, where N is the integral part of x, after which they increase again. Thus, for any fixed x, truncating the series after the N'th term will yield the best approximation to f(x) for that value of x. In general, even when x is only moderately large, truncating the series after only a few terms will provide an excellent approximation to f(x). In the above case, if x = 30 and the series is truncated after only two terms, the magnitude of the error involved when evaluating f(30) is less than 2!/303 = 7.4 × 10−5 . 1.14.2 Definition and Properties of Asymptotic Series 1. Let Sn(z) be the sum of the first (n + 1) terms of the series S (z) = A0 + A1 z + A2 z2 + . . . + An zn + . . . , where, in general, z is a complex number. Let Rn (z) = S(z) − Sn(z). Then the series S(z) is said to be an asymptotic expansion of f(z) if for arg z in some interval α ≤ arg z ≤ β, and for each fixed n, lim \|z\|→∞ zn Rn (z) = 0. 1.15 Basic Results from the Calculus 95 The relationship between f(z) and its asymptotic expansion S(z) is indicated by writing f(z) ∼ S(z). 2. The operations of addition, subtraction, multiplication, and raising to a power can be performed on asymptotic series just as on absolutely convergent series. The series obtained as a result of these operations will also be an asymptotic series. 3. Division of asymptotic series is permissible and yields another asymptotic series provided the first term A0 of the divisor is not equal to zero. 4. Term-by-term integration of an asymptotic series is permissible and yields another asymptotic series. 5. Term-by-term differentiation of an asymptotic series is not, in general, permissible. 6. For arg z in some interval α ≤ arg z ≤ β, the asymptotic expansion of a function f(x) is unique. 7. A series can be the asymptotic expansion of more than one function. 1.15 BASIC RESULTS FROM THE CALCULUS 1.15.1 Rules for Differentiation 1.15.1.1 Let u, v be differentiable functions of their arguments and let k be a constant. 1. dk dx = 0 2. d (ku) dx = k du dx 3. d (u + v) dx = du dx + dv dx (differentiation of a sum) 4. d (uv) dx = u dv dx + v du dx (differentiation of a product) 5. d dx (u/v) = v du dx − u dv dx v2 [v = 0] (differentiation of a quotient) 6. Let v be differentiable at some point x, and u be differentiable at the point v(x); then the composite function (u ◦ v) (x) is differentiable at the point x. In particular, if z = u(y) and y = v(x), so that z = u (v (x)) = (u ◦ v) (x), then dz dx = dz dy · dy dx or, equivalently, dz dx = u (y) v (x) = u (v (x)) v (x) . (chain rule) 1.15 Basic Results from the Calculus 97 3. f(x) dx = F(x) + C, where C is an arbitrary constant. The expression on the right-hand side of 1.15.2.1.3 is called an indefinite integral. The term indefinite is used because of the arbitrariness introduced by the arbitrary addi- tive constant of integration C. 4. u dv dx dx = uv − v du dx dx. (formula for integration by parts) This is often abbreviated to u dv = uv − v du. 5. A definite integral involves the integration of f(x) from a lower limit x = a to an upper limit x = b, and it is written b a f(x) dx. The first fundamental theorem of calculus asserts that if f(x) is an antiderivative of f(x), then b a f(x) dx = F(b) − F(a). Since the variable of integration in a definite integral does not enter into the final result it is called a dummy variable, and it may be replaced by any other symbol. Thus b a f(x) dx = b a f(t) dt = · · · = b a f(s) ds = F(b) − F(a). 6. If F(x) = x a f(t) dt, then dF/dx = f(x) or, equivalently, d dx x a f(t) dt = f(x) . (second fundamental theorem of calculus) 7. b a f(x) dx = − a b f(x) dx. (reversal of limits changes the sign) 8. Let f(x) and g(x) be continuous in the interval [a, b], and let g (x) exist and be continuous in this same interval. Then, with the substitution u = g(x), 1.15 Basic Results from the Calculus 99 + ϕ(α) ψ(α) ∂ ∂α [f(x, α)] dx. (differentiation of a definite integral with respect to a parameter) 13. If f(x) is integrable over the interval [a, b], then b a f(x) dx ≤ b a \|f(x)\| dx. (absolute value integral inequality) 14. If f(x) and g(x) are integrable over the interval [a, b] and f(x) ≤ g(x), then b a f(x) dx ≤ b a g (x) dx. (comparison integral inequality) 15. The following are mean-value theorems for integrals. (i) If f(x) is continuous on the closed interval [a, b], there is a number ξ, with a < ξ < b, such that b a f(x) dx = (b − a) f(ξ) . (ii) If f(x) and g(x) are continuous on the closed interval [a, b] and g(x) is monotonic (either decreasing on increasing) on the open interval [a, b], there is a number ξ, with a < ξ < b, such that b a f(x) g (x) dx = g (a) ξ a f(x) dx + g (b) b ξ f(x) dx. (iii) If in (ii), g(x) > 0 on the open interval [a, b], there is a number ξ, with a < ξ < b, such that when g(x) is monotonic decreasing b a f(x) g (x) dx = g(a) ξ a f(x) dx, and when g(x) is monotonic increasing b a f(x) g (x) dx = g(b) b ξ f(x) dx. 1.15.3 Reduction Formulas 1.15.3.1 When, after integration by parts, an integral containing one or more parameters can be expressed in terms of an integral of similar form involving parameters with reduced values, the result is called a reduction formula (recursion relation). Its importance derives from the 1.15 Basic Results from the Calculus 101 so we arrive at the Wallis infinite product (see 1.9.2.15) π 2 = ∞ k = 1 2k 2k − 1 2k 2k + 1 . 1.15.4 Improper Integrals An improper integral is a definite integral that possesses one or more of the following properties: (i) the interval of integration is either semi-infinite or infinite in length; (ii) the integrand becomes infinite at an interior point of the interval of integration; (iii) the integrand becomes infinite at an end point of the interval of integration. An integral of type (i) is called an improper integral of the first kind, while integrals of types (ii) and (iii) are called improper integrals of the second kind. The evaluation of improper integrals. I. Let f(x) be defined and finite on the semi-infinite interval [a, ∞). Then the improper integral ∞ a f(x) dx is defined as ∞ a f(x) dx = lim R→∞ R a f(x) dx. The improper integral is said to converge to the value of this limit when it exists and is finite. If the limit does not exist, or is infinite, the integral is said to diverge. Corresponding definitions exist for the improper integrals a −∞ f(x) dx = lim R→∞ a −R f(x) dx, and ∞ −∞ f(x) dx = lim R1→∞ a −R1 f(x) dx + lim R2→∞ R2 a f(x) dx, [arbitrary a], where R1 and R2 tend to infinity independently of each other. II. Let f(x) be defined and finite on the interval [a, b] except at a point ξ interior to [a, b] at which point it becomes infinite. The improper integral b a f(x) dx is then defined as b a f(x) dx = lim ε→0 ξ−ε a f(x) dx + lim δ→0 b ξ+δ f(x) dx, where ε > 0, δ > 0 tend to zero independently of each other. The improper integral is said to converge when both limits exist and are finite, and its value is then the sum of 102 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus the values of the two limits. The integral is said to diverge if at least one of the limits is undefined, or is infinite. III. Let f(x) be defined and finite on the interval [a, b] except at an end point, say, at x = a, where it is infinite. The improper integral b a f(x) dx is then defined as b a f(x) dx = lim ε→0 b a+ε f(x) dx, where ε > 0. The improper integral is said to converge to the value of this limit when it exists and is finite. If the limit does not exist, or is infinite, the integral is said to diverge. A corresponding definition applies when f(x) is infinite at x = b. IV. It may happen that although an improper integral of type (i) is divergent, modifying the limits in I by setting R1 = R2 = R gives rise to a finite result. This is said to define the Cauchy principal value of the integral, and it is indicated by inserting the letters PV before the integral sign. Thus, when the limit is finite, PV ∞ −∞ f(x) dx = lim R→∞ R −R f(x) dx. Similarly, it may happen that although an improper integral of type (ii) is divergent, modifying the limits in II by setting ε = δ, so the limits are evaluated symmetrically about x = ξ, gives rise to a finite result. This also defines a Cauchy principal value, and when the limits exist and are finite, we have PV b a f(x) dx = lim ε→0 ξ−ε a f(x) dx + b ξ+ε f(x) dx . When an improper integral converges, its value and the Cauchy principal value coincide. Typical examples of improper integrals are: ∞ 0 xe−x dx = 1, ∞ 1 dx 1 + x2 = π 4 , 2 0 dx (x − 1) 2/3 = 6, 1 0 dx √ 1 − x2 = π 2 , ∞ 0 sin x dx diverges, PV ∞ −∞ cos x a2 − x2 dx = π a sin a. Convergence tests for improper integrals. 1. Let f(x) and g(x) be continuous on the semi-infinite interval [a, ∞), and let g (x) be continuous and monotonic decreasing on this same interval. Suppose (i) lim x→∞ g (x) = 0, 1.15 Basic Results from the Calculus 105 Area under a curve. The definite integral 1. A = b a f(x) dx may be interpreted as the algebraic sum of areas above and below the x-axis that are bounded by the curve y = f(x) and the lines x = a and x = b, with areas above the x-axis assigned positive values and those below it negative values. Volume of revolution. (i) Let f(x) be a continuous and nonnegative function defined on the interval a ≤ x ≤ b, and A be the area between the curve y = f(x), the x-axis, and the lines x = a and x = b. Then the volume of the solid generated by rotating A about the x-axis is 2. V = π b a [f(x)] 2 dx. (ii) Let g(y) be a continuous and nonnegative function defined on the interval c ≤ y ≤ d, and A be the area between the curve x = g(y), the y-axis, and the lines y = c and y = d. Then the volume of the solid generated by rotating A about the y-axis is 3. V = π d c [g (y)] 2 dy. 106 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus (iii) Let g(y) be a continuous and nonnegative function defined on the interval c ≤ y ≤ d with c ≥ 0, and A be the area between the curve x = g(y), the y-axis, and the lines y = c and y = d. Then the volume of the solid generated by rotating A about the x-axis is. 4. V = 2π d c yg (y) dy. (iv) Let f(x) be a continuous and nonnegative function defined on the interval a ≤ x ≤ b with a ≥ 0, and A be the area between the curve y = f(x), the x-axis, and the lines x = a and x = b. Then the volume of the solid generated by rotating A about the y-axis is 5. V = 2π b a xf(x) dx. (v) Theorem of Pappus Let a closed curve C in the (x, y)-plane that does not intersect the x-axis have a cir- cumference L and area A, and let its centroid be at a perpendicular distance ¯y from the x-axis. Then the surface area S and volume V of the solid generated by rotating the area within the curve C about the x-axis are given by 6. S = 2π¯yL and V = 2π¯yA. Length of an arc. (i) Let f(x) be a function with a continuous derivative that is defined on the interval a ≤ x ≤ b. Then the length of the arc along y = f(x) from x = a to x = b is 1.15 Basic Results from the Calculus 107 7. s = b a 1 + [f (x)] 2 dx, δs2 = δx2 + δy2 ; in the limit ds dx 2 = 1 + dy dx 2 = 1 + [f (x)]2 (ii) Let g(y) be a function with a continuous derivative that is defined on the interval c ≤ y ≤ d. Then the length of the arc along x = g(y) from y = c to y = d is 8. s = d c 1 + [g (y)] 2 dy. δs2 = δy2 + δx2 ; in the limit ds dy 2 = 1 + dx dy 2 = 1 + [g (y)]2 Area of surface of revolution. (i) Let f(x) be a nonnegative function with a continuous derivative that is defined on the interval a ≤ x ≤ b. Then the area of the surface of revolution generated by rotating the curve y = f(x) about the x-axis between the planes x = a and x = b is 9. S = 2π b a f(x) 1 + [f (x)] 2 dx. (see also Pappus's theorem 1.15.6.1.5) (ii) Let g(y) be a nonnegative function with a continuous derivative that is defined on the interval c ≤ y ≤ d. Then the area of the surface generated by rotating the curve x = g(y) about the y-axis between the planes y = c and y = d is 10. S = 2π d c g (y) 1 + [g (y)] 2 dy. Center of mass and moment of inertia. (i) Let a plane lamina in a region R of the (x, y)-plane within a closed plane curve C have the continuous mass density distribution ρ(x, y). Then the center of mass (gravity) of the lamina is located at the point G with coordinates (¯x, ¯y), where 108 Chapter 1 Numerical, Algebraic, and Analytical Results for Series and Calculus 11. ¯x = R xρ (x, y) dA M , ¯y = R yρ (x, y) dA M , with dA the element of area in R and 12. M = R ρ (x, y) dA. (mass of lamina) When this result is applied to the area R within the plane curve C in the (x, y)-plane, which may be regarded as a lamina with a uniform mass density that may be taken to be ρ(x, y) ≡ 1, the center of mass is then called the centroid. (ii) The moments of inertia of the lamina in (i) about the x-, y-, and z-axes are given, respectively, by 13. Ix = R y2 ρ (x, y) dA, Iy = R x2 ρ (x, y) dA Iz = R x2 + y2 ρ (x, y) dA. The radius of gyration of a body about an axis L denoted by kL is defined as 14. k2 L = IL/M, where IL is the moment of inertia of the body about the axis L and M is the mass of the body. Chapter 2 Functions and Identities 2.1 COMPLEX NUMBERS AND TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS 2.1.1 Basic Results 2.1.1.1 Modulus-Argument Representation In the modulus-argument (r, θ) representation of the complex number z = x + iy, located at a point P in the complex plane, r is the radial distance of P from the origin and θ is the angle measured from the positive real axis to the line OP. The number r is called the modulus of z (see 1.1.1.1), and θ is called the argument of z, written arg z, and it is chosen to lie in the interval 1. −π < θ ≤ π. By convention, θ = arg z is measured positively in the counterclockwise sense from the positive real axis, so that 0 ≤ θ ≤ π, and negatively in the clockwise sense from the positive real axis, so that −π < θ ≤ 0. Thus, 2. z = x + iy = r cos θ + ir sin θ or 3. z = r (cos θ + i sin θ). The connection between the Cartesian representation z = x + iy and the modulus- argument form is given by 4. x = r cos θ, y = r sin θ 5. r = x2 + y2 1/2 . The periodicity of the sine and cosine functions with period 2π means that for given r and θ, the complex number z in 2.1.1.1.3 will be unchanged if θ = arg z is replaced by 109 2.1 Complex Numbers and Trigonometric and Hyperbolic Functions 113 Cardano Formula for the Roots of a Cubic Given the cubic x3 + Ax2 + Bx + C = 0, the substitution x = y − 1 3 A reduces it to a cubic of the form y3 + ay + b = 0. Setting p = − 1 2 b + d 1/3 and q = − 1 2 b − d 1/3 , where d is the positive square root d = 1 3 a 3 + 1 2 b 2 , the Cardano formulas for the roots x1, x2, and x3 are x1 = p + q − 1 3 A, x2 = − 1 2 (p + q) − 1 3 A + 1 2 (p − q) i √ 3, x3 = − 1 2 (p + q) − 1 3 A − 1 2 (p − q) i √ 3 2.1.1.4 Relationship Between Roots Once a root w∗ of wn = z has been found for a given z and integer n, so that w∗ = z1/n is one of the n'th roots of z, the other n−1 roots are given by w·w1, w·w2, . . . , w·wn−1, where the wk are the n'th roots of unity defined in 2.1.1.3.3. 2.1.1.5 Roots of Functions, Polynomials, and Nested Multiplication If f(x) is an arbitrary function of x, a number x0 such that f(x0) = 0 is said to be a root of the function or, equivalently, a zero of f(x). The need to determine a root of a function f(x) arises frequently in mathematics, and when it cannot be found analytically it becomes necessary to make use of numerical methods. In the special case in which the function is the polynomial of degree n 1. Pn (x) ≡ a0xn + a1xn−1 + · · · + an, with the coefficients a0, a1, . . . , an, it follows from the fundamental theorem of algebra that Pn(x) = 0 has n roots x1, x2, . . . , xn, although these are not necessarily all real or distinct. If xi is a root of Pn(x) = 0 then (x − xi) is a factor of the polynomial and it can be expressed as 2. Pn (x) ≡ a0 (x − x1) (x − x2) · · · (x − xn) . If a root is repeated m times it is said to have multiplicity m. 114 Chapter 2 Functions and Identities The important division algorithm for polynomials asserts that if P(x) and Q(x) are polynomials with the respective degrees n and m, where 1 ≤ m ≤ n, then a polynomial R(x) of degree n−m and a polynomial S(x) of degrees m−1 or less, both of which are unique, can be found such that P (x) = Q (x) R (x) + S (x) . Polynomials in which the coefficients a0, a1, . . . , an are real numbers have the following special properties, which are often of use: (i) If the degree n of Pn(x) is odd it has at least one real root. (ii) If a root z of Pn(x) is complex, then its complex conjugate ¯z is also a root. The quadratic expression (x − z) (x − ¯z) has real coefficients, and is a factor of Pn(x). (iii) Pn(x) may always be expressed as a product of real linear and quadratic factors with real coefficients, although they may occur with a multiplicity greater than unity. The following numerical methods for the determination of the roots of a function f(x) are arranged in order of increasing speed or convergence. The bisection method. The bisection method, which is the most elementary method for the location of a root of f(x) = 0, is based on the intermediate value theorem. The theorem asserts that if f(x) is continuous on the interval a ≤ x ≤ b, with f(a) and f(b) of opposite signs, then there will be at least one value x = c, strictly intermediate between a and b, such that f(c) = 0. The method, which is iterative, proceeds by repeated bisection of intervals containing the required root, which after each bisection the subinterval that contains the root forms the interval to be used in the next bisection. Thus the root is bracketed in a nested set of intervals I0, I1, I2, . . . , where I0 = [a, b], and after the n'th bisection the interval In is of length (b − a)/2n . The method is illustrated in Figure 2.1. Figure 2.1. 2.1 Complex Numbers and Trigonometric and Hyperbolic Functions 115 The bisection algorithm Step 1. Find the numbers a0, b0 such that f(a0) and f(b0) are of opposite sign, so that f(x) has at least one root in the interval I0 = [a0, b0]. Step 2. Starting from Step 1, construct a new interval In+1 = [an+1, bn+1] from interval In = [an, bn] by setting 3. kn+1 = an + 1 2 (bn − an) and choosing 4. an+1 = an, bn+1 = kn+1 if f(an)f(kn+1) < 0 and 5. an+1 = kn+1, bn+1 = bn if f(an)f(kn+1) > 0. Step 3. Terminate the iteration when one of the following conditions is satisfied: (i) For some n = N, the number kN is an exact root of f(x), so f(kN ) = 0. (ii) Take kN as the approximation to the required root if, for some n = N and some preassigned error bound ε > 0, it follows that \|kN − kN−1\| < ε. To avoid excessive iteration caused by round-off error interfering with a small error bound ε it is necessary to place an upper bound M on the total number of iterations to be performed. In the bisection method the number M can be estimated by using M > log2 b−a ε . The convergence of the bisection method is slow relative to other methods, but it has the advantage that it is unconditionally convergent. The bisection method is often used to determine a starting approximation for a more sophisticated and rapidly convergent method, such as Newton's method, which can diverge if a poor approximation is used. The method of false position (regula falsi). The method of false position, also known as the regula falsi method, is a bracketing technique similar to the bisection method, although the nesting of the intervals In within which the root of f(x) = 0 lies is performed differently. The method starts as in the bisection method with two numbers a0, b0 and the interval I0 = [a0, b0] such that f(a0) and f(b0) are of opposite signs. The starting approxi- mation to the required root in I0 is taken to be the point k0 at which the chord joining the points (a0, f(a0)) and (b0, f(b0)) cuts the x-axis. The interval I0 is then divided into the two subintervals [a0, k0] and [k0, b0], and the interval I1 is chosen to be the subinterval at the ends of which f(x) has opposite signs. Thereafter, the process continues iteratively until, for some n = N, \|kN − kN−1\| < ε, where ε > 0 is a preassigned error bound. The approximation to the required root is taken to be kN . The method is illustrated in Figure 2.2. The false position algorithm Step 1. Find two numbers a0, b0 such that f(a0) and f(b0)are of opposite signs, so that f(x) has at least one root in the interval I0 = [a0, b0]. 116 Chapter 2 Functions and Identities Figure 2.2. Step 2. Starting from Step 1, construct a new interval In+1 = [an+1, bn+1] from the interval In = [an, bn] by setting 6. kn+1 = an − f(an) (bn − an) f(bn) − f(an) and choosing 7. an+1 = an, bn+1 = kn+1 if f(an) f(kn+1) < 0 or 8. an+1 = kn+1, bn+1 = bn if f(an) f(kn+1) > 0. Step 3. Terminate the iterations if either, for some n = N, kN is an exact root so that f(kN ) = 0, or for some preassigned error bound ε > 0, \|kN − kN−1\| < ε, in which case kN is taken to be the required approximation to the root. It is necessary to place an upper bound M on the total number of iterations N to prevent excessive iteration that may be caused by round-off errors interfering with a small error bound ε. The secant method. Unlike the previous methods, the secant method does not involve bracketing a root of f(x) = 0 in a sequence of nested intervals. Thus the covergence of the secant method cannot be guaranteed, although when it does converge the process is usually faster than either of the two previous methods. The secant method is started by finding two approximations k0 and k1 to the required root of f(x) = 0. The next approximation k2 is taken to be the point at which the secant drawn 2.1 Complex Numbers and Trigonometric and Hyperbolic Functions 117 Figure 2.3. through the points through the points (k0, f(k0)) and (k1, f(k1)) cuts the x-axis. Thereafter, iteration takes place with secants drawn as shown in Figure 2.3. The secant algorithm. Starting from the approximation k0, k1 to the required root of f(x) = 0, the approximation kn+1 is determined from the approximations kn and kn−1 by using 9. kn+1 = kn − f(kn) (kn − kn−1) f(kn) − f(kn−1) . The iteration is terminated when, for some n = N and a preassigned error bound ε > 0, \|kN − kN−1\| < ε. An upper bound M must be placed on the number of iterations N in case the method diverges, which occurs when \|kN \| increases without bound. Newton's method. Newton's method, often called the Newton–Raphson method, is based on a tangent line approximation to the curve y = f(x) at an approximate root x0 of f(x) = 0. The method may be deduced from a Taylor series approximation to f(x) about the point x0 as follows. Provided f(x) is differentiable, then if x0 + h is an exact root, 0 = f(x0 + h) = f(x0) + hf (x0) + h2 2! f (x0) + · · ·. Thus, if h is sufficiently small that higher powers of h may be neglected, an approximate value h0 to h is given by h0 = − f(x0) f (x0) , and so a better approximation to x0 is x1 = x0 + h0. This process is iterated until the required accuracy is attained. However, a poor choice of the starting value x0 may cause the method to diverge. The method is illustrated in Figure 2.4. 118 Chapter 2 Functions and Identities Figure 2.4. The Newton algorithm. Starting from an approximation x0 to the required root of f(x) = 0, the approximation xn+1 is determined from the approximation xn by using 10. xn+1 = xn − f(x) f (xn) . The iteration is terminated when, for some n = N and a preassigned error bound ε > 0, \|xN − xN−1\| < ε. An upper bound M must be placed on the number of iterations N in case the method diverges. Notice that the secant method is an approximation to Newton's method, as can be seen by replacing the derivative f (xn) in the above algorithm with a difference quotient. Newton's algorithm—two equations and two unknowns The previous method extends immediately to the case of two simultaneous equations in two unknowns x and y f(x, y) = 0 and g(x, y) = 0, for which x∗ and y∗ are required, such that f(x∗ , y∗ ) = 0 and g(x∗ , y∗ ) = 0. In this case the iterative process starts with an approximation (x0, y0) close to the required solution, and the (n + 1)th iterates (xn+1, yn+1) are found from the nth iterates (xn, yn) by means of the formulas 11. xn+1 = xn − f∂g/∂y − g∂f/∂y ∂f/∂x∂g/∂y − ∂f/∂y∂g/∂x (xn,yn) yn+1 = yn − g∂f/∂x − f∂g/∂x ∂f/∂x∂g/∂y − ∂f/∂y∂g/∂x (xn,yn) . The iterations are terminated when, for some n = N and a preassigned error bound ε > 0, \|xN − xN+1\| < ε and \|yN − yN+1\| < ε. As in the one variable case, an upper bound M must be placed on the number of iterations N to terminate the algorithm if it diverges, in which case a better starting approximation (x0, y0) must be used. 2.1 Complex Numbers and Trigonometric and Hyperbolic Functions 119 Nested multiplication. When working with polynomials in general, or when using them in conjunction with one of the previous root-finding methods, it is desirable to have an efficient method for their evaluation for specific arguments. Such a method is provided by the technique of nested multiplication. Consider, for example, the quartic P4 (x) ≡ 3x4 + 2x3 − 4x2 + 5x − 7. Then, instead of evaluating each term of P4 (x) separately for some x = c, say, and summing them to find P4 (c) , fewer multiplications are required if P4 (x) is rewritten in the nested form: P4 (x) ≡ x{x [x (3x + 2) − 4] + 5} − 7. When repeated evaluation of polynomials is required, as with root-finding methods, this economy of multiplication becomes significant. Nested multiplication is implemented on a computer by means of the simple algorithm given below, which is based on the division algorithm for polynomials, and for this reason the method is sometimes called synthetic division. The nested multiplication algorithm for evaluating Pn(c). Suppose we want to evaluate the polynomial 12. Pn (x) = a0xn + a1xn−1 + · · · + an for some x = c. Set b0 = a0, and generate the sequence b1, b2, . . . , bn by means of the algorithm 13. bi = cbi−1 + ai, for 1 ≤ i ≤ n; then 14. Pn (c) = bn. The argument that led to the nested multiplication algorithm for the evaluation of Pn (c) also leads to the following algorithm for the evaluation of Pn (c) = [d Pn (x) /dx]x=c . These algorithms are useful when applying Newton's method to polynomials because all we have to store is the sequence b0, b1, . . . , bn. Algorithm for evaluating Pn(c). Suppose we want to evaluate Pn (x) when x = c, where Pn (x) is the polynomial in the nested multiplication algorithm, and b0, b1, . . . , bn is the sequence generated by that algorithm. Set d0 = b0 and generate the sequence d1, d2, . . . , dn−1 by the means of the algorithm 15. di = cdi−1 + bi, for 1 ≤ i ≤ n−1; then 16. Pn (c) = dn−1. Chapter 12 Elliptic Integrals and Functions 12.1 ELLIPTIC INTEGRALS 12.1.1 Legendre Normal Forms 12.1.1.1 An elliptic integral is an integral of the form R(x, P(x))dx, in which R is a rational function of its arguments and P(x) is a third- or fourth-degree polynomial with distinct zeros. Every elliptic integral can be reduced to a sum of integrals expressible in terms of algebraic, trigonometric, inverse trigonometric, logarithmic, and exponential functions (the elementary functions), together with one or more of the following three special types of integral: Elliptic integral of the first kind. 1. dx (1 − x2)(1 − k2x2) k2 < 1 Elliptic integral of the second kind. 2. √ 1 − k2x2 √ 1 − x2 dx k2 < 1 Elliptic integral of the third kind. 3. dx (1 − nx2) (1 − x2)(1 − k2x2) [k2 < 1] 241 254 Chapter 13 Probability Distributions and Integrals, and the Error Function 5. P(x) = 1 √ 2π x −∞ e−t2 /2 dt. Related functions used in statistics are 6. Q (x) = 1 √ 2π ∞ x e−t2 /2 dt = 1 − P(x) = P(−x) 7. A (x) = 1 √ 2π x −x e−t2 /2 dt = 2P(x) − 1. Special values of P(x), Q(x), and A(x) are P(−∞) = 0, P(0) = 0.5, P(∞) = 1, Q(−∞) = 1, Q(0) = 0.5, Q(∞) = 0, A(0) = 0, A(∞) = 1. An abbreviated tabulation of P(x), Q(x), and A(x) is given in Table 13.1. 13.1.1.2 The binomial distribution, also called the Bernoulli distribution, is a discrete distribution that describes the behavior of n independent random variables, x1, x2, . . . , xn, often called experiments, that can assume one of two different states, say A and B. Let the probability of occurrence of state A be p, in which case the probability of occurrence of state B will be 1 − p. Then in a test of n trials, the probability P(k) that state A will occur precisely k times is P (k) = n k pk (1 − p) n−k = n! k! (n − k)! pk (1 − p) n−k (k = 1, 2, . . . , n) The mean value µ of the binomial distribution, also called the mathematical expectation and denoted by E(xn), is given by µ = E(xn) = np and the variance of the binomial distri- bution is given by σ2 = np (1 − p). A set of n independent random trials with the probability that the occurrence of event A is given by P(k) is said to satisfy a binomial distribution. A typical example of a set of trials described by a binomial distribution occurs when a coin is flipped n times. In this case if state A is the occurrence of "heads" and state B is the occurrence of "tails," the probability of the occurrence of "heads" will only be p = 0.5 if the coin is unbiased; otherwise, p = 0.5. When n is large, provided p and 1 − p are not too small, the binomial distribution can be approximated by the normal distribution in 13.1.1.1, using as the mean µ = np and as the variance σ2 = np(1 − p) to arrive at the approximation lim n→∞ P xn − µ σ ≤ κ = 1 √ 2π κ −∞ exp − 1 2 s2 ds. This provides a good approximation when p ≈ 0.5 and n is large, though the approximation is also used when p is not close to 0.5, provided both np and n (1 − p) exceed 4. 16.11 Fourier Series and Discontinuous Functions 285 16.11 FOURIER SERIES AND DISCONTINUOUS FUNCTIONS 16.11.1 Periodic Extensions and Convergence of Fourier Series 16.11.1.1 Let f(x) be defined on the interval −π ≤ x ≤ π. Then since each function in the Fourier series of f(x) is periodic with a period that is a multiple of 2π, the Fourier series itself will be periodic with period 2π. Thus, irrespective of how f(x) is defined outside the fundamental interval −π ≤ x ≤ π, the Fourier series will replicate the behavior of f(x) in the intervals (2n − 1)π ≤ x ≤ (2n + 1)π, for n = ±1, ±2, . . . . Each of these intervals is called a periodic extension of f(x). At a point x0, a Fourier series, which is piecewise continuous with a finite number of jump discontinuities, converges to the number 1. f(x0 + 0) + f(x0 − 0) 2 . Thus, if f(x) is continuous at x0, the Fourier series converges to the number f(x0), while if it is discontinuous it follows from 16.11.1.1.1 that it converges to the average of f(x0 − 0) and f(x0 + 0). The periodicity of the Fourier series for f(x) implies that these properties that are true in the fundamental interval −π ≤ x ≤ π are also true in every periodic extension of f(x). In particular, if f(−π) = f(π), there will be a jump discontinuity (a saltus) at each end of the fundamental interval (and at the ends of each periodic extension) where the Fourier series will converge to the value 1 2 [f(π) + f(−π)]. These same arguments apply to Fourier series defined on an arbitrary interval a ≤ x ≤ b, and not only to the interval −π ≤ x ≤ π. 16.11.2 Applications to Closed-Form Summations of Numerical Series 16.11.2.1 The implications of 16.11.1.1 are best illustrated by means of examples that show how closed- form summations may be obtained for certain numerical series. 1. Application to f (x) = x, −π ≤ x ≤ π. From 16.10.1.1 the Fourier series of f(x) = x, − π ≤ x ≤ π is known to be 2 ∞ n=1 (−1)n+1 n sin nx. A convenient choice of x will reduce this Fourier series to a simple numerical series. Let us choose x = π/2, at which point f(x) is continuous and the Fourier series simplifies. At x = π/2 the Fourier series converges to f(π/2) = π/2. Using this result and setting x = π/2 in the Fourier series gives 288 Chapter 16 Different Forms of Fourier Series Had the value x = 0 been chosen, at which point f(x) is continuous, the Fourier series would have converged to f(0) = 1, and setting x = 0 in the Fourier series would have yielded 1 = sinh πa πa + 2 sinh πa π ∞ n=1 (−1)n a a2 + n2 , from which it follows that 1 2a [1 − πa cosech πa] = ∞ n=1 (−1)n+1 a a2 + n2 . Chapter 17 Bessel Functions 17.1 BESSEL'S DIFFERENTIAL EQUATION 17.1.1 Different Forms of Bessel's Equation 17.1.1.1 In standard form, Bessel's equation is either written as 1. x2 d2 y dx2 + x dy dx + (x2 − ν2 )y = 0 or as 2. d2 y dx2 + 1 x dy dx + 1 − ν2 x2 y = 0, where the real parameter ν determines the nature of the two linearly independent solutions of the equation. By convention, ν is understood to be any real number that is not an integer, and when integral values of this parameter are involved ν is replaced by n. Two linearly independent solutions of 17.1.1.1.1 are the Bessel functions of the first kind of order ν, written Jν(x) and J−ν(x). Another solution of 17.1.1.1.1, to which reference will be made later, is the Bessel function of the second kind of order ν, written Yν(x). The general solution of 17.1.1.1.1 is 3. y = AJν (x) + BJ−ν(x) [ν not an integer]. 289 296 Chapter 17 Bessel Functions Two linearly independent solutions of 17.6.1.1.1 are the modified Bessel functions of the first kind of order ν, written Iν(x) and I−ν(x). Another solution of 17.6.1.1.1, to which reference will be made later, is the modified Bessel function of the second kind of order ν, written Kν(x). The general solution of 17.6.1.1.1 is 3. y = AIν(x) + BI−ν(x) [ν not an integer]. When the order is an integer (ν = n), the modified Bessel functions In(x) and I−n(x) cease to be linearly independent because 4. I−n(x) = In(x). A second solution of 17.6.1.1.1 that is always linearly independent of Iν(x) is Kν(x), irrespective of the value of ν. Thus, the general solution of 17.6.1.1.1 can always be written 5. y = AIν (x) + BKν(x). The modified Bessel function of the second kind Kν(x) is defined as 6. Kν(x) = π 2 I−ν(x) − Iν(x) sin(νπ) , and when ν is an integer n or zero, 7. Kn(x) = lim ν→n Kν(x). A more general form of Bessel's modified equation that arises in many applications is 8. x2 d2 y dx2 + x dy dx − λ2 x2 + ν2 y = 0 or, equivalently, 9. d2 y dx2 + 1 x dy dx − λ2 + ν2 x2 y = 0. These forms may be derived from 17.6.1.1.1 by first making the change of variable x = λu, and then replacing u by x. Bessel's modified equations 17.6.1.1.8–9 always have the general solution 10. y = AIν (λx) + BKν (λx). 11. Figure 17.3 shows the behavior of In (x) and Kn (x) for n = 0, 1 and 0 ≤ x ≤ 4. 17.15 Fourier-Bessel Expansions 307 17.14.6 Asymptotic Expansions of jn(x) and yn(x) When the Order n Is Large 1. jn (x) ∼ 1 x cos x − 1 2 (n + 1)π 2. yn (x) ∼ 1 x sin x − 1 2 (n + 1)π 17.15 FOURIER-BESSEL EXPANSIONS A function f(r) can be expanded over the interval 0 ≤ x ≤ R in terms of the Bessel function Jn, with fixed n, to obtain a Fourier-Bessel expansion of the form 1. f (r) = ∞ k=1 akJn (jn,kr/R) = a1Jn (jn,1r/R) + a2Jn (jn,2r/R) + · · · , where jn,k is the kth zero of Jn(x), the first few of which are listed in 17.5.1, and the coefficients ak are given by 2. ak = 2 R2 R 0 Jn (jn,k r/R) f(r) dr R 0 Jn+1 (jn,k) 2 , [k = 1, 2, . . .] Example When a stretched uniformly thin circular elastic drumhead of radius R is dis- turbed from its equilibrium position, its oscillatory displacement u (r, θ, t) perpendicular to its equilibrium position at (r, θ) is determined by the wave equation in plane polar coordinates utt = c2 (urr + (1/r) ur + (1/r2 ) uθθ), (1) where r is the radial distance from the center, θ is an angle in the drumhead measured from a fixed line in the equilibrium position, and t is the time. When the variables are separated by setting u = R(r)Θ(θ)T(t) the normal displacement u (r, θ, t) at time t can be written as u(r, θ, t) = U(r, θ) T(t), where U describes the spatial displacement and satisfies the Helmholtz equation ∇2 U + λ2 U = 0, with λ a separation constant (see 25.2.1) and ∇2 is the Laplacian in (1). As U (r, Θ) = R (r) Θ (θ), each term Jn(jn,k r/R) in the Fourier-Bessel expansion in equation 1 is a term in R(r), and the displacement at the point (r, θ) is of the form, Unm (r, θ) = Jn(jn,m r/R) [A sin nθ + b cos nθ], with n = 0, 1, 2, . . . and m = 1, 2, . . . . This spatial mode is modulated by a sinusoidal time variation with frequency jn,m c/(2π) cycles in a unit time. When the initial configuration of the drumhead is given by specifying f(r), the displacement of the drumhead U(r, θ) follows once results of equation 2 are used with the Fourier-Bessel expansion in equation 1. If the expansion is performed in terms of J1, a representative mode U12(r, θ) = J1(j1,2 r/R) cos θ, for which j1,2 = 7.01559, and typically A = 1, B = 0, is shown in Figure 17.4(a), while Figure 17.4(b) shows a contour plot of this mode. 308 Chapter 17 Bessel Functions (a) (b) Figure 17.4. The time variation of this mode is obtained by modulating it with a sinusoid of frequency j1,2c/(2π), and when f(r) is specified, the corresponding term in the expansion in equation 1 is multiplied by the coefficient a2 found from 2. Unless f(r) is particularly simple, it becomes necessary to determine the coefficients ak by means of numerical integration. Chapter 18 Orthogonal Polynomials 18.1 INTRODUCTION 18.1.1 Definition of a System of Orthogonal Polynomials 18.1.1.1 Let {Φn(x)} be a system of polynomials defined for a ≤ x ≤ b such that Φn(x) is of degree n, and let w(x) > 0 be a function defined for a ≤ x ≤ b. Define the positive numbers Φn 2 as 1. Φn 2 = b a [Φn(x)] 2 w(x) dx. Then the system of polynomials {Φn(x)} is said to be orthogonal over a ≤ x ≤ b with respect to the weight function w (x) if 2. b a Φm(x)Φn(x)w(x) dx = 0, m = n Φn 2 , m = n [m, n = 0, 1, 2, . . .]. The normalized system of polynomials [φn(x)], where φn(x) = Φn(x)/ Φn , is said to be orthonormal over a ≤ x ≤ b with respect to the weight function w(x), where Φn is called the norm of Φn(x) and it follows from 18.1.1.1.1–2 that 3. b a φm(x)φn(x)w(x) dx = 0, m = n 1, m = n [m, n = 0, 1, 2, . . .]. Orthogonal polynomials are special solutions of linear variable coefficient second-order dif- ferential equations defined on the interval a ≤ x ≤ b, in which n appears as a parameter. 309 318 Chapter 18 Orthogonal Polynomials 18.2.11 Spherical Harmonics When solutions u for the Laplace and Helmholtz equations are required in spherical regions, it becomes necessary to express the equations in terms of the spherical coordinates (r, θ, φ)(see Fig. 24.3). In this coordinate system, r is the radius, θ is the polar angle with 0 ≤ θ ≤ π, and φ is the azimuthal angle with 0 ≤ φ ≤ 2π. After separating the variables by setting u(r, θ, φ) = R(r)Θ(θ)Φ(φ), the dependence of the solution u on the polar angle θ and the azimuthal angle φ is found to satisfy the equation 1. 1 Θ(θ) sin θ d dθ sin θ dΘ dθ + 1 Φ(φ) sin2 θ d2 Φ(φ) dφ2 + n(n + 1) = 0, where the integer n is a separation constant introduced when the variable r was separated. The first term in this equation is a function only of θ, and the expression (1/Φ(φ))d2 Φ/dφ2 in the second term is a function only of φ, so as the functions Φ(φ) and Θ(θ) are independent for all φ and θ, for this result to be true, the terms in θ and in φ must each be equal to a constant. Setting the φ dependence equal to the separation constant −m2 , with m an integer, shows that the azimuthal dependence must obey the equation 2. d2 Φ(φ) dφ2 + m2 Φ(φ) = 0. So the two linearly independent azimuthal solutions are given by the two complex conjugate functions Φ1(φ) = eimφ and Φ2(φ) = e−imφ . These two complex solutions could, of course, have been replaced by the two linearly independent real solutions sin mφ and cos mφ, but for what is to follow, the complex representations will be retained. When the φ dependence is removed from equation 1 and replaced with −m2 , the equation for the θ dependence that remains becomes the associated Legendre equation with solutions Pm n (θ), namely 3. 1 sin θ d dθ sin θ dPm n (θ) dθ + n(n + 1) − m2 sin2 θ Pm n (θ) = 0. Solutions of equation 1, called spherical harmonics, involve the product of Φ1(φ) or Φ2(φ) and Pm n (θ), but unfortunately, there is no standard notation for spherical harmonics, nor is there agreement on whether Pm n (θ) should be normalized when it is used in the definition of spherical harmonics. The products of these functions are called spherical harmonics because solutions of the Laplace equation are called harmonic functions, and these are harmonic func- tions on the surface of a sphere. As spherical harmonics are mainly used in physics, the notation and normalization in what is to follow are the ones used in that subject. However, whereas in mathematics a complex conjugate is denoted by an overbar, as is the case elsewhere in this book, in physics it is usually denoted by an asterisk . The spherical harmonic Y m n (θ, φ) defined as 4. Y m n (θ, φ) = (−1)m 2n + 1 2 (n − m)! (n + m)! Pm n (cos θ)eimφ , where −n ≤ m ≤ n. Chapter 19 Laplace Transformation 19.1 INTRODUCTION 19.1.1 Definition of the Laplace Transform 19.1.1.1 The Laplace transform of the function f(x), denoted by F(s), is defined as the improper integral 1. F(s) = ∞ 0 f(x)e−sx dx [Re {s} > 0]. The functions f(x) and F(x) are called a Laplace transform pair, and knowledge of either one enables the other to be recovered. If f can be integrated over all finite intervals, and there is a constant c for which 2. ∞ 0 f(x)e−cx dx, is finite, then the Laplace transform exists when s = σ + iτ is such that σ ≥ c. Setting 3. F(s) = L[f(x); s], to emphasize the nature of the transform, we have the symbolic inverse result 4. f(x) = L−1 [F(s); x]. 337 340 Chapter 19 Laplace Transformation 19.1.3 The Dirac Delta Function δ(x) 19.1.3.1 The Dirac delta function δ(x), which is particularly useful when working with the Laplace transform, has the following properties: 1. δ(x − a) = 0 [x = a] 2. ∞ −∞ δ(x − a) dx = 1 3. x −∞ δ(ζ − a)dξ = H(x − a), where H(x − a) is the Heaviside step function defined in 19.1.2.1.5. 4. ∞ −∞ f(x)δ(x − a) dx = f(a) The delta function, which can be regarded as an impulse function, is not a function in the usual sense, but a generalized function or distribution. 19.1.4 Laplace Transform Pairs Table 19.1 lists Laplace transform pairs, and it can either be used to find the Laplace transform F(s) of a function f(x) shown in the left-hand column or, conversely, to find the inverse Laplace transform f(x) of a given Laplace transform F(s) shown in the right-hand column. To assist in the task of determining inverse Laplace transforms, many commonly occurring Laplace transforms F(s) of an algebraic nature have been listed together with their more complicated inverse Laplace transforms f(x). The list of both Laplace transforms and inverse transforms may be extended by appeal to the theorems listed in 19.1.2. 19.1.5 Solving Initial Value Problems by the Laplace Transform The Laplace transform is a powerful technique for solving initial value problems for linear ordi- nary differential equations involving an unknown function y(t), provided the initial conditions are specified at t = 0. The restriction on the value of t at which the initial conditions must be specified is because, when transforming the differential equation, the Laplace transform of each derivative is found by using result 19.1.2.1. 2. This result leads to the introduction of the Laplace transform Y (s) = L{y(t)} and also to terms involving the untransformed initial conditions at t = 0. The Laplace transform method is particularly straightforward when solving constant coeffi- cient differential equations. This is because it replaces by routine algebra the usual process of first determining the general solution and then matching its arbitrary constants to the given initial conditions. The steps involved when solving a linear constant coefficient differential equation by means of the Laplace transform are as follows: Chapter 20 Fourier Transforms 20.1 INTRODUCTION 20.1.1 Fourier Exponential Transform 20.1.1.1 Let f(x) be a bounded function such that in any interval (a, b) it has only a finite number of maxima and minima and a finite number of discontinuities (it satisfies the Dirichlet conditions). Then if f(x) is absolutely integrable on (−∞, ∞), so that ∞ −∞ \|f(x)\| dx < ∞, the Fourier transform of f(x), also called the exponential Fourier transform, is the F (ω) defined as 1. F(ω) = 1 √ 2π ∞ −∞ f(x) eiωx dx. The functions f(x) and F(ω) are called a Fourier transform pair, and knowledge of either one enables the other to be recovered. Setting 2. F(ω) = F[f(x); ω], where F is used to denote the operation of finding the Fourier transform, we have the symbolic inverse result 353 Chapter 21 Numerical Integration 21.1 CLASSICAL METHODS 21.1.1 Open- and Closed-Type Formulas The numerical integration (quadrature) formulas that follow are of the form 1. I = b a f(x)dx = n k=0 w (n) k f(xk) + Rn, with a ≤ x0 < xl < · · · < xn ≤ b. The weight coefficients w (n) k and the abscissas xk, with k = 0, 1, . . . , n are known numbers independent of f(x) that are determined by the numerical integration method to be used and the number of points at which f(x) is to be evaluated. The remainder term, Rn, when added to the summation, makes the result exact. Although, in general, the precise value of Rn is unknown, its analytical form is usually known and can be used to determine an upper bound to \|Rn\| in terms of n. An integration formula is said to be of the closed type if it requires the function values to be determined at the end points of the interval of integration, so that x0 = a and xn = b. An integration formula is said to be of open type if x0 > a and xn < b, so that in this case the function values are not required at the end points of the interval of integration. Many fundamental numerical integration formulas are based on the assumption that an interval of integration contains a specified number of abscissas at which points the function must be evaluated. Thus, in the basic Simpson's rule, the interval a ≤ x ≤ b is divided into two subintervals of equal length at the ends of which the function has to be evaluated, so that f(x0), f(x1), and f(x2) are required, with x0 = a, x1 = 1 2 (a + b), and x2 = b. To control the 363 368 Chapter 21 Numerical Integration stage. The process may be continued until the result is accurate to the required number of decimal places, provided that at the nth stage the derivative f(n) (x) is nowhere singular in the interval of integration. Romberg integration is based on the composite trapezoidal rule and an extrapolation pro- cess (Richardson extrapolation), and is well suited to implementation on a computer. This is because of the efficient use it makes of the function evaluations that are necessary at each successive stage of the computation, and its speed of convergence, which enables the numerical estimate of the integral to be obtained to the required degree of precision relatively quickly. The Romberg Method At the mth stage of the calculation, the interval of integration a ≤ x ≤ b is divided into 2m intervals of length (b − a)/2m . The corresponding composite trapezoidal estimate for I is then given by 1. I0,0 = (b − a) 2 [f(a) + f(b)] and I0,m = (b − a) 2m 1 2 f(a) + 1 2 f(b) + 2m −1 r=1 f(xr) , where xr = a + r[(b − a)/2m ] for r = 1, 2, . . . , 2m − 1. Here the initial suffix represents the computational step reached at the m'th stage of the calculation, with the value zero indicating the initial trapezoidal estimate. The second suffix starts with the number of subintervals on which the initial trapezoidal estimate is based and the steps down to zero at the end of the mth stage. Define 2. Ik,m = 4k Ik−1,m+1 − Ik−1,m 4k − 1 , for k = 1, 2, . . . and m = 1, 2, . . . . For some preassigned error ε > 0 and some integer N, IN,0 is the required estimate of the integral to within an error ε if 3. \|IN,0 − IN−1,1\| < ε. The pattern of the calculation proceeds as shown here. Each successive entry in the r'th row provides an increasingly accurate estimate of the integral, with the final estimate at the end of the r'th stage of the calculation being provided by Ir,0: 21.1 Classical Methods 369 To illustrate the method consider the integral I = 5 1 ln(1 + √ x) 1 + x2 dx. Four stages of Romberg integration lead to the following results, inspection of which shows that the approximation I4,0 = 0.519256 has converged to five decimal places. Notice that the accuracy achieved in I4,0 was obtained as a result of only 17 function evaluations at the end of the 16 subintervals involved, coupled with the use of relation 21.1.7.2. To obtain comparable accuracy using only the trapezoidal rule would involve the use of 512 subintervals. I0,m I1,m I2,m I3,m I4,m 0.783483 0.592752 0.529175 0.537275 0.518783 0.518090 0.523633 0.519086 0.519106 0.519122 0.520341 0.519243 0.519254 0.519255 0.519256 Chapter 22 Solutions of Standard Ordinary Differential Equations 22.1 INTRODUCTION 22.1.1 Basic Definitions 22.1.1.1 An nth-order ordinary differential equation (ODE) for the function y(x) is an equation defined on some interval I that relates y(n) (x) and some or all of y(n−1) (x), y(n−2) (x), . . . , y(1) (x), y(x) and x, where y(r) (x) = dr y/dxr . In its most general form, such an equation can be written 1. F x, y(x), y(1) (x) , . . . , y(n) (x) = 0, where y(n) (x) /≡ 0 and F is an arbitrary function of its arguments. The general solution of 22.1.1.1.1 is an n times differentiable function Y (x), defined on I, that contains n arbitrary constants, with the property that when y = Y (x) is substituted into the differential equation reduces it to an identity in x. A particular solution is a special case of the general solution in which the arbitrary constants have been assigned specific numerical values. 22.1.2 Linear Dependence and Independence 22.1.2.1 A set of functions y1(x), y2(x), . . . , yn(x) defined on some interval I is said to be linearly dependent on the interval I if there are constants c1, c2, . . . , cn, not all zero, such that 371 22.3 Linear First-Order Equations 373 which are defined for all x. For x < 0 we have W[y1, y2] = x 0 1 0 = 0, whereas for x ≥ 0 W[y1, y2] = 0 x 0 1 = 0, so for all x W[y1, y2] = 0. However, despite the vanishing of the Wronskian for all x, the functions y1 and y2 are not linearly dependent, as can be seen from 22.1.2.1.1, because there exist no nonzero constants c1 and c2 such that c1y1 + c2y2 = 0 for all x. This is not a failure of the Wronskian test, because although y1 and y2 are continuous, their first derivatives are not continuous as required by the Wronskian test. 22.2 SEPARATION OF VARIABLES 22.2.1 A differential equation is said to have separable variables if it can be written in the form 1. F(x)G(y) dx + f(x)g(y) dy = 0. The general solution obtained by direct integration is 2. F(x) f(x) dx + g(y) G(y) dy = const. 22.3 LINEAR FIRST-ORDER EQUATIONS 22.3.1 The general linear first-order equation is of the form 1. dy dx + P(x)y = Q(x). This equation has an integrating factor 374 Chapter 22 Solutions of Standard Ordinary Differential Equations 2. µ(x) = e R P (x)dx , and in terms of µ(x) the general solution becomes 3. y(x) = c µ(x) + 1 µ(x) Q(x) µ(x) dx, where c is an arbitrary constant. The first term on the right-hand side is the complementary function yc(x), and the second term is the particular integral of yp(x) (see 22.7.1). Example 22.4 Find the general solution of dy dx + 2 x y = 1 x2 (1 + x2) . In this case P(x) = 2/x and Q(x) = x−2 1 + x2 −1 , so µ(x) = exp 2 x dx = x2 and y(x) = c x2 + 1 x2 dx 1 + x2 so that y(x) = c x2 + 1 x2 arctan x. Notice that the arbitrary additive integration constant involved in the determination of µ(x) has been set equal to zero. This is justified by the fact that, had this not been done, the constant factor so introduced could have been incorporated into the arbitrary constant c. 22.4 BERNOULLI'S EQUATION 22.4.1 Bernoulli's equation is a nonlinear first-order equation of the form 1. dy dx + p(x)y = q(x)yα , with α = 0 and α = 1. Division of 22.4.1.1 by yα followed by the change of variable 2. z = y1−α converts Bernoulli's equation to the linear first-order equation 376 Chapter 22 Solutions of Standard Ordinary Differential Equations where c2 is an arbitrary constant. Thus, the required solution is x3 y − 1 3 y3 = c, where c = − (c1 + c2) is an arbitrary constant. 22.6 HOMOGENEOUS EQUATIONS 22.6.1 In its simplest form, a homogeneous equation may be written 1. dy dx = F y x , where F is a function of the single variable 2. u = y/x. More generally, a homogeneous equation is of the form 3. P(x, y) dx + Q(x, y) dy = 0, where P and Q are both algebraically homogeneous functions of the same degree. Here, by requiring P and Q to be algebraically homogeneous of degree n, we mean that 4. P(kx, ky) = kn P(x, y) and Q(kx, ky) = kn Q(x, y), with k a constant. The solution of 22.6.1.1 is given in implicit form by 5. ln x = du F(u) − u . The solution of 22.6.1.3 also follows from this same result by setting F = −P/Q. 22.7 LINEAR DIFFERENTIAL EQUATIONS 22.7.1 An nth-order variable coefficient differential equation is linear if it can be written in the form 1. ˜a0(x)y(n) + ˜a1(x)y(n−1) + · · · + ˜an(x)y = ˜f(x), where ˜a0, ˜a1, . . . , ˜an and ˜f are real-valued functions defined on some interval I. Provided ˜a0 = 0 this equation can be rewritten as 22.10 Linear Differential Equations—Inhomogeneous Case and the Green's Function 385 22.10.2 The Green's Function Another way of solving initial and boundary value problems for inhomogeneous linear differen- tial equations is in terms of an integral using a specially constructed function called a Green's function. In what follows, only the solution of a second order equation will be considered, though the method extends in a straightforward manner to linear inhomogeneous equations of any order. The Solution of a Linear Inhomogeneous Equation Using a Green's Function When expressed in terms of the Green's function G(x, t), the solution of an initial value problem for the linear second order inhomogeneous equation 1. p(x) d2 y dx2 + q(x) dy dx + r(x)y = f(x), subject to the homogeneous initial conditions 2. y(0) = 0 and y (0) = 0 can be written in the form 3. y(x) = x 0 G(x, t)f(t) dt, where G(x, t) is the Green's function. The Green's function can be found by solving the initial value problem 4. p(x) d2 y dx2 + q(x) dy dx + r(x)y = 0, subject to the initial conditions 5. y(t) = 0 and y (t) = 1 for t < x. This way of finding the Green's function is equivalent to solving equation 1 subject to homogeneous initial conditions, with a Dirac delta function δ(x − t) as the inhomogeneous term. The linearity of the equation then allows solution 3 to be found by weighting the Green's function response by the inhomogeneous term at x = t and integrating the result with respect to t over the interval 0 ≤ t ≤ x. The Green's function can be defined as 6. G(x, t) = − 1 p(x) ϕ1(x) ϕ2(x) ϕ1(t) ϕ2(t) ϕ1(t) ϕ2(t) ϕ1(t) ϕ2(t) , for 0 ≤ t ≤ x, where ϕ1(x) and ϕ2(x) are two linearly independent solutions of the homogeneous form of equation 1 386 Chapter 22 Solutions of Standard Ordinary Differential Equations 7. p(x) d2 y dx2 + q(x) dy dx + r(x)y = 0. The advantage of the Green's function approach is that G(x, t) is independent of the inho- mogeneous term f(x), so once G(x, t) has been found, result 3 gives the solution for any inhomogeneous term f(x). If the initial conditions for equation 1 are not homogeneous, all that is necessary to modify result 3 is to find a solution of the homogeneous form of the differential equation that satisfies the inhomogeneous initial conditions and then to add it to the result in 3. Example 22.12 Use the Green's function to solve (a) the initial value problem for y + y = cos x, subject to the homogeneous initial conditions y(0) = 0 and y (0) = 0 and (b) the same equation subject to the inhomogeneous initial conditions y(0) = 2 and y (0) = 0. Solution (a) A solution set {ϕ1(x), ϕ2(x)} for the homogeneous form of this equation is given by ϕ1(x) = cos x and ϕ2(x) = sin x. As p(x) = 1, and the inhomogeneous term f(x) = cos x, the Green's function becomes G(x, t) = − cos x sin x cos t sin t cos t sin t − sin t cos t = sin x cos t − cos x sin t = sin(x − t). Substituting for G(x, t) in 3 and setting f(t) = cos t, gives y(x) = x 0 sin(x − t) cos t dt, so the solution subject to homogeneous initial conditions is y(x) = 1 2 x sin x. (b) The general solution of the homogeneous form of the equation is yC(x) = A sin x + B cos x, so if this is to satisfy the inhomogeneous initial conditions yC(0) = 2 and yC(0) = 0, we must set A = 0 and B = 2, giving yC(x) = 2 cos x. Thus the required solution subject to the inhomogeneous initial conditions becomes y(x) = yC(x) + x 0 sin(x − t) cos t dt, and so y(x) = 2 cos x + 1 2 x sin x. 22.10 Linear Differential Equations—Inhomogeneous Case and the Green's Function 387 When two-point boundary value problems over the interval a ≤ x ≤ b are considered for equations with inhomogeneous terms, where the solution must satisfy homogeneous boundary conditions, it is necessary to modify the definition of a Green's function. To be precise, the solution is required for a two-point boundary value problem over the interval a ≤ x ≤ b, for the inhomogeneous equation 8. p(x) d2 y dx2 + q(x) dy dx + r(x)y = f(x), that satisfies the homogeneous two-point boundary conditions 9. α1y(a) + β1y (a) = 0 and α2y(b) + β2y (b) = 0, where the constants α, β1, α2, and β2 are such that α2 1 + β2 1 > 0 and α2 2 + β2 2 > 0. The Solution of a Two-Point Boundary Value Problem Using a Green's Function The solution of boundary value problem for equation 8, subject to the boundary conditions 9, will only exist if the problem is properly set, in the sense that it is possible for the solution to satisfy the boundary conditions. The condition for this is that the homogeneous form of the equation 10. p(x) d2 y dx2 + q(x) dy dx + r(x)y = 0, subject to the boundary conditions 9, only has a trivial solution—that is, the only solution is the identically zero solution. Let φ1(x) be a solution of the homogeneous equation 10 that satisfies boundary conditions 9 at x = a, and let φ2(x) be a linearly independent solution of equation 10 that satisfies boundary conditions 9 at x = b. Then the Green's function for the homogeneous equation 10 is defined as 11. G(x, t) =    φ1(t)φ2(x) p(t)W[φ1(t), φ2(t)] for a ≤ x ≤ t, φ1(x)φ2(t) p(t)W[φ1(t), φ2(t)] for t ≤ x ≤ b. The solution of the two-point boundary value problem for equation 8, subject to the boundary conditions 9, is 12. y(x) = b a G(x, t)b(t)dt, or equivalently 13. y(x) = φ2(x) x a φ1(t)b(t) p(t)W[φ1(t), φ2(t)] dt + φ1(x) b x φ2(t)b(t) p(t)W[φ1(t), φ2(t)] dt. 388 Chapter 22 Solutions of Standard Ordinary Differential Equations If required, this form of the solution can be derived by modifying the method of variation of parameters. Example 22.13 Verify that the two-point boundary value problem y + y = 1, y(0) = 0, y(π/2) = 0 has a solution, and find it with the aid of a Green's function. Solution The solution of the homogeneous form of the equation y + y = 0, subject to the boundary conditions y(0) = 0, y(π/2) = 0, only has the trivial solution y(x) ≡ 0, so the Green's function method may be used to find the solution y(x). The function φ1(x) must be constructed from a linear combination of the solutions of the homogeneous form of the equation, namely y + y = 0, with the solution set {ϕ1(x), ϕ2(x)}, where ϕ1(x) = cos x and ϕ2(x) = sin x. So we set φ1(x) = c1 cos x + c2 sin x and require φ1(x) to satisfy the left boundary condition φ1(0) = 0. This shows we must set φ1(x) = c2 sin x. However, the differential equation is homogeneous, so as this solution can be scaled arbitrarily, for simplicity we choose to set c2 = 1, when φ1(x) = sin x. The function φ2(x) must also be constructed from a linear combination of solutions of the homogeneous form of the equation y + y = 0, so now we set φ1(x) = d1 cos x + d2 sin x and require φ2(x) to satisfy the right boundary condition φ2(π/2) = 0. This shows that φ2(x) = d2 cos x, but again, as the differential equation is homogeneous, this solution can also be scaled arbitrarily. So, for simplicity, we choose to set d1 = 1 when φ2(x) = cos x. In this case p(x) = 1, and the Wronskian W[φ1(x), φ2(x)] = sin x cos x cos x − sin x = −1, so the Green's function G(x, t) = − sin t cos x, 0 ≤ x ≤ t − sin x cos t, t ≤ x ≤ π/2. As the inhomogeneous term f(x) = 1, we must set f(t) = 1, showing the solution of the boundary value problem is given by y(x) = − x 0 sin t cos x dt − π/2 x sin x cos t dt, so the required solution is y(x) = 1 − cos x − sin x. To illustrate the necessity for the homogeneous form of the equation to have a trivial solution if the solution of the two-point boundary value problem is to exist, we need only con- sider the equation y + y = 1 subject to the boundary conditions y(0) = 0 and y(π) = 0. The 390 Chapter 22 Solutions of Standard Ordinary Differential Equations where 7. α = − 1 2 a and β = 1 2 √ 4b − a2. 22.12 DETERMINATION OF PARTICULAR INTEGRALS BY THE METHOD OF UNDETERMINED COEFFICIENTS 22.12.1 An alternative method may be used to find a particular integral when the inhomogeneous term f(x) is simple in form. Consider the linear inhomogeneous n'th-order constant coefficient differential equation 1. y(n) + a1y(n−1) + a2y(n−2) + · · · + any = f(x), in which the inhomogeneous term f(x) is a polynomial, an exponential, a product of a power of x, and a trigonometric function of the form xs cos qx or xs sin qx or a sum of any such terms. Then the particular integral may be obtained by the method of undetermined coefficients (constants). To arrive at the general form of the particular integral it is necessary to proceed as follows: (i) If f(x) = constant, include in yp(x) the undetermined constant term C. (ii) If f(x) is a polynomial of degree r then: (a) if L [y(x)] contains an undifferentiated term y, include in yp(x) terms of the form A0xr + A1xr−1 + · · · + Ar, where A0, A1, . . . , Ar are undetermined constants; (b) if L [y(x)] does not contain an undifferentiated y, and its lowest order derivative is y(s) , include in yp(x) terms of the form A0xr+s + A1xr+s−1 + · · · + Arxs , where A0, A1, . . . , Ar are undetermined constants. (iii) If f(x) = eαx then: (a) if eax is not contained in the complementary function (it is not a solution of the homogeneous form of the equation), include in yp(x) the term Beax , where B is an undetermined constant; (b) if the complementary function contains the terms eax , xeax , . . . , xm eax , include in yp(x) the term Bxm+1 eax , where B is an undetermined constant. 22.12 Determination of Particular Integrals by the Method of Undetermined Coefficients 391 (iv) If f(x) = cos qx and/or sin qx then: (a) if cos qx and/or sin qx are not contained in the complementary function (they are not solutions of the homogeneous form of the equation), include in yp(x) terms of the form C cos qx and D sin qx, where C and D are undetermined constants; (b) if the complementary function contains terms xs cos qx and/or xs sin qs with s = 0, 1, 2, . . . , m, include in yp(x) terms of the form xm+1 (C cos qx + D sin qx), where C and D are undetermined constants. (v) The general form of the particular integral is then the sum of all the terms generated in (i) to (iv). (vi) The unknown constant coefficients occurring in yp(x) are found by substituting yp(x) into 22.12.1.1, and then choosing them so that the result becomes an identity in x. Example 22.14 Find the complementary function and particular integral of d2 y dx2 + dy dx − 2y = x + cos 2x + 3ex , and solve the associated initial value problem in which y(0) = 0 and y(1) (0) = 1. The characteristic polynomial P2 (λ) = λ2 + λ − 2 = (λ − 1) (λ + 2), so the linearly independent solutions of the homogeneous form of the equation are y1(x) = ex and y2(x) = e−2x , and the complementary function is yc (x) = c1ex + c2e−2x . Inspection of the inhomogeneous term shows that only the exponential ex is contained in the complementary function. It then follows from (ii) that, corresponding to the term x, we must include in yp(x) terms of the form A + Bx. 22.15 Bessel's Equations 395 subject to the boundary conditions y(0) = 2 and y(1) (a) = 0. The equation is Bessel's equation of order 0, so the general solution is y(x) = c1J0(λx) + c2Y0(λx). The boundary condition y(0) = 2 requires the solution to be finite at the origin, but Y0(0) is infinite (see 17.2.2.2.1) so we must set c2 = 0 and require that 2 = c1J0(0), so c1 = 2 because J0(0) = 1 (see 17.2.1.2.1). Boundary condition y(a) = 0 then requires that 0 = 2J0(λa), but the zeros of J0(x) are j0,1 j0,2, . . . , j0,m, . . . (see Table 17.1), so λa = j0,m, or λm = j0,m/a, for m = 1, 2, . . . . Consequently, the required solutions are ym(x) = 2J0 j0,m x a [m = 1, 2, . . .]. The numbers λ1, λ2, . . . are the eigenvalues of the problem and the functions y1, y2, . . . , are the corresponding eigenfunctions. Because any constant multiple of ym(x) is also a solution, it is usual to omit the constant multiplier 2 in these eigenfunctions. Example 22.16 Solve the two-point boundary value problem x2 d2 y dx2 + x dy dx − x2 + 4 y = 0, subject to the boundary conditions y(a) = 0 and y(b) = 0 with 0 < a < b < ∞. The equation is Bessel's modified equation of order 2, so the general solution is y(x) = c1I2(λx) + c2K2(λx). Since 0 < a < b < ∞, I2, and K2 are finite in the interval a ≤ x ≤ b, so both must be retained in the general solution (see 17.7.1.2 and 17.7.2.2). Application of the boundary conditions leads to the conditions: (y(a) = 0) c1I2(λa) + c2K2(λa) = 0 (y(b) = 0) c1I2(λb) + c2K2(λb) = 0, and this homogeneous system only has a nontrivial solution (c1, c2 not both zero) if I2(λa) K2(λa) I2(λb) K2(λb) = 0. 396 Chapter 22 Solutions of Standard Ordinary Differential Equations Thus, the required values λ1, λ2, . . . of λ must be the zeros of the transcendental equation I2(λa)K2(λb) − I2(λb)K2(λa) = 0. For a given a, b it is necessary to determine λ1, λ2, . . . , numerically. Here also the numbers λ1, λ2, . . . , are the eigenvalues of the problem, and the functions ym(x) = c1 I2 (λmx) − I2 (λma) K2 (λma) K2 (λmx) or, equivalently, ym(x) = c1 I2 (λmx) − I2 (λmb) K2 (λmb) K2 (λmx) , are the corresponding eigenfunctions. As in Example 22.15, because any multiple of ym(x) is also a solution, it is usual to set c1 = 1 in the eigenfunctions. 22.16 POWER SERIES AND FROBENIUS METHODS 22.16.1 To appreciate the need for the Frobenius method when finding solutions of linear variable coefficient differential equations, it is first necessary to understand the power series method and the reason for its failure in certain circumstances. To illustrate the power series method, consider seeking a solution of d2 y dx2 + x dy dx + y = 0 in the form of a power series about the origin y(x) = ∞ n=0 anxn . Differentiation of this expression to find dy/dx and d2 y/dx2 , followed by substitution into the differential equation and the grouping of terms leads to the result (a0 + 2a2) + ∞ n=1 [(n + 1)(n + 2)an+2 + (n + 1)an]xn = 0. For y(x) to be a solution, this expression must be an identity for all x, which is only possible if a0 + 2a2 = 0 and the coefficient of every power of x vanishes, so that (n + 1)(n + 2)an+2 + (n + 1)an = 0 [n = 1, 2, . . .]. 22.16 Power Series and Frobenius Methods 397 Thus, a2 = − 1 2 a0, and in general the coefficients an are given recursively by an+2 = − 1 n + 2 an. from which it follows that a2, a4, a6, . . . are all expressible in terms of an arbitrary nonzero constant a0, while a1, a3, a5, . . . are all expressible in terms of an arbitrary nonzero constant a1. Routine calculations then show that a2m = (−1)m 2mm! a0 [m = 1, 2, . . .], while a2m+1 = (−1)m 1.3.5 · · · (2m + 1) a1 [m = 1, 2, . . .]. After substituting for the an in the power series for y(x) the solution can be written y(x) = a0 ∞ m=0 (−1) m 2mm! x2m + a1 ∞ m=0 (−1) m 1.3.5 · · · (2m + 1) x2m+1 = a0 1 − 1 2 x + 1 8 x2 − · · · + a1 x − 1 3 x3 + 1 15 x5 · · · . Setting y1(x) = 1 − 1 2 x + 1 8 x2 − · · · and y2(x) = x − 1 3 x3 + 1 15 x5 − · · ·, it follows that y1 and y2 are linearly independent, because y1 is an even function and y2 is an odd function. Thus, because a0, a1 are arbitrary constants, the general solution of the differential equation can be written y(x) = a0y1(x) + a2y2(x). The power series method was successful in this case because the coefficients of d2 y/dx2 , dy/dx and y in the differential equation were all capable of being expressed as Maclaurin series, and so could be combined with the power series expression for y(x) and its derivatives that led to y1(x) and y2(x). (In this example, the coefficients 1, x, and 1 in the differential equation are their own Maclaurin series.) The method fails if the variable coefficients in a differential equation cannot be expressed as Maclaurin series (they are not analytic at the origin). The Frobenius method overcomes the difficulty just outlined for a wide class of variable coefficient differential equations by generalizing the type of solution that is sought. To proceed further, it is first necessary to define regular and singular points of a differential equation. 398 Chapter 22 Solutions of Standard Ordinary Differential Equations The second-order linear differential equation with variable coefficients 1. d2 y dx2 + p(x) dy dx + q(x)y = 0 is said to have a regular point at the origin if p(x) and q(x) are analytic at the origin. If the origin is not a regular point it is called a singular point. A singular point at the origin is called a regular singular point if 2. lim x→0 {xp(x)} is finite and 3. lim x→0 {x2 q(x)} is finite. Singular points that are not regular are said to be irregular. The behavior of a solution in a neighborhood of an irregular singular point is difficult to determine and very erratic, so this topic is not discussed further. There is no loss of generality involved in only considering an equation with a regular singular point located at the origin, because if such a point is located at x0 the transformation X = x − x0 will shift it to the origin. If the behavior of a solution is required for large x (an asymptotic solution), making the transformation x = 1/z in the differential equation enables the behavior for large x to be determined by considering the case of small z. If z = 0 is a regular point of the transformed equation, the original equation is said to have regular point at infinity. If, however, z = 0 is a regular singular point of the transformed equation, the original equation is said to have a regular singular point at infinity. The Frobenius method provides the solution of the differential equation 4. d2 y dx2 + p(x) dy dx + q(x)y = 0 in a neighborhood of a regular singular point located at the origin. Remember that if the regular singular point is located at x0, the transformation X = x − x0 reduces the equation to the preceding case. A solution is sought in the form 5. y(x) = xλ ∞ n=0 anxn [a0 = 0], where the number λ may be either real or complex, and is such that a0 = 0. Expressing p(x) and q(x) as their Maclaurin series 6. p(x) = p0 + p1(x) + p2x2 + · · · and q(x) = q0 + q1(x) + q2x2 + · · ·, 22.16 Power Series and Frobenius Methods 399 and substituting for y(x), p(x) and q(x) in 22.16.1.4, as in the power series method, leads to an identity involving powers of x, in which the coefficient of the lowest power xλ is given by [λ(λ − 1) + p0λ + q0]a0 = 0. Since, by hypothesis, a0 = 0, this leads to the indicial equation 7. λ(λ − 1) + p0λ + q0 = 0, from which the permissible values of the exponent λ may be found. The form of the two linearly independent solutions y1(x) and y2(x) of 22.16.1.4 is determined as follows: Case 1. If the indicial equation has distinct roots λ1 and λ2 that do not differ by an integer, then for x > 0 8. y1(x) = xλ1 ∞ n=0 anxn and 9. y2(x) = xλ2 ∞ n=0 bnxn , where the coefficients an and bn are found recursively in terms of a0 and b0, as in the power series method, by equating to zero the coefficient of the general term in the identity in powers of x that led to the indicial equation, and first setting λ = λ1 and then λ = λ2. Here the coefficients in the solution for y2(x) have been denoted by bn, instead of an, to avoid confusion with the coefficients in y1(x). The same convention is used in Cases 2 and 3, which follow. Case 2. If the indicial equation has a double root λ1 = λ2 = λ, then 10. y1(x) = xλ ∞ n=0 anxn and 11. y2(x) = y1(x) ln \|x\| + xλ ∞ n=1 bnxn . Case 3. If the indicial equation has roots λ1 and λ2 that differ by an integer and λ1 > λ2, then 12. y1(x) = xλ1 ∞ n=0 anxn 400 Chapter 22 Solutions of Standard Ordinary Differential Equations and 13. y2(x) = Ky1(x) ln \|x\| + xλ2 ∞ n=0 bnxn , where K may be zero. In all three cases the coefficients an in the solutions y1(x) are found recursively in terms of the arbitrary constant a0 = 0, as already indicated. In Case 1 the solution y2(x) is found in the same manner, with the coefficients bn being found recursively in terms of the arbitrary constant b0 = 0. However, the solutions y2(x) in Cases 2 and 3 are more difficult to obtain. One technique for finding y2(x) involves using a variant of the method of variation of parameters (see 22.10.1). If x < 0 replace xλ by \|x\|λ in results 5 to 13. If y1(x) is a known solution of 14. d2 y dx2 + p(x) dy dx + q(x)y = 0, a second linearly independent solution can be shown to have the form 15. y2(x) = y1(x)υ(x), where 16. υ(x) = exp − p(x)dx [y1(x)] 2 dx. This is called the integral method for the determination of a second linearly independent solution in terms of a known solution. Often the integral determining υ(x) cannot be evaluated analytically, but if the numerator and denominator of the integrand are expressed in terms of power series, the method of 1.11.1.4 may be used to determine their quotient, which may then be integrated term by term to obtain υ(x), and hence y2(x) in the form y2(x) = y1(x)υ(x). This method will generate the logarithmic term automatically when it is required. Example 22.17 (distinct roots λ1, λ2 not differing by an integer). The differential equation x d2 y dx2 + 1 2 − x dy dx + 2y = 0 has a regular singular point at the origin. The indicial equation is λ λ − 1 2 = 0, which corresponds to Case 1 because λ1 = 0, λ2 = 1 2 . 404 Chapter 22 Solutions of Standard Ordinary Differential Equations where for c = 0, 1, 2, . . . , 11. y1(x) = F(b, c; x) = 1 + b c x + b(b + 1) c(c + 1) x2 2! + b(b + 1)(b + 2) c(c + 1)(c + 2) x3 3! + · · · 12. y2(x) = x1−c F(b − c + 1, 2 − c; x). 22.18 NUMERICAL METHODS 22.18.1 When numerical solutions to initial value problems are required that cannot be obtained by analytical means it is necessary to use numerical methods. From the many methods that exist we describe in order of increasing accuracy only Euler's method, the modified Euler method, the fourth-order Runge–Kutta method and the Runge–Kutta–Fehlberg method. The section concludes with a brief discussion of how methods for the solution of initial value problems can be used to solve two-point boundary value problems for second-order equations. Euler's method. The Euler method provides a numerical approximation to the solution of the initial value problem 1. dy dx = f(x, y), which is subject to the initial condition 2. y(x0) = y0. Let each increment (step) in x be h, so that at the nth step 3. xn = x0 + nh. Then the Euler algorithm for the determination of the approximation yn+1 to y (xn+1) is 4. yn+1 = yn + f(xn, yn)h. The method uses a tangent line approximation to the solution curve through (xn, yn) in order to determine the approximation yn+1 to y (xn+1). The local error involved is O h2 . The method does not depend on equal step lengths at each stage of the calculation, and if the solution changes rapidly the step length may be reduced in order to control the local error. Modified Euler method. The modified Euler method is a simple refinement of the Euler method that takes some account of the curvature of the solution curve through (xn, yn) 22.18 Numerical Methods 405 when estimating yn+1. The modification involves taking as the gradient of the tangent line approximation at (xn, yn) the average of the gradients at (xn, yn) and (xn+1, yn+1) as determined by the Euler method. The algorithm for the modified Euler method takes the following form: If 5. dy dx = f(x, y), subject to the initial condition 6. y (x0) = y0, and all steps are of length h, so that after n steps 7. xn = x0 + nh, then 8. yn+1 = yn + h 2 [f(xn, yn) + f(xn + h, yn) + f(xn, yn)h]. The local error involved when estimating yn+1 from yn is O h3 . Here, as in the Euler method, if required the step length may be changed as the calculation proceeds. Runge–Kutta Fourth-Order Method. The Runge–Kutta (R–K) fourth-order method is an accurate and flexible method based on a Taylor series approximation to the function f(x, y) in the initial value problem 9. dy dx = f(x, y) subject to the initial condition 10. y(x0) = y0. The increment h in x may be changed at each step, but it is usually kept constant so that after n steps 11. xn = x0 + nh. The Runge–Kutta algorithm for the determination of the approximation yn+1 to y(xn+1) is 12. yn+1 = yn + 1 6 (k1 + 2k2 + 2k3 + k4), 22.18 Numerical Methods 407 and 20. K1 = hg(xn, yn, zn) K2 = hg xn + 1 2 h, yn + 1 2 k1, zn + 1 2 K1 K3 = hg xn + 1 2 h, yn + 1 2 k2, zn + 1 2 K2 K4 = hg(xn + h, yn + k3, zn + K3). As with the Runge–Kutta method, the local error involved in the determination of yn+1 from yn and zn+1 from zn is O h5 . This method may also be used to solve the second-order equation 21. d2 y dx2 = H x, y, dy dx subject to the initial conditions 22. y(x0) = a and y (x0) = b, by setting z = dy/dx and replacing the second-order equation by the system 23. dy dx = z 24. dz dx = H(x, y, z), subject to the initial conditions 25. y (x0) = a, z (x0) = b. Runge–Kutta–Fehlberg Method. The Runge–Kutta–Fehlberg (R–K–F) method is an adaptive technique that uses a Runge-Kutta method with a local error of order 5 in order to estimate the local error in the Runge–Kutta method of order 4. The result is then used to adjust the step length h so the magnitude of the global error is bounded by some given tolerance ε. In general, the step length is changed after each step, but high accuracy is attained if ε is taken to be suitably small, though this may be at the expense of (many) extra steps in the calculation. Because of the computation involved, the method is only suitable for implementation on a computer, though the calculation is efficient because only six evaluations of the f(x, y) are required at each step compared with the four required for the Runge–Kutta method of order 4. The R–K–F algorithm for the determination of the approximation ˜yn+1 to y(xn+1) is as follows: It is necessary to obtain a numerical solution to the initial value problem 22.18 Numerical Methods 411 Figure 22.1. where γ = δ are chosen arbitrarily. Let the two different solutions y1(x) and y2(x) correspond, respectively, to the initial conditions (I) and (II). Typical solutions are illustrated in Figure 22.1, in which y1(b) and y2(b) differ from the required result y(b) ≡ β. The approach used to obtain the desired solution is called the shooting method because if each solution curve is considered as the trajectory of a particle shot from the point x = a at the elevation y(a) = α, provided there is a unique solution, the required terminal value y(b) = β will be attained when the trajectory starts with the correct gradient y (a). When 22.18.1.31 is a linear equation, and so can be written 34. y = p(x)y + q(x)y + r(x) [a ≤ x ≤ b], the appropriate choice for y (a) is easily determined. Setting 35. y(x) = k1y1(x) + k2y2(x), with 36. k1 + k2 = 1, and substituting into 22.18.1.35 leads to the result 37. d2 dx2 (k1y1 + k2y2) = p(x) d dx (k1y1 + k2y2) + q(x)(k1y1 + k2y2) + r(x), which shows that y(x) = k1y1(x) + k2y2(x) is itself a solution. Furthermore, y(x) = k1y1(x) + k2y2(x) statisfies the left-hand boundary condition y(a) = α. Setting x = b, y(b) = β in 22.18.1.35 gives β = k1y1(b) + k2y2(b), 412 Chapter 22 Solutions of Standard Ordinary Differential Equations which can be solved in conjunction with k1 + k2 = 1 to give 38. k1 = β − y2(b) y1(b) − y2(b) and k2 = 1 − k1. The solution is then seen to be given by 39. y(x) = β − y2(b) y1(b) − y2(b) y1(x) + y1(b) − β y1(b) − y2(b) y2(x) [a ≤ x ≤ b]. A variant of this method involves starting from the two quite different initial value problems; namely, the original equation 40. y = p(x)y + q(x)y + r(x), with the initial conditions (III) y(a) = α and y (a) = 0 and the corresponding homogeneous equation 41. y = p(x)y + q(x)y, with the initial conditions (IV) y(a) = 0 and y (a) = 1. Using the fact that adding to the solution of the homogeneous equation any solution of the inhomogeneous equation will give rise to the general solution of the inhomogeneous equation, a similar form of argument to the one used above shows the required solution to be given by 42. y(x) = y3(x) + β − y3(b) y4(b) y4(x), where y3(x) and y4(x) are, respectively, the solutions of 22.18.1.40 with boundary conditions (III) and 22.18.1.41 with boundary conditions (IV). The method must be modified when 22.18.1.31 is nonlinear, because solutions of homoge- neous equations are no longer additive. It then becomes necessary to use an iterative method to adjust repeated estimates of y (a) until the terminal value y(b) = β is attained to the required accuracy. We mention only the iterative approach based on the secant method of interpolation, because this is the simplest to implement. Let the two-point boundary value problem be 43. y = f(x, y, y ) [a ≤ x ≤ b], 22.18 Numerical Methods 413 subject to the boundary conditions 44. y(a) = α and y(b) = β. Let k0 and k1 be two estimates of the initial gradient y (a), and denote the solution of y = f(x, y, y ), with y(a) = α and y (a) = k0, by y0(x), and the solution of y = f(x, y, y ), with y(a) = α and y (a) = k1, by y1(x). The iteration then proceeds using successive values of the gradient ki, for i = 2, 3, . . . , and the corresponding terminal values yi(b), in the scheme ki = ki−1 − (yi−1(b) − β) (ki−1 − ki−2) yi−1(b) − yi−2(b) , starting from k0 and k1, until for some i = N, \|yN (b) − yN−1(b)\| < ε, where ε is a preassigned tolerance. The required approximate solution is then given by yN (x), for a ≤ x ≤ b. In particular, it is usually necessary to experiment with the initial estimates k0 and k1 to ensure the convergence of the iterative scheme. Chapter 23 Vector Analysis 23.1 SCALARS AND VECTORS 23.1.1 Basic Definitions 23.1.1.1 A scalar quantity is completely defined by a single real number (positive or negative) that measures its magnitude. Examples of scalars are length, mass, temperature, and electric potential. In print, scalars are represented by Roman or Greek letters like r, m, T, and φ. A vector quantity is defined by giving its magnitude (a nonnegative scalar), and its line of action (a line in space) together with its sense (direction) along the line. Examples of vectors are velocity, acceleration, angular velocity, and electric field. In print, vector quantities are represented by Roman and Greek boldface letters like ν, a, , and E. By convention, the magnitudes of vectors ν, a, and are usually represented by the corresponding ordinary letters ν, a, and , etc. The magnitude of a vector r is also denoted by \|r\|, so that 1. r = \|r\| . A vector of unit magnitude in the direction of r, called a unit vector, is denoted by er, so that 2. r = rer. The null vector (zero vector) 0 is a vector with zero magnitude and no direction. A geometrical interpretation of a vector is obtained by using a straight-line segment parallel to the line of action of the vector, whose length is equal (or proportional) to the magnitude of the vector, with the sense of the vector being indicated by an arrow along the line segment. 415 416 Chapter 23 Vector Analysis The end of the line segment from which the arrow is directed is called the initial point of the vector, while the other end (toward which the arrow is directed) is called the terminal point of the vector. A right-handed system of rectangular cartesian coordinate axes 0{x, y, z} is one in which the positive direction along the z-axis is determined by the direction in which a right-handed screw advances when rotated from the x- to the y-axis. In such a system the signed lengths of the projections of a vector r with initial point P(x0, y0, z0) and terminal point Q(x1, y1, z1) onto the x-, y-, and z-axes are called the x, y, and z components of the vector. Thus the x, y, and z components of r directed from P to Q are x1 − x0, y1 − y0, and z1 − z0, respectively (Figure 23.1(a)). A vector directed from the origin 0 to the point P(x0, y0, z0) has x0, y0, and z0 as its respective x, y, and z components [Figure 23.1(b)]. Special unit vectors directed along the x-, y-, and z-axes are denoted by i, j, and k, respectively. The cosines of the angles α, β, and γ between r and the respective x-, y-, and z-axes shown in Figure 23.2 are called the direction cosines of the vector r. If the components of r are x, y, and z, then the respective direction cosines of r, denoted by l, m, and n, are 3. l = x r , m = y r , n = z r with r = x2 + y2 + z2 1/2 . The direction cosines are related by 4. l2 + m2 + n2 = 1. Numbers u, v, and w proportional to l, m, and n, respectively, are called direction ratios. Figure 23.1. 23.1 Scalars and Vectors 417 Figure 23.2. 23.1.2 Vector Addition and Subtraction 23.1.2.1 Vector addition of vectors a and b, denoted by a + b, is performed by first translating vector b, without rotation, so that its initial point coincides with the terminal point of a. The vector sum a + b is then defined as the vector whose initial point is the initial point of a, and whose terminal point is the new terminal point of b (the triangle rule for vector addition) (Figure 23.3(a)). The negative of vector c, denoted by −c, is obtained from c by reversing its sense, as in Figure 23.3(b), and so 1. c = \|c\| = \|−c\|. The difference a − b of vectors a and b is defined as the vector sum a + (−b). This corresponds geometrically to translating vector −b, without rotation, until its initial point coincides with the terminal point of a, when the vector a − b is the vector drawn from the initial point of a to the new terminal point of −b (Figure 23.4(a)). Equivalently, a − b is obtained by bringing into coincidence the initial points of a and b and defining a − b as the vector drawn from the terminal point of b to the terminal point of a [Figure 23.4(b)]. Vector addition obeys the following algebraic rules: 2. a + (−a) = a − a = 0 3. a + b + c = a + c + b = b + c + a (commutative law) 4. (a + b) + c = a + (b + c) (associative law) 23.1 Scalars and Vectors 419 Figure 23.5. 23.1.4 Vectors in Component Form 23.1.4.1 If a, b, and c are any three noncoplanar vectors, an arbitrary vector r may always be written in the form 1. r = λ1a + λ2b + λ3c, where the scalars λ1, λ2, and λ3 are the components of r in the triad of reference vectors a, b, c. In the important special case of rectangular Cartesian coordinates 0{x, y, z}, with unit vectors i, j, and k along the x-, y-, and z-axes, respectively, the vector r drawn from point P(x0, y0, z0) to point Q(x1, y1, z1) can be written (Figure 23.1(a)) 2. r = (x1 − x0)i + (y1 − y0)j + (z1 − z0)k. Similarly, the vector drawn from the origin to the point P(x0, y0, z0) becomes (Figure 23.1(b)) 3. r = x0i + y0j + z0k. 428 Chapter 23 Vector Analysis while for three points P1, P2 and P3 on C, 5. P2 P1 F · dr = P3 P1 F · dr + P2 P3 F · dr. A special case of a line integral occurs when F is given by 6. F = grad φ = ∇φ, where in rectangular Cartesian coordinates 7. grad φ = i ∂φ ∂x + j ∂φ ∂y + k ∂φ ∂z , for 8. C F · dr = P2 P1 F · dr = φ(P2) − φ(P1), and the line integral is independent of the path C, depending only on the initial point P1 and terminal point P2 of C. A vector field of the form 9. F = grad φ is called a conservative field, and φ is then called a scalar potential. For the definition of grad φ in terms of other coordinate systems see 24.2.1 and 24.3.1. In a conservative field, if C is a closed curve, it then follows that 10. C F · dr = C F · dr = 0, where the symbol indicates that the curve (contour) C is closed. 23.10 VECTOR INTEGRAL THEOREMS 23.10.1 Let a surface S defined by z = f(x, y) that is bounded by a closed space curve C have an element of surface area dσ, and let n be a unit vector normal to S at a representative point P (Figure 23.11). Then the vector element of surface area dS of surface S is defined as 1. dS = dσ n. The surface integral of a vector function F(x, y, z) over the surface S is defined as 23.10 Vector Integral Theorems 429 Figure 23.11. 2. S F · d S = S F · n dσ. The Gauss divergence theorem states that if S is a closed surface containing a volume V with volume element dV, and if the vector element of surface area dS = n dσ, where n is the unit normal directed out of V and dσ is an element of surface area of S, then 3. V div F dV = S F· dS = S F· n dσ. The Gauss divergence theorem relates the volume integral of div F to the surface integral of the normal component of F over the closed surface S. In terms of the rectangular Cartesian coordinates 0{x, y, z}, the divergence of the vector F = F1i + F2j + F3k, written div F, is defined as 4. div F = ∂F1 ∂x + ∂F2 ∂y + ∂F3 ∂z . For the definitions of div F in terms of other coordinate systems see 24.2.1 and 24.3.1. Stoke's theorem states that if C is a closed curve spanned by an open surface S, and F is a vector function defined on S, then 5. C F · dr = S (∇ × F) · dS = S (∇ × F) · n dσ. In this theorem the direction of unit normal n in the vector element of surface area d S = dσ n is chosen such that it points in the direction in which a right-handed screw would advance 430 Chapter 23 Vector Analysis when rotated in the sense in which the closed curve C is traversed. A surface for which the normal is defined in this manner is called an oriented surface. The surface S shown in Figure 23.11 is oriented in this manner when C is traversed in the direction shown by the arrows. In terms of rectangular Cartesian coordinates 0{x, y, z}, the curl of the vector F = F1i + F2j + F3k, written either ∇ × F, or curl F, is defined as 6. ∇ × F = i ∂ ∂x + j ∂ ∂y + k ∂ ∂z × F = ∂F3 ∂y − ∂F2 ∂z i + ∂F1 ∂z − ∂F3 ∂x j + ∂F2 ∂x − ∂F1 ∂y k. For the definition of ∇ × F in terms of other coordinate systems see 24.2.1 and 24.3.1. Green's first and second theorems (identities). Let U and V be scalar functions of posi- tion defined in a volume V contained within a simple closed surface S, with an outward-drawn vector element of surface area d S. Suppose further that the Laplacians ∇2 U and ∇2 V are defined throughout V , except on a finite number of surfaces inside V , across which the second- order partial derivatives of U and V are bounded but discontinuous. Green's first theorem states that 7. (U∇V ) · d S = V U∇2 V + (∇U) · (∇V ) dV, where in rectangular Cartesian coordinates 8. ∇2 U = ∂2 U ∂x2 + ∂2 U ∂y2 + ∂2 U ∂z2 . The Laplacian operator ∇2 is also often denoted by , or by n if it is necessary to specify the number n of space dimensions involved, so that in Cartesian coordinates 2U = ∂2 U/∂x2 + ∂2 U/∂y2 . For the definition of the Laplacian in terms of other coordinate systems see 24.2.1 and 24.3.1. Green's second theorem states that 9. V U∇2 V − V ∇2 U dV = S (U∇V − V ∇U) · dS. In two dimensions 0{x, y}, Green's theorem in the plane takes the form 10. C (P dx + Q dy) = A ∂Q ∂x − ∂P ∂y dx dy, where the scalar functions P(x, y) and Q(x, y) are defined and differentiable in some plane area A bounded by a simple closed curve C except, possibly, across an arc γ in A joining two distinct points of C, and the integration is performed counterclockwise around C. Chapter 25 Partial Differential Equations and Special Functions 25.1 FUNDAMENTAL IDEAS 25.1.1 Classification of Equations 25.1.1.1 A partial differential equation (PDE) of order n, for an unknown function Φ of the m independent variables x1, x2, . . . , xm(m ≥ 2), is an equation that relates one or more of the nth-order partial derivatives of Φ to some or all of Φ, x1, x2, . . . , xm and the partial derivatives of Φ or order less than n. The most general second-order PDE can be written 1. F x1, x2, . . . , xm, Φ, ∂Φ ∂x1 , . . . , ∂Φ ∂xm , ∂2 Φ ∂x2 1 , . . . , ∂2 Φ ∂xi∂xj , . . . , ∂2 Φ ∂x2 m = 0, where F is an arbitrary function of its arguments. A solution of 25.1.1.1.1 in a region R of the space to which the independent variables belong is a twice differentiable function that satisfies the equation throughout R. A boundary value problem for a PDE arises when its solution is required to satisfy conditions on a boundary in space. If, however, one of the independent variables is the time t and the solution is required to satisfy certain conditions when t = 0, this leads to an initial value problem for the PDE. Many physical situations involve a combination of both of these situations, and they then lead to an initial boundary value problem. 447 448 Chapter 25 Partial Differential Equations and Special Functions A linear second-order PDE for the function Φ(x1, x2, . . . , xm) is an equation of the form 2. m i, j=1 Aij ∂2 Φ ∂xi∂xj + m i=1 Bi ∂Φ ∂xi + CΦ = f, where the Aij, Bi, C, and f are functions of the m independent variables x1, x2, . . . , xm. The equation is said to be homogeneous if f ≡ 0, otherwise it is inhomogeneous (non- homogeneous). The most general linear second-order PDE for the function Φ(x, y) of the two independent variables x and y is 3. A(x, y) ∂2 Φ ∂x2 + 2B(x, y) ∂2 Φ ∂x∂y + C(x, y) ∂2 Φ ∂y2 + d(x, y) ∂Φ ∂x + e(x, y) ∂Φ ∂y + f(x, y)Φ = g(x, y) , where x, y may be two spatial variables, or one space variable and the time (usually denoted by t). An important, more general second-order PDE that is related to 25.1.1.1.1 is 4. A ∂2 Φ ∂x2 + 2B ∂2 Φ ∂x∂y + C ∂2 Φ ∂y2 = H x, y, Φ, ∂Φ ∂x , ∂Φ ∂y , where A, B, and C, like H, are functions of x, y, Φ, ∂Φ/∂x, and ∂Φ/∂y. A PDE of this type is said to be quasilinear (linear in its second (highest) order derivatives). Linear homogeneous PDEs such as 25.1.1.1.2 and 25.1.1.1.3 have the property that if Φ1 and Φ2 are solutions, then so also is c1Φ1 + c2Φ2, where c1 and c2 are arbitrary constants. This behavior of solutions of PDEs is called the linear superposition property, and it is used for the construction of solutions to initial or boundary value problems. The second-order PDEs 25.1.1.1.3 and 25.1.1.1.4 are classified throughout a region R of the (x, y)-plane according to certain of their mathematical properties that depend on the sign of ∆ = B2 − AC. The equations are said to be of hyperbolic type (hyperbolic) whenever ∆ > 0, to be of parabolic type (parabolic) whenever ∆ = 0, and to be of elliptic type (elliptic) whenever ∆ < 0. The most important linear homogeneous second-order PDEs in one, two, or three space dimensions and time are: 5. 1 c2 ∂2 Φ ∂t2 = ∇2 Φ (wave equation: hyperbolic) 6. k ∂Φ ∂t = ∇2 Φ [diffusion (heat) equation: parabolic] 7. ∇2 Φ = 0 (Laplace's equation: elliptic), 25.1 Fundamental Ideas 449 where c and k are constants, and the form taken by the Laplacian ∇2 Φ is determined by the coordinate system that is used. Laplace's equation is independent of the time and may be regarded as the steady-state form of the two previous equations, in the sense that they reduce to it if, after a suitably long time, their time derivatives may be neglected. Only in exceptional cases is it possible to find general solutions of PDEs, so instead it becomes necessary to develop techniques that enable them to be solved subject to auxiliary conditions (initial and boundary conditions) that identify specific problems. The most fre- quently used initial and boundary conditions for second-order PDEs are those of Cauchy, Dirichlet, Neumann, and Robin. For simplicity these conditions are now described for second- order PDEs involving two independent variables, although they can be extended to the case of more independent variables in an obvious manner. When the time t enters as an independent variable, a problem involving a PDE is said to be a pure initial value problem if it is completely specified by describing how the solution starts at time t = 0, and no spatial boundaries are involved. If only a first-order time derivative ∂Φ/∂t of the solution Φ occurs in the PDE, as in the heat equation, the initial condition involving the specification of Φ at t = 0 is called a Cauchy condition. If, however, a second-order time derivative ∂2 Φ/∂t2 occurs in the PDE, as in the wave equation, the Cauchy conditions on the initial line involve the specification of both Φ and ∂Φ/∂t at t = 0. In each of these cases, the determination of the solution Φ that satisfies both the PDE and the Cauchy condition(s) is called a Cauchy problem. More generally, Cauchy conditions on an arc Γ involve the specification of Φ and ∂Φ/∂n on Γ, where ∂Φ/∂n is the derivative of Φ normal to Γ. In other problems only the spatial variables x and y are involved in a PDE that con- tains both the terms ∂2 Φ/∂x2 and ∂2 Φ/∂y2 and governs the behavior of Φ in a region D of the (x, y)-plane. The region D will be assumed to lie within a closed boundary curve Γ that is smooth at all but a finite number of points, at which sharp corners occur (region D is bounded). A Dirichlet condition for such a PDE involves the specification of Φ on Γ. If ∂Φ/∂n = n · grad Φ, with n the inward drawn normal to Γ, a Neumann condition involves the specification of ∂Φ/∂n on Γ. A Robin condition arises when Φ is required to satisfy the condition α (x, y) Φ + β (x, y) ∂Φ/∂n = h (x, y) on Γ, where h may be identically zero. The determinations of the solutions Φ satisfying both the PDE and Dirichlet, Neumann, or Robin conditions on Γ are called boundary value problems of the Dirichlet, Neumann, or Robin types, respectively. When a PDE subject to auxiliary conditions gives rise to a solution that is unique (except possibly for an arbitrary additive constant), and depends continuously on the data in the auxiliary conditions, it is said to be well posed, or properly posed. An ill-posed problem is one in which the solution does not possess the above properties. Well-posed problems for the Poisson equation (the inhomogeneous Laplace equation) 8. ∇2 Φ (x, y) = H(x, y) involving the above conditions are as follows: 450 Chapter 25 Partial Differential Equations and Special Functions The Dirichlet problem 9. ∇2 Φ (x, y) = H(x, y) in D with Φ = f (x, y) on Γ yields a unique solution that depends continuously on the inhomogeneous term H(x, y) and the boundary data f(x, y). The Neumann problem 10. ∇2 Φ (x, y) = H(x, y) in D with ∂Φ/∂n = g (x, y) on Γ yields a unique solution, apart from an arbitrary additive constant, that depends continuously on the inhomogeneous term H(x, y) and the boundary data g(x, y), provided that H and g satisfy the compatibility condition 11. D H(x, y)dA = Γ gdσ, where dA is the element of area in D and dσ is the element of arc length along Γ. No solution exists if the compatibility condition is not satisfied. The Robin problem 12. ∇2 Φ (x, y) = H(x, y) in D with α Φ + β ∂Φ ∂n = h on Γ yields a unique solution that depends continuously on the inhomogeneous term H(x, y) and the boundary data α(x, y), β(x, y), and h(x, y). If the PDE holds in a region D that is unbounded, the above conditions that ensure well-posed problems for the Poisson equation must be modified as follows: Dirichlet conditions Add the requirement that Φ is bounded in D. Neumann conditions Delete the compatibility condition and add the requirement that Φ (x, y) → 0 as x2 + y2 → ∞. Robin conditions Add the requirement that Φ is bounded in D. The variability of the coefficients A(x, y), B(x, y), and C(x, y) in 25.1.1.1.3 can cause the equation to change its type in different regions of the (x, y)-plane. An equation exhibiting this property is said to be of mixed type. One of the most important equations of mixed type is the Tricomi equation y ∂2 Φ ∂x2 + ∂2 Φ ∂y2 = 0, which first arose in the study of transonic flow. This equation is elliptic for y > 0, hyperbolic for y < 0, and degenerately parabolic along the line y = 0. Such equations are difficult to study because the appropriate auxiliary conditions vary according to the type of the equation. When a solution is required in a region within which the parabolic degeneracy occurs, the matching of the solution across the degeneracy gives rise to considerable mathematical difficulties. 25.2 Method of Separation of Variables 451 25.2 METHOD OF SEPARATION OF VARIABLES 25.2.1 Application to a Hyperbolic Problem 25.2.1.1 The method of separation of variables is a technique for the determination of the solution of a boundary value or an initial value problem for a linear homogeneous equation that involves attempting to separate the spatial behavior of a solution from its time variation (temporal behavior). It will suffice to illustrate the method by considering the special linear homoge- neous second-order hyperbolic equation 1. div (k∇φ) − hφ = ρ ∂2 φ ∂t2 , in which k > 0, h ≥ 0, and φ(x, t) is a function of position vector x, and the time t. A typical homogeneous boundary condition to be satisfied by φ on some fixed surface S in space bounding a volume V is 2. k1 ∂φ ∂n + k2φ S = 0, where k1, k2 are constants and ∂/∂n denotes the derivative of φ normal to S. The appropriate initial conditions to be satisfied by φ when t = 0 are 3. φ (S, 0) = φ1(S) and ∂φ ∂t (S, 0) = φ2 (S). The homogeneity of both 25.2.1.1.1 and the boundary condition 25.2.1.1.2 means that if ˜φ1 and ˜φ2 are solutions of 25.2.1.1.1 by satisfying 25.2.1.1.2, then the function c1 ˜φ1 + c2 ˜φ2 with c1, c2 being arbitrary constants will also satisfy the same equation and boundary condition. The method of separation of variables proceeds by seeking a solution of the form 4. φ (x, t) = U(x)T(t), in which U(x) is a function only on the spatial position vector x, and T(t) is a function only of the time t. Substitution of 25.2.1.1.4 into 25.2.1.1.1, followed by division by ρU(x)T(t), gives 5. L [U] ρU = T T , where T = d2 T/dt2 , and we have set 6. L [U] = div (k∇U) − hU. The spatial vector x has been separated from the time variable t in 25.2.1.1.5, so the left-hand side is a function only of x, whereas the right-hand side is a function only of t. It is only possible for these functions of x and t to be equal for all x and t if 7. L [U] ρU = T T = −λ, 452 Chapter 25 Partial Differential Equations and Special Functions with λ an absolute constant called the separation constant. This result reduces to the equation 8. L [U] + λρU = 0, with the boundary condition 9. k1 ∂U ∂n + k2U S = 0 governing the spatial variation of the solution, and the equation 10. T + λT = 0, with 11. T(0) = φ1 and T (0) = φ2, governing the time variation of the solution. Problem 25.2.1.1.8 subject to boundary condition 25.2.1.1.9 is called a Sturm-Liouville problem, and it only has nontrivial solutions (not identically zero) for special values λ1, λ2, . . . of λ called the eigenvalues of the problem. The solutions U1, U2, . . . , corresponding to the eigenvalues λ1, λ2, . . . , are called the eigenfunctions of the problem. The solution of 25.2.1.1.10 for each λ1, λ2, . . . , subject to the initial conditions of 25.2.1.1.11, may be written 12. Tn(t) = Cn cos √ λnt + Dn sin √ λnt [n = 1, 2, . . .] so that Φn(x, t) = Un(x) Tn(t) . The solution of the original problem is then sought in the form of the linear combination 13. φ (x, t) = ∞ n=1 Un(x) [Cn cos λnt + Dn sin λnt]. To determine the constants Cn and Dn it is necessary to use a fundamental property of eigen- functions. Setting 14. Un 2 = D ρ(x) U2 n(x) dV, and using the Gauss divergence theorem, it follows that the eigenfunctions Un(x) are orthog- onal over the volume V with respect to the weight function ρ(x), so that 15. D ρ (x) Um(x) Un(x) dV = 0, m = n, Un 2 , m = n. The constants Cn follow from 25.2.1.1.13 by setting t = 0, replacing φ(x, t), by φ1(x), multiplying by Um(x), integrating over D, and using 25.2.1.1.15 to obtain 25.3 The Sturm–Liouville Problem and Special Functions 453 16. Cm = 1 Um 2 D ρ(x) φ1(x) Um(x) dV . The constants Dm follow in similar fashion by differentiating φ(x, t) with respect to t and then setting t = 0, replacing ∂φ/∂t by φ2(x) and proceeding as in the determination of Cm to obtain 17. Dm = 1 Um 2 D ρ (x) φ2 (x) Um(x) dV. The required solution to 25.2.1.1.1 subject to the boundary condition 25.2.1.1.2 and the initial conditions 25.2.1.1.3 then follows by substituting for Cn, Dn in 25.2.1.1.13. If in 25.2.1.1.1 the term ρ∂2 φ/∂t2 is replaced by ρ∂φ/∂t the equation becomes parabolic. The method of separation of variables still applies, though in place of 25.2.1.1.10, the equation governing the time variation of the solution becomes 18. T + λT = 0, and so 19. T (t) = e−λt . Apart from this modification, the argument leading to the solution proceeds as before. Finally, 25.2.1.1.1 becomes elliptic if ρ ≡ 0, for which only an appropriate boundary condi- tion is needed. In this case, separation of variables is only performed on the spatial variables, though the method of approach is essentially the same as the one already outlined. The bound- ary conditions lead first to the eigenvalues and eigenfunctions, and then to the solution. 25.3 THE STURM–LIOUVILLE PROBLEM AND SPECIAL FUNCTIONS 25.3.1 In the Sturm–Liouville problem 25.2.1.1.8, subject to the boundary condition 25.2.1.1.9, the operator L[φ] is a purely spatial operator that may involve any number of space dimensions. The coordinate system that is used determines the form of L[φ] (see 24.2.1) and the types of special functions that enter into the eigenfunctions. To illustrate matters we consider as a representative problem 25.2.1.1.1 in cylindrical polar coordinates (r, θ, z) with k = const., ρ = const., and h ≡ 0, when the equation takes the form 1. ∂2 φ ∂r2 + 1 r ∂φ ∂r + 1 r2 ∂2 φ ∂θ2 + ∂2 φ ∂z2 = 1 c2 ∂2 φ ∂t2 , where c2 = k/ρ. The first step in separating variables involves separating out the time variation by writing 2. φ (r, θ, z, t) = U (r, θ, z) T (t). Substituting for φ in 25.3.1.1 and dividing by UT gives 25.3 The Sturm–Liouville Problem and Special Functions 455 To proceed further we must make use of the boundary condition for the problem which, as yet, has not been specified. It will be sufficient to consider the solution of 25.3.1.1 in a circular cylinder of radius a with its axis coinciding with the z-axis, inside of which the solution φ is finite, while on its surface φ = 0. Because the solution φ will be of the form φ(r, θ, z, t) = R (r)Θ(θ)T(t), it follows directly from this that the boundary condition φ(a, θ, z, t) = 0 corresponds to the condition 12. R(a) = 0. The solution of 25.3.1.10 is 13. Θ (θ) = ˜A cos µθ + ˜B sin µθ, where ˜A, ˜B are arbitrary constants. Inside the cylinder r = a the solution must be invariant with respect to a rotation through 2π, so that Θ(θ + 2π) = Θ(θ), which shows that µ must be an integer n = 0, 1, 2, . . . . By choosing the reference line from which θ is measured 25.3.1.3 may be rewritten in the more convenient form 14. Θn(θ) = An cos nθ. The use of integer values of n in 25.3.1.11 shows the variation of the solution φ with r to be governed by Bessel's equation of integer order n 15. r2 d2 R dr2 + r dR dr + λ2 r2 − n2 R = 0, which has the general solution (see 17.1.1.1.8) 16. R (r) = BJn(λr) + CYn(λr). The condition that the solution φ must be finite inside the cylinder requires us to set C = 0, because Yn(λr) is infinite at the origin (see 17.2.22.2), while the condition R(a) = 0 in 25.3.1.12 requires that 17. Jn(λa) = 0, which can only hold if λa is a zero of Jn(x). Denoting the zeros of Jn(x) by jn, 1, jn, 2, . . . (see 17.5.11) it follows that 18. λn,m = jn,m/a [n = 0, 1, . . . ; m = 1, 2, . . . ]. The possible spatial modes of the solution are thus described by the eigenfunctions 19. Un,m(r, θ) = Jn jn,mr a cos nθ [n = 0, 1, . . . ; m = 1, 2, . . .]. 456 Chapter 25 Partial Differential Equations and Special Functions The solution φ is then found by setting 20. φ(r, θ, z, t) = ∞ n=0 m=1 Jn jn,mr a cos nθ[Anm cos(λn,mct) + Bnm sin(λn,mct)], and choosing the constants Anm, Bnm so that specified initial conditions are satisfied. In this result the multiplicative constant An in 25.3.1.14 has been incorporated into the arbitrary constants Anm, Bnm introduced when 25.3.1.4 was solved. The eigenfunctions 25.3.1.19 may be interpreted as the possible spatial modes of an oscil- lating uniform circular membrane that is clamped around its circumference r = a. Each term in 25.3.1.20 represents the time variation of a possible mode, with the response to specified initial conditions comprising a suitable sum of such terms. Had 25.2.1.1 been expressed in terms of spherical polar coordinates (r, θ, φ), with k = const., h ≡ 0, and ρ = 0, the equation would have reduced to Laplace's equation ∇2 φ = 0. In the case where the solution is required to be finite and independent of the azimuthal angle φ, separation of variables would have led to eigenfunctions of the from Un(r, θ) = Arn Pn(cos θ), where Pn(cos θ) is the Legendre polynomial of degree n. Other choices of coordinate systems may lead to different and less familiar special functions, the properties of many of which are only to be found in more advanced reference works. 25.4 A FIRST-ORDER SYSTEM AND THE WAVE EQUATION 25.4.1 Although linear second-order partial differential equations govern the behavior of many physi- cal systems, they are not the only type of equation that is of importance. In most applications that give rise to second-order equations, the equations occur as a result of the elimination of a variable, sometimes a vector, between a coupled system of more fundamental first-order equations. The study of such first-order systems is of importance because when variables can- not be eliminated in order to arrive at a single higher order equation the underlying first-order system itself must be solved. A typical example of the derivation of a single second-order equation from a coupled system of first-order equations is provided by considering Maxwell's equations in a vacuum. These comprise a first-order linear system of the form 1. ∂E ∂t = curl H, ∂H ∂t = −curl E, with div E = 0. Here E = (E1, E2, E3) is the electric vector, H = (H1, H2, H3) is the magnetic vector, and the third equation is the condition that no distributed charge is present. Differentiation of the first equation with respect to t followed by substitution for ∂H/∂t from the second equation leads to the following second-order vector differential equation for E: 2. ∂2 E ∂t2 = −curl (curl E). 25.5 Conservation Equations (Laws) 457 Using vector identity 23.1.1.4 together with the condition div E = 0 shows E satisfies the vector wave equation 3. ∂2 E ∂t2 = ∇2 E. The linearity of this equation then implies that each of the components of E separately must satisfy this same equation, so 4. ∂2 U ∂t2 = ∇2 U, where U may be any component of E. An identical argument shows that the components of H satisfy this same scalar wave equa- tion. Thus the study of solutions of the first-order system involving Maxwell's equations reduces to the study of the single second-order scalar wave equation. An example of a first-order quasilinear system of equations, the study of which cannot be reduced to the study of a single higher order equation, is provided by the equations governing compressible gas flow 5. ∂ρ ∂t + div(ρu) = 0 6. ∂u ∂t + u · grad u + 1 ρ grad p = 0 7. p = f(ρ), where ρ is the gas density, u is the gas velocity, p is the gas pressure, and f(ρ) is a known function of ρ (it is the constitutive equation relating the pressure and density). Only in the case of linear acoustics, in which the pressure variations are small enough to justify linearization of the equations, can they be reduced to the study of the scalar wave equation. 25.5 CONSERVATION EQUATIONS (LAWS) 25.5.1 A type of first-order equation that is of fundamental importance in applications is the con- servation equation, sometimes called a conservation law. In the one-dimensional case let u = u(x, t) represent the density per unit volume of a quantity of physical interest. Then in a cylindrical volume of cross-sectional area A normal to the x-axis and extending from x = a to x = b, the amount present at time t is 1. Q = A b a u (x, t) dx. Let f(x, t) at position x and time t be the amount of u that is flowing through a unit area normal to the x-axis per unit time. The quantity f(x, t) is called the flux of u at position x and time t. Considering the flux at the ends of the cylindrical volume we see that 458 Chapter 25 Partial Differential Equations and Special Functions 2. Q = A [f(a, t) − f(b, t)], because in a one-dimensional problem there is no flux normal to the axis of the cylinder (through the curved walls). If there is an internal source for u distributed throughout the cylinder it is necessary to take account of its effect on u before arriving at a final balance equation (conservation law) for u. Suppose u is created (or removed) at a rate h(x, t, u) at position x and time t. Then the rate of production (or removal) of u throughout the volume = A b a h(x, t, u) dx. Balancing all three of these results to find the rate of change of Q with respect to t gives 3. d dt b a u(x, t) dx = f(a, t) − f(b, t) + b a h(x, t, u) dx. This is the integral form of the conservation equation for u. Provided u(x, t) and f(x, t) are differentiable, this may be rewritten as 4. b a ∂u ∂t + ∂f ∂x − h(x, t, u) dx = 0, for arbitrary a and b. The result can only be true for all a and b if 5. ∂u ∂t + ∂f ∂x = h(x, t, u), which is the differential equation form of the conservation equation for u. In more space dimensions the differential form of the conservation equation becomes the equation in divergence form 6. ∂u ∂t + div f = h(x, t, u). 25.6 THE METHOD OF CHARACTERISTICS 25.6.1 Because the fundamental properties of first-order systems are reflected in the behavior of sin- gle first-order scalar equations, the following introductory account will be restricted to this simpler case. Consider the single first-order quasilinear equation 1. a(x, t, u) ∂u ∂t + b(x, t, u) ∂u ∂x = h(x, t, u), subject to the initial condition u(x, 0) = g(x). Let a curve Γ in the (x, t)-plane be defined parametrically in terms of σ by t = t(σ), x = x(σ). The tangent vector T to Γ has components dx/dσ, dt/dσ, so the directional derivative of u with respect to σ along Γ is 25.6 The Method of Characteristics 459 2. du dσ = T · grad u = dx dσ ∂u ∂x + dt dσ ∂u ∂t . Comparison of this result with the left-hand side of 25.6.1.1 shows that it may be rewritten in the first characteristic form 3. du dσ = h(x, t, u), along the characteristic curves in the (x, t)-plane obtained by solving 4. dt dσ = a(x, t, u) and dx dσ = b(x, t, u). On occasion it is advantageous to retain the parameter σ, but it is often removed by multi- plying du/dσ and dx/dσ by dσ/dt to obtain the equivalent second characteristic form for the partial differential equation 25.6.1.1. 5. du dt = h(x, t, u) a(x, t, u) , along the characteristic curves obtained by solving 6. dx dt = b(x, t, u) a(x, t, u) . This approach, called solution by the method of characteristics, has replaced the original partial differential equation by the solution of an ordinary differential equation that is valid along each member of the family of characteristic curves in the (x, t)-plane. If a characteristic curve C0 intersects the initial line at (x0, 0), it follows that at this point the initial condition for u along C0 must be u(x0, 0) = g(x0). This situation is illustrated in Figure 25.1, which shows typical members of a family of characteristics in the (x, t)-plane together with the specific curve C0. Figure 25.1. 460 Chapter 25 Partial Differential Equations and Special Functions When the partial differential equation is linear, the characteristic curves can be found independently of the solution, but in the quasilinear case they must be determined simultane- ously with the solution on which they depend. This usually necessitates the use of numerical methods. Example 25.1 This example involves a constant coefficient first-order equation. Consider the equation 7. ∂u ∂t + c ∂u ∂x = 0 [c > 0 a const.], sometimes called the advection equation, and subject to the initial condition u(x, 0) = g(x). When written in the second characteristic form this becomes 8. du dt = 0 along the characteristic curves given by integrating dx/dt = c. Thus the characteristic curves are the family of parallel straight lines x = ct + ξ that intersects the initial line t = 0 (the x-axis) at the point (ξ, 0). The first equation shows that u = const. along a characteristic, but at the point (ξ, 0) we have u(ξ, 0) = g(ξ), so u(x, t) = g(ξ) along this characteristic. Using the fact that ξ = x − ct it then follows that the required solution is 9. u(x, t) = g(x − ct). The solution represents a traveling wave with initial shape g(x) that moves to the right with constant speed c without change of shape. The nature of the solution relative to the characteristics is illustrated in Figure 25.2. Figure 25.2. 462 Chapter 25 Partial Differential Equations and Special Functions 17. ∂u ∂t + f(u) ∂u ∂x = 0, subject to the initial condition u(x, 0) = g(x), where f(u) and g(x) are known continuous and differentiable functions. Proceeding as before, the second characteristic form for the equation becomes 18. du dt = 0 along the characteristic curves given by integrating dx/dt = f(u). The first equation shows u = constant along a characteristic curve. Using this result in the second equation and inte- grating shows the characteristic curves to be the family of straight lines 19. x = t f(u) + ξ, where (ξ, 0) is the point on the initial line from which the characteristic emanates. From the initial condition it follows that the value of u transported along this characteristic must be u = g(ξ). Because ξ = x − t f(u), if follows that the solution is given implicitly by 20. u(x, t) = g [x − t f(u)]. The implicit nature of this solution indicates that the solution need not necessarily always be unique. This can also be seen by computing ∂u/∂x, which is 21. ∂u ∂x = g (x − t f(u)) 1 + tg (x − t f(u)) f (u) . If the functions f and g are such that the denominator vanishes for some t = tc > 0, the derivative ∂u/∂x becomes unbounded when t = tc. When this occurs, the differential equation can no longer govern the behavior of the solution, so it ceases to have meaning and the solution may become nonunique. The development of the solution when f(u) = u and g(x) = sin x is shown in Figure 25.3. The wave profile is seen to become steeper due to the influence of nonlinearity until t = tc, where the tangent to part of the solution profile becomes infinite. Subsequent to time tc the solution is seen to become many valued (nonunique). This result illustrates that, unlike the linear case in which f(u) = c (const.), a quasilinear equation of this type cannot describe traveling waves of constant shape. 25.7 DISCONTINUOUS SOLUTIONS (SHOCKS) 25.7.1 To examine the nature of discontinuous solutions of first-order quasilinear hyperbolic equations, it is sufficient to consider the scalar conservation equation in differential form 25.7 Discontinuous Solutions (Shocks) 463 Figure 25.3. 1. ∂u ∂t + div f = 0, where u = u(x, t) and f = f(u). Let V (t) be an arbitrary volume bounded by a surface S(t) moving with velocity ν. Provided u is differentiable in V (t), it follows from 23.11.1.1 that the rate of change of the volume integral of u is 2. d dt V(t) u dV = V(t) ∂u ∂t + div (uν) dV. Substituting for ∂u/∂t gives 3. d dt V(t) u dV = V(t) [ div(uν) − div f ] dV , and after use of the Gauss divergence theorem 23.9.1.3 this becomes 4. d dt V(t) u dV = S(t) (uν − f ) · dσ, where dσ is the outward drawn vector element of surface area of S(t) with respect to V (t). Now suppose that V (t) is divided into two parts V1(t) and V2(t) by a moving surface Ω(t) across which u is discontinuous, with u = u1 in V1(t) and u = u2 in V2 (t). Subtracting from this last result the corresponding results when V (t) is replaced first by V1(t) and then by V2(t) gives 5. 0 = Ω(t) (uν − f )1 · dΩ1 + Ω(t) (uν − f )2 · dΩ2, where now ν is restricted to Ω(t), and so is the velocity of the discontinuity surface, while dΩi is the outwardly directed vector element of surface area of Ω(t) with respect to Vi(t) for 464 Chapter 25 Partial Differential Equations and Special Functions i = 1, 2. As dΩ1 = −dΩ2 = ndΩ, say, where n is the outward drawn unit normal to Ω1(t) with respect to V1(t), it follows that 6. 0 = Ω(t) [(uν − f )1 · n − (uν − f )2 · n] dΩ = 0. The arbitrariness of V (t) implies that this result must be true for all Ω(t), which can only be possible if 7. (uν − f )1 · n = (uν − f )2 · n. The speed of the discontinuity surface along the normal n is the same on either side of Ω(t), so setting ν1 · n = ν2 · n = s, leads to the following algebraic jump condition that must be satisfied by a discontinuous solution 8. (u1 − u2)s =(f1 − f2) · n. This may be written more concisely as 9. [[u]] s = [[f ]] · n, where [[α]] = α1 − α2 denotes the jump in α across Ω(t). In general, f = f(u) is a nonlinear function of u, so for any given s, specifying u on one side of the discontinuity and using the jump condition to find u on the other side may lead to more than one value. This possible nonuniqueness of discontinuous solutions to quasilinear hyperbolic equations is typical, and in physical problems a criterion (selection principle) must be developed to select the unique physically realizable discontinuous solution from among the set of all mathematically possible but nonphysical solutions. In gas dynamics the jump condition is called the Rankine–Hugoniot jump condition, and a discontinuous solution in which gas flows across the discontinuity is called a shock, or a shock wave. In the case in which a discontinuity is present, but there is no gas flow across it, the discontinuity is called a contact discontinuity. In gas dynamics two shocks are possible mathematically, but one is rejected as nonphysical by appeal to the second law of thermodynamics (the selection principle used there) because of the entropy decrease that occurs across it, so only the compression shock that remains is a physically realizable shock. To discuss discontinuous solutions, in general it is necessary to introduce more abstract selec- tion principles, called entropy conditions, which amount to stability criteria that must be satisfied by solutions. The need to work with conservation equations (equations in divergence form) when consid- ering discontinuous solutions can be seen from the preceding argument, for only then can the Gauss divergence theorem be used to relate the solution on one side of the discontinuity to that on the other. 25.8 Similarity Solutions 465 25.8 SIMILARITY SOLUTIONS 25.8.1 When characteristic scales exist for length, time, and the dependent variables in a physical problem it is advantageous to free the equations from dependence on a particular choice of scales by expressing them in nondimensional form. Consider, for example, a cylinder of radius ρ0 filled with a viscous fluid with viscosity ν, which is initially at rest at time t = 0. This cylinder is suddenly set into rotation about the axis of the cylinder with angular velocity ω. Although this is a three-dimensional problem, because of radial symmetry about the axis, the velocity u at any point in the fluid can only depend on the radius ρ and the time t if gravity is neglected. Natural length and time scales are ρ0,which is determined by the geometry of the problem, and the period of a rotation τ = 2π/ω, which is determined by the initial condition. Appro- priate nondimensional length and time scales are thus r = ρ/ρ0 and T = t/τ. The dependent variable in the problem is the fluid velocity u(r, t), which can be made nondimensional by selecting as a convenient characteristic speed u0 = ωρ0, which is the speed of rotation at the wall of the cylinder. Thus, an appropriate nondimensional velocity is U = u/u0. The only other quantity that remains to be made nondimensional is the viscosity ν. It proves most convenient to work with 1/ν and to define the Reynolds number Re = u0ρ0/ν, which is a dimensionless quantity. Once the governing equation has been reexpressed in terms of r, T, U, and Re, and its dimensionless solution has been found, the result may be used to provide the solution appropriate to any choice of ρ0, ω, and ν for which the governing equations are valid. Equations that have natural characteristic scales for the independent variables are said to be scale-similar. The name arises from the fact that any two different problems with the same numerical values for the dimensionless quantities involved will have the same nondimensional solution. Certain partial differential equations have no natural characteristic scales for the indepen- dent variables. These are equations for which self-similar solutions can be found. In such problems the nondimensionalization is obtained not by introducing characteristic scales, but by combining the independent variables into nondimensional groups. The classic example of a self-similar solution is provided by considering the unsteady heat equation 1. ∂2 T ∂x2 = 1 κ ∂T ∂t applied to a semi-infinite slab of heat conducting material with thermal diffusivity κ, with x measured into the slab normal to the bounding face x = 0. Suppose that for t < 0 the material is all at the initial temperature T0, and that at t = 0 the temperature of the face of the slab is suddenly changed to T1. Then there is a natural characteristic temperature scale provided by the temperature difference T1 − T0, so a convenient nondimensional temperature is τ = (T − T0)/(T1 − T0). However, in this example, no natural length and time scales can be introduced. 466 Chapter 25 Partial Differential Equations and Special Functions If a combination η, say, of variables x and t is to be introduced in place of the separate variables x and t themselves, the one-dimensional heat equation will be simplified if this change of variable reduces it to a second-order ordinary differential equation in terms of the single independent variable η. The consequence of this approach will be that instead of there being a different temperature profile for each time t, there will be a single profile in terms of η from which the temperature profile at any time t can be deduced. It is this scaling of the solution on itself that gives rise to the name self-similar solution. Let us try setting 2. η = Dx/tn and τ = f(η) [D = const.] in the heat equation, where a suitable choice for n has still to be made. Routine calculation shows that the heat equation becomes 3. d2 f dη2 + n κD2 t2n−1 η df dη = 0. For this to become an ordinary differential equation in terms of the independent variable η, it is necessary for the equation to be independent of t, which may be accomplished by setting n = 1 2 to obtain 4. d2 f dη2 + 1 2κD2 η df dη = 0. It is convenient to choose D so that 2κD2 = 1, which corresponds to 5. D = 1/ √ 2κ. The heat equation then reduces to the variable coefficient second-order ordinary differential equation 6. d2 f dη2 + η df dη = 0, with η = x/ √ 2κt. The initial condition shows that f(0) = 1, while the temperature variation must be such that for all time t, T → T0 as η → ∞, so another condition on f is f(η) → 0 as η → ∞. Integration of the equation for f subject to these conditions can be shown to lead to the result 7. T = T0 + (T1 − T0)erfc x 2 √ κt . Sophisticated group theoretic arguments must be used to find the similarity variable in more complicated cases. However, this example illustrates the considerable advantage of this approach when a similarity variable can be found, because it reduces by one the number of independent variables involved in the partial differential equation. 25.9 Burgers's Equation, the KdV Equation, and the KdVB Equation 467 As a final simple example of self-similar solutions, we mention the cylindrical wave equation 8. ∂2 Φ ∂r2 + 1 r ∂Φ ∂r = 1 c2 ∂2 Φ ∂t2 , which has the self-similar solution 9. Φ(r, t) = rf(η), with η = r ct , where f is a solution of the ordinary differential equation 10. η(1 − η2 )f (η) + 3 − 2η2 f + f/η = 0. If the wave equation is considered to describe an expanding cylindrically symmetric wave, the radial speed of the wave vr can be shown to be 11. vr = −[f(η) + ηf (η)] , which in turn can be reduced to 12. vr = A 1 − η2 /η, η ≤ 1, 0, η > 1, where A is a constant of integration. Other solutions of 25.8.1.8 can be found if appeal is made to the fact that solutions are invariant with respect to a time translation, so that in the solution t may be replaced by t − t∗ , for some constant t∗ . Result 25.8.1.12 then becomes 13. vr = A(t∗ )[c2 (t − t∗ ) 2 − r2 ]1/2 /r, t∗ ≤ t − r/c, 0, t∗ > t − r/c, and different choices for A(t) will generate different solutions. 25.9 BURGERS'S EQUATION, THE KdV EQUATION, AND THE KdVB EQUATION 25.9.1 In time-dependent partial differential equations, certain higher order spatial derivatives are capable of interpretation in terms of important physical effects. For example, in Burgers's equation 468 Chapter 25 Partial Differential Equations and Special Functions 1. ∂u ∂t + u ∂u ∂x = ν ∂2 u ∂x2 [ν > 0] . the term on the right may be interpreted as a dissipative effect; namely, as the removal of energy from the system described by the equation. In the Korteweg–de Vries (KdV) equation 2. ∂u ∂t + u ∂u ∂x + µ ∂3 u ∂x3 = 0, the last term on the left represents a dispersive effect; namely, a smoothing effect that causes localized disturbances in waves that are propagated to spread out and disperse. Burgers's equation serves to model a gas shock wave in which energy dissipation is present (ν > 0). The steepening effect of nonlinearity in the second term on the left can be balanced by the dissipative effect, leading to a traveling wave of constant form, unlike the case examined earlier corresponding to ν = 0 in which a smooth initial condition evolved into a discontinuous solution (shock). The steady traveling wave solution for Burgers's equation that describes the so-called Burgers's shock wave is 3. u(ζ) = 1 2 (u− ∞ + u+ ∞) − 1 2 (u− ∞ − u+ ∞) tanh u− ∞ − u+ ∞ 4ν ζ , where ζ = x − ct, with the speed of propagation c = 1 2 (u− ∞ + u+ ∞) , u− ∞ > u+ ∞, and u− ∞ and u+ ∞ denote, respectively, the solutions at ζ → −∞ and ζ → +∞. This describes a smooth transition from u− ∞ to u+ ∞. The Burgers's shock wave profile is shown in Figure 25.4. The celebrated KdV equation was first introduced to describe the propagation of long waves in shallow water, but it has subsequently been shown to govern the asymptotic behavior of many other physical phenomena in which nonlinearity and dispersion compete. In the KdV equation, the smoothing effect of the dispersive term can balance the steepening effect of the nonlinearity in the second term to lead to a traveling wave of constant shape in the form of Figure 25.4. 25.9 Burgers's Equation, the KdV Equation, and the KdVB Equation 469 a pulse called a solitary wave. The solution for the KdV solitary wave in which u → u∞ as ζ → ± is 4. u(ζ) = u∞ + a sech2 ζ a 12µ 1/2 [u∞ > 0], where ζ = x − ct, with the speed of propagation c = u∞ + 1 3 a. Notice that, relative to u∞, the speed of propagation of the solitary wave is proportional to the amplitude a. It has been shown that these solitary wave solutions of the KdV equation have the remarkable property that, although they are solutions of a nonlinear equation, they can interact and preserve their identity in ways that are similar to those of linear waves. However, unlike linear waves, during the interaction the solutions are not additive, though after it they have interchanged their positions. This property has led to these waves being called solitons. Interaction between two solitons occurs when the amplitude of the one on the left exceeds that of the one on the right, for then overtaking takes place due to the speed of the one on the left being greater than the speed of the one on the right. This interaction is illustrated in Figure 25.5; the waves are unidirectional since only a first-order time derivative is present in the KdV equation. The Korteweg–de Vries–Burgers's (KdVB) equation 5. ∂u ∂t + u ∂u ∂x − ν ∂2 u ∂x2 + µ ∂3 u ∂x3 = 0 [ν > 0], describes wave propagation in which the effects of nonlinearity, dissipation, and dispersion are all present. In steady wave propagation the combined smoothing effects of dissipation and dis- persion can balance the steepening effect of nonlinearity and lead to a traveling wave solution moving to the right given by 6. u(ζ) = 3ν2 100µ sech2 (ζ/2) + 2 tanh(ζ/2) + 2 , with 7. ζ = −ν 5µ x − 6ν2 25µ t and speed c = 6ν2 /(25µ), or to one moving to the left given by Figure 25.5. 470 Chapter 25 Partial Differential Equations and Special Functions 8. u(ζ) = 3ν2 100µ sech2 (ζ/2) − 2 tanh(ζ/2) − 2 , with 9. ζ = ν 5µ x + 6ν2 25µ t and speed c = −6ν2 /(25µ). The wave profile for a KdVB traveling wave is very similar to that of the Burgers's shock wave. 25.10 THE POISSON INTEGRAL FORMULAS There are two fundamental boundary value problems for a solution u of the Laplace equation in the plane that can be expressed in terms of integrals. The first involves finding the solution (a harmonic function) in the upper half of the (x, y)-plane when Dirichlet conditions are imposed on the x-axis. The second involves finding the solution (a harmonic function) inside a circle of radius R centered on the origin when Dirichlet conditions are imposed on the boundary of the circle. The two results are called the Poisson integral formulas for solutions of the Laplace equation in the plane, and they take the following forms: The Poisson integral formula for the half-plane Let f(x) be a real valued function that is bounded and may be either continuous or piecewise continuous for −∞ < x < ∞. Then, when the integral exists, the function 1. u(x, y) = y π ∞ −∞ f(s)ds (x − s) 2 + y2 is harmonic in the half-plane y > 0 and on the x-axis assumes the boundary condition u(x, 0) = f(x) wherever f(x) is continuous. The Poisson integral formula for a disk Let f(θ) be a real valued function that is bounded and may be either continuous or piecewise continuous for −π < θ ≤ π. Then, when the integral exists, the function 2. u(r, θ) = 1 2π 2π 0 R2 − r2 f(ψ)dψ R2 − 2rR cos(ψ − θ) + r2 is harmonic inside the disk 0 ≤ r < R, and on the boundary of the disk r = R, it assumes the boundary condition u(R, θ) = f(θ) wherever f(θ) is continuous. These integrals have many applications, and they are often used in conjunction with confor- mal mapping when a region of complicated shape is mapped either onto a half-plane or onto a disk when these formulas will give the solution. However, unless the boundary conditions are simple, the integrals may be difficult to evaluate, though they may always be used to provide numerical results. 25.11 The Riemann Method 471 Example 25.4 Find the harmonic function u in the half-plane y > 0 when u(x, 0) = −1 for −∞ < x < 0 and u(x, 0) = 1 for 0 < x < ∞. Setting f(x) = −1 for − ∞ < x < 0 and f(x) = 1 for 0 < x < ∞ in formula 1 and integrating gives u(x, y) = − 1 2 (π − 2 arctan(x/y)) π for − ∞ < x < 0 and u(x, y) = 1 2 (π + 2 arctan(x/y)) π for 0 < x < ∞. Notice that in this case the boundary condition f(x) is discontinuous at the origin. These solutions do not assign a value for u at the origin, but this is to be expected because none exists, and as the results are derived from formula 1 on the assumption that y = 0, the formula vanishes when y = 0. Example 25.5 Find the harmonic function u inside the unit disk centered on the origin when on its boundary u(1, θ) = sin2 θ. Setting R = 1 and f(θ) = sin2 θ in formula 2 leads to the integrand (1 − r2 ) sin2 ψ 1 − 2r cos(ψ − θ) + r2 . The change of variable to z = eiψ converts the integral into a complex integral with a pole of order 2 and also a simple pole inside the unit circle. An application of the residue theorem then shows the required solution to be u(r, θ) = 1 2 (1 − r2 cos 2θ). 25.11 THE RIEMANN METHOD The Riemann method of solution applies to linear hyperbolic equations when Cauchy conditions are prescribed along a finite arc ΓQR in the (x, y)-plane. It gives the solution u(x0, y0) at an arbitrary point (x0, y0) in terms of an integral representation involving a func- tion v(x, y; x0, y0) called the Riemann function, where u(x, y) satisfies the equation 1. ∂2 u ∂x∂y + a(x, y) ∂u ∂x + b(x, y) ∂u ∂y + c(x, y)u = f(x, y). In this representation x0 and y0 are considered to be parameters of a point P in the (x, y)-plane at which the solution is required, and their relationships to the points Q and R are shown in Figure 25.6. The Riemann function v is a solution of the homogeneous adjoint equation associated with 1 2. ∂2 v ∂x∂y − ∂(av) ∂x − ∂(bv) ∂y + cv = 0, 472 Chapter 25 Partial Differential Equations and Special Functions P (x0, y0) Q R y 0 x y0 x0 GQR Area PQR Figure 25.6. Points P, Q, and R and the arc ΓQR along which the Cauchy conditions for u are specified. subject to the conditions 3. v(x, η; x0, y0) = exp x x0 b(σ, y0) dσ and v(x0, y; x0, y0) = exp y y0 a(x0, σ) dσ. When equation 1 is written as L[u] = f(x, y) and equation 2 as M[v] = 0, equation 1 will be self-adjoint when the operators L and M are such that L ≡ M. In general, solving the adjoint equation for v is as difficult as solving the original equation for u, so a significant simplification results when the equation is self-adjoint. The Riemann integral formula for the solution u(x0, y0) at a point P in the (x, y)-plane terms of the Riemann function v is 4. u(x0, y0) = 1 2 (uv)R + 1 2 (uv)Q + P QR Fvdxdy − ΓQR buv + 1 2 (vux − uvx) dx − auv + 1 2 (vuy − uvy) dy, where P QR Fvdxdy denotes the integral over the region PQR. This method only gives closed form solutions when the equations involved are simple, so its main use is in deriving information about domains of dependence and influence for equation 1 of a rather general type. Chapter 26 Qualitative Properties of the Heat and Laplace Equation 26.1 THE WEAK MAXIMUM/MINIMUM PRINCIPLE FOR THE HEAT EQUATION Let u(x, t) satisfy the heat (diffusion) equation ut = k2 uxx in the space-time rectangle 0 ≤ x ≤ L, 0 ≤ t ≤ T. Then the maximum value of u(x, t) occurs either initially when t = 0, or on the sides of the rectangle x = 0 or x = L for 0 ≤ t ≤ T. Similarly, the minimum value of u(x, t) occurs either initially when t = 0, or on the sides of the rectangle x = 0 or x = L for 0 ≤ t ≤ T. 26.2 THE MAXIMUM/MINIMUM PRINCIPLE FOR THE LAPLACE EQUATION Let D be a connected bounded open region in two or three space dimensions, and let the function u be harmonic and continuous throughout D and on its boundary ∂D. Then if u is not constant it attains its maximum and minimum values only on the boundary ∂D. 26.3 GAUSS MEAN VALUE THEOREM FOR HARMONIC FUNCTIONS IN THE PLANE If u is harmonic in a region D of the plane, the value of u at any interior point P of D is the average of the values of u around any circle centered on P and lying entirely inside D. 473 474 Chapter 26 Qualitative Properties of the Heat and Laplace Equation 26.4 GAUSS MEAN VALUE THEOREM FOR HARMONIC FUNCTIONS IN SPACE If u is harmonic in a region D of space, the value of u at any interior point P of D is the average of the values of u around the surface of any sphere centered on P and lying entirely inside D. Chapter 28 The z-Transform 28.1 THE z-TRANSFORM AND TRANSFORM PAIRS The z-transform converts a numerical sequence x[n] into a function of the complex variable z, and it takes two different forms. The bilateral or two-sided z-transform, denoted here by Zb{x[n]}, is used mainly in signal and image processing, while the unilateral or one-sided z-transform, denoted here by Zu{x[n]}, is used mainly in the analysis of discrete time systems and the solution of linear difference equations. The bilateral z-transform Xb(z) of the sequence x[n] = {xn}∞ n = −∞ is defined as Zb{x [n]} = ∞ n = −∞ xnz−n = Xb(z) , and the unilateral z-transform Xu(z) of the sequence x[n] = {xn}∞ n=0 is defined as Zu{x [n]} = ∞ n=0 xnz−n = Xu(z) , where each series has its own domain of convergence (DOC). The series Xb(z) is a Laurent series, and Xu(z) is the principal part of the Laurent series of Xb(z). When xn = 0 for n < 0, 493 494 Chapter 28 The z-Transform the two z-transforms Xb(z) and Xu(z) are identical. In each case the sequence x[n] and its associated z-transform is called a z-transform pair. The inverse z-transformation x[n] = Z−1 {X(z)} is given by x[n] = 1 2πi Γ X(z) zn−1 dz, where X(z) is either Xb(z) or Xu(z), and Γ is a simple closed contour containing the origin and lying entirely within the domain of convergence of X(z). In many practical situations the z-transform is either found by using a series expansion of X(z) in the inversion integral or, if X(z) = N(z)/D(z) where N(z) and D(z) are polynomials in z, by means of partial fractions and the use of an appropriate table of z-transform pairs. In order that the inverse z-transform is unique it is necessary to specify the domain of convergence, as can be seen by comparison of entries 3 and 4 of Table 28.2. Table 28.1 lists general properties of the bilateral z-transform, and Table 28.2 lists some bilateral z-transform pairs. In what follows, use is made of the unit integer function h(n) = 0, n < 0 1, n ≥ 0 that is a generalization of the Heaviside step function, and the unit integer pulse function ∆(n − k) = 1, n = k 0, n = k , that is a generalization of the delta function. The relationship between the Laplace transform of a continuous function x(t) sampled at t = 0, T, 2T, . . . and the unilateral z-transform of the function ˆx(t) = ∞ n = 0 x(nT)δ(t − nT) follows from the result L{ˆx(t)} = ∞ 0 ∞ k=0 x(kT)δ(t − kT) e−st dt = ∞ k=0 x(kT) e−ksT . Setting z = esT this becomes L{ˆx(t)} = ∞ k=0 x(kT)z−k = X(z) , showing that the unilateral z-transform Xu(z) can be considered to be the Laplace transform of a continuous function x(t) for t ≥ 0 sampled at t = 0, T, 2T, . . . . Table 28.3 lists some general properties of the unilateral z-transform, and Table 28.4 lists some unilateral z-transform pairs. As an example of the use of the unilateral z-transform when solving a linear constant coefficient difference equation, consider the difference equation xn+2 − 4xn+1 + 4xn = 0 with the initial conditions x0 = 2, x1 = 3. 28.1 The z-Transform and Transform Pairs 495 Table 28.1. General Properties of the Bilateral z-Transform Xb(n) = ∞ n=−∞ xnz−n Term in sequence z-Transform Xb(z) Domain of convergence 1. αxn + βyn αXb(z) + βYb(z) Intersection of DOCs of Xb(z) and Yb(z) with α, β constants 2. xn −N z−N Xb(z) DOC of Xb(z) to which it may be necessary to add or delete the origin or the point at infinity 3. nxn −zd Xb(z)/dz DOC of Xb(z) to which it may be necessary to add or delete the origin and the point at infinity 4. zn 0 xn Xb(z/z0) DOC of Xb(z) scaled by \|z0\| 5. nzn 0 xn −zd Xb(z/z0)/dz DOC of Xb(z) scaled by \|z0\| to which it may be necessary to add or delete the origin and the point at infinity 6. x−n Xb(1/z) DOC of radius 1/R, where R is the radius of convergence of DOC of Xb(z) 7. nx−n −zd Xb(1/z)/dz DOC of radius 1/R, where R is the radius of convergence of DOC of Xb(z) 8. xn Xb(z) The same DOC as xn 9. Re xn 1 2 Xb(z) + Xb(z) DOC contains the DOC of xn 10. Im xn 1 2i Xb(z)−Xb(z) DOC contains the DOC of xn 11. ∞ k=−∞ xkyn−k Xb(z)Yb(z) DOC contains the intersection of the DOCs of Xb(z) and Yb(z) (convolution theorem) 12. xnyn 1 2πi Γ Xb(ζ) Yb(z/ζ)ζ−1 dζ DOC contains the DOCs of Xb(z) and Yb(z), with Γ inside the DOC and containing the origin (convolution theorem) 13. Parseval ∞ n=−∞ xn yn = 1 2πi Γ Xb(ζ) Y b 1/ζ ζ−1 dζ DOC contains the intersection of formula DOCs of Xb(z) and Yb(z), with Γ inside the DOC and containing the origin 14. Initial value x0 = lim z→∞ Xb(z) theorem for xnh(n) Chapter 29 Numerical Approximation 29.1 INTRODUCTION The derivation of numerical results usually involves approximation. Some of the most common reasons for approximation are round-off error, the use of interpolation, the approximate values of elementary functions generated by computer sub-routines, the numerical approximation of definite integrals, and the numerical solution of both ordinary and partial differential equations. This section outlines some of the most frequently used methods of approximation, rang- ing from linear interpolation, spline function fitting, the economization of series and the Pad´e approximation of functions, to finite difference approximations for ordinary and partial deriva- tives. More detailed information about these methods, their use, and suitability in varying circumstances can be found in the numerical analysis references at the end of the book. 29.1.1 Linear Interpolation Linear interpolation is the simplest way to determine the value of a function f(x) at a point x = c in the interval x0 ≤ x ≤ x1 when it is known only at the data points x = x0 and x = x1 at the ends of the interval, where it has the respective values f(x0) and f(x1). This involves fitting a straight line L through the data points (x0, f(x0)) and (x1, f(x1)), an
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Math Central is maintained by the math and education departments at the Canadian University of Regina. Possibly the most valuable section of the site is the Resource Room, which has an impressive database that is... Production functions, such as the Cobb-Douglas and CES, underly cost schedules used to model firms in micro-economics. The models illustrate this link and show the relation between long and short run cost schedules.... A program that challenges participants to combine knowledge learned in school with an awareness of current events, a curiosity about the business and economic environment, and an interest in managing money. Participants... This unit discusses crystals, offering an opportunity to apply math to real-life objects: growing crystals, the history and theory of crystals, group theory, Miller indices, and mechanical drawing. It develops the... "In the Classroom" highlights how some schools and organizations use Mathematica extensively in their curricula. The section on "Collaborative Initiatives" illustrates how businesses have teamed up with Wolfram Research...
Course Details MATH 096 Intermediate Algebra and Geometry 5 hours lecture, 5 units Letter Grade or Pass/No Pass Option Description: Intermediate Algebra and Geometry is the second of a two-semester integrated sequence in algebra and geometry. This course covers systems of equations and inequalities; radical and quadratic equations; quadratic functions and their graphs; complex numbers; nonlinear inequalities; exponentials and logarithmic functions; conic sections; sequences and series; and solid geometry. The course will also include application problems involving the topics covered. This course is the prerequisite for numerous collegiate level/transfer level mathematics courses. Degree Link This course can help you earn the following degree(s) or certificate(s):
Mathematics at Merit Academy High School At Merit Academy, all students are required to take four years of Mathematics. Freshmen begin by taking Geometry, followed by Algebra II, then Trigonometry, then Pre-Calculus, and then Calculus. With the individualized support of our Math teachers, the students learn to apply mathematical concepts to real-world problems. A strong background in Mathematics (Calculus or beyond) is especially important for students entering the fields of Engineering, Medicine, and Computer Science. At Merit, all students take the Calculus 1 AB course, while other students may advance beyond that level to take Calculus 1 BC. Merit's Math program is designed to be taken concurrently with our Science program, so that students understand the close relationship between mathematical concepts and scientific phenomenon. Students simultaneously take the Kumon Math program, which is designed to give students the repetition needed to learn complex concepts. First, each student tests into the appropriate level. Students then work on a packet of math problems. Before moving on to the next level, the student must show proficiency. The Kumon program is a perfect complement to our Math course—Kumon builds a solid foundation, while the other stimulates and challenges the students to solve problems, analyze data, and apply new strategies and concepts.
Beginning Algebra - 8th edition Summary: This text teaches solid mathematical skills while supporting the student with careful pedagogy. Gustafson and Frisk effectively prepare students for their next mathematics course as it presents all the topics associated with a first course in algebra. Mathematical concepts are motivated by need and illustrated through discussion, examples, and real-world applications Heavy wear along the spine and cover, pages are clean and in tact
Learn to write programs to solve linear algebraic problems The Second Edition of this popular textbook provides a highly accessible introduction to the numerical solution of linear algebraic problems. Readers gain a solid theoretical foundation for all the methods discussed in the text and learn to write FORTRAN90 and MATLAB(r) programs to solve problems.... more... Collecting results scattered throughout the literature into one source, An Introduction to Quasigroups and Their Representations shows how representation theories for groups are capable of extending to general quasigroups and illustrates the added depth and richness that result from this extension. To fully understand representation theory, the first... more... This book gives a coherent and detailed description of analytical methods devised to study random matrices. These methods are critical to the understanding of various fields in in mathematics and mathematical physics, such as nuclear excitations, ultrasonic resonances of structural materials, chaotic systems, the zeros of the Riemann and other zeta... more... The research results published in this book range from pure mathematical theory (semigroup theory, discrete mathematics, etc.) to theoretical computer science, in particular formal languages and automata. The papers address issues in the algebraic and combinatorial theories of semigroups, words and languages, the structure theory of automata, the classification... more... This book is a carefully written exposition of Coxeter groups, an area of mathematics which appears in algebra, geometry, and combinatorics. In this book, the combinatorics of Coxeter groups has mainly to do with reduced expressions, partial order of group elements, enumeration, associated graphs and combinatorial cell complexes, and more. more... Series overview MathsWorks for Teachers has been developed to provide a coherent and contemporary framework for conceptualising and implementing aspects of middle and senior mathematics curricula. more... Boost Your grades with this illustrated quick-study guide. You will use it from college all the way to graduate school and beyond. FREE chapters on Linear equations, Determinant, and more in the trial version. Clear and concise explanations. Difficult concepts are explained in simple terms. Illustrated with graphs and diagrams. Table of Contents. I.... more... Drawn from the literature on the asymptotic behavior of random permanents and random matchings, this book presents a connection between the problem of an asymptotic behavior for a certain family of functionals on random matrices and the asymptotic results in the classical theory of the U-statistics. more...
Intermediate Algebra: Connecting Concepts through Applications 9780534496364 ISBN: 0534496369 Edition: 1 Pub Date: 2011 Publisher: Brooks Cole Summary: INTERMEDIATE ALGEBRA: CONNECTING CONCEPTS THROUGH APPLICATIONS shows students how to apply traditional mathematical skills in real-world contexts. The emphasis on skill building and applications engages students as they master concepts, problem solving, and communication skills. It modifies the rule of four, integrating algebraic techniques, graphing, the use of data in tables, and writing sentences to communicate so...lutions to application problems. The authors have developed several key ideas to make concepts real and vivid for students. First, the authors integrate applications, drawing on real-world data to show students why they need to know and how to apply math. The applications help students develop the skills needed to explain the meaning of answers in the context of the application. Second, they emphasize strong algebra skills. These skills support the applications and enhance student comprehension. Third, the authors use an eyeball best-fit approach to modeling. Doing models by hand helps students focus on the characteristics of each function type. Fourth, the text underscores the importance of graphs and graphing. Students learn graphing by hand, while the graphing calculator is used to display real-life data problems. In short, INTERMEDIATE ALGEBRA: CONNECTING CONCEPTS THROUGH APPLICATIONS takes an application-driven approach to algebra, using appropriate calculator technology as students master algebraic concepts and skills. Clark, Mark is the author of Intermediate Algebra: Connecting Concepts through Applications, published 2011 under ISBN 9780534496364 and 0534496369. Seven hundred fifteen Intermediate Algebra: Connecting Concepts through Applications textbooks are available for sale on ValoreBooks.com, two hundred twenty three used from the cheapest price of $28.36, or buy new starting at $170 INSTRUCTORS EDITION534496371
Calculus II For Dummies [NOOK Book]... More About This Book use them, approximate integration, and improper integrals. This hands-on guide also covers sequences and series, with introductions to multivariable calculus, differential equations, and numerical analysis. Best of all, it includes practical exercises designed to simplify and enhance understanding of this complex subject. Related Subjects Meet the Author Mark Zegarelli, a math tutor and writer with 25 years of professional experience, delights in making technical information crystal clear — and fun — for average readers. He is the author of Logic For Dummies and Basic Math & Pre-Algebraourtlandt Posted September 19, 2009 Calculus II for Dummies About 5/8ths of the book is a review of Calculus I for Dummies... however, they are different authors so I will not complain too much. Included are tear out cheat sheets on front and back.... however, I did notice one small error on page 1. It states that " d/dx arcsin(x) = -1/sqrt(1-x^2) " and that is wrooooooongggg!!!! hopefully alot of students will notice that and not memorize the wrong thing.... that little "-" sign can be the difference in an A and a B. kind of a minor but big screw up Wiley Publishing. The error was fixed everywhere else in the book but the problem is that the error IS on the tear out cheat sheet that is designed to be studied and memorized. anywho... not a bad book, definitely resourceful. 4 out of 5 stars! 3 out of 5 people found this review helpful. Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged. chibiseiya Posted April 10, 2010 I wish I had this book last quarter!! This book is a lifesaver! After taking calculus 2 and failing, I decided to look for books that dealt with integration (the concept of calculus 2). I was thrilled when I purchased this book. It is full of excellent explainations and examples that really helped me get a better understanding of calculus 2. I would recommend this book to all current and future calculus students! 2 out of 2 people found this review helpful. Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged.
The Matrix Algebra Tutor: Learning by Example DVD Series teaches students about matrices and explains why they're useful in mathematics. This episode teaches students about Gaussian Elimination and Gauss-Jordan Elimination. These methods are used to solve a system of equations using matrix math, and are two methods that are similar to simplying matrix equations. Grades 9-College. 59 minutes on DVD.
Intermediate Algebra - 8th edition ISBN13:978-0495108405 ISBN10: 0495108405 This edition has also been released as: ISBN13: 978-0495384977 ISBN10: 0495384976 Summary: Algebra is accessible and engaging with this popular text from Charles ''Pat'' McKeague! hel...show morep you to move through each new concept with ease. Real-world applications in every chapter of this user-friendly book highlight the relevance of what you are learning. And studying is easier than ever with the book's multimedia learning resources, including ThomsonNOW for INTERMEDIATE ALGEBRA, a personalized online learning companion10897.77 +$3.99 s/h VeryGood AlphaBookWorks Alpharetta, GA 0495108405, but text is in tact. Slight water damage, text is completely in tact.. This book needs to be rebound. ...show less $200.00 +$3.99 s/h New TextbookBarn Woodland Hills, CA 0495108405 .May contain school stamp or sticker for inventory purposes. $234.91
Hitchcock, TX CalculusThomas and received an A in the course. Linear Algebra is the study of matrices and their properties. The applications for linear algebra are far reaching whether you want to continue studying advanced algebra or computer science use Linux several hours a day. I currently have a Linux server as my personal home computer, currently Arch Linux x86_64 SMP preempt 3.11.4. I was first exposed to UNIX system administration in 1991 as the custodian of my workgroup's computing workstation, I have continued to maintain my skills.
Beginning Algebra - With CD - 4th edition Summary: For college-level courses in beginning or elementary algebra. Elayn Martin-Gay's success as a developmental math author and teacher starts with a strong focus on mastering the basics through well-written explanations, innovative pedagogy and a meaningful, integrated program of learning resources. The revisions provide new pedagogy and resources to build student confidence, help students develop basic skills and understand concepts, and provide the highest ...show morelevel of instructor and adjunct support. Martin-Gay's series is well known and widely praised for an unparalleled ability to: Relate to students through real-life applications that are interesting, relevant, and practical. Martin-Gay believes that every student can: Test better: The new Chapter Test Prep Video shows Martin-Gay working step-by-step video solutions to every problem in each Chapter Test to enhance mastery of key chapter content. Study better: New, integrated Study Skills Reminders reinforce the skills introduced in section 1.1, "Tips for Success in Mathematics" to promote an increased focus on the development of all-important study skills. Learn better: The enhanced exercise sets and new pedagogy, like the Concept Checks, mean that students have the tools they need to learn successfully. Martin-Gay believes that every student can succeed, and with each successive edition enhances her pedagogy and learning resources to provide evermore relevant and useful tools to help students and instructors achieve success. ...show less 0131444441 Please read before purchase: This book has extreme wear and is not pretty; still usable. No Disk(s) Included. Meets the acceptable condition guidelines. Has wear. Five star seller - Ships Q...show moreuickly - Buy with confidence! ...show less $116
Math You are here Information and resources about personal finance and money, algebra, geometry, elementary math, pre-calculus and calculus, and trigonometry. Includes videos to help explain math and Google search tips This year's Nobel prize Economics in economics was giving to three American economists. Together their work tells us more about how markets work what the Stock Markets will do in the future. Even if...
Moorestown Trigonometry Math. One of the advantages of learning higher level math (such as differential equations and real analysis) is that it gives you a much better perspective of the basic math and the underlying structure and foundation. I have worked with Microsoft Excel for over 10 yearsYou can either buy a computer that is made specifically to use Windows/Microsoft technology or Macintosh/Apple technology. Computers that use Windows technology are referred to as PCs (a term that I personally do not like because PC stands for Personal Computer) or Mac for computers that use the...
what does pre-algebra mean?? Pre-Algebra is basically preparing you for Algebra. Pre-Algebra teaches you Order of Operations, Properties of Numbers, Rational and Irrational Numbers, Exponents, PEMDAS, ect. Monday, August 25, 2008 at 9:11pm by Delilah what does pre-algebra mean?? pre algebra is like a bunch of math that comes before algebra in middle school. Monday, August 25, 2008 at 9:11pm by Grace what does pre-algebra mean?? i have to draw things that have to do with pre algebra but...i dont understand the meaning of it Monday, August 25, 2008 at 9:11pm by damainmind pre-algebra write an inequality for the sentence. The total t is greater than five. Well your answer is going to be T and the sign for greater than is simply >. So the answer would be T > 5 ok again my name is above. i was just wondering what pre-algebra was? i mean i take the ... Friday, March 9, 2007 at 8:50am by Joe pre-algebra kk my teach told me to describe and draw things that are about pre-algebra and she told me to write what pre-algebra is... so can anyone help me?? Monday, August 25, 2008 at 9:03pm by damainmind pre algebra With a mean of 72, and SD of 10 of High temprature: Temprature of 93 is how many SD above mean? Temprature of 54 is how many SD below mean? Show your calculations. Sunday, June 20, 2010 at 6:45pm by jenny pre- algebra what does grams mean?? Thursday, December 18, 2008 at 7:47pm by Taylor Pre Algebra Thank you so much. I have another question :) In my Pre Algebra book, I have a question like this: Draw a line from one vertex to a point on another side to create a triangle. Cut along the line. What do they exactly mean by 'cut along the line'? Thursday, March 20, 2008 at 6:36pm by Lily Pre- Algebra What does the prefix centi- mean??? Thursday, December 18, 2008 at 7:44pm by Taylor pre algebra I suspect you mean =0 (r-8)(r-3) = 0 that is true if r = 8 or if r = 3 Friday, April 23, 2010 at 3:14pm by Damon Pre-algebra Does the dot mean multiply? Wednesday, May 25, 2011 at 4:21pm by Ms. Sue pre algebra what you mean scroll down..??? tell me the answer!! Wednesday, June 16, 2010 at 9:01pm by lavena 7th grade Pre-Algebra That's ok; however, you still are not writing it clearly. e^-4f can mean (e^-4)*f or it can mean e^(-4f). If the former, it's the first answer I gave you. If the latter, it's the second answer. Also, E and e aren't the same. Tuesday, September 22, 2009 at 9:52pm by DrBob222 Pre-Algebra What does P-E-M-D-A-S mean? I know they are the orders of operation, but I forgot what each letter stand for??? HELP! Wednesday, November 28, 2007 at 10:38pm by Ellen what does pre-algebra mean?? so am i dumb for learning this in 8th grade?? well um basically its elementary math Monday, August 25, 2008 at 9:11pm by damainmind Pre-Algebra the world wobbl e contests? i dunoo pre-algebra sucks Tuesday, September 15, 2009 at 11:25pm by marina pre algebra i dont have a math text. i had to take it back to the teacher and plus she never thought me how to do this im in pre algebra not trigonometry Sunday, June 20, 2010 at 9:38pm by lavena pre-algebra so the 15 - 9=4 okay thanks ms. sue for helping me with my 7th grade pre algebra hw I will probably need more help I will jut post my other questions too. Sunday, September 16, 2012 at 4:23pm by emily pre algebra check I don't know what you mean by the "percent equation." However 1.2% of 130 = 1.56 Did you misplace the decimal point? Tuesday, May 29, 2012 at 4:35pm by Ms. Sue Pre-Algebra-math Sorry I am normally great at Pre-Algebra, but this one i cannot figure out. Sorry, Kim! Good Luck! Tuesday, September 25, 2012 at 3:51am by Delilah Pre Algebra What does the -y- mean? here: 9y cm or 5y cm This is for finding the area. Thursday, March 20, 2008 at 6:38pm by Lily Pre Algebra do you mean x2 + 2x + 4 ? it does not factor using rationals, so that is all you can do. Tuesday, June 10, 2008 at 6:48pm by Reiny Pre-Algebra ?? English?? Do you think either of these sentences is biased? What does biased mean? Monday, February 2, 2009 at 7:31pm by Ms. Sue Pre-Algebra Can someone show me how to turn this into an equation? I mean like 'y=x-z' -2x+ y > or equal to 2 Did I say what I needed to get right? -MC Thursday, May 14, 2009 at 7:10pm by mysterychicken Pre-Algebra This is about Reasoning Strategy, I really need help on this, I have a test tomorrow on it! Please help... Ok, here's what it says: 3a. What is the y-intercept of the trend line? (What do they mean???) 3b. Locate one other point on the trend line, then find the slope of the ... Wednesday, February 20, 2008 at 3:13pm by Mathilde Pre-Algebra Did you add them up and divide by 8? Wednesday, March 20, 2013 at 1:53pm by Writeacher pre algebra IM guessing by Plots the mean the rectangle garden, so find the peremiter, which will give you the entire lenght of the rectangle the multyply that by 5 Wednesday, October 9, 2013 at 9:50pm by Flare pre-calc I really do not understand what you mean by picking a series of x values from -10 to 10? Do you mean the window? And how do you calculate the corresponding values of a^2-x^2? And do you mean that I'm supposed to just type in a^2-x^2 into my graphing calculator because when I ... Sunday, September 14, 2008 at 11:30am by Freddie by Amber shall PreAlgebra The only t table I know about is the statistical t and that doesn't fit a pre-algebra problem. So I assume you mean to substitute -3 for x and you want to know y. Then substitute 4 for y and you want to know x. If so, y = 3x+7 y = 3(-3)+7 y = -9+7 y = -2 etc. Check my work. Thursday, May 21, 2009 at 4:35pm by DrBob222 pre-algebra um as in for what is pre-algebra?? Monday, August 25, 2008 at 9:03pm by damainmind Pre-Algebra its my pre-algebra homework Tuesday, January 18, 2011 at 9:23pm by TiffanyJ Pre Algebra random question: i have a pre-algebra unscramble and i need help to unscramble the word the word they gave me was: Seproropit can any of u help me? Thursday, June 5, 2008 at 5:21pm by coco pre algebra It depends on what you mean by "solve". Do you mean solve for y? The first is not an equation, there is no equal sign. The second add 5y to each side -25=-x+5y add x to each side 5y=x-25 divide each side by 5 y= x/5 - 5 Thursday, January 20, 2011 at 9:15pm by bobpursleyPre-Algebra Evidently you haven't mastered pre-Algebra since your answer is WRONG. It takes him 4 hours to catch her. While she does ride for 5 hours, it only takes him 4 to catch her. Wednesday, August 18, 2010 at 6:28pm by Joe Mama Pre-Algebra: Help I have to do a webquest for pre-algebra and we have to make up a fundraiser thing, I need a list of things that one needs for a charity fundraiser, such as chairs, buffet tables, tables, tents, tablecloths. HELPP Thursday, March 3, 2011 at 12:20pm by Cheryl oh no Vitaliy, this person is in 7th grade pre-agebra or 8th grade pre-algebra. Thursday, February 5, 2009 at 6:58pm by haha PRE-CALCULUS THIS IS FOR ALGEBRA/PRE-CALCULUS MATH. I POSTED IT JUST AS THE TEACHER WROTE IT. Tuesday, September 13, 2011 at 10:41pm by aLVIN pre-algebra i don't get them Thursday, November 14, 2013 at 11:07am by pre-algbra pre calculus You need to write out the question. I have no idea what "pre calculus" this could be, maybe zeroes, however, that is standard fare for an algebra course. Tuesday, December 10, 2013 at 6:26pm by bobpursley pre calculus You need to write out the question. I have no idea what "pre calculus" this could be, maybe zeroes, however, that is standard fare for an algebra course. Tuesday, December 10, 2013 at 6:26pm by bobpursley SHAY THE MATH QUESTION IS IT ALGEBRA , PRE ALGEBRA' OR GEOMETRY? Quadriatic functions. so algebra i think. Wednesday, January 17, 2007 at 7:09pm by ROSA Pre-Algebra Sorry, you are PRE-algebra, so you may not know how to solve for x. 8x = 10x - 10 Subtract 10x from both sides. -2x = -10 Divide both sides by -2. x = 5 Wednesday, August 18, 2010 at 6:28pm by PsyDAG pre-calculus Do you mean "Can y be expressed as a function of x?". If so then the answer is yes it can, if by "x^2y" you mean "x²y" and not "x raised to the power of 2y". The equation x²y - x² + 4y = 0 can be written as (x²+4)y = x², from which you can easily express either x in terms of y... Saturday, September 13, 2008 at 6:13pm by David Q Pre Algebra This is about Reasoning Strategy... It says this: 3a. What is the y-intercept of the trend line? 3c. Write an equation for the trend line in slope-intercept form. I don't really know what they mean? Please help me. Thursday, February 21, 2008 at 12:42am by Mathilde algebra 2 thats part of algebra 2 thats easy were doing that now in pre algebra Wednesday, April 29, 2009 at 1:14pm by Tanisha Statistics In Professor White's statistics course the correlation between the students' total scores before the final examination and their final examination scores is r = 0.9. The pre-exam totals for all students in the course have mean 275 and standard deviation 50. The final exam ... Sunday, October 21, 2012 at 11:08pm by Kevin math (pre clac) What do you mean? Friday, April 25, 2008 at 4:55pm by Jordan pre calc asap what do you mean? Thursday, May 7, 2009 at 10:57pm by peter Pre-calculus what does sqrt mean? Tuesday, November 27, 2012 at 6:01pm by John Pre-Algebra since almost all the values are greater than 15, it is unlikely that the mean is 15. Add up all the values and divide by 8, since there are 8 numbers. Now what do you get? Wednesday, March 20, 2013 at 3:16pm by Steve pre calculus Pre-Calculus? I wonder what the question is. In a basic algebra course, you would be looking for zeroes, and probably get those by factoring. 2x^2+11x-21=0 (2x-3)(x+7)=0 x= 1.5, or x=-7 Tuesday, December 10, 2013 at 12:39pm by bobpursley Pre-calc Pre calculus? This is stock Algebra II. put the equations in the form of ax^2+bx+c=0 then factor. for instance, c is already in that form. 2x^2+x-6=0 (2x-3)(x+2)=0 x= 3/2 x=-2 do the others the same method. Monday, January 25, 2010 at 5:18pm by bobpursley PRE-CALCULUS You need a set of data to have an arithmetic mean. Tuesday, September 13, 2011 at 10:41pm by Ms. Sue Pre-Cal Do you mean f(g)? f(g(x))=x^3/(x^3+1) which has a domain of x for all x, except x=-1 Tuesday, February 10, 2009 at 1:58am by bobpursley Pre Cal - incomplete and? while you're at it, what does "93 x" mean? Tuesday, October 2, 2012 at 11:29pm by Steve pre- calculus Do you mean 3x + (2y/3y), or (3x + 2y)/3y ? The first is 3x + (2/3) The second is (x/y) + 2/3 Until you learn to master the importance and use of parentheses in algebra, you will not master the subject. Thursday, August 12, 2010 at 5:46pm by drwls Math - Pre-Algebra 44) sampled... Monday, August 30, 2010 at 9:29pm by BC Math (Algebra/Pre-Algebra) That is one of the choices so my guess is that that is correct. Thank you! =] Thursday, June 4, 2009 at 8:48pm by Samantha Math (Algebra/Pre-Algebra) in finding the x-intercept let y = 0 in the equation and solve for x so what do you get for x? Thursday, June 4, 2009 at 9:13pm by Reiny Pre-Algebra it is about math in algebra i dont know why but it is my teacher is wierd Tuesday, September 15, 2009 at 11:25pm by Mikayla pre algebra You have to decide you're going to learn how to do algebra. Wednesday, June 16, 2010 at 10:05pm by Ms. Sue
books.google.com - In the international research community, the teaching and learning of algebra have received a great deal of interest. The difficulties encountered by students in school algebra show the misunderstandings that arise in learning at different school levels and raise important questions concerning the functioning... to Algebra Approaches to Algebra: Perspectives for Research and Teaching In the international research community, the teaching and learning of algebra have received a great deal of interest. The difficulties encountered by students in school algebra show the misunderstandings that arise in learning at different school levels and raise important questions concerning the functioning of algebraic reasoning, its characteristics, and the situations conducive to its favorable development. This book looks more closely at some options that aim at giving meaning to algebra, and which are considered in contemporary research: generalization, problem solving, modeling, and functions. Salient research on these four perspectives addressed the question of the emergence and development of algebraic thinking by a dual focus on epistemological (via the history of the development of algebra) and didactic concerns. References from web pages Building up the notion of Dependence Relationship between ... Approaches to Algebra. Perspectives for research and teaching. Kluwer Academic Publishers, Netherlands. Janvier, C.: 1996. Modeling and the Initiation into ... meta/ p117670_index.html?PHPSESSID=62a2d4248508793753383a59b0b314b0
Almost any workflow involves computing results, and that's what Mathematica does—from building a hedge fund trading website or publishing interactive engineering textbooks to developing embedded image recognition algorithms or teaching calculus. MatDescription: A fairly detailed training video course on how to shoot a video camera (for beginners). Creators - Blue Crane Digital - the ones who create movies using Nikon cameras and Canon (corresponding torrent is on the tracker). The movie is in English, but those who are not "owned" will still be clear. The same video - everything is visible. The disc is filled completely, so that the menu and all additional. materials remained in the original sound. The menu has a choice of scenes, if that remains unclear - you can go back and look over the menu.
Introductory physics courses are mathematically demanding, even those for non-physics science majors. Students must become adept at solving a wide variety of quantitative problems. However, even students with calculus experience often lack facility with basic pre-calculus skills. A large contributing factor to the problem is the students' generally poor retention of working math skills, but they may also be struggling to transfer their math knowledge to unfamiliar problem domains. In either case, these students should benefit from early intervention that continues to scaffold throughout the term. We report on our efforts to create math-related, online formative assessment modules for first-semester introductory physics. These online tutorials target specific mathematical skills that are essential to success in physics, and are designed to progress from a purely math-centered review of each basic skill, to problems of increasing generality and complexity, and ultimately toward a transfer of these skills to physics problem domains.
This is a free textbook by Boundless that is offered by Amazon for reading on a Kindle. Anybody can read Kindle books—even... see more This is a free textbook by designedVisually searchable database of algebra 1 videos. Click on a problem to see the solution worked out on YouTube. The... see more Visually searchable database of algebra 1 videos. Click on a problem to see the solution worked out on YouTube. The solutions are meant to accompany the free and open textbook Elementary Algebra that can be found on the flat world knowledge website. Visually searchable collection of algebra 2 videos. Click on a problem to see the solution worked out on YouTube. These... see more Visually searchable collection of algebra 2 videos. Click on a problem to see the solution worked out on YouTube. These videos are meant to accompany the free and open textbook Intermediate Algebra that can be found on the flat world knowledge website. " Algebra for College Students is designed to be used as an intermediate level text for students who have had some prior... see more " Algebra for College Students is designed to be used as an intermediate level text for students who have had some prior exposure to beginning algebra in either high school or college. This text explains the why's of algebra, rather than simply expecting students to imitate examples is a free, online textbook offered in conjunction with MIT's OpenCourseWare. "Over the last 100 years, the mathematical... see more This is a free, online textbook offered in conjunction with MIT's OpenCourseWare. "Over״ This is a free, online textbook that is a wikibook. "This book will help you learn how to do mathematics using Algebra. It... see more This is a free, online textbook that is a wikibook. "This book will help you learn how to do mathematics using Algebra. It has chapters (parts of the book) with lessons (parts of the chapter about one idea). A lesson has five parts: 1.Vocabulary - gives special words you need for the lesson. 2.Lesson - gives a new idea and how to use this idea. 3.Example Problems - gives the steps to do problems using the new idea. 4.Practice Games - gives places for amusement where you do problems. 5.Practice Problems - You do problems.״ This is a free, onlne textbook. According to the authors, "We are two college mathematics professors who grew weary of... see more This is a free, onlne textbook. According to the authors, "We are two college mathematics professors who grew weary of forcing our students to purchase expensive college algebra textbooks whose mathematical content has slowly degraded over the years. Our solution? Write our own. The twist? We made our college algebra book free and we distribute it as a .pdf file under the Creative Commons License. What's more, the LaTeX source code is also available under the same license.״
Phase II is a 42,000 square foot addition that will provide a new home for Physics, Math and Computer Sciences as well as the new Science Center for Integrated Learning. Ground breaking for Phase 2 is scheduled for March 2014, with a completion date of May 2015. This is the second of a three-phase science building initiative repesenting an investment of more than $30 million. Combining logic and precision with intuition and imagination At Earlham, we strive to teach our students mathematical fundamentals and problem-solving skills that they can apply in a variety of disciplines or in further study of mathematics. Mathematics students may participate in weekly "mathophiles" seminars and informal lunches, attend regional meetings of professional mathematicians, and participate in mathematically related off-campus programs during the academic year or the summer. The Association for Women in Mathematics (Careers That Count) describes mathematics as "… a powerful tool for solving practical problems and a highly creative field of study, combining logic and precision with intuition and imagination. The basic goal of mathematics is to reveal and explain patterns — whether the pattern appears as electrical impulses in an animal's nervous system, as fluctuations in stock market prices, or as fine detail of an abstract geometric figure." Jacob LaChance says math is simple, and the key is not to think about math but rather to think with math. As a sophomore, he was part of the Earlham duo that placed first in the Michigan Autumn Take Home Challenge. It is easy for Earlham students to design and participate in projects that explore connections between math and other interests as a class project, independent study or as a double major. Earlham students have participated in the Budapest Semesters in Mathematics, studying math in one of the world's great centers of mathematical research and discovering Hungarian culture
This chapter discusses the design, development, and evaluation of a set of domain-independent instructional strategies for teaching problem-solving outcomes in elementary algebra through Intelligent Computer-Aided Instruction (ICAI). The elementary algebra course is one of the most important for all entry-level freshman students of any major at most universities and 2-year colleges in the United States. Many students have a difficult time understanding the facts, concepts, and principles of algebra. As a result, the procedural portion in subsequent problem solving becomes a difficult task. Many students eventually fail the course. Our long-term goal was to develop a computer-based tutor that will be as effective in algebra instruction as a human tutor. The purpose of this chapter is an attempt to combine artificial intelligence technology with instructional theories of problem solving to achieve an effective teaching environment in algebra. The design implemented various instructional strategies from various design theories, and developed a set of transformations in the context of linear algebra to improve an existing ICAI system. INTELLIGENT COMPUTER-ASSISTED INSTRUCTION In recent years there has been increased emphasis on individualized instruction and computer technology specially applying intelligent learning systems that facilitate learning at all levels of education and training ( O'Neil, 1981). The intelligent learning systems are known as intelligent tutoring systems
34,"ASIN":"0816051240","isPreorder":0},{"priceBreaksMAP":null,"buyingPrice":85.53,"ASIN":"0691118809","isPreorder":0}],"shippingId":"0816051240::G4D%2F7sWDtpeFxLZYcVelLUlsJ578eXAO7X3o3gVoNyqoZAfXC7A7K5j7OxOf0hIffb21y10dedz6G2h0Vg6HzenXpZLcRPmxwyLoh79b7DI%3D,0691118809::HX914JRgBY25z6jlFIdTXiBIvoyZnh5wlaLh3cq12TxQu3RBORrJMmeHtn7bKq7xRdR%2BWUzPi0riKWO2ZecBHvegEt5kViZNSyL0zKeEBearcher, author, and educator Tanton has compiled this encyclopedia to share his enthusiasm for thinking about and doing mathematics. More than 800 alphabetically arranged entries present a wide variety of mathematical definitions, theorems, historical figures, formulas, examples, charts, and pictures. Many cross-references serve to connect concepts or extend a concept further. A mathematical time line listing major accomplishments is available following the entries, along with a list of current mathematics organizations. The bibliography contains print and Web resources, and the index is helpful in locating terms and concepts. Each entry varies in length depending on the term, concept, or person being described. Six longer essays describe the history of the branches of mathematics. The writing style is straightforward and readable and sometimes contains parenthetical notes that add background or context. If an entry contains a word or words in capital letters, that term or person is also an entry in the encyclopedia. This remarkable book is not just a collection of facts about mathematics, but is a fairly detailed treatment (within the limits imposed by space considerations) of various mathematical terms and topics. It does not restrict itself to simple mathematics and devotes full attention to several advanced concepts, but is always clearly written. I really commend the author for including proofs for some of the more important theorems and results (e.g. proof of the fundamemtal theorem of algebra, derivation of the least-squares method, and many more). And yes, you will learn tons and tons of things from this excellent book. It is a must read for anybody interested in mathematics! Given these four, there is hardly a topic from among the current 495 math fields of study that isn't at least explained in enough detail to save LOTS of time on link expeditions. At minimum, these give head starts on alphabetized keywords that will quickly fill holes in any research project, class, or syllabus. Looking for divisibility rules for numbers that you didn't think had divisibility rules? Looking for names of symbols you didn't think had names? Dr. Tanton provides the facts and the explanations along with the stories behind the topics.
Introduction Introduction Basic Math and Pre-Algebra is the first book in the Master Math series. The series also includes Algebra, Pre-Calculus, Geometry, Trigonometry, and Calculus. The Master Math series presents the general principles of mathematics from grade school through college including arithmetic, algebra, geometry, trigonometry, pre-calculus, and introductory calculus. Basic Math and Pre-Algebra is a comprehensive arithmetic book that explains the subject matter in a way that makes sense to the reader. It begins with the most basic fundamental principles and progresses through more advanced topics to prepare a student for algebra. Basic Math and Pre-Algebra explains the principles and operations of arithmetic, provides step-by-step procedures and solutions and presents examples and applications.
Teaching just became a lot easier thanks to Wolfram Research Institute and the resources they have put online. Called the 'Wolfram Education Portal', it combines the power of Wolfram research's best computation engines with other teaching aids like lesson plans to make learning as pleasurable for both teachers and students. Hold on tight as we introduce to you the different features of this wonderful portal. Wolfram had already demonstrated the power of interactive computational techniques by developing the Computable Document Format or CDF, where it is possible to spice up documents with interactive graphs and figures. The present development seems to be an even bigger jump. Introducing the Wolfram Education Portal The Education Portal has a lot of material in the algebra and calculus section, but it will soon expand into other sections as well. Instructors will benefit greatly by being able to easily present the methods of calculation, like finding the slope of a curve, the meaning of discontinuity and numerical integration. It also aims to stress the inculcation of Wolfram's wonderful web-based mathematical software-cum-database Wolfram\|Alpha. There are also a number of introductions to different Wolfram products like Mathematica and CDF. Exploring it myself I decided to explore what the big fuss was and was quite impressed. You'll have to log in with a certain Wolfram ID. If you don't have one, creating one is extremely easy and it's free. Once that is done, you can access everything that has been put out there. Algebra Let's first start off with the Algebra section. In the so-called Library view, you can see that there are currently 90 textbook sections, 68 lesson plans, 15 demonstrations and 10 widgets. I especially liked the widgets; they do simple things quickly and without fuss. I checked out several sections, 'Multi-Step Equations', 'Graphs of Quadratic Functions' and 'The Pythagorean Theorem and its Converse'. They contain textbook material, which provides direct, easy-to-understand-and-present material, deliciously sprinkled with a healthy dose of problems. Instructors might be more interested in the Lesson Plans. It draws up a list of things that the instructor is supposed to teach and the students are supposed to work out. Examples are nicely provided and stress has been laid to the use of Wolfram\|Alpha in classrooms. There are also widgets provided in between the examples. Calculus Next comes the calculus section. I loved this section more for the simple reason that it is richer in content. This section has demonstrations and widgets. The demonstrations are brilliant and spent quite some time fiddling around with them, even though I knew every technique being shown here. It's great fun, and it makes you love the things you already know. It will definitely be a great help for students, more as a visual aid than as a computational technique. I loved the demonstration of numerical integration, using the three different techniques – rectangular (not so accurate), trapezoidal rule (more accurate) and Simpson's rule (quite accurate). You can easily see the comparison and judge which method works best for different functions. What method are you supposed to use for functions which are discontinuous at certain points? Use the different functions and different methods interactively to find out! It's a fun way to learn. Image 2: The derivative of a polynomial. You can change the polynomial (blue) by changing the position of the black knobs. The purple bar, showing the derivative, changes automatically. I had to mention the demonstration on the squeeze theorem and taking derivatives of polynomials. Can you draw the derivative of any given polynomial by simply looking at it? No? Then give this a try, fiddle around with it and you'll know how you do that! Wolfram\|Alpha and the classroom Lastly, I cannot but mention the Wolfram\|Alpha and how Wolfram wants teachers to use it for instruction. Wolfram has this comprehensive step-by-step-math guide for Wolfram\|Alpha. Give Wolfram\|Alpha something to solve and then ask it to show the steps as well. If it can, it will. Image 3: Wolfram\|Alpha showing the steps of the calculation. The 'Show Steps' button is placed right where the 'Hide Steps' button (arrow) is now placed. The equation is quadratic. Image 4: No steps shown for the cubic equation. But do notice the graphical solution shown. I found out that it can easily show steps for quadratic equations (image 3), but not so for cubic equations (image 4). I think the method of intersection of curves to solve equations is something that is not given its due importance in classrooms, so it was quite refreshing to see Wolfram\|Alpha displaying that as a primary technique. There you have it! Oh, you can also give suggestions, share material and inform Wolfram about any novel teaching methods that you might have thought of by clicking the give feedback link at the top of the page
You are here Math & Stats Help The Math/Stats Help service of the University Learning Centre is located in the Murray (Main) Library, Room 144 (map). The service is free of charge to all University of Saskatchewan students for help with University of Saskatchewan courses. Services Math/Stats help staff help students with a variety of mathematical or statistical topics, with a focus on first-year or introductory courses. Our service is primarily a drop-in service: students are welcome to drop in and work on homework, and ask questions when needed. Our service is designed for current University of Saskatchewan students looking for help with U of S courses. We also offer workshops or review sessions on certain topics in mathematics. These will be announced on this page and on PAWS. Unfortunately, we do not have the resources to provide help with research-level statistics at this time. Please contact the Department of Mathematics and Statistics for advice on research-level statistics. You may also find the resources at the Murray Library page on Numeric Data helpful. Announcements We will be open for Term 2 of Regular Session 2013-2014 on Monday, January 6, 2014. See below ("Hours of Operation") for schedule. Math 104 session: see below. Math Readiness course in term 2: If you are interested in taking Math Readiness in term 2, please email Holly Fraser at holly.fraser at usask.ca . (Replace ' at ' with the @ symbol.) Update (Jan. 30) * Organizational meeting and introductory session * If you are interested in the Math Readiness course, please attend this organizational session. We will discuss the course, and go over some basic math. Date/time: Thursday, Feb. 6, 7pm Location: Room 102 Murray Library (one floor up from the ground floor, across from the elevators) - Math Placement Test (for students registered in certain 100-level mathematics courses): For more information about the placement test, go to . If you still have questions, please contact Amos Lee in the Department of Mathematics and Statistics (lee at math.usask.ca - replace the 'at' with @). To try a sample version of the placement test that you will write in September, please go to: - Math Readiness evening course - See above under "Announcements." - Math help for physics - If you are looking for help with the mathematics used in physics, you may want to consider registering in the "Math for Physics" course being offered by the Department of Physics and Engineering Physics through CCDE (the Centre for Continuing and Distance Education). For more information, please visit the "Math for Physics" website here: . Alternatively, you may want to seek help at your physics tutorial, at the Structured Study Sessions for physics, or at Math/Stats Help. Please note that Math/Stats Help tutors are not necessarily prepared to help with physics concepts, but can help with math calculations involved. Hours of Operation Hours during Term 2 of Winter Session, 2013-2014 (January-April) (see below for schedule during Reading Week) Regular Schedule Monday 10am-6pm & 7-9pm Tuesday 10am-6pm & 7-9pm Wednesday 10am-8pm Thursday 10:30am-6pm Friday 10:30am-4pm Saturday closed Sunday 1:30-4:30pm Schedule during Reading Week Saturday Feb. 15: closed Sunday Feb. 16: 1:30-4:30pm Monday Feb. 17: closed (holiday) Tuesday Feb. 18: 11am-4pm Wednesday Feb. 19: 11am-4pm Thursday Feb. 20: 11am-4pm Friday Feb. 21: 12noon-3pm Saturday Feb. 22: closed Sunday Feb. 23: closed Regular hours will resume on Monday, Feb. 24. This schedule is subject to change. If these hours are not compatible with your schedule, please email Holly at holly.fraser at usask.ca (replace the 'at' with the appropriate symbol). Schedule: Math/Stats Help is generally open whenever the U of S is in session; that is, during Regular Session (September-April) and Spring and Summer Session (mid-May until mid-August). We are not able to open on statutory holidays. We often hold exra hours during the final exam periods. U of S students seeking help with non-U of S courses should be aware that students needing help with U of S courses have priority. Unfortunately, we usually do not have the capacity at this time to answer questions from people who are not University of Saskatchewan students.
History of Mathematics An Introduction 9780073051895 ISBN: 0073051896 Edition: 6 Pub Date: 2005 Publisher: McGraw-Hill College Summary: David Burton covers the history behind the topics typically covered in an undergraduate maths curriculum or in elementary or high schools. He illuminates the people, stories, and social context behind mathematics' greatest historical advances, while maintaining appropriate focus on the mathematical concepts themselves. Burton, David M. is the author of History of Mathematics An Introduction, published 2005 u...nder ISBN 9780073051895 and 0073051896. Twenty seven History of Mathematics An Introduction textbooks are available for sale on ValoreBooks.com, twenty two used from the cheapest price of $15.06, or buy new starting at $183.05.[read more
and Intermediate Algebra Developmental mathematics is the gateway to success in academics and in life. George Woodbury strives to provide his students with a complete ...Show synopsisDevelopmental mathematics is the gateway to success in academics and in life. George Woodbury strives to provide his students with a complete learning package that empowers them for success in developmental mathematics and beyond. The Woodbury suite consists of a combined text written from the ground up to minimize overlap between elementary and intermediate algebra, a new workbook that helps students make connections between skills and concepts, and a robust set of MyMathLab resources. Note: this item is for the textbook only; supplements are available separately.Hide synopsis Description:Good. Hardcover. May include moderately worn cover, writing,...Good. Hardcover. May include moderately worn cover, writing, markings or slight discoloration. SKU: 97803216654851665485. Description:New in new dust jacket. Brand New as listed. ISBN 9780321771940....New in new dust jacket. Brand New as listed. ISBN 97803217719401760203....New in new dust jacket. Brand New as listed. ISBN 9780321760203. Clean! Out of sight Shipping & Customer Service! We process all orders same day
Precise Calculator has arbitrary precision and can calculate with complex numbers, fractions, vectors and matrices. Has more than 150 mathematical functions and statistical functions and is programmable (if, goto, print, return, for).
0387950605 9780387950600 Understanding Analysis (Undergraduate Texts in Mathematics):This book outlines an elementary, one-semester course which exposes students to both the process of rigor, and the rewards inherent in taking an axiomatic approach to the study of functions of a real variable. The aim of a course in real analysis should be to challenge and improve mathematical intuition rather than to verify it. The philosophy of this book is to focus attention on questions which give analysis its inherent fascination. Back to top Rent Understanding Analysis (Undergraduate Texts in Mathematics) 1st edition today, or search our site for Stephen textbooks. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Pearson.
Expands upon the topics of Algebra I including rational expressions, radicals and exponents, quadratic equations, systems of equations, and applications. Develops the mathematical proficiency necessary for selected curriculum entrance. Credits not applicable toward graduation. Prerequisites: a placement recommendation for MTH 04 and Algebra I or equivalent. (5 cr.) Rational exponents, evaluating and estimating radicals, simplifying radical expressions, addition, subtraction, multiplication and division, rationalizing the denominator with one or two terms, radical equations, definition of complex numbers, simplifying complex numbers in the form of solutions of quadratic equations. Applications and problem solving including right triangles, Pythagorean theorem and the distance formula. (1 cr) Solving quadratic equations by the square root method and the quadratic formula. Graphing parabolas, finding the vertex of a parabola or center of a circle by completing the square, naming the intercepts. Applications and problem solving including mathematical modeling with quadratic functions. Solving systems of linear equations by graphing, elimination and substitution, applications including geometry problems. Graphing systems of inequalities in two variables. (1 cr)
Theory and Problems of Linear Algebra 9780071362009 ISBN: 0071362002 Edition: 3 Pub Date: 2000 Publisher: McGraw-Hill Companies, The Summary: Master linear algebra with Schaum'¬"s'¬"the high-performance study guide. It will help you cut study time, hone problem-solving skills, and achieve your personal best on exams and projects! Students love Schaum'¬"s Outlines because they produce results. Each year, hundreds of thousands of students improve their test scores and final grades with these indispensable study guides. Get the edge on your classmates. Use Sc...haum'¬"s! If you don't have a lot of time but want to excel in class, this book helps you: * Use detailed examples to solve problems * Brush up before tests * Find answers fast * Study quickly and more effectively * Get the big picture without poring over lengthy textbooks Schaum'¬"s Outlines give you the information your teachers expect you to know in a handy and succinct format'¬"without overwhelming you with unnecessary jargon. You get a complete overview of the subject. Plus, you get plenty of practice exercises to test your skill. Compatible with any classroom text, Schaum'¬"s let you study at your own pace and remind you of all the important facts you need to remember'¬"fast! And Schaum'¬"s are so complete, they'¬"re perfect for preparing for graduate or professional exams. Inside, you will find: * A bridge between computational calculus and formal mathematics * Clear explanations of eigenvalues, eigenvectors, linear transformations, linear equations, vectors, and matrices * Solved problems that relate to the field you are studying * Easy-to-understand information, perfect for pre-test review If you want top grades and a thorough understanding of linear algebra, this powerful study tool is the best tutor you can have! Chapters include: Vectors in Rn and Cn * Matrix Algebra * Linear Equations * Vector Spaces * Linear Mappings * Linear Mapings and Matrices * Inner Product Spaces, Orthogonality * Determinants * Diagonalization: Eigenvalues and Eigenvectors * Canonical Forms * Linear Functionals and the Dual Space * Bilinear, Quadratic, and Hermitian Forms * Linear Operators on Inner Product Spaces * Polynomials
This is a page of links to several different learning/teaching materials, including: an extensive laboratory manual of... see more This is a page of links to several different learning/teaching materials, including: an extensive laboratory manual of computer activities for calculus in pdf format, a page of animations in various calculus subjects produced by Mathcad 6.0 but viewable in most browsers, an interactive tutorial in infinite series and an interactive test in integration techniques. While this site is described here as primarily "Lecture/Presentation," unquestionably parts of it would serve well for "tutorial" purposes and even "drill and practice.״ This is an online supplement designed for a calculus course taught at the University of British Columbia. The site contains... see more This is an online supplement designed for a calculus course taught at the University of British Columbia. The site contains course notes, interactive labs, and in-class demonstrations. The topics covered include differential and integral calculus, differential equations, and series. A high-content, award-winning puzzle site that is divided into six main categories: puzzles & tests, optical illusions,... see more A high-content, award-winning puzzle site that is divided into six main categories: puzzles & tests, optical illusions, custom puzzles, teachers' resource, curiosities, art & language. Among the myriad of fun things, visitors will find number games, riddles, puzzles with downloadable pieces, oxymorons, and even a gift shop. A principal objective throughout the site is the enhancement of critical thinking skills. Here's a quote from Scientific American (May 27, 2003): "This virtual lab borrows the empirical spirit and creative curiosity that Archimedes brought to his work and invites visitors to explore with the same expectations for mind-blowing discovery." This site contains the learning materials for eleven online courses that are currently taught at Carnegie Mellon... see more This site contains the learning materials for eleven online courses that are currently taught at Carnegie Mellon University. Students anywhere may use these materials as learning resouces. Instructors at other learning institutions may use and even base their own courses on these materials free of charge. An instructor can create an account on OLI, select and sequence course materials, and take advantage of OLI's tracking of student progress. While all courses come in the "open and free" version, at least one course (Logic and Proofs) also comes in a version that requires payment of a fee to OLI. The available couses at this time are: Biology, Calculus, Causal Reasoning, Chemistry, Economics, Emperical Research Methods, French, Logic and Proofs, Physics, Statistics, and Statics. This is a subsite associated with the parent site called IDEA (Internet Differential Equation Activities). The... see more This is a subsite associated with the parent site called IDEA (Internet Differential Equation Activities). The activity on this page explores the diffusion of pollutants in landfills using the steady state diffusion model. This extensive site is primarily intended as an online supplement to the text Finite Mathematics by Stefan Waner and Steven... see more This extensive site is primarily intended as an online supplement to the text Finite Mathematics by Stefan Waner and Steven R. Costenoble. However, as the authors explain on their home page, this material is pertinent and freely available to anyone studying in this subject area. Of particular interest will be the online tutorials, interactive quizzes and exercises, and various Java, Javascript, and Excel spreadsheet utilities. The following chapter headings indicate the topics covered: Linear Functions and Models, Systems of Linear Equations and Matrices, Matrix Algebra, Linear Programming, Mathematics of Finance, Sets and Counting, Probability, Statistics, Markov Systems, Game Theory. There is also a review of Algebra. The author offers reflections on specific questions mathematicians and philosophers have asked about the infinite over the... see more The author offers reflections on specific questions mathematicians and philosophers have asked about the infinite over the centuries. He examines why explorers of the infinite, even in its strictly mathematical forms, often find it to be sublime. The purpose of the examples on this Web site is to illustrate the use of LiveMath software in the solution of problems that... see more A collection of LiveMath notebooks in various topics including elementary algebra, linear algebra, and calculus. This... see more A collection of LiveMath notebooks in various topics including elementary algebra, linear algebra, and calculus. This collection is intended for both classroom demonstrations and individual student use. Fractals, chaos theory, and juggling examples are also explored on this site.
About the Actor Smith Show Media Group Inc. Product Description Join Mrs. Franklin as she tackles math problems by using systems of linear equations! See how these systems can have zero, one or many common solutions. In the process of finding these solutions -- by graphing, substitution or elimination -- students will discover the three possible situations associated with graphs of these systems, be warned of some common pitfalls and determine when to choose substitution over elimination. With the support of graphs and tables, see how algebra comes to life in this video program. Practice test examples, a section on concentrating during tests and a teacher's guide is included on disc two.
absolutely phenomenal respect... respectively covered fractions, decimals, and negative numbers. The preceding topics were thoroughly familiar to my son thus unnecessary and should be to anyone who is looking to begin algebra. We began our endeavor 1/05 and it is now 6/05 and we have completed chapters 2 and 5-7 and my son has a complete understanding of all the covered topics. The topics we covered thus far were variables, exponents and roots, factoring (quadratics and many other aspects of factoring), and linear equations. Each topic is put in such a step by step fashion that it could almost be miscontrued as repetitive. It is so thought out that anyone who knows all their basic math (+,-,*,/,percents,decimals,fractions)can ease through this book. At first when I looked ahead to Exponents and roots, factoring and quadratics I thought it might be a challenge that was above us, I took one step at a time and my boy eased through it. Now I find myself looking ahead to chapter 8 (word problems), at this time it looks like a mountain to climb but I am comforted with the fact that when I looked ahead to roots I had thoughts that I might have to skip that section but on the contrary he took to it like a fish to water. I'd like to thank the author for this wonderfully thought out book and would certainly recommend this book for beginners and for review. I will write my final assessment of this book upon completion of chapters 8-11. posted by Anonymous on June 17, 2005 Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged. Most Helpful Critical Review 2 out of 2 people found this review helpful reinforc...posted by kls525 on January 6, 2011 Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged2 out of 2 people found this review helpful. Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged. Anonymous Posted October 23, 2002 Not for beginners If you are using this book, you should have already taken Algebra, and are looking for a refresher. It's a good refresher with lots and lots of examples and practice problems, but the explanations leave a lot to be desired, and some are just plain confusing. The cartoon on the front of the book is intesting, as the equation has an error in it, and had I spotted it while perusing the book I probably would have been reluctant to buy it. However, I didn't find a lot of errors in the book itself. If you are reviewing, then this could be a good book, but if you have no Algebra background, this is not the book for you. 1 out of 2 people found this review helpful. Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged.
The JOMA Global Positioning System and Imagery Collection is a growing library of data, how-tos, and materials for learning mathematics, science, and engineering using data collected with GPS units and both digital still and movie cameras. In this Editor's note, I discuss how to use HTML, CSS style sheets, and JavaScript to produce mathematical documents that are well-structured, conform to best practices, and look good. I also provide a number of sample documents and links to resources. This is a Developer's Area article that describes how to use a Java applet called LiveGraphics3D to speed up the process of creating interactive graphics by removing the need to create a graphics engine in Java, Flash, or other programming languages. This article argues for pedagogical reasons that the dot and cross products should be defined by their geometric properties, from which algebraic representations can be derived, rather than the other way around.
Foundations of Mathematical & Computational Economics 9780324235838 ISBN: 0324235836 Edition: 1 Pub Date: 2006 Publisher: Thomson Learning Summary: Economics doesn't have to be a mystery anymore. FOUNDATIONS OF MATHEMATICAL AND COMPUTATION ECONOMICS shows you how mathematics impacts economics and econometrics using easy-to-understand language and plenty of examples. Plus, it goes in-depth into computation and computational economics so you'll know how to handle those situations in your first economics job. Get ready for both the test and the workforce with this ...economics textbook. Dadkhah, Kamran is the author of Foundations of Mathematical & Computational Economics, published 2006 under ISBN 9780324235838 and 0324235836. Four hundred eighty five Foundations of Mathematical & Computational Economics textbooks are available for sale on ValoreBooks.com, one hundred thirty used from the cheapest price of $12.96, or buy new starting at $99
Mathematics Endorsement Why Elementary Education Mathematics endorsement? Students preparing to teach either in K-6 elementary school settings or grades 5-8 middle school mathematics settings will gain valuable knowledge by adding this endorsement to their Elementary Education major. Elementary Education (K-6) majors wishing to add an endorsement in Mathematics will complete the following courses (see "Degree Requirements" below). This course, followed by M149, provides a two-semester sequence that covers the material of a traditional Calculus I course along with built-in coverage of precalculus topics. Topics in M148 include: solving equations, functions, classes of functions (polynomial, rational, algebraic, exponential, logarithmic), right triangle trigonometry, angle measure, limits and continuity, derivatives, rules for derivatives. Credit is not granted for this course and M151 or courses equivalent to college algebra and college trigonometry. This course completes the two-semester sequence that begins with M148, and together with M148 provides a two-semester sequence that covers the material of a traditional Calculus I course along with built-in coverage of precalculus topics. Topics in M149 include: trigonometric and inverse trigonometric functions, rules for derivatives, applications of derivatives, and definite and indefinite integrals. Credit is not granted for this course and M151. This course provides an introduction to the differential and integral calculus. Topics include: the concepts of function, limit, continuity, derivative, definite and indefinite integrals, and an introduction to transcendental functions. Credit is not granted for this course and M148 and M149. Prerequisites: departmental placement or courses equivalent to college algebra and college trigonometry. This course is designed to strengthen the mathematical background of students in elementary education. It is required for the endorsement in mathematics for elementary education. The course consists of a selection of mathematical topics of wide interest and applicability. Topics include: graph models, linear programming, scheduling and packing problems, allocation problems, and social decision problems. This course may not be used as an upper-division elective for the mathematics major or minor or the mathematics education major. This course is designed to develop student facility in the use of statistical methods and the understanding of statistical concepts. The course takes a practical approach based on statistical examples taken from everyday life. Topics include: descriptive and inferential statistics, an intuitive introduction to probability, estimation, hypothesis testing, chi-square tests, regression and correlation. Appropriate technology is used to perform the calculations for many applications, and correspondingly an emphasis is placed on interpreting the results of statistical procedures. Credit is not granted for this course and any of the following: BU215, B392 or ST232.
Gain a solid understanding of the principles of trigonometry and how these concepts apply to real life with McKeague/Turner's best-selling ...Show synopsisGain a solid understanding of the principles of trigonometry and how these concepts apply to real life with McKeague/Turner's best-selling TRIGONOMETRY 6e, International Edition. This book's proven approach presents contemporary concepts in brief, manageable sections using current, detailed examples and high-interest applications. Captivating illustrations drawn from Lance Armstrong's cycling success, the Ferris wheel, and even the human cannonball show trigonometry in action. Unique Historical Vignettes offer a fascinating glimpse at how many of the central ideas in trigonometry began. TRIGONOMETRY 6e, International Edition, uses a standard right-angle approach with an emphasis on the study skills most important for success both now and in advanced courses, such as calculus. The book's proven blend of exercises, fresh applications, and projects is combined with a simplified approach to graphing and the convenience of new Enhanced WebAssign--a leading, time-saving online homework tool--and the innovative CengageNOW teaching system. With TRIGONOMETRY 6e, International Edition, you'll find everything you need for a thorough understand of trigonometry concepts now and the solid foundation you need for future coursework and career success.Hide synopsis Description:New. 1111826854 ANNOTATED INSTRUCTOR'S EDITION contains the...New. 1111826854 500 p. Contains: Illustrations. Description:NEW, Perfect! -INSTRUCTORS EDITION: hardcover; same content as...NEW, Perfect! -INSTRUCTORS EDITION: hardcover; same content as student's text with answers to ALL problems in back of book. See also ISBN 0495382590 for the Instructor's Solutions Manual, has worked-out solutions to... TEACHER EDITION
I'm an adult student attending college in Littleton, Colorado (yes, = barely a mile away from Columbine High School). We also utilize the FOIL = method for multiplying binomials. IMHO, it really didn't need to have = it's own little acronym since it is simply the most logical way to = perform the operation. I have looked at several other algebra textbooks = besides the one my class uses (Beginning Algebra, K. Elayn = Martin-Gay)whenever I was having difficulty understanding the way our = text/professor explained something. They all include FOIL in their = chapter on binomials. --=20 Miss Dee ________________________________ Most people don't care what you know, they just want to know that you care. JIM HARMON <SXBG@grove.iup.edu> wrote in message = news://01JAIVVGGV4I91XMHD@grove.iup.edu... I am doing a lesson on binomials and was just wondering what students = or teachers thought of the FOIL method and whether it is used everywhere = or not.=20 A reply would be appreciated. Thanks.
IDEA: Internet Differential Equations Activities IDEA is Internet Differential Equations Activities, an interdisciplinary effort to provide students and teachers around the world with computer based activities for differential equations in a wide variety of disciplines. IDEA is sponsored by the National Science Foundation with a grant from the Division of Undergraduate Education. The Idea behind IDEA As with every site on the Web, IDEA is evolving. IDEA contains a database of computer activities illustrating both mathematical concepts and the application of these concepts in a wide variety of disciplines. The aim is to show differential equations where they live, rather than in a purely mathematical setting. In addition to the exercises, it provides a variety of software for solving and describing differential equations. These packages include DynaSys, a package for Microsoft operating system that can be used in a very flexible way to create displays of solutions of differential equations; Java software that may be used over the Web to develop your own activities, or to work on the ones presented here; and Flash components that again may be used to develop differential equations activities. CODEE, the Consortium for Differential Equations Experiments, has been revitalized. CODEE was quite active in the 1990s in spreading differential equations activities, information, and software tools. In particular, CODEE formed the organization for the ODE Architect software. Recently, an NSF project headed by Darryl Young of Harvey Mudd College has reinvigorated CODEE. New activities! There are two new activities concerning the Idaho plan for wolf management, and a model for an insurgency. See the project page for details.
Numerical Computing in C# In this lesson I will show how to numerically solve algebraic and ordinary differential equations, and perform numerical integration with Simpson method. I will start with the solution of algebraic equations. The secant method is one of the simplest methods for solving algebraic equations. It is usually used as a part of a larger algorithm to improve convergence. As in any numerical algorithm, we need to check that the method is converging to a given precision in a certain number of steps. This is a precaution to avoid an infinite loop. Our second example is a Simpson integration algorithm. The Simpson algorithm is more precise the naive integration algorithm I have used there. The basic idea of the Simpson algorithm is to sample the integrand in a number of points to get a better estimate of its variations in a given interval. Finally, let me show a simple code for solving first order ordinary differential equations. The code uses a Runge-Kutta method. The simplest method to solve ODE is to do a Taylor expansion, which is called Euler's method. Euler's method approximates the solution with the series of consecutive secants. The error in Euler's method is O(h) on every step of size h. The Runge-Kutta method has an error O(h^4) Runge-Kutta methods with a variable step size are often used in practice since they converge faster than fixed size methods
About the Author Stan Gibilisco is an electronics engineer, researcher, and mathematician who has authored a number of titles for the McGraw-Hill Demystified series, along with more than 30 other books and dozens of magazine articles. His work has been published in several languages. Pre-Calculus Know-It-ALL Stan Gibilisco New York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto Preface This book is intended to complement standard pre-calculus texts at the high-school, trade-school, and college undergraduate levels. It can also serve as a self-teaching or home-schooling supplement. Prerequisites include beginning and intermediate algebra, geometry, and trigonometry. Pre-Calculus Know-It-ALL forms an ideal "bridge" between Algebra Know-It-ALL and Calculus Know-It-ALL. This course is split into two major sections. Part 1 (Chapters 1 through 10) deals with coordinate systems and vectors. Part 2 (Chapters 11 through 20) is devoted to analytic geometry. Chapters 1 through 9 and 11 through 19 end with practice exercises. They're "open-book" quizzes. You may (and should) refer to the text as you work out your answers. Detailed solutions appear in Appendices A and B. In many cases, these solutions don't represent the only way a problem can be figured out. Feel free to try alternatives! Chapters 10 and 20 contain question-and-answer sets that finish up Parts 1 and 2, respectively. These chapters aren't tests. They're designed to help you review the material, and to strengthen your grasp of the concepts. A multiple-choice Final Exam concludes the course. It's a "closed-book" test. Don't look back at the chapters, or use any other external references, while taking it. You'll find these questions more general (and easier) than the practice exercises at the ends of the chapters. The exam is meant to gauge your overall understanding of the concepts, not to measure how fast you can perform calculations or how well you can memorize formulas. The correct answers are listed in Appendix C. I've tried to introduce "mathematicalese" as the book proceeds. That way, you'll get used to the jargon as you work your way through the examples and problems. If you complete one chapter a week, you'll get through this course in a school year with time to spare, but don't hurry. Proceed at your own pace. Stan Gibilisco xi This page intentionally left blank Acknowledgments I extend thanks to my nephew Tony Boutelle, a technical writer based in Minneapolis, Minnesota, who offered insights and suggestions from the viewpoint of the intended audience, and found a few arithmetic errors before they got into print! I'm also grateful to Andrew A. Fedor, M.B.A., P.Eng (afedor@look.ca), a freelance consultant from Hampton, Ontario, Canada, for his proofreading help. Andrew has often provided suggestions for my existing publications and ideas for new ones. xiii This page intentionally left blank Pre-Calculus Know-It-ALL This page intentionally left blank PART 1 Coordinates and Vectors 1 This page intentionally left blank CHAPTER 1 Cartesian Two-Space If you've taken a course in algebra or geometry, you've learned about the graphing system called Cartesian (pronounced "car-TEE-zhun") two-space, also known as Cartesian coordinates or the Cartesian plane. Let's review the basics of this system, and then we'll learn how to calculate distances in it. How It's Assembled We can put together a Cartesian plane by positioning two identical real-number lines so they intersect at their zero points and are perpendicular to each other. The point of intersection is called the origin. Each number line forms an axis that can represent the values of a mathematical variable. The variables Figure 1-1 shows a simple set of Cartesian coordinates. One variable is portrayed along a horizontal line, and the other variable is portrayed along a vertical line. The number-line scales are graduated in increments of the same size. Figure 1-2 shows how several ordered pairs of the form (x,y) are plotted as points on the Cartesian plane. Here, x represents the independent variable (the "input"), and y represents the dependent variable (the "output"). Technically, when we work in the Cartesian plane, the numbers in an ordered pair represent the coordinates of a point on the plane. People sometimes say or write things as if the ordered pair actually is the point, but technically the ordered pair is the name of the point. Interval notation In pre-calculus and calculus, we'll often want to express a continuous span of values that a variable can attain. Such a span is called an interval. An interval always has a certain minimum value and a certain maximum value. These are the extremes of the interval. Let's be sure that 3 How It's Assembled 5 you're familiar with standard interval terminology and notation, so it won't confuse you later on. Consider these four situations: 0<x<2 −1 ≤ y < 0 4<z≤8 −p ≤ q ≤ p These expressions have the following meanings, in order: • • • • The value of x is larger than 0, but smaller than 2. The value of y is larger than or equal to −1, but smaller than 0. The value of z is larger than 4, but smaller than or equal to 8. The value of q is larger than or equal to −p, but smaller than or equal to p. The first case is an example of an open interval, which we can write as x ∈ (0,2) which translates to "x is an element of the open interval (0,2)." Don't mistake this open interval for an ordered pair! The notations look the same, but the meanings are completely different. The second and third cases are examples of half-open intervals. We denote this type of interval with a square bracket on the side of the included value and a rounded parenthesis on the side of the non-included value. We can write y ∈ [−1,0) which means "y is an element of the half-open interval [−1,0)," and z ∈ (4,8] which means "z is an element of the half-open interval (4,8]." The fourth case is an example of a closed interval. We use square brackets on both sides to show that both extremes are included. We can write this as q ∈ [−p,p] which translates to "q is an element of the closed interval [−p,p]." Relations and functions Do you remember the definitions of the terms relation and function from your algebra courses? (If you read Algebra Know-It-All, you should!) These terms are used often in pre-calculus, so it's important that you be familiar with them. A relation is an operation that transforms, or maps, values of a variable into values of another variable. A function is a relation in which there is never more than one value of the dependent variable for any value of the independent variable. In other words, there can't be more than one output for any input. (If a particular input 6 Cartesian Two-Space produces no output, that's okay.) The Cartesian plane gives us an excellent way to illustrate relations and functions. The axes In a Cartesian plane, both axes are linear, and both axes are graduated in increments of the same size. On either axis, the change in value is always directly proportional to the physical displacement. For example, if we travel 5 millimeters along an axis and the value changes by 1 unit, then that fact is true everywhere along that axis, and it's also true everywhere along the other axis. The quadrants Any pair of intersecting lines divides a plane into four parts. In the Cartesian system, these parts are called quadrants, as shown in Fig. 1-3: • In the first quadrant, both variables are positive. • In the second quadrant, the independent variable is negative and the dependent variable is positive. • In the third quadrant, both variables are negative. • In the fourth quadrant, the independent variable is positive and the dependent variable is negative. y 6 II I 4 Second quadrant 2 First quadrant x –6 –4 –2 Third quadrant III 2 –2 –4 4 6 Fourth quadrant IV –6 Figure 1-3 The Cartesian plane is divided into quadrants. The first, second, third, and fourth quadrants are sometimes labeled I, II, III, and IV, respectively. How It's Assembled The quadrants are sometimes labeled with Roman numerals, so that • • • • Quadrant I is at the upper right Quadrant II is at the upper left Quadrant III is at the lower left Quadrant IV is at the lower right If a point lies on one of the axes or at the origin, then it is not in any quadrant. Are you confused? Why do we insist that the increments be the same size on both axes in a Cartesian two-space graph? The answer is simple: That's how the Cartesian plane is defined! But there are other types of coordinate systems in which this exactness is not required. In a more generalized system called rectangular coordinates or the rectangular coordinate plane, the two axes can be graduated in divisions of different size. For example, the value on one axis might change by 1 unit for every 5 millimeters, while the value on the other axis changes by 1 unit for every 10 millimeters. Here's a challenge! Imagine an ordered pair (x,y), where both variables are nonzero real numbers. Suppose that you've plotted a point (call it P) on the Cartesian plane. Because x ≠ 0 and y ≠ 0, the point P does not lie on either axis. What will happen to the location of P if you multiply x by −1 and leave y the same? If you multiply y by −1 and leave x the same? If you multiply both x and y by −1? Solution If you multiply x by −1 and do not change the value of y, P will move to the opposite side of the y axis, but will stay the same distance away from that axis. The point will, in effect, be "reflected" by the y axis, moving to the left if x is positive to begin with, and to the right if x is negative to begin with. • • • • If P starts out in the first quadrant, it will move to the second. If P starts out in the second quadrant, it will move to the first. If P starts out in the third quadrant, it will move to the fourth. If P starts out in the fourth quadrant, it will move to the third. If you multiply y by −1 and leave x unchanged, P will move to the opposite side of the x axis, but will stay the same distance away from that axis. In a sense, P will be "reflected" by the x axis, moving straight downward if y is initially positive and straight upward if y is initially negative. • • • • If P starts out in the first quadrant, it will move to the fourth. If P starts out in the second quadrant, it will move to the third. If P starts out in the third quadrant, it will move to the second. If P starts out in the fourth quadrant, it will move to the first. 7 8 Cartesian Two-Space If you multiply both x and y by −1, P will move diagonally to the opposite quadrant. It will, in effect, be "reflected" by both axes. • • • • If P starts out in the first quadrant, it will move to the third. If P starts out in the second quadrant, it will move to the fourth. If P starts out in the third quadrant, it will move to the first. If P starts out in the fourth quadrant, it will move to the second. If you have trouble envisioning these point maneuvers, draw a Cartesian plane on a piece of graph paper. Then plot a point or two in each quadrant. Calculate how the x and y values change when you multiply either or both of them by −1, and then plot the new points. Distance of a Point from Origin On a straight number line, the distance of any point from the origin is equal to the absolute value of the number corresponding to the point. In the Cartesian plane, the distance of a point from the origin depends on both of the numbers in the point's ordered pair. An example Figure 1-4 shows the point (4,3) plotted in the Cartesian plane. Suppose that we want to find the distance d of (4,3) from the origin (0,0). How can this be done? We can calculate d using the Pythagorean theorem from geometry. In case you've forgotten that principle, here's a refresher. Suppose we have a right triangle defined by points P, Q, and R. Suppose the sides of the triangle have lengths b, h, and d as shown in Fig. 1-5. Then b2 + h2 = d 2 We can rewrite this as d = (b 2 + h 2)1/2 where the 1/2 power represents the nonnegative square root. Now let's make the following point assignments between the situations of Figs. 1-4 and 1-5: • The origin in Fig. 1-4 corresponds to the point Q in Fig. 1-5. • The point (4,0) in Fig. 1-4 corresponds to the point R in Fig. 1-5. • The point (4,3) in Fig. 1-4 corresponds to the point P in Fig. 1-5. Continuing with this analogy, we can see the following facts: • The line segment connecting the origin and (4,0) has length b = 4. • The line segment connecting (4,0) and (4,3) has height h = 3. • The line segment connecting the origin and (4,3) has length d (unknown). Distance of a Point from Origin 9 y 6 4 2 (4, 3) d x –6 –4 –2 What's the distance d ? 6 2 –2 –4 (4, 0) –6 Figure 1-4 We can use the Pythagorean theorem to find the distance d of the point (4,3) from the origin (0,0) in the Cartesian plane. The side of the right triangle having length d is the longest side, called the hypotenuse. Using the Pythagorean formula, we can calculate d = (b 2 + h 2)1/2 = (42 + 32)1/2 = (16 + 9)1/2 = 251/2 = 5 We've determined that the point (4,3) is 5 units distant from the origin in Cartesian coordinates, as measured along a straight line connecting (4,3) and the origin. Figure 1-5 The Pythagorean theorem for right triangles. 10 Cartesian Two-Space The general formula We can generalize the previous example to get a formula for the distance of any point from the origin in the Cartesian plane. In fact, we can repeat the explanation of the previous example almost verbatim, only with a few substitutions. Consider a point P with coordinates (xp,yp). We want to calculate the straight-line distance d of the point P from the origin (0,0), as shown in Fig. 1-6. Once again, we use the Pythagorean theorem. Turn back to Fig. 1-5 and follow along by comparing with Fig. 1-6: • The origin in Fig. 1-6 corresponds to the point Q in Fig. 1-5. • The point (xp,0) in Fig. 1-6 corresponds to the point R in Fig. 1-5. • The point (xp,yp) in Fig. 1-6 corresponds to the point P in Fig. 1-5. The following facts are also visually evident: • The line segment connecting the origin and (xp,0) has length b = xp. • The line segment connecting (xp,0) and (xp,yp) has height h = yp. • The line segment connecting the origin and (xp,yp) has length d (unknown). The Pythagorean formula tells us that d = (b 2 + h 2)1/2 = (xp2 + yp2)1/2 y Point P (xp, yp) d x What's the distance d ? Figure 1-6 (xp, 0) Using the Pythagorean theorem, we can derive a formula for the distance d of a generalized point P = (xp,yp) from the origin. Distance of a Point from Origin 11 That's it! The point (xp,yp) is (xp2 + yp2)1/2 units away from the origin, as we would measure it along a straight line. Are you confused? You might ask, "Can the distance of a point from the origin ever be negative?" The answer is no. If you look at the formula and break down the process in your mind, you'll see why this is so. First, you square xp, which is the x coordinate of P. Because xp is a real number, its square must be a nonnegative real. Next, you square yp, which is the y coordinate of P. This result must also be a nonnegative real. Next, you add these two nonnegative reals, which must produce another nonnegative real. Finally, you take the nonnegative square root, getting yet another nonnegative real. That's the distance of P from the origin. It can't be negative in a Cartesian plane whose axes represent real-number variables. Here's a challenge! Imagine a point P = (xp,yp) in the Cartesian plane, where xp ≠ 0 and yp ≠ 0. Suppose that P is d units from the origin. What will happen to d if you multiply xp by −1 and leave y unchanged? If you multiply yp by −1 and leave x unchanged? If you multiply both xp and yp by −1? Solution This is a three-part challenge. Let's break each part down into steps and apply the distance formula in each case. In the first situation, we change the x coordinate of P to its negative. Let's call the new point Px−. Its coordinates are (−xp,yp). Let dx− be the distance of Px− from the origin. Plugging the values into the formula, we obtain dx− = [(−xp)2 + yp2]1/2 = [(−1)2xp2 + yp2]1/2 = (xp2 + yp2)1/2 = d In the second situation, we change the y coordinate of P to its negative. This time, let's call the new point Py−. Its coordinates are (xp,−yp). Let dy− represent the distance of Py− from the origin. Plugging the values into the formula, we obtain dy− = [(xp)2 + (−yp)2]1/2 = [xp2 + (−1)2yp2]1/2 = (xp2 + yp2)1/2 = d In the third case, we change both the x and y coordinates of P to their negatives. We can call the new point Pxy− with coordinates (−xp,−yp). If we let dxy− represent the distance of Pxy− from the origin, we have dxy− = [(−xp)2 + (−yp)2]1/2 = [(−1)2xp2 + (−1)2yp2]1/2 = (xp2 + yp2)1/2 = d We've shown that we can negate either or both of the coordinate values of a point in the Cartesian plane, and although the point's location will usually change, its distance from the origin will always stay the same. 12 Cartesian Two-Space Distance between Any Two Points The distance between any two points on a number line is easy to calculate. We take the absolute value of the difference between the numbers corresponding to the points. In the Cartesian plane, each point needs two numbers to be defined, so the process is more complicated. Setting up the problem Figure 1-7 shows two generic points, P and Q, in the Cartesian plane. Their coordinates are P = (xp,yp) and Q = (xq,yq) Suppose we want to find the distance d between these points. We can construct a triangle by choosing a third point, R (which isn't on the line defined by P and Q) and then connecting P, Q, and R by line segments to get a triangle. The shape of triangle PQR depends on the location of R. If we choose certain coordinates for R, we can get a right triangle with the right angle at vertex R. With the help of Fig. 1-7, it's easy to see what the coordinates of R should be. If I travel "straight down" (parallel to the y axis) from P, and if you travel "straight to the right" (parallel to the Figure 1-7 We can find the distance d between two points P = (xp,yp) and Q = (xq,yq) by choosing point R to get a right triangle, and then applying the Pythagorean theorem. Distance between Any Two Points 13 x axis) from Q, our paths will cross at a right angle when we reach the point whose coordinates are (xp,yq). Those are the coordinates that R must have if we want the two sides of the triangle to be perpendicular there. Are you confused? "Wait!" you say. "Isn't there another point besides R that we can choose to create a right triangle along with points P and Q?" Yes, there is. The situation is shown in Fig. 1-8. If I go "straight up" (parallel to the y axis) from Q, and if you go "straight to the left" (parallel to the x axis) from P, we will meet at a right angle when we reach the coordinates (xq,yp). In this case, we might call the right-angle vertex point S. We won't use this geometry in the derivation that follows. But we could, and the final distance formula would turn out the same. Figure 1-8 Alternative geometry for finding the distance between two points. In this case, the right angle appears at point S. Dimensions and "deltas" Mathematicians use the uppercase Greek letter delta (Δ) to stand for the phrase "the difference in" or "the difference between." Using this notation, we can say that • The difference in the x values of points R and Q in Fig. 1-7 is xp − xq, or Δx. That's the length of the base of a right triangle. • The difference in the y values of points P and R is yp − yq, or Δy. That's the height of a right triangle. 14 Cartesian Two-Space We can see from Fig. 1-7 that the distance d between points Q and P is the length of the hypotenuse of triangle PQR. We're ready to find a formula for d using the Pythagorean theorem. The general formula Look back once more at Fig. 1-5. The relative positions of points P, Q, and R here are similar to their positions in Fig. 1-7. (I've set things up that way on purpose, as you can probably guess.) We can define the lengths of the sides of the triangle in Fig. 1-7 as follows: • The line segment connecting points Q and R has length b = Δx = xp − xq. • The line segment connecting points R and P has height h = Δy = yp − yq. • The line segment connecting points Q and P has length d (unknown). The Pythagorean formula tells us that d = (b 2 + h 2)1/2 = (Δx 2 + Δy 2)1/2 = [(xp − xq)2 + (yp − yq)2]1/2 An example Let's find the distance d between the following points in the Cartesian plane, using the formula we've derived: P = (−5,−2) and Q = (7,3) Plugging the values xp = −5, yp = −2, xq = 7, and yq = 3 into our formula, we get d = [(xp − xq)2 + (yp − yq)2]1/2 = [(−5 − 7)2 + (−2 − 3)2]1/2 = [(−12)2 + (−5)2]1/2 = (144 + 25)1/2 = 1691/2 = 13 Here's a challenge! It's reasonable to suppose that the distance between two points shouldn't depend on the direction in which we travel. But if you're a "show-me" person (as a mathematician should be), you might demand proof. Let's do it! Solution When we derived the distance formula previously, we traveled upward and to the right in Fig. 1-7 (from Q to P). When we work with directional displacement, it's customary to subtract the starting-point coordinates from the finishing-point coordinates. That's how we got Δx = xp − xq Finding the Midpoint 15 and Δy = yp − yq If we travel downward and to the left (from P to Q), we get Δx = xq − xp and Δy = yq − yp when we subtract the starting-point coordinates from the finishing-point coordinates. These new "star deltas" are the negatives of the original "plain deltas" because the subtractions are done in reverse. If we plug the "star deltas" straightaway into the derivation for d we worked out a few minutes ago, we can maneuver to get d = (Δx 2 + Δy 2)1/2 = [(−Δx)2 + (−Δy)2]1/2 = [(−1)2Δx 2 + (−1)2Δy 2]1/2 = (Δx 2 + Δy 2)1/2 = [(xp − xq)2 + (yp − yq)2]1/2 That's the same distance formula we got when we went from Q to P. This proves that the direction of travel isn't important when we talk about the simple distance between two points in Cartesian coordinates. (When we work with vectors later in this book, the direction will matter. Directional distance is known as displacement.) Finding the Midpoint We can find the midpoint between two points on a number line by calculating the arithmetic mean (or average value) of the numbers corresponding to the points. In Cartesian xy coordinates, we must make two calculations. First, we average the x values of the two points to get the x value of the point midway between. Then, we average the y values of the points to get the y value of the point midway between. A "mini theorem" Once again, imagine points P and Q in the Cartesian plane with the coordinates P = (xp,yp) and Q = (xq,yq) Suppose we want to find the coordinates of the midpoint. That's the point that bisects a straight line segment connecting P and Q. As before, we start out by choosing the point R "below and 16 Cartesian Two-Space y Point P (xp, yp) Point M lies midway between points P and Q Point M Point My x Point Q (xq, yq) Point R (xp, yq) Point Mx What are the coordinates of M ? Figure 1-9 We can calculate the coordinates of the midpoint of a line segment whose endpoints are known. to the right" that forms a right triangle PQR, as shown in Fig. 1-9. Imagine a movable point M that we can slide freely along line segment PQ. When we draw a perpendicular from M to side QR, we get a point Mx. When we draw a perpendicular from M to side RP, we get a point My. Consider the three right triangles MQMx, PMMy, and PQR. The laws of basic geometry tell us that these triangles are similar, meaning that the lengths of their corresponding sides are in the same ratios. According to the definition of similarity for triangles, we know the following two facts: • Point Mx is midway between Q and R if and only if M is midway between P and Q. • Point My is midway between R and P if and only if M is midway between P and Q. Now, instead of saying that M stands for "movable point," let's say that M stands for "midpoint." In this case, the x value of Mx (the midpoint of line segment QR) must be the x value of M, and the y value of My (the midpoint of line segment RP) must be the y value of M. The general formula We've reduced our Cartesian two-space midpoint problem to two separate number-line midpoint problems. Side QR of triangle PQR is parallel to the x axis, and side RP of triangle PQR is parallel to the y axis. We can find the x value of Mx by averaging the x values of Q and R. When we do this and call the result xm, we get xm = (xp + xq)/2 Finding the Midpoint 17 In the same way, we can calculate the y value of My by averaging the y values of R and P. Calling the result ym, we have ym = (yp + yq)/2 We can use the "mini theorem" we finished a few moments ago to conclude that the coordinates of point M, the midpoint of line segment PQ, are (xm,ym) = [(xp + xq)/2,(yp + yq)/2] An example Let's find the coordinates (xm,ym) of the midpoint M between the same two points for which we found the separation distance earlier in this chapter: P = (−5,−2) and Q = (7,3) When we plug xp = −5, yp = −2, xq = 7, and yq = 3 into the midpoint formula, we get (xm,ym) = [(xp + xq)/2,(yp + yq)/2] = [(−5 + 7)/2,(−2 + 3)/2] = (2/2,1/2) = (1,1/2) Are you a skeptic? It seems reasonable to suppose the midpoint between points P and Q should not depend on whether we go from P to Q or from Q to P. We can prove this by showing that for all real numbers xp, yp, xq, and yq, we have [(xp + xq)/2,(yp + yq)/2] = [(xq + xp)/2,(yq + yp)/2] This demonstration is easy, but let's go through it step-by-step to completely follow the logic. For the x coordinates, the commutative law of addition tells us that xp + xq = xq + xp Dividing each side by 2 gives us (xp + xq)/2 = (xq + xp)/2 For the y coordinates, the commutative law says that yp + yq = yq + yp 18 Cartesian Two-Space Again dividing each side by 2, we get (yp + yq)/2 = (yq + yp)/2 We've shown that the coordinates in the ordered pair on the left-hand side of the original equation are equal to the corresponding coordinates in the ordered pair on the right-hand side. The ordered pairs are identical, so the midpoint is the same in either direction. Are you confused? To find a midpoint of a line segment in Cartesian two-space, you simply average the coordinates of the endpoints. This method always works if the midpoint lies on a straight line segment between the two endpoints. But you might wonder, "How can we find the midpoint between two points along an arc connecting those points?" In a situation like that, we must determine the length of the arc. Depending on the nature of the arc, that can be fairly hard, very hard, or almost impossible! Arc-length problems are beyond the scope of this book, but you'll learn how to solve them in Calculus Know-It-All. Here's a challenge! Consider two points in the Cartesian plane, one of which is at the origin. Show that the coordinate values of the midpoint are exactly half the corresponding coordinate values of the point not on the origin. Solution We can plug in (0,0) as the coordinates of either point in the general midpoint formula, and work things out from there. First, let's suppose that point P is at the origin and the coordinates of point Q are (xq,yq). Then xp = 0 and yp = 0. If we call the coordinates of the midpoint (xm,ym), we have (xm,ym) = [(xp + xq)/2,(yp + yq)/2] = [(0 + xq)/2,(0 + yq)/2] = (xq /2,yq /2) Now, let Q be at the origin and let the coordinates of P be (xp,yp). In that case, we have (xm,ym) = [(xp + xq)/2,(yp + yq)/2] = [(xp + 0)/2,(yp + 0)/2] = (xp /2,yp /2 20 Cartesian Two-Space 1. What are the x and y coordinates of the points shown in Fig. 1-10? 2. Determine the distance of the point (−4,5) from the origin in Fig. 1-10. Using a calculator, round off the answer to three decimal places. 3. Determine the distance of the point (−5,−3) from the origin in Fig. 1-10. Using a calculator, round it off to three decimal places. 4. Determine the distance of the point (1,−6) from the origin in Fig. 1-10. Using a calculator, round it off to three decimal places. 5. Determine the distance between the points (−4,5) and (−5,−3) in Fig. 1-10. Using a calculator, round it off to three decimal places. 6. Determine the distance between the points (−5,−3) and (1,−6) in Fig. 1-10. Using a calculator, round it off to three decimal places. 7. Determine the distance between the points (1,−6) and (−4,5) in Fig. 1-10. Using a calculator, round it off to three decimal places. 8. Determine the coordinates of the midpoint of line segment L in Fig. 1-11. Express the values in fractional and decimal form. 9. Determine the coordinates of the midpoint of line segment M in Fig. 1-11. Express the values in fractional and decimal form. 10. Determine the coordinates of the midpoint of line segment N in Fig. 1-11. Express the values in fractional and decimal form. CHAPTER 2 A Fresh Look at Trigonometry Trigonometry (or "trig") involves the relationships between angles and distances. Traditional texts usually define the trigonometric functions of an angle as ratios between the lengths of the sides of a right triangle containing that angle. If you've done trigonometry with triangles, get ready for a new perspective! Circles in the Cartesian Plane In Cartesian xy coordinates, circles are represented by straightforward equations. The equation for a particular circle depends on its radius, and also on the location of its center point. The unit circle In trigonometry, we're interested in the circle whose center is at the origin and whose radius is 1. This is the simplest possible circle in the xy plane. It's called the unit circle, and is represented by the equation x2 + y2 = 1 The unit circle gives us an elegant way to define the basic trigonometric functions. That's why these functions are sometimes called the circular functions. Before we get into the circular functions themselves, let's be sure we know how to define angles, which are the arguments (or inputs) of the trig functions. Naming angles Mathematicians often use Greek letters to represent angles. The italic, lowercase Greek letter theta is popular. It looks like an italic numeral 0 with a horizontal line through it (q). When writing about two different angles, a second Greek letter is used along with q. Most often, it's the italic, lowercase letter phi. This character looks like an italic lowercase English letter o with a forward slash through it (f). 21 22 A Fresh Look at Trigonometry Sometimes the italic, lowercase Greek letters alpha, beta, and gamma are used to represent angles. These, respectively, look like the following symbols: a, b, and g. When things get messy and there are a lot of angles to talk about, numeric subscripts may be used with Greek letters, so don't be surprised if you see text in which angles are denoted q1, q2, q3, and so on. If you read enough mathematical papers, you'll eventually come across angles that are represented by other lowercase Greek letters. Angle variables can also be represented by more familiar characters such as x, y, or z. As long as we know the context and stay consistent in a given situation, it really doesn't matter what we call an angle. Radian measure Imagine two rays pointing outward from the center of a circle. Each ray intersects the circle at a point. Suppose that the distance between these points, as measured along the arc of the circle, is equal to the radius of the circle. In that case, the measure of the angle between the rays is one radian (1 rad). There are always 2p rad in a full circle, where p (the lowercase, non-italic Greek letter pi) stands for the ratio of a circle's circumference to its diameter. The number p is irrational. Its value is approximately 3.14159. Mathematicians prefer the radian as a standard unit of angular measure, and it's the unit we'll work with in this course. It's common practice to omit the "rad" after an angle when we know that we're working with radians. Based on that convention: • • • • An angle of p /2 represents 1/4 of a circle An angle of p represents 1/2 of a circle An angle of 3p /2 represents 3/4 of a circle An angle of 2p represents a full circle An acute angle has a measure of more than 0 but less than p /2, a right angle has a measure of exactly p /2, an obtuse angle has a measure of more than p /2 but less than p, a straight angle has a measure of exactly p, and a reflex angle has a measure of more than p but less than 2p. Degree measure The angular degree (°), also called the degree of arc, is the unit of angular measure familiar to lay people. One degree (1°) is 1/360 of a full circle. You probably know the following basic facts: • • • • An angle of 90° represents 1/4 of a circle An angle of 180° represents 1/2 of a circle An angle of 270° represents 3/4 of a circle An angle of 360° represents a full circle An acute angle has a measure of more than 0 but less than 90°, a right angle has a measure of exactly 90°, an obtuse angle has a measure of more than 90° but less than 180°, a straight angle has a measure of exactly 180°, and a reflex angle has a measure of more than 180° but less than 360°. Primary Circular Functions 23 Are you confused? If you're used to measuring angles in degrees, the radian can seem unnatural at first. "Why," you might ask, "would we want to divide a circle into an irrational number of angular parts?" Mathematicians do this because it nearly always works out more simply than the degree-measure scheme in algebra, geometry, trigonometry, pre-calculus, and calculus. The radian is more natural than the degree, not less! We can define the radian in a circle without having to quote any numbers at all, just as we can define the diagonal of a square as the distance from one corner to the opposite corner. The radian is a purely geometric unit. The degree is contrived. (What's so special about the fraction 1/360, anyhow? To me, it would have made more sense if our distant ancestors had defined the degree as 1/100 of a circle.) Here's a challenge! The measure of a certain angle q is p /6. What fraction of a complete circular rotation does this represent? What is the measure of q in degrees? Solution A full circular rotation represents an angle of 2p. The value p /6 is equal to 1/12 of 2p. Therefore, the angle q represents 1/12 of a full circle. In degree measure, that's 1/12 of 360°, which is 30°. Primary Circular Functions Let's look again at the equation of a unit circle in the Cartesian xy plane. We get it by adding the squares of the variables and setting the sum equal to 1: x2 + y2 = 1 Imagine that q is an angle whose vertex is at the origin, and we measure this angle in a counterclockwise sense from the x axis, as shown in Fig. 2-1. Suppose this angle corresponds to a ray that intersects the unit circle at a point P, where P = (x0, y0) We can define the three basic circular functions, also called the primary circular functions, of q in a simple way. But before we get into that, let's extend our notion of angles to include negative values, and also to deal with angles larger than 2p. Offbeat angles In trigonometry, any direction angle, no matter how extreme, can always be reduced to something that's nonnegative but less than 2p. Even if the ray OP in Fig. 2-1 makes more than one complete revolution counterclockwise from the x axis, or if it turns clockwise instead, its 24 A Fresh Look at Trigonometry y O P ( x0, y0) x Each axis division is 1/4 unit Unit circle Figure 2-1 The unit circle, whose equation is x2 + y2 = 1, can serve as the basis for defining trigonometric functions. In this graph, each axis division represents 1/4 unit. direction can always be defined by some counterclockwise angle of least 0 but less than 2p relative to the x axis. Think of this situation another way. The point P must always be somewhere on the circle, no matter how many times or in what direction the ray OP rotates to end up in a particular position. Every point on the circle corresponds to exactly one nonnegative angle less than 2p counterclockwise from the x axis. Conversely, if we consider the continuous range of angles going counterclockwise over the half-open interval [0,2p), we can account for every point on the circle. Any offbeat direction angle such as −9p /4 can be reduced to a direction angle that measures at least 0 but less than 2p by adding or subtracting some whole-number multiple of 2p. But we must be careful about this. A direction angle specifies orientation only. The orientation of the ray OP is the same for an angle of 3p as for an angle of p, but the larger value carries with it the idea that the ray (also called a vector) OP has rotated one and a half times around, while the smaller angle implies that it has undergone only half of a rotation. For our purposes now, this doesn't matter. But in some disciplines and situations, it does! Negative angles are encountered in trigonometry, especially in graphs of functions. Multiple revolutions of objects are important in physics and engineering. So if you ever hear or read about an angle such as −p /2 or 5p, you can be confident that it has meaning. The negative value indicates clockwise rotation. An angle larger than 2p indicates more than one complete rotation counterclockwise. An angle of less than −2p indicates more than one complete rotation clockwise. Primary Circular Functions 25 The sine function Look again at Fig. 2-1. Imagine that ray OP points along the x axis, and then starts to rotate counterclockwise at steady speed around its end point O, as if that point is a mechanical bearing. The point P, represented by coordinates (x0, y0), therefore revolves around O, following the unit circle. Imagine what happens to the value of y0 (the ordinate of point P ) during one complete revolution of ray OP. The ordinate of P starts out at y0 = 0, then increases until it reaches y0 = 1 after P has gone 1/4 of the way around the circle (that is, the ray has turned through an angle of p /2). After that, y0 begins to decrease, getting back to y0 = 0 when P has gone 1/2 of the way around the circle (the ray has turned through an angle of p ). As P continues in its orbit, y0 keeps decreasing until the value of y0 reaches its minimum of −1 when P has gone 3/4 of the way around the circle (the ray has turned through an angle of 3p /2). After that, the value of y0 rises again until, when P has gone completely around the circle, it returns to y0 = 0 for q = 2p. The value of y0 is defined as the sine of the angle q. The sine function is abbreviated as sin, so we can write sin q = y0 Circular motion Imagine that you attach a "glow-in-the-dark" ball to the end of a string, and then swing the ball around and around at a steady rate of one revolution per second. Suppose that you make the ball circle your head so the path of the ball lies in a horizontal plane. Imagine that you are in the middle of a flat, open field at night. The ball describes a circle as viewed from high above, as shown in Fig. 2-2A. If a friend stands far away with her eyes exactly in the plane of the ball's orbit, she sees a point of light that oscillates back and forth, from right-to-left and left-to-right, along what appears to be a straight-line path (Fig. 2-2B). Starting from its rightmost apparent position, the glowing point moves toward the left for 1/2 second, speeding up and then slowing down; then it reverses direction; then it moves toward the right for You Ball A Top view String Ball B Side view Figure 2-2 Orbiting ball and string. At A, as seen from above; at B, as seen edge-on. 26 A Fresh Look at Trigonometry 1/2 second, speeding up and then slowing down; then turns around again. As seen by your friend, the ball reaches its extreme rightmost position at 1-second intervals, because its orbital speed is one revolution per second. The sine wave If you graph the apparent position of the ball as seen by your friend with respect to time, the result is a sine wave, which is a graphical plot of a sine function. Some sine waves "rise higher and lower" (corresponding to a longer string), some are "flatter" (the equivalent of a shorter string), some are "stretched out" (a slower rate of revolution), and some are "squashed" (a faster rate of revolution). But the characteristic shape of the wave, known as a sinusoid, is the same in every case. You can whirl the ball around faster or slower than one revolution per second, thereby altering the frequency of the sine wave: the number of times a complete wave cycle repeats within a specified interval on the independent-variable axis. You can make the string longer or shorter, thereby adjusting the amplitude of the wave: the difference between the extreme values of its dependent variable. No matter what changes you might make of this sort, the sinusoid can always be defined in terms of a moving point that orbits a central point at a constant speed in a perfect circle. If we want to graph a sinusoid in the Cartesian plane, the circular-motion analogy can be stated as y = a sin bq where a is a constant that depends on the radius of the circle, and b is a constant that depends on the revolution rate. The angle q is expressed counterclockwise from the positive x axis. Figure 2-3 illustrates a graph of the basic sine function; it's a sinusoid for which a = 1 and b = 1, and for which the angle is expressed in radians. sin q 3 2 1 –3p q 3p –1 –2 –3 Figure 2-3 Graph of the sine function for values of q between −3p and 3p. Each division on the horizontal axis represents p /2 units. Each division on the vertical axis represents 1/2 unit. Primary Circular Functions 27 The cosine function Look again at Fig. 2-1. Imagine, once again, a ray OP running outward from the origin through point P on the circle. Imagine that at first, the ray points along the x axis, and then it rotates steadily in a counterclockwise direction. Now let's think about what happens to the value of x0 (the abscissa of point P ) during one complete revolution of ray OP. It starts out at x0 = 1, then decreases until it reaches x0 = 0 when q = p /2. Then x0 continues to decrease, getting down to x0 = −1 when q = p. As P continues counterclockwise around the circle, x0 increases. When q = 3p /2, we get back up to x0 = 0. After that, x0 increases further until, when P has gone completely around the circle, it returns to x0 = 1 for q = 2p. The value of x0 is defined as the cosine of the angle q. The cosine function is abbreviated as cos, so we can write cos q = x0 The cosine wave Circular motion in the Cartesian plane can be defined in terms of the cosine function by means of the equation y = a cos bq where a is a constant that depends on the radius of the circle, and b is a constant that depends on the revolution rate, just as is the case with the sine function. The angle q is measured or defined counterclockwise from the positive x axis, as always. The shape of a cosine wave is exactly the same as the shape of a sine wave. Both waves are sinusoids. But the entire cosine wave is shifted to the left by 1/4 of a cycle with respect to the sine wave. That works out to an angle of p /2. Figure 2-4 shows a graph of the basic cosine cos q 3 2 1 3p q –3p –1 –2 –3 Figure 2-4 Graph of the cosine function for values of q between −3p and 3p. Each division on the horizontal axis represents p /2 units. Each division on the vertical axis represents 1/2 unit. 28 A Fresh Look at Trigonometry function; it's a cosine wave for which a = 1 and b = 1. Because the cosine wave in Fig. 2-4 has the same frequency but a difference in horizontal position compared with the sine wave in Fig. 2-3, the two waves are said to differ in phase. For those of you who like fancy technical terms, a phase difference of 1/4 cycle (or p /2) is known in electrical engineering as phase quadrature. The tangent function Once again, refer to Fig. 2-1. The tangent (abbreviated as tan) of an angle q can be defined using the same ray OP and the same point P = (x0,y0) as we use when we define the sine and cosine functions. The definition is tan q = y0 /x0 We've seen that sin q = y0 and cos q = x0, so we can express the tangent function as tan q = sin q /cos q The tangent function is interesting because, unlike the sine and cosine functions, it "blows up" at certain values of q. This is shown by a graph of the function (Fig. 2-5). Whenever x0 = 0, the denominator of either quotient above becomes 0, so the tangent function is not defined for any angle q such that cos q = 0. This happens whenever q is a positive or negative odd-integer multiple of p /2. tan q 3 2 1 3p q –3p –1 –2 –3 Figure 2-5 Graph of the tangent function for values of q between −3p and 3p. Each division on the horizontal axis represents p /2 units. Each division on the vertical axis represents 1/2 unit. Primary Circular Functions 29 Singularities When a function "blows up" as the tangent function does at all the odd-integer multiples of p /2, we say that the function is singular for the affected values of the input variable. Such a "blow-up point" is called a singularity. If you've read books or watched movies about space travel and black holes, maybe you've seen or heard the term space-time singularity. That's a place where all the familiar rules of the universe break down. In a mathematical singularity, things aren't quite so dramatic, but the output value of a function becomes meaningless. In Fig. 2-5, the singularities are denoted by vertical dashed lines. The dashed lines themselves are known as asymptotes. Inflection points Midway between the singularities, the graph of the tangent function crosses the q axis, and the sense of the curvature changes. Below the q axis, the curves are always concave to the right and convex to the left. Above the q axis, the curves are always concave to the left and convex to the right. Whenever we have a point on a curve where the sense of the curvature reverses, we call that point an inflection point or a point of inflection. (Some texts spell the word "inflexion.") Lots of graphs have inflection points. If you're astute, you'll look back in this chapter and notice that the sine and cosine waves also have them. From your algebra courses, you might also remember that the graphs of many higher-degree polynomial functions have inflection points. Are you confused? Some students wonder if there's a way to define a function at a singularity. If you scrutinize Fig. 2-5 closely, you might be tempted to say that tan (p /2) = ±∞ where the symbol ±∞ means positive or negative infinity. The graph suggests that the output of the tangent function might attain values of infinity at the singular input points, doesn't it? It's an interesting notion; the problem is that we don't have a formal definition for infinity as a number. Mathematicians have found it difficult, over the generations, to make up a rigorous, workable definition for infinity as a number. Some mathematicians have grappled with the notion of infinity and come up with a way of doing arithmetic with it. Most notable among these people was Georg Cantor, a German mathematician who lived from 1845 to 1918. He discovered the apparent existence of "multiple infinities," which he called transfinite numbers. If you're interested in studying transfinite numbers, try searching the Internet using that term as a phrase. Here's a challenge! Figure out the value of tan (p /4). Don't do any calculations. You should be able to infer this on the basis of geometry alone. 30 A Fresh Look at Trigonometry Solution Draw a diagram of a unit circle, such as the one in Fig. 2-1, and place ray OP so that it subtends an angle of p /4 with respect to the x axis. (That's exactly "northeast" if the positive x axis goes "east" and the positive y axis goes "north.") Note that the ray OP also subtends an angle of p /4 with respect to the y axis, because the x and y axes are mutually perpendicular (oriented at an angle of exactly p /2 with respect to each other), and p /4 is half of p /2. Every point on the ray OP is equally distant from the x and y axes, including the point (x0,y0) where the ray intersects the circle. It follows that x0 = y0. Neither of them is equal to 0, so you know that y0/x0 = 1. According to the definition of the tangent function, you can conclude that tan (p /4) = y0 /x0 = 1 Secondary Circular Functions The three primary circular functions, as already defined, form the cornerstone of trigonometry. Three more circular functions exist. Their values represent the reciprocals of the values of the primary circular functions. The cosecant function Imagine the ray OP in Fig. 2-1, oriented at a certain angle q with respect to the x axis, pointing outward from the origin, and intersecting the unit circle at P = (x0,y0). The reciprocal of the ordinate, 1/y0, is defined as the cosecant of the angle q. The cosecant function is abbreviated as csc, so we can write csc q = 1/y0 Because y0 is the value of the sine function, the cosecant is the reciprocal of the sine. For any angle q, the following equation is always true as long as sin q ≠ 0: csc q = 1/sin q The cosecant of an angle q is undefined when q is any integer multiple of p. That's because the sine of any such angle is 0, which would make the cosecant equal to 1/0. Figure 2-6 is a graph of the cosecant function for values of q between −3p and 3p. The vertical dashed lines denote the singularities. There's also a singularity along the y axis. The secant function Consider the reciprocal of the abscissa, that is, 1/x0, in Fig. 2-1. This value is the secant of the angle q. The secant function is abbreviated as sec, so we can write sec q = 1/x0 The secant of an angle is the reciprocal of the cosine. When cos q ≠ 0, the following equation is true: sec q = 1/cos q The secant is undefined for any positive or negative odd-integer multiple of p /2. Figure 2-7 is a graph of the secant function for values of q between −3p and 3p. Note the input values for which the function is singular (vertical dashed lines). 32 A Fresh Look at Trigonometry The cotangent function Now let's think about the value of x0 /y0 at the point P where the ray OP crosses the unit circle. This ratio is called the cotangent of the angle q. The cotangent function is abbreviated as cot, so we can write cot q = x0 /y0 Because we already know that cos q = x0 and sin q = y0, we can express the cotangent function in terms of the cosine and the sine: cot q = cos q/sin q The cotangent function is also the reciprocal of the tangent function: cot q = 1/tan q Whenever y0 = 0, the denominators of all three quotients above become 0, so the cotangent function is not defined. Singularities occur at all integer multiples of p. Figure 2-8 is a graph of the cotangent function for values of q between −3p and 3p. Singularities are, as in the other examples here, shown as vertical dashed lines. cot q 3 2 1 q 3p –3p –1 –2 –3 Figure 2-8 Graph of the cotangent function for values of q between −3p and 3p. Each division on the horizontal axis represents p /2 units. Each division on the vertical axis represents 1/2 unit. Pythagorean Extras 33 Are you confused? Now that you know how the six circular functions are defined, you might wonder how you can determine the output values for specific inputs. The easiest way is to use a calculator. This approach will usually give you an approximation, not an exact value, because the output values of trigonometric functions are almost always irrational numbers. Remember to set the calculator to work for inputs in radians, not in degrees! The values of the sine and cosine functions never get smaller than −1 or larger than 1. The values of the other four functions can vary wildly. Put a few numbers into your calculator and see what happens when you apply the circular functions to them. When you input a value for which a function is singular, you'll get an error message on the calculator. Here's a challenge! Figure out the value of cot (5p /4). As in the previous challenge, you should be able to solve this problem entirely with geometry. Solution As you did before, draw a unit circle on a Cartesian coordinate grid. This time, orient the ray OP so that it subtends an angle of 5p /4 with respect to the x axis. (That's exactly "southwest" if the positive x axis goes "east" and the positive y axis goes "north.") Every point on OP is equally distant from the x and y axes, including (x0,y0) where the ray intersects the circle. You can see that x0 = y0 and both of them are negative, so the ratio x0/y0 must be equal to 1. According to the definition of the cotangent function, you can therefore conclude that cot (5p /4) = 1 Pythagorean Extras The Pythagorean theorem for right triangles, which we reviewed in Chap. 1, can be extended to cover three important identities (equations that always hold true) involving the circular functions. Pythagorean identity for sine and cosine The square of the sine of an angle plus the square of the cosine of the same angle is always equal to 1. We can write this fact as (sin q)2 + (cos q)2 = 1 When the value of a trigonometric function is squared, the exponent 2 is customarily placed after the abbreviation of the function and before the input variable, so the parentheses can be eliminated from the expression. In that format, the above equation is written as sin2 q + cos2 q = 1 34 A Fresh Look at Trigonometry Pythagorean identity for secant and tangent The square of the secant of an angle minus the square of the tangent of the same angle is always equal to 1, as long as the angle is not an odd-integer multiple of p /2. We write this as sec2 q − tan2 q = 1 Pythagorean identity for cosecant and cotangent The square of the cosecant of an angle minus the square of the cotangent of the same angle is always equal to 1, as long as the angle is not an integer multiple of p. We write this as csc2 q − cot2 q = 1 Are you confused? You've probably seen the above formula for the sine and cosine in your algebra or trigonometry courses. If you haven't seen the other two formulas, you might wonder where they come from. They can both be derived from the first formula using simple algebra along with the facts we've reviewed in this chapter. You'll get a chance to work them out in Problems 9 and 10, later. Here's a challenge! Use a drawing of the unit circle to show that sin2 q + cos2 q = 1 for angles q greater than 0 and less than p /2. (Here's a hint: A right triangle is involved.) Solution Figure 2-9 shows the unit circle with q defined counterclockwise between the x axis and a ray emanating from the origin. When the angle is greater than 0 but less than p /2, a right triangle y Length = 1 unit sin q q x cos q Unit circle Figure 2-9 This drawing can help show that sin2 q + cos2 q = 1 when 0 < q < p /2. Pythagorean Extras 35 is formed, with a segment of the ray as the hypotenuse. The length of this segment is equal to the radius of the unit circle. This radius, by definition, is 1 unit. According to the Pythagorean theorem for right triangles, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. It is easy to see that the lengths of these other two sides are sin q and cos q. Therefore, sin2 q + cos2 q = 1 Here's another challenge! Use another drawing of the unit circle to show that sin2 q + cos2 q = 1 for angles q greater than 3p /2 and less than 2p. (Here's a hint: This range of angles is equivalent to the range of angles greater than −p /2 and less than 0.) Solution Figure 2-10 shows how this can be done. Draw a mirror image of Fig. 2-9, with the angle q defined clockwise instead of counterclockwise. Again, you get a right triangle with a hypotenuse 1 unit long, while the other two sides have lengths of sin q and cos q. This triangle, like all right triangles, obeys the Pythagorean theorem. As in the previous challenge, you end up with sin2 q + cos2 q = 1 y Unit circle cos q x q sin q Length = 1 unit Figure 2-10 This drawing can help show that sin2 q + cos2 q = 1 when 3p /2 < q < 2p. 36 A Fresh Look at Trigonometry Approximately how many radians are there in 1°? Use a calculator and round the answer off to four decimal places, assuming that p ≈ 3.14159. 2. What is the angle in radians representing 7/8 of a circular rotation counterclockwise? Express the answer in terms of p, not as a calculator-derived approximation. 3. What is the angle in radians corresponding to 120° counterclockwise? Express the answer in terms of p, not as a calculator-derived approximation. 4. Suppose that the earth is a perfectly smooth sphere with a circumference of 40,000 kilometers (km). Based on that notion, what is the angular separation (in radians) between two points 1000 /p km apart as measured over the earth's surface along the shortest possible route? 5. Sketch a graph of the function y = sin x as a dashed curve in the Cartesian xy plane. Then sketch a graph of y = 2 sin x as a solid curve. How do the two functions compare? 6. Sketch a graph of the function y = sin x as a dashed curve in the Cartesian xy plane. Then sketch a graph of y = sin 2x as a solid curve. How do the two functions compare? 7. The secant of an angle can never be within a certain range of values. What is that range? 8. The cosecant of an angle can never be within a certain range of values. What is that range? 9. The Pythagorean formula for the sine and cosine is sin2 q + cos2 q = 1 From this, derive the fact that sec2 q − tan2 q = 1 10. Once again, consider the formula sin2 q + cos2 q = 1 From this, derive the fact that csc2 q − cot2 q = 1 CHAPTER 3 Polar Two-Space The Cartesian plane isn't the only tool for graphing on a flat surface. Instead of moving right-left and up-down from the origin, we can travel in a specified direction straight outward from the origin to reach a desired point. The direction angle is expressed in radians with respect to a reference axis. The outward distance is called the radius. This scheme gives us polar two-space or the polar coordinate plane. The Variables Figure 3-1 shows the basic polar coordinate plane. The independent variable is portrayed as an angle q relative to a ray pointing to the right (or "east"). That ray is the reference axis. The dependent variable is portrayed as the radius r from the origin. In this way, we can define points in the plane as ordered pairs of the form (q,r). The radius In the polar plane, the radial increments are concentric circles. The larger the circle, the greater the value of r. In Fig. 3-1, the circles aren't labeled in units. We can imagine each concentric circle, working outward, as increasing by any number of units we want. For example, each radial division might represent 1, 5, 10, or 100 units. Whatever size increments we choose, we must make sure that they stay the same size all the way out. That is, the relationship between the radius coordinate and the actual radius of the circle representing it must be linear. The direction As pure mathematicians, we express polar-coordinate direction angles in radians. We go counterclockwise from a reference axis pointing in the same direction as the positive x axis normally goes in the Cartesian xy plane. The angular scale must be linear. That is, the physical angle on the graph must be directly proportional to the value of q. 37 38 Polar Two-Space Figure 3-1 The polar coordinate plane. Angular divisions are straight lines passing through the origin. Each angular division represents p/6 units. Radial divisions are circles. Strange values In polar coordinates, it's okay to have nonstandard direction angles. If q ≥ 2p, it represents at least one complete counterclockwise rotation from the reference axis. If the direction angle is q < 0, it represents clockwise rotation from the reference axis rather than counterclockwise rotation. We can also have negative radius coordinates. If we encounter some point for which we're told that r < 0, we can multiply r by −1 so it becomes positive, and then add or subtract p to or from the direction. That's like saying "Proceed 10 km due east" instead of "Proceed −10 km due west." Which variable is which? If you read a lot of mathematics texts and papers, you'll sometimes see ordered pairs for polar coordinates with the radius listed first, and then the angle. Instead of the form (q,r), the ordered pairs will take the form (r,q). In this scheme, the radius is the independent variable, and the direction is the dependent variable. It works fine, but it's easier for most people to imagine that the radius depends on the direction. Think of an old-fashioned radar display like the ones shown in war movies made in the middle of the last century. A bright radial ray rotates around a circular screen, revealing targets at various distances. The rotation continues at a steady rate; it's independent. Target distances are functions of the direction. Theoretically, a radar display could work in the opposite sense with an expanding bright circle instead of a rotating ray, and all of the targets would show up The Variables 39 in the same places. But that geometry wasn't technologically practical when radar sets were first designed, and it was never used. Let's use the (q,r) format for ordered pairs, where q is the independent variable and r is the dependent variable. Are you confused? You ask, "How we can write down relations and functions intended for polar coordinates as opposed to those meant for Cartesian coordinates?" It's simple. When we want to denote a relation or function (call it f ) in polar coordinates where the independent variable is q and the dependent variable is r, we write r = f (q) We can read this out loud as "r equals f of q." When we want to denote a relation or function (call it g) in Cartesian coordinates where the independent variable is x and the dependent variable is y, we can write y = g (x) We can read this out loud as "y equals g of x." Here's a challenge! Provide an example of a graphical object that represents a function in polar coordinates when q is the independent variable, but not in Cartesian xy coordinates when x is the independent variable. Solution Consider a polar function that maps all inputs into the same output, such as f (q) = 3 Because f (q) is another way of denoting r, this function tells us that r = 3. The graph is a circle with a radius of 3 units. In Cartesian coordinates, the equation of the circle with radius of 3 units is x2 + y2 = 9 (Note that 9 = 32, the square of the radius.) If we let y be the dependent variable and x be the independent variable, we can rearrange this equation to get y = ±(9 − x 2)1/2 We can't claim that y = g (x) where g is a function of x in this case. There are values of x (the independent variable) that produce two values of y (the dependent variable). For example, if x = 0, then y = ±3. If we want to say that g is a relation, that's okay; but g is not a function. 40 Polar Two-Space Three Basic Graphs Let's look at the graphs of three generalized equations in polar coordinates. In Cartesian coordinates, all equations of these forms produce straight-line graphs. Only one of them does it now! Constant angle When we set the direction angle to a numerical constant, we get a simple polar equation of the form q=a where a is the constant. As we allow the value of r to range over all the real numbers, the graph of any such equation is a straight line passing through the origin, subtending an angle of a with respect to the reference axis. Figure 3-2 shows two examples. In these cases, the equations are q = p /3 and q = 7p /8 p /2 q = p /3 q = 7p/8 p 0 3p/2 Figure 3-2 When we set the angle constant, the graph is a straight line through the origin. Here are two examples. Three Basic Graphs 41 Constant radius Imagine what happens if we set the radius to a numerical constant. This gives us a polar equation of the form r=a where a is the constant. The graph is a circle centered at the origin whose radius is a, as shown in Fig. 3-3, when we allow the direction angle q to rotate through at least one full turn of 2p. If we allow the angle to span the entire set of real numbers, we trace around the circle infinitely many times, but that doesn't change the appearance of the graph. Angle equals radius times positive constant Now let's investigate a more interesting situation. Figure 3-4 shows an example of what happens in polar coordinates when we set the radius equal to a positive constant multiple of the angle. We get a pair of "mirror-image spirals." To see how this graph arises, imagine a ray pointing from the origin straight out toward the right along the reference axis (labeled 0). The angle is 0, so the radius is 0. Now suppose the ray starts to rotate counterclockwise, like the sweep on an old-fashioned military radar screen. The angle increases positively at a constant rate. Therefore, the radius also increases at a constant rate, because the radius is a positive constant multiple of the angle. The resulting p/ 2 p 0 r=a 3p /2 Figure 3-3 When we set the radius constant, the graph is a circle centered at the origin. In this case, the radius is an arbitrary value a. 42 Polar Two-Space p /2 Positive angle, positive radius p Negative angle, negative radius 0 3p /2 Figure 3-4 When we set the radius equal to a positive constant multiple of the angle, we get a pair of spirals. graph is the solid spiral. The pitch (or "tightness") of the spiral depends on the value of the constant a in the equation r = aq Small positive values of a produce tightly curled-up spirals. Larger positive values of a produce more loosely pitched spirals. Now suppose that the ray starts from the reference axis and rotates clockwise. At first, the angle is 0, so the radius is 0. As the ray turns, the angle increases negatively at a constant rate. That means the radius increases negatively at a constant rate, too, because we're multiplying the angle by a positive constant. We must plot the points in the exact opposite direction from the way the ray points. When we do that, we get the dashed spiral in Fig. 3-4. The pitch is the same as that of the heavy spiral, because we haven't changed the value of a. The entire graph of the equation consists of both spirals together. Angle equals radius times negative constant Figure 3-5 shows an example of what happens in polar coordinates when we set the radius equal to a negative constant multiple of the angle. As in the previous case, we get a pair of spirals, but they're "upside-down" with respect to the case when the constant is positive. To see Three Basic Graphs Positive angle, negative radius 43 p /2 p 0 3p /2 Negative angle, positive radius Figure 3-5 When we set the radius equal to a negative constant multiple of the angle, we get a pair of spirals "upside-down" relative to those for a positive constant multiple of the angle. Illustration for Problem 4. how this works, you can trace around with rotating rays as we did in Fig. 3-4. Be careful with the signs and directions! Remember that negative angles go clockwise, and negative radii go in the opposite direction from the way the angle is defined. Are you confused? Look back at Fig. 3-2. If you ponder this graph for awhile, you might suspect that the indicated equations aren't the only ones that can represent these lines. You might ask, "If we allow r to range over all the real numbers, both positive and negative, can't the line for q = p /3 also be represented by other equations such as q = 4p /3 or q = −2p /3? Can't the line representing the q = 7p /8 also be represented by q = 15p /8 or q = −p /8?" The answers to these questions are "Yes." When we see an equation of the form q = a representing a straight line through the origin in polar coordinates, we can add any integer multiple of p to the constant a, and we get another equation whose graph is the same line. In more formal terms, a particular line q = a through the origin can be represented by q = kp a where k is any integer and a is a real-number constant. 44 Polar Two-Space Here's a challenge! What's the value of the constant, a, in the function shown by the graph of Fig. 3-4? What's the equation of this pair of spirals? Assume that each radial division represents 1 unit. Solution Note that if q = p, then r = 2. You can solve for a by substituting this number pair in the general equation for the pair of spirals. Plugging in the numbers (q,r) = (p,2), proceed as follows: r = aq 2 = ap 2 /p = a Therefore, a = 2 /p, and the equation you seek is r = (2/p )q If you don't like parentheses, you can write it as r = 2q /p Here's another challenge! What is the polar equation of a straight line running through the origin and ascending at an angle of p /4 as you move to the right, with the restriction that 0 ≤ q < 2p ? If you drew this line on a standard Cartesian xy coordinate grid instead of the polar plane, what equation would it represent? Solution Two equations will work here. They are q = p /4 and q = 5p /4 Keep in mind that the value of r can be any real number: positive, negative, or zero. First, look at the situation where q = p /4. When r > 0, you get a ray in the p /4 direction. When r < 0, you get a ray in the 5p /4 direction. When r = 0, you get the origin point. The union of these two rays and the origin point forms the line running through the origin and ascending at an angle of p /4 as you move toward the right. Now examine events with the equation q = 5p /4. When r > 0, you get a ray in the 5p /4 direction. When r < 0, you get a ray in the p /4 direction. When r = 0, you get the origin point. The union of the two rays and the origin point forms the same line as in the first case. In the Cartesian xy plane, this line would be the graph of the equation y = x. Coordinate Transformations 45 Coordinate Transformations We can convert the coordinates of any point from polar to Cartesian systems and vice versa. Going from polar to Cartesian is easy, like floating down a river. Getting from Cartesian to polar is more difficult, like rowing up the same river. As you read along here, refer to Fig. 3-6, which shows a point in the polar grid superimposed on the Cartesian grid. Polar to Cartesian Suppose we have a point (q,r) in polar coordinates. We can convert this point to Cartesian coordinates (x,y) using the formulas x = r cos q and y = r sin q p /2 P y r q p 0 3p /2 x Figure 3-6 A point plotted in both polar and Cartesian coordinates. Each radial division in the polar grid represents 1 unit. Each division on the x and y axes of the Cartesian grid also represents 1 unit. The shaded region is a right triangle x units wide, y units tall, and having a hypotenuse r units long. 46 Polar Two-Space To understand how this works, imagine what happens when r = 1. The equation r = 1 in polar coordinates gives us a unit circle. We learned in Chap. 2 that when we have a unit circle in the Cartesian plane, then for any point (x,y) on that circle x = cos q and y = sin q Suppose that we double the radius of the circle. This makes the polar equation r = 2. The values of x and y in Cartesian coordinates both double, because when we double the length of the hypotenuse of a right triangle (such as the shaded region in Fig. 3-6), we also double the lengths of the other two sides. The new triangle is similar to the old one, meaning that its sides stay in the same ratio. Therefore x = 2 cos q and y = 2 sin q This scheme works no matter how large or small we make the circle, as long as it stays centered at the origin. If r = a, where a is some positive real number, the new right triangle is always similar to the old one, so we get x = a cos q and y = a sin q If our radius r happens to be negative, these formulas still work. (For "extra credit," can you figure out why?) An example Consider the point (q,r) = (p,2) in polar coordinates. Let's find the (x,y) representation of this point in Cartesian coordinates using the polar-to-Cartesian conversion formulas x = r cos q and y = r sin q Plugging in the numbers gives us x = 2 cos p = 2 × (−1) = −2 and y = 2 sin p = 2 × 0 = 0 Therefore, (x,y) = (−2,0). Coordinate Transformations 47 Cartesian to polar: the radius Figure 3-6 shows us that the radius r from the origin to our point P = (x,y) is the length of the hypotenuse of a right triangle (the shaded region) that's x units wide and y units tall. Using the Pythagorean theorem, we can write the formula for determining r in terms of x and y as r = (x 2 + y 2 )1/2 That's straightforward enough. Now it's time to work on the more difficult conversion: finding the polar angle for a point that's given to us in the Cartesian xy plane. The Arctangent function Before we can find the polar direction angle for a point that's given to us in Cartesian coordinates, we must be familiar with an inverse trigonometric function known as the Arctangent, which "undoes" the work of the tangent function. (The capital "A" is not a typo. We'll see why in a minute.) Consider, for example, the fact that tan (p /4) = 1 A true function that "undoes" the tangent must map an input value of 1 in the domain to an output value of p /4 in the range, but to no other values. In fact, no matter what we input to the function, we must never get more than one output. To ensure that the inverse of the tangent behaves as a true function, we must restrict its range (output) to an open interval where we don't get any redundancy. By convention, mathematicians specify the open interval (−p /2,p /2) for this purpose. When mathematicians make this sort of restriction in an inverse trigonometric function, they capitalize the "first letter" in the name of the function. That's a "code" to tell us that we're working with a true function, and not a mere relation. Some texts use the abbreviation tan−1 instead of Arctan to represent the inverse of the tangent function. We won't use this symbol here because some readers might confuse it with the reciprocal of the tangent, which is the cotangent, not the Arctangent! If you're curious as to what the Arctangent function looks like when graphed, check out Fig. 3-7. This graph consists of the principal branch of the tangent function, tipped on its side and then flipped upside-down. Compare Fig. 3-7 with Fig. 2-5 on page 28. The principal branch of the tangent function is the one that passes through the origin. Once we've made sure we won't run into any ambiguity, we can state the above fact using the Arctangent function, getting Arctan 1 = p /4 For any real number u except odd-integer multiples of p /2 (for which the tangent function is undefined), we can always be sure that Arctan (tan u) = u Going the other way, for any real number v, we can be confident that tan (Arctan v) = v 48 Polar Two-Space y p p /2 x –3 –2 1 –1 2 3 –p /2 –p Figure 3-7 A graph of the Arctangent function. The domain extends over all the real numbers. The range is restricted to values larger than −p /2 and smaller than p /2. Each division on the y axis represents p /6 units. Cartesian to polar: the angle We now have the tools that we need to determine the polar angle q for a point on the basis of its Cartesian coordinates x and y. We already know that x = r cos q and y = r sin q As long as x ≠ 0, it follows that y /x = (r sin q)/(r cos q) = (r /r)(sin q)/(cos q) = (sin q)/(cos q) = tan q Simplifying, we get tan q = y /x Coordinate Transformations 49 If we take the Arctangent of both sides, we obtain Arctan (tan q ) = Arctan ( y /x) which can be rewritten as q = Arctan (y /x) Suppose the point P = (x,y) happens to lie in the first or fourth quadrant of the Cartesian plane. In this case, we have −p /2 < q < p /2 so we can directly use the conversion formula q = Arctan ( y /x) If P = (x,y) is in the second or third quadrant, then we have p /2 < q < 3p /2 That's outside the range of the Arctangent function, but we can remedy this situation if we subtract p from q. When we do this, we bring q into the allowed range but we don't change its tangent, because the tangent function repeats itself every p radians. (If you look back at Fig. 2-5 again, you will notice that all of the branches in the graph are identical, and any two adjacent branches are p radians apart.) In this situation, we have q − p = Arctan ( y /x) which can be rewritten as q = p + Arctan ( y /x) Now we're ready to derive specific formulas for q in terms of x and y. Let's break the scenario down into all possible general locations for P = (x,y), and see what we get for q in each case: P at the origin. If x = 0 and y = 0, then q is theoretically undefined. However, let's assign q a default value of 0 at the origin. By doing that, we can "fill the hole" that would otherwise exist in our conversion scheme. P on the +x axis. If x > 0 and y = 0, then we're on the positive x axis. We can see from Fig. 3-6 that q = 0. P in the first quadrant. If x > 0 and y > 0, then we're in the first quadrant of the Cartesian plane where q is larger than 0 but less than p /2. We can therefore directly apply the conversion formula q = Arctan ( y /x) 50 Polar Two-Space P on the +y axis. If x = 0 and y > 0, then we're on the positive y axis. We can see from Fig. 3-6 that q = p /2. P in the second quadrant. If x < 0 and y > 0, then we're in the second quadrant of the Cartesian plane where q is larger than p /2 but less than p . In this case, we must apply the modified conversion formula q = p + Arctan ( y /x) P on the -x axis. If x < 0 and y = 0, then we're on the negative x axis. We can see from Fig. 3-6 that q = p. P in the third quadrant. If x < 0 and y < 0, then we're in the third quadrant of the Cartesian plane where q is larger than p but less than 3p /2, so we apply the modified conversion formula q = p + Arctan ( y /x) P on the -y axis. If x = 0 and y < 0, then we're on the negative y axis. We can see from Fig. 3-6 that q = 3p /2. P in the fourth quadrant. If x > 0 and y < 0, then we're in the fourth quadrant of the Cartesian plane where q is larger than 3p /2 but smaller than 2p. That's the same thing as saying that −p /2 < q < 0. We'll get an angle in that range if we apply the original conversion formula q = Arctan ( y /x) In the interest of elegance, we'd like the angle in the polar representation of a point to always be nonnegative but less than 2p. We can make this happen by adding in a complete rotation of 2p to the basic conversion formula, getting q = 2p + Arctan ( y /x) We have taken care of all the possible locations for P. A summary of the nine-part conversion formula that we've developed is given in the following table. q=0 At the origin q=0 On the +x axis q = Arctan ( y /x) In the first quadrant q = p /2 On the +y axis q = p + Arctan ( y /x) In the second quadrant q=p On the −x axis q = p + Arctan ( y /x) In the third quadrant q = 3p /2 On the −y axis q = 2p + Arctan ( y /x) In the fourth quadrant Coordinate Transformations 51 An example Let's convert the Cartesian point (−5,−12) to polar form. Here, x = −5 and y = −12. When we plug these numbers into the formula for r, we get r = [(−5)2 + (−12)2]1/2 = (25 + 144)1/2 = 1691/2 = 13 Our point is in the third quadrant of the Cartesian plane. To find the angle, we should use the formula q = p + Arctan ( y /x) When radians (not degrees) tells us that Arctan (12/5) ≈ 1.1760 rounded off to four decimal places. (Remember that the "wavy" equals sign means "is approximately equal to.") If we let p ≈ 3.1416, also rounded off to four decimal places, we get q ≈ 3.1416 + 1.1760 ≈ 4.3176 The polar equivalent of (x,y) = (−5,−12) is therefore (q,r) ≈ (4.3176,13), where q is approximated to four decimal places and r is exact. Are you confused? If the foregoing angle-conversion formula derivation baffles you, don't feel bad. It's complicated! If you don't grasp it to your satisfaction right now, set it aside for awhile. Read it again tomorrow, or the day after that. You might want to make up some problems with points in all four quadrants of the Cartesian plane, and then use these formulas to convert them to polar form. As you work out the arithmetic, you'll gain a better understanding of how (and why) the formulas work. Here's a challenge! Find the distance d in radial units between the points P = (p,3) and Q = (p /2,4) in polar coordinates, where a radial unit is equal to the radius of a unit circle centered at the origin. Solution Let's convert the polar coordinates of P and Q to Cartesian coordinates, and then employ the Cartesian distance formula to determine how far apart the two points are. Let's call the Cartesian versions of the points P = (xp,yp) 52 Polar Two-Space and Q = (xq,yq) For P, we have xp = 3 cos p = 3 × (−1) = −3 and yp = 3 sin p = 3 × 0 = 0 The Cartesian coordinates of P are therefore (xp,yp) = (−3,0). For Q, we have xq = 4 cos p /2 = 4 × 0 = 0 and yq = 4 sin p /2 = 4 × 1 = 4 The Cartesian coordinates of Q are therefore (xq,yq) = (0,4). Using the Cartesian distance formula, we obtain d = [(xp − xq)2 + ( yp − yq)2]1/2 = [(−3 − 0)2 + (0 − 4)2]1/2 = [(−3)2 + (−4)2]1/2 = (9 + 16)1/2 = 251/2 = 5 We've found that the points P = (p,3) and Q = (p /2,4) are precisely 5 radial units apart in the polar coordinate plane Is the relation q = p /4 a function in polar coordinates, where q is the independent variable and r is the dependent variable? Why or why not? Is q = p /2 a function in the same polar system? Why or why not? 2. Suppose that we draw the lines representing the polar relations q = p /4 and q = p /2 directly onto the Cartesian xy plane, where x is the independent variable and y is the dependent variable. Do either of the resulting graphs represent functions in the Cartesian coordinate system? Why or why not? Practice Exercises 53 3. Imagine a circle centered at the origin in polar coordinates. The equation for the circle is r = a, where a is a real-number constant. What other equation, if any, represents the same circle? 4. In Fig. 3-5 on page 43, suppose that each radial increment is p units. What's the value of the constant a in this case? What's the equation of the pair of spirals? (Here are a couple of reminders: The radial increments are the concentric circles. The value of a in this situation turns out negative.) 5. Figure 3-8 shows a line L and a circle C in polar coordinates. Line L passes through the origin, and every point on L is equidistant from the horizontal and vertical axes. Circle C is centered at the origin. Each radial division represents 1 unit. What's the polar equation representing L when we restrict the angles to positive values smaller than 2p? What's the polar equation representing C? (Here's a hint: Both equations can be represented in two ways.) 6. When we examine Fig. 3-8, we can see that L and C intersect at two points P and Q. What are the polar coordinates of P and Q, based on the information given in Problem 5? (Here's a hint: Both points can be represented in two ways.) Intersection point P p /2 Line L Circle C p 0 3p /2 Intersection point Q Figure 3-8 Illustration for Problems 5 through 10. Each radial division is 1 unit. 54 Polar Two-Space 7. Solve the system of equations from the solution to Problem 5, verifying the polar coordinates of points P and Q in Fig. 3-8. 8. Based on the information given in Problem 5, what are the Cartesian xy-coordinate equations of line L and circle C in Fig. 3-8? 9. Solve the system of equations from the solution to Problem 8 to determine the Cartesian coordinates of the intersection points P and Q in Fig. 3-8. 10. Based on the polar coordinates of points P and Q in Fig. 3-8 (the solutions to Problems 6 and 7), use the conversion formulas to derive the Cartesian coordinates of those two points. CHAPTER 4 Vector Basics We can define the length of a line segment that connects two points, but the direction is ambiguous. If we want to take the direction into account, we must make a line segment into a vector. Mathematicians write vector names as bold letters of the alphabet. Alternatively, a vector name can be denoted as a letter with a line or arrow over it. The "Cartesian Way" In diagrams and graphs, a vector is drawn as a directed line segment whose direction is portrayed by putting an arrow at one end. When working in two-space, we can describe vectors in Cartesian coordinates or in polar coordinates. Let's look at the "Cartesian way" first. Endpoints, locations, and notations Figure 4-1 shows four vectors drawn on a Cartesian coordinate grid. Each vector has a beginning (the originating point) and an end space (the terminating point). In this situation, any of the four vectors can be defined according to two independent quantities: • The length (magnitude) • The way it points (direction) It doesn't matter where the originating or terminating points actually are. The important thing is how the two points are located with respect to each other. Once a vector has been defined as having a specific magnitude and direction, we can "slide it around" all over the coordinate plane without changing its essential nature. We can always think of the originating point for a vector as being located at the coordinate origin (0,0). When we place a vector so that its originating point is at (0,0), we say that the vector is in standard form. The standard form is convenient in Cartesian 55 56 Vector Basics y 6 4 c b 2 x –6 –4 –2 2 –2 4 6 d a –4 –6 Figure 4-1 Four vectors in the Cartesian plane. In each case, the magnitude corresponds to the length of the line segment, and the direction is indicated by the arrow. coordinates, because it allows us to uniquely define any vector as an ordered pair corresponding to • The x coordinate of its terminating point (x component) • The y coordinate of its terminating point (y component) Figure 4-2 shows the same four vectors as Fig. 4-1 does, but all of the originating points have been moved to the coordinate origin. The magnitudes and directions of the corresponding vectors in Figs. 4-1 and 4-2 are identical. That's how we can tell that the vectors a, b, c, and d in Fig. 4-2 represent the same mathematical objects as the vectors a, b, c, and d in Fig. 4-1. Cartesian magnitude Imagine an arbitrary vector a in the Cartesian xy plane, extending from the origin (0,0) to the point (xa,ya) as shown in Fig. 4-3. The magnitude of a (which can be denoted as ra, as \|a\|, or as a) can be found by applying the formula for the distance of a point from the origin. We learned that formula in Chap. 1. Here it is, modified for the vector situation: ra = (xa2 + ya2)1/2 58 Vector Basics Figure 4-4 The direction of a vector can be defined as its angle, in radians, going counterclockwise from the positive x axis in the Cartesian plane. Cartesian direction Now let's think about the direction of a, as shown in Fig. 4-4. We can denote it as an angle qa or by writing dir a. To define qa in terms of its terminating-point coordinates (xa,ya), we must go back to the polar-coordinate direction-finding system in Chap. 3. The following table has those formulas, modified for our vector situation. qa = 0 qa = 0 qa = Arctan ( ya /xa) qa = p /2 qa = p + Arctan ( ya /xa) qa = p qa = p + Arctan ( ya /xa) qa = 3p /2 qa = 2p + Arctan ( ya /xa) When xa = 0 and ya = 0 so a terminates at the origin When xa > 0 and ya = 0 so a terminates on the +x axis When xa > 0 and ya > 0 so a terminates in the first quadrant When xa = 0 and ya > 0 so a terminates on the +y axis When xa < 0 and ya > 0 so a terminates in the second quadrant When xa < 0 and ya = 0 so a terminates on the −x axis When xa < 0 and ya < 0 so a terminates in the third quadrant When xa = 0 and ya < 0 so a terminates on the −y axis When xa > 0 and ya < 0 so a terminates in the fourth quadrant The "Cartesian Way" 59 Cartesian vector sum Let's consider two arbitrary vectors a and b in the Cartesian plane, in standard form with terminating-point coordinates a = (xa,ya) and b = (xb,yb) We calculate the sum vector a + b by adding the x and y terminating-point coordinates separately and then combining the sums to get a new ordered pair. When we do that, we get a + b = [(xa + xb),(ya + yb)] This sum can be illustrated geometrically by constructing a parallelogram with the two vectors a and b as adjacent sides, as shown in Fig. 4-5. The sum vector, a + b, corresponds to the directional diagonal of the parallelogram going away from the coordinate origin. y [(x a+ xb ) , ( ya +y b)] (x b, y b) b Sum of a and b runs along diagonal of parallelogram a+b qb a ( xa, ya) qa x Figure 4-5 We can determine the sum of two vectors a and b by finding the directional diagonal of a parallelogram with a and b as adjacent sides. 60 Vector Basics An example Consider two vectors in the Cartesian plane. Suppose they're both in standard form. (From now on, let's agree that all vectors are in standard form so they "begin" at the coordinate origin, unless we specifically state otherwise.) The vectors are defined according to the ordered pairs a = (4,0) and b = (3,4) In this case, we have xa = 4, xb = 3, ya = 0, and yb = 4. We find the sum vector by adding the corresponding coordinates to get a + b = [(xa + xb),(ya + yb)] = [(4 + 3),(0 + 4)] = (7,4) Cartesian negative of a vector To find the Cartesian negative of a vector, we take the additive inverses (that is, the negatives) of both coordinate values. Given the vector a = (xa,ya) its Cartesian negative is −a = (−xa,−ya) The Cartesian negative of a vector always has the same magnitude as the original, but points in the opposite direction. Cartesian vector difference Suppose that we want to find the difference between the two vectors a = (xa,ya) and b = (xb,yb) by subtracting b from a. We can do this by finding the Cartesian negative of b and then adding −b to a to get a − b = a + (−b) = {[(xa + (−xb)],[(ya + (−yb)]} = [(xa − xb),(ya − yb)] The "Cartesian Way" 61 We can skip the step where we find the negative of the second vector and directly subtract the coordinate values, but we must be sure we keep the vectors and coordinate values in the correct order if we take that shortcut. An example Let's look again at the same two vectors for which we found the Cartesian sum a few moments ago: a = (4,0) and b = (3,4) As before, we have xa = 4, xb = 3, ya = 0, and yb = 4. We can find a − b by taking the differences of the corresponding coordinates, as long as we keep the vectors in the correct order. Then we get a − b = [(xa − xb),(ya − yb)] = [(4 − 3),(0 − 4)] = (1,−4) Are you confused? If you have trouble with the notion of a vector, here are three real-world examples of vector quantities in two dimensions. If you like, draw diagrams to help your mind's eye envision what's happening in each case: • When the wind blows at 5 meters per second from east to west, you can say that the magnitude of its velocity vector is 5 and the direction is toward the west. In Cartesian coordinates where the +x axis goes east, the +y axis goes north, the −x axis goes west, and the −y axis goes south, you would assign this vector the ordered pair (−5,0). • When you push on a rolling cart with a force of 10 newtons toward the north, you're applying a force vector to the cart with a magnitude of 10 and a direction toward the north. In Cartesian coordinates where the +x axis goes east, the +y axis goes north, the −x axis goes west, and the −y axis goes south, you would assign this vector the ordered pair (0,10). • When you accelerate a car at 5 feet per second per second in a direction somewhat to the east of north, the magnitude of the car's acceleration vector is 5 and the direction is somewhat to the east of north. A "neat" situation of this sort occurs when the x (or eastward) component is 3 and the y (or northward) component is 4, so you get the ordered pair (3,4). These components form the two shorter sides of a 3:4:5 right triangle whose hypotenuse measures 5 units (the magnitude). Here's a challenge! Show that Cartesian vector addition is commutative. That is, show that for any two vectors a and b expressed as ordered pairs in the Cartesian plane, a+b=b+a 62 Vector Basics Solution This fact is easy, although rather tedious, to demonstrate rigorously. (In pure mathematics, the term rigor refers to the process of proving something in a series of absolutely logical steps. It has nothing to do with the physical condition called rigor mortis.) We must define the two vectors by coordinates, and then work through the arithmetic with those coordinates. Let's call the two vectors a = (xa,ya) and b = (xb,yb) As defined earlier in this chapter, the Cartesian sum a + b is a + b = [(xa + xb),(ya + yb)] Using the same definition, the Cartesian sum b + a is b + a = [(xb + xa),(yb + ya)] All four of the coordinate values xa, xb, ya, and yb are real numbers. We know from basic algebra that addition of real numbers is commutative. Therefore, we can reverse both of the sums in the elements of the ordered pair above, getting b + a = [(xa + xb),(ya + yb)] That's the ordered pair that defines a + b. We have just shown that a+b=b+a for any two Cartesian vectors a and b. The "Polar Way" In the polar coordinate plane, we draw a vector as a ray going straight outward from the origin to a point defined by a specific angle and a specific radius. Figure 4-6 shows two vectors a and b with originating points at (0,0) and terminating points at (qa,ra) and (qb,rb), respectively. Polar magnitude and direction The magnitude and direction of a vector a = (qa,ra) in the polar coordinate plane are defined directly by the coordinates. The magnitude is ra, the straight-line distance of the terminating point from the origin. The direction angle is qa, the angle that the ray subtends in a The "Polar Way" (q b, rb) 63 p /2 qb b (qa, ra) qa a p 0 3p /2 Figure 4-6 Vectors in the polar plane are defined by ordered pairs for their terminating points, denoting the direction angle (relative to the reference axis marked 0) and the radius (the distance from the origin). counterclockwise sense from the reference axis (labeled 0 here). By convention, we restrict the vector magnitude and direction to the ranges ra ≥ 0 and 0 ≤ qa < 2p If a vector's magnitude is 0, then the direction angle doesn't matter; the usual custom is to set it equal to 0. Special constraints When defining polar vectors, we must be more particular about what's "legal" and what's "illegal" than we were when defining polar points in Chap. 3. With polar vectors: • We don't allow negative magnitudes • We don't allow negative direction angles • We don't allow direction angles of 2p or larger These constraints ensure that the set of all polar-plane vectors can be paired off in a one-to-one correspondence (also called a bijection) with the set of all Cartesian-plane vectors. 64 Vector Basics Polar vector sum If we have two vectors in polar form, their sum can be found by following these steps, in order: 1. Convert both vectors to Cartesian coordinates 2. Add the vectors the Cartesian way 3. Convert the Cartesian vector sum back to polar coordinates Let's look at the situation in more formal terms. Suppose we have two vectors expressed in polar form as a = (qa,ra) and b = (qb,rb) To convert these vectors to Cartesian coordinates, we can use formulas adapted from the polar-to-Cartesian conversion we learned in Chap. 3. The modified formulas are (xa,ya) = [(ra cos qa),(ra sin qa)] and (xb,yb) = [(rb cos qb),(rb sin qb)] Once we have obtained the Cartesian ordered pairs, we add their elements individually to get a + b = [(xa + xb),(ya + yb)] Let's call this Cartesian sum vector c, and say that c = a + b = [(xa + xb),(ya + yb)] = (xc,yc) To convert c from Cartesian coordinates into polar coordinates, we can use the formulas given earlier in this chapter for the magnitude and direction angle of a vector in the xy plane. If we call the magnitude rc and the direction angle qc, we can write down the polar coordinates of sum vector as c = (qc, rc) An example Let's find the polar sum of the vectors a = (p /4,2) 66 Vector Basics Putting these coordinates into an ordered pair, we derive our final answer as a + b = [0,(2 × 21/2)] That's the polar sum of our original two polar vectors. The first coordinate is the angle in radians. The second coordinate is the magnitude in linear units. Polar vector difference When we want to subtract a polar vector from another polar vector, we follow these steps in order: 1. 2. 3. 4. Convert both vectors to Cartesian coordinates Find the Cartesian negative of the second vector Add the first vector to the negative of the second vector the Cartesian way Convert the resultant back to polar coordinates Once again, imagine that we have a = (qa,ra) and b = (qb,rb) The Cartesian equivalents are (xa,ya) = [(ra cos qa),(ra sin qa)] and (xb,yb) = [(rb cos qb),(rb sin qb)] We find the Cartesian negative of b as −b = (−xb,−yb) The difference vector a − b is therefore a − b = a + (−b) = {[(xa + (−xb)],[(ya + (−yb)]} = [(xa − xb),(ya − yb)] Let's call this difference vector d. We can say that d = a − b = [(xa − xb),(ya − yb)] = (xd,yd) We can skip the step where we find the Cartesian negative of the second vector and directly subtract the coordinate values, but we must take special care to keep the vectors and coordinate The "Polar Way" 67 values in the correct order if we do it that way. To convert d from Cartesian coordinates into polar coordinates, we can take advantage of the same formulas that we use to complete the process of polar vector addition. Polar negative of a vector Once in awhile, we'll want to find the negative of a vector the polar way. To do that, we reverse its direction and leave the magnitude the same. We can do this by adding p to the angle if it's at least 0 but less than p to begin with, or by subtracting p if it's at least p but less than 2p to begin with. In formal terms, suppose we have a polar vector a = (qa,ra) If 0 ≤ qa < p, then the polar negative is −a = [(qa + p ),ra] If p ≤ qa < 2p, then the polar negative is −a = [(qa − p ),ra] An example Let's find the polar difference a − b between the vectors a = (p /4,2) and b = (7p /4,2) In the addition example we finished a few minutes ago, we found that the Cartesian ordered pairs for these vectors are a = (21/2,21/2) and b = (21/2,−21/2) The negative of b is −b = (−21/2,21/2) When we add a to −b, we get a + (−b) = {[21/2 + (−21/2)],(21/2 + 21/2)} = [0,(2 × 21/2)] 68 Vector Basics That's the same as a − b. Let's call this Cartesian difference vector d = (xd,yd). Then xd = 0 and yd = 2 × 21/2 Using the Cartesian-to-polar conversion table, we can see that qd = p /2 and rd = (xd2 + yd2)1/2 = [02 + (2 × 21/2)2]1/2 = 2 × 21/2 The polar ordered pair is therefore a − b = [(p /2),(2 × 21/2)] The first coordinate is the angle in radians. The second coordinate is the magnitude in linear units. Are you confused? By now you might wonder, "What's the difference between a polar vector sum and a Cartesian vector sum? Or a polar vector negative and a Cartesian vector negative? Or a polar vector difference and a Cartesian vector difference? If we start with the same vector or vectors, shouldn't we get the same vector when we're finished calculating, whether we do it the polar way or the Cartesian way?" That's an excellent question. The answer is yes. The mathematical methods differ, but the resultant vectors are equivalent whether we work them out the polar way or the Cartesian way. Here's a challenge! Draw polar coordinate diagrams of the vector addition and subtraction facts we worked out in this section. Solution The original two polar vectors were a = (qa,ra) = (p /4,2) and b = (qb,rb) = (7p /4,2) The "Polar Way" 69 We found their polar sum to be a + b = [0,(2 × 21/2)] and their polar difference to be a − b = [(p /2),(2 × 21/2)] When we converted the two vectors to Cartesian form, we got a = (21/2,21/2) and b = (21/2,−21/2) We found their Cartesian sum to be a + b = [(2 × 21/2),0] and their Cartesian difference to be a − b = [0,(2 × 21/2)] We can illustrate the original vectors, the vector sum, the negative of the second vector, and the vector difference in four diagrams: • Figure 4-7 shows the polar sum, including a, b, and a + b. Figure 4-7 Polar sum of two vectors. Each radial division represents 1/2 unit. Practice Exercises 71 • Figure 4-10 shows the Cartesian difference, including a, b, −b, and a − b. y a – b = a + (–b) = [0, (2 × 21/2)] a = (21/2, 21/2) –b = (–21/2, 21/2) x Each axis division is 1/2 unit Figure 4-10. b = (21/2, –21/2) Cartesian difference between two vectors. Each axis division represents 1/2 unit vectors a and b in the Cartesian plane, with coordinates defined as follows: a = (−3,6) and b = (2,5) Work out, in strict detail, the Cartesian vector sums a + b, b + a, a − b, and b − a. 2. A vector is defined as the zero vector (denoted by a bold numeral 0) if and only if its magnitude is equal to 0. In the Cartesian plane, the zero vector is expressed as the ordered pair (0,0). Show that when a vector is added to its Cartesian negative in either order, the result is the zero vector. 72 Vector Basics 3. Imagine two arbitrary vectors a and b in the Cartesian plane, with coordinates defined as follows: a = (xa,ya) and b = (xb,yb) Show that the vector b − a is the Cartesian negative of the vector a − b. 4. Find the Cartesian sum of the vectors a = (4,5) and b = (−2,−3) Compare this with the sum of their negatives −a = (−4,−5) and −b = (2,3) 5. Prove that Cartesian vector negation distributes through Cartesian vector addition. That is, show that for two Cartesian vectors a and b, it's always true that −(a + b) = −a + (−b) 6. Find the polar sum of the vectors a = (p /2,4) and b = (p,3) 7. Find the polar negative of the vector a + b from the solution to Problem 6. 8. Find the polar negatives −a and −b of the vectors stated in Problem 6. 9. Find the polar sum of the vectors −a and −b from the solution to Problem 8. Compare this with the solution to Problem 7. 10. Find the polar differences a − b and b − a between the vectors stated in Problem 6. CHAPTER 5 Vector Multiplication We've seen how vectors add and subtract in two dimensions. In this chapter, we'll learn how to multiply a vector by a real number. Then we'll explore two different ways in which vectors can be multiplied by each other. Product of Scalar and Vector The simplest form of vector multiplication involves changing the magnitude by a real-number factor called a scalar. A scalar is a one-dimensional quantity that can be positive, negative, or zero. If the scalar is positive, the vector direction stays the same. If the scalar is negative, the vector direction reverses. If the scalar is zero, the vector disappears. Cartesian vector times positive scalar Imagine a standard-form vector a in the Cartesian xy plane, defined by an ordered pair whose coordinates are xa and ya, so that a = (xa,ya) Suppose that we multiply a positive scalar k+ by each of the vector coordinates individually, getting two new coordinates. Mathematically, we write this as k+a = (k+ xa,k+ ya) This vector is called the left-hand Cartesian product of k+ and a. If we multiply both original coordinates on the right by k+ instead, we get a k+ = (xak+,yak+) 73 74 Vector Multiplication That's the right-hand Cartesian product of a and k+. The individual coordinates of k+a and ak+ are products of real numbers. We learned in pre-algebra that real-number multiplication is commutative, so it follows that k+a = (k+ xa,k+ya) = (xak+,yak+) = a k+ We've just shown that multiplication of a Cartesian-plane vector by a positive scalar is commutative. We don't have to worry about whether we multiply on the left or the right; we can simply talk about the Cartesian product of the vector and the positive scalar. An example Figure 5-1 illustrates the Cartesian vector (−1,−2) as a solid, arrowed line segment. If we multiply this vector by 3 on the left, we get 3 × (−1,−2) = {[3 × (−1)],[3 × (−2)]} = (−3,−6) If we multiply the original vector by 3 on the right, we get (−1,−2) × 3 = [(−1 × 3)],(−2 × 3)] = (−3,−6) The new vector is shown as a dashed, gray, arrowed line segment pointing in the same direction as the original vector, but 3 times as long. Figure 5-1 Cartesian products of the scalars 3 and −3 with the vector (−1,−2). Product of Scalar and Vector 75 Cartesian vector times negative scalar Now suppose we want to multiply a by a negative scalar instead of a positive scalar. Let's call the scalar k−. The left-hand Cartesian product of k− and a is k−a = (k−xa,k−ya) The right-hand Cartesian product is a k− = (xak−,yak−) As with the positive constant, the commutative property of real-number multiplication tells us that k−a = (k− xa,k−ya) = (xak−,yak−) = ak− We don't have to worry about whether we multiply on the left or the right. We get the same result either way. An example Once again, look at Fig. 5-1 with the vector (−1,−2) shown as a solid, arrowed line segment. When we multiply it by the scalar −3 on the left, we obtain −3 × (−1,−2) = {[−3 × (−1)],[ −3 × (−2)]} = (3,6) Multiplying by the scalar on the right, we get (−1,−2) × (−3) = {[−1 × (−3)],[−2 × (−3)]} = (3,6) This result is shown as a dashed, gray, arrowed line segment pointing in the opposite direction from the original vector, and 3 times as long. Polar vector times positive scalar Imagine some vector a in the polar-coordinate plane whose direction angle is qa and whose magnitude is ra. If it's in standard form, we can express it as the ordered pair a = (qa,ra) When we multiply a on the left by a positive scalar k+, the angle remains the same, but the magnitude becomes k+ra. This gives us the left-hand polar product of k+ and a, which is k+a = (qa,k+ra) If we multiply a on the right by k+, we get the right-hand polar product of a and k+, which is a k+ = (qa,rak+) 76 Vector Multiplication Because real-number multiplication is commutative, we know that k+a = (qa,k+ra) = (qa,rak+) = a k+ As in the Cartesian case, we don't have to worry about whether we multiply on the left or the right. The polar product of the vector and the positive scalar is the same either way. An example In Fig. 5-2, the polar vector (7p /4,3/2) is shown as a solid, arrowed line segment. When we multiply this vector by 3 on the left, we get 3 × (7p /4,3/2) = [7p /4,(3 × 3/2)] = (7p /4,9/2) Multiplying by 3 on the right yields (7p /4,3/2) × 3 = {7p /4,[(3/2) × 3]} = (7p /4,9/2) This polar product vector is represented by a dashed, gray, arrowed line segment pointing in the same direction as the original vector, but 3 times as long. Polar vector times negative scalar Again, consider our polar vector a = (qa,ra). Suppose that we want to multiply a on the left by a negative scalar k−. It's tempting to suppose that we can leave the angle the same and make Figure 5-2 Polar products of the scalars 3 and −3 with the vector (7p /4,3/2). Each radial division represents 1 unit. Product of Scalar and Vector 77 the magnitude equal to k−ra. But that gives us a negative magnitude, which is forbidden by the rules we've accepted for polar vectors. The proper approach is to multiply the original vector magnitude ra by the absolute value of k−. In this situation, that's −k−. Then we reverse the direction of the vector by either adding or subtracting p to get a direction angle that's nonnegative but smaller than 2p. We define the result as the left-hand polar product of k− and a, and write it as; therefore −k−ra is positive, which ensures that our scalar-vector product has positive magnitude. If we multiply a on the right by k−, we get the right-hand polar product of a and k−, which is ak− = [(qa + p ),ra(−k−)] if 0 ≤ qa < p, and ak− = [(qa − p ),ra(−k−)] if p ≤ qa < 2p. As before, k− is negative so −k− is positive; that means ra(−k−) is positive, ensuring that our vector-scalar product has positive magnitude. The commutative law assures us that for any negative scalar k− and any polar vector a, it's always true that k−a = ak− As before, we can leave out the left-hand and right-hand jargon, and simply talk about the polar product of the vector and the scalar. An example Look again at Fig. 5-2. When we multiply the original polar vector (7p /4,3/2) by −3 on the left, we get −3 × (7p /4,3/2) = [(7p /4 − p),(3 × 3/2)] = (3p /4,9/2) Multiplying the original polar vector by −3 on the right yields (7p /4,3/2) × (−3) = {(7p /4 − p),[(3/2) × 3]} = (3p /4,9/2) This result is shown as a dashed, gray, arrowed line segment pointing in the opposite direction from the original vector, and 3 times as long. 78 Vector Multiplication Are you confused? You ask, "What happens when our positive scalar k+ is between 0 and 1? What happens when our negative scalar k− is between −1 and 0? What do we get if the scalar constant is 0?" If 0 < k+ < 1, the product vector points in the same direction as the original, but it's shorter. If −1 < k− < 0, the product vector points in the opposite direction from the original, and it's shorter. If we multiply a vector by 0, we get the zero vector. In all of these cases, it doesn't matter whether we work in the Cartesian plane or in the polar plane. Here's a challenge! Prove that the multiplication of a Cartesian-plane vector by a positive scalar is left-hand distributive over vector addition. That is, if k+ is a positive constant, and if a and b are Cartesian-plane vectors, then k+(a + b) = k+a + k+b Solution At first glance, this might seem like one of those facts that's intuitively obvious and difficult to prove. But all we have to do is work out some arithmetic with fancy characters. Let's start with k+(a + b) where k+ is a positive real number, a = (xa,ya), and b = (xb,yb). We can expand the vector sum into an ordered pair, writing the above expression as k+(a + b) = k+[(xa + xb),(ya + yb)] The definition of left-hand scalar multiplication of a Cartesian vector tells us that we can rewrite this as k+(a + b) = {[k+(xa + xb)],[k+(ya + yb)]} In pre-algebra, we learned that real-number multiplication is left-hand distributive over real-number addition, so we can morph the above equation to get k+(a + b) = [(k+xa + k+xb),(k+ya + k+yb)] Let's set this equation aside for a little while. We shouldn't forget about it, however, because we're going to come back to it shortly. Now, instead of the product of the scalar and the sum of the vectors, let's start with the sum of the scalar products k+a + k+b We can expand the individual vectors into ordered pairs to get k+a + k+b = k+(xa,ya) + k+(xb,yb) Dot Product of Two Vectors 79 The definition of left-hand scalar multiplication lets us rewrite this equation as k+a + k+b = (k+xa,k+ya) + (k+xb,k+yb) According to the definition of the Cartesian sum of vectors, we can add the elements of these ordered pairs individually to get a new ordered pair. That gives us k+a + k+b = [(k+xa + k+xb),(k+ya + k+yb)] Take a close look at the right-hand side of this equation. It's the same as the right-hand side of the equation we put into "brain memory" a minute ago. That equation was k+(a + b) = [(k+xa + k+xb),(k+ya + k+yb)] Taken together, the above two equations show us that k+(a + b) = k+a + k+b Dot Product of Two Vectors Mathematicians define two ways in which a vector can be multiplied by another vector. The simpler operation is called the dot product and is symbolized by a large dot (•). Sometimes it's called the scalar product because the end result is a scalar. Some texts refer to it as the inner product. Cartesian dot product Suppose we're given two standard-form vectors a and b in Cartesian coordinates, defined by the ordered pairs a = (xa,ya) and b = (xb,yb) The Cartesian dot product a • b is the real number we get when we multiply the x values by each other, multiply the y values by each other, and then add the two results. The formula is a • b = xa xb + ya yb An example Consider two standard-form vectors in the Cartesian xy plane, given by the ordered pairs a = (4,0) 80 Vector Multiplication and b = (3,4) In this case, xa = 4, xb = 3, ya = 0, and yb = 4. We calculate the dot product by plugging the numbers into the formula, getting a • b = (4 × 3) + (0 × 4) = 12 + 0 = 12 Polar dot product Now let's work in the polar-coordinate plane. Imagine two vectors defined by the ordered pairs a = (qa,ra) and b = (qb,rb) Let q b − qa be the angle between vectors a and b, expressed in a rotational sense starting at a and finishing at b as shown in Fig. 5-3. We calculate the polar dot product a • b by multiplying the magnitude of a by the magnitude of b, and then multiplying that result by the cosine of q b − qa to get a • b = rarb cos (q b − qa) p /2 (q a, ra) qb– qa a p 0 b (q b, rb) 3p /2 Figure 5-3 To find the polar dot product of two vectors, we must know the angle between them as we rotate from the first vector (in this case a) to the second vector (in this case b). Dot Product of Two Vectors 81 An example Suppose that we're given two vectors a and b in the polar plane, and told that their coordinates are a = (p /6,3) b = (5p /6,2) In this situation, ra = 3, rb = 2, qa = p /6, and qb = 5p /6. We have qb − qa = 5p /6 − p /6 = 2p /3 Therefore, the dot product is a • b = rarb cos (qb − qa) = 3 × 2 × cos (2p /3) = 3 × 2 × (−1/2) = −3 Are you confused? Do you wonder if the dot product of two polar-plane vectors is always equal to the dot product of the same vectors in the Cartesian plane when expressed in standard form? The answer is yes. Let's find out why. Here's a challenge! Prove that for any two vectors a and b in two-space, the polar dot product a • b is the same as the Cartesian dot product a • b when both vectors are in standard form. Solution We will start with the polar versions of the vectors, calling them a = (qa,ra) and b = (qb,rb) Let's convert these vectors to Cartesian form. We can use the formulas for conversion of points from polar to Cartesian coordinates (from Chap. 3). When we apply them to vector a, we get xa = ra cos qa and ya = ra sin qa so the standard Cartesian form of the vector is a = [(ra cos qa),(ra sin qa)] 82 Vector Multiplication When we apply the same conversion formulas to b, we obtain xb = rb cos qb and yb = rb sin qb so the standard Cartesian form is b = [(rb cos qb),(rb sin qb)] The Cartesian dot product of the two vectors is a • b = xa xb + ya yb Substituting the values we found for the individual vector coordinates, we get a • b = (ra cos qa)(rb cos qb) + (ra sin qa)(rb sin qb) = rarb (cos qa cos qb + sin qa sin qb) As we think back to our trigonometry courses, we recall that there's a trigonometric identity telling us how to expand the cosine of the difference between two angles. When we name the angles so they apply to our situation here, that formula becomes cos (qb − qa) = cos qa cos qb + sin qa sin qb We can substitute the left-hand side of this identity in the last part of the long equation we got a minute ago for the dot product, obtaining a • b = rarb cos (qb − qa) This is the formula for the polar dot product! We've taken the polar versions of a and b, found their Cartesian dot product, and then found that it's identical to the polar dot product. We can now say, Quod erat demonstradum. That's Latin for "Which was to be proved." Some mathematicians write the abbreviation for this expression, "QED," when they've finished a proof. Cross Product of Two Vectors The more complicated (and interesting) way to multiply two vectors by each other gives us a third vector that "jumps" out of the coordinate plane. This operation is known as the cross product. Some mathematicians call it the vector product. The cross product of two vectors a and b is written as a × b. Cross Product of Two Vectors 83 Polar cross product Imagine two arbitrary vectors in the polar-coordinate plane, expressed in standard form as ordered pairs a = (qa,ra) and b = (qb,rb) The magnitude of a × b is always nonnegative by default, and is easy to define. When a and b are in standard form, the originating point of a × b is at the coordinate origin, so all three vectors "start" at the same spot. The direction of a × b is always along the line passing through the origin at a right angle to the plane containing a and b. But it's quite a trick to figure out in which direction the cross vector product points along this line! Suppose that the difference qb − qa between the direction angles is positive but less than p, as shown in the example of Fig. 5-4. If we start at vector a and rotate until we get to vector b, we turn through an angle of qb − qa. To calculate the magnitude of a × b (which we will denote (q b, rb) p /2 0 < q b – qa < p b (q a, ra) a p 0 Cross product a ë b points ... 3p /2 ... straight toward us Figure 5-4 If qa < qb and the two angles differ by less than p, then a × b points straight toward us as we look down on the plane containing a and b. 84 Vector Multiplication as ra×b), we multiply the original vector magnitudes by each other, and then multiply by the sine of the difference angle. Mathematically, ra×b = rarb sin (q b − qa) In a situation of the sort shown in Fig. 5-4, the vector a × b points from the coordinate origin straight out of the page toward us. If qb − qa is larger than p, then things get a little bit complicated. To be sure that we assign the correct direction to the vector a × b, we must always rotate counterclockwise, and we're never allowed to turn through more than a half circle. Figure 5-5 shows an example. We rotate through one full circular turn minus qb − qa, so the difference angle is 2p − (qb − qa) which can be more simply written as 2p + qa − qb In a situation like this, a × b points straight away from us. p < q b – qa < 2 p p /2 (q a, ra) a p 0 2p + qa –q b b (q b, rb) Cross product a ë b points ... 3p /2 ... straight away from us Figure 5-5 If qa < qb and the two angles differ by more than p, then a × b points straight away from us as we look down on the plane containing a and b. 86 Vector Multiplication Are you confused? We haven't discussed how to directly calculate the cross product of two Cartesian-plane vectors. There's a way to do it, but we must know how to work with vectors in Cartesian three-space. We'll learn those techniques in Chap. 8. Meanwhile, we can indirectly find the cross product of two Cartesian-plane vectors by converting them both to polar form and then finding their cross product the polar way. Are you still confused? Here's a game that can help you find the direction of the cross product a × b (in that order) between two vectors a and b. It involves some maneuvers with your right hand. Some mathematicians, engineers, and physicists call this the right-hand rule for cross products. If 0 < qb − qa < p (as in Fig. 5-4), point your right thumb out as if you're making a thumbs-up sign. Curl your fingers in the counterclockwise rotational sense from a to b. Your thumb will point in the general direction of a × b. If the page on which the vectors are printed is horizontal, your thumb should point straight up. If p < qb − q a < 2p (as in Fig. 5-5), curl your right-hand fingers in the clockwise rotational sense from a to b. If the page on which the vectors are printed is horizontal, you'll have to twist your wrist in a clumsy fashion so that your thumb points straight down in the general direction of a × b. Remember that a × b always comes out of the origin precisely perpendicular to the plane containing a and b. Here's a challenge! A few moments ago, it was mentioned that if two vectors point in the same direction or in opposite directions, then their cross product is the zero vector. Prove it! Solution First, consider two vectors a and b that have the same direction angle q but different magnitudes ra and rb, so that a = (q,ra) and b = (q,rb) The magnitude of a × b is ra×b = rarb sin (q − q) = rarb sin 0 = rarb × 0 = 0 Whenever a vector has a magnitude of zero, then it's the zero vector by definition, so a×b=0 Cross Product of Two Vectors 87 Now look at the case where a and b have angles that differ by p, so they point in opposite directions. As before, you can assign the coordinates a = (q,ra) Two possibilities exist for the direction angle of a. You can have 0≤q<p or p ≤ q < 2p If 0 ≤ q < p, then b = [(q + p),rb] and the magnitude of a × b is ra×b = rarb sin [(q + p) − q] = rarb sin p = rarb × 0 = 0 Therefore a×b=0 If p ≤ q < 2p, then b = [(q − p),rb] In this case, the magnitude of a × b is ra×b = rarb sin [(q − p) − q] = rarb sin (−p) = rarb × 0 = 0 so again, a×b=0 Here's an "extra credit" challenge! Prove that the cross product of two vectors is anticommutative. That is, show that for any two polar-plane vectors a and b, the magnitudes of a × b and b × a are the same, but they point in opposite directions. Solution You're on your own. That's what makes this is an "extra credit" problem! 88 Vector Multiplication standard-form vectors a and b in the Cartesian plane, represented by the ordered pairs a = (5,−5) and b = (−5,5) Calculate and compare the Cartesian products 4a and −4b. 2. Convert the original two vectors from Problem 1 into polar form. Then calculate and compare the polar products 4a and −4b. 3. Prove that the multiplication of a standard-form vector by a positive scalar is right-hand distributive over Cartesian-plane vector subtraction. That is, if k+ is a positive constant, and if a and b are vectors in the xy plane, then (a − b)k+ = ak+ − bk+ 4. Consider two standard-form vectors a and b in the Cartesian plane, represented by a = (4,4) and b = (−7,7) Calculate and compare the Cartesian dot products a • b and b • a. 5. Convert the original two vectors from Problem 4 into polar form. Then calculate and compare the polar dot products a • b and b • a. 6. Prove that the dot product is commutative for standard-form vectors in the Cartesian plane. 7. Prove that the dot product is commutative for vectors in the polar plane. 8. Prove that if k+ is a positive constant, and if a and b are standard-form vectors in Cartesian or polar two-space, then k+a • k+b = k+2(a • b) Demonstrate the Cartesian case first, and then the polar case. CHAPTER 6 Complex Numbers and Vectors If you've had a comprehensive algebra course such as the predecessor to this book, Algebra Know-It-All, then you've been exposed to imaginary numbers and complex numbers. In this chapter, we'll take a closer look at how these quantities behave. Numbers with Two Parts A complex number consists of two components, the real part and the imaginary part. Complex numbers can be defined as ordered pairs and mapped one-to-one onto the points of a coordinate plane. They can also be represented as vectors. The unit imaginary number The set of imaginary numbers arises when we ask, "What is the square root of a negative real number?" This question poses a mystery to anyone who is familiar only with the real numbers. Unless we come up with some new sort of quantity, we have to say, "It's undefined." In order to define the square root of a negative real number, mathematicians invented the unit imaginary number, called it i, and defined it on the basis of the equation i 2 = −1 Once they had set down this rule, mathematicians explored how this strange new number behaved, and a new branch of number theory evolved. Engineers and physicists use j instead of i to denote the unit imaginary number. That's what we'll use, because the lowercase italic i is found in other mathematical contexts, particularly in sequences and series. The unit imaginary number j is equal to the positive square root of −1. That is, j = (−1)1/2 When we use the symbol j to represent the unit imaginary number, we can also call it the j operator, a term commonly used by engineers. 90 Numbers with Two Parts 91 The set of imaginary numbers We can multiply j by any real number, known as a real-number coefficient, and the result is an imaginary number. The real coefficient is customarily written after j if it is positive or 0, and after −j if it is negative. Examples are j3 = j × 3 = 3 × j −j5 = j × (−5) = −5 × j −j 2 /3 = j × (−2 /3) = −2 /3 × j j0 = j × 0 = 0 × j = 0 The set of all possible real-number multiples of j composes the set of imaginary numbers. For practical purposes, the elements of this set can be depicted along a number line corresponding one-to-one with the real-number line. By convention, the imaginary-number line is oriented vertically, as shown in Fig. 6-1. When either j or −j is multiplied by 0, the result is equal to the real number 0. Therefore, the intersection of the sets of imaginary and real numbers contains one element, namely, 0. j8 j6 j4 j2 j0 –j 2 –j 4 –j 6 –j 8 Center of continuous line Negative-imaginary numbers can be depicted as points on a vertical line. As we go upward, we get more positiveimaginary numbers; as we go downward, we get more negativeimaginary numbers. Positive-imaginary numbers Figure 6-1 Imaginary numbers 92 Complex Numbers and Vectors Complex numbers When we add a real number to an imaginary number, we get a complex number. The general form for a complex number is a + jb where a and b are real numbers. If the real-number coefficient of j happens to be negative, then its absolute value is written following j, and a minus sign is used instead of a plus sign in the composite expression. So instead of a + j(−b) we should write a − jb Individual complex numbers can be depicted as points on a Cartesian coordinate plane as shown in Fig. 6-2. The intersection point between the real- and imaginary-number lines corresponds Real part negative, imaginary part positive Real part positive, imaginary part positive j8 j6 j4 j2 –8 –6 –4 –2 –j 2 2 4 6 8 –j 4 Real part negative, imaginary part negative –j 6 –j 8 Real part positive, imaginary part negative Figure 6-2 Complex numbers can be depicted as points on a plane, which is defined by the intersection of perpendicular real- and imaginary-number lines. 94 Complex Numbers and Vectors Are you confused? We have learned that (−1)1/2 = j. You might now ask, "What about the square root of a negative real number other than −1, such as −4 or −100?" The positive square root of any negative real number is equal to j times the positive square root of the absolute value of that real number. For example, (−4)1/2 = j × 41/2 = j2 and (−100)1/2 = j × 1001/2 = j10 We can also have negative square roots of negative reals. That's because −j is not the same quantity as j. (You'll get a chance to prove this fact in Problem 1 at the end of this chapter.) Negating the above examples, we get −(−4)1/2 = −j × 41/2 = −j2 and −(−100)1/2 = −j × 1001/2 = −j10 Here's a challenge! Demonstrate what happens when −j is raised to successively higher positive-integer powers. Solution Keep in mind that −j is the negative square root of −1, which is −(−1)1/2. By definition, we know that j 2 = −1, so we can calculate the square of −j as (−j )2 = (−1 × j )2 = (−1)2 × j 2 = 1 × j2 = 1 × (−1) = −1 Now for the cube: (−j )3 = (−j )2 × (−j ) = −1 × (−j ) =j How Complex Numbers Behave 97 The square brackets, while technically superfluous, are included to visually set apart the real and imaginary parts of the result. We have just derived a general complex-number ratio formula that we can always use: (a + jb)/(c + jd ) = [(ac + bd )/(c + d 2)] + j [(bc − ad )/(c 2 + d 2)] 2 For this formula to work, the denominator must not be equal to 0 + j0. That means we cannot have both c = 0 and d = 0. If both of these coefficients are 0, then we end up dividing by 0. That operation, unlike the square root of a negative real, remains undefined, at least as far as this book is concerned! Complex number raised to positive-integer power If a + jb is a complex number and n is a positive integer, then (a + jb)n is the result of multiplying (a + jb) by itself n times. Complex conjugates Suppose we encounter two complex numbers that have the same coefficients, but opposite signs between the real and imaginary parts, as in a + jb and a − jb We call any two such quantities complex conjugates. They have some interesting properties. When we add a complex number to its conjugate, we get twice the real coefficient. In general, we have (a + jb) + (a − jb) = 2a When we multiply a complex number by its conjugate, we get the sum of the squares of the coefficients. In general, we have (a + jb)(a − jb) = a2 + b2 Complex conjugates are often encountered in engineering. They're especially useful in alternatingcurrent (AC) circuit, radio-frequency (RF) antenna, and transmission-line theories. Sum example Let's find the sum of the two complex numbers 5 + j4 and 2 − j3. When we add the real parts, we get 5+2=7 100 Complex Numbers and Vectors The cube can be expressed directly as (2 − j3)3 = −46 − j 9 Are you confused? When working with complex numbers, you should pay close attention to whether or not a numeral after the j operator is a superscript. The two notations are perilously similar! For example, if you see 5 + j2 it means 5 plus twice j, which is a complex number that's neither pure real nor pure imaginary. But if you see 5 + j2 it means 5 plus j squared, which can be simplified to 5 + (−1) or 4, which is pure real. Here's a challenge! Prove that the square of a complex number is equal to the square of the negative of that complex number. That is, show that (a + jb)2 = (−a − jb)2 for all real-number coefficients a and b. Solution First, let's work out the square of a + jb. We get (a + jb)2 = (a + jb)(a + jb) = a2 + jab + jba + j 2b2 = a2 + j 2ab − b2 = a2 − b2 + j 2ab Note that in the final term j2ab, the numeral 2 is a multiplier, not an exponent! Now let's find the square of −a − jb. This is a "nightmare of negatives," so we must be careful with the signs. We have (−a − jb)2 = (−a − jb)(−a − jb) = (−a)2 + (−a)(−jb) + (−jb)(−a) + (−jb)2 102 Complex Numbers and Vectors Polar model Any vector in the Cartesian plane can also be represented as a vector in the polar coordinate plane. Figure 6-5 shows the vectors from Fig. 6-4 plotted on the polar plane. The radial increments (shown as concentric circles) are the same size as the horizontal- and vertical-axis increments in the Cartesian plane of Fig. 6-4 (that is, 1 unit). The polar scheme is not as common as the Cartesian scheme. But it's equally valid if we restrict the direction angles to positive values less than 2p, and if we forbid negative vector magnitudes. The vectors in Fig. 6-5 theoretically represent the same complex numbers as those in Fig. 6-4. But the polar coordinates for a complex number differ from the Cartesian coordinates. The polar coordinates reflect the direction angle and magnitude of a vector, not the real and imaginary components. We can calculate the direction angle and magnitude of the polar vector if we know the real and imaginary parts of the equivalent complex number. We can also go the other way, and figure out the real and imaginary parts of the complex number if we know the polar vector direction angle and magnitude. Cartesian-to-polar complex vector conversion Imagine a complex number t = a + jb, represented as a vector tc in the Cartesian complexnumber plane, extending from the origin to the point (a,jb). We can derive the magnitude r of the equivalent polar vector tp by applying the Pythagorean distance formula to get r = (a2 + b2 )1/2 Figure 6-5 Complex numbers can be portrayed as vectors in the polar plane. Each radial division represents 1 unit. Cartesian coordinates are shown here. The polar coordinates are entirely different! Complex Vectors 103 To determine the direction angle q of the polar vector tp, we modify the polar-coordinate direction-finding system. Here's what we get. As we did in the Cartesian-to-polar coordinateconversion scheme, we define q = 0 by default when we're at the origin. That way, we get a one-to-one correspondence between the set of Cartesian vectors and the set of polar vectors. (Keep that in mind, because we'll keep doing this whenever the situation comes up!) q=0 by default When a = 0 and jb = j0 that is, at the origin q=0 When a > 0 and jb = j0 q = Arctan (b/a) When a > 0 and jb > j0 q = p /2 When a = 0 and jb > j0 q = p + Arctan (b/a) When a < 0 and jb > j0 q=p When a < 0 and jb = j0 q = p + Arctan (b/a) When a < 0 and jb < j0 q = 3p /2 When a = 0 and jb < j0 q = 2p + Arctan (b/a) When a > 0 and jb < j0 Polar-to-Cartesian complex vector conversion We can always convert a polar complex vector tp into a Cartesian complex vector tc that portrays a complex number a + jb in the familiar form. If we have tp = (q,r) then the Cartesian vector equivalent is tc = [(r cos q ), j(r sin q )] which represents the complex number a + jb = r cos q + j(r sin q ) The parentheses are not strictly necessary here, but they keep the real and imaginary components clearly separated. Absolute value We can find the absolute value of a complex number a + jb, written \| a + jb \|, by calculating the magnitude of its vector. In the Cartesian complex plane, going from the origin (0,0) to the point (a,jb), we have \| a + jb \| = (a 2 + b 2 )1/2 as shown in Fig. 6-6. In the polar plane, the absolute value of a complex vector is the vector radius r. 104 Complex Numbers and Vectors Figure 6-6 The absolute value of a complex number is the magnitude of its vector. jy a + jb b x a \| a + jb \| = ( a 2 + b 2 ) 1/2 = vector magnitude Complex vector sum and difference When we want to add or subtract two complex vectors, we can work on the Cartesian real and imaginary parts separately. If the vectors are presented to us in polar form, we should convert them to Cartesian form and then add. We can always convert the resultant back to polar form after we're done with the addition process. To find the difference between two complex vectors, we must be sure they're both in Cartesian form before we do any calculations. Once the vectors are in the Cartesian form, we take the negative of the second vector by negating both of its coordinates. Then we add the two resulting vectors. Again, if we want, we can convert the resultant back to polar form. Are you confused? You may ask, "Why isn't the addition and subtraction of polar coordinates directly done when we want to add or subtract complex vectors?" That's a good question. We can try to define vector sums and differences this way (adding or subtracting the polar angles and radii separately, for example), and we'll get output numbers when we grind out the arithmetic. But those numbers don't coincide with the geometric definitions of vector addition and subtraction. They don't give us the correct complex-number sums or differences. It's hard to say what those output numbers really mean, even though the idea is interesting! We should use Cartesian coordinates when we add or subtract complex vectors. We should use polar coordinates when we want to multiply or divide them. Polar complex vector product When we want to multiply two complex-number vectors, neither the dot product nor the cross product will give us the proper results. We must invent a new vector operation! Here's how it works. 1. We add the direction angles of the original two vectors to get the direction angle of the product vector. 2. If we end up with a direction angle larger than 2p, then we subtract 2p to get the correct angle for the product vector. Complex Vectors 105 3. We multiply the original vector magnitudes by each other to get the magnitude of the product vector. Polar complex vector ratio When we want to find the ratio of two complex numbers, we can go through the complex vector product process "inside-out." Again, there are three steps. 1. We subtract the direction angle of the denominator vector from the direction angle of the numerator vector to get the direction angle of the ratio vector. 2. If we end up with a negative direction angle, then we add 2p to get the correct angle for the ratio vector. 3. We divide the magnitude of the numerator vector by the magnitude of the denominator vector to get the magnitude of the ratio vector. Polar complex vector power When we want to raise a complex number to a positive-integer power, we multiply the polar angle by that positive integer, and then take the power of the magnitude. If the angle of our resulting vector is 2p or larger, we subtract whatever multiple of 2p is necessary to bring the angle into the range where it's positive but less than 2p. Absolute-value vector example There are infinitely many vectors that represent complex numbers having an absolute value of 6. All the vectors have magnitudes of 6, and they all point outward from the origin. Figure 6-7 shows a few such vectors. jy Set of all points corresponding to \| x + jy \| = 6 j4 j2 x –4 –2 2 4 –j 2 –j 4 Figure 6-7 There are infinitely many complex numbers with an absolute value of 6. They all terminate on a circle of radius 6, centered at the origin. Complex Vectors 107 p /2 Ratio = (3p /4, 4) p (p, 2) 0 (7p /4, 8) Subtract the angles ... 3p /2 ... and divide the magnitudes Figure 6-9 Ratio of the polar complex vector (7p /4,8) to the polar complex vector (p,2). Each radial division represents 1 unit. When we divide the magnitudes, we get 8/2 = 4 so the ratio vector is (3p /4,4). De Moivre's theorem The above schemes for finding products, ratios, and powers of polar complex numbers can be summarized in a famous theorem attributed to the French mathematician Abraham De Moivre, (pronounced "De Mwahvr"), who lived during the late 1600s and early 1700s. This theorem can be found in two different versions, depending on which text you consult. 108 Complex Numbers and Vectors where r1 and r2 are real-number polar magnitudes, and q1 and q2 are real-number polar direction angles in radians. Then the product of c1 and c2 is c1c2 = r1r2 cos (q1 + q2) + j [r1r2 sin (q1 + q2)] If r2 is nonzero, the ratio of c1 to c2 is c1/c2 = (r1/r2) cos (q1 − q2) + j [(r1/r2) sin (q1 − q2)] The second, and more commonly known, version of De Moivre's theorem can be derived from the first version. Suppose that we have a complex number c such that c = r cos q + j(r sin q) where r is the real-number polar magnitude and q is the real-number polar direction angle. Also suppose that n is an integer. Then c to the nth power is cn = rn cos (nq) + j[r n sin (nq)] I recommend that you enter this version of De Moivre's theorem into your "brain storage," and save it there forever! Are you confused? Do you wonder why we haven't described how to find a root of a complex vector? You might think, "It ought to be simple, just like finding a power backward. Can't we divide the polar angle by the index of the root, and then take the root of the magnitude?" That's a good question. Doing that will indeed give us a root. But there are often two or more complex roots for any given complex number. We're about to see an example of this. Here's a challenge! Cube the polar complex vectors (2p /3,1) and (4p /3,1). Here's a warning: The solution might come as a surprise! What do you suppose these results imply? Solution To cube a polar complex vector, we multiply the direction angle by 3 (the value of the exponent) and cube the magnitude. Let's do this with the vectors we've been given here. In the case of (2p /3,1)3, we get an angle of (2p /3) × 3 = 2p That's outside the allowed range of angles, but if we subtract 2p, we get 0, which is okay. We get a magnitude of 13 = 1. Now we know that (2p /3,1)3 = (0,1) Practice Exercises 109 where the first coordinate represents the direction angle in radians, and the second coordinate represents the magnitude. If we draw this vector on a polar graph, we can see that this is the polar representation of the complex number 1 + j0, which is equal to the pure real number 1. (If you like, you can use the conversion formulas to prove it.) In the case of (4p /3,1)3, we get the direction angle (4p /3) × 3 = 4p That's outside the allowed range of angles, but if we subtract 2p twice, then we get an angle of 0, and that's allowed. As before, we get a magnitude of 1 3 = 1. Now we know that (4p /3,1)3 = (0,1) where, again, the first coordinate represents the direction angle in radians, and the second coordinate represents the magnitude. This is the same as the previous result. It's the polar representation of 1 + j0, which is the pure real number 1. We've found two cube roots of 1 in the realm of the complex numbers. Neither of these roots show up when we work with pure real numbers exclusively. There are three different complex cube roots of 1! They are • The pure real number 1 • The complex number corresponding to the polar vector (2p /3,1) • The complex number corresponding to the polar vector (4p /3,1 Prove that −j is not equal to j, even though, when squared, they both give us −1. Here's a hint: Use the tactic of reductio ad absurdum, where a statement is proved by assuming its opposite and then deriving a contradiction from that assumption. 2. Show that the reciprocal of j is equal to its negative; that is, j −1 = −j. 3. Find the sum and difference of the complex numbers −3 + j4 and 1 + j5. 4. Find the ratio of the generalized complex conjugates a + jb and a − jb. That is, work out a general formula for (a + jb) / (a − jb) where a and b are both nonzero real numbers. 5. Prove that if we take any two complex conjugates and square them individually, the results are complex conjugates. In other words, show that for all real-number coefficients a and b, (a + jb)2 is the complex conjugate of (a − jb)2. 110 Complex Numbers and Vectors 6. Find the polar product of the polar complex vectors (p /4,21/2) and (3p /4,21/2). Then convert this product vector to Cartesian form and write down the "real-plus-imaginary" complex number that it represents. 7. Convert the polar complex vectors (p /4,21/2) and (3p /4,21/2) to the complex numbers they represent in "real-plus-imaginary" form. Multiply these numbers and compare with the solution to Problem 6. 8. Look at the results of the last "challenge," where we found these three cube roots of 1: • The pure real number 1 • The complex number corresponding to the polar vector (2p /3,1) • The complex number corresponding to the polar vector (4p /3,1) Convert the polar vectors (2p /3,1) and (4p /3,1) to their "real-plus-imaginary" complex-number forms. 9. Graph the three cube roots of 1 as polar complex vectors. Label them as ordered pairs in the form (q,r), where q is the direction angle and r is the magnitude. 10. Graph the three cube roots of 1 as Cartesian complex vectors. Label them as complex numbers in the form a + jb, where a and b are real numbers. Also graph the unit circle, and note that the vectors all terminate on that circle. CHAPTER 7 Cartesian Three-Space We can create three-dimensional graphs by adding a third axis perpendicular to the familiar x and y axes of the Cartesian plane. The new axis, usually called the z axis, passes through the xy plane at the origin, giving us Cartesian three-space or Cartesian xyz space. How It's Assembled Cartesian three-space has three real-number lines positioned so they all intersect at their zero points, and so each line is perpendicular to the other two. The point where the axes intersect constitutes the origin. Each axis portrays a real-number variable. Axes and variables Figure 7-1 is a perspective drawing of a Cartesian xyz space coordinate system. In a true-to-life three-dimensional portrayal, the positive x axis would run to the right, the negative x axis would run to the left, the positive y axis would run upward, the negative y axis would run downward, the positive z axis would project out from the page toward us, and the negative z axis would project behind the page away from us. In Cartesian three-space, the axes are all linear, and they're all graduated in increments of the same size. For any axis, the change in value is always directly proportional to the physical displacement. If we move 3 millimeters along an axis and the value changes by 1 unit, then that's true all along the axis, and it's also true everywhere along both of the other axes. If the divisions differ in size between the axes, then we have rectangular three-space, but not true Cartesian three-space. Cartesian three-space is often used to graph relations and functions having two independent variables. When this is done, x and y are usually the independent variables, and z is the dependent variable, whose value depends on both x and y. 111 112 Cartesian Three-Space +y –z –x +x +z –y Figure 7-1 A pictorial rendition of Cartesian three-space. In this view, the x axis increases positively from left to right, the y axis increases positively from the bottom up, and the z axis increases positively from far to near. Biaxial planes Cartesian three-space contains three flat biaxial (two-axis) planes that intersect along the coordinate axes. • The xy plane contains the axes for the variables x and y. • The xz plane contains the axes for the variables x and z. • The yz plane contains the axes for the variables y and z. You'll see three rectangles in Fig. 7-2, one parallel to each of the three biaxial planes. Look closely at how these rectangles are oriented. They can help you envision the orientations of the three biaxial planes in space. Each of the three biaxial planes is perpendicular to both of the others. Are you astute? Figure 7-2 shows an alternative perspective on Cartesian three-space, in which we're looking "up" toward the xz plane from somewhere near the negative y axis. There's a difference between the apparent positions of the axes in Fig. 7-2 as compared with their positions in Fig. 7-1, but the orientations of the three axes are the same with respect to each other. You should get used to seeing Cartesian three-space from various points of view. I'll switch points of view often to keep you thinking! How It's Assembled +z 113 Rectangle parallel to xz plane Rectangle parallel to yz plane +y –x +x Rectangle parallel to xy plane –y –z Figure 7-2 Cartesian three-space contains the xy, xz, and yz planes. This drawing shows rectangles parallel to each of these three biaxial planes. Note the difference in the point of view between this illustration and Figure 7-1. Points and ordered triples Figure 7-3 shows two specific points P and Q, plotted in Cartesian three-space. We've returned to the perspective of Fig. 7-1, with the positive z axis coming out of the page toward us. A point can always be denoted as an ordered triple in the form (x,y,z), according to the following scheme: • The x coordinate represents the point's projection onto the x axis. • The y coordinate represents the point's projection onto the y axis. • The z coordinate represents the point's projection onto the z axis. We get the projection of a point onto an axis by drawing a line from that point to the axis, and making sure that the line intersects that axis at a right angle. If this notion gives you trouble, you can think of the x, y, and z values for a particular point in the following way 114 Cartesian Three-Space Q (3, 5, –2) +y Each axis increment is 1 unit –z –x +x +z P (–5, –4, 3) –y Figure 7-3 Two points in Cartesian three-space, along with the corresponding ordered triples of the form (x,y,z). On all three axes, each increment represents 1 unit. Here, we've gone back to the point of view shown in Figure 7-1. In Fig. 7-3, the coordinates of point P are (–5,–4,3), and the coordinates of point Q are (3,5,–2). As the system is portrayed here, we can get to point P from the origin by making the following moves in any order: • Go 5 units in the negative x direction (straight to the left). • Go 4 units in the negative y direction (straight down). • Go 3 units in the positive z direction (straight out of the page). We can get from the origin to point Q by doing the following moves in any order: • Go 3 units in the positive x direction (straight to the right). • Go 5 units in the positive y direction (straight up). • Go 2 units in the negative z direction (straight back behind the page). If we were looking at the coordinate grid from a different viewpoint (that of Fig. 7-2, for example), our movements would look different, but the points and their coordinates would be the same. A note for the picayune An ordered triple represents the coordinates of a point in three-space, not the geometric point itself. But we may talk or write as if an ordered triple actually is a point, just as we sometimes think of a certain person when we read a name. That's okay, as long as we're aware of the semantical difference between the name and the point. How It's Assembled 115 Are you confused? Some people have trouble envisioning three-dimensional situations "in the mind's eye." If you're having problems understanding exactly how the three axes should relate in Cartesian three-space, here's a "pool rule" for the orientation of the axes. Imagine the origin of the system resting on the surface of a swimming pool. Suppose that we align the positive x axis so that it runs along the water surface, pointing due east. Once we've done that, the other axes are oriented as follows: • • • • •You can look at the coordinate axes from any point you want, whether on the surface, in the sky, or under the water. No matter how your view of the system changes, the actual orientation of the axes with respect to each other always stays the same. This relative axis orientation is important. If it's not strictly followed, we'll get into trouble when we work with graphs and vectors in Cartesian three-space. Here's a challenge! Imagine an ordered triple (x,y,z) where all three variables are nonzero real numbers. Suppose that you've plotted a point P in xyz space. Because x ≠ 0, y ≠ 0, and z ≠ 0, the point P doesn't lie on any of the axes. What will happen to the location of P if you • Multiply x by −1 and leave y and z the same? • Multiply y by −1 and leave x and z the same? • Multiply z by −1 and leave x and y the same? Solution Here's what will take place in each of these three situations. You can use Fig. 7-2 as a visual aid. If you're a computer whiz, maybe you can program your machine to create an animated display for each of these three processes: • If you multiply x by −1 and do not change the values of y or z, then point P will move parallel to the x axis to the opposite side of the yz plane, but P will end up at the same distance from the yz plane as it was before. • If you multiply y by −1 and do not change the values of x or z, then point P will move parallel to the y axis to the opposite side of the xz plane, but P will end up at the same distance from the xz plane as it was before. • If you multiply z by −1 and do not change the values of x or y, then point P will move parallel to the z axis to the opposite side of the xy plane, but P will end up at the same distance from the xy plane as it was before. 116 Cartesian Three-Space Distance of Point from Origin In Cartesian three-space, the distance of a point from the origin depends on all three of the coordinates in the ordered triple representing the point. The formula for this distance resembles the formula for the distance of a point from the origin in Cartesian two-space. The general formula It's not difficult to derive a general formula for the distance of a point from the origin in Cartesian three-space, as long as we're willing to use our "spatial mind's eye." Suppose we name the point P, and assign it the coordinates P = (xp,yp,zp) Figure 7-4A shows this situation, along with a point P* = (xp,yp,0), which is the projection of P onto the xy plane. We've moved again back to the perspective of Fig. 7-2, looking in toward the origin from somewhere far out in space near the negative y axis. To find the distance of P* from the origin, we can work entirely in the xy plane. This gives us a two-dimensional distance problem, which we learned how to handle in Chap. 1. Let's call the distance of P* from the origin by the name a. Using the formula we learned in Chap. 1 for the distance of a point from the origin in Cartesian xy plane, we have a = (xp2 + yp2)1/2 First, we find the distance from the origin to P, and call it a +z P (xp, yp, 0) +y –x +x a –y –z P (xp, yp, zp) Figure 7-4A Finding the distance of point P from the origin: step 1. Distance of Point from Origin 117 This completes the first step in a three-phase process. Figure 7-4B shows the second step. Here, we find the distance between P* and P. Let's call that distance b. It's the perpendicular distance of P from the xy plane, which is simply the coordinate value zp. Therefore, we have b = zp That's the end of the second step. In Fig. 7-4C, the distance from the origin to P is labeled c. Note that we now have a right triangle with sides of lengths a, b, and c. The right angle is between the sides whose lengths are a and b. The Pythagorean theorem therefore allows us to make the claim that a2 + b2 = c2 Substituting the previously determined values for a and b into this formula gives us [(xp2 + yp2)1/2]2 + zp2 = c2 which simplifies to xp2 + yp2 + zp2 = c2 +z Second, we find the distance from P* to P, and call it b P* (xp, yp, 0) +y –x +x a b –y –z P (xp, yp, zp) Figure 7-4B Finding the distance of point P from the origin: step 2. 118 Cartesian Three-Space Third, we call the distance from the origin to P by the name c ... +z P* (xp, yp, 0) +y –x +x a Right angle c b –y –z ... and note that c is the length of the hypotenuse of a right triangle! P (xp, yp, zp) Figure 7-4C Finding the distance of point P from the origin: step 3. When we switch the right-hand and left-hand sides of this equation and then take the 1/2 power of both sides, we get the formula we've been looking for, which is c = (xp2 + yp2 + zp2)1/2 An example Let's find the distance from the origin to the point P = (−5,−4,3) as shown in Fig. 7-3. We have xp = −5, yp = −4, and zp = 3. If we call the distance c, then c = (xp2 + yp2 + zp2)1/2 = [(−5)2 + (−4)2 + 32]1/2 = (25 + 16 + 9)1/2 = 501/2 Another example Now let's find the distance from the origin to Q = (3,5,−2) as shown in Fig. 7-3. This time, the coordinates are xq = 3, yq = 5, and zq = −2. We can again call the distance c, so c = (xq2 + yq2 + zq2)1/2 = [32 + 52 + (−2)2]1/2 = (9 + 25 + 4)1/2 = 381/2 Distance of Point from Origin 119 Are you confused? You might ask, "Can the distance of a point from the origin in Cartesian three-space ever be undefined? Can it ever be negative?" The answers are no, and no! Imagine a point P in Cartesian three-space— anywhere you want—with the coordinates (xp,yp,zp). To find the distance of P from the origin, you start by squaring xp, which is the x coordinate of P. Because xp is a real number, its square is a nonnegative real. Then you square yp, which is the y coordinate of P. This result must also be a nonnegative real. Then you square zp, which is the z coordinate of P. This square, too, is a nonnegative real. Next, you add the three nonnegative reals xp2, yp2, and zp2. That sum must be another nonnegative real. Finally, you take the nonnegative square root of the sum of the squares. The nonnegative square root of a nonnegative real number is always defined; and it's never negative itself, of course! Are you still confused? The formula we derived here is based on the idea that we start at the origin and go outward to point P. If we go inward from P to the origin, the distance is exactly the same. (If we were working with vectors, the vector displacements would be negatives of each other, but we're not there yet.) Here's a challenge! Suppose we're given a point P = (xp,yp,zp) in Cartesian three-space. Prove that if we negate any one, any two, or all three of the coordinates, the resulting point is the same distance from the origin as P. Solution For the point P, the distance c from the origin is c = (xp2 + yp2 + zp2)1/2 The square of any real number is always the same as the square of its negative. That tells us three things: (−xp)2 = xp2 (−yp)2 = yp2 (−zp)2 = zp2 By substitution, all these quantities are identical: (xp2 + yp2 + zp2)1/2 [(−xp)2 + yp2 + zp2]1/2 [xp2 + (−yp)2 + zp2]1/2 [xp2 + yp2 + (−zp)2]1/2 [(−xp)2 + (−yp)2 + zp2]1/2 [(−xp)2 + yp2 + (−zp)2]1/2 [xp2 + (−yp)2 + (−zp)2]1/2 [(−xp)2 + (−yp)2 + (−zp)2]1/2 120 Cartesian Three-Space These quantities represent the distances of the following points from the origin, respectively: (xp,yp,zp) (−xp,yp,zp) (xp,−yp,zp) (xp,yp,−zp) (−xp,−yp,zp) (−xp,yp,−zp) (xp,−yp,−zp) (−xp,−yp,−zp) That's all the points we can get, in addition to P itself, by negating any one, any two, or all three of the coordinates of P. They're all the same distance c from the origin, where c = (xp2 + yp2 + zp2)1/2 Distance between Any Two Points When we want to determine the distance between any two points in Cartesian three-space, we can expand the formula from Cartesian two-space that we learned in Chap. 1 into an extra dimension. The general formula Imagine two different points in Cartesian three-space, after the fashion of Fig. 7-5. Let's call the points and their coordinates P = (xp,yp,zp) and Q = (xq,yq,zq) where each coordinate can range over the entire set of real numbers. The distance d between these points, as we follow a straight-line path from P to Q, is d = [(xq − xp)2 + (yq − yp)2 + (zq − zp)2]1/2 If we start at Q and finish at P, we reverse the orders of subtraction, so the formula becomes d = [(xp − xq)2 + (yp − yq)2 + (zp − zq)2]1/2 We always subtract "starting coordinates" from "finishing coordinates." 122 Cartesian Three-Space Are you confused? You are probably not surprised that the distance between P and Q is the same in either direction. But you might ask, "Are there any situations where the distance between two points is different in one direction than in the other?" The answer is no, such a thing can never happen—as long as we always follow the same straight-line path through three-space to get from point to point. Let's prove that the direction doesn't matter when we want to express the distance between two points in space. Here's a challenge! Show that the distance between any two points in Cartesian three-space is the same, whichever direction we go. Solution It's sufficient to prove that for all real numbers xp, yp, zp, xq, yq, and zq, it's always the case that [(xq − xp)2 + (yq − yp)2 + (zq − zp)2]1/2 = [(xp − xq)2 + (yp − yq)2 + (zp − zq)2]1/2 Because xq − xp and xp − xq are negatives of each other, their squares are equal: (xq − xp)2 = (xp − xq)2 Because yq − yp and yp − yq are negatives of each other, their squares are equal: (yq − yp)2 = (yp − yq)2 Because zq − zp and zp − zq are negatives of each other, their squares are equal: (zq − zp)2 = (zp − zq)2 Based on these three facts, we know that the squared differences on both sides of the original equation are equal, no matter what the values of the coordinates might be (as long as they're all real numbers). This tells us that the distance between any two points is the same in either direction. Finding the Midpoint We can find the midpoint along a straight-line path between two points in Cartesian three-space by averaging the corresponding coordinates. Finding the Midpoint 123 The general formula Suppose that we want to find the midpoint M along a straight-line segment connecting two points P and Q as shown in Fig. 7-6. We can assign the points the ordered triples P = (xp,yp,zp) and Q = (xq,yq,zq) Let's say that the coordinates of the midpoint M are M = (xm,ym,zm) We find xm by averaging xp and xq, getting xm = (xp + xq)/2 We find ym by averaging yp and yq, getting ym = (yp + yq)/2 +z Q (xq, yq, zq) Point M is midway between points P and Q M +y –x +x What are the coordinates of point M ? –y P (xp, yp, zp) –z Figure 7-6 Midpoint of line segment connecting two points in Cartesian three-space. Finding the Midpoint 125 Are you a skeptic? Does it seem obvious that the midpoint between two points, say P and Q, doesn't depend on whether we go from P to Q or from Q to P? That's indeed the case; but if we demand proof, we must show that for real numbers xp, yp, zp, xq, yq, and zq, it's always true that [(xp + xq)/2,(yp + yq)/2,(zp + zq)/2] = [(xq + xp)/2,(yq + yp)/2,(zq + zp)/2] This proof is almost trivial, but it's good mental exercise to put it down in rigorous form. The commutative law for addition of real numbers tells us that xp + xq = xq + xp Dividing each side by 2 gives us (xp + xq)/2 = (xq + xp)/2 Using the same logic with the y and z coordinates, we get (yp + yq)/2 = (yq + yp)/2 and (zp + zq)/2 = (zq + zp)/2 Based on these facts, we know that the coordinates on both sides of the original equation are identical. It follows that the midpoint along a straight-line segment connecting any two points in Cartesian three-space is the same, regardless of which way we go. Here's a challenge! Imagine two points in Cartesian three-space where corresponding coordinates are negatives of each other. Show that the midpoint is exactly at the origin. Solution We can choose any point P whose coordinates are all real numbers. Let's suppose that P = (xp,yp,zp) Then the coordinates of Q are Q = (−xp,−yp,−zp) The coordinates of the midpoint M are (xm,ym,zm) = {[(xp + (−xp)]/2,[(yp + (−yp)]/2,[(zp + (−zp)/2]} = [(xp − xp)/2,(yp − yp)/2,(zp − zp)/2] = (0/2,0/2,0/2) = (0,0,0) The point (0,0,0) is, of course, the origin of the coordinate system. 126 Cartesian Three-Space What are the individual x, y, and z coordinates of the three points P, Q, and R shown in Fig. 7-7? +y Q (–5, 4, 0) Origin = (0, 0, 0) N L –z –x +x R (0, 0, 6) Each axis increment is 1 unit +z P (3, –3, 4) M –y Figure 7-7 Illustration for Problems 1 through 10. Each axis division represents 1 unit. 2. Determine the distance of the point P from the origin in Fig. 7-7. Using a calculator, approximate the answer by rounding off to three decimal places. 3. Determine the distance of the point Q from the origin in Fig. 7-7. Using a calculator, approximate the answer by rounding off to three decimal places. 4. Determine the distance of the point R from the origin in Fig. 7-7. This should come out exact, so you won't need a calculator! 5. Determine the length of the line segment L in Fig. 7-7. Using a calculator, approximate the answer by rounding off to three decimal places. 6. Determine the length of the line segment M in Fig. 7-7. Using a calculator, approximate the answer by rounding off to three decimal places. Practice Exercises 127 7. Determine the length of the line segment N in Fig. 7-7. Using a calculator, approximate the answer by rounding off to three decimal places. 8. Determine the coordinates of the midpoint of line segment L in Fig. 7-7. 9. Determine the coordinates of the midpoint of line segment M in Fig. 7-7. 10. Determine the coordinates of the midpoint of line segment N in Fig. 7-7. CHAPTER 8 Vectors in Cartesian Three-Space We've learned how to work with Cartesian coordinates in two and three dimensions, and we've learned about vectors in two dimensions. Now it's time to explore how vectors behave in Cartesian xyz space. How They're Defined Imagine two vectors a and b in three-dimensional space. We have infinitely many more direction possibilities now than we did in two-space! We can denote our vectors as arrowed line segments, "starting" at the origin (0,0,0) and "ending" at points (xa,ya,za) and (xb,yb,zb), as shown in Fig. 8-1. Cartesian standard form In Cartesian xyz space, vectors don't have to "start" at the coordinate origin, but there are advantages to putting them in that form. Any vector in this coordinate system, no matter where it "starts" and "ends," has an equivalent vector whose originating point is at (0,0,0). Such a vector is in Cartesian standard form. Suppose that we have a vector a′ that "starts" at a point P1 and "ends" at another point P2, with coordinates as P1 = (x1,y1,z1) and P2 = (x2,y2,z2) as shown in Fig. 8-2. The standard form of a′, denoted a, is defined by the terminating point Pa such that Pa = (xa,ya,za) = [(x2 − x1),(y2 − y1),(z2 − z1)] 128 +y a (xa, ya, za) –z –x +z +x b (xb, yb, zb) –y Figure 8-1 Two vectors in Cartesian xyz space. This is a perspective drawing (as are all three-space renditions in this book). In "real life," both vectors in this particular case would project generally toward us. +y Pa (xa, ya, za) a P2 (x2, y2, z2) –z +x –x a' P1 (x1, y1, z1) +z –y Figure 8-2 Two vectors in Cartesian xyz space. Vector a is in standard form because it "begins" at the origin (0,0,0). Vector a′ is equivalent to a, because both vectors are equally long, and they both point in the same direction. 129 130 Vectors in Cartesian Three-Space The two vectors a and a′ are equivalent, because they're equally long and they point in the same direction. Left-hand scalar multiplication Imagine the vector a in standard form, defined by (xa,ya,za) as shown in Fig. 8-3. Suppose that we want to multiply a positive real scalar k+ by the vector a. To do this, we multiply each coordinate by k+, getting k+a = k+(xa,ya,za) = (k+xa,k+ya,k+za) The direction of our vector a does not change, but it becomes k+ times as long. If we want to multiply a negative real scalar k− by a, then we follow the same procedure with that constant, obtaining k−a = k−(xa,ya,za) = (k−xa,k−ya,k−za) We've just described how to get the left-hand Cartesian product of a vector and a scalar in Cartesian xyz space. Whenever we multiply a negative scalar by a vector, we reverse the direction in which the vector points. We also change its length by a factor equal to the absolute value of the scalar. Vector a times positive constant greater than 1 +y Vector a (xa , ya, za) –z –x +x Vector a times negative constant less than –1 +z –y Figure 8-3 Multiplication of a standard-form vector by positive and negative real scalars in xyz space. How They're Defined 131 Right-hand scalar multiplication Now suppose that we multiply all three of the original vector coordinates on the right by a scalar k+. In this case, we get ak+ = (xak+,yak+,zak+) That's the right-hand Cartesian product of a and k+. If we multiply the original vector coordinates on the right by a negative constant k−, we get ak− = (xak−,yak−,zak−) That's the right-hand Cartesian product of a and k−. As you might guess, it doesn't matter whether we multiply a vector by a constant on the left or the right; we get the same result either way. Scalar multiplication of a vector is commutative. Magnitude Let's keep thinking about our vector a = (xa,ya,za) in Cartesian xyz space. If we make sure that a is in standard form, we can calculate its magnitude, which we'll denote as ra, by finding the distance of its terminating point from the origin. We learned how to do that in Chap. 7. We get ra = (xa2 + ya2 + za2)1/2 Here, the r stands for "radius." In some texts, vector magnitude is denoted by surrounding its name with absolute-value signs, or by changing the bold letter to a nonbold italic letter. Instead of ra, you might see the magnitude of a written as \|a\| or a. Direction The x, y, and z coordinates contain all the information we need to fully and uniquely define the direction of a vector in Cartesian three-space, as long as the vector is in standard form. But there's a more explicit way to do it. We can define the direction of a Cartesian three-space vector if we know the measures of the angles qx, qy, and qz that the vector subtends relative to the +x, +y, and +z axes, respectively, as shown in Fig. 8-4. These angles, expressed as an ordered triple (qx,qy,qz), are called direction angles. For any nonzero vector in xyz space, the direction angles are always nonnegative, and they're never larger than p. That means 0 ≤ qx ≤ p 0 ≤ qy ≤ p 0 ≤ qz ≤ p When we restrict the angles this way, we don't have to worry about whether we go clockwise or counterclockwise from the axes to the vectors. An example Imagine a nonstandard vector c′ in Cartesian three-space. Suppose that the originating point is (2,3,−7), and the terminating point is (−1,4, −1). Let's convert it to standard form, and call 132 Vectors in Cartesian Three-Space Figure 8-4 The direction of a vector in Cartesian xyz space is defined by the angles that the vector subtends with respect to each of the three positive axes. the resulting vector c. To get the terminating points of c, we must individually subtract the originating coordinates of c′ from the terminating coordinates of c′. The x coordinate of c is xc = −1 − 2 = −3 The y coordinate of c is yc = 4 − 3 = 1 The z coordinate of c is zc = −1 − (−7) = −1 + 7 = 6 Therefore, the standard form of c′ is c = (xc,yc,zc) = (−3,1,6) Another example Imagine the standard-form vector a = (2,3,4) in Cartesian xyz space. Suppose that we want to find the magnitude of this vector, accurate to three decimal places. We can assign it the coordinates xa = 2, ya = 3, and za = 4. Plugging these values into the magnitude formula, we get \|a\| = (xa2 + ya2 + za2)1/2 = (22 + 32 + 42)1/2 = (4 + 9 + 16)1/2 = 291/2 How They're Defined 133 Are you confused? You might wonder, "When we want to do operations with vectors that aren't in standard form, must we always convert them to standard form first?" Not always. Sometimes we'll get a valid result from a vector operation if we leave the vector or vectors in nonstandard form. But sometimes our answer will turn out wrong, and sometimes we won't be able to figure out what to do at all. The safest course of action is to do operations on vectors in xyz space only after they've been converted to standard form. Here's a challenge! Imagine three standard-form vectors a, b, and c in xyz space, defined by ordered triples as a = (4,0,0) b = (0,−5,0) c = (0,0,3) What are the direction angles of these vectors? Solution Figure 8-5 shows this situation. We can see that a lies along the positive x axis, b lies along the negative y axis, and c lies along the positive z axis. In Cartesian three-space, each of the coordinate Figure 8-5 Three standard-form vectors and their direction angles. Each vector lies along one of the coordinate axes. 134 Vectors in Cartesian Three-Space axes is perpendicular to the other two. This fact tells us that a subtends an angle of 0 with respect to the +x axis, an angle of p /2 with respect to the +y axis, and an angle of p /2 with respect to the +z axis. The direction angles of a are therefore (qxa,qya,qza) = (0,p /2,p /2) We know that b subtends an angle of p /2 relative to the +x axis, and an angle of p /2 with respect to the +z axis. Because b points along the negative y axis, its angle is p relative to the +y axis. The direction angles of b are therefore (qxb,qyb,qzb) = (p /2,p,p /2) Finally, vector c subtends an angle of p /2 against the +x axis, p /2 against the +y axis, and 0 against the +z axis, so the direction angles of c are (qxc,qyc,qzc) = (p /2,p /2,0) Remember that these ordered triples contain angle data in radians, not the x, y, and z coordinates for the terminating points! Sum and Difference When we want to add or subtract two vectors in Cartesian xyz space, we should make certain that they're both in standard form before we do anything else. Once we've gotten the vectors into standard form so that they both start at the origin, we can simply add or subtract the x, y, and z coordinates. Cartesian vector sum Suppose we have two generic three-space vectors in standard form, represented by ordered triples as a = (xa,ya,za) and b = (xb,yb,zb) Their vector sum is a + b = [(xa + xb),(ya + yb),(za + zb)] This sum can be found geometrically by constructing a parallelogram with vectors a and b as adjacent sides. The sum vector a + b is the diagonal of the parallelogram. An example is shown in Fig. 8-6. The figure doesn't look like a parallelogram because we're looking at it in Sum and Difference 135 +y a –z +x The four points lie at the vertices of a parallelogram (believe it or not!) a+b b +z –y Figure 8-6 Vector addition in Cartesian xyz space. The terminating points of the three vectors a, b, and a + b, along with the origin, lie at the vertices of a parallelogram. Perspective distorts the view. perspective, and from an oblique angle. All three vectors project generally in our direction; that is, they're all "coming out of the page." Cartesian negative of a vector To find the Cartesian negative of a standard-form vector in xyz space, we take the negatives of all three coordinate values. For example, if we have a = (xa,ya,za) then the Cartesian negative vector is −a = (−xa,−ya,−za) As in two-space, the Cartesian negative of a three-space vector always has the same magnitude as the original, but points in the opposite direction. Cartesian vector difference Let's look again at the two generic vectors a = (xa,ya,za) 136 Vectors in Cartesian Three-Space and b = (xb,yb,zb) Suppose we want to subtract b from a. We can do this by finding the Cartesian negative of b and then adding that result to a, getting a − b = a + (−b) = {[(xa + (−xb)],[(ya + (−yb)],[(za + (−zb)]} = [(xa − xb),(ya − yb),(za − zb)] We can skip the "find-the-negative" step and simply subtract the coordinate values, but we must be sure to keep the coordinates in the correct order if we do it that way. An example Let's look again at the three standard-form vectors that we worked with a few minutes ago. They are a = (4,0,0) b = (0,−5,0) c = (0,0,3) Suppose we want to find the sum vector a + b. We add the x, y, and z coordinates individually to get a + b = (4,0,0) + (0,−5,0) = {(4 + 0),[0 + (−5)],(0 + 0)} = (4,−5,0) If we add c to the right-hand side of this sum, we get (a + b) + c = (4,−5,0) + (0,0,3) = [(4 + 0),(−5 + 0),(0 + 3)] = (4,−5,3) Another example Continuing with the same three vectors as previously, let's find the sum b + c. We add the x, y, and z coordinates individually to get b + c = (0,−5,0) + (0,0,3) = [(0 + 0),(−5 + 0),(0 + 3)] = (0,−5,3) Adding a to the left-hand side of this sum, we obtain a + (b + c) = (4,0,0) + (0,−5,3) = {(4 + 0),[0 + (−5)],(0 + 3)} = (4,−5,3) Sum and Difference 137 Are you confused? The previous example might lead you to ask, "Is vector addition associative in xyz space, just as real-number addition is associative in ordinary algebra?" The answer is yes. The following proof will show you why. Here's a challenge! Show that if a, b, and c are standard-form vectors in Cartesian xyz space, then addition among them is associative. That is (a + b) + c = a + (b + c) Solution Let's begin by assigning generic names to the coordinates of each vector. Using the same style as we've been working with all along, we can say that a = (xa,ya,za) b = (xb,yb,zb) c = (xc,yc,zc) When we add a and b using the formula we've learned, we get a + b = [(xa + xb),(ya + yb),(za + zb)] Adding c to this sum on the right, again using the formula we've learned, we obtain (a + b) + c = {[(xa + xb) + xc],[(ya + yb) + yc],[(za + zb) + zc]} The associative law for addition of real numbers allows us to regroup each of the three coordinates in the ordered triple to get (a + b) + c = {[xa + (xb + xc)],[ya + (yb + yc)],[za + (zb + zc)]} By definition, we know that {[xa + (xb + xc)],[ya + (yb + yc)],[za + (zb + zc)]} = a + (b + c) By substitution, we have (a + b) + c = a + (b + c) 138 Vectors in Cartesian Three-Space Some Basic Properties Here are some fundamental laws that apply to vectors and real-number scalars in xyz space. We won't delve into the proofs. Most of these facts are intuitive, and resemble similar laws in algebra. We've already seen a couple of them, but they're repeated here so you can use this section for reference in the future. Keep in mind that all of these rules assume that the vectors are in standard form. Commutative law for vector addition When we add any two vectors in xyz space, it doesn't matter in which order the addition is done. The resultant vector is the same either way. If a and b are vectors, then a+b=b+a Commutative law for vector-scalar multiplication When we find the product of a vector and a scalar in xyz space, it doesn't matter which way we do it. If a is a vector and k is a scalar, then ka = ak Associative law for vector addition When we add up three vectors in xyz space, it makes no difference how we group them. If a, b, and c are vectors, then (a + b) + c = a + (b + c) Associative law for vector-scalar multiplication Suppose that we have two scalars k1 and k2, along with some vector a in Cartesian xyz space. If we want to find the product k1k2a, it makes no difference how we group the quantities. We can write this rule mathematically as k1k2a = (k1k2)a = k1(k2a) Distributive laws for scalar addition Imagine that we have some vector a in xyz space, along with two real-number scalars k1 and k2. We can always be sure that a(k1 + k2) = ak1 + ak2 and (k1 + k2)a = k1a + k2a The first rule is called the left-hand distributive law for multiplication of a vector by the sum of two scalars. The second law is called the right-hand distributive law for multiplication of the sum of two scalars by a vector. Some Basic Properties 139 Distributive laws for vector addition Suppose we have two vectors a and b in xyz space, along with a real-number scalar k. We can always be certain that k(a + b) = ka + kb and (a + b)k = ak + bk The first rule is called the left-hand distributive law for multiplication of a scalar by the sum of two vectors. The second law is called the right-hand distributive law for multiplication of the sum of two vectors by a scalar. Unit vectors Let's take a close look at the "structures" of two different vectors a and b in xyz space, both of which are expressed in the standard form. Suppose that their coordinates can be written as the familiar generic ordered triples a = (xa,ya,za) and b = (xb,yb,zb) Either of these vectors can be split up into a sum of three component vectors, each of which lies along one of the coordinate axes. The component vectors are scalar multiples of mutually perpendicular vectors with magnitude 1. We haveand b = (xb,yb,zb) = (xb,0,0) + (0,yb,0) + (0,0,zb) = xb(1,0,0) + yb(0,1,0) + zb(0,0,1) The three vectors (1,0,0), (0,1,0), and (0,0,1) are called standard unit vectors. (We can call them SUVs for short.) It's customary to name them i, j, and k, such that i = (1,0,0) j = (0,1,0) k = (0,0,1) 140 Vectors in Cartesian Three-Space Figure 8-7 The three standard unit vectors i, j, and k in Cartesian xyz space. Figure 8-7 illustrates the coordinates and direction angles of the three SUVs in Cartesian three-space, where each axis division represents 1/5 of a unit. Note that each SUV is perpendicular to the other two. A generic example Let's see what happens when we add two generic vectors component-by-component. Again, suppose we have a = (xa,ya,za) and b = (xb,yb,zb) Expressed as sums of multiples of the SUVs, these two vectors are a = (xa,ya,za) = xai + yaj + zak Dot Product 141 and b = (xb,yb,zb) = xbi + ybj + zbk When we add these components straightaway, we get a + b = xai + yaj + zak + xbi + ybj + zbk The commutative law for vector addition allows us to rearrange the addends on the righthand side of this equation to get a + b = xai + xbi + yaj + ybj + zak + zbk Now let's use the right-hand distributive law for multiplication of the sum of two scalars by a vector to morph the previous equation into a + b = (xa + xb)i + (ya + yb)j + (za + zb)k That's the sum of the original vectors, expressed as a sum of multiples of SUVs. A specific example Suppose we're given a vector b = (−2,3,−7), and we're told to break it into a sum of multiples of i, j, and k. We can imagine i as going 1 unit "to the right," j as going 1 unit "upward," and k as going 1 unit "toward us." The breakdown proceeds as follows: b = (−2,3,−7) = −2 × (1,0,0) + 3 × (0,1,0) + (−7) × (0,0,1) = −2i + 3j + (−7)k = −2i + 3j − 7k Are you confused? By now you might wonder, "Must I memorize all of the rules mentioned in this section?" Not necessarily. You can always come back to these pages for reference. But honestly, I recommend that you do memorize them. If you take a lot of physics or engineering courses later on, you'll be glad that you did. Dot Product As we've been doing throughout this chapter, let's revisit our generic standard-form vectors in xyz space, defined as a = (xa,ya,za) 142 Vectors in Cartesian Three-Space and b = (xb,yb,zb) We can calculate the dot product a and b as determined in the plane containing them both, rotating from a to b. An example Let's find the dot product of the two Cartesian vectors a = (2,3,4) and b = (−1,5,0) We can call the coordinates xa = 2, ya = 3, za = 4, xb = −1, yb = 5, and zb = 0. Plugging these values into the formula, we get a • b = xaxb + yayb + zazb = 2 × (−1) + 3 × 5 + 4 × 0 = −2 + 15 + 0 = 13 Another example Suppose we want to find the dot product of the two Cartesian vectors a = (−4,1,−3) and b = (−3,6,6) This time, we have xa = −4, ya = 1, za = −3, xb = −3, yb = 6, and zb = 6. When we substitute these coordinates into the formula, we have a • b = xaxb + yayb + zazb = −4 × (−3) + 1 × 6 + (−3) × 6 = 12 + 6 + (−18) = 0 Dot Product 143 Are you confused? Do you wonder how two nonzero vectors can have a dot product of 0? If we look closely at the alternative formula for the dot product, we can figure it out. That formula, once again, is a • b = rarb cos qab The right-hand side of this equation will attain a value of 0 if at least one of the following is true: • The magnitude of a is equal to 0 • The magnitude of b is equal to 0 • The cosine of the angle between a and b is equal to 0 Neither of the vectors in the preceding example has a magnitude of 0, so we must conclude that cos qab = 0. That can happen only when a and b are perpendicular to each other, so qab is either p /2 or 3p /2. In the preceding example, the two vectors a = (−4,1,−3) and b = (−3,6,6) are mutually perpendicular. That's not obvious from the ordered triples, is it? Here's a challenge! Show that for any two vectors pointing in the same direction, their dot product is equal to the product of their magnitudes. Then show that for any two vectors pointing in opposite directions, their dot product is equal to the negative of the product of their magnitudes. Solution Imagine two vectors a and b that point in the same direction. In this situation, the angle qab between the vectors is equal to 0. If the magnitude of a is ra and the magnitude of b is rb, then the dot product is a • b = rarb cos qab = rarb cos 0 = rarb × 1 = rarb Now think of two vectors c and d that point in opposite directions. The angle qcd between the vectors is equal to p. If the magnitude of c is rc and the magnitude of d is rd, then the dot product is c • d = rcrd cos qcd = rcrd cos p = rcrd × (−1) = −rcrd 144 Vectors in Cartesian Three-Space Cross Product The cross product a ë b of two vectors a and b in three-dimensional space can be found according to the same rules we learned for finding a cross product in polar two-space. We get a vector perpendicular to the plane containing a and b, and whose magnitude ra×b is given by ra×b = rarb sin qab where ra is the magnitude of a, rb is the magnitude of b, and qab is the angle between a and b, expressed in the rotational sense going from a to b. When we want to figure out a cross product, it's always best to keep the angle between the vectors nonnegative, but not larger than p. That is, we should restrict the angle to the following range: 0 ≤ qab ≤ p If we look at vectors a and b from some vantage point far away from the plane containing them, and if qab turns through a half circle or less counterclockwise as we go from a to b, then a ë b points toward us. If qab turns through a half circle or less clockwise as we go from a to b, then a ë b points away from us. In any case, the cross product vector is precisely perpendicular to both the original vectors. An example Consider two vectors a and b in three-space. Imagine that they both have magnitude 2, but their directions differ by p /6. We can plug the numbers into the formula for the magnitude of the cross product of two vectors, and calculate as follows: ra×b = rarb sin qab = 2 × 2 × sin (p /6) = 4 × 1/2 = 2 If the p /6 angular rotation from a to b goes counterclockwise as we observe it, then a ë b points toward us. If the p /6 angular rotation from a to b goes clockwise as we see it, then a ë b points away from us. Another example Now think about two vectors c and d, represented by ordered triples as c = (1,1,1) and d = (−2,−2,−2) Let's find the cross product c ë d. From the information we've been given, we can see immediately that d = -2c. That means the magnitude of d is twice the magnitude of c, and the two vectors point in opposite directions. We can calculate the magnitude rc of vector c as rc = (12 + 12 + 12)1/2 = (1 + 1 + 1)1/2 = 31/2 Cross Product 145 and the magnitude rd of vector d as rd = [(−2)2 + (−2)2 + (−2)2]1/2 = (4 + 4 + 4)1/2 = 121/2 When two vectors point in opposite directions, the angle between them is p, whether we go clockwise or counterclockwise. We now have all the information we need to figure out the magnitude rc×d of the cross product c ë d using the formula rc×d = rcrd sin qcd = 31/2 × 121/2 × sin p = 31/2 × 121/2 × 0 = 0 The cross product c ë d is the zero vector, because its magnitude is 0. Although we don't yet have a formula for figuring out cross products directly from ordered triples in xyz space, we can infer from this result that (1,1,1) ë (−2,−2,−2) = (0,0,0) where the bold times sign (ë) denotes the cross product, not ordinary multiplication. Are you confused? The preceding result might make you wonder, "If two vectors point in exactly the same direction or in exactly opposite directions is their cross product always the zero vector?" The answer is yes, and it doesn't depend on the magnitudes of the original two vectors. Let's prove this fact now. Here's a challenge! Show that the cross product of any two vectors that point in the same direction or in opposite directions, regardless of their magnitudes, is the zero vector. Solution When two vectors a and b point in the same direction, the angle qab between them is 0. In such a situation, the magnitude ra×b of the cross product is ra×b = rarb sin qab = rarb sin 0 = rarb × 0 = 0 Therefore, a ë b = 0, because if a vector has a magnitude of 0, then it's the zero vector by definition. When two vectors c and d point in opposite directions, the angle qcd between them is p, so the magnitude rc×d of the cross product is rc×d = rcrd sin qcd = rcrd sin p = rcrd × 0 = 0 Again, we have c ë d = 0 by definition. 146 Vectors in Cartesian Three-Space Some More Vector Laws Here are some more rules involving vectors. You'll find these useful for future reference if you get serious about higher mathematics, physical science, or engineering. Commutative law for dot product When we figure out the dot product of two vectors, it doesn't matter in which order we work it. The result is the same either way. If a and b are vectors in three-space, then a•b=b•a Reverse-directional commutative law for cross product Suppose qab is the angle between two vectors a and b as defined in the plane containing a and b, such that 0 ≤ qab ≤ p, and such that we're allowed to rotate in either direction. The magnitude of the cross-product vector is a nonnegative real number, and is independent of the order in which the operation is performed. This can be proven on the basis of the commutative property for multiplication of real numbers. We have ra×b = rarb sin qab and rb×a = rbra sin qab = rarb sin qab The direction of b ë a in space is exactly opposite that of a ë b. Figure 8-8 can help us see why this is true when we apply the right-hand rule for cross products (from Chap. 5) both ways. Figure 8-8 The vector b ë a has the same magnitude as vector a ë b, but points in the opposite direction. Some More Vector Laws 147 Distributive laws for dot product over vector addition Imagine that we have three vectors a, b, and c in three-space. We can always be sure that a • (b + c) = (a • b) + (a • c) This fact is called the left-hand distributive law for a dot product over the sum of two vectors. It's also true that (a + b) • c = (a • c) + (b • c) which, as you can probably guess, is the right-hand distributive law for the sum of two vectors over a dot product. Distributive laws for cross product over vector addition Suppose that a, b, and c are vectors in three-space. Then we can always be sure that a × (b + c) = (a × b) + (a × c) This property is known as the left-hand distributive law for a cross product over the sum of two vectors. A similar rule exists when we cross multiply a sum of vectors on the right. The right-hand distributive law for the sum of two vectors over a cross product tells us that (a + b) × c = (a × c) + (b × c) We can expand these rules to pairs of polynomial vector sums, each having n addends (where n = 2, n = 3, n = 4, etc.), in the same way as multiplication is distributive with respect to addition for polynomials in algebra. For example, for n = 2, we have the cross product of two binomial vector sums, getting (a + b) × (c + d) = (a × c) + (a × d) + (b × c) + (b × d) In the case of n = 3, the cross product of two trinomial vector sums expands as (a + b + c) × (d + e + f ) = (a × d) + (a × e) + (a × f ) + (b × d) + (b × e) + (b × f ) + (c × d) + (c × e) + (c × f ) Dot product of cross products Imagine that we have four vectors a, b, c, and d in three-space. We can rearrange a dot product of cross products as (a × b) • (c × d) = (a • c)(b • d) − (a • d)(b • c) We always end up with a scalar quantity (that is, a real number). 148 Vectors in Cartesian Three-Space Dot product of mixed vectors and scalars Suppose that t and u are real numbers, and we have two three-space vectors a and b. We can rearrange a dot product of scalar multiples as ta • ub = tu(a • b) The result is always a scalar. Cross product of mixed vectors and scalars Once again, imagine that t and u are real numbers, and we have two three-space vectors a and b. We can rearrange a cross product of scalar multiples as ta ë ub = tu(a ë b) The result is always a vector quantity. Here's a challenge! Imagine two vectors in Cartesian xyz space whose coordinates are expressed as a = (xa,ya,za) and b = (xb,yb,zb) Derive a general expression for a ë b in the form of an ordered triple. Solution Let's go back to the concept of SUVs that we learned earlier in this chapter. These vectors are i = (1,0,0) j = (0,1,0) k = (0,0,1) Now let's evaluate and list all the cross products we can get from these vectors. Using the righthand rule for cross products (from Chap. 5) along with the formula for the magnitude of the cross product of vectors, we can deduce, along with the help of Fig. 8-7 on page 140, that i×j=k j × i = −k i × k = −j k×i=j j×k=i k × j = −i Some More Vector Laws 149 We can write the cross product a ë b as a ë b = (xa,ya,za) ë (xb,yb,zb) = (xai + yaj + zak) ë (xbi + ybj + zbk) Using the left-hand distributive law for the cross product over vector addition as it applies to trinomials, we can expand this to a ë b = (xai ë xbi) + (xai ë ybj) + (xai ë zbk) + (yaj ë xbi) + (yaj ë ybj) + (yaj ë zbk) + (zak ë xbi) + (zak ë ybj) + (zak ë zbk) With our newfound knowledge of how scalar multiplication and cross products can be mixed (see "Cross product of mixed vectors and scalars"), we can morph each of the terms after the equals sign to get a ë b = xaxb(i ë i) + xayb(i ë j) + xazb(i ë k) + yaxb(j ë i) + yayb(j ë j) + yazb(j ë k) + zaxb(k ë i) + zayb(k ë j) + zazb(k ë k) A few moments ago, we proved that we always get the zero vector if we take the cross product of any vector with another vector pointing in the same direction. That means the cross product of any vector with itself is the zero vector. Because the zero vector has zero magnitude, we get the zero vector if we multiply it by any scalar. With all this information in mind, we can rewrite the previous equation as a × b = 0 + xayb(i × j) + xazb(i × k) + yaxb(j × i) + 0 + yazb(j × k) + zaxb(k × i) + zayb(k × j) + 0 Looking back at the six "factoids" involving pairwise cross products of i, j, and k, and getting rid of the zero vectors in the previous equation, we can simplify it to a ë b = xaybk + xazb(−j) + yaxb(−k) + yazbi + zaxbj + zayb(−i) Rearranging the signs, we obtain a ë b = xaybk − xazbj − yaxbk + yazbi + zaxbj − zaybi This can be morphed a little more, based on rules we've learned in this chapter, getting a ë b = (yazb − zayb)i + (zaxb − xazb)j + (xayb − yaxb)k This SUV-based equation tells us three things: • The x coordinate of a ë b is yazb − zayb • The y coordinate of a ë b is zaxb − xazb • The z coordinate of a ë b is xayb − yaxb 150 Vectors in Cartesian Three-Space Knowing these three facts, we can write the x, y, and z coordinates of a ë b as an ordered triple to get a ë b = [(yazb − zayb),(zaxb − xazb),(xayb − yaxb)] We've found a formula that allows us to directly calculate the cross product of two vectors in xyz space when we're given both vectors as ordered triples. Here's an extra-credit challenge! The formulas for the seven laws in this section were stated straightaway. We didn't show how they are derived. If you're ambitious (and you have a good pen along with plenty of blank sheets of paper), derive these seven laws by working out the general arithmetic step by step. Following are the names of those laws again, for reference: • • • • • • • Commutative law for dot product Reverse-directional commutative law for cross product Distributive laws for dot product over vector addition Distributive laws for cross product over vector addition Dot product of cross products Dot product of mixed vectors and scalars Cross product of mixed vectors and scalars Solution You're on your own. That's why you get extra credit! Here's a hint: The work is rather tedious, but it's straightforward Find the magnitude ra of the standard-form vector a = (8,−1,−6) in Cartesian xyz space. Assume the values given are exact. Using a calculator, round off the answer to three decimal places. 2. Imagine a nonstandard vector a′ that originates at (−2,0,4) and terminates at the origin. Convert a′ to standard form. 3. What's the standard form of the product 4b′, where b′ originates at (2,3,4) and terminates at (6,7,8)? Here's a hint: Convert b′ to its standard form b first, and then multiply that vector by 4. Practice Exercises 151 4. Consider the two standard-form vectors a = (−7,−10,0) and b = (8,−1,−6) in xyz space. What is their dot product? 5. Consider the two standard-form vectors a = (2,6,0) and b = (7,4,3) in xyz space. What is their cross product? 6. Imagine two standard-form vectors f and g that point in the same direction in threespace. Suppose that the magnitude rf of f is equal to 4, and the magnitude rg of g is equal to 7. What is f • g? 7. Imagine two standard-form vectors f and g that point in opposite directions in threespace. Suppose that the magnitude rf of vector f is equal to 4, and the magnitude rg of vector g is equal to 7. What is f • g? 8 • g? What is g • f ? 9 ë g? What is g ë f ? 10. Consider two standard-form vectors a and b that both lie in the xy plane within Cartesian xyz space. Suppose that a = (2,0,0), so it points along the +x axis. Suppose that b has magnitude 2 and rotates counterclockwise in the xy plane, starting at (2,0,0), then going around through (0,2,0), (−2,0,0), and (0,−2,0), finally ending up back at (2,0,0). Now imagine that we watch all this activity from somewhere high above the xy plane, near the +z axis. Describe what happens to the cross product vector a ë b as vector b goes through a complete counterclockwise rotation. What will we see if b keeps rotating counterclockwise indefinitely? CHAPTER 9 Alternative Three-Space We can define the locations of points in three dimensions by methods other than the Cartesian system. In this chapter, we'll learn about the two most common alternative coordinate schemes for three-space. Cylindrical Coordinates Figure 9-1 is a functional diagram of a system of cylindrical coordinates. It's basically a polar coordinate plane of the sort we learned about in Chap. 3, with the addition of a height axis to define the third dimension. How it works To set up a cylindrical coordinate system, we "paste" a polar plane onto a Cartesian xy plane, creating a reference plane. We call the positive Cartesian x axis the reference axis. Imagine a point P in three-space, along with its projection point P ′ onto the reference plane. In this context, the term "projection" means that P ′ is directly above or below P, so a line connecting the two points is perpendicular to the reference plane. We define three coordinates: • The direction angle, which we call q, is the angle in the reference plane as we turn counterclockwise from the reference axis to the ray that goes out from the origin through P ′. • The radius, which we call r, is the straight-line distance from the origin to P ′. • The height, which we call h, is the vertical displacement (positive, negative, or zero) from P ′ to P. These three coordinates give us enough information to uniquely define the position of P as shown in Fig. 9-1. We express the cylindrical coordinates as an ordered triple P = (q,r,h) 152 Cylindrical Coordinates 153 Figure 9-1 Cylindrical coordinates define points in three dimensions according to an angle, a radial distance, and a vertical displacement. Strange values We can have nonstandard direction angles in cylindrical coordinates, but it's best to add or subtract whatever multiple of 2p to bring the angle into the preferred range of 0 ≤ q < 2p. If q ≥ 2p, then we're making at least one complete counterclockwise rotation from the reference axis. If q < 0, then we're rotating clockwise from the reference axis rather than counterclockwise. We can have negative radii, but it's best to reverse the direction angle if necessary to keep the radius nonnegative. We can multiply a negative radius coordinate by −1 so it becomes positive, and then add or subtract p to or from the direction angle to ensure that 0 ≤ q < 2p. The height h can be any real number. We have h > 0 if and only if P is above the reference plane, h < 0 if and only if P is below the reference plane, and h = 0 if and only if P is in the reference plane. An example In the situation shown by Fig. 9-1, the direction angle q appears to be somewhat more than p (half of a rotation from the reference axis) but less than 3p /2 (three-quarters of a rotation). The radius r is positive, but we can't tell how large it is because there are no coordinate increments for reference. The height h is also positive, but again, we don't know its exact value because there are no reference increments. 153 154 Alternative Three-Space Cylinder extends upward forever +z r=k Constant radius +y –x +x Reference plane –y Cylinder extends downward forever Figure 9-2 –z When we set the radius equal to a constant in cylindrical coordinates, we get an infinitely tall vertical cylinder whose axis corresponds to the vertical axis. Another example In Chap. 3, we learned that the equation of a circle in polar two-space is simple; all we have to do is specify a radius. If we do the same thing in cylindrical three-space, we get a vertical cylinder that's infinitely tall, with an axis that corresponds to the vertical coordinate axis. Figure 9-2 shows what we get when we graph the equation r=k in cylindrical three-space, where k is a nonzero constant. Still another example If we set the height equal to a nonzero constant in cylindrical coordinates, we get the set of all points at a specific distance either above or below the reference plane. That's always a plane parallel to the reference plane. Figure 9-3 is an example of the generic situation where h=k In this case, k is a positive real-number constant, but we don't know the exact value because the graph doesn't show us any reference increments for the height coordinate. Cylindrical Coordinates +z h=k Constant height 155 Plane extends forever in all directions +y –x +x Reference plane –y –z Figure 9-3 When we set the height equal to a constant in cylindrical coordinates, we get a plane parallel to the reference plane. Are you confused? Some texts will tell you that the cylindrical coordinates of a point are listed in an ordered triple with the radius first, then the angle, and finally the height, as P = (r,q,h) Don't let this notational inconsistency baffle you. For any particular set of coordinate values, we're talking about the same point, regardless of the order in which we list them. In this book, we indicate the angle before the radius to be consistent with the polar-coordinate system described in Chap. 3. When "traveling" from the origin out to some point P in space in the cylindrical system, most people find it easiest to think of the reference-plane angle q first (as in "face northwest"), then the radius r (as in "walk 40 meters"), and finally the height h (as in "dig down 2 meters to find the treasure"). That's why, in this book, we use the form P = (q,r,h) Here's a challenge! What do we get if we set the direction angle q equal to a constant in cylindrical coordinates? As an example, draw a diagram showing the graph of the equation q = p /2 156 Alternative Three-Space Solution Let's think back again to Chap. 3. In polar coordinates, if we set the direction angle equal to a constant, we get a line passing through the origin. Cylindrical coordinates are simply a vertical extension of polar coordinates, going infinitely upward and infinitely downward. If we hold q constant in cylindrical coordinates but allow the other coordinates to vary at will, we get a vertical plane, which is an infinite vertical extension of a horizontal line. If k is any real-number constant, then the graph of q=k is a plane that passes through the vertical axis. In the case where q = p /2, that plane also contains the ray for the direction angle p /2, as shown in Fig. 9-4. +z Plane extends forever in all directions Constant angle +y –x +x Reference plane –y q = p /2 Figure 9-4 –z When we set the angle equal to a constant in cylindrical coordinates, we get a plane that contains the vertical axis. Cylindrical Conversions Conversion of coordinate values between cylindrical and Cartesian three-space is just as easy as conversion between polar and Cartesian two-space. The only difference is that in threespace, we add the vertical dimension. In xyz space, it's z; in cylindrical three-space, it's h. Cylindrical to Cartesian Let's look at the simplest conversions first. These transformations are like going down a river; we can simply "get into the boat" (sharpen our pencils) and make sure we don't "run aground" Cylindrical Conversions 157 (make an arithmetic error). Suppose we have a point (q,r,h) in cylindrical coordinates. We can find the Cartesian x value of this point using the formula x = r cos q The Cartesian y value is y = r sin q The Cartesian z value is z=h An example Consider the point (q,r,h) = (p,2,−3) in cylindrical coordinates. Let's find the (x,y,z) representation in Cartesian three-space using the preceding formulas. Plugging in the numbers gives us x = 2 cos p = 2 × (−1) = −2 y = 2 sin p = 2 × 0 = 0 z = h = −3 Therefore, we have the Cartesian equivalent point (x,y,z) = (−2,0,−3) Cartesian to cylindrical: finding q Going from Cartesian to cylindrical coordinates is like navigating up a river. We not only have to "go against the current" (do some hard work), but we have to be sure we "take the right tributary" (use the correct angle values). Cartesian-to-cylindrical angle conversion is the same as the Cartesian-to-polar angle conversion process that we learned in Chap. 3. That was messy, because we had to break the situation down into nine different ranges for q. In the cylindrical context, the angle-conversion process works as follows: / /2 When x = 0 and y < 0 q = 2p + Arctan ( y /x) When x > 0 and y < 0 158 Alternative Three-Space If you've forgotten what the Arctangent function is, and why we use a capital "A" to denote it, you can check in Chap. 3 to refresh your memory. Notice that the Cartesian z value is irrelevant when we want to find the direction angle in cylindrical coordinates. Cartesian to cylindrical: finding r When we want to calculate the r coordinate in cylindrical three-space on the basis of a point in Cartesian xyz space, we use the Cartesian two-space distance formula, exactly as we would in the polar plane. The radius depends only on the values of x and y; the z coordinate is irrelevant. The r coordinate is therefore equal to the distance between the projection point P ′ and the origin in the xy plane, which is r = (x2 + y2)1/2 Cartesian to cylindrical: finding h When we want to change the Cartesian z value to the cylindrical h value in three-space, we can make the direct substitution h=z An example Let's convert the Cartesian point (x,y,z) = (1,1,1) to cylindrical three-space coordinates. In this situation, x = 1 and y = 1. To find the angle, we should use the formula q = Arctan ( y /x) because x > 0 and y > 0. When we plug in the values for x and y, we get q = Arctan (1/1) = Arctan 1 = p /4 When we input the values for x and y to the formula for r, we get r = (12 + 12)1/2 = 21/2 Because z = 1, we know that h=z=1 We've just found that the cylindrical equivalent point is (q,r,h) = (p /4,21/2,1) Spherical Coordinates 159 Are you confused? We must pay close attention to the meaning of the radius in cylindrical coordinates. The cylindrical radius goes from the origin to the reference-plane projection of the point whose coordinates we're interested in. It does not go straight through space to the point of interest, which is usually outside the reference plane. Here's a challenge! Convert the Cartesian point (x,y,z) = (−5,−12,8) to cylindrical coordinates. Using a calculator, approximate all irrational values to four decimal places. Solution We have x = −5 and y = −12. To find the angle, we should use the formula q = p + Arctan ( y /x) because x < 0 and y < 0. When radians (not degrees) allows us to approximate this to four decimal places as q ≈ 4.3176 When we input x = −5 and y = −12 to the formula for r, we get r = [(−5)2 + (−12)2]1/2 = (25 + 144)1/2 = 1691/2 = 13 Because z = 8, we know that h=z=8 We've found that the cylindrical equivalent point is (q,r,h) ≈ (4.3176,13,8) The value of q is approximate to four decimal places, while r and h are exact values. Spherical Coordinates Figure 9-5 illustrates a system of spherical coordinates for defining points in three-space. Instead of one angle and two displacements as in cylindrical coordinates, we now use two angles and one displacement. 160 Alternative Three-Space +z Reference axis f P +y r –x q +x P¢ Reference plane –y –z Figure 9-5 Spherical coordinates define points in threespace according to a horizontal angle, a vertical angle, and a radius. How it works In the spherical coordinate arrangement, we start with a horizontal Cartesian reference plane, just as we do when we set up cylindrical coordinates. The positive Cartesian x axis forms the reference axis. Suppose that we want to define the location of a point P. Consider its projection, P ′, onto the reference plane: • The horizontal angle, which we call q, turns counterclockwise in the reference plane from the reference axis to the ray that goes out from the origin through P ′. • The vertical angle, which we call f, turns downward from the vertical axis to the ray that goes out from the origin through P. • The radius, which we call r, is the straight-line distance from the origin to P. These three coordinates, taken all together, provide us with sufficient information to uniquely define the location of P in three-space. We can express the spherical coordinates as an ordered triple P = (q,f,r) Strange values In spherical three-space, we can have nonstandard horizontal direction angles, but it's always best to add or subtract whatever multiple of 2p will keep us within the preferred range of 0 ≤ q < 2p. If q ≥ 2p, it represents at least one complete counterclockwise rotation from the reference axis. If q < 0, it represents clockwise rotation from the reference axis. Spherical Coordinates 161 We can have nonstandard vertical angles, although things are simplest if we keep them nonnegative but no larger than p. Theoretically, all possible locations in space can be covered if we restrict the vertical angle to the range 0 ≤ f ≤ p. If it's outside this range, such as −p < f < 0 or p < f < 2p, we can multiply the radius r by −1, and then add or subtract p to or from f, and we'll end up at the point we want. But those are confusing ways to get there! The radius r can be any real number, but things are simplest if we keep it nonnegative. If our horizontal and vertical direction angles put us on a ray that goes from the origin through P, then r > 0. If our direction angles put us on a ray that goes from the origin away from P, then r < 0. We have r = 0 if and only if P is at the origin. If we find ourselves working with a negative radius, we should reverse the direction by adding or subtracting p to or from both angles, keeping 0 ≤ q < 2p and 0 ≤ f ≤ p. Then we can take the absolute value of the negative radius and use it as the radius coordinate. An example In the situation of Fig. 9-5, the horizontal direction angle q appears to be somewhere between p and 3p /2. The vertical direction angle f appears to be roughly 1 radian. We can't be sure of the exact values of these angles, because we don't have any reference lines to compare them with. The radius r is positive, but we have no idea how large it is because there are no radial coordinate increments. Another example Imagine that we set the horizontal direction angle q equal to a constant in spherical coordinates. For example, let's say that we have the equation q = 7p /5 When we work in polar coordinates and set the direction angle equal to a constant, we get a line passing through the origin. In spherical coordinates, the horizontal angle in the reference plane is geometrically identical to the polar direction angle. Therefore, if k is any real-number constant, the graph of q=k is a plane that passes through the vertical axis. When k = 7p /5, that vertical plane also contains the ray for the direction angle 7p /5, as shown in Fig. 9-6. Still another example If we set the radius equal to a constant in spherical coordinates, we get the set of all points at some fixed distance from the origin. That's a sphere centered at the origin. Figure 9-7 shows what happens when we graph the following equation: r=k in spherical three-space, where k is a nonzero constant. 162 Alternative Three-Space Figure 9-6 When we set the horizontal angle equal to a constant in spherical coordinates, we get a plane that contains the vertical axis. +z Constant radius +y –x +x r=k Reference plane –y –z Figure 9-7 When we set the radius equal to a constant in spherical coordinates, we get a sphere centered at the origin. Spherical Coordinates 163 Are you confused? Don't get the wrong idea about the meaning of the radius in spherical coordinates. It's not the same as the cylindrical-coordinate radius! In spherical coordinates, the radius follows a straightline path from the origin to the point whose coordinates we're interested in. This line almost never lies in the reference plane. In cylindrical coordinates, the radius goes from the origin to the projection of the point in the reference plane. You can see the difference if you compare Fig. 9-1 with Fig. 9-5. Are you still confused? If you've read a lot of other pre-calculus texts (and I recommend that you do), you might notice that the order in which we list spherical coordinates is different from the way it's done in some of those other texts. You might see the spherical coordinates of a point P go with the radius first, then the horizontal angle, and finally the vertical angle, as P = (r,q,f) Theoretically, it doesn't matter in which order we list the coordinates. For any particular values, we're always working with the same point. When we want to get from the origin to a point in spherical three-space, most people find it easiest to think of the horizontal angle q first (as in "face southeast"), then the vertical angle f (as in "fix your gaze at an angle that's p /6 radian from the zenith"), and finally the radius r (as in "follow the string for 150 meters to reach the kite"). That's why we use the form P = (q,f,r) Here's a challenge! What sort of graph do we get if we set the vertical angle f equal to a constant in spherical coordinates? As an example, draw a diagram showing the graph of the following equation: f = p /4 Solution This situation doesn't resemble anything we've seen so far in Cartesian, polar, or cylindrical coordinates. If we hold the vertical angle constant in a spherical coordinate system, we get the set of points formed by a line passing through the origin and rotated with respect to the vertical axis. If k is a real-number constant, then the graph of f=k is a double cone whose axis corresponds to the vertical axis and whose apex is at the origin, as shown in Fig. 9-8. 164 Alternative Three-Space Constant vertical angle +z Cone extends upward forever Reference plane +y –x +x –y f = p /4 Figure 9-8 –z Cone extends downward forever When we set the vertical angle equal to a constant in spherical coordinates, we get a double cone whose axis corresponds to the vertical axis. Spherical Conversions Converting coordinates between xyz space and spherical three-space is a little tricky, but not too difficult. Let's think about a point P whose spherical coordinates are (q,f,r) and whose Cartesian coordinates are (x,y,z). Spherical to Cartesian: finding x In spherical coordinates, the radius is usually outside of the reference plane, so we can't use it directly in the same formulas as the cylindrical radius. But we can construct a projection radius identical to the cylindrical radius: the distance from the origin to the projection point P ′ in the reference plane. In Fig. 9-9, the projection radius is called r ′. From this geometry, we can see that r ′ is equal to the true spherical radius times the sine of the vertical angle. As an equation, we have r ′ = r sin f Spherical Conversions 165 The x value conversion formula from cylindrical coordinates, which we learned earlier in this chapter, tells us that x = r ′ cos q where q is the horizontal direction angle, which is the same in spherical and cylindrical coordinates. Substituting the quantity (r sin f) for r ′ gives us x = r sin f cos q Spherical to Cartesian: finding y When we found the cylindrical equivalent of the Cartesian y value, we took the radius in the reference plane and multiplied by the sine of the direction angle in that plane. In the spherical-coordinate situation of Fig. 9-9, that translates to y = r ′ sin q where q is the horizontal direction angle. We can substitute (r sin f) for r ′ to get y = r sin f sin q Spherical to Cartesian: finding z Let's look again at Fig. 9-9, and locate the projection point P ∗ on the z axis, such that the z values of P ∗ and P are equal. We can see that P ∗, P, P ′, and the origin form the vertices of a +z Reference axis P* P r f +y q –x P¢ +x r¢ Reference plane –y –z Figure 9-9 Conversion between spherical and Cartesian three-space coordinates involves several geometric variables. 166 Alternative Three-Space rectangle perpendicular to the reference plane. It follows that P ∗, P, and the origin are at the vertices of a right triangle. By trigonometry, the z value of P ∗ is equal to the spherical radius r times the cosine of the vertical angle f. Because the z values of P and P ∗ are the same, we can deduce that the z value of P is given by z = r cos f Cartesian to spherical: finding r Now let's figure out how to get from Cartesian xyz space to spherical three-space. The radius is the easiest coordinate to find, so let's do it first. Recall that the spherical radius of a point is its distance from the origin. Therefore, when we want to find the spherical radius r for point P in terms of its xyz space coordinates, we can apply the Cartesian three-space distance formula to get r = (x2 + y2 + z2)1/2 Cartesian to spherical: finding q The horizontal angle in spherical coordinates is identical to its counterpart in cylindrical coordinates, so we can use the conversion table from earlier in this chapter.//2 When x = 0 and y < 0 q = 2p + Arctan ( y /x) When x > 0 and y < 0 The Arccosine Before we can find the vertical spherical angle for a point that's given to us in Cartesian coordinates, we must be familiar with the arccosine relation. It's abbreviated arccos (cos−1 in some texts), and it "undoes" the work of the cosine function. For example, we know that cos (p /3) = 1/2 and cos p = −1 Spherical Conversions 167 For things to work without ambiguity when we go the other way, we want the arccosine to be a true function. To do that, we must restrict its range (output) to an interval where we don't get into trouble with ambiguity. By convention, mathematicians specify the closed interval [0,p] for this purpose. That happens to be the ideal range of values for our vertical angle f in spherical coordinates. When we make this restriction, we capitalize the "A" and write Arccosine or Arccos to indicate that we're working with a true function. Then we can state the above facts "in reverse" using the Arccosine function, getting Arccos 1/2 = p /3 and Arccos (−1) = p For any real number u, we can be sure that Arccos (cos u) = u Going the other way, for any real number v such that −1 ≤ v ≤ 1, we know that cos (Arccos v) = v We restrict v because the Arccosine function is not defined for input values less than −1 or larger than 1. Cartesian to spherical: finding e We've learned how to find the vertical angle on the basis of the Cartesian coordinate z. That formula is z = r cos f We can use algebra to rearrange this, getting cos f = z /r provided r ≠ 0. When we examine Fig. 9-9, we can see that for any given point P, the absolute value of z can never exceed r, so we can be sure that −1 ≤ z /r ≤ 1. Therefore, we can take the Arccosine of both sides of the preceding equation, getting Arccos (cos f) = Arccos (z /r) Simplifying, we obtain f = Arccos (z /r) Spherical Conversions 169 Plugging in the values, we get r = [(−1)2 + (−1)2 + 12]1/2 = (1 + 1 + 1)1/2 = 31/2 To find the horizontal angle, we use the formula q = p + Arctan ( y /x) because x < 0 and y < 0. When we plug in the values for x and y, we get q = p + Arctan [−1/(−1)] = p + Arctan 1 = p + p /4 = 5p /4 To find the vertical angle, we can use the formula f = Arccos (z /r) We already know that r = 31/2, so f = Arccos (1/31/2) = Arccos 3−1/2 Our spherical ordered triple, listing the coordinates in the order P = (q,f,r), is P = [5p /4,(Arccos 3−1/2),31/2] Are you confused? When you come across a messy ordered triple like this, you might ask, "Is there any way to make it look simpler?" Sometimes there is. In this case, there isn't. You can get rid of the grouping symbols if you're willing to use a calculator to approximate the values. But even if you do that, you'll have to remember that in spherical coordinates, the first two values represent angles in radians, and the third value represents a linear distance. Here's a challenge! Suppose we're given the coordinates of a point P in spherical three-space as P = (q,f,r) = (3p /4,p /4,31/2) Find the coordinates of P in cylindrical coordinates. Solution We haven't learned any formulas for direct conversion between spherical and cylindrical coordinates, so we must convert to Cartesian coordinates first, and then to cylindrical coordinates from there. The Cartesian x value is x = r sin f cos q = 31/2 sin (p /4) cos (3p /4) = 31/2 × 21/2/2 × (−21/2/2) = −31/2/2 Practice Exercises 171 Describe the graphs of the following equations in cylindrical coordinates. What would they look like in Cartesian xyz space? q=0 r=0 h=0 2. Plot the point (q,r,h) = (3p /4,6,8) in the cylindrical coordinate system. 3. Consider the point (q,r,h) = (p /4,0,1) in cylindrical coordinates. Find the equivalent of this point in Cartesian xyz space. 4. Consider the point (−4,1,0) in xyz space. Find the equivalent of this point in cylindrical three-space. First, find the exact coordinates. Then, using a calculator, approximate the irrational coordinates to four decimal places. 5. In the chapter text, we used the conversion formulas to find that the cylindrical equivalent of (x,y,z) = (1,1,1) is (q,r,h) = (p /4,21/2,1). Convert these coordinates back to Cartesian xyz coordinates to verify that the result we got was correct and unambiguous. 6. Describe the graphs of the following equations in spherical coordinates. What would they look like in Cartesian xyz space? q=0 f=0 r=0 7. Plot the point (q,f,r) = (3p /4,p /4,8) in the spherical coordinate system. 8. Consider the point (q,f,r) = (p /4,0,1) in spherical coordinates. Find the equivalent of this point in Cartesian xyz space. 9. Consider the point (−4,1,0) in xyz space. Find the equivalent of this point in spherical three-space. First, find the exact coordinates. Then, using a calculator, approximate the irrational coordinates to four decimal places. 10. Work the final "challenge" backward to verify that we did our calculations correctly. Consider the point P in cylindrical three-space given by P = (q,r,h) = [3p /4,61/2/2,61/2/2] Find the coordinates of P in Cartesian coordinates, and from there, convert to spherical coordinates. CHAPTER 10 Review Questions and Answers Part One This is not a test! It's a review of important general concepts you learned in the previous nine chapters. Read it through slowly and let it sink in. If you're confused about anything here, or about anything in the section you've just finished, go back and study that material some more. Chapter 1 Question 1-1 What's the difference between an open interval, a half-open interval, and a closed interval? Answer 1-1 All three types of intervals are continuous spans of values that a variable can attain between a specific minimum and a specific maximum, which are called the extremes. But there are subtle differences between the three types as listed below: • In an open interval, neither extreme is included. • In a half-open interval, one extreme is included, but not the other. • In a closed interval, both extremes are included. Question 1-2 Imagine two real numbers a and b, such that a < b. These numbers can be the extremes of four different intervals: one open, two half-open, and one closed. How can we denote these four intervals for a variable x ? Answer 1-2 If we include neither a nor b, we have an open interval where a < x < b. We can write x ∈ (a,b) 172 Part One 173 which means "x is an element of the open interval (a,b)." If we include a but not b, we have a half-open interval where a ≤ x < b. We can write x ∈ [a,b) which translates to "x is an element of the half-open interval [a,b)." If we include b but not a, we have an open interval where a < x ≤ b. We write x ∈ (a,b] which means "x is an element of the half-open interval (a,b]." If we include both a and b, we have a closed interval where a ≤ x ≤ b. We can write x ∈ [a,b] which means "x is an element of the closed interval [a,b]." Question 1-3 What point of confusion must we avoid when working with interval notation? Answer 1-3 We must never confuse an open interval with an ordered pair, which uses the same notation. If we pay close attention to the context in which the expression appears, we shouldn't have trouble. Question 1-4 Relations and functions are operations that map specific values of a variable into specific values of another variable. There's an important distinction between a relation and a function. What is it? Answer 1-4 In a relation, we can have more than one value of the dependent (or output) variable for a single value of the independent (or input) variable. In a function, we're allowed no more than one output for any given input. All functions are relations, but not all relations are functions. Question 1-5 The Cartesian plane can be used for graphing relations and functions between an independent variable and a dependent variable. The plane is divided into four sections, called quadrants. How do we identify them? Answer 1-5 In the first quadrant (usually the upper right), both variables are positive. In the second quadrant (usually the upper left), the independent variable is negative and the dependent variable is positive. In the third quadrant (usually the lower left), both variables are negative. In the fourth quadrant (usually the lower right), the independent variable is positive and the dependent variable is negative. 174 Review Questions and Answers Question 1-6 Suppose we have a point S in Cartesian two-space that is represented by the ordered pair (xs,ys). We can write this as S = (xs,ys) What's the straight-line distance ds between S and the coordinate origin? What's the minimum possible distance between S and the origin? Can the distance be negative? What's the maximum possible distance? Answer 1-6 We can find the distance using the formula that we derived from the Pythagorean theorem in geometry. In this situation, the formula is ds = (xs2 + ys2)1/2 The minimum possible distance between S and the origin is zero, which occurs if and only if xs = 0 and ys = 0, so that S = (xs,ys) = (0,0) We can never have a negative distance. There is no maximum possible distance between S and the origin. We can make it as large as we want by making xs or ys (or both) huge positively or huge negatively. Question 1-7 Imagine two points in Cartesian two-space, called S and T, such that S = (xs,ys) and T = (xt,yt) What's the straight-line distance dst going from S to T ? What's the straight-line distance dts going from T to S ? Does it make any difference which way we go? Answer 1-7 If we go from S to T, the distance between the points is dst = [(xt − xs)2 + ( yt − ys)2]1/2 If we go from T to S, the distance is dts = [(xs − xt)2 + ( ys − yt)2]1/2 Part One 175 Figure 10-1 Illustration for Question and Answer 1-8. It doesn't matter which way we go when we want to determine the straight-line distance between two points. Therefore, dst = dts. Question 1-8 In Fig. 10-1, what do the expressions Δx and Δy mean? What's the straight-line distance d between the two points, based on the values of Δx and Δy? Answer 1-8 We read Δx as "delta x," which means "the difference in x." We read Δy as "delta y," which means "the difference in y." The straight-line distance d between the points can be found by squaring Δx and Δy individually, adding the squares, and then taking the nonnegative square root of the result, getting d = (Δx2 + Δy2)1/2 Question 1-9 Suppose we want to find the midpoint of a line segment connecting two known points in the Cartesian xy plane. How can we do this? Answer 1-9 We average the x coordinates of the endpoints to get the x coordinate of the midpoint, and we average the y coordinates of the endpoints to get the y coordinate of the midpoint. 176 Review Questions and Answers Question 1-10 Once again, imagine two points S and T in the Cartesian plane with the coordinates S = (xs,ys) and T = (xt,yt) What are the coordinates of the point B that bisects the line segment connecting S and T ? Answer 1-10 The point B is the midpoint of the line segment. When we follow the procedure described in Answer 1-9, we obtain the coordinates (xb,yb) of point B as (xb,yb) = [(xs + xt)/2,( ys + yt)/2] Chapter 2 Question 2-1 What is a radian? Answer 2-1 A radian is the standard unit of angular measure in mathematics. If we have two rays pointing out from the center of a circle, and those rays intersect the circle at the endpoints of an arc whose length is equal to the circle's radius, then the smaller (acute) angle between the rays measures one radian (1 rad ). Question 2-2 How many radians are there in a full circle? In 1/4 of a circle? In 1/2 of a circle? In 3/4 of a circle? Answer 2-2 There are 2p rad in a full circle. Therefore, 1/4 of a circle is p/2 rad, 1/2 of a circle is p rad, and 3/4 of a circle is 3p/2 rad. Question 2-3 Suppose we have an angle whose radian measure is 7p/6. What fraction of a complete circular rotation does this represent? Answer 2-3 Remember that an angle of 2p represents a full rotation. The quantity p/6 is 1/12 of 2p, so an angle of p/6 represents 1/12 of a rotation. Therefore, an angle of 7p/6 represents 7/12 of a rotation. Part One 177 Figure 10-2 Illustration for Questions and Answers 2-4 through 2-9. Each axis division represents 1/4 unit. Question 2-4 In Fig. 10-2, the gray circle is a graph of the equation x2 + y2 = 1. The point (x0,y0) lies on this circle. A ray from the origin through (x0,y0) subtends an angle q going counterclockwise from the positive x axis. How can we define the sine of the angle q ? Answer 2-4 The sine of q as shown in Fig. 10-2 is equal to y0. Mathematically, we write this as sin q = y0 Question 2-5 How can we define the cosine of the angle q in Fig. 10-2? Answer 2-5 The cosine of q is equal to x0. Mathematically, we write this as cos q = x0 Question 2-6 How can we define the tangent of the angle q in Fig. 10-2? 178 Review Questions and Answers Answer 2-6 The tangent of q is equal to y0 divided by x0, as long as x0 is nonzero. If x0 = 0, then the tangent of the angle is not defined. Mathematically, we have tan q = y0 /x0 ⇔ x0 ≠ 0 The double-headed, double-shafted arrow (⇔) is the logical equivalence symbol. It translates to the words "if and only if." We can also define the tangent as tan q = sin q /cos q ⇔ cos q ≠ 0 Question 2-7 How can we define the cosecant of the angle q in Fig. 10-2? Answer 2-7 The cosecant of q is equal to the reciprocal of y0, as long as y0 is nonzero. If y0 = 0, then the cosecant is not defined. Mathematically, we have csc q = 1/y0 ⇔ y0 ≠ 0 We can also define the cosecant as csc q = 1/sin q ⇔ sin q ≠ 0 Question 2-8 How can we define the secant of the angle q in Fig. 10-2? Answer 2-8 The secant of q is equal to the reciprocal of x0, as long as x0 is nonzero. If x0 = 0, then the secant is not defined. Mathematically, we have sec q = 1/x0 ⇔ x0 ≠ 0 We can also define the secant as sec q = 1/cos q ⇔ cos q ≠ 0 Question 2-9 How can we define the cotangent of the angle q in Fig. 10-2? Answer 2-9 The cotangent of q is equal to x0 divided by y0, as long as y0 is nonzero. If y0 = 0, then the cotangent is not defined. Mathematically, we have cot q = x0/y0 ⇔ y0 ≠ 0 Part One 179 We can also define the cotangent as cot q = 1/tan q ⇔ tan q ≠ 0 or as cot q = cos q /sin q ⇔ sin q ≠ 0 Question 2-10 What are the Pythagorean identities for trigonometric functions? Which, if any, of these should be memorized? Answer 2-10 The Pythagorean identities are the three formulas sin2 q + cos2 q = 1 sec2 q – tan2 q = 1 csc2 q – cot2 q = 1 The first of these is worth memorizing, because it comes up quite often in applied mathematics and engineering. The second and third identities can be derived from the first one. Chapter 3 Question 3-1 How are variables and points portrayed on the polar-coordinate plane? Answer 3-1 The independent variable is rendered as a direction angle q, expressed counterclockwise from a reference axis. This reference axis normally goes outward from the origin toward the right (or "due east"), in the same direction as the positive x axis in the Cartesian xy plane. The dependent variable is rendered as a radius r, expressed as the straight-line distance from the origin. Points in the plane are expressed as ordered pairs of the form (q,r), as shown in Fig. 10-3. In some texts, the ordered pair is written as (r,q). Question 3-2 Can a point in polar coordinates have a negative direction angle, or an angle that represents a full rotation or more? Answer 3-2 Yes. If q < 0, it represents clockwise rotation from the reference axis. If q ≥ 2p, it represents at least one complete counterclockwise rotation from the reference axis. Question 3-3 Can a point in polar coordinates have a negative radius? 180 Review Questions and Answers p /2 Angle is q p 0 Radius is r (r, q) 3p /2 Figure 10-3 Illustration for Question and Answer 3-1. Answer 3-3 Yes. If r < 0, we can multiply r by –1 so it becomes positive, and then add or subtract p to or from the direction angle, keeping it within the preferred range 0 ≤ q < 2p. Question 3-4 How we can we portray a relation or function in polar coordinates when the independent variable is q and the dependent variable is r ? Answer 3-4 We can write down an equation with r on the left-hand side and the name of the function followed by q in parentheses on the right-hand side. For example, if our function is g, we write r = g (q) and read it as "r equals g of q." Question 3-5 If we set the polar-coordinate angle equal to a constant, say k, what graph do we get? Answer 3-5 The graph is a straight line passing through the origin. The line appears at an angle of k radians with respect to the reference axis. Question 3-6 If we set the polar-coordinate radius equal to a constant, say m, what graph do we get? Part One 181 Answer 3-6 The graph a circle centered at the origin, so that every point on the circle is m units from the origin. Question 3-7 Suppose we have a point (q,r) in polar coordinates. How can we convert this to coordinates in the Cartesian xy plane? Answer 3-7 We can convert the polar point (q,r) to Cartesian (x,y) using the formulas x = r cos q and y = r sin q Question 3-8 Suppose we have a point (x,y) in the Cartesian plane. What's the polar radius r of this point? Answer 3-8 The polar radius of a point is its distance from the origin. We can use the formula for the distance of a point from the origin to find that r = (x2 + y2)1/2 This gives us a positive value for the radius, which is preferred. Question 3-9 Suppose we have a point (x,y) in the Cartesian plane. What's the polar angle q of this point? Answer 3-9 This problem breaks down into following nine cases, depending on where in the Cartesian plane our point (x,y) liesIf we follow this process carefully, we always get an angle in the range 0 ≤ q < 2p, which is preferred. 182 Review Questions and Answers Question 3-10 What does "Arctan" mean in the conversions listed in Answer 3-9? Answer 3-10 It stands for "Arctangent." That's the function that undoes the work of the trigonometric tangent function. The domain of the Arctangent function is the entire set of real numbers. The range is the open interval (−p /2,p /2). For any real number u within this interval, we have Arctan (tan u) = u Conversely, for any real number v, we have tan (Arctan v) = v Chapter 4 Question 4-1 What is a vector? Answer 4-1 A vector is a quantity with two independent properties: magnitude and direction. A vector can also be defined as a directed line segment having an originating point (beginning) and a terminating point (end ). Question 4-2 What's the standard form of a vector in the xy plane? What's the standard form of a vector in the polar plane? What's the advantage of putting a vector into its standard form? Answer 4-2 In any coordinate system, a vector is in standard form if and only if its originating point is at the coordinate origin. The standard form allows us to uniquely define a vector as an ordered pair that represents the coordinates of its terminating point alone. Question 4-3 How can we find the magnitude of a standard-form vector b in the xy plane whose terminating point has the coordinates (xb,yb)? Answer 4-3 The magnitude of b, which we can write as rb, is found by using the formula for the distance of the terminating point from the origin. In this case, we get rb = (xb2 + yb2)1/2 In some texts, the magnitude of b would be denoted as \|b\| or b. Question 4-4 How can we find the direction of a standard-form vector b in the xy plane whose terminating point has the coordinates (xb,yb)? Part One 183 Answer 4-4 We find the polar direction angle of the point (xb,yb). If we call this angle qb, the process can be broken down into the following nine possible cases: • • • • • • • • • If xb = 0 and yb = 0, then qb = 0 by default. If xb > 0 and yb = 0, then qb = 0. If xb > 0 and yb > 0, then qb = Arctan ( yb/xb). If xb = 0 and yb > 0, then qb = p /2. If xb < 0 and yb > 0, then qb = p + Arctan ( yb/xb). If xb < 0 and yb = 0, then qb = p. If xb < 0 and yb < 0, then qb = p + Arctan ( yb/xb). If xb = 0 and yb < 0, then qb = 3p /2. If xb > 0 and yb < 0, then qb = 2p + Arctan ( yb/xb). In some texts, the direction of b is denoted as dir b. Question 4-5 Imagine two vectors a and b in the xy plane, in standard form with terminating-point coordinates a = (xa,ya) and b = (xb,yb) How can we find the sum of these vectors? Answer 4-5 We calculate the sum vector a + b using the formula a + b = [(xa + xb),( ya + yb)] Question 4-6 How can we calculate the Cartesian negative of a vector that's in standard form? How does the Cartesian negative compare with the original vector? Answer 4-6 We take the negatives of both coordinate values. For example, if we have b = (xb,yb) then its Cartesian negative is -b = (−xb,−yb) The Cartesian negative has the same magnitude as the original vector, but points in the opposite direction. 184 Review Questions and Answers Question 4-7 Imagine two Cartesian vectors a and b, in standard form with terminating-point coordinates a = (xa,ya) and b = (xb,yb) How can we find a − b? How can we find b − a? How do these two vectors compare? Answer 4-7 We calculate the difference vector a − b using the formula a − b = [(xa − xb),( ya − yb)] We find difference vector b − a by reversing the order of subtraction for each coordinate, getting b − a = [(xb − xa),( yb − ya)] In the Cartesian plane, the difference vector b − a is always equal to the negative of the difference vector a − b. Question 4-8 Suppose we have a vector expressed in polar form as c = (qc,rc) where qc is the direction angle of c, and rc is the magnitude of c. How can we convert c to a standard-form vector (xc,yc) in the Cartesian plane? Answer 4-8 We use formulas adapted from the polar-to-Cartesian conversion. We get (xc,yc) = [(rc cos qc),(rc sin qc)] Question 4-9 What restrictions apply when we work with vectors in the polar-coordinate plane? Answer 4-9 A polar vector is not allowed to have a negative radius, a negative direction angle, or a direction angle of 2p or more. These constraints prevent ambiguities, so we can be confident that the set of all polar-plane vectors can be paired off in a one-to-one correspondence with the set of all Cartesian-plane vectors. Part One 185 Question 4-10 Suppose we're given two vectors in polar coordinates. What's the best way to find their sum and difference? What's the best way to find the negative of a vector in polar coordinates? Answer 4-10 The best way to add or subtract polar vectors is to convert them to Cartesian vectors in standard form, then add or subtract those vectors, and finally convert the result back to polar form. The best way to find the negative of a polar vector is to reverse its direction and leave the magnitude the same. Suppose we have a = (qa,ra) If 0 ≤ qa < p, then the polar negative is -a = [(qa + p ),ra] If p ≤ qa < 2p, then the polar negative is -a = [(qa − p ),ra] Chapter 5 Question 5-1 What's the left-hand Cartesian product of a scalar and a vector? What's the right-hand Cartesian product of a vector and a scalar? How do they compare? Answer 5-1 Consider a real-number constant k, along with a standard-form vector a defined in the xy plane as a = (xa,ya) The left-hand Cartesian product of k and a is ka = (kxa,kya) The right-hand Cartesian product of a and k is ak = (xak,yak) The left- and right-hand products of a scalar and a Cartesian vector are always the same. For all real numbers k and all Cartesian vectors a, we can be sure that ka = ak Question 5-2 What's the left-hand polar product of a positive scalar and a vector? What's the right-hand polar product of a vector and a positive scalar? How do they compare? 186 Review Questions and Answers Answer 5-2 Imagine a polar vector a with angle qa and radius ra, such that a = (qa,ra) When we multiply a on the left by a positive scalar k+, we get k+a = (qa,k+ra) When we multiply a on the right by k+, we get ak+ = (qa,rak+) The left- and right-hand polar products of a positive scalar and a polar vector are always the same. For all positive real numbers k+ and all polar vectors a, we can be sure that k+a = ak+ Question 5-3 What's the left-hand polar product of a negative scalar and a vector? What's the right-hand polar product of a vector and a negative scalar? How do they compare? Answer 5-3 Once again, suppose we have a polar vector a with angle qa and radius ra, such that a = (qa,ra) When we multiply a on the left by a negative scalar k−, we get, so −k−ra is positive, ensuring that we get a positive radius for the resultant vector. If we multiply a on the right by k−, we get ak− = [(qa + p ),ra(−k−)] if 0 ≤ qa < p, and ak− = [(qa − p ),ra(−k−)] Part One 187 if p ≤ qa < 2p. Because k− is negative, −k− is positive, so ra(−k−) is positive, ensuring that we get a positive radius for the resultant vector. For all negative real numbers k− and all polar vectors a, k−a = ak− Question 5-4 Suppose we're given two standard-form vectors a and b, defined by the ordered pairs a = (xa,ya) and b = (xb,yb) What's the Cartesian dot product a • b? What's the Cartesian dot product b • a? How do they compare? Answer 5-4 The Cartesian dot product a • b is a real number given by a • b = xaxb + yayb and the Cartesian dot product b • a is a real number given by b • a = xbxa + ybya The Cartesian dot product is commutative, so for any two vectors a and b in the xy plane, we can be confident that a•b=b•a Question 5-5 Imagine a polar vector a with angle qa and radius ra, such that a = (qa,ra) and a polar vector b with angle qb and radius rb, such that b = (qb,rb) What's the polar dot product a • b? Answer 5-5 Let qb − qa be the angle as we rotate from a to b. The polar dot product a • b is given by the formula a • b = rarb cos (qb − qa) 188 Review Questions and Answers Question 5-6 Consider the same two polar vectors as we worked with in Question and Answer 5-5. What's the polar dot product b • a? Answer 5-6 We can define this dot product by reversing the roles of the vectors in the previous problem. Let qa − qb be the angle going from b to a. The polar dot product b • a is given by the formula b • a = rbra cos (qa − qb) Question 5-7 How do the polar dot products a • b and b • a, as defined in Answers 5-5 and 5-6, compare? Answer 5-7 For any two vectors a and b, the polar dot product is commutative. That is a•b=b•a Question 5-8 Imagine a polar vector c with angle qc and radius rc, such that c = (qc,rc) and a polar vector d with angle qd and radius rd, such that d = (qd,rd ) What's the polar cross product c ë d? Answer 5-8 Imagine that we start at vector c and rotate counterclockwise until we get to vector d, so we turn through an angle of qd − qc. Suppose that 0 < qd − qc < p. To calculate the magnitude rc×d of the cross-product vector c × d, we use the formula rc×d = rcrd sin (qd − qc) In this situation, c ë d points toward us. If p < qd − qc < 2p, we can consider the difference angle to be 2p + qc − qd. Then the magnitude of c ë d is rc×d = rcrd sin (2p + qc − qd ) and it points away from us. Part One 189 Question 5-9 What's the right-hand rule for cross products? Answer 5-9 Consider again the two vectors c and d that we defined in Question 5-8, and their difference angle qd − qc that we defined in Answer 5-8. If 0 < qd − qc < p, point your right thumb out, and curl your fingers counterclockwise from c to d. If p < qd − qc < 2p, point your right thumb out, and curl your right-hand fingers clockwise from c to d. Your thumb will then point in the general direction of c ë d. The vector c ë d is always perpendicular to the plane defined by c and d. Question 5-10 How do the polar cross products of two vectors c ë d and d ë c compare? Answer 5-10 They have identical magnitudes, but they point in opposite directions. Chapter 6 Question 6-1 What's the unit imaginary number? What's the j operator? Answer 6-1 These expressions both refer to the positive square root of −1. If we denote it as j, then j = (−1)1/2 and j 2 = −1 Question 6-2 How is the set of imaginary numbers "built up"? How do we denote such numbers? Answer 6-2 If we multiply j by a nonnegative real number a, we get a nonnegative imaginary number. If we multiply j by a negative real number −a, we get a negative imaginary number. We denote nonnegative imaginary numbers by writing j followed by the real-number coefficient. If a ≥ 0, then j × a = a ë j = ja We denote negative imaginary numbers as −j followed by the absolute value of the real-number coefficient. If −a < 0, then j ë (−a) = −a ë j = −ja 190 Review Questions and Answers Question 6-3 How is the set of complex numbers "built up"? How do we denote such numbers? Answer 6-3 A complex number is the sum of a real number and an imaginary number. If a is a real number and b is a nonnegative real number, then the general form for a complex number is a + jb If a is a real number and −b is a negative real number, then we have a + j(−b) but it's customary to write the absolute value of −b after j, and use a minus sign instead of a plus sign in the expression. That gives us the general form a − jb Question 6-4 How do the complex number 0 + j0, the pure real number 0, and the pure imaginary number j0 compare? Answer 6-4 They are all identical. Question 6-5 How do we find the sum of two complex numbers a + jb and c + jd? How do we find their difference? How do we find their product? How do we find their ratio? Answer 6-5 When we want to add, we use the formula (a + jb) + (c + jd ) = (a + c) + j(b + d ) When we want to subtract, we use the formula (a + jb) − (c + jd ) = (a − c) + j(b − d ) When we want to multiply, we use the formula (a + jb)(c + jd ) = (ac − bd ) + j(ad + bc) When we want to find the ratio, we use the formula (a + jb) / (c + jd ) = [(ac + bd ) / (c2 + d 2)] + j [(bc − ad ) / (c2 + d 2)] In a complex-number ratio, the denominator must not be equal to 0 + j0. Part One 191 Question 6-6 What are complex conjugates? What happens when we add a complex number to its conjugate? What happens when we multiply a complex number by its conjugate? Answer 6-6 Complex conjugates have identical coefficients, but opposite signs between the real and imaginary parts, as in a + jb and a − jb When we add a complex number to its conjugate, we get (a + jb) + (a − jb) = 2a When we multiply a complex number by its conjugate, we get (a + jb)(a − jb) = a2 + b2 Question 6-7 What's the Cartesian complex-number plane? What's the polar complex-number plane? How are complex vectors defined in these planes? Answer 6-7 Figure 10-4 shows a Cartesian complex-number plane. The horizontal axis portrays the realnumber part, and the vertical axis portrays the imaginary-number part. A Cartesian complex vector is rendered in standard form, going from the origin to the terminating point corresponding to the complex number. Figure 10-5 shows a polar complex-number plane. Polar complex vectors are defined in terms of their direction angle and magnitude, instead of their real and imaginary parts. Assuming that the axis divisions in Fig. 10-4 are the same size as the radial divisions in Fig. 10-5, the vectors in both drawings represent the same complex number. Question 6-8 How can we convert a Cartesian complex vector to a polar complex vector? Answer 6-8 Imagine a complex number a + jb in the Cartesian complex plane, whose vector extends from the origin to the point (a,jb). We can derive the magnitude r of the equivalent polar vector by applying the distance formula to get r = (a2 + b2)1/2 Part One 193 To determine the direction angle q of the polar vector, we modify the polar-coordinate direction-finding system. Here's what happens: • • • • • • • • • When a = 0 and jb = j0, we have q = 0 by default. When a > 0 and jb = j0, we have q = 0. When a > 0 and jb > j0, we have q = Arctan (b /a). When a = 0 and jb > j0, we have q = p /2. When a < 0 and jb > j0, we have q = p + Arctan (b /a). When a < 0 and jb = j0, we have q = p. When a < 0 and jb < j0, we have q = p + Arctan (b /a). When a = 0 and jb < j0, we have q = 3p /2. When a > 0 and jb < j0, we have q = 2p + Arctan (b /a). Question 6-9 How can we convert a polar complex vector to a Cartesian complex vector? Answer 6-9 Imagine a complex vector (q,r) in the polar complex plane, whose direction angle is q and whose radius is r. The Cartesian vector equivalent is (a,jb) = [(r cos q), j(r sin q)] which represents the complex number a + jb = r cos q + j(r sin q) Question 6-10 What are the two versions of De Moivre's theorem? How are they used? Answer 6-10where r1 and r2 are real-number polar magnitudes, and q1 and q2 are real-number polar angles in radians. Then c1c2 = r1r2 cos (q1 + q2) + j [r1r2 sin (q1 + q2)] and, as long as r2 is nonzero, c1/c2 = (r1/r2) cos (q1 − q2) + j [(r1/r2) sin (q1 − q2)] 194 Review Questions and Answers The second version of De Moivre's theorem involves integer powers. Suppose that c is a complex number, where c = r cos q + j(r sin q) where r is the real-number polar magnitude and q is the real-number polar angle. Also suppose that n is an integer. Then cn = rn cos (nq) + j[rn sin (nq)] Chapter 7 Question 7-1 How are the axes and variables defined in Cartesian xyz space? Answer 7-1 We construct Cartesian xyz space by placing three real-number lines so that they all intersect at their zero points, and they're all mutually perpendicular. One number line represents the variable x, another represents the variable y, and the third represents the variable z. Figure 10-6 shows two perspective drawings of the typical system. Although the point of Figure 10-6 Illustration for +y Question and Answer 7-1. –z A –x +x +z –y +z +y B –x +x –y –z Part One 195 view differs between illustrations A and B, the relative axis orientation is the same in both cases. When we graph relations and functions having two independent variables in Cartesian xyz space, x and y are usually the independent variables, and z is usually the dependent variable. Question 7-2 What's the difference between Cartesian xyz space and rectangular xyz space? Answer 7-2 In Cartesian xyz space, the axes are all linear, and they're all graduated in increments of the same size. In rectangular xyz space, the divisions can differ in size between the axes, although each axis must be linear along its entire length. Question 7-3 What's the "pool rule" for the relative axis orientation and coordinate values in Cartesian xyz space? Answer 7-3 We can imagine that the origin of the coordinate grid rests on the surface of a swimming pool. We orient the positive x axis horizontally along the pool surface, pointing due east. Once we've done that, the coordinate values can be generalized as follows: • • • • • • Positive values of x are east of the origin.Question 7-4 What are the biaxial planes in Cartesian xyz space? Answer 7-4 The biaxial planes are the xy plane, the xz plane, and the yz plane. Each plane is perpendicular to the other two, and all three intersect at the origin. The biaxial planes are defined by pairs of axes as follows: • The xy plane contains the axes for variables x and y. • The xz plane contains the axes for variables x and z. • The yz plane contains the axes for variables y and z. Question 7-5 In Cartesian xyz space, a point can always be denoted as an ordered triple in the form (x,y,z). What do the x, y, and z coordinates represent geometrically? 196 Review Questions and Answers Answer 7-5 We can think of this situation in two different ways. First, we can use the notion of a point's projection. We get the projection of a point onto an axis by drawing a line from the point to the axis, and making sure that the line intersects that axis at a right angle. That way, the coordinates and projection points are related as follows: • The x coordinate represents the point's projection onto the x axis. • The y coordinate represents the point's projection onto the y axis. • The z coordinate represents the point's projection onto the z axis. We can also think of the x, y, and z values for a particular point in terms of perpendicular displacements from the biaxial planes as followsQuestion 7-6 What semantical distinction should we keep in mind when we talk about points in terms of ordered triples? Answer 7-6 An ordered triple represents the coordinates of a point in three-space, not the geometric point itself. Informally, the ordered triple is the name of the point. We can talk about the ordered triple as if it were the actual point, as long as we're aware of the technical difference between the object and its name. Question 7-7 How can we find the distance of a point from the origin in Cartesian xyz space? Answer 7-7 Suppose we name the point Q, and assign it the coordinates Q = (xq,yq,zq) If we call the distance between Q and the origin by the name dq, then dq = (xq2 + yq2 + zq2)1/2 This distance is always defined, it's always unique (unambiguous), it's never negative, and it doesn't depend on whether we go from the origin to the point or from the point to the origin. Question 7-8 How can we find the distance between two points in Cartesian xyz space? Part One 197 Answer 7-8 Let's call the points and their coordinates S = (xs,ys,zs) and T = (xt,yt,zt) where each coordinate can range over the entire set of real numbers. If we go from S to T, the distance between the points is dst = [(xt − xs)2 + ( yt − ys)2 + (zt − zs)2]1/2 If we go from T to S, the distance is dts = [(xs − xt)2 + ( ys − yt)2 + (zs − zt)2]1/2 This distance is always defined and unique. It's never negative, and it doesn't depend on which direction we go. Therefore dst = dts Question 7-9 How can we find the midpoint of a line segment connecting two points in Cartesian xyz space? Answer 7-9 Let's call the points and their coordinates P = (xp,yp,zp) and Q = (xq,yq,zq) We can call the midpoint M, and say that its coordinates are M = (xm,ym,zm) Given this information, the coordinates of M in terms of the coordinates of P and Q are (xm,ym,zm) = [(xp + xq)/2,( yp + yq)/2,(zp + zq)/2] This midpoint is always defined, it's always unique, and it doesn't depend on which direction we go. 198 Review Questions and Answers Question 7-10 Suppose that we have two points in Cartesian xyz space where all three pairs of corresponding coordinates are negatives of each other. Where is the midpoint of a line segment connecting these two points? Answer 7-10 It's always at the origin. Chapter 8 Question 8-1 What's the Cartesian standard form for a vector in xyz space? Answer 8-1 Any vector in xyz space, no matter where its originating and terminating points are located, has an equivalent standard-form vector whose originating point is at (0,0,0). Consider a vector c′ whose originating point is Q1 and whose terminating point is Q2, such that Q1 = (x1,y1,z1) and Q2 = (x2,y2,z2) The standard form of c′, denoted c, has the originating point (0,0,0) and the terminating point Qc such that Qc = (xc,yc,zc) = [(x2 − x1),( y2 − y1),(z2 − z1)] The two vectors c and c′ have identical direction angles and identical magnitudes. That's why we say they're equivalent. Question 8-2 What's the advantage of putting a three-space vector into its standard form? Answer 8-2 The standard form allows us to uniquely define a vector as an ordered triple that represents the coordinates of its terminating point alone. We don't have to worry about the originating point. Question 8-3 How can we find the magnitude rb of a standard-form vector b in xyz space whose terminating point has the coordinates (xb,yb,zb)? Answer 8-3 We can do it by calculating the distance of the terminating point from the origin. In this case, the formula is rb = (xb2 + yb2 + zb2)1/2 Part One 199 Question 8-4 How can we define the direction of a standard-form vector in xyz space whose terminating point has the coordinates (xb,yb,zb)? Answer 8-4 The x, y, and z coordinates implicitly contain all the information we need to define the direction of a standard-form vector in Cartesian three-space. But this information is "indirect." Alternatively, we can define the vector's direction if we know the measures of the angles qx, qy, and qz that the vector subtends relative to the +x, +y, and +z axes, respectively. These angles are never negative, and they're never larger than p. There is a one-to-one correspondence between all possible vector orientations and all possible values of the ordered triple (qx,qy,qz). Question 8-5 Imagine two Cartesian xyz space vectors a and b, in standard form with terminating-point coordinates a = (xa,ya,za) and b = (xb,yb,zb) How can we find the sum a + b? How can we find the difference a - b? How can we find the difference b - a? How can we calculate the Cartesian xyz space negative of a vector that's in standard form? How does the Cartesian negative compare with the original vector? How do the differences a - b and b - a compare? Answer 8-5 We can calculate the sum vector a + b using the formula a + b = [(xa + xb),( ya + yb),(za + zb)] We can calculate the difference vector a − b using the formula a - b = [(xa − xb),( ya − yb),(za − zb)] We can find the difference vector b - a using the formula b - a = [(xb − xa),( yb − ya),(zb − za)] To find the Cartesian xyz space negative of a vector that's in standard form, we take the negatives of all three terminating-point coordinate values. For example, if we have b = (xb,yb,zb) then its Cartesian negative is -b = (−xb,−yb,−zb) 200 Review Questions and Answers The Cartesian negative has the same magnitude as the original vector, but points in the opposite direction. In xyz space, the difference vector b - a is always equal to the Cartesian negative of the difference vector a - b. Question 8-6 What's the left-hand Cartesian product of a scalar and a vector in xyz space? What's the righthand Cartesian product of a vector and a scalar in xyz space? How do they compare? Answer 8-6 Consider a real-number constant k, along with a standard-form vector a defined in xyz space as a = (xa,ya,za) The left-hand Cartesian product of k and a is ka = (kxa,kya,kza) The right-hand Cartesian product of a and k is ak = (xak,yak,zak) For all real numbers k and all Cartesian xyz space vectors a, we can be sure that ka = ak Question 8-7 What are the three standard unit vectors (SUVs) in Cartesian xyz space? Answer 8-7 The three SUVs in Cartesian xyz space are defined as the standard-form vectors i = (1,0,0) j = (0,1,0) k = (0,0,1) Any Cartesian xyz space vector in standard form can be split up into a sum of scalar multiples of the three SUVs. The scalar multiples are the coordinates of the ordered triple representing the vector. For example, suppose we have a = (xa,ya,za) Part One 201 We can break the vector a up in the following manner:= xai + yaj + zak Question 8-8 Suppose we have two standard-form vectors in Cartesian xyz space, defined as a = (xa,ya,za) and b = (xb,yb,zb) How can we calculate the dot product a • b? How can we calculate the dot product b • a? How do they compare? Answer 8-8 We can calculate the vectors as determined in the plane containing them both, rotating from a to b. In the same fashion, we can calculate b • a using the formula b • a = xbxa + ybya + zbza Alternatively, it is b • a = rbra cos qba where rb is the magnitude of b, ra is the magnitude of a, and qba is the angle between the vectors as determined in the plane containing them both, rotating from b to a. The dot product is commutative. In other words, for all vectors a and b in Cartesian xyz space, we can be sure that a•b=b•a 202 Review Questions and Answers Question 8-9 How can we find the cross product of two standard-form vectors a and b in three-space if we know their magnitudes and the angle between them? Answer 8-9 The cross product a ë b is a vector perpendicular to the plane containing both a and b, and whose magnitude ra×b is given by ra×b = rarb sin qab where ra is the magnitude of a, rb is the magnitude of b, and qab is the angle between a and b, expressed in the rotational sense going from a to b. We should define the angle so that it's always within the range 0 ≤ qab ≤ p If we look at a and b from some point far outside of the plane containing them, and if qab turns through a half circle or less counterclockwise as we go from a to b, then the crossproduct vector a ë b points toward us. If qab turns through a half circle or less clockwise as we go from a to b, then a ë b points away from us. Question 8-10 Imagine that we have two vectors in xyz space whose coordinates are a = (xa,ya,za) and b = (xb,yb,zb) How can we express a ë b as an ordered triple? Answer 8-10 We can plug in the coordinate values directly into the formula a ë b = [( yazb − zayb),(zaxb − xazb),(xayb − yaxb)] Chapter 9 Question 9-1 How do we determine the cylindrical coordinates of a point in three-space? Answer 9-1 We "paste" a polar plane onto a Cartesian xy plane, creating a reference plane. The positive Cartesian x axis is the reference axis. To determine the cylindrical coordinates of a point P, we first locate its projection point, P ′ on the reference plane: Part One 203 • The direction angle q is expressed counterclockwise from the reference axis to the ray that goes out from the origin through P ′. • The radius r is the distance from the origin to P ′. • The height h is the vertical displacement (positive, negative, or zero) from P ′ to P. The basic scheme is shown in Fig. 10-7. We express the cylindrical coordinates of our point of interest as an ordered triple: P = (q,r,h) Question 9-2 Can we have nonstandard direction angles in cylindrical coordinates? Can we have negative radii? Are there any restrictions on the values of the height coordinate? Answer 9-2 Theoretically, we can have a nonstandard direction angle. But if we come across that situation, it's best to add or subtract whatever multiple of 2p will bring the direction angle into the preferred range 0 ≤ q < 2p. If q ≥ 2p, it represents at least one complete counterclockwise rotation from the reference axis. If q < 0, it represents clockwise rotation from the reference axis. We can have a negative radius in theoretical terms. However, if we come across that sort of situation, it's best to reverse the direction angle and then consider the radius positive. If r < 0, we can take the absolute value of the negative radius and use it as the radius coordinate. Then we must add or subtract p to or from q to reverse the direction, while also making sure that the new angle is larger than 0 but less than 2p. +z Reference axis P +y h q –x +x P¢ r Reference plane –y –z Figure 10-7 Illustration for Question and Answer 9-1. 204 Review Questions and Answers The height h can be any real number. There are no restrictions on it whatsoever. We have h > 0 if and only if P is above the reference plane, h < 0 if and only if P is below the reference plane, and h = 0 if and only if P is in the reference plane. Question 9-3 Consider a point P = (q,r,h) in cylindrical coordinates. How can we determine the coordinates of P in Cartesian xyz space? Answer 9-3 The Cartesian x value of P is x = r cos q The Cartesian y value is y = r sin q The Cartesian z value is z=h Question 9-4 Consider a point P = (x,y,z) in Cartesian three-space. How can we find the direction angle q of the point P in cylindrical coordinates? Answer 9-4 Cartesian-to-cylindrical angle conversion is the same as Cartesian-to-polar angle conversionQuestion 9-5 Consider a point P = (x,y,z) in Cartesian three-space. How can we find the radius r of the point P in cylindrical coordinates? How can we find the height h of the point P in cylindrical coordinates? Part One 205 Answer 9-5 To find the cylindrical radius coordinate of P, we find the distance between its projection point P ′ and the origin in the xy plane. The z value is irrelevant, so the formula is r = (x2 + y2)1/2 The cylindrical height is simply equal to z. The x and y values are irrelevant, so the formula is h=z Question 9-6 How do we determine the spherical coordinates of a point in three-space? Answer 9-6 We start with a Cartesian reference plane. The positive Cartesian x axis forms the reference axis. To determine the spherical coordinates of a point P, we first locate its projection point, P′, on the reference plane: • The horizontal angle q turns counterclockwise in the reference plane from the reference axis to the ray that goes out from the origin through P ′. • The vertical angle f turns downward from the vertical axis to the ray that goes out from the origin through P. • The radius r is the straight-line distance from the origin to P. The basic scheme is shown in Fig. 10-8. We express the spherical coordinates as an ordered triple P = (q,f,r) +z Reference axis f P +y r –x q +x P¢ Reference plane –y –z Figure 10-8 Illustration for Question and Answer 9-6. 206 Review Questions and Answers Question 9-7 Are their any restrictions on the horizontal or vertical angles in spherical coordinates? Are there any restrictions on the radius? Answer 9-7 Theoretically, we can have a nonstandard horizontal direction angle, but it's best to add or subtract whatever multiple of 2p will bring it into the preferred range 0 ≤ q < 2p. If q ≥ 2p, it represents at least one complete counterclockwise rotation from the reference axis. If q < 0, it represents clockwise rotation from the reference axis. Theoretically, we can have a nonstandard vertical angle, but it's best to restrict it to the range 0 ≤ f ≤ p. We can do that by making sure that we traverse the smallest possible angle between the positive z axis and the ray connecting the origin with P. The radius can be any real number, but things are simplest if we keep it nonnegative. If we find ourselves working with a negative radius, we should reverse the direction by adding or subtracting p to or from both angles, making sure that we end up with 0 ≤ q < 2p and 0 ≤ f ≤ p. Then we must take the absolute value of the negative radius and use it as the radius coordinate. Question 9-8 Consider a point P = (q,f,r) in spherical coordinates. How can we determine the coordinates of P in Cartesian xyz space? Answer 9-8 The Cartesian x value of P is x = r sin f cos q The Cartesian y value is y = r sin f sin q The Cartesian z value is z = r cos f Question 9-9 Consider a point P = (x,y,z) in Cartesian three-space. How can we find the horizontal angle coordinate q of the point P in spherical coordinates? Answer 9-9 The Cartesian-to-spherical horizontal-angle conversion process is identical to the Cartesianto-cylindrical direction-angle conversion process: • If x = 0 and y = 0, then q = 0 by default. • If x > 0 and y = 0, then q = 0. • If x > 0 and y > 0, then q = Arctan ( y /x). CHAPTER 11 Relations in Two-Space If you've taken the course Algebra Know-It-All, you've already had some basic training on relations and functions. They've been mentioned a few times in this book as well. Let's look more closely at how relations and functions behave in two-space. What's a Two-Space Relation? A relation is a special way of assigning, or mapping, the elements of a "source" set to the elements of a "destination" set. In two-space, both the source and destination sets usually consist of numbers. The sets might be identical, partially overlapping, or entirely disjoint. For example, we might have a relation between the set of negative integers and the set of positive integers, or the set of positive real numbers and the set of all real numbers, or the set of all real numbers and itself. Ordered pairs Any point in the Cartesian plane or the polar plane can be uniquely represented by an ordered pair in which a value of the independent variable (an element of the source set) is listed first, followed by a value of the dependent variable (an element of the destination set). The domain is the set of all values of the independent variable for which the relation produces defined values of the dependent variable. The range is the set of all values of the dependent variable that come from the elements of the domain. Here's an example of a relation written as a set of ordered pairs: {(3,2),(4,3),(5,4),(6,5)} The domain of this particular relation (let's call it set D) is the set of first numbers in the ordered pairs. Therefore D = {3,4,5,6} 211 212 Relations in Two-Space The range (let's call it set R) of the relation is the set of second numbers in the ordered pairs, so R = {2,3,4,5} Injection, surjection, and bijection Imagine a relation between numbers x in a set X and numbers y in a set Y. Suppose that each number x in set X corresponds to one, but only one, number y in set Y. Also suppose that no number in Y has more than one "mate" in X. (There might be some numbers in Y without any "mate" in X.) A relation of this type is called an injection. In some older texts, it's called one-to-one. Now imagine a relation that assigns the elements of set X to the elements of set Y so that every element of Y has at least one "mate" in X. This type of relation is called a surjection. Set Y is completely "spoken for." A surjection is sometimes called an onto relation, because it maps (assigns) the values from set X completely onto the entire set Y. Finally, imagine a relation that is both an injection and a surjection. This type of relation is called a bijection. In older texts, you might see it referred to as a one-to-one correspondence (not to be confused with one-to-one, which means an injection). A bijection assigns every value of x in set X to a unique value of y in set Y. Conversely, every y in set Y corresponds to a unique value of x in set X. In this context, "a unique value" means "one and only one value" or "exactly one value." Example 1 Relations are commonly represented by equations. Here's an example of a simple two-space relation that subtracts 1 from every value in the domain to generate values in the range: y=x−1 This relation could describe a one-to-one correspondence between the elements of the domain X = {3,4,5,6} and the elements of the range Y = {2,3,4,5} which we saw a few moments ago. If we allow the domain of the relation to extend over the entire set of real numbers, then the range also covers the entire set of real numbers. When we put specific values of x into the equation, we get results such as the following: • • • • • • If x = −13, then (x,y) = (−13,−14). If x = −1.6, then (x,y) = (−1.6,−2.6). If x = 0, then (x,y) = (0,−1). If x = 1, then (x,y) = (1,0). If x = 3/2, then (x,y) = (3/2,1/2). If x = 81/2, then (x,y) = [81/2,(81/2 − 1)]. What's a Two-Space Relation? 213 For every value of x, the relation assigns one and only one value of y. The converse is also true; for every value of y, there is one and only one corresponding value of x. Example 2 Next, let's consider a real-number relation that squares each element in the domain to produce values in the range. We can write this relation as the following equation: y = x2 In the set of real numbers, this relation is defined for all possible values of x, but we never get any negative values of y. The range is the set of all y such that y ≥ 0. When we plug specific numbers into this equation, we get results such as the following: • • • • • • • If x = −4, then (x,y) = (−4,16). If x = −1, then (x,y) = (−1,1). If x = −1/2, then (x,y) = (−1/2,1/4). If x = 0, then (x,y) = (0,0). If x = 1/2, then (x,y) = (1/2,1/4). If x = 1, then (x,y) = (1,1). If x = 4, then (x,y) = (4,16). For every value of x, the relation assigns a unique value of y, but for every assigned value of y except y = 0 in the range, the domain contains two values of x. Example 3 Now let's look at a real-number relation that takes the positive or negative square root of elements in the domain to get elements in the range. We can write it as the equation y = ±(x1/2) When we plug in some numbers here, we get results like the following: • • • • • • If x = 1/9, then (x,y) = (1/9,1/3) or (1/9,−1/3). If x = 1/4, then (x,y) = (1/4,1/2) or (1/4,−1/2). If x = 1, then (x,y) = (1,1) or (1,−1). If x = 4, then (x,y) = (4,2) or (4,−2). If x = 9, then (x,y) = (9,3) or (9,−3). If x = 0, then (x,y) = (0,0). In the set of real numbers, the domain of this relation is confined to nonnegative values of x. That is, the domain is the set of all x such that x ≥ 0. For every positive value of x in the domain, there are two values of y in the range. If x = 0, then y = 0. The range encompasses all possible real-number values of y. For any value of y in the range, there exists one and only one corresponding value of x in the domain. 214 Relations in Two-Space Example 4 Finally, let's examine a real-number relation that takes the nonnegative square root of values in the domain to get values in the range. We can denote it as y = x1/2 The domain is the set of all real numbers x such that x ≥ 0, and the range is the set of all real numbers y such that y ≥ 0. Following are a few examples of what happens when we input values of x into this equation: If x = 1/9, then (x,y) = (1/9,1/3). If x = 1/4, then (x,y) = (1/4,1/2). If x = 1, then (x,y) = (1,1). If x = 4, then (x,y) = (4,2). If x = 9, then (x,y) = (9,3). If x = 0, then (x,y) = (0,0). • • • • • • For every x in the domain, there is one and only one y in the range. The converse is also true. For every y in the range, there is one and only one x in the domain. Are you confused? Sometimes a relation fails to take all of the elements of the source or destination sets into account. Figure 11-1 illustrates a generic example of a situation of this sort using a graphical scheme called a Venn diagram: • • • • The entire source set is called the maximal domain. The entire destination set is called the co-domain. The domain of a relation is a subset of its maximal domain. The range of a relation is a subset of its co-domain. Here's a challenge! Classify each of the relations in Examples 1 through 4 as an injection, a surjection, a bijection, or "none of them" from the set of real numbers to itself. Solution In each of these relations, our source set is the entire set of real numbers, and that's not necessarily the domain. Also, our destination set is the entire set of reals, and that's not necessarily the range: • In Example 1, we subtract 1 from each value of the independent variable to get a value of the dependent variable. This operation produces a one-to-one correspondence between the set of real numbers and itself. For every value we input, we get a unique output. Also, every output value is the result of one and only one input value. It follows that this relation is a bijection. What's a Two-Space Relation? 215 Domain Maximal domain Mapping Co-domain Range Figure 11-1 The domain of a relation is a subset of the maximal domain. The range is a subset of the co-domain. • In Example 2, we square each value of the independent variable to get a value of the dependent variable. For every value of the dependent variable except 0, two different values of the independent variable are assigned to it. The relation is not an injection, because it's not one-to-one. It can't be a bijection, then, either. The independent variable can attain any real value, but the dependent variable can never be negative, so this relation is not a surjection onto the set of real numbers. We must therefore classify this relation as "none of them." • In Example 3, we take the positive or negative square root of each value of the independent variable to get a value of the dependent variable. The independent variable can't be negative, but the dependent variable can be any real number. The relation is therefore a surjection onto the set of real numbers. But it's not an injection, because most values of the independent variable map to two values of the dependent variable. It's not a bijection then, either. • In Example 4, we take the nonnegative square root of each value of the independent variable to get a value of the dependent variable. As in Example 3, the independent variable can never be negative. Neither can the dependent variable. In this case we don't have an injection, because some real numbers in the source set don't have any counterparts in the destination set. We don't have a surjection either, because the range fails to cover the entire set of real numbers. We must categorize this as a "none of them" relation. 216 Relations in Two-Space What's a Two-Space Function? In two-space, a function is a relation that never maps any value of the independent variable to more than one value of the dependent variable. All functions are relations, but not all relations are functions. Figure 11-2 shows Venn diagrams of a "legal" assignment for a function (left) and an "illegal" assignment (right). The vertical-line test In the Cartesian xy plane, suppose that x is the independent variable, and we plot it against the horizontal axis. Also suppose that y is the dependent variable, and we plot it against the vertical axis. When we see the graph of a simple relation, it usually appears as a line or curve. More complicated relations may graph as groups of lines and/or curves. We can test the graph of any relation in the Cartesian xy plane to see if it represents a function of x. Imagine an infinitely long, movable vertical line that's always parallel to the dependent-variable axis (the y axis). Suppose that we're free to move the line to the left or right, so it intersects the independent-variable axis (the x axis) wherever we want. If the graph is a function of x, then the movable vertical line never intersects the graph of our relation at more than one point. If, in any position, the vertical line intersects the graph at more than one point, then the relation is not a function of x. We call this exercise the vertical-line test. Domain Range A function can do this ... Domain Range ... but not this! Figure 11-2 A true function never assigns any element in its domain to more than one element in its range. What's a Two-Space Function? 217 Example 1 revisited Let's take another look at the relation given by Example 1 in the previous section. We described it using the following equation: y=x−1 Figure 11-3 is a graph of this equation in the Cartesian xy plane. It's a straight line with a slope of 1 and a y intercept of −1. If we imagine an infinitely long, movable vertical line sweeping back and forth, it's easy to see that the vertical line never intersects our graph at more than one point. Therefore, the relation is a function. Example 2 revisited The relation in Example 2 in the previous section has a graph that's a parabola opening upward, as shown in Fig. 11-4. The equation is y = x2 The vertex of the parabola represents the absolute minimum value of the relation, and it coincides with the coordinate origin (0,0). The curve rises symmetrically on either side of the y axis. It's not difficult to see that a movable vertical line never intersects the parabola at more than one point. This fact tells us that the relation is a function of x. y 6 4 2 x –6 –4 –2 2 –4 –6 4 6 Movable vertical line Figure 11-3 Cartesian graph of the relation y = x − 1. The vertical-line test reveals that it's a function of x. 218 Relations in Two-Space y 6 4 2 x –6 –4 –2 2 4 6 –2 –4 –6 Movable vertical line Figure 11-4 Cartesian graph of the relation y = x2. The vertical-line test reveals that it's a function of x. Example 3 revisited Figure 11-5 is a graph of the relation we saw in Example 3 in the previous section. The equation for that relation was stated as y = ±(x1/2) In this case, the graph is a parabola that opens to the right. The vertex coincides with the coordinate origin, but there is no absolute minimum or maximum for the dependent variable. When we construct a movable vertical line in this situation, we find that it doesn't intersect the graph when x < 0. When x = 0, the vertical line intersects the graph at the single point (0,0). When x > 0, the vertical line intersects the graph at two points. Therefore, this relation is not a function of x. Example 4 revisited Figure 11-6 is a graph of the relation we saw in Example 4 in the previous section. It's the upper half of the parabola of Fig. 11-5, with the point (0,0) included. The equation is y = x1/2 The vertical-line test tells us that this relation is a function of x. No matter where we position the vertical line, it never intersects the graph more than once. 220 Relations in Two-Space Are you confused? By now you might wonder, "When we have a relation where the independent variable is represented by the polar angle q and the dependent variable is represented by the polar radius r, how can we tell if the relation is a function of q?" It's easy, but there's a little trick involved. We can draw the graph of the relation in a Cartesian plane with q on the horizontal axis and r on the vertical axis. We must allow both q and r to attain all possible real-number values. Once we've drawn the graph of the polar relation the Cartesian way, we can use the Cartesian vertical-line test to see whether or not the relation is a function of q. Here's a challenge! Consider the relation between an independent variable x and a dependent variable y such that x2 − y2 = 1 Sketch a graph of this relation in the Cartesian xy plane. Use the vertical-line test to determine, on the basis of the graph, whether or not this relation is a function of x. Solution Figure 11-7 is a graph of this relation. It's a geometric figure called a hyperbola. The vertical-line test tells us that the relation is not a function of x. y 6 4 Movable vertical line 2 x –6 –4 –2 2 4 6 –2 –4 –6 Figure 11-7 Cartesian graph of the relation x2 − y2 = 1. The vertical-line test reveals that it isn't a function of x. What's a Two-Space Function? 221 Here's another challenge! Consider the relation between an independent variable q and a dependent variable r defined by the equation r = 2q /p Sketch a graph of this equation in the polar plane. Then redraw it in the Cartesian plane with q on the horizontal axis and r on the vertical axis. Use the vertical-line test to determine, on the basis of the graph, whether or not the relation is a function of q. Solution Figure 11-8 is a graph of the equation in the polar plane. It's a pair of "dueling spirals." When we draw the graph of the equation in a Cartesian plane with q on the horizontal axis and r on the vertical axis, we get a straight line that passes through the origin with a slope of 2 /p, as shown in Fig. 11-9. The Cartesian vertical-line test indicates that the relation is a function of q. Figure 11-8 Polar graph of the relation r = 2q /p. Each radial division represents 1 unit. 222 Relations in Two-Space r 3 2 1 q –3p –2p p –p –2 2p 3p Movable vertical line –3 Figure 11-9 Cartesian graph of the relation r = 2q /p. The vertical-line test shows that it's a function of q. Algebra with Functions Functions can always be written as equations. Therefore, when we want to add, subtract, multiply, or divide two functions, we can use ordinary algebra to add, subtract, multiply, or divide both sides of the equations representing the functions. Cautions There are three "catches" in the algebra of functions. Whenever we add, subtract, multiply, or divide one function by another, we must watch out for these potential pitfalls. Otherwise, we might get misleading or incorrect results: • The independent variables of the two functions must match. That is, they must describe the same parameters or phenomena. We can't algebraically combine functions of two different variables in an attempt to get a new function in a single variable. If we try to do that, we won't know which variable the resultant function should operate on. • The domain of the resultant function is the intersection of the domains of the two functions we combine. Any element in the domain of a sum, difference, product, or ratio function must belong to the domains of both of the constituent functions. The domain of a ratio function may, however, be restricted even further if the denominator function becomes 0 anywhere in its domain. Algebra with Functions 223 • If we divide a function by another function, the resultant function is undefined for any value of the independent variable where the denominator function becomes 0. This can, and often does, restrict the domain of the resultant function to a proper subset of the domain we would get if we were to add, subtract, or multiply the same two functions. New names for old functions So far in this chapter, we've encountered four different functions. Three of them are functions of x; the fourth is a function of q. Following are the equations of the functions once again, for reference: y=x−1 y = x2 y = x1/2 r = 2q /p Let's assign these functions specific names, so that we can write them in the conventional function notation. These are f1 (x) = x − 1 f2 (x) = x2 f3 (x) = x1/2 f4 (q) = 2q /p Sum of two functions When we want to find the sum of two functions, we add both sides of their equations. This can be done in either order, producing identical results. For f1 and f2, we have ( f1 + f2)(x) = f1 (x) + f2 (x) = (x − 1) + x2 = x2 + x − 1 and ( f2 + f1)(x) = f2 (x) + f1 (x) = x2 + (x − 1) = x2 + x − 1 For f1 and f3, we have ( f1 + f3)(x) = f1 (x) + f3 (x) = (x − 1) + x1/2 = x + x1/2 − 1 and ( f3 + f1)(x) = f3 (x) + f1 (x) = x1/2 + (x − 1) = x + x1/2 − 1 For f2 and f3, we have ( f2 + f3)(x) = f2 (x) + f3 (x) = x2 + x1/2 224 Relations in Two-Space and ( f3 + f2)(x) = f3 (x) + f2 (x) = x1/2 + x2 = x2 + x1/2 It's customary to write polynomials (sums or differences of a variable raised to powers) with the largest power first, and then descending powers after that. That's why some of the sums and differences have been rearranged in the above examples. What about f4? The independent variable in f4 doesn't match the independent variable in any of the other three functions, so we can't combine and manipulate the equations as if the variables did match. We can add the equations straightaway, but that doesn't tell us much. For example, we can say that f2 (x) + f4 (q) = x2 + 2q /p but that's all we can do with it. It's like trying to add minutes to millimeters. We can't get a resultant function of a single variable. The same problem occurs if we try to subtract, multiply, or find a ratio involving f4 and any of the other three functions. Difference between two functions When we want to find the difference between two functions, we subtract both sides of their equations. This can be done in either order, usually producing different results. For f1 and f2, we have ( f1 − f2)(x) = f1 (x) − f2 (x) = (x − 1) − x2 = −x2 + x − 1 and ( f2 − f1)(x) = f2 (x) − f1 (x) = x2 − (x − 1) = x2 − x + 1 For f1 and f3, we have ( f1 − f3)(x) = f1 (x) − f3 (x) = (x − 1) − x1/2 = x − x1/2 − 1 and ( f3 − f1)(x) = f3 (x) − f1 (x) = x1/2 − (x − 1) = −x + x1/2 + 1 For f2 and f3, we have ( f2 − f3)(x) = f2 (x) − f3 (x) = x2 − x1/2 and ( f3 − f2)(x) = f3 (x) − f2 (x) = x1/2 − x2 = −x2 + x1/2 226 Relations in Two-Space Are you confused? You ask, "Do the commutative, associative, distributive, and other rules of arithmetic and algebra work with functions in the same ways as they do with numbers and variables?" The answer is a qualified yes. All the rules of addition, subtraction, multiplication, and division of functions are identical to the rules for arithmetic or algebra involving numbers or variables, as long as we heed the cautions outlined earlier in this section. Here's a challenge! Define the real-number domains of all the sum, difference, product, and ratio functions we've found in this section. Solution We found the real-number domains (which we can call the real domains for short) of the functions f1, f2, and f3 earlier in this chapter. Here they are again, for reference: • The real domain of f1 (x), which subtracts 1 from x, is the set of all reals. • The real domain of f2 (x), which squares x, is the set of all reals. • The real domain of f3 (x), which takes the nonnegative square root of x, is the set of all nonnegative reals. The real domains of the sum, difference, and product functions are the intersections of these. Let's list them: • • • • • • The real domains of ( f1 + f2), ( f1 − f2), and ( f1 × f2) are the set of all real numbers. The real domains of ( f2 + f1), ( f2 − f1), and ( f2 × f1) are the set of all real numbers. The real domains of ( f1 + f3), ( f1 − f3), and ( f1 × f3) are the set of all nonnegative real numbers. The real domains of ( f3 + f1), ( f3 − f1), and ( f3 × f1) are the set of all nonnegative real numbers. The real domains of ( f2 + f3), ( f2 − f3), and ( f2 × f3) are the set of all nonnegative real numbers. The real domains of ( f3 + f2), ( f3 − f2), and ( f3 × f2) are the set of all nonnegative real numbers. The real domains of the ratio functions are subsets of the real domains for the sum, product, and difference functions. We have to look at each ratio function and check to see where the denominators are equal to 0: • • • • • • The denominator of ( f1 / f2) becomes 0 when x = 0. The denominator of ( f2 / f1) becomes 0 when x = 1. The denominator of ( f1 / f3) becomes 0 when x = 0. The denominator of ( f3 / f1) becomes 0 when x = 1. The denominator of ( f2 / f3) becomes 0 when x = 0. The denominator of ( f3 / f2) becomes 0 when x = 0. On the basis of these observations, we can create one final list: • The real domain of ( f1 / f2) is the set of all real numbers except 0. • The real domain of ( f2 / f1) is the set of all real numbers except 1. Practice Exercises • • • • 227 The real domain of ( f1 / f3) is the set of all strictly positive real numbers. The real domain of ( f3 / f1) is the set of all nonnegative real numbers except 1. The real domain of ( f2 / f3) is the set of all strictly positive real numbers. The real domain of ( f3 / f2) is the set of all strictly positive real numbers Examine the relation illustrated in Fig. 11-10. Suppose that for every element x in set X, there exists at most one element y in set Y. Is this relation an injection? Is it a surjection? Is it a bijection? Note that the range of the relation is the entire co-domain. Is that true of all relations? As described here, is this relation a function? Explain each answer. Maximal domain Relation Set X Set Y coincides with co-domain Figure 11-10 Illustration for Problem 1. 2. Imagine a relation in which the domain X is the set of all positive rational numbers, while the range Y is the set of all positive integers. Let's call the independent variable x and the dependent variable y. Suppose that for any x in set X, the relation rounds x up to the next larger integer to obtain the corresponding element y in set Y. Is this relation an injection? Is it a surjection? Is it a bijection? Is it a function of x? Explain each answer. 3. Suppose that we reverse the action of the relation described in Problem 2. Let the domain X be the set of all positive integers, while the range Y is the set of all positive rationals. Suppose that for any value of the independent variable x in set X, the relation 228 Relations in Two-Space maps to the set of all rationals in the half open interval (x − 1,x]. Is this relation an injection? Is it a surjection? Is it a bijection? Is it a function of x? Explain each answer. 4. Can a relation whose graph is a circle or ellipse in the Cartesian xy plane ever be a function of x? Why or why not? 5. Can a relation whose graph is a circle in the polar qr plane ever be a function of q ? Why or why not? 6. Find all the sums, differences, products, and ratios of f (x) = x + 2 and g (x) = 3 7. Find all the sums, differences, products, and ratios of f (x) = x + 1 and g (x) = x − 1 8. Find all the sums, differences, products, and ratios of f (x) = x−1 and g (x) = x−2 9. Find all the sums, differences, products, and ratios of f (x) = sin2 q and g (x) = cos2 q 10. What are the real-number domains of all the original and derived functions in Problems 6 through 9? CHAPTER 12 Inverse Relations in Two-Space Any relation in two-space has a unique inverse relation, which can be called simply the inverse if we understand that we're dealing with a relation. We denote the fact that a relation is an inverse by writing a superscript −1 after its name. For example, if we have a relation f (x), then its inverse is f −1(x). Finding an Inverse Relation A relation's inverse does the opposite of whatever the original relation does. To find the inverse of a relation, we can manipulate the equation so that the independent and dependent variables switch roles. We must therefore transpose the domain and range. The algebraic way Suppose we have a relation f (x). The inverse of f, which we call f −1, is another relation such that f −1[ f (x)] = x for all possible values of x in the domain of f, and f [ f −1( y)] = y for all possible values of y in the range of f. When we talk or write about an inverse relation, it's customary to swap the names of the variables so the inverse relation calls the independent and dependent variables by their original names. That means the preceding equation can be rewritten as f [ f −1(x)] = x for all possible values of x in the domain of f −1. 229 230 Inverse Relations in Two-Space An example Let's find the inverse of the following relation: f (x) = x + 2 If we call the dependent variable y, then we can rewrite our relation as y=x+2 Swapping the names of the variables, we get x=y+2 which can be manipulated with algebra to obtain y=x−2 If we replace the new variable y by the relation notation f −1(x), we get f −1(x) = x − 2 The domain and range of the original relation f both span the entire set of real numbers. Therefore, the domain and range of the inverse relation f −1 also both span the entire set of reals. Another example Let's find the inverse of the following relation: g(x) = ±(x1/2) If we call the dependent variable y, then we can rewrite the relation as y = ±(x1/2) When we switch the names of the variables, we get x = ±( y1/2) Squaring both sides produces x2 = y Reversing the left- and right-hand sides gives us y = x2 Finding an Inverse Relation 231 Replacing y by g −1(x), we get g −1(x) = x 2 The domain of the original relation g spans the set of nonnegative reals, and the range of g spans the set of all reals. Therefore, the domain of g −1 includes all reals, while the range of g −1 is confined to the set of nonnegative reals. Still another example Let's find the inverse of the relation h(x) = x 1/2 When we write the 1/2 power of a quantity without including any sign, we mean the nonnegative square root of that quantity. If we call the dependent variable y, then we have y = x1/2 Swapping the names of the variables, we get x = y1/2 Squaring both sides, we obtain x2 = y Reversing the left- and right-hand sides of this equation yields y = x2 Replacing y by h −1(x), we get h −1(x) = x 2 Is the inverse of h identical to the inverse of g we obtained a few moments ago? It looks that way "on the surface," but it's not so simple when we examine the situation more closely. The domain of h spans the set of nonnegative reals, just as the domain of g does. But the range of h spans the set of nonnegative reals only (not the set of all reals, as the range of g does). Transposing, we must conclude that the domain and range of h −1 are both confined to the set of nonnegative reals. The relations h and h −1 are therefore restricted versions of g and g −1. The graphical way Imagine the line represented by the equation y = x in the Cartesian xy plane as a "point reflector." For any point that's part of the graph of the original relation, we can locate its 232 Inverse Relations in Two-Space 12-1 Any point on the graph of the inverse of a relation is the point's image on the opposite side of a point reflector line. The new coordinates are obtained by reversing the sequence of the ordered pair representing the original point. counterpart in the graph of the inverse relation by going to the opposite side of the point reflector, exactly the same distance away. Figure 12-1 shows how this works. The line connecting a point in the original graph and its "mate" in the inverse graph is perpendicular to the point reflector. The point reflector is a perpendicular bisector of every point-connecting line. Mathematically, we can do a point transformation of the sort shown in Fig. 12-1 by reversing the sequence of the ordered pair representing the point. For example, if (4,6) represents a point on the graph of a certain relation, then its counterpoint on the graph of the inverse relation is represented by (6,4). When we want to graph the inverse of a relation, we flip the whole graph over along a "hinge" corresponding to the point reflector line y = x. That moves every point in the graph of the original relation to its new position in the graph of the inverse. Figures 12-2, 12-3, and 12-4 show how this process works with the three relations we dealt with a few moments ago. The positions of the x and y axes haven't changed, but the values of the variables, as well as the domain and range, have been reversed. 236 Inverse Relations in Two-Space Are you confused? It's reasonable for you to wonder, "Can any relation be its own inverse?" The answer is yes. There are plenty of examples. Consider the following equation: x2 + y2 = 25 The Cartesian graph of this equation is a circle centered at the origin and having a radius of 5 units (Fig. 12-5). If we transpose the variables, we get y2 + x2 = 25 which is equivalent to the original relation. If we perform the graphical transformation by mirroring the circle around the line y = x, we get another circle having the same radius and the same center. Theoretically, all but two of points on the new circle are in different places than the points on the original circle, but the graph looks the same as the one shown in Fig. 12-5. Here's a challenge! Consider the following relation where the independent variable is x and the dependent variable is y: x 2/9 + y2/25 = 1 y Circle centered at the origin is symmetrical ... "Point reflector" line 6 4 2 x –6 –4 –2 2 4 6 –2 –4 –6 ... with respect to the "point reflector" line Figure 12-5 Cartesian graph of the relation x2 + y2 = 25. This relation is its own inverse. Finding an Inverse Relation 237 y 6 4 2 x –6 –4 –2 2 4 6 –2 –4 Ellipse centered at origin –6 Figure 12-6 Cartesian graph of the relation x2/9 + y2/25 = 1. Figure 12-6 is a graph of this relation in Cartesian coordinates. It's an ellipse centered at the origin. The distance from the center to the extreme right- or left-hand point on the ellipse measures 3 units, which is the square root of 9. The distance from the center to the uppermost or lowermost point on the ellipse measures 5 units, which is the square root of 25. Determine the inverse of this relation, and graph it. Solution We can obtain the inverse of this relation by swapping the variables. That gives us the equation y 2/9 + x 2/25 = 1 which can be rewritten as x 2/25 + y 2/9 = 1 Figure 12-7 illustrates the graphs of the original relation and its inverse in Cartesian coordinates. The new graph is another ellipse having the same shape as the original one, and centered at the origin just like the original one. But the horizontal and vertical axes of the ellipse have been transposed. The distance from the center to the extreme right- or left-hand point on the "inverse ellipse" measures 5 units, which is the square root of 25. The distance from the center to the uppermost or lowermost point on the "inverse ellipse" measures 3 units, which is the square root of 9. 238 Inverse Relations in Two-Space y Original relation "Point reflector" line 6 4 2 x –6 –4 –2 2 4 6 –2 –4 Inverse relation –6 Figure 12-7 Cartesian graph of the relation x2/25 + y2/9 = 1, the inverse of the relation graphed in Fig. 12-6. Finding an Inverse Function If a function is a bijection (that is, a perfect one-to-one correspondence) over a certain domain and range, then we can transpose the domain and range, and the resulting inverse relation will always be a function. If a function is many-to-one, then its inverse relation is one-to-many, so it's not a function. "Undoing" the work Suppose that f and f −1 are both true functions that are inverses of each other. Then for all x in the domain of either function, we have f −1[ f (x)] = x and f [ f −1(x)] = x An inverse function undoes the work of the original function in an unambiguous manner when the domains and ranges are restricted so that the original function and the inverse are both bijections. Finding an Inverse Function 239 Sometimes we can simply turn f "inside-out" to get an inverse relation, and the inverse will be a true function for all the values in the domain and range of f. But often, when we seek the inverse of a function f, we get a relation that's not a true function, because some elements in the range of f map from more than one element in the domain of f. When this happens, we must restrict f to define an inverse f −1 that's a true function. We can usually (but not always) find a way to "force" f −1 to behave as a true function by excluding all values of either variable that map to more than one value of the other variable. Once we've done that, we get a bijection, ensuring that there is no ambiguity or redundancy either way. Making a relation behave as a function A little while ago, we looked at a relation whose graph is a circle with a radius of 5 units (Fig. 12-5). The equation of that relation, once again, is x 2 + y 2 = 25 which can be rewritten as y 2 = 25 − x 2 and then morphed to y = ±(25 − x 2)1/2 If we use relation notation to express this equation and name the relation f, we have f (x) = ±(25 − x 2)1/2 The vertical-line test tells us that f is not a true function of x. We can modify it so that it becomes a function of x if we restrict the range to nonnegative values. Graphically, that eliminates the lower half of the circle, so that for every input value in the domain, we get only one output in the range. Figure 12-8A is an illustration of this function, which we can call f+ and define as f+(x) = (25 − x 2)1/2 Once again, we mustn't forget that when we take the 1/2 power of a quantity without including any sign, we mean, by default, the nonnegative square root of that quantity. The solid dots indicate that the plotted points are part of the range of f+. Now suppose that we eliminate the top half of the circle including the points (−5,0) and (5,0), getting the graph shown in Fig. 12-8B. The vertical-line test indicates that this is a true function of x. If we call this function f−, we can write f−(x) = −(25 − x2)1/2 The white dots (small open circles) tell us that the plotted points are not part of the range of f−. We can restrict the range further, say to values strictly larger than 1 or values smaller than or Finding an Inverse Function 241 equal to −2, and we'll get more true functions. You can doubtless imagine other restrictions we can impose on the range of the original relation f to get true functions of x. What about the inverse of f+ ? Let's manipulate f+ algebraically to find its inverse. If we call the dependent variable y, then y = (25 − x 2)1/2 Swapping the names of the variables, we get x = (25 − y 2)1/2 Squaring both sides, we obtain x 2 = 25 − y 2 Subtracting 25 from each side yields x 2 − 25 = −y 2 When we multiply through by −1 and transpose the left-hand and right-hand sides of the equation, we obtain y 2 = 25 − x 2 Taking the complete square root of both sides gives us y = ±(25 − x 2)1/2 Replacing y by f+−1(x) to indicate the inverse of f+, we get f+−1(x) = ±(25 − x 2)1/2 Does this look like the same thing as the inverse of the original relation f ? Don't be fooled; it isn't the same! We haven't quite finished our work. We must transpose the domain and range of f+ to get the domain and range of the inverse relation f+−1. The domain of f+ is the closed interval [−5,5], and the range of f+ is the closed interval [0,5]. Therefore, the domain of f+−1 is the closed interval [0,5], and the range of f+−1 is the closed interval [−5,5]. Figure 12-9A is a graph of this inverse relation. It's easy to see that f+−1 fails the vertical-line test, so it's not a true function of x. What about the inverse of f-? Now let's go through the algebra to figure out the inverse of the function f−. This process is almost identical to the work we just finished, but it's a good practice to carry it out step by step anyway. If we call the dependent variable y, then y = −(25 − x2)1/2 242 Inverse Relations in Two-Space y 6 (0, 5) is part of the graph 4 2 x A –6 –4 –2 2 4 6 –2 –4 (0, –5) is part of the graph –6 y 6 (0, 5) is not part of the graph 4 2 B x –6 –4 –2 2 4 6 –2 –4 (0, –5) is not part of the graph –6 Figure 12-9 At A, Cartesian graph of the inverse of the function y = (25 − x 2)1/2. At B, Cartesian graph of the inverse of the function y = −(25 − x 2)1/2. Vertical-line tests indicate that neither of these inverse relations is a function. Finding an Inverse Function 243 Switching the names of the variables, we get x = −(25 − y2)1/2 Squaring both sides, we obtain x2 = 25 − y2 Subtracting 25 from each side gives us x2 − 25 = −y2 When we multiply through by −1 and transpose the left- and right-hand sides of the equation, we get y2 = 25 − x2 Taking the complete square root of both sides, we have y = ±(25 − x2)1/2 Replacing y by f−−1(x) to indicate the inverse of f−, we get f−−1(x) = ±(25 − x2)1/2 We transpose the domain and range of f− to get the domain and range of f−−1. Things get a little tricky here. Refer again to Fig. 12-8B. The domain of f− is the open interval (−5,5), and the range of f− is the half-open interval [−5,0). Transposing, we can see that the domain of f−−1 is the half-open interval [−5,0), and the range of f−−1 is the open interval (−5,5). Figure 12-9B is a graph of f−−1. This inverse relation fails the vertical-line test, so it's not a true function of x. Making an inverse behave as a function Do you get the idea that we can't make the relation graphed in Fig. 12-5 behave as a function whose inverse is another function, no matter what limitations we impose on the domain and range? Don't give up. There are plenty of ways. For example, we can restrict both the domain and the range of the original relation x2 + y2 = 25 to values that all show up in the first quadrant of the Cartesian plane. When we do that, the domain and range are both narrowed down to the open interval (0,5). The relation becomes a true function of x, and its inverse also becomes a true function. Similar things happen if we restrict both the domain and the range to values that show up entirely in the second quadrant, entirely in the third quadrant, or entirely in the fourth quadrant. Feel free to draw the graphs, put a point reflector line to work, and see for yourself. 244 Inverse Relations in Two-Space Are you confused? You might ask, "We've seen an example of a relation that's its own inverse. Can a function be its own inverse?" The answer is yes. The function f (x) = x is its own inverse; the domain and range both span the entire set of real numbers. It's the ultimate in simplicity. The function's graph coincides with the point reflector line, so it's identical to its own reflection! We have f −1[ f (x)] = f −1(x) = x so therefore f [ f −1(x)] = f (x) = x Another example Consider g(x) = 1/x, with the restriction that the domain and range can attain any real-number value except zero. This function is its own inverse. We have g −1[ g(x)] = g −1(1/x) = 1/(1/x) = x so therefore g [g −1(x)] = g (1/x) = 1/(1/x) = x Still another example Consider the function h(x) = 3 for all real numbers x. Figure 12-10 shows its graph. When we transpose the variables, domain, and range, we must set x = 3 for h −1(x) to mean anything. Then we end up with all the real numbers at the same time. This relation fails the vertical-line function test in the worst possible way, because the graph is itself a vertical line (Fig. 12-11). Here's a challenge! Consider the following three functions: f (x) = x − 11 g(x) = x2/4 h(x) = −32x5 The inverse of one of these functions is not a function. Which one? 246 Inverse Relations in Two-Space Solution The inverse of g is not a function. If we call the dependent variable y, we get y = x2/4 The domain is the entire set of reals, and the range is the set of non-negative reals. If we swap the names of the independent and dependent variables, we get x = y2/4 which is the same as y2 = 4x Taking the complete square root of both sides gives us y = ± 2x1/2 The plus-or-minus symbol indicates that for every nonzero value of the independent variable x that we input to this relation, we get two values of the dependent variable y, one positive and the other negative. We can also write g −1(x) = ± 2x1/2 The original function g is two-to-one (except when x = 0). That's okay. But the inverse relation is one-to-two except when x = 0. That prevents g −1 from qualifying as a true function. The other two functions, f and h, have inverses that are also functions. Both f and h are one-toone, so their inverses are also one-to-one. We have f (x) = x − 11 and f −1(x) = x + 11 We also have h(x) = −32x5 and h −1(x) = (−x)1/5/2 Practice Exercises 247 Use algebra to find the inverse of the relation f (x) = 2x + 4 2. Use algebra to find the inverse of the relation g(x) = x2 − 4x + 4 3. Use algebra to find the inverse of the relation h(x) = x3 − 5 4. Determine the real-number domains and ranges of the relations and inverses from the statements and solutions of Problems 1, 2, and 3. 5. Consider the two-space relation x2/4 − y2/9 = 1 Figure 12-12 is a graph of this relation in Cartesian coordinates. It's a hyperbola centered at the origin, opening to the right and left, and crossing the x axis at (2,0) and (−2,0). y 6 4 2 x –6 –4 4 –2 –4 –6 Figure 12-12 Illustration for Problem 5. 6 248 Inverse Relations in Two-Space Call the independent variable x and the dependent variable y. Call the relation f. Determine f (x) and f −1(x) mathematically. State them both using relation notation. 6. What is the real-number domain of the relation f (x) that you determined when you solved Problem 5? What is its real-number range? 7. Sketch a graph of the inverse relation you found when you solved Problem 5. What is its real-number domain? What is its real-number range? 8. The relation described and graphed in Problem 5 can be modified by restricting its domain to the set of reals greater than or equal to 2. Show graphically, by means of the vertical-line test, that this restriction makes the inverse f −1 into a function. 9. The relation described and graphed in Problem 5 can be modified by restricting its domain to the set of reals smaller than or equal to −2. Show graphically, by means of the vertical-line test, that this restriction makes the inverse f −1 into a function. 10. The relation described and graphed in Problem 5 can be modified by restricting its range to the set of nonnegative reals. Show graphically, and by means of verticalline tests, that this restriction makes f into a function, but does not make f −1 into a function. CHAPTER 13 Conic Sections In this chapter, we'll learn the fundamental properties of curves called conic sections. These curves include the circle, the ellipse, the parabola, and the hyperbola. The conic sections can always be represented in the Cartesian plane as equations that contain the squares of one or both variables. Geometry Imagine a double right circular cone with a vertical axis that extends infinitely upward and downward. Also imagine a flat, infinitely large plane that can be moved around so that it slices through the double cone in various ways, as shown in Fig. 13-1. The intersection between the plane and the double cone is always a circle, an ellipse, a parabola, or a hyperbola, as long as the plane doesn't pass through the point where the apexes of the cones meet. Geometry of a circle and an ellipse Figure 13-1A shows what happens when the plane is perpendicular to the axis of the double cone. In that case, we get a circle. In Fig. 13-1B, the plane is not perpendicular to the axis of the cone, but it isn't tilted very much. The curve is closed, but it isn't a perfect circle. Instead, it's an "elongated circle" or ellipse. Geometry of a parabola As the plane tilts farther away from a right angle with respect to the double-cone axis, the ellipse becomes increasingly elongated. Eventually, we reach an angle of tilt where the curve is no longer closed. At precisely this threshold angle, the intersection between the plane and the cone is a parabola (Fig. 13-1C). Geometry of a hyperbola So far, the plane has only intersected one half of the double cone. If we tilt the plane beyond the angle at which the intersection curve is a parabola, the plane intersects both halves of the cone. In that case, we get a hyperbola. If we tilt the plane as far as possible so that it becomes parallel to the cone's axis, we still get a hyperbola (Fig. 13-1D). 249 250 Conic Sections Double circular cone Flat plane Double circular cone Flat plane B A Flat plane Flat plane Double circular cone Double circular cone C D Figure 13-1 The conic sections can be defined by the intersection of a flat plane with a double right circular cone. At A, a circle. At B, an ellipse. At C, a parabola. At D, a hyperbola. Are you confused? You might ask, "We haven't mentioned the flare angle of the double cone (the measure of the angle between the axis of the cone and its surface). Does the size of this angle make any difference?" Quantitatively, it does. As the flare angle increases (the cones become "fatter"), we get ellipses less often and hyperbolas more often. As the flare angle decreases (the cones get "slimmer"), we obtain ellipses more often and hyperbolas less often. However, we can always get a circle, an ellipse, a parabola, or a hyperbola by manipulating the plane to the desired angle, regardless of the flare angle. Here's a challenge! Imagine that you're standing on a frozen lake at night, holding a flashlight that throws a coneshaped beam with a flare angle of p /10; in other words, the outer face of the light cone subtends an angle of p /10 with respect to the beam center. How can you aim the flashlight so that the edge of the light cone forms a circle on the ice? An ellipse? A parabola? Solution The edge of the light cone is a circle if and only if the flashlight is pointed straight down, so the center of the beam is perpendicular to the surface of the ice (Fig. 13-2A). The edge of the region Geometry 251 of light is an ellipse if and only if the beam axis subtends an angle of more than p /10 with respect to the ice, but the entire light cone still lands on the surface (Fig. 13-2B). The edge of the region of light is a parabola if and only if the beam axis subtends an angle of exactly p /10 with respect to the ice, so the top edge of the light cone is parallel to the surface (Fig. 13-2C). Here's another challenge! Imagine the scenario described above with the flashlight. How can you aim the flashlight so the edge of the light cone forms a half-hyperbola on the ice? Flashlight A Edge of bright region is a circle Flashlight B Edge of bright region is an ellipse Flashlight C Edge of bright region is a parabola Figure 13-2 At A, the edge of the light cone creates a circle on the surface. At B, the edge of the light cone creates an ellipse on the surface. At C, the edge of the light cone creates a parabola on the surface. The dashed lines show the edges of the light cones. The dotted-anddashed lines show the central axes of the light cones. 252 Conic Sections Solution The edge of the region of light is a half-hyperbola if and only if one of the following conditions is met: • The beam's central axis intersects the lake at an angle of less than p /10 with respect to the surface of the ice (Fig. 13-3A). • The beam's central axis is aimed horizontally (Fig. 13-3B). • The beam's central axis is aimed into the sky at an angle of less than p /10 above the horizon (Fig. 13-3C). Flashlight A Edge of bright region is a half-hyperbola Flashlight B Edge of bright region is a half-hyperbola Flashlight C Edge of bright region is a half-hyperbola Figure 13-3 At A, B, and C, the edge of the light cone creates a half-hyperbola on the surface. The uppermost part of the light cone is above the horizon in all three cases. The dashed lines show the edges of the light cones. The dotted-and-dashed lines show the central axes of the light cones. Basic Parameters 253 Basic Parameters Figure 13-4 illustrates generic examples of a circle (at A), an ellipse (at B), and a parabola (at C) in the Cartesian xy plane. The circle and ellipse are closed curves, while the parabola is an open curve. In the circle, r is represents the radius. In the ellipse, a and b represent the semi-axes. The longer of the two is called the major semi-axis. The shorter of the two is called the minor semi-axis. In these examples, the circle and the ellipse are centered at the origin, and the parabola's vertex (the extreme point where the curvature is sharpest) is at the origin. Specifications for a parabola Suppose that we're traveling in a geometric plane along a course that has the contour of a parabola. At any given time, our location on the curve is defined by the ordered pair (x,y). To follow a parabolic path, we must always remain equidistant from a point called the focus and y y b r a x x A B y x C Figure 13-4 Three basic conic sections in the Cartesian xy plane. At A, a circle centered at the origin with radius r. At B, an ellipse centered at the origin with semi-axes a and b. At C, a parabola with the vertex at the origin. 254 Conic Sections y Point (x, y) u Focus x u f f Vertex u = 2f + y Directrix Figure 13-5 All the points on a parabola are at equal distances u from the focus and the directrix. The focus and the directrix are at equal distances f from the vertex of the curve. a line called the directrix as shown in Fig. 13-5, where the focus and the directrix both lie in the same plane as the parabola. Let's call this distance u. In this illustration, the focus of the parabola is at the coordinate origin (0,0). Now imagine a straight line passing through the focus and intersecting the directrix at a right angle. This line forms the axis of the parabola. In Fig. 13-5, the parabola's axis happens to coincide with the coordinate system's y axis. Along the axis line, the distance u is called the focal length, which mathematicians and scientists usually call f. (Be careful here! Don't confuse this f with the name of a relation or a function.) By drawing a line through the focus parallel to the directrix and perpendicular to the axis, we can divide u, measuring our distance from the directrix, into two line segments, one having length 2f and the other having length y. Therefore u = 2f + y The focus is at the point (0,0). Therefore, the distance u is the length of the hypotenuse of a right triangle whose base length is x and whose height is y. The Pythagorean theorem tells us that x2 + y2 = u2 If we divide the distance from the focus to point (x,y) on the curve by the distance from (x,y) to the directrix, we get a figure called the eccentricity of the curve. The eccentricity is Basic Parameters 255 symbolized e. (Don't confuse this with the exponential constant, which is also symbolized e.) In the case of a parabola, these distances are both equal to u, so e = u /u = 1 Specifications for an ellipse and a circle Suppose that we want to construct a curve in which the eccentricity is positive but less than 1. We can use a geometric arrangement similar to the one we used with the parabola, but the distance from the focus is eu instead of u, as shown in Fig. 13-6. In this situation we get an ellipse. The focus is at the origin (0,0). The ellipse has two vertices (points where the curvature is sharpest), both of which lie on the y axis, and the ellipse is taller than it is wide. When we draw an ellipse this way, its variables and parameters are related according to the equations u = f + f /e + y and x2 + y2 = (eu)2 y Point (x, y) Focus eu x f u = f + f/e + y Vertex u f/e Directrix Figure 13-6 Construction of an ellipse based on a defined focus and directrix. The eccentricity e is an expression of the elongation of the ellipse. 256 Conic Sections As the eccentricity e approaches 0, the focus gets farther from the directrix, and the ellipse gets less elongated. When e reaches 0, then f /e becomes meaningless, the directrix vanishes ("runs away to infinity"), and we have a circle where f is equal to the radius r. A circle is actually an ellipse whose major and minor semi-axes are the same length. Going the other way, as the eccentricity e approaches 1, the focus gets closer to the directrix, and the ellipse gets more elongated. When e reaches 1, the ellipse "breaks open" at one end and becomes a parabola. Summarizing the above we can say • For a circle, e = 0 • For an ellipse, 0 < e < 1 • For a parabola, e = 1 The ellipse has another focus besides the one shown in Fig. 13-6. It's located at the same distance from the upper vertex of the curve as the coordinate origin is from the lower vertex. We can flip the ellipse in Fig. 13-6 upside-down, putting the upper focus in place of the lower focus and vice versa, and we'll get a diagram that looks exactly the same. The center of the ellipse is midway between the two foci. How the foci, directrix, and eccentricity relate Let's look at the circle, the ellipse, and the parabola in terms of the parameters we've just described. The circle has a single focus, which is at the center. The directrix is "at infinity." The ellipse has two foci separated by a finite distance. The curve is symmetrical with respect to a straight line that goes through the two foci. The curve is also symmetrical with respect to a straight line equidistant from the foci. The ellipse has two directrixes at finite distances from the vertices. We can think of a parabola as having two foci: one "within reach" and the other "at infinity." Its single directrix is at a distance from the vertex equal to the focal length. There's an alternative way to define the eccentricity of an ellipse. Suppose we know the distance d between the foci, and we also know the length s of the major semi-axis. The eccentricity can be found by taking the ratio e = d /(2s) Specifications for a hyperbola If we construct a conic section for which e > 1, we get a curve called a hyperbola. Figure 13-7 shows an example. The hyperbola looks like two parabolas back-to-back, but there's an important difference in the shape of a hyperbola compared with the shape of a double parabola. The parameters that help define hyperbolas are straight lines called asymptotes. Hyperbolas always have asymptotes, but parabolas never do. In the scenario of Fig. 13-7, the hyperbola has two asymptotes that happen to pass through the origin. In this case, the equations of the asymptotes are y = (b /a) x and y = −(b /a) x The curve approaches the asymptotes as we move away from the center of the hyperbola, but the curves never quite reach the asymptotes, no matter how far from the center we go. Basic Parameters 257 y Asymptotes b a x Asymptotes Figure 13-7 A basic hyperbola in the Cartesian xy plane. The eccentricity is greater than 1. The distances a and b are the semi-axes. Are you confused? You might ask, "Is it possible to have a conic section with negative eccentricity?" For our purposes in this course, the answer is no. Negative eccentricity involves the notion of negative distances. If we allow the eccentricity of a noncircular conic section to become negative, we get an "inside-out" ellipse, parabola, or hyperbola. In ordinary geometry, such a curve is the same as a "real-world" ellipse, parabola, or hyperbola. That said, it's worth noting that in certain high-level engineering and physics applications, negative distances sometimes behave differently than positive distances. In those special situations, an inside-out conic section might represent an entirely different phenomenon from a real-world conic section. Keep that in the back of your mind if you plan on becoming an astronomer, cosmologist, or high-energy physicist someday! Here's a challenge! Using the alternative formula for the eccentricity of an ellipse, show that if we have an ellipse in which e = 0, then that ellipse is a circle. 258 Conic Sections Solution First, we should realize that a circle is a special sort of ellipse in which the two semi-axes have identical length. With that in mind, let's plug e = 0 into the alternative equation for the eccentricity of an ellipse. That gives us 0 = d /(2s) where d is the distance between the foci, and s is the length of the major semi-axis. We can multiply the above formula by 2s to obtain 0=d which tells us that the two foci are located at the same point, so there's really only one focus. A circle is the only type of ellipse that has a single focus. Standard Equations When we graph a conic section in the Cartesian xy plane, we can find a unique equation that represents that curve. These equations are always of the second degree, meaning that the equation must contain the square of one or both variables. Equations for circles We can write the standard-form general equation for a circle in terms of its center point and its radius as (x − x0)2 + ( y − y0)2 = r2 where x0 and y0 are real constants that tell us the coordinates (x0,y0) of the center of the circle, and r is a positive real constant that tells us the radius (Fig. 13-8). When a circle is centered at the origin, the equation is simpler because x0 = 0 and y0 = 0. Then we have x2 + y2 = r2 The simplest possible case is the unit circle, centered at the origin and having a radius equal to 1. Its equation is x 2 + y2 = 1 Equations for ellipses The standard-form general equation of an ellipse in the Cartesian xy plane, as shown in Fig. 13-9, is (x − x0)2/a2 + ( y − y0)2/b2 = 1 260 Conic Sections where x0 and y0 are real constants representing the coordinates (x0,y0) of the center of the ellipse, a is a positive real constant that represents the distance from (x0,y0) to the curve along a line parallel to the x axis, and b is a positive real constant that tells us the distance from (x0,y0) to the curve along a line parallel to the y axis. When we plot x on the horizontal axis and y on the vertical axis (the usual scheme), a is the length of the horizontal semi-axis or horizontal radius of the ellipse, and b is the length of the vertical semi-axis or vertical radius. For ellipses centered at the origin, we have x0 = 0 and y0 = 0, so the general equation is x2/a2 + y2/b2 = 1 If a = b, then the ellipse is a circle. Remember that a circle is an ellipse for which the eccentricity is 0. Equations for parabolas Figure 13-10 is an example of a parabola in the Cartesian xy plane. The standard-form general equation for this curve is y = ax2 + bx + c The vertex is at the point (x0,y0). We can find these values according to the formulas x0 = −b /(2a) y x Vertex x0 = –b/(2a) y0 = –b 2/(4a) + c Figure 13-10 Graph of the parabola for y = ax2 + bx + c. Standard Equations 261 and y0 = ax02 + bx0 + c = −b2/(4a) + c If a > 0, the parabola opens upward, and the vertex represents the absolute minimum value of y. If a < 0, the parabola opens downward, and the vertex represents the absolute maximum value of y. In the graph of Fig. 13-10, the parabola opens upward, so we know that a > 0 in its equation. Equations for hyperbolas The standard-form general equation of a hyperbola in the Cartesian xy plane, as shown in Fig. 13-11, is (x − x0)2/a2 − ( y − y0)2/b2 = 1 where x0 and y0 are real constants that tell us the coordinates (x0,y0) of the center. y D b a x (x0, y0) Figure 13-11 Graph of the hyperbola for (x − x0)2/a2 − (y − y0)2/b2 = 1. 262 Conic Sections The dimensions of a hyperbola are harder to define than the dimensions of a circle or an ellipse. Suppose that D is a rectangle whose center is at (x0,y0), whose vertical edges are tangent to the hyperbola, and whose corners lie on the asymptotes. When we define D this way, then a is the distance from (x0,y0) to D along a line parallel to the x axis, and b is the distance from (x0,y0) to D along a line parallel to the y axis. We call a the width of the horizontal semi-axis, and we call b the height of the vertical semi-axis. For hyperbolas centered at the origin, we have x0 = 0 and y0 = 0, so the general equation becomes x2/a2 − y2/b2 = 1 The simplest possible case is the unit hyperbola whose equation is x 2 − y2 = 1 Are you astute? You might imagine that the above-mentioned standard forms are not the only ways that the equations of conic sections can present themselves. If that's what you're thinking, you're right! However, you can always convert the equation of a conic section to its standard form. For example, suppose you encounter 49x2 + 25y2 = 1225 You say, "This looks like it might be the equation for an ellipse, but it's not in the standard form for any known conic section." Then you notice that 1225 is the product of 49 and 25. When you divide the whole equation through by 1225, you get 49x2/1225 + 25y2/1225 = 1225 / 1225 which simplifies to x2/25 + y2/49 = 1 which can also be written as x2/52 + y2/72 = 1 Now you know that the equation represents an ellipse centered at the origin whose horizontal semi-axis is 5 units wide, and whose vertical semi-axis is 7 units tall. Here's a challenge! Whenever we have an equation that can be reduced to the standard form y = ax2 + bx + c Practice Exercises 263 we get a parabola that opens either upward or downward, and that represents a true function of x. How can we write the standard-form general equation of a parabola that opens to the right or the left? Does such a parabola represent a true function of x? Solution We can simply switch the variables to get x = ay2 + by + c If a > 0, we have a parabola that opens to the right. If a < 0, we have a parabola that opens to the left. If we define x as the independent variable and y as the dependent variable as is usually done in Cartesian xy coordinates, then vertical-line tests reveal that these parabolas do not represent true functions of x At the beginning of this chapter, we learned that the intersection between a plane and a double right circular cone is always a circle, an ellipse, a parabola, or a hyperbola, as long as the plane doesn't pass through the point where the apexes of the two cones meet. What happens if the plane does pass through that point? 2. Figure 13-12 shows an ellipse in the Cartesian xy plane with some dimensions labeled. The lower focus is at the origin (0,0). The lower vertex is at (0,−2). Both foci and both vertices lie on the y axis. The ellipse is taller than it is wide. What is its eccentricity? 3. Recall the formulas relating the parameters of an ellipse when plotted in the manner of Fig. 13-12: u = f + f /e + y and x2 + y2 = (eu)2 Based on these formulas, the information provided in the figure, and the solution you worked out to Problem 2, determine a relation between x and y that describes our ellipse. The equation should include only the variables x and y, but it doesn't have to be in the standard form. 4. What are the coordinates of the upper vertex of the ellipse shown in Fig. 13-12? What are the coordinates of the upper focus of the ellipse shown in Fig. 13-12? 264 Conic Sections y Upper vertex Upper focus Point (x, y) Lower focus eu x f f /e u = f + f /e + y Lower vertex is at (0, –2) u Equation of directrix is y = –6 Figure 13-12 Illustration for Problems 2 through 5. 5. What are the coordinates of the center of the ellipse shown in Fig. 13-12? What is the length of the vertical semi-axis? What is the length of the horizontal semi-axis? Based on these results, write down the standard-form equation for the ellipse. 6. Determine the type of conic section the following equation represents, and then draw its graph: x2 + 9y2 = 9 7. Determine the type of conic section the following equation represents, and then draw its graph: x2 + y2 + 2x − 2y + 2 = 4 8. Determine the type of conic section the following equation represents, and then draw its graph: x2 − y2 + 2x + 2y = 4 Practice Exercises 265 9. Determine the type of conic section the following equation represents, and then draw its graph: x2 − 3x − y + 3 = 1 10. Following is an equation in the standard form for a hyperbola: (x − 1)2/4 − ( y + 2)2/9 = 1 First, find the coordinates (x0,y0) of the center point. Then determine the length a of the horizontal semi-axis and the length b of the vertical semi-axis. Next, sketch a graph of the curve. Finally, work out the equations of the lines representing the asymptotes. Here's a hint: Use the point-slope form of the equation for a straight line in the xy plane. If it has slipped your memory, the general form is y − y0 = m(x − x0) where m is the slope of the line, and (x0,y0) represents a known point on the line. CHAPTER 14 Exponential and Logarithmic Curves In your algebra courses, you learned about exponential functions and logarithmic functions. If you need a refresher, the basics are covered in Chap. 29 of Algebra Know-It-All. Let's look some graphs that involve these functions. Graphs Involving Exponential Functions An exponential function of a real variable x is the result of raising a positive real constant, called the base, to the xth power. The base is usually e (an irrational number called Euler's constant or the exponential constant) or 10. The value of e is approximately 2.71828. Exponential: example 1 When we raise e to the xth power, we get the natural exponential function of x. When we raise 10 to the xth power, we get the common exponential function of x. Figure 14-1 shows graphs of these functions. At A, we see the graph of y = ex over the portion of the domain between −2.5 and 2.5. At B, we see the graph of y = 10x over the portion of the domain between −1 and 1. The curves have similar contours. When we "tailor" the axis scales in a certain relative way (as we do here), the two curves appear almost identical. In the overall sense, both of these functions have domains that include all real numbers, because we can raise e or 10 to any real-number power and always get a real-number as the result. However, the ranges of both functions are confined to the set of positive reals. No matter what real-number exponent we attach to e or 10, we can never produce an output that's equal to 0, and we can never get an output that's negative. 266 Graphs Involving Exponential Functions 267 y Figure 14-1 Graphs of the natural exponential function (at A) and the common exponential function (at B). 10 A 5 x –2 0 –1 1 2 y 10 B 5 x –1 0 1 Exponential: example 2 Let's see what happens to the graphs of the foregoing functions when we take their reciprocals and then graph them over the same portions of their domains as we did before. Figure 14-2A is a graph of y = 1/ex Figure 14-2B is a graph of y = 1/10x 268 Exponential and Logarithmic Curves y Figure 14-2 Graphs of the 10 reciprocals of the natural exponential (at A) and the common exponential (at B). A 5 x –2 0 –1 1 2 y 10 B 5 x –1 0 1 These curves are exactly reversed left-to-right from those in Fig. 14-1. The above reciprocal functions can be rewritten, respectively, as y = e−x and y = 10−x When we negate x before taking the power of the exponential base, we "horizontally mirror" all of the function values. The y axis acts as a "point reflector." The overall domains and ranges of these reciprocal functions are the same as the domains and ranges of the original functions. Graphs Involving Exponential Functions 269 Exponential: example 3 Now that we have two pairs of exponential functions, let's create two new functions by adding them, and see what their graphs look like. The solid black curve in Fig. 14-3A is a graph of y = ex + 1/ex = ex + e−x The solid black curve in Fig. 14-3B is a graph of y = 10x + 1/10x = 10x + 10−x The domains of these sum functions both encompass all the real numbers. The ranges are limited to the reals greater than or equal to 2. Figure 14-3 Graphs of the natural exponential plus its reciprocal (solid black curve at A) and the common exponential plus its reciprocal (solid black curve at B). The dashed gray curves are the graphs of the original functions. y 10 A 5 1/e x ex x –2 0 –1 1 2 y 10 B 5 1/10x 10x x –1 0 1 270 Exponential and Logarithmic Curves Here's a "heads up"! In many of the graphs to come, you'll see two dashed gray curves representing functions to be combined in various ways, as is the case in Fig. 14-3. But the constituent functions won't be labeled as they are in Fig. 14-3. The absence of labels will keep the graphs from getting too cluttered, so you'll be able to see clearly how they relate. Also, the lack of labels will force you to think! Based on your knowledge of the way the functions behave, you should be able to tell which graph is which without having them labeled. Exponential: example 4 Figure 14-4 shows what happens when we subtract the reciprocal of the natural exponential function from the original function and then graph the result. The solid black curve is the graph of y = ex − 1/ex = ex − e−x Figure 14-4 Graph of the natural y exponential minus its reciprocal (solid black curve). The dashed gray curves are the graphs of the original functions. 10 5 x –2 1 –1 –5 –10 2 Graphs Involving Exponential Functions 271 y Figure 14-5 Graph of the 10 common exponential minus its reciprocal (solid black curve). The dashed gray curves are the graphs of the original functions. 5 x 1 –1 –5 –10 In Fig. 14-5, we do the same thing with the common exponential function and its reciprocal. Here, the solid black curve represents y = 10x − 1/10x = 10x − 10−x In both of these figures, the dashed gray curves represent the original functions. The domains and ranges of both difference functions include all real numbers. Are you confused? Do you wonder how we arrived at the graphs in Figs. 14-3 through 14-5? We can plot sum and difference functions in two ways. We can graph the original functions separately, and then add or subtract their values graphically (that is, geometrically) by moving vertically upward or downward at various points within the spans of the domains shown. Alternatively, we can, with the help of a calculator, plot several points for each sum or difference function after calculating the outputs for several different input values. Once we have enough points for the sum or difference function, we can draw an approximation of the graph for that function directly. 272 Exponential and Logarithmic Curves Here's a challenge! Plot a graph of the function we get when we raise e to the power of 1/x. In rectangular xy coordinates, the curve is represented by the equation y = e(1/x) What is the domain of this function? What is its range? Solution We can use a calculator to determine the values of y for various values of x. Figure 14-6 is the resulting graph for values of x ranging from −10 to 10. When we input x = 0, we get e1/0, which is undefined. For any other real value of x, the output value y is a positive real number. Therefore, the domain of this function is the set of all nonzero reals. No matter how large we want y to be when y > 1, we can always find some value of x that will give it to us. Similarly, no matter how small we want y to be when 0 < y < 1, we can always find some value of x that will do the job. However, we can't find any value for x that will give us y = 1. For that to happen, we must raise e to the 0th power, meaning that we must find some x such that 1/x = 0. That's impossible! Therefore, the range of our function is the set of all positive reals except 1. The graph has a horizontal asymptote whose equation is y = 1, and a vertical asymptote corresponding to the y axis. The open circle at the point (0,0) indicates that it's not part of the graph. y 10 Asymptote along y axis Range includes all positive real numbers except 1 5 Asymptote at y = 1 x –10 –5 0 5 10 Domain includes all nonzero real numbers Figure 14-6 Graph of the function y = e(1/x). Note the "hole" in the domain at x = 0 and the "hole" in the range at y = 1. Graphs Involving Logarithmic Functions 273 Graphs Involving Logarithmic Functions A logarithm (sometimes called a log) of a quantity is a power to which a positive real constant is raised to get that quantity. As with exponential functions, the constant is called the base, and it's almost always equal to either e or 10. The base-e log function, also called the natural logarithm, is usually symbolized by writing "ln" or "log e" followed by the argument (the quantity on which the function operates). The base-10 log function, also called the common logarithm, is usually symbolized by writing "log10" or "log" followed by the argument. They're inverses! A logarithmic function is the inverse of the exponential function having the same base. The natural logarithmic function "undoes" the work of the natural exponential function and vice versa, as long as we restrict the domains and ranges so that both functions are bijections. The common logarithmic and exponential functions also behave this way, so we can say that ln ex = x = e(ln x) and log 10x = x = 10(log x) For these formulas to work, we must restrict x to positive real-number values, because the logarithms of quantities less than or equal to 0 are not defined. Logarithm: example 1 Figure 14-7 illustrates graphs of the two basic logarithmic functions operating on a variable x. At A, we see the graph of the base-e logarithmic function, over the portion of the domain from 0 to 10. The equation is y = ln x At B in Fig. 14-7, we see the graph of the base-10 logarithmic function, over the portion of the domain from 0 to 10. The equation is y = log10 x As with the exponential graphs, these curves have similar contours, and they look almost identical if we choose the axis scales as we've done here. The domains of the natural and common log functions both span the entire set of positive reals. When we try to take a logarithm of 0 or a negative number, however, we get a meaningless quantity (or, at least, something outside the set of reals!). By inputting just the right positive real value to a log function, we can get any real-number output we want. The ranges of the log functions therefore include all real numbers. 274 Exponential and Logarithmic Curves y 2 1 A 0 x 5 10 5 10 –1 –2 y 1 B x 0 –1 Figure 14-7 Graphs of the natural logarithmic function (at A) and the common logarithmic function (at B). Logarithm: example 2 Let's take the reciprocal of the independent variable x before performing the natural or common log, and then plot the graphs. Figure 14-8 shows the results. At A, we see the graph of the function y = ln (1/x) and at B, we see the graph of the function y = log10 (1/x) Graphs Involving Logarithmic Functions 275 y 2 1 A 0 x 5 10 5 10 –1 –2 y 1 B x 0 –1 Figure 14-8 Graphs of the natural log of the reciprocal (at A) and the common log of the reciprocal (at B). These functions can also be written as y = ln (x−1) and y = log10 (x−1) Based on our knowledge of logarithms from algebra, we can rewrite these functions, respectively, as y = −1 ln x = −ln x 276 Exponential and Logarithmic Curves and y = −1 log10 x = −log10 x When we raise x to the −1 power before taking the logarithm, we negate all the function values, compared to what they'd be if we left x alone. The x axis acts as a point reflector. The domains and ranges of these reciprocal functions are the same as the domains and ranges of the original functions. Logarithm: example 3 We can create two interesting functions by multiplying the functions defined in the previous two paragraphs. Let's do that, and see what the graphs look like. We want to graph the functions y = (ln x) [ln (x−1)] and y = (log10 x) [log10 (x−1)] Our knowledge of logarithms allows us to rewrite these functions, respectively, as y = −(ln x)2 and y = −(log10 x)2 The results are shown in Figs. 14-9 and 14-10. The domains of both product functions span the entire set of positive reals. The ranges of both functions are confined to the set of nonpositive reals (that is, the set of all reals less than or equal to 0). Figure 14-9 Graph of the natural log times the natural log of the reciprocal (solid black curve). The dashed gray curves are the graphs of the original functions. y 6 3 0 –3 –6 5 x 10 Graphs Involving Logarithmic Functions 277 y Figure 14-10 Graph of the common log times the common log of the reciprocal (solid black curve). The dashed gray curves are the graphs of the original functions. 1 5 x 10 –1 Logarithm: example 4 Finally, let's take the log functions we've been working with and find their ratios, as follows: y = (ln x) / [ ln (x−1)] and y = (log10 x) / [ log10 (x−1)] Our knowledge of logarithms allows us to simplify these, respectively, to y = (ln x) / (−ln x) = −1 and y = (log10 x) / (−log10 x) = −1 These functions are defined only if 0 < x < 1 or x > 1. The domains have "holes" at x = 1 because when we input 1 to either quotient, we end up dividing by 0. (Try it and see!) The ranges are confined to the single value −1. 278 Exponential and Logarithmic Curves Don't let them confuse you! In some texts, natural (base-e) logs are denoted by writing "log" without a subscript, followed by the argument. In other texts and in most calculators, "log" means the common (base-10) log. To avoid confusion, you should include the base as a subscript whenever you write "log" followed by anything. For example, write "log 10" or "log e" instead of "log" all by itself, unless it's impractical to write the subscript. You don't need a subscript when you write "ln" for the natural log. If you aren't sure what the "log" key on a calculator does, you can do a test to find out. If your calculator says that the "log" of 10 is equal to 1, then it's the common log. If the "log" of 10 turns out to be an irrational number slightly larger than 2.3, then it's the natural log. Here's a challenge! Draw graphs of the ratio functions we found in "Logarithm: example 4." Be careful! They're a little tricky. Solution Figure 14-11 is a graph of the function y = (ln x) / [ln (x−1)] Figure 14-11 Graph of the natural log divided by the natural log of the reciprocal (solid black line with "holes"). The dashed gray curves are the graphs of the original functions. y 6 3 0 –3 –6 5 Points (0, –1) and (1, –1) are not part of the graph of the ratio function! x 10 Logarithmic Coordinate Planes Figure 14-12 Graph of the common log divided by the common log of the reciprocal (solid black line with "holes"). The dashed gray curves are the graphs of the original functions. 279 y 1 5 x 10 –1 Points (0, –1) and (1, –1) are not part of the graph of the ratio function! Figure 14-12 is a graph of the function y = (log10 x) / [log10 (x−1)] In both graphs, the original numerator and denominator functions are graphed as dashed gray curves. The ratio functions are graphed as solid black lines with "holes." The small open circles at the points (0,−1) and (1,−1) indicate that those points are not part of either graph. That's the trick I warned you about. Without the open circles, these graphs would be wrong. Logarithmic Coordinate Planes Engineers and scientists sometimes use coordinate systems in which one or both axes are graduated according to the common (base-10) logarithm of the displacement. Let's look at the three most common variants. Semilog (x -linear) coordinates Figure 14-13 shows semilogarithmic (semilog) coordinates in which the independent-variable axis is linear, and the dependent-variable axis is logarithmic. The values that can be depicted on the y axis are restricted to one sign or the other (positive or negative). The graphable intervals in this example are –1 ≤ x ≤ 1 and 0.1 ≤ y ≤ 10 280 Exponential and Logarithmic Curves y Figure 14-13 The semilog 10 coordinate plane with a linear x axis and a logarithmic y axis. 3 1 0.3 x –1 0 1 The y axis in Fig. 14-13 spans two orders of magnitude (powers of 10). The span could be increased to encompass more powers of 10, but the y values can never extend all the way down to 0. Semilog (y -linear) coordinates Figure 14-14 shows semilog coordinates in which the independent-variable axis is logarithmic, and the dependent-variable axis is linear. The values that can be depicted on the x axis are restricted to one sign or the other (positive or negative). The graphable intervals in this illustration are 0.1 ≤ x ≤ 10 and –1 ≤ y ≤ 1 The x axis in Fig. 14-14 spans two orders of magnitude. The span could cover more powers of 10, but in any case the x values can't extend all the way down to 0. Log-log coordinates Figure 14-15 shows log-log coordinates. Both axes are logarithmic. The values that can be depicted on either axis are restricted to one sign or the other (positive or negative). In this example, the graphable intervals are 0.1 ≤ x ≤ 10 and 0.1 ≤ y ≤ 10 Both axes in Fig. 14-15 span two orders of magnitude. The span of either axis could cover more powers of 10, but neither axis can be made to show values down to 0. Logarithmic Coordinate Planes y Figure 14-14 The semilog coordinate plane with a logarithmic x axis and a linear y axis. 281 1 x 0 0.3 1 3 0.3 1 3 10 –1 Figure 14-15 The log-log coordinate plane. The x and y axes are both logarithmic. y 10 3 1 0.3 0.1 0.1 x 10 Are you confused? Semilog and log-log coordinates distort the graphs of relations and functions because the axes aren't linear. Straight lines in Cartesian or rectangular coordinates usually show up as curves in semilog or log-log coordinates. Some functions whose graphs appear as curves in Cartesian or rectangular coordinates turn out to be straight lines in semilog or log-log coordinates. Try plotting some linear, logarithmic, and exponential functions in Cartesian, semilog, and log-log coordinates. See for yourself what happens! Use a calculator, plot numerous points, and then "connect the dots" for each function you want to graph. 282 Exponential and Logarithmic Curves Here's a challenge! Plot graphs of each of the following three functions in x-linear semilog coordinates, y-linear semilog coordinates, and log-log coordinates (use the templates from Figs. 14-13 through 14-15): y=x y = ln x y = ex Solution Use a scientific calculator and input various values of x. Plot several points for each function and then draw curves through them, interpolating as you go. Be sure that your calculator is set for the natural logarithmic and exponential functions (that is, base e), not common logarithm or common exponential functions (base 10). You should get graphs that look like those shown in Fig. 14-16. In two cases, only a single point of the function shows up in the coordinate spans portrayed here. You'd have to expand the linear axis (the x axis) at A beyond 1 to see any of the graph for y = ln x. You'd have to expand the linear axis (the y axis) at B beyond 1 to see any of the graph for y = ex. y y 10 y = ex 1 3 1 A 10 1 x 0 3 0.3 y = ln x 0.3 –1 x B 0 –1 1 y 10 y=x 3 y = ln x C 1 y = ex 0.3 0.1 0.1 x 0.3 1 3 10 Figure 14-16 Simple functions in x-linear semilog coordinates (at A), y-linear semilog coordinates (at B), and log-log coordinates (at C). Practice Exercises 283 When plotting graphs here, feel free to use a calculator, locate numerous points, and "connect the dots." 1. When we discussed the range of the natural exponential function, we claimed that no real-number power of e is equal to 0. Prove it. Here's a hint: Use the technique of reductio ad absurdum, in which we assume the truth of a statement and then derive something obviously false or contradictory from that assumption. 2. Set up a rectangular coordinate system like the one in Fig. 14-1A, where the values of x are portrayed from −2.5 to 2.5, and the values of y are portrayed from 0 to 10. Sketch the graphs of the following functions on this coordinate grid: y = ex y = e−x y = ex e−x 3. Set up a rectangular coordinate system like the one in Fig. 14-1B, where the values of x are portrayed from −1 to 1, and the values of y are portrayed from 0 to 10. Sketch the graphs of the following functions on this coordinate grid: y = 10x y = 10−x y = 10x/10−x 4. Draw the graphs of the three functions from Problem 3 on an x-linear semilog coordinate grid, where the values of x are portrayed from −1 to 1, and the values of y are portrayed over the three orders of magnitude from 0.1 to 100. 5. Plot a rectangular-coordinate graph of the function we get when we raise 10 to the power of 1/x. The curve is represented by the following equation: y = 10(1/x) What is the domain of this function? What is its range? Include all values of the domain from −10 to 10. 6. We've claimed that the natural log of 0 isn't a real number. Prove it. Here's a hint: Use reductio ad absurdum, and use the solution to Problem 1 as a lemma (a theorem that helps in the proof of another theorem). 7. Plot a rectangular-coordinate graph of the sum of the natural log function and the common log function. The curve is represented by the following equation: y = ln x + log10 x What is the domain of this function? What is its range? Include all values of the domain from 0 to 10. Include all values of the range from −5 to 5. 284 Exponential and Logarithmic Curves 8. Plot a rectangular-coordinate graph of the product of the natural log function and the common log function. The curve is represented by the following equation: y = (ln x) (log10 x) What is the domain of this function? What is its range? Include all values of the domain from 0 to 10, and all values of the range from 0 to 5. 9. Draw the graphs of the three functions from Fig. B-18 (the illustration for the solution to Problem 7) on a y-linear semilog coordinate grid. Portray values of x over the two orders of magnitude from 0.1 to 10. Portray values of y from −5 to 5. 10. Draw the graphs of the three functions from Fig. B-19 (the illustration for the solution to Problem 8) on a y-linear semilog coordinate grid. Portray values of x over the single order of magnitude from 1 to 10. Portray values of y from 0 to 2.5. CHAPTER 15 Trigonometric Curves If you've taken basic algebra and geometry, you're familiar with the trigonometric functions. You also got some experience with them in Chap. 2 of this book. Now we'll graph some algebraic combinations of these functions. Graphs Involving the Sine and Cosine Let's find out what happens when we add, multiply, square, and divide the sine and cosine functions. Sine and cosine: example 1 Figure 15-1 shows superimposed graphs of the sine and cosine functions along with a graph of their sum. You can follow along by inputting numerous values of q into your calculator, determining the output values, plotting the points corresponding to the input/output ordered pairs, and then filling in the curve by "connecting the dots." In Fig. 15-1, the dashed gray curves are the individual sine and cosine waves. The solid black curve is the graph of the sum function: f (q) = sin q + cos q The sum-function wave has the same period (distance between the corresponding points on any two adjacent waves) as the sine and cosine waves. In this situation, that period is 2p. The new wave also has the same frequency as the originals. The frequency of any regular, repeating wave is always equal to the reciprocal of its period. The peaks (recurring maxima and minima) of the sine and cosine waves attain values of ±1. The peaks of the new wave attain values of ±21/2, which occur at values of q where the graphs of the sine and cosine cross each other. By definition, the peak amplitude of the new function is 21/2 times the peak amplitude of either original function. The new wave appears "sine-like," but we can't be sure that it's a true sinusoid on the basis of its appearance in this graph. The domain of our function f includes all real numbers. The range of f is the set of all reals in the closed interval [−21/2,21/2]. 285 286 Trigonometric Curves Figure 15-1 Graphs of the sine and cosine functions (dashed gray curves) and the graph of their sum (solid black curve). Each division on the horizontal axis represents p /2 units. Each division on the vertical axis represents 1/2 unit. Sine and cosine: example 2 In Fig. 15-2, we see graphs of the sine and cosine functions along with a graph of their product. The dashed gray curves are the superimposed sine and cosine waves. The solid black curve is the graph of the product function: f (q) = sin q cos q The new function's graph has a period of p, which is half the period of the sine wave, and half the period of the cosine wave. The peaks of the new wave are ±1/2, which occur at values of q where the graphs of the sine and cosine intersect. As in the previous example, the new wave looks like a sinusoid, but we can't be sure about that by merely looking at it. The domain of the product function spans the entire set of reals. The range is the set of all reals in the closed interval [−1/2,1/2]. Sine and cosine: example 3 Figure 15-3 shows the graphs of the sine function (at A) and the cosine function (at B) along with their squares. The dashed gray curve at A is the sine wave; the dashed gray curve at B is the cosine wave. In illustration A, the solid black curve is the graph of f (q) = sin2 q In illustration B, the solid black curve is the graph of g (q) = cos2 q Figure 15-2 Graphs of the sine and cosine functions (dashed gray curves) along with the graph of their product (solid black curve). Each horizontal division represents p/2 units. Each vertical division represents 1/4 unit. Each horizontal division is p/2 units f (q ) q Each vertical division is 1/4 unit Figure 15-3 The dashed gray curves are graphs of the sine function (at A) and the cosine function (at B). The solid black curves are graphs of the square of the sine function (at A) and the square of the cosine function (at B). Each division on the horizontal axes represents p /2 units. Each division on the vertical axes represents 1/4 unit. Each horizontal division is p /2 units f (q ) q A Each vertical division is 1/4 unit Each horizontal division is p /2 units B g(q) q Each vertical division is 1/4 unit 288 Trigonometric Curves The squared-function waves have periods of p, half the periods of the original functions. Therefore, the frequencies of the squared functions are twice those of the original functions. The waves for the squared functions are displaced upward relative to the waves for the original functions. The squared functions attain repeated minima of 0 and repeated maxima of 1. In other words, the positive peak amplitudes of the squared-function waves are equal to 1, while the minimum peak amplitudes are equal to 0. We define the peak-to-peak amplitude of a regular, repeating wave as the difference between the positive peak value and the negative peak value. In this example, the original waves have peak-to-peak amplitudes of 1 − (−1) = 2, while the squared-function waves have peak-to-peak amplitudes of 1 − 0 = 1. The waves representing the squared functions f and g look like sinusoids, but we can't be certain about that on the basis of their appearance alone. The domains of f and g include all real numbers. The ranges of f and g are confined to the set of all reals in the closed interval [0,1]. Sine and cosine: example 4 Let's add the squared functions from the previous example and graph the result. The solid black line in Fig. 15-4 is a graph of the sum of the squares of the sine and the cosine functions, which are shown as superimposed dashed gray curves. We have f (q) = sin2 q + cos2 q In this case, the function has a constant value. The domain includes all of the real numbers. The range is the set containing the single real number 1. Figure 15-4 Graph of the sum of the squares of the sine and cosine functions (solid black line). The dashed gray curves are the graphs of the original squared functions. Each horizontal division represents p/2 units. Each vertical division represents 1/4 unit. f (q ) q Each horizontal division is p /2 units Each vertical division is 1/4 unit Graphs Involving the Sine and Cosine 289 Are you confused? You might wonder, "How can we be certain that the graph in the previous example is actually a straight, horizontal line? When I input values into my calculator, I always get an output of 1, but I've learned that even a million examples can't prove a general truth in mathematics." Your skepticism shows that you're thinking! But let's remember one of the basic trigonometric identities that we learned in Chap. 2. For all real numbers q, the following equation holds true: sin2 q + cos2 q = 1 This fact assures us that in the previous example, we have f (q) = 1 so the graph of the sum-of-squares function is indeed the horizontal, solid black line portrayed in Fig. 15-4. Here's a challenge! Sketch a graph of the ratio of the square of the sine function to the square of the cosine function. That is, graph f (q) = (sin2 q)/(cos2 q) Solution The solid black complex of curves in Fig. 15-5 is the graph of the ratio of the square of the sine to the square of the cosine. The superimposed gray curves are graphs of the original sine-squared and cosine-squared functions. Figure 15-5 Graph of the ratio of the square of the sine function to the square of the cosine function (solid black curves). Each horizontal division represents p /2 units. Each vertical division represents 1/2 unit. The dashed gray curves are the graphs of the original sinesquared and cosinesquared functions. The vertical dashed lines are asymptotes of f. f (q ) q Each horizontal division is p /2 units Each vertical division is 1/2 unit 290 Trigonometric Curves This ratio function f is singular (that is, it "blows up") when q is any odd-integer multiple of p /2. That's because cos2 q (the denominator) equals 0 at those points, while sin2 q (the numerator) equals 1. The function attains values of 0 at all integer multiples of p because at those points, sin2 q (the numerator) equals 0, while cos2 q (the denominator) equals 1. The period of f is p, the distance between the asymptotes; the graph repeats itself completely between each adjacent pair of asymptotes. The peak amplitude and the peak-to-peak amplitude are both undefined. (It's tempting to call them "infinite," but let's not go there!) The domain includes all reals except the odd-integer multiples of p /2. The range is the set of all nonnegative reals. Graphs Involving the Secant and Cosecant In Chap. 2, we saw graphs of the basic secant and cosecant functions, which are the reciprocals of the cosine and sine, respectively. Let's combine these two functions after the fashion of the previous section, and see what the resulting graphs look like. Secant and cosecant: example 1 The dashed gray curves in Fig. 15-6 are the superimposed graphs of the secant and cosecant functions. The complex of solid black curves is a graph of their sum. As always, you can reproduce this graph by inputting a sufficient number of values into your calculator, plotting the Figure 15-6 Graph of the sum of f(q ) the sec an asymptote of f. q Each horizontal division is p /2 units Each vertical division is 1 unit Graphs Involving the Secant and Cosecant 291 output points, and then "connecting the dots." You'll have to take some time to "investigate" this function before you can accurately plot this graph, but be patient! We have f (q) = sec q + csc q The graph of f has asymptotes that pass through every point where the independent variable is an integer multiple of p/2. If you examine Fig. 15-6 closely, you'll see that the graph is regular and it repeats with a period of 2p, but we can't call it a wave. The domain includes all real numbers except the integer multiples of p/2 because, whenever q attains one of those values, either the secant or the cosecant is undefined. The range spans the set of all real numbers. Secant and cosecant: example 2 Figure 15-7 shows graphs of the secant and cosecant functions along with their product. The dashed gray curves are graphs of the original functions superimposed on each other; the solid black curves show the graph of f (q) = sec q csc q This function f has a period of p, which is half that of the secant and cosecant functions. Like the sum-function graph, this graph has asymptotes that pass through every point where the independent variable is an integer multiple of p /2. The domain is the set of all reals except the integer multiples of p /2. The range spans the set of all real numbers except those in the open interval (−2,2). Alternatively, we can say that the range includes all reals y such that y ≥ 2 or y ≤ −2. Figure 15-7 Graph of the product of the sec an asymptote of f. f(q ) q Each horizontal division is p /2 units Each vertical division is 1 unit 292 Trigonometric Curves Secant and cosecant: example 3 The dashed gray curves in Fig. 15-8 are the graphs of the secant function (at A) and the cosecant function (at B). At A, the solid black curve is the graph of f (q) = sec2 q At B, the solid black curve is the graph of g (q) = csc2 q Figure 15-8 The solid black curves are the graphs of the squares of the secant function (at A) and the cosecant function (at B). The dashed gray curves are the graphs of the original functions. Each division on the horizontal axes represents p/2 units. Each division on the vertical axes represents 1 unit. The vertical dashed lines are asymptotes of f and g. At B, the dependentvariable axis is also an asymptote of g. f(q ) q A Each horizontal division is p /2 units Each vertical division is 1 unit g(q ) B q Graphs Involving the Secant and Cosecant 293 The squared functions have periods of p, which are half the periods of the original functions. Therefore, the frequencies of the squared functions are double those of the originals. Singularities occur at the same points on the independent-variable axes as they do for the original functions. The domain of the secant-squared function is the set of all reals except odd-integer multiples of p/2. The domain of the cosecant-squared function is the set of all reals except integer multiples of p. The ranges in both cases are confined to the set of reals y such that y ≥ 1. Secant and cosecant: example 4 Figure 15-9 shows what happens when we add the secant-squared function to the cosecantsquared function. The solid black curves compose the graph of f (q) = sec2 q + csc2 q The dashed gray curves are superimposed graphs of the original functions. This sum function has a period equal to half that of the original functions, or p/2. The domain includes all reals except the integer multiples of p/2. The range is the set of reals y such that y ≥ 4. Figure 15-9 f (q ) Graph of the sum of the squares of the secant and cosecant functions (solid black curves). The dashed gray curves are the graphs of the original squared functions. Each horizontal division represents p/2 units. Each vertical division represents 1 unit. The vertical dashed lines are asymptotes of f. The positive dependent-variable axis is also an asymptote of f. q Each horizontal division is p /2 units Each vertical division is 1 unit 294 Trigonometric Curves Are you confused? You're bound to wonder, "How do we know that the range of the sum-of-squares function in the previous example is the set of all reals greater than or equal to 4?" Another way of stating this fact is that the minima of the solid black curves in Fig. 15-9 have dependent-variable values equal to 4. These minima occur at values of q where the graphs of the secant-squared and cosecant-squared functions (dashed gray curves) intersect. Every one of those points occurs where q is an odd-integer multiple of p /4. With the help of your calculator, you can determine that whenever q is an odd-integer multiple of p/4, the secant squared and cosecant squared are both 2, so their sum is 4. If you move slightly to the right or left of any of these points, the value of the sum-of-squares function increases (a fact that you can, again, check out with your calculator). It follows that the sum-ofsquares function can never attain any real-number value less than 4. However, there's no limit to how large the value of the function can get. One or the other of the original functions "blows up positively" at every point where q attains an integer multiple of p /2. Here's a challenge! Sketch a graph of the ratio of the square of the secant function to the square of the cosecant function. That is, graph f (q) = (sec2 q)/(csc2 q) Determine the domain and range of f. Be careful! Both the domain and the range have some tricky restrictions. Solution We can simplify this problem by remembering a few basic facts in trigonometry, and by applying a little algebra. First, let's remember that the cosecant is equal to the reciprocal of the sine, so the converse is also true. We have 1/(csc q) = sin q When we square both sides, we get 1/(csc2 q) = sin2 q Substituting in the equation for our function gives us f (q) = (sec2 q) (sin2 q) We've learned that the secant is equal to the reciprocal of the cosine. We have sec q = 1/(cos q) so we can square both sides to get sec2 q = 1/(cos2 q) Graphs Involving the Secant and Cosecant 295 Substituting again in the equation for our original function, we obtain f (q) = (sin2 q)/(cos2 q) = [(sin q)/(cos q)]2 The sine over the cosine is equal to the tangent, so we can substitute again to conclude that our original function is f (q) = tan2 q with the restriction that we can't define it for any input value where either the secant or the cosecant become singular. The solid black curves in Fig. 15-10 show the result of squaring all the values of the tangent function, noting the additional undefined values as open circles. At the points shown by the open circles, the cosecant function is singular so its square is undefined. That means we can't define our ratio function f at any such point. At the asymptotes (dashed vertical lines), the secant function is singular so its square is undefined, making it impossible to define the ratio function f for those values of q. Our function f has a period of p. The domain of f includes all real numbers except integer multiples of p /2, where one or the other of the original squared functions is singular. The range is the set of all positive reals. f (q ) q Each horizontal division is p /2 units Each vertical division is 1 unit Figure 15-10 Graph of the ratio of the square of the secant function to the square of the cosecant function (solid black curves). The dashed gray curves are the graphs of the original squared functions. Each horizontal division represents p /2 units. Each vertical division represents 1 unit. The vertical dashed lines are asymptotes of f. 296 Trigonometric Curves Graphs Involving the Tangent and Cotangent You were introduced to graphs of the basic tangent and cotangent functions in Chap. 2. The tangent is the ratio of the sine to the cosine, and the cotangent is the ratio of the cosine to the sine. Now we'll see what happens when we alter or combine these functions in a few different ways. Tangent and cotangent: example 1 In Fig. 15-11, the dashed gray curves are superimposed graphs of the tangent and cotangent functions. The solid black curves portray the graph of f (q) = tan q + cot q The graph of f has asymptotes that pass through every point where the independent variable attains an integer multiple of p /2. The period is p. The domain is the set of all real numbers except the integer multiples of p /2. The range is the set of reals larger than or equal f (q ) q Each horizontal division is p /2 units Each vertical division is 1 unit Figure 15-11 Graph of the sum of the tangent and cotangent functions (solid black curves). The dashed gray curves are the graphs of the original functions. Each division on the horizontal axis represents p /2 units. Each vertical division represents 1 unit. The vertical dashed lines are asymptotes of f. The dependent-variable axis is also an asymptote of f. Graphs Involving the Tangent and Cotangent 297 to 2 or smaller than or equal to −2. We can also say that the range spans the set of all reals except those in the open interval (−2,2). Tangent and cotangent: example 2 Figure 15-12 shows superimposed graphs of the tangent and cotangent functions (dashed gray curves) along with their product (black line with "holes" in it). We have f (q) = tan q cot q We can simplify the calculations to graph this function when we recall that the cotangent and the tangent are reciprocals of each other, so we have cot q = 1/(tan q) This equation is valid as long as both functions are defined and tan q ≠ 0. By substitution, the equation for our function f becomes f (q) = (tan q)/(tan q) = 1 f (q ) q Each horizontal division is p /2 units Each vertical division is 1 unit Figure 15-12 Graph of the product of the tangent and cotangent functions (solid black curve). The dashed gray curves are the graphs of the original functions. Each division on the horizontal axis represents p /2 units. Each vertical division represents 1 unit. 298 Trigonometric Curves The graph of f is a horizontal, straight line with infinitely many "holes," with each hole located at a point where q is an integer multiple of p /2. If we want to get creative with our terminology, we can say that the graph of f consists of infinitely many open-ended line segments, each of length p /2, placed end-to-end in a collinear arrangement. The domain of f spans the set of all reals except the integer multiples of p /2. The range is the set containing the single real number 1. Tangent and cotangent: example 3 The dashed gray curves in Fig. 15-13 are the graphs of the tangent function (at A) and the cotangent function (at B). The solid black curves in drawing A compose the graph of f (q) = tan2 q Figure 15-13 The solid black curves are graphs of the squares of the tangent function (at A) and the cotangent function (at B). The dashed gray curves are graphs of the original functions. Each division on the horizontal axes represents p /2 units. Each division on the vertical axes represents 1 unit. The vertical dashed lines are asymptotes of f and g. At B, the positive dependentvariable axis is also an asymptote of g. f (q ) q A Each horizontal division is p /2 units Each vertical division is 1 unit g(q ) B q Graphs Involving the Tangent and Cotangent 299 The solid black curves in drawing B compose the graph of g (q) = cot2 q Both f and g have periods of p, the same as the periods of the tangent and cotangent functions. Therefore, the frequencies of the squared functions are the same as those of the originals. Singularities occur in f and g at the same points on the independent-variable axis as they do for the original functions. The domain of f is the set of all reals except odd-integer multiples of p /2. The domain of g is the set of all reals except integer multiples of p. The ranges of both f and g span the set of nonnegative real numbers. Tangent and cotangent: example 4 Figure 15-14 is a graph of the sum of the tangent-squared function and the cotangent-squared function. The solid black curves compose the graph of f (q) = tan2 q + cot2 q The dashed gray curves are superimposed graphs of the original squared functions. This sum function f has a period equal to half that of the original functions, or p /2. The domain includes all reals except the integer multiples of p /2. The range is the set of reals y such that y ≥ 2. Figure 15-14 Graph of the sum of the squares of the tangent and cotangent functions (solid black curves). The dashed gray curves are the graphs of the original squared functions. Each division on the horizontal axis represents p /2 units. Each vertical division represents 1 unit. The vertical dashed lines are asymptotes of f. The positive dependentvariable axis is also an asymptote of f. f(q ) q Each horizontal division is p /2 units Each vertical division is 1 unit 300 Trigonometric Curves Are you confused? You might wonder how we can be sure that the range of the sum-of-squares function graphed in Fig. 5-14 is the set of all reals greater than or equal to 2. To understand this, we can use the same reasoning as we did when we added the squares of the secant and the cosecant functions. All the minima on the solid black curves in Fig. 15-14 correspond to dependent-variable values of 2, because they occur where the graphs of the dashed gray curves intersect. At all such points, the tangent squared and cotangent squared are both 1, so their sum is 2. If you move slightly on either side of any such point, the value of the sum-of-squares function increases. Here's a challenge! Sketch a graph of the ratio of the square of the tangent function to the square of the cotangent function. That is, graph f (q) = (tan2 q)/(cot2 q) State the domain and range of f. Be careful! There are some tricky restrictions in the domain. Solution Let's use our knowledge of trigonometry to break this ratio down into sines and cosines. We recall that tan q = (sin q)/(cos q) as long as q isn't an odd-integer multiple of p /2, and cot q = (cos q)/(sin q) provided q isn't an integer multiple of p. Therefore, tan2 q = (sin2 q)/(cos2 q) and cot2 q = (cos2 q)/(sin2 q) with the same restrictions. By substitution, our ratio function becomes f (q) = [(sin2 q)/(cos2 q)]/[(cos2 q)/(sin2 q)] as long as q isn't an integer multiple of p /2. The above equation can be rewritten as f (q) = [(sin2 q)/(cos2 q)] [(sin2 q)/(cos2 q)] which simplifies to f (q) = (sin4 q)/(cos4 q)] Graphs Involving the Tangent and Cotangent 301 and finally to f (q) = tan4 q with, once again, the important restriction that q cannot be any integer multiple of p /2. If we input any integer multiple of p /2 to the original function, we can't define the output because either the numerator or the denominator function encounters a singularity. The black curves with the holes in Fig. 15-15 show the result of raising all the values of the tangent function to the fourth power, noting the additional undefined values as open circles. The dashed gray curves are the original tangent-squared and cotangent-squared functions. Our function f has a period of p. The domain includes all real numbers except integer multiples of p /2. The range includes all positive real numbers. f (q ) q Each horizontal division is p /2 units Figure 15-15 Each vertical division is 1 unit Graph of the ratio of the square of the tangent function to the square of the cotangent function (solid black curves). The dashed gray curves are the graphs of the original squared functions. Each division on the horizontal axis represents p /2 units. Each vertical division represents 1 unit. The vertical dashed lines represent asymptotes of f. 302 Trigonometric Curves Look back at Fig. 15-1, which shows the graphs of the sine and cosine waves (dashed gray curves) along with their sum (solid black curve). Sketch a graph, and state the domain and the range, of the difference function: h (q) = sin q − cos q 2. Look back at Fig. 15-4, which shows the sine-squared and cosine-squared waves (dashed gray curves) along with their sum (solid black horizontal line). Sketch a graph, and state the domain and the range, of the difference-of-squares function: h (q) = sin2 q − cos2 q 3. Look back at Fig. 15-5, which shows the graphs of the sine-squared and cosine-squared waves (dashed gray curves) along with the graph of the ratio-of-squares function: f (q) = (sin2 q)/(cos2 q) Sketch a graph, and state the domain and the range, of the ratio-of-squares function going the other way. That function is h (q) = (cos2 q)/(sin2 q) 4. Look back at Fig. 15-6, which shows the graphs of the secant and cosecant functions (dashed gray curves) along with their sum (solid black curves). Sketch a graph, and state the domain and the range, of the difference function: h (q) = sec q − csc q 5. Look back at Fig. 15-9, which shows the graphs of the secant-squared and cosecantsquared functions (dashed gray curves) along with their sum (solid black curves). Sketch a graph, and state the domain and the range, of the difference function: h (q) = sec2 q − csc2 q 6. Look back at Fig. 15-10, which shows the graphs of the secant-squared and cosecantsquared functions (dashed gray curves) along with the graph of f (q) = (sec2 q)/(csc2 q) Sketch a graph, and state the domain and the range, of the ratio-of-squares function going the other way. That function is h (q) = (csc2 q)/(sec2 q) 7. Look back at Fig. 15-11, which shows the graphs of the tangent and cotangent functions (dashed gray curves) along with the graph of their sum (solid black curves). Sketch a graph, and state the domain and the range, of the difference function: h (q) = tan q − cot q Practice Exercises 303 8. Look back at Fig. 15-14, which shows the graphs of the tangent-squared and cotangentsquared functions (dashed gray curves) along with the graph of their sum (solid black curves). Sketch a graph, and state the domain and the range, of the difference-ofsquares function: h (q) = tan2 q − cot2 q 9. In Chap. 2, we learned that square of the secant of an angle minus the square of the tangent of the same angle is always equal to 1, as long as the angle is not an odd-integer multiple of p /2. That is, sec2 q – tan2 q = 1 Sketch a graph that illustrates this principle, which is sometimes called the Pythagorean theorem for the secant and tangent. 10. In Chap. 2, we learned that the square of the cosecant of an angle minus the square of the cotangent of the same angle is always equal to 1, as long as the angle is not an integer multiple of p. That is, csc2 q – cot2 q = 1 Sketch a graph that illustrates this principle, which is sometimes called the Pythagorean theorem for the cosecant and cotangent. CHAPTER 16 Parametric Equations in Two-Space In the two-space relations and functions we've seen so far, the value of one variable depends on the value of the other variable. In this chapter, we'll learn how to express two-space relations and functions in which both variables depend on an external factor called a parameter. What's a Parameter? In a two-space relation or function, a parameter acts as a "master controller" for one or both variables. When there exists a relation between x and y, for example, we don't have to say that x depends on y or vice versa. Instead, we can say that a parameter, which we usually call t, independently governs the values of x and y. We use parametric equations to describe how this happens. A rectangular-coordinate example Here's an example of a pair of parametric equations that produce a straight line in the Cartesian xy plane. Consider x = 2t and y = 3t To generate the graph of this system, we can input various values of t to both of the parametric equations, and then plot the ordered pairs (x,y) that come out. Following are some examples: • • • • • 304 When t = −2, we have x = 2 × (−2) = −4 and y = 3 × (−2) = −6. When t = −1, we have x = 2 × (−1) = −2 and y = 3 × (−1) = −3. When t = 0, we have x = 2 × 0 = 0 and y = 3 × 0 = 0. When t = 1, we have x = 2 × 1 = 2 and y = 3 × 1 = 3. When t = 2, we have x = 2 × 2 = 4 and y = 3 × 2 = 6. What's a Parameter? 305 When we plot the (x,y) ordered pairs based on the above list as points on a Cartesian plane and then "connect the dots," we get a line passing through the origin with a slope of 3/2, as shown in Fig. 16-1. From our knowledge of the slope-intercept form of a line in the xy plane, we can write down the equation in that form as y = (3/2)x Alternatively, we can use algebra to derive the equation of our system in terms of x and y alone, without t. Let's take the first parametric equation x = 2t and multiply it through by 3/2 to get (3/2)x = (3/2)(2t) = 3t Deleting the middle portion in the above three-way equation gives us (3/2)x = 3t The second parametric equation tells us that 3t = y, so we can substitute directly in the above equation to obtain (3/2)x = y y t=2 6 4 t=1 t=0 2 x –6 2 4 6 –2 t = –1 –4 t = –2 Figure 16-1 Cartesian-coordinate graph of the parametric equations x = 2t and y = 3t. 306 Parametric Equations in Two-Space which is identical to the slope-intercept equation y = (3/2)x The polar-coordinate counterpart Let's see what happens if we change our pair of parametric equations to polar form. We'll put q in place of x, and put r in place of y. Now we have q = 2t and r = 3t To create the polar graph, we can input various values of t, just as we did before. To keep things from getting messy, let's restrict ourselves to values of t such that we see only the part of the graph corresponding to the first full counterclockwise rotation of the direction angle, so 0 ≤ q ≤ 2p. Consider the following cases: • • • • • When t = 0, we have q = 2 × 0 = 0 and r = 3 × 0 = 0. When t = p /4, we have q = 2 × p /4 = p /2 and r = 3p /4. When t = p /2, we have q = 2 × p /2 = p and r = 3p /2. When t = 3p /4, we have q = 2 × 3p /4 = 3p /2 and r = 3 × 3p /4 = 9p /4. When t = p, we have q = 2p and r = 3p. Our graph is a spiral, as shown in Fig. 16-2. Its equation can be derived with algebra exactly as we did in the Cartesian plane, substituting q for x and r for y to get r = (3/2)q Figure 16-2 Polar-coordinate graph of the parametric equations q = 2t and r = 3t. Each radial division represents p units. For simplicity, we restrict q to the interval [0,2p ]. What's a Parameter? 307 Are you confused? If you're having trouble understanding the concept of a parameter, imagine the passage of time. In science and engineering, elapsed time t is the parameter on which many things depend. In the Cartesian situation described above, as time flows from the past (t < 0) through the present moment (t = 0) and into the future (t > 0), a point moves along the line in Fig. 16-1, going from the third quadrant (lower left, in the past) through the origin (right now) and into the first quadrant (upper right, in the future). In the polar case, as time flows from the present (t = 0) into the future (t > 0), a point travels along the spiral in Fig. 16-2, starting at the center (right now) and going counterclockwise, arriving at the outer end when t = p (a little while from now). Here's a challenge! Find a pair of parametric equations that represent the line shown in Fig. 16-3. Solution We're given two points on the line. One of them, (0,3), tells us that the y-intercept is 3. We can deduce the slope from the coordinates of the other point. When we move 4 units to the right from (0,3), we must go downward by 3 units (or upward by −3 units) to reach (4,0). The "rise over run" ratio is −3 to 4, so the slope is −3/4. The slope-intercept form of the equation for our line is y = (−3/4)x + 3 y 6 (0, 3) 4 2 (4, 0) x –6 –4 –2 2 4 6 –2 –4 –6 Figure 16-3 How can we represent this line as a pair of parametric equations? 308 Parametric Equations in Two-Space We can let x vary directly with the parameter t. We describe that relation simply as x=t That's one of our two parametric equations. We can substitute t for x into the point-slope equation to get y = (−3/4)t + 3 That's the other parametric equation. Here's an experiment! Do you suspect that the pair of equations x=t and y = (−3/4)t + 3 isn't the only parametric way we can represent the line in Fig. 16-3? If so, maybe you're right. Let x = −2t, or x = t + 1, or x = −2t + 1, and see what happens when you generate the equation for y in terms of t on that basis. When you put the two parametric equations together, do you get the same line as the one shown in Fig. 16-3? From Equations to Graph Parametric equations allow us to define complicated curves in an elegant, and often simpler, way than we can do with ordinary equations. Let's look at a couple of examples, and plot their graphs in rectangular and polar coordinates. Rectangular-coordinate graph: example 1 Suppose that x varies directly with the square of t, and y varies directly with the cube of t. In this situation, we have the parametric equations x = t2 and y = t3 From Equations to Graph 309 Let's construct a graph of this equation by inputting several values of t to the system and then plotting the points. We can break the situation down as follows: • • • • • • • • • When t = −4, we have x = (−4)2 = 16 and y = (−4)3 = −64. When t = −3, we have x = (−3)2 = 9 and y = (−3)3 = −27. When t = −2, we have x = (−2)2 = 4 and y = (−2)3 = −8. When t = −1, we have x = (−1)2 = 1 and y = (−1)3 = −1. When t = 0, we have x = 02 = 0 and y = 03 = 0. When t = 1, we have x = 12 = 1 and y = 13 = 1. When t = 2, we have x = 22 = 4 and y = 23 = 8. When t = 3, we have x = 32 = 9 and y = 33 = 27. When t = 4, we have x = 42 = 16 and y = 43 = 64. We can plot the points for these nine xy-plane coordinates and then connect them by curve fitting to get the graph of Fig. 16-4. To keep the picture clean, the points aren't labeled. In this illustration, we have a rectangular-coordinate graph, but not a true Cartesian graph. That's because the divisions on the y axis represent different increments than those on the x axis. The result is a curve that's "vertically squashed" compared to the way it would look if plotted on a true Cartesian coordinate grid, but we can fit more of the curve into the available space. Polar-coordinate graph: example 1 Figure 16-5 illustrates what happens when we substitute q for x and r for y in the above example, and then graph the result in polar coordinates. For simplicity, let's restrict the graph Figure 16-4 Rectangularcoordinate graph of the parametric equations x = t2 and y = t3. y 60 30 x –16 –8 8 –30 –60 16 From Equations to Graph 311 y 2 1 x –8 4 –4 8 –1 –2 Figure 16-6 Rectangular-coordinate graph of the parametric equations x = t−1 and y = ln t. We can construct a rectangular-coordinate graph of the relation between x and y by tabulating the values for several points, based on various values of t. Let's break things down into the following cases: • • • • • • • When t ≤ 0, ln t is undefined, so there are no points to plot. When t = e−2 ≈ 0.14, we have x = (e−2)−1 = e2 ≈ 7.39 and y = ln (e−2) = −2. When t = e−1 ≈ 0.37, we have x = (e−1)−1 = e ≈ 2.72 and y = ln (e−1) = −1. When t = 1, we have x = 1−1 = 1 and y = ln 1 = 0. When t = 2, we have x = 2−1 = 1/2 and y = ln 2 ≈ 0.69. When t = e ≈ 2.72, we have x = e−1 ≈ 0.37 and y = ln e = 1. When t = e2 ≈ 7.39, we have x = (e2)−1 = e−2 ≈ 0.14 and y = ln (e2) = 2. Figure 16-6 shows the curve we obtain when we plot these points and "connect the dots." To keep the picture clean, the points aren't labeled. As in Fig. 16-4, we use distorted rectangular coordinates to help us fit more of the curve on the page than we could with a true Cartesian grid. Polar-coordinate graph: example 2 We can directly substitute q for x and r for y in the above example, tabulate some values, graph the results, and get the curve shown in Fig. 16-7. Let's restrict t to keep q within the closed interval [0,2p] so we see only the first full positive revolution. The situation breaks down as follows: • When t = e5 ≈ 148, we have q = (e5)−1 = e−5 ≈ 0.0067 and r = ln (e5) = 5. • When t = e2 ≈ 7.39, we have q = (e2)−1 = e−2 ≈ 0.14 and r = ln (e2) = 2. 312 Parametric Equations in Two-Space Each radial division ... p /2 p 0 3p /2 Figure 16-7 • • • • • ... is 1 unit Polar-coordinate graph of the parametric equations q = t−1 and r = ln t. Each radial division represents 1 unit. When t = 2, we have q = 2−1 = 1/2 and r = ln 2 ≈ 0.69. When t = 1, we have q = 1−1 = 1 and r = ln 1 = 0. When t = 1/2, we have q = (1/2)−1 = 2 and r = ln (1/2) ≈ −0.69. When t = p −1 ≈ 0.32, we have q = (p −1)−1 = p and r = ln (p −1) ≈ −1.14. When t = (2p )−1 ≈ 0.16, we have q = [(2p )−1]−1 = 2p and r = ln [(2p )−1] ≈ −1.84. As we plot the points to obtain this graph, we must remember that when we have a negative radius in polar coordinates, we go outward from the origin by a distance equal to \|r\|, but in the opposite direction from that indicated by q. Are you confused? The polar graph in Fig. 16-7 can be baffling. Imagine that we start out facing east, in the direction q = 0. Our graph is infinitely far away in this direction. As we turn counterclockwise, the curve approaches us as r becomes finite and decreases. When we have turned counterclockwise through an angle of 1 rad (approximately 57º), the graph has come all the way in and reached the origin. As we continue to turn counterclockwise, the radius becomes negative, so the graph is behind us. As we rotate farther counterclockwise, r increases negatively. When we've rotated all the way around through a complete circle, the graph is approximately 1.84 units to our rear. From Equations to Graph 313 Here's a challenge! Plot a rectangular-coordinate graph of the pair of parametric equations where x varies directly with et and y varies directly with t2. Then plot a polar-coordinate graph of the pair of parametric equations where q varies directly with et and r varies directly with t2. For simplicity, restrict the polar graph to values of t such that 0 ≤ q ≤ 2p. Solution The parametric equations for plotting the system in the rectangular xy plane are x = et and y = t2 Let's tabulate the x and y values for several points, based on various values of t: • • • • • • When t = −2, we have x = e−2 ≈ 0.14 and y = (−2)2 = 4. When t = −1, we have x = e−1 ≈ 0.37 and y = (−1)2 = 1. When t = 0, we have x = e0 = 1 and y = 02 = 0. When t = 1, we have x = e1 = e ≈ 2.72 and y = 12 = 1. When t = 3/2, we have x = e3/2 ≈ 4.48 and y = (3/2)2 = 9/4 = 2.25. When t = 2, we have x = e2 ≈ 7.39 and y = 22 = 4. Figure 16-8 shows the graph we obtain by plotting the points in the xy plane. This is a true Cartesian graph; the divisions on the x and y axes are the same size. Figure 16-8 Cartesian-coordinate graph of the parametric equations x = et and y = t2. y 6 4 2 x –4 2 –2 –2 –4 –6 4 6 8 314 Parametric Equations in Two-Space Now let's tabulate some values for a polar graph. We substitute q for x and r for y, input values of t to get a good sampling of polar angles in the output, and restrict t to keep q within the closed interval [0,2p]. The situation breaks down into the following cases: • • • • • • When t = −2, we have q = e−2 ≈ 0.14 and r = (−2)2 = 4. When t = −1, we have q = e−1 ≈ 0.37 and r = (−1)2 = 1. When t = 0, we have q = e0 = 1 and r = 02 = 0. When t = 1, we have q = e1 = e ≈ 2.72 and r = 12 = 1. When t = 3/2, we have q = e3/2 ≈ 4.48 and r = (3/2)2 = 9/4 = 2.25. When t = ln 2p, we have q = e(ln 2p ) = 2p and r = (ln 2p )2 ≈ 3.38. Figure 16-9 shows the resulting curve in the polar plane. If the above tabulation doesn't generate enough points to satisfy you, feel free to work out a few more. As you gain experience in plotting graphs like this, you'll learn to get a sense of where the curves go without having to calculate very many discrete values. Each radial division ... p /2 p 0 3p /2 Figure 16-9 ... is 1 unit Polar-coordinate graph of the parametric equations q = et and r = t2. Each radial division represents 1 unit. From Graph to Equations We've seen how we can go from parametric equations to graphs. Now we'll do an exercise going from a graph to a pair of parametric equations. From Graph to Equations 315 Cartesian-coordinate graph to equations Consider a circle of radius a, centered at the origin in the Cartesian xy plane as shown in Fig. 16-10. From trigonometry, we remember that x = a cos f and y = a sin f where f is the angle going counterclockwise from the positive x axis. Both x and y depend on the value of f. Let's rename f and call it t, so our equations become x = a cos t and y = a sin t This is a pair of parametric equations representing a circle of radius a, centered at the origin in the Cartesian xy plane. For any particular circle, a is a constant (not a variable), so the parameter t is the only variable on the right-hand side of either equation. y x = a cos f f x a (x, y) y = a sin f Figure 16-10 Cartesian-coordinate graph of a circle with radius a, centered at the origin. We can let f = t to describe this circle as a pair of parametric equations. 316 Parametric Equations in Two-Space Figure 16-11 Polar-coordinate graph of a circle with radius a, centered at the origin. We can let f = t to describe this circle as a pair of parametric equations. p /2 r=a f p 0 a (f, r) 3p /2 f=t Polar-coordinate graph to equations Now let's convert the circle in the previous example to a pair of polar-form parametric equations. Suppose the polar direction angle is f, and the polar radius is r. The equation of a circle having radius a as shown in Fig. 16-11 is r=a Let's call the angle f our parameter t, just as we did in the xy-plane situation. Then we can write the parametric equations of our circle as f=t and r=a Are you confused? Does the above pair of parametric equations seem strange to you? The second equation doesn't contain the parameter! That's not a problem in this situation. The parameter has no effect because the polar radius r is always the same. Here's a challenge! Suppose that we come across a pair of parametric equations similar to the one in the Cartesiancoordinate example above, except that the cosine and sine of the parameter are multiplied by different nonzero real-number constants a and b, like this: x = a cos t From Graph to Equations 317 and y = b sin t What sort of curve should we expect to get if we graph the relation defined by this pair of parametric equations? Solution We've been told that a and b are both nonzero real numbers. Therefore, we can divide the equations through by their respective constants to get x /a = cos t and y /b = sin t If we square both sides of both equations, we obtain (x /a)2 = cos2 t and (y /b)2 = sin2 t When we add these two equations, left-to-left and right-to-right, we obtain the new equation (x /a)2 + (y /b)2 = cos2 t + sin2 t From trigonometry, we remember that for any real number t, it's always true that cos2 t + sin2 t = 1 Therefore, the preceding equation can be rewritten as (x /a)2 + (y /b)2 = 1 Expanding the squared ratios on the left-hand side gives us x2 /a2 + y2/b2 = 1 which is the equation of an ellipse centered at the origin. The horizontal (x-coordinate) semi-axis is a units wide, and the vertical (y-coordinate) semi-axis is b units high. 318 Parametric Equations in Two-Space In "Rectangular-coordinate graph: example 1" (Fig. 16-4), the parametric equations are x = t2 and y = t3 Find an equation for this relation that expresses x in terms of y without the parameter t. Then find an equation that expresses y in terms of x without the parameter t. 2. Is the relation defined in your second answer to Problem 1 a function of x? 3. In "Polar-coordinate graph: example 2" (Fig. 16-7), the parametric equations are q = t −1 and r = ln t Find an equation for this relation that expresses q in terms of r without the parameter t. Then find an equation that expresses r in terms of q without the parameter t. 4. Is the relation defined in your second answer to Problem 3 a function of q? 5. In "Cartesian-coordinate graph to equations" (Fig. 16-10), the parametric equations are x = a cos t and y = a sin t where a is a nonzero constant. Find an equation for this relation in terms of x and y only, without the parameter t. 6. Express the solution to Problem 5 as a relation in which x is the independent variable and y is the dependent variable. You should end up with y alone on the left-hand side of the equals sign, and an expression containing x (but not y) on the right-hand side. Is this relation a function of x? 7. Suppose that we come across the pair of parametric equations x = sec t and y = tan t Find an equation for this relation in terms of x and y only, without the parameter t. What sort of curve should we expect to get if we graph this relation in the Cartesian xy plane? Practice Exercises 319 8. Consider the pair of parametric equations x = a csc t and y = b cot t where a and b are nonzero real-number constants. Find an equation for this relation in terms of x and y only, without the parameter t. What sort of curve should expect to get if we graph this relation in the Cartesian xy plane? 9. Express the relation x = sin (cos y) as a pair of parametric equations. 10. Manipulate the equation stated in Problem 9 so that y appears all by itself on the left-hand side of the equals sign, and operations involving x appear on the right-hand side. Then manipulate your answer to Problem 9 to get the same equation. CHAPTER 17 Surfaces in Three-Space Three-space can contain an infinite variety of surfaces, all of which can be defined as equations in terms of three variables. In this chapter, we'll examine a few basic surfaces and their equations in Cartesian three-space. Planes An intuitive way to express the equation for a plane in Cartesian xyz space is to define the direction of a vector normal (perpendicular) to the plane, and then to identify the coordinates of a point in the plane. We don't have to know the magnitude of the vector, and the point in the plane doesn't have to be the one where the vector originates. General equation of plane Figure 17-1 shows a plane W in Cartesian three-space, a point P = (x0,y0,z0) in the plane W, and a vector (a,b,c) = ai + bj + ck that's normal to plane W. The vector (a,b,c) originates at a point Q that differs from P, and which is also located away from the coordinate origin. The values x = a, y = b, and z = c for the vector are nevertheless based on the vector's standard form, as if it originated at (0,0,0). The point and the vector give us enough information to uniquely define the plane and write its equation in standard form as a(x − x0) + b(y − y0) + c(z − z0) = 0 This equation can also be written as ax + by + cz + d = 0 where d is a stand-alone constant. With a little algebra, we can work out its value in terms of the other constants and coefficients as d = −ax0 − by0 − cz0 320 Planes Vector (a, b, c) normal to W at point Q +y 321 Point P (x0, y0 , z0) in plane W Point Q in plane W x +x Plane W +z y Figure 17-1 A plane W can be uniquely defined on the basis of a point P in the plane and a vector (a,b,c) normal to the plane. Plotting a plane When we want to construct a plane in Cartesian xyz space based on its equation, we can do it by figuring out the coordinates of points where the plane crosses each of the three coordinate axes. These points are the x-intercept, the y-intercept, and the z-intercept. When we plot these intercept points on the axes, we can envision the position and orientation of the plane. There's a potential "hangup" with this scheme for plane-graphing. Not all planes cross all three axes in Cartesian xyz-space. If a plane is parallel to one of the axes, then it does not cross that axis, although must cross at least one of the other two. If a plane is parallel to the plane formed by two coordinate axes, then that plane crosses only the axis with respect to which it is not parallel. An example Suppose that a plane contains the point (3,−6,2), and the standard form of a vector normal to the plane is 4i + 3j + 2k. Let's find the plane's equation in the standard form given above. To begin, we know that the vector 4i + 3j + 2k is equivalent to the ordered triple (a,b,c) = (4,3,2) 322 Surfaces in Three-Space We've been told that (x0,y0,z0) = (3,−6,2) and that this point lies in the plane. The general formula for the plane is a(x − x0) + b(y − y0) + c(z − z0) = 0 Plugging in the known values for a, b, c, x0, y0, and z0, we get 4(x − 3) + 3[y − (−6)] + 2(z − 2) = 0 which simplifies to 4x + 3y + 2z + 2 = 0 Are you confused? The standard-form equation of a plane in xyz space looks like an extrapolation of the standardform equation of a straight line in the xy plane. This can confuse some people. Don't let it baffle you! An equation of the form ax + by + cz + d = 0 where a, b, c, and d are constants represents a plane, not a line. In Chap. 18, you'll learn how to describe straight lines in Cartesian xyz space. Here's a challenge! Draw a graph of the plane represented by the following equation: −2x − 4y + 3z − 12 = 0 Solution Let's work out the graph by finding the coordinate-axis intercepts. The x-intercept, or the point where the plane intersects the x axis, can be found by setting y = 0 and z = 0, and then solving the resultant equation for x. Let's call this point P. We have −2x − 4 × 0 + 3 × 0 − 12 = 0 Solving step-by-step, we get −2x − 12 = 0 −2x = 12 x = 12/(−2) = −6 Therefore P = (−6,0,0) Planes 323 The y-intercept, or the point where the plane intersects the y axis, can be found by setting x = 0 and z = 0, and then solving the resultant equation for y. Let's call this point Q. We have −2 × 0 − 4y + 3 × 0 − 12 = 0 Solving, we get −4y − 12 = 0 − 4y = 12 y = 12/(−4) = −3 Therefore Q = (0,−3,0) The z-intercept, or the point where the plane intersects the z axis, can be found by setting x = 0 and y = 0, and then solving the resultant equation for z. Let's call this point R. We have −2 × 0 − 4 × 0 + 3z − 12 = 0 Solving, we get 3z − 12 = 0 3z = 12 z = 12/3 = 4 Therefore R = (0,0,4) These three points are shown in Fig. 17-2. We can now envision the plane because, as we recall from our courses in spatial geometry, a plane in three dimensions can be uniquely defined on the basis of three points. +y Each axis division is 1 unit P = (–6, 0, 0) –z –x +x Q = (0, –3, 0) +z R = (0, 0, 4) –y Figure 17-2 Here's the graph of a plane, based on the locations of the three axis intercept points P, Q, and R. 324 Surfaces in Three-Space Spheres A spherical surface is defined as the set of all points that lie at a fixed distance from a known central point in three dimensions. When we recall the formula for the distance between a point and the origin, it's easy to work out equations for spheres in Cartesian xyz space. Center at the origin Imagine a sphere whose center lies at the origin (0,0,0), as shown in Fig. 17-3. Any point on the sphere's surface is at the same distance from the origin as any other point on the sphere's surface. Suppose that P is one such point whose coordinates are given by P = (xp,yp,zp) In Chap. 7, we learned that the distance r of the point P from the origin in Cartesian xyz space is r = (xp2 + yp2 + zp2)1/2 We can square both sides of the above equation to get r2 = xp2 + yp2 + zp2 +y Center of sphere is at (0, 0, 0) –x +x Radius = r +z –y Figure 17-3 A sphere of radius r in Cartesian xyz space, centered at the origin. All points on the sphere's surface are at distance r from the center point (0,0,0). Spheres 325 Transposing the left- and right-hand sides, we have xp2 + yp2 + zp2 = r2 Every point on the sphere's surface is the same distance from the origin as P, so we can generalize the above equation to get x2 + y2 + z2 = r2 which defines the set of all points in three dimensions that lie at a fixed distance r from the origin. That's all there is to it! We've found the standard-form equation for a sphere of radius r, centered at the origin in Cartesian xyz space. Center away from the origin Consider a sphere whose center is somewhere other than the origin in Cartesian xyz space. Suppose that the coordinates of the center point are (x0,y0,z0), as shown in Fig. 17-4. Whatever point P that we choose on the sphere's surface, the distance between P and the center is equal to the sphere's radius r. Adapting the distance-between-points formula for Cartesian xyz space from Chap. 7, we get r = [(xp − x0)2 + (yp − y0)2 + (zp − z0)2]1/2 Center of sphere is at (x0, y0, z0) +y –z –x +x Radius = r +z –y Figure 17-4 A sphere of radius r in Cartesian xyz space, centered away from the origin. All points on the sphere's surface are at distance r from the center point (x0,y0,z0). 326 Surfaces in Three-Space Squaring both sides of this equation and then transposing the left- and right-hand sides, we obtain (xp − x0)2 + (yp − y0)2 + (zp − z0)2 = r2 Every point on the sphere's surface is the same distance from P as (x0,y0,z0), so we can generalize to get (x − x0)2 + (y − y0)2 + (z − z0)2 = r2 This is the standard-form equation for a sphere of radius r, centered at the point (x0,y0,z0) in Cartesian xyz space. An example Suppose we have a sphere whose center is at the origin, and whose radius is 7 units. If we let r = 7 in the general equation for a sphere centered at the origin, then we have x2 + y2 + z2 = 72 which can be simplified to x2 + y2 + z2 = 49 Another example Consider a sphere centered at the point (−2,4,−1) with a radius of 5 units in Cartesian xyz space. We can let x0 = −2 y0 = 4 z0 = −1 r=5 in the general equation for a sphere centered at a point other than the origin. When we plug in the numbers, we obtain [x − (−2)]2 + (y − 4)2 + [z − (−1)]2 = 52 which simplifies to (x + 2)2 + (y − 4)2 + (z + 1)2 = 25 Spheres 327 Are you confused? The radius of a sphere is usually defined as a positive real number. If we define the radius of a particular sphere as a negative real number, we get the same equation as we would if we defined the radius as the absolute value of that number. That's because we square the radius when we work out the formula. For example, if we have a sphere centered at the origin with a radius of 4 units, then its equation is x2 + y2 + z2 = 42 which simplifies to x2 + y2 + z2 = 16 If we find a companion "antisphere" centered at the origin with radius −4 units, then its equation is x2 + y2 + z2 = (−4)2 which also simplifies to x2 + y2 + z2 = 16 In physics and engineering, it's possible to come up with spheres having negative radii, as well as negative dimensions for other physical objects. These results are usually mere artifacts of the calculation process, and don't have any significance in the real world. However, if you ever encounter a sphere whose radius is represented by an imaginary number such as j4, then you have good reason to be confused until you know what sort of object or phenomenon the equation describes! Here's a challenge! Suppose we're told that the following equation represents a sphere in Cartesian xyz space: x2 − 6x + y2 − 2y + z2 + 4z = 86 We're also informed that the sphere has a radius of 10 units. What are the coordinates of the center of the sphere? Solution To solve this problem, we need some intuition. We know that the radius r of the sphere is 10 units. Therefore, r2 = 100. If we add 14 to both sides of the original equation, we get r2 on the right-hand side: x2 − 6x + y2 − 2y + z2 + 4z + 14 = 100 We can split the stand-alone constant 14 into the sum of 9, 1, and 4, getting x2 − 6x + y2 − 2y + z2 + 4z + 9 + 1 + 4 = 100 328 Surfaces in Three-Space Rearranging the addends on the left-hand side produces the following equation: x2 − 6x + 9 + y2 − 2y + 1 + z2 + 4z + 4 = 100 When we group the terms on the left-hand side by threes, we obtain (x2 − 6x + 9) + (y2 − 2y + 1) + (z2 + 4z + 4) = 100 This is a sum of three perfect squares! Factoring them individually gives us (x − 3)2 + (y − 1)2 + (z + 2)2 = 100 The coordinates of the center point are therefore x0 = 3 y0 = 1 z0 = −2 Expressed as an ordered triple, it's (3,1,−2). Distorted Spheres Spheres can be made "out of the round" by increasing or decreasing the axial radii in the x, y, and z directions individually. Alternative equation for a sphere centered at the origin Once again, consider the general equation of a perfect sphere centered at the origin. That equation, in standard form, is x2 + y2 + z2 = r2 where r is the radius. If we divide through by r2, we get x2/r2 + y2/r2 + z2/r2 = 1 This equation tells us that the radius is always the same, whether we measure it in the direction of the x axis, y axis, or z axis. To emphasize the fact that we can, if desired, change any of all of these axial radii, let's rewrite the above equation as x2/a2 + y2/b2 + z2/c2 = 1 where a, b, and c are positive real numbers representing the radii along the x, y, and z axes, respectively. In the case of a perfect sphere, we have a=b=c If these three positive real-number constants a, b, and c are not all the same, then we have a distorted sphere. Distorted Spheres 329 Oblate sphere centered at the origin Suppose that we take a perfect sphere and then shorten one of the three axial radii. This process gives us an object called an oblate sphere. It's flattened, like a soft rubber ball when pressed between our hands. Figure 17-5 shows an example. This is what we get if we take the sphere from Fig. 17-3 and reduce the axial radius b (the one that goes along the y axis), while leaving the axial radii a and c unchanged. The center of the object is still at the origin, but we can no longer say that all the points on its surface are equidistant from the origin. The general equation for an oblate sphere holds true among them: a<b=c b<a=c c<a=b Alternative equation for a sphere centered away from the origin Earlier in this chapter, we learned that the general equation of a sphere centered at some point other than the origin in Cartesian xyz space is (x − x0)2 + (y − y0)2 + (z − z0)2 = r2 +y Center is at (0, 0, 05 An oblate sphere in Cartesian xyz space, centered at the origin. 330 Surfaces in Three-Space where r is the radius, and (x0,y0,z0) are the coordinates of the center. Dividing through by r2, we obtain (x − x0)2/r2 + (y − y0)2/r2 + (z − z0)2/r2 = 1 As we did with the sphere centered at the origin, we can rewrite this equation, getting (x − x0)2/a2 + (y − y0)2/b2 + (z − z0)2/c2 = 1 where a, b, and c are the radii parallel to the x, y, and z axes, respectively. As before, with a perfect sphere, we have a=b=c If a, b, and c are not all the same, then the sphere is distorted. Oblate sphere centered away from the origin If we take a sphere that's centered at (x0,y0,z0) and shorten one of the axial radii, we get an oblate sphere<b=c b<a=c c<a=b Figure 17-6 should give you a general idea of what happens in a case like this. Imagine a sphere centered at (x0,y0,z0) that has been squashed in the direction defined by a line parallel to the y axis. Ellipsoid centered at the origin Again, imagine that we have a perfect sphere centered at the origin in Cartesian xyz space. Let's lengthen one of the axial radii while leaving the other two unchanged. This stretching process produces an ellipsoid. It's elongated, like a football with blunted ends. Figure 17-7 shows an example. Imagine that we take the sphere from Fig. 17-3 and then stretch it in the z direction. The general equation for an ellipsoid is true: a>b=c b>a=c c>a=b 332 Surfaces in Three-Space Ellipsoid centered away from the origin Consider a sphere centered at (x0,y0,z0). If we make one of the axial radii longer while leaving the other two unchanged, we get an ellipsoid>b=c b>a=c c>a=b Figure 17-8 portrays a situation in which a sphere centered at (x0,y0,z0) has been stretched along a line parallel to the z axis to obtain an ellipsoid. Oblate ellipsoid centered at the origin One more time, imagine a sphere centered at the origin. We start out with all three axial radii equal in measure. Then we lengthen one of them, shorten another, and leave the third one unchanged. This process gives us an oblate ellipsoid. Figure 17-9 shows an example where we take the sphere from Fig. 17-3, squash the radius in the y direction, stretch the radius in the z direction, and leave the radius unchanged in the x direction. The general equation for an oblate ellipsoid centered at the origin is x2/a2 + y2/b2 + z2/c2 = 1 +y Center is at (x0, y0, z08 An ellipsoid in Cartesian xyz space, centered at (x0,y0,z0). Distorted Spheres 333 +y Center is at (0, 0, 0) Radius in y direction =b –x +x +z Radius in x direction =a Radius in z direction =c –y Figure 17-9 An oblate ellipsoid in Cartesian xyz space, centered at the origin. where a is the x-axial radius, b is the y-axial radius, c is the z-axial radius, and all of the following are true: a≠b b≠c a≠c Oblate ellipsoid centered away from the origin Finally, imagine a sphere that's centered at (x0,y0,z0). If we lengthen one of the axial radii, shorten another, and leave the third one unchanged, we get an oblate ellipsoid defined by (x − x0)2/a2 + (y − y0)2/b2 + (z − z0)2/c2 = 1 where all of the following are true: a≠b b≠c a≠c Figure 17-10 shows an example of what happens when we move the center of the oblate ellipsoid from Fig. 17-9 away from the origin. 334 Surfaces in Three-Space +y Radius in y direction =b Center is at (x0, y0, z0) –z –x +x Radius in x direction =a Radius in z direction =c +z –y Figure 17-10 An oblate ellipsoid in Cartesian xyz space, centered at (x0,y0,z0). An example Suppose that the coordinates of the center of a certain oblate sphere in Cartesian xyz space are (1,2,3). The axial radius in the x direction is 4, the axial radius in the y direction is 4, and the axial radius in the z direction is 2. The general equation b is the axial radius in the y direction, and c is the axial radius in the z direction. We know that (x0,y0,z0) = (1,2,3) a=4 b=4 c=2 Plugging these values into the general equation, we conclude that our oblate sphere can be represented by the following equation: (x − 1)2/42 + (y − 2)2/42 + (z − 3)2/22 = 1 which simplifies to (x − 1)2/16 + (y − 2)2/16 + (z − 3)2/4 = 1 Distorted Spheres 335 Another example The coordinates of the center of an ellipsoid are (−3,−2,−6). The axial radius in the x direction is 3, the axial radius in the y direction is 7, and the axial radius in the z direction is 3. The general equation is (x − x0)2/a2 + (y − y0)2/b2 + (z − z0)2/c2 = 1 This time, we have (x0,y0,z0) = (−3,−2,−6) a=3 b=7 c=3 Plugging these values into the general equation, we obtain [x − (−3)]2/32 + [y − (−2)]2/72 + [z − (−6)]2/32 = 1 which simplifies to (x + 3)2/9 + (y + 2)2/49 + (z + 6)2/9 = 1 Still another example The coordinates of the center of an oblate ellipsoid are (0,−3,11). The axial radius in the x direction is 5, the axial radius in the y direction is 8, and the axial radius in the z direction is 1. The general equation is (x − x0)2/a2 + (y − y0)2/b2 + (z − z0)2/c2 = 1 In this case, we have (x0,y0,z0) = (0,−3,11) a=5 b=8 c=1 Plugging these values into the general equation gives us [x − 0]2/52 + [y − (−3)]2/82 + (z − 11)2/12 = 1 which simplifies to x2/25 + (y + 3)2/64 + (z − 11)2 = 1 336 Surfaces in Three-Space Are you astute? So far, we've described various surfaces by adding squared binomials to each other. You have every right to ask, "What will happen if we subtract any of the squared binomials in equations like these?" We'll do that shortly, and you'll see a few examples of what can take place. When we add squared binomials, the graphs always turn out to be spheres, oblate spheres, ellipsoids, or oblate ellipsoids in Cartesian xyz space. These are closed surfaces. They're "air-tight." If we subtract one or more of the squared binomials, we get open surfaces that "can't hold air." Such surfaces can take diverse, interesting forms. Here's a challenge! Consider a distorted sphere represented by the following equation: 12x2 + 72x + 20y2 − 80y + 15z2 − 30z = −143 What are the coordinates of the center? What are the axial radii? Is the object an oblate sphere, an ellipsoid, or an oblate ellipsoid? Solution This problem requires a lot of insight to solve! Let's begin by adding 203 to each side of the equation to obtain 12x2 + 72x + 20y2 − 80y + 15z2 − 30z + 203 = 60 The number we've added, 203, happens to be the sum of 108, 80, and 15. Let's add these three numbers into the above equation just after the terms 72x, 80y, and −30z, respectively. The equation then becomes 12x2 + 72x + 108 + 20y2 − 80y + 80 + 15z2 − 30z + 15 = 60 Grouping the addends on the left-hand side by threes gives us (12x2 + 72x + 108) + (20y2 − 80y + 80) + (15z2 − 30z + 15) = 60 which is equivalent to 12(x2 + 6x + 9) + 20(y2 − 4y + 4) + 15(z2 − 2z + 1) = 60 The three trinomials factor into perfect squares, so we can further morph the equation to obtain 12(x + 3)2 + 20(y − 2)2 + 15(z − 1)2 = 60 Dividing through by 60, we get (x + 3)2/5 + (y − 2)2/3 + (z − 1)2/4 = 1 Other Surfaces 337 We recall that general formula for a distorted sphere in Cartesian xyz space axial radius in the x direction, b is the axial radius in the y direction, and c is the axial radius in the z direction. In this situation, we have (x0,y0,z0) = (−3,2,1) a = 51/2 b = 31/2 c = 41/2 = 2 Our object is an oblate ellipsoid centered at (−3,2,1). The radius in the x direction is 51/2. The radius in the y direction is 31/2. The radius in the z direction is 2. Other Surfaces Let's look at three general objects that arise in Cartesian xyz space from equations with sums and differences of terms containing x2, y2, and z2. Hyperboloid of one sheet Figure 17-11 shows a generic example of a hyperboloid of one sheet. In this context, the term sheet refers to an unbroken surface. We get this type of object when we graph an equation of the form x2/a2 + y2/b2 − z2/c2 = 1 where a, b, and c are positive real-number constants. This equation is like the one for a distorted sphere, except that one of the plus signs has been replaced by a minus sign. That sign change makes a huge difference! Instead of a closed surface centered at the origin, we get an infinitely tall, pinched cylinder whose axis lies along the coordinate z axis, and whose center coincides with the origin. The dimensions and shape of the hyperboloid depend on the values of a, b, and c. The perpendicular cross sections are always circles or ellipses. If we move the minus sign so that it's in front of the term containing y2 instead of the term containing z2, we get the general equation x2/a2 − y2/b2 + z2/c2 = 1 Again, we get a hyperboloid of one sheet, but its axis is along the coordinate y axis, and its center is at the origin. If we move the minus sign one more place to the left, putting it in front of the term containing x2, the general equation becomes −x2/a2 + y2/b2 + z2/c2 = 1 This is the general form of the equation for a hyperboloid of one sheet whose axis coincides with the coordinate x axis, and whose center is at the origin. 338 Surfaces in Three-Space +z –x –y +y +x –z Figure 17-11 A hyperboloid of one sheet in Cartesian xyz space, centered at the origin. Are you astute? Figure 17-11 shows a perspective on Cartesian xyz space that we haven't seen before. We're looking "down" on the yz plane from somewhere near the positive x axis. Nevertheless, the axes are correctly oriented with respect to each other, as you can verify by referring back to Chap. 7. Let's stay with this axis orientation as we look at the next couple of objects. Hyperboloid of two sheets Figure 17-12 shows a hyperboloid of two sheets, which is the graph in Cartesian xyz space of an equation having the form −x2/a2 + y2/b2 − z2/c2 = 1 where a, b, and c are positive real-number constants. Here, we have two surfaces that resemble bowls facing in opposite directions. In theory, the bowls extend infinitely toward the left and the right in this illustration. Both surfaces share a common straight-line axis that coincides with the coordinate y axis, and the two sheets are exact mirror images of each other. The center of the entire hyperboloid is at the origin. The contours of the surfaces depend on the values of a, b, and c. Other Surfaces 339 +z –x –y +y +x –z Figure 17-12 A hyperboloid of two sheets in Cartesian xyz space, centered at the origin. If we make the term containing x2 positive instead of the term containing y2, we get the general equation x2/a2 − y2/b2 − z2/c2 = 1 which produces a hyperboloid of two sheets whose axis lies along the coordinate x axis, and whose center is at the origin. If we move the plus sign so it's in front of the term containing z2, the general equation becomes −x2/a2 − y2/b2 + z2/c2 = 1 This maneuver gives us a hyperboloid of two sheets whose axis lies along the coordinate z axis, and whose center is at the origin. Elliptic cone Figure 17-13 shows an elliptic cone. It's what we get when we graph an equation of the form x2/a2 + y2/b2 − z2/c2 = 0 where a, b, and c are positive real-number constants. The perpendicular cross sections of the cone are always circles or ellipses. The cone's axis coincides with the coordinate z axis, and the cone's vertex coincides with the origin. The flare angles, as well as the eccentricity of the crosssectional ellipses, depend on the values of a, b, and c. 340 Surfaces in Three-Space +z –x –y +y +x –z Figure 17-13 An elliptic cone in Cartesian xyz space, centered at the origin. If we move the minus sign so it's in front of the term containing y2, we get the general equation x2/a2 − y2/b2 + z2/c2 = 0 whose graph is an elliptic cone with the axis along the coordinate y axis, and whose center is at the origin. If we move the minus sign so that it's in front of the term containing x2, the general equation becomes −x2/a2 + y2/b2 + z2/c2 = 0 and the graph becomes an elliptic cone whose axis lies along the coordinate x axis, and whose center is at the origin. An example Consider the object in Cartesian xyz space represented by 36x2 − 16y2 + 36z2 = 0 We can divide through by 144 to obtain x2/4 − y2/9 + z2/4 = 0 This is the equation for an elliptic cone whose vertex is at the origin, and whose axis coincides with the coordinate y axis. Other Surfaces 341 Another example Consider the object in Cartesian xyz space represented by −x2 + y2 + z2 = −7 When we divide through by −7, we get −x2/(−7) + y2/(−7) + z2/(−7) = 1 which simplifies to x2/7 − y2/7 − z2/7 = 1 This equation describes a hyperboloid of two sheets whose center is at the origin, and whose axis lies along the coordinate x axis. Still another example Consider the object in Cartesian xyz space represented by 15x2 + 10y2 = 6z2 + 30 We can subtract 6z2 from each side, getting 15x2 + 10y2 − 6z2 = 30 Dividing through by 30 gives us x2/2 + y2/3 − z2/5 = 1 This is the equation for a hyperboloid of one sheet whose center is at the origin, and whose axis lies along the coordinate z axis. Are you confused? It's reasonable to ask, "What if the center of a hyperboloid, or the vertex of an elliptic cone, lies somewhere other than the origin, say at (x0,y0,z0)? What happens to the equation in that case?" If you're willing to exercise your mathematical intuition, you can probably guess the answer. Make the following substitutions in the equation: • Replace every occurrence of x with x − x0 • Replace every occurrence of y with y − y0 • Replace every occurrence of z with z − z0 Consider a hyperboloid of two sheets such as the one in Fig. 17-12. The straight-line axis of the "bowls" lies along the coordinate y axis, and the center of the entire object is at the origin. If a = 2, b = 3, and c = 4, the equation is −x2/22 + y2/32 − z2/42 = 1 342 Surfaces in Three-Space which simplifies to −x2/4 + y2/9 − z2/16 = 1 Now suppose that you change the equation to −(x − 7)2/4 + (y + 1)2/9 − (z − 5)2/16 = 1 You've moved the entire hyperboloid, without altering its overall shape or orientation. It has a new center whose coordinates are (7,−1,5) instead of (0,0,0). The straight-line axis of the two bowls is parallel to, but no longer coincides with, the coordinate y axis. If you want to disguise this equation, you can multiply it through by the product of the denominators on the left-hand side, getting −144(x − 7)2 + 64(y + 1)2 − 36(z − 5)2 = 576 Here's a challenge! Consider the object in Cartesian xyz space represented by 3x2 + 6x − 4y2 − 16y + 2z2 − 4z = 35 What do we get when we graph this equation in Cartesian xyz space? Where is the center of the object? How is its axis oriented? Solution As with some of the examples we've seen, we need lot of intuition to solve this problem. Let's subtract 11 from both sides of the equation. That gives us 3x2 + 6x − 4y2 − 16y + 2z2 − 4z − 11 = 24 When we subtract 11, we in effect add −11, which happens to be the sum of 3, −16, and 2. That means we can rewrite the above equation as 3x2 + 6x − 4y2 − 16y + 2z2 − 4z + 3 − 16 + 2 = 24 which can be rearranged to get 3x2 + 6x + 3 − 4y2 − 16y − 16 + 2z2 − 4z + 2 = 24 Grouping the terms on the left-hand side into trinomials, and paying special attention to the signs associated with the variable y as we group the second three terms, we get (3x2 + 6x + 3) − (4y2 + 16y + 16) + (2z2 − 4z + 2) = 24 which morphs to 3(x2 + 2x + 1) − 4(y2 + 4y + 4) + 2(z2 − 2z + 1) = 24 Practice Exercises 343 and further to 3(x + 1)2 − 4(y + 2)2 + 2(z − 1)2 = 24 Dividing through by 24, we get (x + 1)2/8 − (y + 2)2/6 + (z − 1)2/12 = 1 This is the equation of a hyperboloid of one sheet whose center is at (−1,−2,1), and whose axis is oriented along a line parallel to the coordinate y axis Suppose that a plane contains the point (0,0,0), and the standard form of a vector normal to the plane is −4i + 4j − 4k. Find the plane's equation in standard form. 2. Suppose that a plane contains the point (4,5,6), and the standard form of a vector normal to the plane is −2i + 0j + 0k. Find the plane's equation in standard form. 3. Consider a sphere whose equation is x2 + 2x + 1 + y2 − 2y + 1 + z2 + 8z + 16 = 64 What are the coordinates of the center of this sphere? What's its radius? 4. What's the equation of a sphere centered at the point (5,7,−3) and whose radius is equal to the positive square root of 23? 5. Consider the equation 8(x − 1)2 + 8(y + 2)2 + 6(z + 7)2 = 24 What sort of object does this equation describe? Does the object have a center? If so, what are the coordinates of the center point? Does the object have axial radii? If so, what are they? 6. Consider the equation 400(x + 2)2 + 225(y − 4)2 + 144z2 − 3600 = 0 What sort of object does this equation describe? Does the object have a center? If so, what are the coordinates of the center point? Does the object have axial radii? If so, what are they? 344 Surfaces in Three-Space 7. Consider a surface whose equation is x2 + 2x + 1 + y2 − 2y + 1 − z2 + 6z − 9 = 36 What sort of object is this? What are the coordinates of the center? How is the axis oriented? 8. Write down a generalized equation for an elliptic cone whose axis is parallel to the coordinate y axis, and whose vertex is at (−2,3,4). 9. Suppose we slice the elliptic cone described in Problem 8 straight through with the coordinate xz plane. The cone's surface intersects the xz plane in a curve. Derive a generalized equation of that curve in the variables x and z. What sort of curve is it? Here's a hint: At every point in the xz plane, y = 0. 10. Suppose we slice the elliptic cone described in Problem 8 straight through with the coordinate xy plane. The cone's surface intersects the xy plane in a curve. Derive a generalized equation of that curve in the variables x and y. What sort of curve is it? Here's a hint: At every point in the xy plane, z = 0. CHAPTER 18 Lines and Curves in Three-Space In Chap. 16, we learned how parametric equations can define curves that are difficult to portray as conventional relations. "Parametric power" becomes more apparent when we graduate to three dimensions. Straight Lines Finding an equation for a straight line in Cartesian three-space is harder than it is in the Cartesian plane. The extra dimension makes expressing the line's location and orientation more complicated. There are at least two ways we can do it: the symmetric method and the parametric method. Symmetric method A straight line in Cartesian xyz space can be represented by a three-part symmetric-form equation. Suppose that (x0,y0,z0) are the coordinates of a known point on the line, and a, b, and c are nonzero real-number constants. Given this information, we can represent the line as (x − x0)/a = (y − y0)/b = (z − z0)/c If a = 0 or b = 0 or c = 0, then we get a zero denominator somewhere, and the system becomes meaningless. Direction numbers In the symmetric-form equation of a straight line, the constants a, b, and c are known as the direction numbers. Imagine a vector m whose originating point is at the origin (0,0,0) and whose terminating point has coordinates (a,b,c). Under these circumstances, the vector m either lies right along, or is parallel to, the line denoted by the symmetric-form equation. 345 346 Lines and Curves in Three-Space +y Line L and vector m are parallel L m = (a, b, c) z x +x P = (x0, y0, z0) +z y Figure 18-1 We can uniquely define a line L in Cartesian xyz space on the basis of a point P on L and a vector m = (a,b,c) parallel to L. (In three-space, a vector m and a straight line L are parallel if and only if the line containing m occupies the same plane as L but does not intersect L.) We have m = ai + bj + ck where m is the three-dimensional equivalent of the slope of a line in the Cartesian plane. Figure 18-1 shows a generic example. Parametric method Given any particular line L in Cartesian xyz space, we can find infinitely many vectors to play the role of the direction-defining vector m. If t is a nonzero real number, then any vector t m = (ta,tb,tc) = ta i + tb j + tc k works just as well as m = ai + bj + ck for the purpose of defining the direction of L, so we have an alternative way to describe a straight line using the following equations: x = x0 + at y = y0 + bt z = z0 + ct Straight Lines 347 The variable t behaves as a "master controller" for the variables x, y, and z, so the above system is a set of parametric equations for a straight line in Cartesian xyz space. To completely define a straight, infinitely long line this way, we must let t vary throughout the entire set of real numbers, including t = 0 to "fill the hole" at the point (x0,y0,z0). An example Let's find the symmetric-form equation for the line L shown in Fig. 18-2. As indicated in the drawing, L passes through the point P = (−5,−4,3) and is parallel to the vector m = 3i + 5j − 2k The direction numbers of L are the coefficients of the vector m, so we have a=3 b=5 c = −2 +y L m = 3i + 5j – 2k Each axis division equals 1 unit –z –x +x Line L and vector m are parallel +z P= (–5, –4, 3) Figure 18-2 –y What are the symmetric and parametric equations for line L? 348 Lines and Curves in Three-Space We are given a point P on the line L with the coordinates x0 = −5 y0 = −4 z0 = 3 The general symmetric-form equation for a line in Cartesian xyz space is (x − x0)/a = (y − y0)/b = (z − z0)/c When we plug in the known values, we get the three-part equation [x − (−5)]/3 = [y − (−4)]/5 = (z − 3)/(−2) which simplifies to (x + 5)/3 = (y + 4)/5 = (z − 3)/(−2) Another example Let's find a set of parametric equations for the line L shown in Fig. 18-2. In this case, our work is easy. We can take the values of x0, y0, z0, a, b, and c that we already know, and plug them into the generalized set of parametric equations x = x0 + at y = y0 + bt z = z0 + ct The results are x = −5 + 3t y = −4 + 5t z = 3 − 2t Are you confused? For any particular line in Cartesian xyz space, there are infinitely many valid ordered triples that can represent the direction numbers. If a line has the direction numbers (2,3,4), then we can multiply all three entries by a real number other than 0 or 1, and we'll get another valid ordered triple of direction numbers. For example, all of the following ordered triples represent the same line orientation as (2,3,4): (4,6,8) (−2,−3,−4) (20,30,40) (−20,−30,−40) (2p,3p,4p ) (−2p,−3p,−4p ) Straight Lines 349 "That's interesting," you say, "but which direction numbers are the best?" In theory, it doesn't matter; any of the above ordered triples is as "good" as any other. Nevertheless, from an esthetic point of view, it's a good idea to reduce an ordered triple of direction numbers so that the only common divisor is 1, and so that there is at most one negative element. According to that standard, (2,3,4) are the preferred direction numbers. Here's a challenge! Consider the following three-way equation that represents a straight line in Cartesian xyz space: 3x − 6 = 4y − 12 = 6z − 24 Find a point on the line. Determine the preferred direction numbers. Based on that information, write down the direction vector as a sum of multiples of i, j, and k. Solution Before we think about the direction numbers or any specific point on the line, let's try to get the equation into the standard symmetric form. We can multiply the left-hand part of the equation by 4/4, the middle part by 3/3, and the right-hand part by 2/2. That gives us 4(3x − 6)/4 = 3(4y − 12)/3 = 2(6z − 24)/2 Multiplying out the numerators, we get (12x − 24)/4 = (12y − 36)/3 = (12z − 48)/2 We can factor out 12 from each of the numerators to obtain 12(x − 2)/4 = 12(y − 3)/3 = 12(z − 4)/2 Dividing the entire equation through by 12 gives us the standard symmetric form (x − 2)/4 = (y − 3)/3 = (z − 4)/2 We remember that the generalized symmetric equation for a straight line in Cartesian xyz space is (x − x0)/a = (y − y0)/b = (z − z0)/c where (x0,y0,z0) are the coordinates of a specific point on the line, and a, b, and c are the direction numbers. Comparing the symmetric-form equation we derived with the generalized form, we can see that x0 = 2 y0 = 3 z0 = 4 350 Lines and Curves in Three-Space This tells us that (2,3,4) is a point on the line. We can also see that a=4 b=3 c=2 so the line's direction numbers are (4,3,2). We can write down a standard-form direction vector m from these numbers as m = 4i + 3j + 2k Parabolas From algebra, we remember that a quadratic equation in a variable x can always be written in the form a1x2 + a2x + a3 = 0 where a1, a2, and a3 are real-number constants called the coefficients, and a1 ≠ 0. If we replace the 0 on the right-hand side of this equation by another variable and then transpose the sides, we get an expression for a quadratic function. For example, 4x2 + 2x + 1 = 0 is a quadratic equation in x, but y = 4x2 + 2x + 1 is a quadratic function in which the independent variable is x and the dependent variable is y. If we give our function a name (f, for example), then we can denote it as f (x) = 4x2 + 2x + 1 When we graph a quadratic function in Cartesian two-space, we always get a parabola that's fairly easy to graph, because there's only one plane to worry about (the xy plane, if our independent variable is x and our dependent variable is y). In xyz space, the situation is more complicated, because we have an extra variable. There are infinitely many different planes in which a parabola can lie, as well as infinitely many different shapes and orientations for a parabola in any particular plane. Let's look at a few simple cases. Hold x constant Imagine a parameter t that's allowed to wander all over the set of real numbers. Also imagine a generalized quadratic function f of this parameter, such that f (t) = a1t2 + a2t + a3 Parabolas 351 +y Parabola in plane x=c (c, 0, 0) z x +x +z x=c y Figure 18-3 Parabola in a plane where x is held to a constant value c. The plane is perpendicular to the x axis, and intersects that axis at the point (c,0,0). where a1, a2, and a3 are real-number coefficients. Let's go into Cartesian xyz space and restrict ourselves to a single plane in which the value of x is some real-number constant c. This plane is parallel to the yz plane, and it intersects the x axis at the point (c,0,0). Consider a parabola in the plane x = c whose axis is parallel to the y axis, as shown in Fig. 18-3. (The axis of a parabola is a straight line in the same plane as the parabola, and on either side of which the parabola is symmetrical.) In this situation, the value of z tracks right along with the value of t, while the variable y follows f (t). Therefore x=c y = f (t) = a1t2 + a2t + a3 z=t The above set of equations is a parametric description of our parabola. If we want to describe a parabola in the plane x = c whose axis is parallel the z axis instead of the y axis, then y follows t while z follows f (t), and we have x=c y=t z = f (t) = a1t2 + a2t + a3 352 Lines and Curves in Three-Space Hold y constant Now suppose that we restrict our movements to a plane in which the value of y is always equal to a constant c. The equation of the plane is y = c. It's parallel to the xz plane, and it intersects the y axis at (0,c,0). Imagine a parabola in this plane whose axis is parallel to the z axis, as shown in Fig. 18-4. In this situation, x follows t while z follows f (t), and the curve can be described as x=t y=c z = f (t) = a1t2 + a2t + a3 To describe a parabola in the plane y = c whose axis is parallel the x axis, we can let z follow t and let x follow f (t), getting the system x = f (t) = a1t2 + a2t + a3 y=c z=t Hold z constant Finally, let's confine our attention to a single plane in which the value of z is some real-number constant c. The plane z = c is parallel to the xy plane, and it intersects the z axis at (0,0,c). +y Parabola in plane y=c y=c z x +x (0, c, 0) +z y Figure 18-4 Parabola in a plane where y is held to a constant value c. The plane is perpendicular to the y axis, and intersects that axis at the point (0,c,0). Parabolas Parabola in plane z=c 353 +y (0, 0, c) z x +x +z z=c y Figure 18-5 Parabola in a plane where z is held to a constant value c. The plane is perpendicular to the z axis, and intersects that axis at the point (0,0,c). Imagine a parabola in the plane z = c whose axis is parallel to the y axis as shown in Fig. 18-5. Here, x follows t while y follows f (t). We therefore have the parametric system x=t y = f (t) = a1t2 + a2t + a3 z=c For a parabola in the plane z = c whose axis is parallel to the x axis instead of the y axis, the value of y follows t while the value of x follows f (t), so we have x = f (t) = a1t2 + a2t + a3 y=t z=c An example Consider a quadratic function in the plane x = 2. Suppose that the parametric equations are x=2 y=t z = t2 − 3t + 2 354 Lines and Curves in Three-Space Using the knowledge we've gained so far in this chapter, along with our existing knowledge of algebra (such as we got from Algebra Know-It-All or a comparable algebra book), let's draw a graph of this function. Imagine that we're gazing broadside at the plane x = 2 from some distant point on the positive x axis. We've been told that y = t. If we stay in the plane x = 2, we can therefore write the quadratic function by direct substitution as z = y2 − 3y + 2 The coefficient of y2 is positive, so the parabola opens in the positive z direction. The above polynomial equation factors into z = (y − 1)(y − 2) so we can see that z = 0 when y = 1, and also that z = 0 when y = 2. Because x is always equal to 2, we know that the points (2,1,0) and (2,2,0) are on the parabola. The curve opens in the positive z direction, so we know that the parabola must have an absolute minimum. The y-value at the point, ymin, is the average of the y-values of the points where z = 0. Therefore ymin = (1 + 2)/2 = 3/2 To find the z-value at this point, we plug 3/2 into the quadratic function and get zmin = (3/2)2 − 3 × 3/2 + 2 = 9/4 − 9/2 + 2 = 9/4 − 18/4 + 8/4 = (9 − 18 + 8)/4 = −1/4 We've determined that the coordinates of the absolute minimum are (2,3/2,−1/4). We also know that the points (2,1,0) and (2,2,0) lie on the parabola. Figure 18-6 shows these points. They're close together, so it's difficult to get a clear picture of the parabola based on their locations. But we can find another point to help us draw the curve. When we plug in 0 for y, we get z = y2 − 3y + 2 = 02 − 3 × 0 + 2 =0−0+2=2 This tells us that the point (2,0,2) is on the curve. It's also shown in Fig. 18-6. Another example Now let's look at a quadratic function in the plane where y = 5. Suppose that the parametric equations are x=t y=5 z = 2t2 + 4t + 3 Parabolas 355 +z (2, 0, 2) Each axis increment is 1/4 unit (2, 1, 0) +y (2, 2, 0) (2, 3/2, –1/4) Figure 18-6 Graph of a parabola in a plane parallel to the yz plane, such that x has a constant value of 2. On both axes, each increment represents 1/4 unit. Imagine that we're gazing broadside at the plane y = 5 from somewhere on the negative y axis. We have been told that x = t, so we can write the quadratic function as z = 2x2 + 4x + 3 This parabola opens in the positive z direction, because the coefficient of x2 is positive. That means this parabola attains an absolute minimum for some value of x. Let's call it xmin. When x is the independent variable and z is the dependent variable, the general polynomial form for a quadratic function is z = a1x2 + a2x + a3 where a1, a2, and a3 are constants. From our algebra courses, we know that xmin = −a2/(2a1) In this situation, we have xmin = −4/(2 × 2) = (−4)/4 = −1 356 Lines and Curves in Three-Space The z-value at the absolute minimum point is zmin = 2xmin2 + 4xmin + 3 = 2 × (−1)2 + 4 × (−1) + 3 = 2 − 4 + 3 = −2 + 3 = 1 Now we know that the coordinates of the parabola's vertex are (−1,5,1). As the basis for our next point, let's choose x = −3. We can plug it directly into the function to get z = 2x2 + 4x + 3 = 2 × (−3)2 + 4 × (−3) + 3 = 18 − 12 + 3 = 6 + 3 = 9 This gives us (−3,5,9) as the coordinates of a second point on the curve. Finally, let's set x = 1. Plugging it in, we obtain z = 2x2 + 4x + 3 = 2 × 12 + 4 × 1 + 3 =2+4+3=9 The third point on our curve is (1,5,9). We now have three points: (−3,5,9), (−1,5,1), and (1,5,9). Figure 18-7 shows these points, along with a graph of the parabola passing through them, as seen in the plane where y maintains a constant value of 5. +z (–3, 5, 9) (1, 5, 9) Each axis increment is 1 unit +x (–1, 5, 1) Figure 18-7 Graph of a parabola in a plane parallel to the xz plane, such that y has a constant value of 5. On both axes, each increment represents 1 unit. Circles 357 Are you curious and ambitious? Think about the graphs of higher-degree polynomial functions confined to specific planes in Cartesian xyz space. For example, consider the cubic function in the plane where x = 2, such that x=2 y=t z = t3 or the quartic function in the plane where z = −7, such that x = 3t 4 + 6 y=t z = −7 Can you draw graphs of these curves? Circles In Chap. 13, we learned that the equation of a circle centered at the origin in the Cartesian xy plane can be written in the form x2 + y2 = r2 where r is the radius. In Chap. 16, we learned that the parametric equations for such a circle are x = r cos t and y = r sin t where t is the parameter. Let's expand these notions to deal with any circle in xyz space that's centered on, and exists entirely in a plane perpendicular to, one of the three coordinate axes. Hold x constant Consider a plane x = c in Cartesian xyz space, where c is a constant. This plane is parallel to the yz plane, and it intersects the x axis at (c,0,0). Imagine a circle of radius r in the plane x = c that's centered on the x axis as shown in Fig. 18-8. The variable y follows along with 358 Lines and Curves in Three-Space +y Circle in plane x=c x=c (c, 0, 0) z x +x Radius of circle =r +z y Figure 18-8 Circle in a plane where x is held to a constant value c. The plane is perpendicular to the x axis, and intersects that axis at the point (c,0,0). The circle has radius r and is centered at (c,0,0). r cos t, while the variable z follows along with r sin t. Therefore, we can define our circle with the system of parametric equations x=c y = r cos t z = r sin t For the circle to be fully circumscribed, the parameter t must range continuously over a span of values sufficient to ensure that a moving point makes at least one full revolution around the x axis. The smallest such span is any half-open interval that's at least 2p units wide. Hold y constant Now suppose that we restrict ourselves to a plane such that y = c, where c is a constant. This plane is parallel to the xz plane, and it intersects the y axis at (0,c,0). Imagine a circle in the plane y = c that's centered on the y axis, as shown in Fig. 18-9. In this case, the circle is described by the system x = r cos t y=c z = r sin t Circles 359 +y Radius of circle =r y=c z x +z +x (0, c, 0) Circle in plane y=c y Figure 18-9 Circle in a plane where y is held to a constant value c. The plane is perpendicular to the y axis, and intersects that axis at the point (0,c,0). The circle has radius r and is centered at (0,c,0). For a complete circle to be described, the parameter t must range continuously over a span of values sufficient to ensure that a moving point makes at least one full revolution around the y axis. The smallest such span is any half-open interval that's at least 2p units wide. Hold z constant Finally, consider a plane in which z = c. It's parallel to the xy plane, and it intersects the z axis at (0,0,c). Imagine a circle in the plane z = c that's centered on the z axis as shown in Fig. 18-10. Here, we have x = r cos t y = r sin t z=c For a complete circle to be described, the parameter t must range continuously over a span of values sufficient to ensure that a moving point makes at least one full revolution around the z axis. The smallest such span is any half-open interval that's at least 2p units wide. 360 Lines and Curves in Three-Space +y z=c Circle in plane z=c (0, 0, c) z x +x +z Radius of circle =r y Figure 18-10 Circle in a plane where z is held to a constant value c. The plane is perpendicular to the z axis, and intersects that axis at the point (0,0,c). The circle has radius r and is centered at (0,0,c). An example Imagine a circle in the plane x = 3. Suppose that the circle is centered on the x axis, and its parametric equations are x=3 y = 3 cos t z = 3 sin t In the plane x = 3, our circle can be described by the two parametric equations y = 3 cos t and z = 3 sin t When we express this system as a relation between y and z, we have y2 + z2 = 9 Circles 361 +y Each axis increment is 1 unit (3, 0, 0) x +x Radius of circle =3 +z Circle in plane x=3 x=3 y Figure 18-11 Graph of a circle of radius 3 in the plane x = 3, centered on (3,0,0). Each axis increment represents 1 unit. Figure 18-11 is a perspective rendition of this circle's graph in Cartesian xyz space. The radius is 3, and the center is at (3,0,0). Another example Now let's look at a circle having a radius of 3 units, and contained in the plane where y maintains a constant value of −2. The parametric equations are x = 3 cos t y = −2 z = 3 sin t In the plane y = −2, our circle can be described by the parametric system x = 3 cos t and z = 3 sin t As a relation between x and z, this system can be represented by x2 + z2 = 9 362 Lines and Curves in Three-Space +y Radius of circle =3 Each axis increment is 1 unit z +x x y = –2 +z (0, –2, 0) Circle in plane y = –2 y Figure 18-12 Graph of a circle of radius 3 in the plane y = −2, centered on (0,−2,0). Each axis increment represents 1 unit. Figure 18-12 is a perspective graph of this circle in Cartesian xyz space. Are you confused? All of the parabolas and circles described in this chapter are confined to planes parallel to the xy plane, the xz plane, or the yz plane. Finding equations for curves in other planes is sometimes easy, but more often it's difficult. The process can be streamlined by adding a function called a coordinate transformation to the relation describing the curve. That way, any curve that lies in a single plane (no matter how the plane is oriented in space, and no matter where the curve is positioned within the plane) can be described in terms of a curve in the xy plane, the xz plane, or the yz plane. You'll learn how to do coordinate transformations in advanced calculus or analysis courses. Here's a challenge! Consider a curve whose parametric equations are x = 2 cos t y = 3 sin t z = −3 What sort of curve is this? Sketch its graph in Cartesian xyz space. Circular Helixes 363 Solution For a moment, suppose that the coefficient in the first equation was 3 rather than 2. In that case, the set of parametric equations would be x = 3 cos t y = 3 sin t z = −3 and we'd have a circle in the plane z = −3. Figure 18-10, on page 360 is an approximate graph of this circle if we imagine each coordinate axis division to represent 1 unit. However, the coefficient in the first equation is 2, not 3. Therefore, the curve is squashed in the x direction; it's only 2/3 as wide as the above described circle. This squashed circle is an ellipse centered on the point (0,0,−3). Figure 18-13 shows how its graph looks in Cartesian xyz space, from a vantage point far from the origin but close to the positive z axis. +y z = –3 Ellipse in plane z = –3 (0, 0, –3) x +x +z Major semi-axis =3 Minor semi-axis =2 y Figure 18-13 Graph of an ellipse in the plane z = −3, centered on (0,0,−3). Each axis increment is 1 unit. Circular Helixes When we created the generalized circles and graphed them as shown in Figs. 18-8 through 18-10, we held one variable constant and forced the other two variables to follow the parametric equations for a circle in a plane. Now imagine that, instead of holding one variable 364 Lines and Curves in Three-Space constant, we let it change according to a constant multiple of the parameter. When we do this, we get a three-dimensional object called a circular helix. Center on x axis Consider a moving plane x = ct in Cartesian xyz-space, where c is a constant and t is the parameter. This plane is always perpendicular to the x axis, so it's always parallel to the yz plane. It intersects the x axis at a moving point (ct,0,0). Imagine a moving a circle of radius r in the moving plane x = ct that's centered on the x axis. On this circle, the value of y tracks along with r cos t, while the value of z tracks along with r sin t. The complete set of parametric equations is x = ct y = r cos t z = r sin t When we graph the path of a point on this moving circle as t varies, we get a circular helix of uniform pitch (that means its "coil turns" are evenly spaced, like those of a well-designed spring). The pitch depends on c. Small values of c produce tightly compressed helixes, while large values of c produce stretched-out helixes. The helix axis corresponds to the coordinate x axis, so the helix is centered on the x axis. Figure 18-14 is a generic graph of a circular helix oriented in this way. +y Helix is centered on the x axis z x +x Radius of helix =r +z y Figure 18-14 Circular helix of radius r, centered on the x axis. The pitch depends on the constant by which t is multiplied to obtain x. Circular Helixes 365 Center on y axis Now imagine a moving plane y = ct that's perpendicular to the y axis, parallel to the xz plane, and intersects the y axis at a moving point (0,ct,0). The value of x tracks along with r cos t, while the value of z tracks along with r sin t, so we have the system x = r cos t y = ct z = r sin t The graph of this set of parametric equations is a circular helix of uniform pitch, centered on the y axis as shown in Fig. 18-15. Center on z axis Finally, envision a moving plane z = ct that's perpendicular to the z axis, parallel to the xy plane, and intersects the z axis at a moving point (0,0,ct). The value of x follows r cos t, while y follows r sin t. Our parametric equations are therefore x = r cos t y = r sin t z = ct Helix is centered on the y axis +y z x +x +z Radius of helix =r y Figure 18-15 Circular helix of radius r, centered on the y axis. The pitch depends on the constant by which t is multiplied to obtain y. 366 Lines and Curves in Three-Space +y Radius of helix =r Helix is centered on the z axis z x +x +z y Figure 18-16 Circular helix of radius r, centered on the z axis. The pitch depends on the constant by which t is multiplied to obtain z. In Cartesian xyz space, these equations produce a circular helix of uniform pitch, centered on the z axis. Figure 18-16 is a generic graph. An example Consider a circular helix centered on the x axis, described by the parametric equations x = t /(2p ) y = cos t z = sin t Here are some values of x, y, and z that we can calculate as t varies, causing a point on the helix to complete a single revolution in a plane perpendicular to the x axis: • • • • • When t = 0, we have x = 0, y = 1, and z = 0. When t = p /2, we have x = 1/4, y = 0, and z = 1. When t = p, we have x = 1/2, y = −1, and z = 0. When t = 3p /2, we have x = 3/4, y = 0, and z = −1. When t = 2p, we have x = 1, y = 1, and z = 0. Circular Helixes 367 Every time t increases by 2p, our point makes one complete revolution in a moving plane that's always perpendicular to the x axis. Also, every time t increases by 2p, our point gets 1 unit farther away from the yz plane. The pitch of the helix is therefore equal to 1 linear unit per revolution. Another example Consider a circular helix centered on the y axis, described by the parametric equations x = 2 cos t y=t z = 2 sin t Here are some values of x, y, and z that we can calculate as t varies, causing a point on the helix to complete a single revolution in a plane perpendicular to the y axis: • • • • • When t = 0, we have x = 2, y = 0, and z = 0. When t = p /2, we have x = 0, y = p /2, and z = 2. When t = p, we have x = −2, y = p, and z = 0. When t = 3p /2, we have x = 0, y = 3p /2, and z = −2. When t = 2p, we have x = 2, y = 2p, and z = 0. Every time t increases by 2p, our point makes a complete revolution in a moving plane that's always perpendicular to the y axis. Also, every time t increases by 2p, our point moves 2p units farther away from the xz plane. The pitch of the helix is therefore equal to 2p linear units per revolution. Are you confused? You might ask, "When describing a helix with parametric equations, does it make any difference if we multiply t by a positive constant or a negative constant?" That's an excellent question. The answer is yes; it matters a lot! The polarity of the constant affects the sense in which the helix rotates as we move in the positive direction. For example, suppose we have a helix described by the parametric equations x = 3t y = 3 cos t z = 3 sin t In this case, the helix turns counterclockwise as we move in the positive x direction. If we observe the situation from somewhere on the positive x axis while the value of t increases, a point on the 368 Lines and Curves in Three-Space helix will appear to approach us and rotate counterclockwise. If the value of t decreases, a point on the helix will appear to retreat from us and rotate clockwise. Now suppose that we reverse the sign of the constant in the first equation, so our system becomes x = −3t y = 3 cos t z = 3 sin t If we watch this scene from somewhere on the positive x axis while the value of t increases, a point on the helix will appear to retreat from us and rotate counterclockwise. If the value of t decreases, a point on the helix will appear to approach us and rotate clockwise. Are you astute? Imagine yourself at some point far from the origin on the +x axis in Fig. 18-14, or far from the origin on the +y axis in Fig. 18-15, or far from the origin on the +z axis in Fig. 18-16. If you have excellent spatial perception, you'll be able to figure out that in all three of these situations, the constant c is negative! In each case, a retreating point on the helix will appear to revolve counterclockwise, and an approaching point on the helix will appear to revolve clockwise. Here's a challenge! Suppose that we encounter an object in Cartesian xyz space whose parametric equations are x = 2 cos t y = 3 sin t z = −3t What sort of object is this? Solution Let's divide the first two equations through by their respective constants. That gives us x /2 = cos t and y /3 = sin t Squaring both sides of both equations, we obtain (x /2)2 = cos2 t Circular Helixes 369 and (y /3)2 = sin2 t When we add these two equations, left-to-left and right-to-right, we get (x /2)2 + (y /3)2 = cos2 t + sin2 t The rules of trigonometry tell us that cos2 t + sin2 t = 1 so the preceding equation can be rewritten as (x /2)2 + (y /3)2 = 1 and further morphed into x2/4 + y2/9 = 1 This equation describes an ellipse in the xy plane whose horizontal (x-coordinate) semi-axis measures 2 units, and whose vertical (y-coordinate) semi-axis measures 3 units. Now let's consider the z coordinate. The equation for z in terms of t is z = −3t This equation tells us that a point on our object travels in the negative z direction as the value of the parameter t increases. The complete set of three parametric equations therefore describes an elliptical helix centered on the z axis. As we move in the positive z direction, the helix rotates clockwise, because the coefficient of t is negative. It looks something like the helix in Fig. 18-16, except that it's stretched by approximately 50 percent in the positive and negative y directions (vertically in this particular illustration). Here's an extra-credit challenge! Sketch three-dimensional perspective graphs of the helixes described in the foregoing two examples and challenge. Solution You're on your own. That's why you get extra credit! 370 Lines and Curves in Three-Space Practice Exercises This is an open-book quiz. You may (and should) refer to the text as you solve these problems. Don't hurry! You'll find worked-out answers in App. B. The solutions in the appendix may not represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it! 1. Consider the following three-way equation for a straight line in Cartesian xyz space: x−1=y−2=z−4 Find a point on the line, find the preferred direction numbers, and determine the direction vector as a sum of multiples of i, j, and k. 2. Consider the following three-way equation for a straight line in Cartesian xyz space: 4x = 5y = 6z Find a point on the line, find the preferred direction numbers, and determine the direction vector as a sum of multiples of i, j, and k. 3. Consider the following three-way equation for a straight line in Cartesian xyz space: (x − 2)/3 = (4y − 8)/4 = (z + 5)/(−2) Find a point on the line, find the preferred direction numbers, and determine the direction vector as a sum of multiples of i, j, and k. 4. Consider a relation in Cartesian xyz space described by the system of parametric equations x = −4 y=t z = −t2 − 1 Draw a two-dimensional graph of this relation as it appears when we look broadside at the plane containing it. 5. Consider a relation in Cartesian xyz space described by the system of parametric equations x = t2 + 2t y=t z=0 Practice Exercises 371 Draw a two-dimensional graph of this relation as it appears when we look broadside at the plane containing it. 6. Consider a relation in Cartesian xyz space described by the system of parametric equations x=t y = −7 z = t2/2 − 5 Draw a two-dimensional graph of this relation as it appears when we look broadside at the plane containing it. 7. Consider a relation in Cartesian xyz space described by the system of parametric equations x = 4 cos t y = 4 sin t z=1 Draw a two-dimensional graph of this relation as it appears when we look broadside at the plane containing it. 8. Consider a relation in Cartesian xyz space described by the system of parametric equations x = 5 cos t y=0 z = 5 sin t Draw a two-dimensional graph of this relation as it appears when we look broadside at the plane containing it. 9. Consider a relation in Cartesian xyz space described by the system of parametric equations x = 5 cos t y = 3 sin t z=p Draw a two-dimensional graph of this relation as it appears when we look broadside at the plane containing it. 372 Lines and Curves in Three-Space 10. Consider a relation in Cartesian xyz space described by the system of parametric equations x = 2 cos t y = t /(2p ) z = 2 sin t Draw a perspective view of this relation's three-dimensional graph. Here's a hint: You can probably tell that the graph is a circular helix, but as you draw it, pay attention to the orientation, the pitch, and the sense of rotation. CHAPTER 19 Sequences, Series, and Limits Have you ever tried to find the missing number in a list? Have you ever figured out how much money an interest-bearing bank account will hold after 10 years? Have you ever calculated the value that a function approaches but never reaches? If you can answer "Yes" to any of these questions, you've worked with sequences (also called progressions), series, or limits. Repeated Addition A sequence is a list of numbers. Some sequences are finite; others are infinite. The simplest sequences have values that repeatedly increase or decrease by a fixed amount. Here are some examples: A = 1, 2, 3, 4, 5, 6 B = 0, −1, −2, −3, −4, −5 C = 2, 4, 6, 8 D = −5, −10, −15, −20 E = 4, 8, 12, 16, 20, 24, 28, ... F = 2, 0, −2, −4, −6, −8, −10, ... The first four sequences are finite. The last two are infinite, as indicated by an ellipsis (three dots) at the end. Arithmetic sequence In each of the sequences shown above, the values either increase steadily (in A, C, and E ) or decrease steadily (in B, D, and F ). In all six sequences, the spacing between numbers is constant throughout. Here's how each sequence changes as we move along from term to term: • The values in A always increase by 1. • The values in B always decrease by 1. • The values in C always increase by 2. 373 374 Sequences, Series, and Limits • The values in D always decrease by 5. • The values in E always increase by 4. • The values in F always decrease by 2. Each sequence has an initial value. After that, we can easily predict subsequent values by repeatedly adding a constant. If the constant is positive, the sequence increases. If the added constant is negative, the sequence decreases. Suppose that s0 is the first number in a sequence S. Let c be a real-number constant. If S can be written in the form S = s0, (s0 + c), (s0 + 2c), (s0 + 3c), ... then it's an arithmetic sequence or an arithmetic progression. In this context, the word "arithmetic" is pronounced "err-ith-MET-ick." The numbers s 0 and c can be integers, but that's not a requirement. They can be fractions such as 2/3 or −7/5. They can be irrational numbers such as the square root of 2. As long as the separation between any two adjacent terms is the same wherever we look, we have an arithmetic sequence, even in the trivial case S0 = 0, 0, 0, 0, 0, 0, 0, ... Arithmetic series A series is the sum of all the terms in a sequence. For an arithmetic sequence, the corresponding arithmetic series can be defined only if the sequence has a finite number of terms. For the above sequences A through F, let the corresponding series be called A+ through F+. The total sums are as follows. A+ = 1 + 2 + 3 + 4 + 5 + 6 = 21 B+ = 0 + (−1) + (−2) + (−3) + (−4) + (−5) = −15 C+ = 2 + 4 + 6 + 8 = 20 D+ = (−5) + (−10) + (−15) + (−20) = −50 E+ is not defined F+ is not defined Now consider the infinite series S0+ = 0 + 0 + 0 + 0 + 0 + 0 + 0 + ··· We might think of S0+ as "infinity times 0," because it's the sum of 0 added to itself infinitely many times. It's tempting to suppose that S0+ = 0, but we can't prove it. When we add up any finite number of "nothings", we get "nothing", of course. However, when we try to find the sum of infinitely many nothings, we encounter a mystery. The best we can do is say that S0+ is undefined. Repeated Addition 375 Graphing an arithmetic sequence When we plot the values of an arithmetic sequence as a function of the term number in rectangular coordinates, we get a set of discrete points. We can depict the term number along the horizontal axis going toward the right, so the term number plays the role of the independent variable. We can plot the term value along the vertical axis, so it plays the role of the dependent variable. Figure 19-1 illustrates two arithmetic sequences as they appear when graphed in this way. (The dashed lines connect the dots, but they aren't actually parts of the sequences.) One sequence is increasing, and the dashed line connecting this set of points ramps upward as we go toward the right. Because this sequence is finite, the dashed line ends at (6,6). The other sequence is decreasing, and its dashed line ramps downward as we go toward the right. This sequence is infinite, as shown by the ellipsis at the end of the string of numbers, and also by the arrow at the right-hand end of the dashed line. When any arithmetic sequence is graphed according to the scheme shown in Fig. 19-1, its points lie along a straight line. The slope m of the line depends on whether the sequence increases ( positive slope) or decreases (negative slope). In fact, m is equal to the constant c in the general arithmetic series form: S = s0, (s0 + c), (s0 + 2c), (s0 + 3c), ... regardless of how many terms the sequence contains. 6 4 1, 2, 3, 4, 5, 6 Term value 2 Term number 0 1 2 3 4 5 6 –2 6, 3, 0, –3, –6, ... –4 –6 Figure 19-1 Rectangular-coordinate plots of two arithmetic sequences. Repeated Multiplication 379 If the constant is positive, the values either remain positive or remain negative. If the constant is negative, the values alternate between positive and negative. Let t0 be the first number in a sequence T, and let k be a constant. Imagine that T can be written in the general form: T = t 0, t 0k, t 0k 2, t0k 3, t0k 4, ... for as long as the sequence goes. Such a sequence is called a geometric sequence or a geometric progression. If k = 1, the sequence consists of the same number over and over. (In that case, it's also an arithmetic sequence with a constant equal to 0!) If k = −1, the sequence alternates between t0 and its negative. If t 0 is less than −1 or greater than 1, the values get farther from 0 as we move along in the series. If t 0 is between (but not including) −1 and 1, the values get closer to 0. If t0 = 1 or t0 = −1, the values stay the same distance from 0. The numbers t0 and k can be whole numbers, but this is not a requirement. As long as the multiplication factor between any two adjacent terms in a sequence is the same, the sequence is a geometric sequence. In the sequence L above, we have the constant k = 1/2. This is an especially interesting case, as we'll see in a moment. Geometric series In a geometric sequence, the corresponding geometric series, which is the sum of all the terms, can always be defined if the sequence is finite, and can sometimes be defined if the sequence is infinite. For the above sequences G through L, let the corresponding series be called G+ through L+. Then we have G+ = 1 + 2 + 4 + 8 + 16 + 32 = 63 H+ = 1 − 1 + 1 − 1 + 1 − 1 + ··· = ? I+ = 1 + 10 + 100 + 1000 = 1111 J+ = −5 − 15 − 45 − 135 − 405 = −605 K+ is not defined L+ = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ··· = ? The finite series G+, I+, and J+ are straightforward. There's no mystery there! The partial sums of H+ alternate between 0 and 1, but can't settle on either of those values. It's tempting to say that H+ has two values, just as certain equations have solution sets containing two roots. But we're looking for a single, identifiable number, not the solution set of an equation. On that basis, we're forced to conclude that H+ is not definable. The infinite series K+ goes "out of control." It's an example of a divergent series; its values keep getting farther from 0 without ever reaching a limit. Convergence For the above sequences H, K, and L, the sequences of partial sums, which we'll denote using asterisk subscripts, go as follows: Sequences, Series, and Limits H* = 1, 0, 1, 0, 1, 0, ... K* = 3, 12, 39, 120, 363, 1092, 3279, ... L* = 1/2, 3/4, 7/8, 15/16, 31/32, ... The partial sums denoted by H* and K* don't settle down on anything. But the partial sums denoted by L* seem to approach 1. They don't "run away" into uncharted territory, and they don't alternate between or among multiple numbers. The partial sums in L* seem to have a clear destination that they could reach, if only they had an infinite amount of time to get there. It turns out that the complete series L+, representing the sum of the infinite string of numbers in the sequence L, is exactly equal to 1! We can get an intuitive view of this fact by observing that the partial sums approach 1. As the position in the sequence of partial sums, L, gets farther and farther along, the denominators keep doubling, and the numerator is always 1 less than the denominator. In fact, if we want to find the nth number Ln in the sequence of partial sums L, we can calculate it by using the following formula: Ln = (2n − 1)/2n As n becomes large, 2n becomes large much faster, and the proportional difference between 2 − 1 and 2n becomes smaller. When n reaches extremely large positive integer values, the quotient (2n − 1)/2n is almost exactly equal to 1. We can make the quotient as close to 1 as we want by going out far enough in the series of partial sums, but we can never make it equal to or larger than 1. The sequence L* is said to converge on the number 1. The sequence of partial sums L* is an example of a convergent sequence. The series L+ is an example of a convergent series. n Plotting a geometric sequence A geometric sequence, like an arithmetic sequence, appears as a set of points when plotted on a Cartesian plane. Figure 19-2 shows examples of two geometric sequences as they appear 40 40, 20, 10, 5, 2.5, 1.25, ... 30 Term value 380 1, 2, 4, 8, 16, 32 20 10 0 1 Figure 19-2 2 3 4 Term number 5 6 Rectangular-coordinate plots of two geometric sequences. Repeated Multiplication 381 when graphed. Note that the dashed curves, which show the general trends of the sequences (but aren't actually parts of the sequences), aren't straight lines, but they are "smooth." They don't turn corners or make sudden leaps. One of the sequences in Fig. 19-2 is increasing, and the dashed curve connecting this set of points goes upward as we move to the right. Because this sequence is finite, the dashed curve ends at the point (6,32), where the term number is 6 and the term value is 32. The other sequence is decreasing, and the dashed curve goes downward and approaches 0 as we move to the right. This sequence is infinite, as shown by the three dots at the end of the string of numbers, and also by the arrow at the right-hand end of the dashed curve. If a geometric sequence has a negative factor, that is, if k < 0, the plot of the points alternates back and forth on either side of 0. The points fall along two different curves, one above the horizontal axis and the other below. If you want to see what happens in a case like this, try plotting an example. Set t 0 = 64 and k = −1/2, and plot the resulting points. An example Suppose you get a 5-year certificate of deposit (CD) at your local bank for $1000.00, and it earns interest at the annualized rate of exactly 5 percent per year. The CD will be worth $1276.28 after 6 years. To calculate this, multiply $1000 by 1.05, then multiply this result by 1.05, and repeat this process a total of 5 times. The resulting numbers form a geometric sequence: • • • • • After 1 year: $1000.00 × 1.05 = $1050.00 After 2 years: $1050.00 × 1.05 = $1102.50 After 3 years: $1102.50 × 1.05 = $1157.63 After 4 years: $1157.63 × 1.05 = $1215.51 After 5 years: $1215.51 × 1.05 = $1276.28 Another example Is the following sequence a geometric sequence? If so, what are the values t0 (the starting value) and k (the factor of change)? T = 3, −6, 12, −24, 48, −96, ... This is a geometric sequence. The numbers change by a factor of −2. In this case, t 0 = 3 and k = −2. Are you confused? It's reasonable to ask, "Can we categorize all sequences as either arithmetic or geometric?" The answer is no! Consider U = 10, 13, 17, 22, 28, 35, 43, ... This sequence shows a pattern, but it's neither arithmetic nor geometric. The difference between the first and second terms is 3, the difference between the second and third terms is 4, the difference 382 Sequences, Series, and Limits between the third and fourth terms is 5, and so on. The difference keeps increasing by 1 for each succeeding pair of terms. This is a fairly simple example of a nonarithmetic, nongeometric sequence with an identifiable pattern. Here's a challenge! Suppose a particular species of cell undergoes mitosis (splits in two) every half hour, precisely on the half hour. We take our first look at a cell culture at 12:59 p.m., and find three cells. At 1:00 p.m., mitosis occurs for all the cells at the same time, and then there are six cells in the culture. At 1:30 p.m., mitosis occurs again, and we have 12 cells. How many cells are there in the culture at 4:01 p.m.? Solution There are 3 hours and 2 minutes between 12:59 p.m. and 4:01 p.m. This means that mitosis takes place 7 times: at 1:00, 1:30, 2:00, 2:30, 3:00, 3:30, and 4:00. Table 19-1 illustrates the scenario. We look at the culture repeatedly at 1 minute past each half hour. There are 384 cells at 4:01 p.m., just after the mitosis event that occurs at 4:00 p.m. Table 19-1 Cell division as a function of time, assuming mitosis occurs every half hour Time 12:59 1:01 1:31 2:01 2:31 3:01 3:31 4:01 Number of cells 3 6 12 24 48 96 192 384 Limit of a Sequence A limit is a specific, well-defined quantity that a sequence, series, relation, or function approaches. The value of the sequence, series, relation, or function can get arbitrarily close to the limit, but doesn't always reach it. An example Let's look at an infinite sequence A that starts with 1 and then keeps getting smaller. For any positive integer n, the nth term is 1/n, so we have A = 1, 1/2, 1/3, 1/4, 1/5, ..., 1/n, ... Limit of a Sequence 383 This is a simple example of a special type of sequence called a harmonic sequence. In this particular case, the values of the terms approach 0. The hundredth term is 1/100; the thousandth term is 1/1000; the millionth term is 1/1,000,000. If we choose a tiny but positive real number, we can always find a term in the sequence that's closer to 0 than that number. But no matter how much time we spend generating terms, we'll never get 0. We say that "The limit of 1/n, as n approaches infinity, is 0," and write it as Lim 1/n = 0 n→∞ Another example Consider the sequence B in which the numerators ascend one by one through the set of natural numbers, while every denominator is equal to the corresponding numerator plus 1. For any positive integer n, the nth term is (n − 1)/n, so we have B = 0/1, 1/2, 2/3, 3/4, 4/5, ..., (n − 1)/n, ... As n becomes extremely large, the numerator (n − 1) gets closer and closer to the denominator, when we consider the difference in proportion to the value of n. Therefore Lim (n − 1)/n = n/n = 1 n→∞ Still another example Let's see what happens in a sequence C where every numerator is equal to the square of the term number, while every denominator is equal to twice the term number. For any positive integer n, the nth term is n 2/(2n), so we have C = 1/2, 4/4, 9/6, 16/8, 25/10, 36/12, 49/14, ..., n 2 / (2n), ... Note that n 2/(2n) = n /2 This tells us that Lim n 2/(2n) = Lim n /2 n→∞ n→∞ As n grows larger without end, so does n /2. Therefore Lim n /2 n→∞ is undefined, so we know that Lim n 2/(2n) n→∞ is also undefined. Alternatively, we can say that this limit doesn't exist, or that it's meaningless. 384 Sequences, Series, and Limits Are you confused? By now, you should suspect that any given sequence must fall into one or the other of two categories: convergent (meaning that it has a limit) or divergent (meaning that it doesn't have a limit). But what if a sequence alternates between two numbers endlessly? Once again, look at the sequence H = 1, −1, 1, −1, 1, −1, ... We might be tempted to suggest that a sequence of this type "has two different limits," but it doesn't converge on any single number. However, that won't work because a limit must always be a single value that we can specify as a number. In cases like this, it's customary to say that the limit is not defined. Here's a challenge! Consider the sequence D in which the numerators alternate between −1 and 1, while the denominators start at 1 and increase by 1 with each succeeding term. For any positive integer n, the nth term is (−1)n/n, so that D = −1/1, 1/2, −1/3, 1/4, −1/5, ..., (−1)n/n, ... Does this sequence have a limit? If so, what is it? If not, why not? Solution As n becomes extremely large, the absolute value of the numerator is always 1, although the sign alternates. The denominator increases steadily, and without end. If we choose a tiny positive or negative real number, we can always find a term that's closer to 0 than that number, but we'll never actually reach 0 from either the positive side or the negative side. Therefore Lim (−1)n/n = 0 n→∞ Here's another challenge! Consider the following sequence: K = (−1 − 1/1), (1 + 1/2), (−1 − 1/3), (1 + 1/4), ..., [(−1)n + (−1)n/n], ... The parentheses and brackets are not technically necessary here, but they visually isolate the terms from one another. Does K have a limit? If so, what is it? If not, why not? Solution Each term in K is expressed as a sum. The first addend alternates between −1 and 1, endlessly. The second addend is identical to the corresponding term in the sequence D that we evaluated in the previous challenge. We determined that D converges toward 0. The terms in K therefore approach Summation "Shorthand" 385 two different values, −1 and 1, as we generate terms indefinitely. If we want to claim that a sequence has a limit, we must take that expression literally. "A limit" means "one and only one limit." We therefore conclude that Lim (−1)n + (−1)n/n n→∞ is not defined. Summation "Shorthand" Mathematicians have a "shorthand" way to denote long sums. This technique can save a lot of space and writing time. We can even write down an infinite sum in a compact statement. It's called summation notation. Specify the series Imagine a set of constants, all denoted by a with a subscript, such as {a1, a2, a3, a4, a5, a6, a7, a8} Suppose that we add up the elements of this set, and call the sum b. We can write this sum out term by term as a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 = b That's easy because we have only eight terms, but if the set contained 800 elements, writing down the entire sum would be exasperating. We could put an ellipsis in the middle of the sum, calling it c and then writing a1 + a2 + a3 + ··· + a798 + a799 + a800 = c If the series had infinitely many terms, we could use an ellipsis after the first few terms and leave the statement wide open after that, calling it d and then writing a1 + a2 + a3 + a4 + a5 + ··· = d Tag the terms Let's invent a nonnegative-integer variable and call it i. Written as a subscript, i can serve as a counting tag in a series containing a large number of terms. Don't confuse this i with the symbol some texts use to represent the unit imaginary number, which is the positive square root of −1! In the above-described series, we can call each term by the generic name ai. In the first series, we add up eight ai's to get the final sum b, and the counting tag i goes from 1 to 8. In 386 Sequences, Series, and Limits the second series, we add up 800 ai's to get the final sum c, and the counting tag i goes from 1 to 800. In the third series, we add up infinitely many ai's to get the final sum d, and the counting tag i ascends through the entire set of positive integers. Suppose that we have a series with n terms, as follows: a1 + a2 + a3 + ··· + an−2 + an−1 + an = k In this case, we add up n ai's to get the final sum k. The big sigma Let's go back to the series with eight terms. We can write it down in a cryptic but informationdense manner as 8 ∑ ai = b i =1 We read this expression out loud as, "The summation of the terms ai, from i = 1 to 8, is equal to b." The large symbol Σ is the uppercase Greek letter sigma, which stands for summation or sum. Now let's look at the series in which 800 terms are added: 800 ∑ ai = c i =1 We can read this aloud as, "The summation of the terms ai, from i = 1 to 800, is equal to c." In the third example containing infinitely many terms, we can write ∞ ∑ ai = d i =1 This statement can be read as, "The summation of the terms ai, from i = 1 to infinity, is equal to d." Finally, in the general case, we can write n ∑ ai = k i =1 and read it aloud as, "The summation of the terms ai, from i = 1 to n, is equal to k." A more sophisticated example Suppose we want to determine the value of an infinite series starting with 1, then adding 1/2, then adding 1/4, then adding 1/8, and going on forever, each time cutting the value in half. As things work out, we get 1 + 1/2 + 1/4 + 1/8 + ··· = 2 even though the series has infinitely many terms. We can also write 1/20 + 1/21 + 1/22 + 1/23 + ··· = 2 In summation notation, we write ∞ ∑ 1/2i = 2 i =0 Summation "Shorthand" 387 Are you confused? If you're baffled by the idea that we can add up infinitely many numbers and get a finite sum, you can use the "frog-and-wall" analogy. Imagine that a frog sits 8 meters (8 m) away from a wall. Then she jumps halfway to the wall, so she's 4 m away from it. Now imagine that she continues to make repeated jumps toward the wall, each time getting halfway there (Fig. 19-3). No finite number of jumps will allow the frog to reach the wall. To accomplish that goal, she would have to take infinitely many jumps. This scenario can be based on a sequence of partial sums of a series S = 4 + 2 + 1 + 1/2 + 1/4 + 1/8 + ··· A real-world frog cannot reach the wall by jumping halfway to it, over and over. But in the imagination, she can. There are two ways this can happen. First, in the universe of mathematics, we have an infinite amount of time, so an infinite number of jumps can take place. Another way around the problem is to keep halving the length of time in between jumps, say from 4 seconds to 2 seconds, then to 1 second, then to 1/2 second, and so on. This will make it possible for our "cosmic superfrog" to hop an infinite number of times in a finite span of time. Either way, when she has finished her journey and her nose touches the wall, she'll have traveled exactly 8 m. Therefore, the sum total of the lengths of her jumps is S = 4 + 2 + 1 + 1/2 + 1/4 + 1/8 + ··· = 8 Here's a challenge! Consider the series that we dealt with in "A more sophisticated example" a couple of paragraphs ago, but only up to the reciprocal of the nth power of 2. Let Sn be the partial sum of this series up to, and including, that term. Write Sn in summation notation. Wall 1st jump 4th 2nd jump 3rd 4m 2m 1m Initial position of frog (point on surface exactly below her nose) Figure 19-3 1/2 m A frog jumps toward a wall, getting halfway there with each jump. 388 Sequences, Series, and Limits Solution Let's use the letter i as the counting tag. We start at i = 0 and go up to i = n, with each term having the value 1/2i. Therefore, the summation notation is n ∑ 1/2i i =0 Limit of a Series If a series has a limit, we can sometimes figure it out by creating a sequence from the partial sums, and then finding the limit of that sequence. An example Think of the summation in the previous challenge, and imagine what happens as n increases endlessly—that is, as n approaches infinity. As n grows larger, the sequence of partial sums approaches 2. We can plug the summation into a limit template, and then state that Lim n→∞ n ∑ 1/2i = 2 i =0 Another example Let's look once again at the infinite sequence V we saw a little while ago, where the numerators keep alternating between −1 and 1, as follows: V = −1/2, 1/2, −1/2, 1/2, −1/2, 1/2, −1/2 ... Let's replace every comma by a plus sign, creating the infinite series V+ = −1/2 + 1/2 − 1/2 + 1/2 − 1/2 + 1/2 − 1/2 + ··· We can write this series in summation form as ∞ ∑ (−1)i/2 i =1 Now consider the limit of the sequence of partial sums of V+ as the number of terms becomes arbitrarily large. We write this quantity symbolically as Lim n→∞ n ∑ (−1) /2 i i =1 This limit does not exist, because the sequence of partial sums alternates endlessly between two values, −1/2 and 0. Limit of a Series 389 Are you confused? Does the combination of limit and summation notation look intimidating? Besides getting used to the symbology, you have to keep track of two different indexes, i for the sum and n for the limit. It helps if you remember that the two indexes are independent of each other. You're finding the limit of a sum as you keep making that sum longer. Here's a challenge! Find the limit of the partial sums of the infinite series 1/100 + 1/1002 + 1/1003 + 1/1004 + 1/1005 + ··· as the number of terms in the partial sum increases without end. That is, find n Lim n→∞ ∑ 1/100i i =1 Solution In decimal form, 1/100 = 0.01, 1/1002 = 0.0001, 1/1003 = 0.000001, and so on. Let's arrange these numbers in a column with each term underneath its predecessor, and all the decimal points along a vertical line, as the following: 0.01 0.0001 0.000001 0.00000001 0.0000000001 ↓ ⎯⎯⎯⎯⎯⎯⎯ 0.0101010101... When we look at the series this way, we can see that it must ultimately add up to the nonterminating, repeating decimal 0.0101010101.... From our algebra or number theory courses, we recall that this endless decimal number is equal to 1/99. That's the limit of the sequence of partial sums in the series: n ∑ 1/100i i =1 as the positive integer n increases without end. It's also the value of the entire infinite series: ∞ ∑ 1/100i i=1 390 Sequences, Series, and Limits Limits of Functions So far, we've looked at situations where we move from term to term in a sequence or series. Sometimes, such sequences and series have limits (they converge); in other cases they don't have limits (they diverge). Similar phenomena can occur when we have a variable that changes in a smooth, continuous manner, rather than jumping among discrete values. Some functions have limits, and some don't Certain functions increase or decrease without bound, while others reach specific values and stay there. Still others increase or decrease continuously without ever passing, or even reaching, a certain value. It's also possible for a function to "blow up" and have no limit at all. The solid curve in Fig. 19-4 shows the reciprocal function in the first quadrant of the Cartesian plane, where the value of the independent variable is positive. The dashed curve shows the negative reciprocal function in the fourth quadrant, where, again, the value of the independent variable is positive. The functions are f (x) = x −1 Value of function Each axis division is 1 unit f (x) = x –1 for x > 0 x What are the limits? Figure 19-4 –f (x) = –x –1 for x > 0 Graphs of the reciprocal function (solid curve) and its negative (dashed curve) in the first and fourth quadrants of the Cartesian plane, where x > 0. Each axis division represents 1 unit. Limits of Functions 391 and −f (x) = −x −1 As the value of x increases without end, both of these functions approach, but never reach, 0. We can therefore write Lim f (x) = 0 x →∞ and Lim −f (x) = 0 x →∞ As the value of x becomes arbitrarily small but remains positive, both of these functions approach singularity. The reciprocal function "blows up positively" while the negative reciprocal function "blows up negatively." Therefore, we must conclude that neither Lim f (x) x →0 nor Lim −f (x) x →0 exists. It's tempting to claim that Lim f (x) = +∞ x →0 and Lim −f (x) = −∞ x →0 However, we haven't explicitly defined +∞ ("positive infinity") or −∞ ("negative infinity"), so such statements are informal at best. Right-hand limit at a point Consider again the reciprocal function f (x ) = x −1 To specify that we approach 0 from the positive direction, we can refine the limit notation by placing a plus sign after the 0, as follows: Lim f (x ) x →0+ This expression reads, "The limit of f (x ) as x approaches 0 from the positive direction." We can also say, "The limit of f (x ) as x approaches 0 from the right." (In most graphs where x is on the horizontal axis, the value of x becomes more positive as we move toward the right.) 392 Sequences, Series, and Limits This sort of limit is called a right-hand limit. Because f is singular where x = 0, this particular limit is not defined. Left-hand limit at a point Let's expand the domain of f to the entire set of reals except 0, for which f is not defined because 0−1 is meaningless. Suppose that we start out with negative real values of x and approach 0 from the left. As we do this, f decreases endlessly. Another way of saying this is that f increases negatively without limit, or that it "blows up negatively." Therefore, Lim f (x ) x →0− is not defined. We read the above symbolic expression as, "The limit of f (x) as x approaches 0 from the negative direction." We can also say, "The limit of f (x) as x approaches 0 from the left." This sort of limit is called a left-hand limit. An example Let's consider a function g that takes the reciprocal of twice the independent variable. If the independent variable is x, then we have g (x ) = (2x ) −1 Imagine that we allow x to be any positive real number. As x gets arbitrarily large positively, g (x) gets arbitrarily small positively, approaching 0 but never quite getting there. We can say, "The limit of g (x ), as x approaches infinity, is 0," and write Lim g (x) = 0 x →∞ This scenario is similar to what happens with the reciprocal function, except that this function g approaches 0 at a different rate than the reciprocal function as the independent variable becomes arbitrarily large. Now let's see what happens when x gets smaller but stays positive, so that g (x) gets larger. If we make x close enough to 0, we can make g (x) as large as we want. This function, like the reciprocal function, "blows up" as x approaches 0 from the positive direction, but at a different rate. Therefore Lim g (x ) x →0+ is not defined. Another example Suppose that x is a positive real-number variable, and we want to evaluate Lim 1/x 2 x →∞ Let's start out with x at some positive real number for which the function is defined. As we increase the value of x, the value of 1/x 2 decreases, but it always remains positive. If we choose some tiny positive real number r, no matter how close to 0 it might be, we can always find Limits of Functions 393 some large value of x for which 1/x 2 is smaller than r. Therefore, as x grows without bound, 1/x 2 approaches 0, telling us that Lim 1/x 2 = 0 x →∞ Still another example Again, let x be a positive real-number variable. This time, let's evaluate Lim 1/x 2 x →0+ Suppose that we start out with x at some positive real number for which the function is defined and then decrease x, letting it get arbitrarily close to 0 but always remaining positive. As we decrease the value of x, the value of 1/x 2 remains positive and increases. If we choose some large positive real number s, no matter how gigantic, we can always find some small, positive value of x for which 1/x 2 is larger than s. As x becomes arbitrarily small positively, 1/x 2 grows without bound, so Lim 1/x 2 x →0+ is not defined. Are you confused? It's easy to get mixed up by the meanings of negative direction and positive direction, and how these relate to the notions of left hand and right hand. These terms are based on the assumption that we're talking about the horizontal axis in a graph, and that this axis represents the independent variable. In most graphs of this type, the value of the independent variable gets more negative as we move to the left, and it gets more positive as we move to the right. As we travel along the horizontal axis, we might be in positive territory the whole time; we might be in negative territory the whole time; we might cross over from the negative side to the positive side or vice versa. Whenever we come toward a point from the left, we approach from the negative direction, even if that point corresponds to something like x = 567. Whenever we come toward a point from the right, we approach from the positive direction, even if the point is at x = −53,535. The location of the point doesn't matter. The important consideration is the direction from which we approach the point. Here's a challenge! Consider the base-10 logarithm function (symbolized log10). Sketch a graph of the function f (x) = log10 x for values of x from 0.1 to 10, and for values of f from −1 to 1. Then determine Lim log10 x x→5− Solution Figure 19-5 is a graph of the function f (x) = log10 x for values of x from 0.1 to 10, and for values of f from −1 to 1. The function varies smoothly throughout this span. If we start at values of x a little smaller than 5 and work our way toward 5, the value of f approaches log10 5. Therefore, Lim log10 x = log10 5 x→5− 394 Sequences, Series, and Limits We "close in" on this point ... f (x) 1 ... from the negative direction x 0 10 5 Common logarithm function f (x) = log 10 x –1 Figure 19-5 An example of the limit of a function as we approach a point from the negative direction. Memorable Limits of Series Certain limits of series are found often in calculus and analysis. If you plan to go on to Calculus Know-It-All after finishing this book, you're certain to see the three examples that follow! An example Imagine an infinite series where we take a positive integer i and then divide it by the square of another positive integer n. Symbolically, we write this as ∞ ∑ i /n 2 i =1 When we expand this series out, we write it as 1/n 2 + 2/n 2 + 3/n 2 + ··· + n/n 2 + ··· which simplifies to (1 + 2 + 3 + ··· + n + ···)/n 2 Memorable Limits of Series 395 Suppose that we let n grow endlessly larger, increasing the number of terms in the series. Let's consider Lim (1 + 2 + 3 + ··· + n)/n 2 n→∞ As things work out, this limit is equal to 1/2. Therefore n Lim n→∞ ∑ i /n 2 = 1/2 i =1 Another example Now imagine an infinite series where we square a positive integer i and then divide it by the cube of another positive integer n. Symbolically, we write this as ∞ ∑ i 2/n 3 i =1 We can expand it to 12/n 3 + 22/n 3 + 32/n 3 + ··· + n 2/n 3 + ··· which simplifies to (12 + 22 + 32 + ··· + n 2 + ···)/n 3 As n grows endlessly larger, we have Lim (12 + 22 + 32 + ··· + n 2)/n 3 n→∞ This limit turns out to be 1/3. Therefore n Lim n→∞ ∑ i 2/n 3 = 1/3 i =1 Still another example Finally, let's look at an infinite series where we cube a positive integer i and then divide it by the fourth power of another positive integer n. Symbolically, we write this as ∞ ∑ i 3/n 4 i =1 When we write this series out, we obtain 13/n 4 + 23/n 4 + 33/n 4 + ··· + n 3/n 4 + ··· which simplifies to (13 + 23 + 33 + ··· + n 3 + ···)/n 4 396 Sequences, Series, and Limits As n grows endlessly larger, we have Lim (13 + 23 + 33 + ··· + n 3)/n 4 n→∞ This limit turns out to be 1/4. Therefore n Lim n→∞ ∑ i 3/n 4 = 1/4 i =1 Figure 19-6 is a graph of the first few elements of an infinite arithmetic sequence. If we call the sequence S, then S = s 0, (s 0 + c ), (s 0 + 2c ), (s 0 + 3c ), ... where s0 is the initial term value and c is a constant. Based on the information given in this graph, what is s0? What is c ? What is the value of the hundredth term in S ? 6 5, 3, 1, –1, –3, –5, ... 4 Term value 2 4 0 1 2 5 3 –2 –4 –6 Figure 19-6 Illustration for Problem 1. 6 Term number Practice Exercises 397 2. Does the infinite arithmetic sequence described in Problem 1 converge? If so, on what value does it converge? If not, why not? 3. The general form for an infinite geometric sequence T is T = t0, t 0k, t 0k2, t 0k 3, t 0k 4, ... where t0 is the initial value and k is the constant of multiplication. Calculate, and write down, the first seven terms in an infinite geometric sequence T where t 0 = 2 and k = −4. Does this sequence converge? If so, on what value does it converge? If not, why not? 4. Suppose that in the scenario of Problem 3, we change k from −4 to −1/4. Calculate and list the first seven values of the resulting infinite sequence. Does it converge? If so, on what value does it converge? If not, why not? 5. Consider again the sequence we saw earlier in this chapter: B = 0/1, 1/2, 2/3, 3/4, 4/5, ..., (n − 1)/n, ... We determined that the limit of B, as n grows without end, is Lim (n − 1)/n = 1 n→∞ so we know that B converges. Write down the series B+ that we get when we add the elements of B. Then write down the first five terms of the sequence B, which is made up of the partial sums in B+. Does the sequence B converge? If so, to what value does it converge? If not, why not? 6. Express the following series by writing out the first five terms followed by an ellipsis: n S+ = ∑ 1/10i i =1 First, express the terms as fractions. Then express them as powers of 10. Then express them as decimal quantities. Finally, write down the first five terms in the sequence S* of partial sums. 7. Find the following limit if it exists. If no limit exists, explain: n Lim n→∞ ∑ 1/10i i =1 8. Using a calculator, plug in n = 2, n = 6, n = 10, and n = 20 to informally illustrate that Lim (1 + 2 + 3 + ··· + n)/n 2 = 1/2 = 0.5 n→∞ and therefore that n Lim n→∞ ∑ i /n 2 = 1/2 = 0.5 i =1 Work out the partial sums to obtain decimal quantities. Round off your results to five decimal places when you encounter repeating or lengthy decimals. 9. Using a calculator, plug in n = 2, n = 6, n = 10, and n = 20 to informally illustrate that Lim (12 + 22 + 32 + ··· + n 2)/n 3 = 1/3 = 0.33333... n→∞ CHAPTER 20 Review Questions and Answers Part Two This is not a test! It's a review of important general concepts you learned in the previous nine chapters. Read it though slowly and let it sink in. If you're confused about anything here, or about anything in the section you've just finished, go back and study that material some more. Chapter 11 Question 11-1 What's a mathematical relation? Answer 11-1 A relation is a clearly defined way of assigning, or mapping, some or all of the elements of a source set to some or all of the elements of a destination set. Suppose that X is the source set for a relation, and Y is the destination set for the same relation. In that case, the relation can be expressed as a collection of ordered pairs of the form (x,y), where x is an element of set X and y is an element of set Y. Question 11-2 What's an injection, also known as an injective relation? Answer 11-2 Imagine two sets X and Y. Suppose that a relation assigns each element of X to exactly one element of Y. Also suppose that, according to the same relation, an element of Y never has more than one mate in X. (Some elements of Y might have no mates in X.) In a situation like this, the relation is an injection. Question 11-3 What's a surjection, also called an onto relation? 399 400 Review Questions and Answers Answer 11-3 Again, imagine two sets X and Y. Suppose that according to a certain relation, every element of Y has at least one (and maybe more than one) mate in X, so that no element of Y is left out. A relation of this type is a surjection from X onto Y. Question 11-4 What's a bijection, also called a one-to-one correspondence? Answer 11-4 A bijection is a relation that's both an injection and a surjection. Given two sets X and Y, a bijection assigns every element of X to exactly one element of Y, and vice versa. This is why a bijection is sometimes called a one-to-one correspondence. Question 11-5 What's a two-space function? Is every two-space function a relation? Is every two-space relation a function? Answer 11-5 A two-space function is a relation between two sets that never maps any element of the source set to more than one element of the destination set. All two-space functions are relations. However, not all two-space relations are functions. Question 11-6 What's the vertical-line test for the graph of a two-space function? Answer 11-6 The vertical-line test is a quick way to determine, based on the graph of a two-space relation, whether or not the relation is a function. Imagine an infinitely long, movable line that's always parallel to the dependent-variable axis (usually the vertical axis). Suppose that we're free to move the line to the left or right, so it intersects the independent-variable axis (usually the horizontal axis) wherever we want. The graph is a function of the independent variable if and only if the movable vertical line never intersects the graph at more than one point. Question 11-7 Based on the vertical-line test, which of the curves in Fig. 20-1 are functions of x within the span of values for which −6 < x < 6? Answer 11-7 Only f is a function of x. If we construct a movable vertical line (always parallel to the y axis), it never intersects the curve for f at more than one point over the span of values for which −6 < x < 6. However, the movable vertical line intersects the curve for g at more than one point for some values of x where −6 < x < 6. The same is true of the curve for h. Question 11-8 Suppose we're working in the polar coordinate plane, and we encounter the graph of a relation where the independent variable is represented by q (the direction angle) and the dependent Part Two 401 y 6 4 f 2 x –6 –4 –2 2 4 6 –2 –4 g –6 h Figure 20-1 Illustration for Question and Answer 11-7. variable is represented by r (the radial distance from the origin). How can we tell if the relation is a function of q ? Answer 11-8 We can draw the graph of the relation in a Cartesian plane, plotting values of q along the horizontal axis, and plotting values of r along the vertical axis. We can allow both q and r to attain all possible real-number values. Then we can use the Cartesian vertical-line test to see if the relation is a function of q. Question 11-9 How do functions add, subtract, multiply, and divide? Answer 11-9 To add one function to another, we add both sides of their equations. This can be done in either order, producing identical results. If f1 and f2 are functions of x, then ( f1 + f2)(x) = f1(x) + f2(x) and ( f2 + f1)(x) = f2(x) + f1(x) 402 Review Questions and Answers To subtract one function from another, we subtract both sides of their equations. This can be done in either order, usually producing different results. If f1 and f2 are functions of x, then ( f1 − f2)(x) = f1(x) − f2(x) and ( f2 − f1)(x) = f2(x) − f1(x) To multiply one function by another, we multiply both sides of their equations. This can be done in either order, producing identical results. If f1 and f2 are functions of x, then ( f1 × f2)(x) = f1(x) × f2(x) and ( f2 × f1)(x) = f2(x) × f1(x) To divide one function by another, we divide both sides of their equations. This can be done in either order, usually producing different results. If f1 and f2 are functions of x, then ( f1/f2)(x) = f1(x)/f2(x) and ( f2/f1)(x) = f2(x)/f1(x) Question 11-10 When we add, subtract, multiply, or divide functions, we must adhere to three important rules. What are they? Answer 11-10 First, we must be sure that the functions both operate on the same thing. In other words, the independent variables must describe the same parameters or phenomena. Second, we must restrict the domain of the resultant function to only those values that are in the domains of both functions (the intersection of the domains). Third, if we divide a function by another function, we can't define the resultant function for any value of the independent variable where the denominator function becomes 0. Chapter 12 Question 12-1 How can we informally define the inverse of a relation? Part Two 403 Answer 12-1 The inverse of a relation is another relation that undoes whatever the original relation does. Also, the original relation undoes whatever its inverse does. Question 12-2 How can we rigorously define the inverse of a relation? Answer 12-2 Let f be a relation where x is the independent variable and y is the dependent variable. The inverse relation for f is another relation f −1 such that f −1 [ f (x)] = x for all values of x in the domain of f, and f [f −1 (y)] = y for all values of y in the range of f. Question 12-3 Suppose we've drawn the graph of a relation f in the Cartesian xy plane. How can we create the graph of the inverse relation f −1? Answer 12-3 Imagine the line y = x as a "point reflector." For any point on the graph of f, its counterpoint on the graph of f −1 lies on the opposite side of the line y = x but the same distance away, as shown in Fig. 20-2. Mathematically, we can do this transformation by reversing the sequence of the ordered pair representing the point. When we want to obtain the graph of f −1 based on the graph of f, we can "flip the whole graph over in three dimensions" around the line y = x, as if that line were the hinge of a revolving door. Question 12-4 Is it possible for a relation to be its own inverse? Answer 12-4 Yes. The simplest example is the relation described in the Cartesian xy plane by the equation y=x which can also be written as f (x) = x Another, less obvious example, is y = −x 404 Review Questions and Answers 20-2 Illustration for Question and Answer 12-3. which can also be written as f (x) = −x If a relation's graph is a circle centered at the origin, then that relation is its own inverse. Examples include all relations of the form x2 + y2 = r2 where r is the radius of the circle. We can also write such a relation in the form f (x) = ±(r2 − x2)1/2 Question 12-5 How can we tell, simply by looking at the graph of a relation, whether or not that relation is its own inverse? Answer 12-5 Suppose that when we "flip the graph over in three dimensions" along the line y = x as if that line were the hinge of a revolving door, we end up with exactly the same graph as the one we started with. In any case like that, the relation is its own inverse. If we do the "revolving door" transformation and end up with a graph that's different in any way from the one we started Part Two 405 y Graph of inverse relation 6 4 2 x –6 –4 y=x –2 2 4 6 –2 –4 –6 Flip whole graph over by 1/2 revolution along this line Figure 20-3 Graph of original relation Illustration for Question and Answer 12-5. with, then the relation isn't its own inverse. Figure 20-3 shows an example of a graph of the second type, where the inverse obviously differs from the original relation. Question 12-6 Suppose we have a relation that's not a function, because it maps some values of the independent variable x to more than one value of the dependent variable y. Is it possible to modify such a relation so that it becomes a function of x ? Answer 12-6 Yes, in most cases it's possible. If we can restrict the domain or the range to values such that the modified relation never maps any value of x to more than one value of y, then the modified relation is a function of x. Question 12-7 Is the inverse of a function always a function? Answer 12-7 No, not always. Suppose we have a function f that maps values of an independent variable x to values of a dependent variable y. Also imagine that, for any value of x in the domain, there's 406 Review Questions and Answers only one corresponding value of y in the range. On that basis, we know that f is a function of x. However, if some values of y are mapped from two or more values of x, then we don't have a function of y when we consider y as the independent variable and x as the dependent variable. Although the inverse f −1 is a relation, it's not a true function. Question 12-8 Consider a function f that maps values of x to values of y. Suppose that f −1, which maps values of y to values of x, is a relation but not a true function. Is it possible to modify the inverse relation f −1 so that it becomes a function of y ? Answer 12-8 In most cases, yes. If we can restrict the inverse relation's domain (the set of y values for which f −1 is defined) or the inverse relation's range (the set of x values for which f −1 is defined) so that the modified version of f −1 never maps any value of y to more than one value of x, then the modified inverse is a true function of y. Question 12-9 Consider the two functions f (x) = x and g (x) = −x Both f and g are their own inverses, and the inverses are also true functions. Is it possible for any other true function to be its own inverse, with that inverse also constituting a true function? Answer 12-9 Yes, this can happen. Consider the real-number function h (x) = 1/x where x ≠ 0. This function is its own inverse, because h −1[h (x)] = h −1(1/x) = 1/(1/x) = x and h[h −1(x)] = h (1/x) = 1/(1/x) = x Question 12-10 Imagine a function f such that y = f (x) Part Two 407 and whose inverse f −1 is a true function, so that f −1( y) = x for all values of x in the domain of f, and for all values of y in the range of f. Based on this information, what can we conclude about the nature of the mapping that f represents between the elements of its domain and the elements of its range? Answer 12-10 Every element x in the domain maps to exactly one element y in the range, and every element y in the range is mapped from exactly one element x in the domain. Therefore, within the specified domain and range, the mapping that f represents is a one-to-one correspondence, technically known as a bijection. Chapter 13 Question 13-1 What are the four basic types of conic sections? What do they look like in the Cartesian plane? Answer 13-1 The conic sections are geometric curves representing the intersection of a plane with a double cone. There are four types: the circle, the ellipse, the parabola, and the hyperbola. Figure 20-4 shows generic graphs of each type of conic section in the Cartesian plane. Question 13-2 How are the conic sections generated in 3D geometry? Answer 13-2 When the plane is perpendicular to the axis of the double cone, we get a circle, as shown in Fig. 20-5A. When the plane is not perpendicular to the axis of the cone but the intersection curve is closed, we get an ellipse ( Fig. 20-5B). When we tilt the plane just enough to open up the curve, we get a parabola ( Fig. 20-5C). When we tilt the plane still more, we get a hyperbola ( Fig. 20-5D). Question 13-3 What is meant by the term "eccentricity" with respect to a conic section? How do the eccentricity values compare for a circle, an ellipse, a parabola, and a hyperbola? Answer 13-3 Eccentricity (symbolized e) is a nonnegative real number that defines the extent to which a conic section differs from a circle. Here's how the eccentricity values compare for the four types of conic section: • A circle has e = 0 • An ellipse has 0 < e < 1 408 Review Questions and Answers Circle Ellipse Parabola Hyperbola Figure 20-4 Illustration for Question and Answer 13-1. • A parabola has e = 1 • A hyperbola has e > 1 Question 13-4 What's the focus of a parabola? What's the directrix of a parabola? How are they related? Answer 13-4 The focus of a parabola is a point in the same plane as the parabola, and the directrix is a line in that plane that does not pass through the focus. On a parabola, every point is equidistant from a specific focus and a specific directrix, as shown in Fig. 20-6. For any particular focus and directrix in geometric space, there exists exactly one parabola. Conversely, for any particular parabola in space, there exists exactly one focus, and exactly one directrix. Question 13-5 What's the focal length of a parabola? Part Two 409 Ellipse Circle A B Parabola Hyperbola D C Figure 20-5 Illustration for Question and Answer 13-2. For any point on the parabola, these distances are equal Parabola Focus Directrix Figure 20-6 Illustration for Question and Answer 13-4. 410 Review Questions and Answers Answer 13-5 The focal length of a parabola is the distance between the focus and the point on the parabola closest to the focus. The focal length is also equal to half the distance between the focus and the point on the directrix closest to the focus. Question 13-6 What's the standard-form general equation for a circle in the Cartesian xy plane? Answer 13-6 The standard-form general equation is (x − x0)2 + (y − y0)2 = r2 where x0 and y0 are real-number constants that tell us the coordinates (x0,y0) of the center of the circle, and r is a positive real-number constant that tells us the radius of the circle. Question 13-7 What's the standard-form general equation for an ellipse in a Cartesian xy plane where the x axis is horizontal and the y axis is vertical? Answer 13-7 The standard-form general equation is (x − x0)2/a2 + (y − y0)2/b2 = 1 where x0 and y0 are real-number constants representing the coordinates (x0,y0) of the center of the ellipse, a is a positive real-number constant that tells us the length of the horizontal semiaxis, and b is a positive real-number constant that tells us the length of the vertical semi-axis. Question 13-8 What's the standard-form general equation for a parabola that opens upward or downward in a Cartesian xy plane where the x axis is horizontal and the y axis is vertical? Answer 13-8 The standard-form general equation is y = ax2 + bx + c where a, b, and c are real-number constants, and a ≠ 0. If a > 0, the parabola opens upward. If a < 0, the parabola opens downward. Question 13-9 How can we locate the coordinates (x0,y0) of the vertex point on a parabola that opens upward or downward in a Cartesian xy plane where the x axis is horizontal and the y axis is vertical? Part Two 411 How can we tell whether that vertex point represents the absolute minimum value of y or the absolute maximum value of y? Answer 13-9 We can find the coordinates (x0,y0) of the vertex point based on the constants in the standardform equation of the parabola. The x value is x0 = −b/(2a) The y value is y0 = −b2/(4a) + c If a > 0, the parabola opens upward, so the vertex represents the absolute minimum value of y on the curve. If a < 0, the parabola opens downward, so the vertex represents the absolute maximum value of y on the curve. Question 13-10 What's the standard-form general equation for a hyperbola that opens toward the right and left in a Cartesian xy plane where the x axis is horizontal and the y axis is vertical? Answer 13-10 The standard-form general equation is (x − x0)2/a2 − (y − y0)2/b2 = 1 where x0 and y0 are real-number constants that tell us the coordinates (x0,y0) of the center of the hyperbola, a is a positive real-number constant that tells us the length of the horizontal semi-axis, and b is a positive real-number constant that tells us the length of the vertical semi-axis. Chapter 14 Question 14-1 How can we informally describe the graph of the function y = ex in the Cartesian xy plane? Answer 14-1 The graph is a smooth, continually increasing curve that crosses the y axis at the point (0,1). The domain is the set of all real numbers, and the range is the set of all positive real numbers. The curve is entirely contained within the first and second quadrants. As we move to the left (in the negative x direction), the curve approaches, but never reaches, the x axis. As we move to the right (in the positive x direction), the graph rises at an ever-increasing rate. 412 Review Questions and Answers Question 14-2 How can we informally describe the graph of the function y = e−x in the Cartesian xy plane? Answer 14-2 The graph is a smooth, continually decreasing curve that crosses the y axis at (0,1). The domain is the set of all real numbers, and the range is the set of all positive real numbers. The curve is entirely contained within the first and second quadrants. As we move to the right, the curve approaches the x axis but never quite reaches that axis. As we move to the left, the graph rises at an ever-increasing rate. In fact, the curve for the function y = e−x has exactly the same shape as the curve for the function y = ex but is reversed left-to-right around the y axis, so the two graphs are horizontal mirror images of each other. Question 14-3 How can we informally describe the graphs of the functions y = 10x and y = 10−x in the Cartesian xy plane? Answer 14-3 The graphs of these functions are curves that closely resemble the graphs of the functions y = ex and y = e−x respectively. Both base-10 graphs cross the y axis at (0,1), just as the base-e graphs do. However, the contours differ. The base-10 curves are somewhat steeper than the base-e curves. Part Two 413 Question 14-4 How can we visually and qualitatively compare the graphs of the four functions described in Questions 14-1 through 14-3? Answer 14-4 We can graph them all together on a generic rectangular-coordinate grid such as the one shown in Fig. 20-7. Question 14-5 How can we informally describe the graph of the function y = ln x in the Cartesian xy plane? Answer 14-5 The graph is a smooth, continually increasing curve that crosses the x axis at the point (1,0). The domain is the set of positive real numbers, and the range is the set of all real numbers. The curve is entirely contained within the first and fourth quadrants. As we move to the left (in the negative x direction) from the point (1,0), the curve "blows up negatively," approaching the y axis but never reaching it. As we move to the right from (1,0), the graph rises at an everdecreasing rate. +y y = 10 –x y = 10 x y = e –x y = ex (0, 1) –x +x –y Figure 20-7 Illustration for Question and Answer 14-4. 414 Review Questions and Answers Question 14-6 How can we verbally describe the graph of y = ln (1/x) in the Cartesian xy plane? Answer 14-6 The graph is a smooth, continually decreasing curve that crosses the y axis at (1,0). The domain is the set of positive real numbers, and the range is the set of all real numbers. The curve is entirely contained within the first and fourth quadrants. As we move to the left from the point (1,0), the curve "blows up positively," approaching the y axis but never reaching it. As we move to the right from (1,0), the graph falls at an ever-decreasing rate. In fact, the curve representing y = ln (1/x) has exactly the same shape as the curve for y = ln x but is reversed top-to-bottom with respect to the x axis, so the two graphs are vertical mirror images of each other. Question 14-7 How can we informally describe the graphs of the common-log functions y = log10 x and y = log10 (1/x) in the Cartesian xy plane? Answer 14-7 The graphs of these functions closely resemble the graphs of the functions y = ln x and y = ln (1/x) respectively. Both common-log graphs cross the x axis at (1,0), just as the natural-log graphs do. However, the contours differ. The natural-log curves are somewhat steeper than the commonlog curves. Part Two 415 Question 14-8 How can we visually and qualitatively compare the graphs of the four functions described in Questions 14-5 through 14-7? Answer 14-8 We can graph them all together on a generic rectangular-coordinate grid such as the one shown in Fig. 20-8. Question 14-9 How can we plot the graph of the sum of two functions or the difference between two functions? Answer 14-9 There are two ways in which this can be done. First, we can graph the original functions separately, and then add or subtract their values visually to infer the sum or difference graph. Second, we can calculate several outputs for each function using inputs that we've selected to get a good sampling. Then we can add or subtract these outputs arithmetically. Based on that data, we can graph the sum or difference function. Question 14-10 Texts don't always agree in the denotation of logarithmic functions. How can we avoid confusion when we write our own papers? +y y = ln x y = log 10 x (1, 0) –x +x y = log 10 (1/x) y = ln (1/x) –y Figure 20-8 Illustration for Question and Answer 14-8. 416 Review Questions and Answers Answer 14-10 We should always clarify the logarithmic base when we write "log" followed by anything. For example, we should write "log10" or "loge" instead of "log" (unless we can't portray the subscript within the constraints of a text-editing or Web site–building program). We don't need to write a subscript when we write "ln" to denote the natural logarithm, because "ln" means natural log or base-e log all the time. Chapter 15 Special note If you want to see graphical illustrations of the answers to the following 10 questions, feel free to look back at Chap. 15. Try to envision or draw the graphs yourself before you look back! Question 15-1 Consider a function f of a real-number variable q such that f (q) = sin q + cos q What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-1 The period of f is 2p. That's the same as the period of the sine. It's also the same as the period of the cosine. The positive peak amplitude of f is 21/2. The negative peak amplitude of f is −21/2. The domain of f is the set of all real numbers. The range of f is the set of all real numbers f (q) such that −21/2 ≤ f (q) ≤ 21/2 Question 15-2 Consider a function f of a real-number variable q such that f (q) = sin q cos q What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-2 The period of f is p, which is equal to half the period of the sine, and is also half the period of the cosine. The positive peak amplitude of f is 1/2. The negative peak amplitude of f is −1/2. The domain of f is the set of all real numbers. The range of f is the set of all real numbers f (q) such that −1/2 ≤ f (q) ≤ 1/2 Part Two 417 Question 15-3 Consider a function f of a real-number variable q such that f (q) = sin2 q + cos2 q What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-3 In this case, the function f has a constant value of 1. The period is not defined, because the function's output value never changes, and is defined for all inputs. The positive peak amplitude of f is equal to 1. The negative peak amplitude of f is also equal to 1. The domain of f is the set of all real numbers. The range of f is the set containing the single number 1. Question 15-4 Consider a function f of a real-number variable q such that f (q) = sec q + csc q What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-4 The period of f is 2p, which is the same as the period of the secant, and the same as the period of the cosecant. The positive and negative peak amplitudes of f are not defined, because f blows up in both the positive and negative directions whenever q is an integer multiple of p /2. The domain of f is the set of all real numbers except the integer multiples of p /2. The range of f is the set of all real numbers. Question 15-5 Consider a function f of a real-number variable q such that f (q) = sec q csc q What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-5 The period of f is p, which is half the period of the secant, and half the period of the cosecant. The positive and negative peak amplitudes of f are undefined, because f blows up both positively and negatively at all integer multiples of p /2. The domain of f is the set of all real numbers except the integer multiples of p /2. The range is the set of all real numbers f (q) such that f (q) ≥ 2 or f (q) ≤ −2 418 Review Questions and Answers Question 15-6 Consider a function f of a real-number variable q such that f (q) = sec2 q + csc2 q What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-6 The period of f is p /2, which is half the period of the secant squared, and half the period of the cosecant squared. The positive peak amplitude of f is undefined, because f "blows up" positively at all integer multiples of p /2. The negative peak amplitude of f is equal to 4, which occurs whenever q is an odd-integer multiple of p /4. The domain of f is the set of all real numbers except the integer multiples of p /2. The range is of f the set of all real numbers f (q) such that f (q) ≥ 4 Question 15-7 Consider a function f of a real-number variable q such that f (q) = tan q + cot q What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-7 The period of f is p, which is the same as that of the tangent, and the same as that of the cotangent. The positive and negative peak amplitudes of f are both undefined, because f blows up positively and negatively at all integer multiples of p /2. The domain of f is the set of all real numbers except the integer multiples of p /2. The range of f is the set of all real numbers f (q) such that f (q) ≥ 2 or f (q) ≤ −2 Question 15-8 Consider a function f of a real-number variable q such that f (q) = tan q cot q What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-8 This particular function presents a strange situation. The graph of f is a horizontal, straight line with single-point gaps wherever q is an integer multiple of p /2. The period of f is p /2, because the graph consists of infinitely many open line segments placed end to end, each of Part Two 419 length p /2. The positive peak amplitude of f is equal to 1. The negative peak amplitude of f is also equal to 1. The domain of f is the set of all real numbers except the integer multiples of p /2. The range is the set containing the single element 1. Question 15-9 Consider a function f of a real-number variable q such that f (q) = tan2 q + cot2 q What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-9 The period of f is p /2, which is half that of the tangent squared, and is also half that of the cotangent squared. The positive peak amplitude of f is undefined, because the function blows up positively at all integer multiples of p /2. The negative peak amplitude of f is equal to 2, which occurs whenever q is an odd-integer multiple of p /4. The domain of f is the set of all real numbers except the integer multiples of p /2. The range of f is the set of real numbers f (q) such that f (q) ≥ 2 Question 15-10 Consider a function f of a real-number variable q such that f (q) = (tan2 q)/(cot2 q) What are the period, the positive peak amplitude and the negative peak amplitude of f ? What are the domain and range of f ? Answer 15-10 The period of f is p, which is the same as the period of the tangent squared, and is also the same as the period of the cotangent squared. The positive peak amplitude of f is undefined, because the function blows up positively at all odd-integer multiples of p /2. The negative peak amplitude of f is undefined as well, although f (q) approaches 0 whenever q approaches any integer multiple of p from either side. (We can't say that the negative peak amplitude is 0, because the function never actually attains that value.) The domain of f is the set of all real numbers except the integer multiples of p /2. The range of f is the set of all positive real numbers. Chapter 16 Question 16-1 What's a parameter? What's a set of parametric equations? Answer 16-1 A parameter is an independent variable on which other variables depend. A set of parametric equations is a collection of equations, at least one of which has one or more variables that 420 Review Questions and Answers depend on the parameter. The parameter, which is often symbolized t, plays the role of master controller for the other variables in the system. Question 16-2 Consider the pair of parametric equations x = 3t and y = −t where t is the parameter on which both x and y depend. How can we sketch a Cartesian graph of this system? How can we find an equivalent equation in terms of the variables x and y only, based on the graph? Answer 16-2 We can input various values of t to both equations, and plot the ordered pairs (x,y) that come out of those equations. Following are some examples: • • • • • When t = −2, we have x = 3 × (−2) = −6 and y = −1 × (−2) = 2. When t = −1, we have x = 3 × (−1) = −3 and y = −1 × (−1) = 1. When t = 0, we have x = 3 × 0 = 0 and y = −1 × 0 = 0. When t = 1, we have x = 3 × 1 = 3 and y = −1 × 1 = −1. When t = 2, we have x = 3 × 2 = 6 and y = −1 × 2 = −2. When we plot the (x,y) ordered pairs based on this list as points on a Cartesian plane and then "connect the dots," we get a line through the origin with a slope of −1/3, as shown in Fig. 20-9. In slope-intercept form, the line can be represented as y = (−1/3)x Question 16-3 Consider again the pair of parametric equations x = 3t and y = −t How can we use algebra alone (without the help of a graph) to determine the equivalent equation in terms of x and y only? Answer 16-3 We can take the first parametric equation x = 3t 422 Review Questions and Answers and r = −t where t is the parameter on which both q and r depend. How can we sketch a polar graph of this system? Answer 16-4 We can input various values of t, restricting ourselves to values such that we see only the part of the graph corresponding to the first full counterclockwise rotation of the direction angle, where 0 ≤ q ≤ 2p: • • • • When t = 0, we have q = 3 × 0 = 0 and r = −1 × 0 = 0. When t = p /4, we have q = 3p /4 and r = −p /4 ≈ −0.79. When t = p /2, we have q = 3p /2 and r = −p /2 ≈ −1.57. When t = 2p /3, we have q = 3 × 2p /3 = 2p and r = −2p /3 ≈ −2.09. Figure 20-10 illustrates this graph, based on these four points and the intuitive knowledge that the graph must be a spiral, starting at the origin and expanding as we rotate counterclockwise. The graph is a little tricky, because all of the radii are negative! Also, we should remember that the concentric circles represent radial divisions on the polar coordinate grid; the straight lines represent angular divisions. Figure 20-10 Illustration for Question and Answer 16-4. In this coordinate system, each radial division represents p /4 units. Part Two 423 Question 16-5 Consider again the pair of parametric equations q = 3t and r = −t How can we use algebra to determine the equivalent equation in terms of q and r only? Answer 16-5 The equation can be derived using the same algebraic process that we used in the Cartesian situation. We substitute q in place of x, and we substitute r in place of y. When we do that, we get r = (−1/3)q Question 16-6 What are the parametric equations for a circle centered at the origin in the Cartesian xy plane? Answer 16-6 The parametric equations are x = a cos t and y = a sin t where a is the radius of the circle and t is the parameter. Question 16-7 What are the parametric equations for a circle centered at the origin in the polar coordinate plane? Answer 16-7 Let the polar direction angle be q, and let the polar radius be r. The parametric equations of a circle having radius a, and centered at the origin, are q=t and r=a where t is the parameter. 424 Review Questions and Answers Question 16-8 Why does only one of the equations in Answer 16-7 contain the parameter t ? Shouldn't both equations contain it? Answer 16-8 The parameter t has no effect in the second equation, because the polar radius r of a circle centered at the origin is always the same, no matter how anything else varies. Question 16-9 What are the parametric equations for an ellipse centered at the origin in the Cartesian xy plane? Answer 16-9 The parametric equations are x = a cos t and y = b sin t where a is the length of the horizontal (x-coordinate) semi-axis, b is the length of the vertical (y-coordinate) semi-axis, and t is the parameter. Question 16-10 Why is the passage of time a common parameter in science and engineering? Answer 16-10 In the physical world, many effects and phenomena depend on elapsed time. If we find time acting as a mathematical variable, then that variable is almost always independent. We often come across situations where two or more factors fluctuate with the passage of time. An example is the variation of temperature, humidity, and barometric pressure versus time in a specific location. In a situation of this sort, time can be considered as the parameter on which the other three physical variables depend. Chapter 17 Question 17-1 What information do we need to determine the equation of a plane in Cartesian xyz space? Answer 17-1 We can find an equation for a plane in Cartesian xyz space if we know the direction of at least one vector that's perpendicular to the plane, and if we know the coordinates of at least one point in the plane. We don't have to know the magnitude of the vector, but only its direction. The point's coordinates don't have to tell us where the vector begins or ends; the point can be anywhere in the plane. Part Two 425 Question 17-2 Imagine a plane that passes through a point whose coordinates are (x0,y0,z0) in Cartesian xyz space. Also suppose that we've found a vector ai + bj + ck that's normal (perpendicular) to the plane. Based on this information, how can we write down an equation that represents the plane? Answer 17-2 We can write the plane's equation in the standard form a(x − x0) + b(y − y0) + c(z − z0) = 0 We can also write the equation as ax + by + cz + d = 0 where d is a constant that works out to d = −ax0 − by0 − cz0 Question 17-3 What's the general equation for a sphere centered at the origin and having radius r in Cartesian xyz space? Answer 17-3 The equation can be written in the standard form x2 + y2 + z2 = r2 Question 17-4 What's the general equation for a sphere of radius r in Cartesian xyz space, centered at a point whose coordinates are (x0,y0,z0)? Answer 17-4 The equation can be written in the standard form (x − x0)2 + (y − y0)2 + (z − z0)2 = r2 Question 17-5 Can a sphere have a negative radius in Cartesian xyz space? Answer 17-5 Normally, we define a sphere's radius as a positive real number. Nevertheless, spheres with negative radii can exist in theory. If we encounter a sphere whose radius happens to be defined 426 Review Questions and Answers as a negative real number, then that sphere has the same equation as it would if we defined the radius as the absolute value of that number. For all real numbers r, it's always true that r2 = \|r\|2, so the following two equations: (x − x0)2 + (y − y0)2 + (z − z0)2 = r2 and (x − x0)2 + (y − y0)2 + (z − z0)2 = \|r \|2 are equivalent, whether r is positive or negative. Question 17-6 What's the general equation for a distorted sphere in Cartesian xyz space? Answer 17-6 The equation can be written in the standard form b is the axial radius in the y direction, and c is the axial radius in the z direction. Normally, all three of the constants a, b, and c are positive reals. Question 17-7 There are three distinct classifications of distorted sphere. What are they? How can we tell, from the standard-form equation, which of these three types we have? Answer 17-7 We can have an oblate sphere, an ellipsoid, or an oblate ellipsoid. We can tell which of these three types a particular standard-form equation represents by comparing the values of the axial radii a, b, and c. We have an oblate sphere if and only if two of the positive real-number axial radii are equal, and the third is smaller. In that case, one of the following is true: a<b=c b<a=c c<a=b We have an ellipsoid if and only if two of the positive real-number axial radii are equal, and the third is larger. Then one of the following is true: a>b=c b>a=c c>a=b Part Two 427 We have an oblate ellipsoid if and only if no two of the positive real-number axial radii are equal. In that scenario, all of the following are true: a≠b b≠c a≠c Question 17-8 What's the general equation for a hyperboloid of one sheet in Cartesian xyz space? Answer 17-8 The equation can be written in one of the following standard forms: + (z − z0)2/c2 = 1 −9 What's the general equation for a hyperboloid of two sheets in Cartesian xyz space? Answer 17-9 The equation can be written in one of the following standard forms: − − (z − z0)2/c2 = 1 −(x − x0)2/a2 −10 What's the general equation for an elliptic cone in Cartesian xyz space? Answer 17-10 The equation can be written in one of the following standard forms: (x − x0)2/a2 + (y − y0)2/b2 − (z − z0)2/c2 = 0 (x − x0)2/a2 − (y − y0)2/b2 + (z − z0)2/c2 = 0 −(x − x0)2/a2 + (y − y0)2/b2 + (z − z0)2/c2 = 0 428 Review Questions and Answers where (x0,y0,z0) are the coordinates of the point where the apexes of the two halves of the double cone meet, the constants a, b, and c are positive real numbers that define the object's general shape, and the locations of the signs (plus and minus) define the object's orientation with respect to the coordinate axes. Don't get confused by the similarity between these equations and those for hyperboloids of one sheet. The only difference is that the net values are all equal to 1 for the hyperboloids, and all equal to 0 for the cones. Chapter 18 Question 18-1 What's the general symmetric equation for a straight line in Cartesian xyz space? Answer 18-1 Imagine that (x0,y0,z0) are the coordinates of a specific point. Suppose that a, b, and c are nonzero real-number constants. The general symmetric equation of a straight line passing through (x0,y0,z0) is (x − x0)/a = (y − y0)/b = (z − z0)/c The constants a, b, and c are called direction numbers. When considered all together as an ordered pair (a,b,c), these numbers define the direction or orientation of the line with respect to the coordinate axes. Question 18-2 What are the general parametric equations for a straight line in Cartesian xyz space? Answer 18-2 Let (x0,y0,z0) be the coordinates of a specific point, and suppose that a, b, and c are nonzero real-number constants. The general parametric equations for a straight line passing through (x0,y0,z0) are x = x0 + at y = y0 + bt z = z0 + ct where the parameter t can range over the entire set of real numbers. As with the symmetric form, the constants a, b, and c are direction numbers that tell us how the line is orientated relative to the coordinate axes. Question 18-3 What is meant by the expression "preferred direction numbers" when describing the orientation of a straight line in Cartesian xyz space? Answer 18-3 For any line in Cartesian xyz space, there are infinitely many ordered triples that can define its orientation with respect to the coordinate axes. For example, if a line has the direction numbers (a,b,c), then we can multiply all three entries by a real number other than 0 or 1, and we'll get
Prealgebra, by definition is the transition from arithmetic to algebra. Miller/OíNeill/Hyde Prealgebra will introduce algebraic concepts early and repeat them as student would work through a Basic College Mathematics or arithmetic table of contents. Prealegbra is the ground work thatís needed for developmental students to take the next step into a traditional algebra course. As in previous editions, the focus in PREALGEBRA & INTRODUCTORY ALGEBRA remains on the Aufmann Interactive Method (AIM). Students are encouraged to be active participants in the classroom and in their own studies as they work through the How To examples and the paired Examples and You Try It problems. Student engagement is crucial to success. Presenting students with worked examples, and then providing them with the opportunity to immediately solve similar problems, helps them build their confidence and eventually master the concepts. Simplicity is key in the organization of this edition, as in all other editions. All lessons, exercise sets, tests, and supplements are organized around a carefully constructed hierarchy of objectives.
Curriculum & assignment schedule were clear and organized. Lessons from the book were always supported with instructor handouts, guided notes provided by instructor and explained in lectures. All aspects were valuable I feel that the hands on attempt towards class is more helpful but visual is also helpful as well she always makes sure we understand. the group worksheets done in class everything in the class is going to be used throught my career so it is not one particular thing. learning the basic culinary math rules Being very intuitive with students, having examples and worksheets being done in class along with lecture. Juli is a great teacher, but I already had a firm grasp on the concepts being offered in this course. Hence the answer to the above question. Loved the vocab sheets, very helpful. Having a tutor in class to help us. Juli Umetsu provided us with many chapter study guides. This was a big help to me. when we relate the math to real life situatons all of them.. Juli applied everything we learned to the culinary world making it more relatable and applicable The tutoring and the help from my teacher was valuable during or after class. the reason why is because, she the teacher and he the tutor helped the class with problems that we needed help on. So my teacher and tutor was the most valuable overall. The worksheets we did everyday in class because they helped me to understand the material better. The hands on learning was awesome juli is great at explaining things multiple times so everyone can understand! class work, hand outs1 (7%) 14 (93%) 4. The instructor seems to enjoy teaching56713 (100%) 8915 (100%) 10. Comment on the instructor's professional attitude and behavior. Juli is the best! she has a very good attitude and profesional behavior shes a really great teacher. Very helpful and knowledgable. Makes learning easier and fun. i totally admire and adore Juli. she is a great teacher and she explains math problems real good. juli is a role model and just a good person all together Juli makes you feel like she really wants everyone of us to get this. She offers her time after class to help anyone who might need it. She has patience and it is greatly appreciated in this classroom. She loves teaching, she loves the industry and it's plain to see. Julie was very helpful in all areas; i have always had difficulty with math, but she made things very easy to understand and remember. she's a great instructor Excelent very professional yet approachable she loves her job and it shows through her teaching she is great teaching Juli is an excellent teacher and always has a positive attitude. She always encourages students to come to class and keep grades up because she cares for each and everyone of her students. Even though she had a personal problem on a particular day she handled herself with all the professionalism I have come to expect of her. Her ability to adapt and overcome is truly awe inspiring. I truly love her spunk in class it motivates students to learn. She is very thorough in her teaching methods. My memory just sucks. VERY GOOD! always professional Juli is very professional in both her attitude and behavior. Not overly professional however, she still relates to her students and connects with them on a level that makes them feel more comfortable in the classroom Her additude toward the class was very nice because she was always smiling and encouraging students to work hard in class and never give up learning. So her behavior as a proffessional was pretty good. Good teacher THE best teacher EVER She is a very good teacher!! She is the best. She is very kind, helpful person and personality. She teach us very good, organized, make me like/love the math with simple way. She treat all students the same, encourage all students.. she is always fair Great attitude, good instructor, positive energy 11. What did you find most valuable and helpful about the instructor? She is very kind and patient. she made sure everyone was doing there work and talked to students regularly about grades her patience with students her great attitude and the fact that she cares for real Her enthusiasm for teaching is encouraging. She treats everyone equally, and encourages everyone the same. I like that there isn't any favorites! Her willingness to keep going over a specific problem until it was understood by all. that we can go over anything that we may not understand her explanations variations on material so that it became easier to understand given different learning abilities she is a visual learner just like me the way she teaches She went over worksheets with everyone as a class and made it sure that everyone understood She knows that I already knew the criteria of the class and has allowed me to better utalize my time within. While this may seem unacceptable to some teachers it allows me to work on other assignments that require a greater deal of my attention. She offers help when needed. she always checks to see if everyone gets whats going on How friendly and genuinely interested she seemed in the students understanding and achievement in the course I found that asking her questions about work in class was valuable to me, because if I had problem doing a worksheet or textbook assignment she would know the process of the problem without looking at any of her notes. That she is always availble for help Her encouragement Everything that she taught me. open to answer questions 12. What did you find least valuable and helpful about the instructor? Inventory turnover rate section, but that's not her fault, it's part of the book. nothing Nothing. nothing nothing is least about her stupid question She rocks! Nothing that I can think of nothing she's a great instructor nothing nothing.. Juli was awsome Everything was help in its own rite. The heavy dependance on a particular calculator was annoying. Nothing. none She was valuable most of the time because the main thing that is important was getting the help I needed from her to learn something and get through the class without any problems and have a better understanding of my work. Nothing nothing. yield percent experiment 13. The lab sections were a valuable part of this course93 Freq(%) 0 (0%) 0 (0%) 2 (15%) 1 (7%) 8 (62%) 14. My grades accurately represent my performance in the course. Mean N-Size Std Dev Strongly Disagree Disagree Neutral Agree Strongly Agree 4.93 14 0.27 Freq(%) 0 (0%) 0 (0%) 0 (0%) 3 (23%) 10 (77%) 15. Other comments: I was lucky to have her as a teacher! nice and sweet lady with a lot of patience. awesome teacher and very very helpful thank heavens for juli umetsu good job Juli was an awsome teacher Again her vocabulary sheets are very helpful then paired with lecture, hard to get a low grade. Easy to understands and get it. juli rocks juli is great! I would like to say Thanks Mrs. Umetsu for the great experience of culinary math it was a bumpy ride but in the end I got through the class with posibly flying colors.
Lessons Include: Ratios, Unit Rates, Proportions, Solving Proportions, Fractions and Percents, Decimals and Percents, Find the Percent, Percent of a Number (Finding the Part) Finding a Number When the Percent is Known, Discount, Markup, Percent of Increase, Percent of Decrease Lessons Include: Area of a Triangle, Area of a Parallelogram, Area of Similar Figures, Area of a Trapezoid, Area of a Rhombus or Kite, Area of a Circle, Area of a Sector of a Circle, Area of Regular Polygons, Area of an Irregular Shape, Comparing Area and Perimeters, Using Trigonometry to Find the Area of a Triangle, Geometric Probability Graphing Quadratic Functions, Properties of a Graph of a Quadratic Function, Writing a Quadratic Function from its Graph, Quadratic Function in Intercept Form, Solving Quadratic Equations Using Square Roots, Solving a Quadratic Equation by Completing the Square, Quadratic Formula, Solving a Quadratic Equation by Factoring, Using the Discriminant, Methods for Solving Quadratic Equations, Writing an Equation of an Ellipse, Foci of an EllipseLessons Include: Relations and Functions, Types of Functions, Direct Variation, Inverse Variation, Slope-Intercept Form, Point-Slope Form I, Point-Slope Form II, Linear Parametric Equations, Writing a System of Equations as a Matrix, Using Matrices to Solve a System of Two Equations, Cramer's Rule For Beginning & Intermediate ESL & ELL students Features include: Six chapters cover the basic math skills in English Creative illustrations help add visual cues to problem-solving Emphasis given to learning pronunciation of words and numbers in English Multicultural notes and sidebars
Description: Mathematics education in schools has seen a revolution in recent years. In this book, the author steers a simple and well motivated path through the central ideas of real analysis. Each concept is introduced only after its need has ... Description: This self contained text, suitable for advanced undergraduates, provides an extensive introduction to mathematical analysis, from the fundamentals to more advanced material. It begins with the properties of the real numbers and continues with a rigorous treatment of sequences, ...
Introduction To Graph Theory 9780073204161 ISBN: 0073204161 Pub Date: 2004 Publisher: McGraw-Hill College Summary: Written by one of the leading authors in the field, this text provides a student-friendly approach to graph theory for undergraduates. Much care has been given to present the material at the most effective level for students taking a first course in graph theory. Gary Chartrand and Ping Zhang's lively and engaging style, historical emphasis, unique examples and clearly-written proof techniques make it a sound yet acc...essible text that stimulates interest in an evolving subject and exploration in its many applications.This text is part of the Walter Rudin Student Series in Advanced Mathematics. Chartrand, Gary is the author of Introduction To Graph Theory, published 2004 under ISBN 9780073204161 and 0073204161. Two hundred eighty Introduction To Graph Theory textbooks are available for sale on ValoreBooks.com, twenty three used from the cheapest price of $23.79, or buy new starting at $109204161-4-0-3 Orders ship the same or next business day. Expedited shipping within U.S. [more] May include moderately worn cover, writing, markings or slight discoloration. SKU:9780073204
Scienceresents an introduction to trigonometry, providing simplified explanations of such topics as mapping, functions, vectors, the unit-circle paradigm, and polar coordinates, with quizzes at the end of each chapter, along with answers and explanations. more Read more... more
MA 211 Calculus and Analytic Geom II Maldonado, Aldo's philosophy is based on constructivist view of mathematics learning. Students construct their own schema. Instructor provides guidance like lectures, HW assignment and tutoring but learning task is placed entirely on student's shoulders. Full use of technology is one of the ways instructor guides students Learning Outcomes: Core Learning Outcomes Compute simple integrals by guessing the antiderivative Compute the area under a given curve using Riemann sums Compute integrals using the power rule, the sum and difference rules, and simple substitution. State and utilize the Fundamental Theorem of Calculus to compute definite integrals. Differentiate and integrate inverse functions Application of integrals (areas, volumes of revolution, work, etc.) Core Assessment: Periodic assignments Quizzes Tests Class Assessment: 1 midterm, 1 comprehensible final, Quizzes and Homework. Midterm, final, quizzes and HW consist of questions and mostly problems at the level of and based in textbook Grading: HW 15% QUIZZES 15% MIDTERM 30% FINAL 40 % Late Submission of Course Materials: No late submission is allowed Classroom Rules of Conduct: Cell phones and pagers must be turned off to prevent unnecessary disruptions during the class. Disruptive behavior, racist, or sexist speech out of context will not be tolerated LAST DAY TO DROP: Monday, March 24, 2008 LAST DAY TO WITHDRAW: Sunday, April 20, 2008 Course Topic/Dates/Assignments: Class Activities Assignments Week 1 Anti-derivatives sections 4.3-4.5 Week 2 Intro to integration sections 4.6 – 5.2 Week 3 Applications of definite integral sections 5.3-5.5 Week 4 Applications of definite integral sections 5.8-6.3 Week 5 Transcendental functions sections 6.4-6.6 Midterm Week 6 Techniques of integration sections 6.9, 7.1, 7.2,7.3 Week 7 Parametric Equations and Polar Coordinates sections 7.7, 9.1, 9.3, 9.4 Week 8 Review and final Review,
This free and open online course in Elementary Algebra was produced by the WA State Board for Community & Technical... see more This free and open online course in Elementary Algebra was produced by the WA State Board for Community & Technical Colleges [ course is the study of basic algebraic operations and concepts and the structure and use of algebra. This includes the solutions to algebraic equations. factoring algebraic functions, working with rational expressions, free and open online course in Linear Algebra was produced by the WA State Board for Community & Technical Colleges... see more This free and open online course in Linear Algebra was produced by the WA State Board for Community & Technical Colleges [ course is the study of basic algebraic operations and concepts and the structure and use of algebra. This includes the solutions to algebraic equations, factoring algebraic expressions, working with rational expressionsThe study of "abstract algebra" grew out of an interest in knowing how attributes of sets of mathematical objects behave when... see more The study of "abstract algebra" grew out of an interest in knowing how attributes of sets of mathematical objects behave when one or more properties we associate with real numbers are restricted. The student will begin this course by reviewing basic set theory, integers, and functions in order to understand how algebraic operations arise and are used. The student then will proceed to the heart of the course, which is an exploration of the fundamentals of groups, rings, and fields. This free course may be completed online at any time. See course site for detailed overview and learning outcomes. (Mathematics 231) This course is a continuation of Abstract Algebra I: the student will revisit structures like groups, rings, and fields as... see more This course is a continuation of Abstract Algebra I: the student will revisit structures like groups, rings, and fields as well as mappings like homomorphisms and isomorphisms. The student will also take a look at ring factorization, general lattices, and vector spaces. Later, this course presents more advanced topics, such as Galois theory—one of the most important theories in algebra, but one that requires a thorough understanding of much of the content we will study beforehand. This free course may be completed online at any time. See course site for detailed overview and learning outcomes. (Mathematics 232) This is a free online course offered by the Saylor Foundation.'This introductory mathematics course is for you if you have a... see more This is a free online course offered by the Saylor Foundation.'This introductory mathematics course is for you if you have a solid foundation in arithmetic (that is, you know how to perform operations with real numbers, including negative numbers, fractions, and decimals). Numbers and basic arithmetic are used often in everyday life in both simple situations, like estimating how much change you will get when making a purchase in a store, as well as in more complicated ones, like figuring out how much time it would take to pay off a loan under interest.The subject of algebra focuses on generalizing these procedures. For example, algebra will enable you to describe how to calculate change without specifying how much money is to be spent on a purchase–it will teach you the basic formulas and steps you need to take no matter what the specific details of the situation are. Likewise, accountants use algebraic formulas to calculate the monthly loan payments for a loan of any size under any interest rate. In this course, you will learn how to work with formulas that are already known from science or business to calculate a given quantity, and you will also learn how to set up your own formulas to describe various situations by translating verbal descriptions to mathematical language. In the later units of this course, you will discover another tool used in mathematics to describe numbers and analyze relationships: graphing. You will learn that any pair of numbers can be represented by a point on a coordinate plane and that a relationship between two quantities can be represented by a line or a curve.Units 6, 7, and 8 may seem more abstract than the earlier ones, as you will deal with expressions that contain mostly variables and not too many numbers. While the procedures you will master in these units might seem to have little practical application, you have to keep in mind that they result in formulas that describe very real situations in business, accounting, and science. Knowing how to perform various operations with algebraic expressions will eventually enable you to solve quadratic and even more complex equations. You will explore a variety of real-world scenarios that can be described by these kinds of equations. For example, if a ball is thrown up in the air, solving a quadratic equation will help you find out when it will hit the ground. As another example, if you know the area of a rectangular garden, then you can use a quadratic equation to find the length of each side.' This is a free online course offered by the Saylor Foundation.'In this course, you will study basic algebraic operations and... see more This is a free online course offered by the Saylor Foundation.'In this course, you will study basic algebraic operations and concepts, as well as the structure and use of algebra. This includes the solutions to algebraic equations, factoring algebraic expressions, working with rational expressions, and graphing of linear equations. You will apply these skills to solve real world problems (word problems). Each unit will have its own application problems, depending on the concepts you have been exposed to. This course is also intended to provide you with a strong foundation for intermediate algebra and beyond.This course will begin with a review of some math concepts formed in pre-algebra, such as order of operations and simplifying simple algebraic expressions to get your feet wet. You will then build on these concepts by learning more about functions, graphing of functions, evaluation of functions, and factorization. You will spend time on the rules of exponents and their applications in distribution of multiplication over addition/subtraction.' introduces cryptography by addressing topics such as ciphers that were used before World War II, block cipher... see more This course introduces cryptography by addressing topics such as ciphers that were used before World War II, block cipher algorithms, the advanced encryption standard for a symmetric-key encryption adopted by the U.S. government, MD5 and SHA-1 hash functions, and the message authentication code. The course will focus on public key cryptography (as exemplified by the RSA algorithm), elliptic curves, the Diffie-Hellman key exchange, and the elliptic curve discrete logarithm problem. The course concludes with key exchange methods, study signature schemes, and discussion of public key infrastructure. Note: It is strongly recommended that you complete an abstract algebra course (such as the Saylor Foundation's MA231) before taking this course. This free course may be completed online at any time. See course site for detailed overview and learning outcomes. (Computer Science 409)
Transition to Algebra Full Description: Transition to Algebra is a new, full-year algebra support curriculum designed to run concurrently with first-year algebra and raise the competence and confidence of students who may benefit from supports for algebra success. Developed with funding from the National Science Foundation, Transition to Algebra builds students' Algebraic Habits of Mind, several key mathematical ways of thinking aligned with the Common Core Standards for Mathematical Practice. Students explore algebraic logic puzzles that connect to and extend algebra course topics and learn broadly-applicable tools and strategies to help them make sense of what they are learning in algebra. Students also discuss and refine their ideas as they work through mental mathematics activities, written puzzles, spoken dialogues, and hands-on explorations that engage them in cultivating mathematical knowledge, intuition, and skills. Professional development is offered to support use of the curriculum.
More About This Textbook Overview Now in its fifth edition, A Mathematics Sampler presents mathematics as both science and art, focusing on the historical role of mathematics in our culture. It uses selected topics from modern mathematics—including computers, perfect numbers, and four-dimensional geometry—to exemplify the distinctive features of mathematics as an intellectual endeavor, a problem-solving tool, and a way of thinking about the rapidly changing world in which we live. A Mathematics Sampler also includes unique LINK sections throughout the book, each of which connects mathematical concepts with areas of interest throughout the humanities. The original course on which this text is based was cited as an innovative approach to liberal arts mathematics in Lynne Cheney's report, "50 HOURS: A Core Curriculum for College Students", published by the National Endowment for the Humanities. Related Subjects Meet the Author William P. Berlinghoff is visiting professor of mathematics at Colby College. Kerry E. Grant is professor of mathematics at Southern Connecticut State University. Dale Skrien is professor of computer science at Colby
Co-ordinate Geometry(2D): Co-ordinate system, Cartesian & polar, Relation between Cartesian and polar, Distance between two points section Formula. Formula for Area of Triangle problems. Straight Line, Different forms of equations of straight line, angle between two straight lines, conditions of parallelism and perpendicularity of two straight lines, Distance of a point from a given straight line, problems Mensuration: Area and perimeter of regular polygons of n sides, formula and problems. Area and premeter of a circle and an ellipse, applications Evaluation of Volume and surface area of a prism,pyramid and a cone, cylinder, trapezoid. Calculation of area of curvilinear figures by Simpson's One Third rule. Different Calculus: Function : Definition, Even and Odd, explicit and implien, periodic and parametric, Limit,Theorems on limits (statement only), standard limits, evaluation of limits, Definition of continuity of a function, testing of continuity, problems. Differentiation, Differentiation of parametric and implicit functions. Successive differentiation up to second order, problems. Integral Calculus: Integration as the inverse process of differentiation, List of formula for integration, Method of substitution, integration by parts, evaluation of integrals by each of the above methods. Definite integral,rules and properties of definite integral (statement only), Evaluation of definite integrals. Differential Equation: Definition, order and degree of a differential equation, solution of differential equation of 1st order and 1st degree by the method of separation of variables
The Mathematics of Games of Strategy by Melvin Dresher This text offers an exceptionally clear presentation of the mathematical theory of games of strategy and its applications to many fields including economics, military, business, and operations research. Two-Person Game Theory by Anatol Rapoport Clear, accessible treatment of mathematical models for resolving conflicts in politics, economics, war, business, and social relationships. Topics include strategy, game tree and game matrix, and much more. Minimal math background required. 1970 edition. Products in Game Theory Analytical Methods of Optimization by D. F. Lawden Suitable for advanced undergraduates and graduate students, this text surveys the classical theory of the calculus of variations. Topics include static systems, control systems, additional constraints, the Hamilton-Jacobi equation, and the accessory optimization problem. 1975 edition. Our Price:$11.95 Building Models by Games by Wilfrid Hodges This volume covers basic model theory and examines such algebraic applications as completeness for Magidor-Malitz quantifiers, Shelah's recent and sophisticated omitting types theorem for L(Q), and applications to Boolean algebras. Over 160 exercises. 1985 edition. Differential Forms by Henri Cartan The famous mathematician addresses both pure and applied branches of mathematics in a book equally essential as a text, reference, or a brilliant mathematical exercise. "Superb." — Mathematical Review. 1971 edition. Our Price:$14.95 Differential Games by Avner Friedman Graduate-level text surveys games of fixed duration, games of pursuit and evasion, the computation of saddle points, games of survival, games with restricted phase coordinates, and N-person games. 1971 edition. Differential Geometry by William C. Graustein This first course in differential geometry presents the fundamentals of the metric differential geometry of curves and surfaces in a Euclidean space of 3 dimensions, using vector notation and technique. Nearly 200 problems.1935 edition. Game Theory and Politics by Steven J. Brams Many illuminating and instructive examples of the applications of game theoretic models to problems in political science appear in this volume, which requires minimal mathematical background. 1975 edition. 24 figures. Our Price:$17.95Introduction to Stochastic Models: Second Edition by Roe Goodman Newly revised by the author, this undergraduate-level text introduces the mathematical theory of probability and stochastic processes. Features worked examples as well as exercises and solutions. Our Price:$21.95Our Price:$19.95Our Price:$9.95 The Mathematics of Games of Strategy by Melvin Dresher This text offers an exceptionally clear presentation of the mathematical theory of games of strategy and its applications to many fields including economics, military, business, and operations research. Our Price:$10.95 N-Person Game Theory: Concepts and Applications by Anatol Rapoport In this sequel to Two-Person Game Theory, the author introduces the necessary mathematical notation (mainly set theory), presents basic concepts, discusses a variety of models, and provides applications to social situations. 1970 edition. Our Price:$22.95 Nonlinear Potential Theory of Degenerate Elliptic Equations by Juha Heinonen, Tero Kilpeläinen, Olli Martio A self-contained treatment appropriate for advanced undergraduates and graduate students, this text offers a detailed development of the necessary background for its survey of the nonlinear potential theory of superharmonic functions. 1993 edition.
ARKANSAS SCHOOL FOR MATHEMATICS, SCIENCES AND THE ARTS Fall & Spring 2008-2009 MATH MODELING Instructor: Bruce Turkal…….classroom 2407……..ext 5496 Office 2525 th Class Meeting Time: 6 period Prerequisites: Texts: Precalculus none required Students will build confidence in modeling functions from linear to trigonometric. They will use mathematical skills thru precalculus to find solutions for various problems in which the solution may not be unique. Estimation and approximation techniques will be critical in many problems. Students are traditionally taught mathematics by being shown examples and asking them to reproduce what they have been taught. Real application of mathematics occurs when one is confronted with an unfamiliar problem and the student is required to create his own roadmap on the way to a solution. Many times they have hard decisions to make in order to find the proper direction. Teaching techniques in the math modeling class will be used that promote independent creative thought. Learning to simplify complex problems through assumption and estimation play a pivotal role. The COMAP High School Math Modeling competition is a required activity for the class. Discussion, preparation, and participation in the contest in integral to the course. The course will cover the following areas: I. Building Confidence with Estimations & Approximations. a. Volume estimates b. Lake Construction Activity c. Scud Missile Deployment Use Numerical Techniques to find Approximate Solutions. a. Develop sum(seq( technique b. Baseball Park Fencing Problem c. Bridge Expansion II. 1 III. Build Functions to Represent Models a. Piping problem b. Kite c. Big Ben Clock d. Paper Roll e. Sprinkler Design f. Jeep in the Desert g. Car Rebate h. Cuba Fireworks i. Sonic Blast j. String Around Earth Internet Activites a. Great Circle Routes b. GPS c. Coriolis Effect Comap Practice Problems Discrete Topics a. Venn Diagrams b. Vertex Coloring c. Game Theory IV. V. Vl. Course Format: The daily classroom format is student driven activities. Problems are presented to the students and they work in groups finding solutions. They frequently are working at the blackboards showing their completed work. See ASMSA Student Handbook for attendance, late work, makeup work policies and Inclement Weather Policy. Grading/Evaluation: Major tests will count 100 points each with usually one per grading period. Class participation is the major basis for grading in the class. Weekly 100 points are assessed for each student on the basis of how they worked in class and with their groups. Most work is done in class with only limited outside work. Those not understanding the in class activities may be required to do outside work
Davie, FL Statistics diffe... ...They will gain a thorough introduction to functions, the basis of all of algebra and higher mathematics, such as calculus. Students will learn how to solve linear equations, including multistep equations, equations with multiple variables and equations involving decimals, as well as write a line
The Basics of Computer Arithmetic Made Enjoyable and Accessible-with a Special Program Included for Hands-on Learning "The combination of this book and its associated virtual computer is fantastic! Experience over the last fifty years has shown me that there's only one way to truly understand how computers work; and that is to learn one computer and... more... Presents the research in design and analysis of algorithms, computational optimization, heuristic search and learning, modeling languages, parallel and distribution computing, simulation, computational logic and visualization. This book emphasizes a variety of novel applications in the interface of CS, AI, and OR/MS. more... Aims to reinforce the interface between physical sciences, theoretical computer science, and discrete mathematics. This book assembles theoretical physicists and specialists of theoretical informatics and discrete mathematics in order to learn about developments in cryptography, algorithmics, and more. more... Real-world problems and modern optimization techniques to solve them Here, a team of international experts brings together core ideas for solving complex problems in optimization across a wide variety of real-world settings, including computer science, engineering, transportation, telecommunications, and bioinformatics. Part One—covers methodologies... more... About the Book: The book `Fundamental Approach to Discrete Mathematics` is a required part of pursuing a computer science degree at most universities. It provides in-depth knowledge to the subject for beginners and stimulates further interest in the topic. The salient features of this book include: Strong coverage of key topics involving recurrence... more... The field of discrete calculus, also known as 'discrete exterior calculus', focuses on finding a proper set of definitions and differential operators that make it possible to operate the machinery of multivariate calculus on a finite, discrete space. In contrast to traditional goals of finding an accurate discretization of conventional multivariate... more... A Complete Introduction to probability AND its computer Science Applications USING R Probability with R serves as a comprehensive and introductory book on probability with an emphasis on computing-related applications. Real examples show how probability can be used in practical situations, and the freely available and downloadable statistical programming... more... Discrete mathematics and theoretical computer science are closely linked research areas with strong impacts on applications and various other scientific disciplines. Both fields deeply cross fertilize each other. One of the persons who particularly contributed to building bridges between these and many other areas is Laszlo Lovasz, a scholar whose
Algebra E-Course and Homework Information Formulas and Cheat Sheets An Online Algebra Class that Guarantees Your Success In Algebra 1 An affordable online Algebra Class for students who are struggling with Algebra 1 Is this a picture of you as you sit down to complete your Algebra work? Do you feel your blood pressure start to rise even before you turn to the correct page in your book? Do you end up crumpling your paper and feel like you're about to throw your Algebra book out of the window? If you answered "YES" to all of these questions, then join the club! That's right - you are NOT the only one feeling this way! The number one subject in school that causes anxiety in students is Math! It doesn't matter whether you are home schooled, attend public or private school, or whether you are attending college - the answer is always the same..... Math! Like Us on Facebook Recommend Algebra Class Algebra Class E-course Members Check It Out Only $9.99 per month or $49.99 per year Who Should Take This Online Algebra Class? This online algebra class was created with "Any" Algebra student in mind. It's perfect for the home school student who is looking for an easy to understand curriculum that allows for independent study. It's also perfect for a middle or high school student who needs extra help with their Algebra class. Many teachers and tutors use it for students who are falling behind in class and I also have many adults purchase the curriculum so that they can brush up on their skills before taking an introductory Algebra class in college. How Does The Online Algebra Class Work? You will log in to the E-course from any computer that you choose. You will start by watching a video lesson. Each video explains the concept in "bite size pieces". Through several examples, you are guided step by step through the process. You are also given a pre-designed "notes worksheet" to take notes on as you watch the video. In this way, you'll have notes to refer back to when completing your practice problems. You will then have ample practice problems to complete in order to fully master each skill. Each set of practice problems ends with a small formative assessment to make sure that you've mastered the skill. The BEST part is that you will have step-by-step solutions to every single problem! You will never have to spend time trying to figure out where you made your mistake. This allows you to learn from your mistakes and to fully understand the concept. If you are using this online algebra class as your only curriculum (for home schooling or teaching purposes), I've also included chapter quizzes, tests, and a mid-term and final exam. (Yes - with step-by-step solutions!) It Really Does Work... Take a look at this email that I received from a struggling 38 year old mother of five. Dear Karin, I am a 38 year old mother of five who has never done mathematics in high school. I am considering a career change so I need mathematics desperately. In my attempts to learn high school maths, I had tried several programs from the internet. I had bought at least three maths programs locally. Then I Googled something on equations and came across your web page. As I have stated earlier in the previous email, it is the best! I have not seen anything that comes close to your material. You have taken away my frustration and made learning maths very easy and fun for me. For me, your course explains even better than one of the high school teachers whose services I had acquired to help with my studies. I literally came from one of his classes crying, thinking that I had become so dumb that I could not comprehend a thing in what he was trying to explain. Until I used your material, I felt that mathematics was not for "dummies" like me. Since I started using your program less than 4 weeks ago, I already feel like one of the Professors in Mathematics. I am learning from home now without any maths tutor. I love maths. You are a great teacher. Thank you very much. Regards, Ntombifuthi So... What Units Will You Study? Below is a list of all the units and the lessons contained in the units. If you are using this as a curriculum, you can see that this Algebra Course contains a very thorough study of Algebra. If you are using this as a supplement to your studies, you will find many individual lessons guaranteed to suit your needs. Pre-algebra Review Integers Algebraic Expressions Order of Operations Like Terms and Distributive Property Distributive Property Intro to Matrices Using Formulas Solving Equations Intro to Equations One-Step Equations (4 lessons) Mixed Review Practice Two-Step Equations Distributive Property Equations Equations with Fractions Literal Equations Variables on Both Sides Word Problems Absolute Value Absolute Value Pt 2 Graphing Equations Graphing Points Table of Values Calculating Slope Graphing Slope Slope Intercept Form Finding Slope Given 2 Points Rate of Change Standard Form Equations (1) Standard Form Equations (2) Graphing Absolute Value Eq Writing Equations Slope Intercept Form Standard Form Word Problems Slope and a Point Two Points Point-slope Form Parallel and Perpendicular Lines Line of Best Fit Cumulative Test on Units 1-3 Systems of Equations Graphing Systems Substitution Method Linear Combinations Method Addition & Multiplication Methods Systems Word Problems Systems in Three Variables Study of Matrices Intro to Matrices Operations with Matrices Multiplying Matrices Determinants Cramer's Rule Identify and Inverse Matrices Solving Systems with Matrices Inequalities Intro to Inequalities Solving Inequalities in 1 Variable More Solving Inequalities Inequality Word Problems Sets Compound Inequalities Absolute Value Inequalities Graphing Linear Inequalities More Graphing Inequalities Absolute Value Inequalities Systems of Inequalities Relations and Functions Relations Identifying Functions Evaluating Functions Domain and Range Linear Functions Linear Functions (2) Quadratic Functions Quadratic Functions (2) One More Look at Functions Exponents and Monomials Review of Exponents Compound Interest Laws of Exponents Multiplying Monomials Dividing Monomials Complex Expressions Zero and Negative Exponents Scientific Notation Polynomials Adding Polynomials Subtracting Polynomials Multiplying Polynomials Using FOIL Special Binomials More Multiplying Polynomials Factoring Polynomials Introduction to Factoring Using the GCF Factoring Trinomials Factoring Special Trinomials More Factoring Trinomials Cumulative Test on Polynomials and Factoring Quadratic Equations Square Roots Simple Quadratic Equations Pythagorean Theorem Solving By Factoring More Factoring Graphing Quadratic Equations Quadratic Formula Discriminants Discriminants When Graphing Problem Solving You'll have instant access to all tweleve units of the Algebra Class curriculum, plus a mid-term and final exam! Sign up Now for only $9.99 per month OR Save more than 60% and subscribe for a Whole Year for just $49.99 (Plus you can sign up risk free. Should you not be happy at any time within the next 30 days, just send me an email and I will refund your payment promptly.) That's right, all you have to do is choose a username and password and you will have instant access to the entire Algebra Class curriculum! There is absolutely nothing that you have to download! It can't get much easier! Please Note: You will not receive an actual book in the mail. This is an online course, and all materials can be saved or printed from your computer. You will access all worksheets and videos once you login to the secure online course. A home school mom says... Dear Karin, I am a home school mom of five who struggled with finding a solid Algebra program for my older boys until I came upon your online algebra class. Your tutorials are straight forward and detailed. Your assignments are designed in such a way that we can spend extra time on what my children need to and move on when they understand something more easily. I am a math-lover myself and am excited that you have finished Part 2 of this course. My boys are understanding Algebra in a way that they never were able to with other courses due to the detail and care that are in each of your lessons. I also appreciate that you are available to answer questions. Thank You! Jacqui Coleman Another Home school Parent Says... My daughter is a ninth grade student attending a local high school in Toronto, Ontario. I am using the online Algebra Class to help her. She was not well prepared in Pre-Algebra, hence most of the subjects in Algebra 1 are new to her. Algebra 1 is not a well taught subject. Her high school Math teacher does not use the method that you use, consequently, she is not grasping the subject matter thoroughly. Your step-by-step problem solving technique is helping her to understand the subject matter and individual topics. My daughter comments that she likes your video presentations and wishes she could see a picture of you. This e-course has helped my daughter to better understand Algebra in the following ways: The topic (e.g.. Solving Equations) is explained and every detail of one step, two step, etc... is explained in such a clear manner that sometimes she stops the video to work out the problem, and then continue. She can now distinguish between solving equations, writing equations, graphing equations, systems of equations, and word problems because of your video presentations. In subtracting polynomials, she could not grasp it until she began to follow your Keep-Change-Change procedure. We have used other Algebra products, but this differs because Karin is the only teacher who makes my daughter feel as if she is sitting beside the teacher. Thank You! Reggie Clark Every package is backed by my 30 day guarantee. I am so confident that you will find success with Algebra Class, that I will give you 30 full days to use the workbook and video tutorials. If for any reason, you are not satisfied, just contact me and I will promptly refund your money! If you have questions regarding your Algebra Class purchase, you may contact me at: 410-937-8468 or you can contact me via email and I will respond in less than 24 hours. Frequently Asked Questions What if I order and for some reason it doesn't work for me? We know that everyone has different learning styles. If you feel that you are not mastering Algebra 1, simply contact me (within 60 days of signing up) and I will refund your money promptly. (I've only had one return and that was because the customer thought this curriculum covered Algebra 2.) Will I receive the books in the mail that are shown on website? No, the books are simply a graphic. This is an E-course and all of the materials can be printed. You will login and have access to all materials. Nothing will arrive in the mail. Can I try it out before I purchase? Yes. You can sign up for the free Pre-algebra unit. It's a pre-algebra refresher and it will give you an idea of how the E-course is set up. You can sign up here. Does this program work with Mac computers? Yes, you can use a Mac computer, a PC computer, or a tablet. All videos are formatted to be viewed on all computers. What if I'm a teacher and need multiple logins for my students?Contact me and I will give you my group rates for schools or teachers. What if I have trouble logging on or need technical support? Simply contact me through the website and I will respond within 24 hours, usually must sooner. I can also be reached by phone at 410-937-8468. Do I have to use Pay Pal? No, you may use pay pal or you can use your credit card. The box below the Pay Pal payment information is the link for using your credit card. What if I want to cancel my monthly subscription? You can login to your pay pal account at any time to cancel your automatic payments. Once you login, go to Profile and then My Money then click update on My Preapproved Payments. Here you can cancel your automatic payment to Algebra Class. Is your payment processor secure? Yes! I use Pay Pal and it's very secure. You can verify this by checking your address bar on the payment page. It will start with https if the payment page is secure. Can I pay by check or money order?Yes! If you are not comfortable with using your credit card or pay pal, simply contact me and I will give you an address to send the payment. Get rid of that frustration and math anxiety today. Create your user name and password and get started instantly. Are you studying Algebra for a short period of time? You can sign up for just $9.99 per month. Or.. are you studying Algebra for the whole year? Save more than $100 by signing up for a whole year for just $49.99.
The Lone Star College–Tomball Math Center has videotapes, supplemental textbooks, manipulatives for the hands-on (kinesthetic) learner, graphing calculators, math topic handouts, and computer math programs available for students to use in the Math Center.
Tutorial fee-based software for PCs that must be downloaded to the user's computer. It covers topics from pre-algebra through pre-calculus, including trigonometry and some statistics. The software pos... More: lessons, discussions, ratings, reviews,... Teach your geometry students advanced circle concepts with this set of interactive whiteboard mini-movies. Animated flash cards bring the concepts to life while the interactive circle allows you to cr... More: lessons, discussions, ratings, reviews,... An interactive applet and associated web page that define and describe the radius of a regular polygon. The applet has a polygon where the user can change the umber of sides and alter the radiu... More: lessons, discussions, ratings, reviews,... An interactive applet and associated web page that demonstrate the radius of a circle. The applet shows a circle with a radius line. The radius endpoints are draggable and the figure changes ... More: lessons, discussions, ratings, reviews,... An interactive applet and associated web page that describe the radius of an arc and how to derive it from the width and height of the segment defined by that arc. A practical use is described fo... More: lessons, discussions, ratings, reviews,... These instructions, for using the sequence command to generate either a specific sequence, or a series of pseudo-random numbers, include questions and answers for testing students' understanding. More: lessons, discussions, ratings, reviews,... In this activity, students use simulations and graphs to explore the common-sense notion that repeatedly flipping a coin results in "heads up" about half of the time. First, students simulate
A+ National Pre-traineeship Maths and Literacy for Retail by Andrew Spencer Book Description Pre-traineeship Maths and Literacy for Retail is a write-in workbook that helps to prepare students seeking to gain a Retail Traineeship. It combines practical, real-world scenarios and terminology specifically relevant to the Retail Industry, and provides students with the mathematical skills they need to confidently pursue a career in the Retail Trade. Mirroring the format of current apprenticeship entry assessments, Pre-traineeship Maths and Literacy for Retail includes hundreds of questions to improve students' potential of gaining a successful assessment outcome of 75-80% and above. This workbook will therefore help to increase students' eligibility to obtain a Retail Traineeship. Pre-traineeship Maths and Literacy for Retail also supports and consolidates concepts that students studying VET (Vocational Educational Training) may use, as a number of VCE VET programs are also approved pre-traineeships. This workbook is also a valuable resource for older students aiming to revisit basic literacy and maths in their preparation to re-enter the workforce at the apprenticeship level. Buy A+ National Pre-traineeship Maths and Literacy for Retail book by Andrew Spencer from Australia's Online Bookstore, Boomerang Books. You might also like... Charts the history of markets in the British Isles from the small rural fairs selling harvested crops, to the teeming markets of London's Covent Garden, Smithfield and Camden. This book surveys the historic buildings and sites where these markets took place, and describes how they worked and what they sold. In Workbook 10: Promoting Safety, the retail sales associate will learn how to report safety problems in the store or department, how to follow emergency procedures, and how to maintain accurate safety records. By ensuring the store is a safe place to work and shop, the sales associate can become an even more valuable and successful employee. Suitable for people who have considered owning a retail store but just don't know where to start, how much it would cost, whether it would be profitable and, most importantly, whether they would really enjoy it. This title also includes information on small retailer trends, retailing software, and online selling. Once the social and commercial core of the rural community, the village shop has become as much the victim of the accelerating pace of social and economic change as the parish school and pub, and has now almost entirely disappeared from everyday life. This book charts the development and history of the village shop and it's slow demise. Books By Author Andrew Spencer Current approaches to morphology, Andrew Spencer argues, are flawed. He uses intermediate types of lexical relatedness in different languages to develop a morphologically-informed model of the lexical entry. He uses this to build a model of lexical relatedness consistent with paradigm-based models. A book for all morphologists and lexicographers. Helps learners' improve their Maths and English skills and help prepare for Level 1 and Level 2 Functional Skills exams. This title enables learners to improve their maths and English skills and real-life questions and scenarios are written with an automotive context to help learners find essential Maths and English theory understandable Hairdressing context beauty therapy context
an AP Calculus Student This workbook provides an excellent way to practice a variety of basic calculus problems. The disadvantage is that this book is not really tailored to the AP Calculus student. This book is aimed at college students. Finally, the problems are fairly easy to understand, but the explanations are confusing. The authors of this book use so many different variable expressions, it got me confused quickly. Was this review helpful? YesNoThank you for your feedback.Report this reviewThank you, this review has been flagged.
Prealgebra (Hardcover), 6th Edition Description Elayn Martin-Gay firmly believes that every student can succeed, and her developmental math textbooks and video resources are motivated by this belief. Prealgebra, Sixth Edition was written to help readers effectively make the transition from arithmetic to algebra. The new edition offers new resources like the Student Organizer (available separately) and now includes Student Resources in the back of the book to help students on their quest for success. Table of Contents 1. Whole Numbers 1.1 Tips for success in Mathematics 1.2 Place Value, Names for Numbers, and Reading Tables 1.3 Adding and Subtracting Whole Numbers, and Perimeter 1.4 Rounding and Estimating 1.5 Multiplying Whole Numbers and Area 1.6 Dividing Whole Numbers and Area Integrated Review – Operations on Whole Numbers 1.7 Exponents and Order of Operations 1.8 Introduction to Variables, Algebraic Expressions, and Equations 2. Integers and Introduction to Solving Equations 2.1 Introduction to Integers 2.2 Adding Integers 2.3 Subtracting Integers 2.4 Multiplying and Dividing Integers Integrated Review – Integers 2.5 Order of Operations 2.6 Solving Equations: The Addition and Multiplication Properties 3. Solving Equations and Problem Solving 3.1 Simplifying Algebraic Expressions 3.2 Solving Equations: Review of the Addition and Multiplication Properties Integrated Review - Expressions andEquations 3.3 Solving Linear Equations in One Variable 3.4 Linear Equations in One Variable and Problem Solving 4. Fractions and Mixed Numbers 4.1 Introduction to Fractions and Mixed Numbers 4.2 Factors and Simplest Form 4.3 Multiplying and Dividing Fractions 4.4 Adding and Subtracting Like Fractions, Least Common Denominator and Equivalent Fractions 4.5 Adding and Subtracting Unlike Fractions Integrated Review - Summary on Fractions and Operations on Fractions 4.6 Complex Fractions and Review of Order of Operations 4.7 Operations on Mixed Numbers 4.8 Solving Equations Containing Fractions 5. Decimals 5.1 Introduction to Decimals 5.2 Adding and Subtracting Decimals 5.3 Multiplying Decimals and Circumference of a Circle 5.4 Dividing Decimals Integrated Review - Operations on Decimals 5.5 Fractions, Decimals, and Order of Operations 5.6 Solving Equations Containing Decimals 5.7 Decimal Applications: Mean, Median, and Mode 6. Ratio, Proportion, and Triangle Applications 6.1 Ratios and Rates 6.2 Proportions Integrated Review - Ratio, Rate, and Proportions 6.3 Proportions and Problem Solving 6.4 Square Roots and the Pythagorean Theorem 6.5 Congruent and Similar Triangles 7. Percent 7.1 Percents, Decimals, and Fractions 7.2 Solving Percent Problems with Equations 7.3 Solving Percent Problems with Proportions Integrated Review - Percent and Percent Problems 7.4 Application of Percent 7.5 Percent and Problem Solving: Sales Tax, Commission, and Discount 7.6 Percent and Problem Solving: Interest 8. Graphing and Introduction to Statistics 8.1 Reading Pictographs, Bar Graphs, Histograms, and Line Graphs 8.2 Reading Circle Graphs 8.3 The Rectangular Coordinate System and Paired Data Integrated Review - Reading Graphs 8.4 Graphing Linear Equations in Two Variables 8.5 Counting and Introduction to Probability 9. Geometry and Measurement 9.1 Lines and Angles 9.2 Perimeter 9.3 Area, Volume, and Surface Area Integrated Review - Geometry Concepts 9.4 Linear Measurement 9.5 Weight and Mass 9.6 Capacity 9.7 Temperature and Conversions between the U.S. and Metric Systems 10. Exponents and Polynomials 101. Adding and Subtracting Polynomials 10.2 Multiplication Properties of Exponents Integrated Review - Operations on Polynomials 10.3 Multiplying Polynomials 10.4 Introduction to Factoring Polynomials Appendices Appendix A. Tables 1. Geometric Figures 2. Percents, Decimals, and Fraction Equivalents 3. Finding Common Percents of a Number 4. Squares and Square Roots Appendix B. Quotient Rule and Negative Exponents Appendix C. Scientific Notation Appendix D. Geometric Formulas Student Resources Study Skills Builders Bigger Picture—Study Guide Outline Practice Final Exam Answers to Selected Exercises Solutions to Selected Exercises This title is also sold in the various packages listed below. Before purchasing one of these packages, speak with your professor about which one will help you be successful in your course.
The prediction of college mathematics grades has been widely researched. Results from other studies have shown that the prediction of college mathematics grades can vary depending on the set of predictor variables considered. For instance, some
Book Description: This is a well-balanced introduction to topology that stresses geometric aspects. Focusing on historical background and visual interpretation of results, it emphasizes spaces with few dimensions, where visualization is possible, and interaction with combinatorial group theory via the fundamental group. It also present algorithms for topological problems. Most of the results and proofs are known, but some have been simplified or placed in a new perspective. Over 300 illustrations, many interesting exercises, and challenging open problems are included. New in this edition is a chapter on unsolvable problems, which includes the first textbook proof that the main problem of topology, the homeomorphism problem, is unsolvable
A Simple Math Dictionary PDF (Acrobat) Document File Be sure that you have an application to open this file type before downloading and/or purchasing. 1.61 MB \| 33 pages PRODUCT DESCRIPTION A critical and often overlooked part of any math program is the use of proper mathematical language. Did you know that by 5th grade there are 53 geometry terms your students need to comprehend? This 31 page math dictionary uses easy and clear definitions as well as formulas and examples so that students can learn and understand new math words without difficulty or cumbersome language. Most definitions include diagrams and/or illustrations for the visual learner. Over 300 common math terms are organized alphabetically for quick reference. When students are uncertain of a word or its meaning, this simple and straightforward dictionary is a good starting place. Make copies for your students so that they can place it in their math folder at the beginning of the year. It is appropriate for students in grades 4-10. Since it covers most mathematics vocabulary, it is aligned with the CCSS. You are welcome. If you find a word missing, let me know. I'm always updating this product 12, 2012, Buyer said: Love this! I keep it on my thumb drive and it's always ready to pull up and use when my GED students have a question about math terminology. Only one disappointment - I see that it was included in the free items last week, and I had already purchased it! It is a great resource - thanks! Oh my goodness! I've hit the Jackpot today, for sure. This is so thorough and clear... I'm sure it's going to set me on the road to curing my "math phobia" for sure! Will definitely come in handy when tutoring, and providing extra help to kids who are just using my room to catch up on other subjects! THANKS! I currently teach remedial math students on the college level. I found their textbook was lacking on definitions that weren't written in "mathtese" so that is why I took the time to write this dictionary. All my students keep it in their math folders so it easily accessible. I wish I had been given something like this when I was in school. The language is so simple, no matheze. The illustrations are uncluttered and compliment the definitions perfectly. The best thing I have ever seen. So easy to use. Great job Scipi. Product Questions & Answers Be the first to ask Scipi a question about this product. Please log in to do that. $6.75 Digital Download PRODUCT LICENSING3.38.
Other Formats Product Description Make geometry concepts appealing to students with different learning styles-especially visual learners-with these graphical representation strategies! Enriches any geometry class by focusing on the concept as well as the skill. Includes a variety of graphic tools Correlates to the new Principles and Stanard for School Mathematics Reproducible teacher book Grades 6-10
MISCELLANEOUS INFORMATION Preface This handbook has been prepared for the use of mathematics majors at Oberlin College. It contains suggestions for planning a viable major and for starting a career, as well as information about the mathematics program and staff that will be of interest to students. However, no booklet is a substitute for individual advising. The Department strongly encourages students to seek out faculty members on an individual basis to discuss careers and plans. These discussions need not and should not be limited to your advisor; all members of the Department will be happy to discuss the program, as well as career options in mathematics, with anyone who cares to inquire. One of the reasons for coming to Oberlin is the personal attention available from faculty members; students are encouraged to take advantage of this opportunity. INTRODUCTION What Is Mathematics? Mathematics is the study of structure and the way it can be applied to solve specific problems. The mathematics one sees in high school and the first year or so of college––techniques for solving equations, trigonometry, analytic geometry and calculus––represents only a small corner of the discipline. Some of the structures discussed in more advanced courses include algebraic systems such as groups and vector spaces, geometric notions such as surfaces, manifolds and topological spaces, and spaces of differentiable or integrable functions. Such structures are used to construct mathematical models that may explain and predict events in a wide variety of disciplines. Mathematics has been with us since antiquity and is a pervasive force in our society; it is a diverse field encompassing many subjects that are, unfortunately, largely unknown outside mathematics. The major branches of mathematics and a few of their applications are briefly described below. Pure mathematics has as its main purpose the search for a deeper understanding of mathematics itself. As a result pure mathematics seems at first far removed from everyday life. However, many important applications have been the results of advances in pure mathematics. This subject is traditionally divided into four main areas: algebra, analysis, geometry, and logic. However, some of the most exciting developments throughout the history of mathematics have resulted from the interaction between different areas. Algebra is the study of abstract mathematical systems. These systems, such as groups, rings, and fields, generalize properties of familiar structures such as integers, polynomials, and matrices. The general, abstract approach of algebra has been fruitful in solving many problems in both mathematics and other disciplines. For example, using algebraic techniques, one can show that it is impossible to trisect an angle using only a straight-edge and compass. Group theory has been employed liberally in quantum mechanical physics and physical chemistry. Other recent applications of algebra include cryptography and coding theory. Analysis is the study of infinite processes. As such, it concerns itself with phenomena that are continuous as opposed to discrete. Starting with the fundamental notions of function and limit, it builds differential and integral calculus, which is the mathematics of continuous change. Analysis in turn gives rise to a deeper and more general study of functions of both real and complex variables. The area is pervasive, and it finds rich and varied applications in almost every field of pure and applied mathematics. Geometry is the study of curves, surfaces, their higher-dimensional analogues, and the properties they possess under various types of transformations. Geometry and topology frequently make use of techniques and notions from algebra and analysis. Geometry is an important subject for our understanding of the nature and structure of spatial relations. Logic is at the very foundation of mathematics. In this field one studies the formulation of mathematical statements, the meaning and nature of mathematical truth and proof, and what can possibly be proved in a mathematical system. For example, there is a famous theorem due to Kurt Goedel that says that in any logical system rich enough to contain arithmetic there are true statements that can neither be proved nor disproved. Logic has found many important applications in the study of computability in computer science. Applied Mathematics Applied mathematics is the development and use of mathematical concepts and techniques to solve problems in many other disciplines. Unlike pure mathematics, the areas of applied mathematics fall under no simple classification. Nonetheless, the following topics cover many of the important applications. Applied analysis involves the study of techniques for analyzing continuous processes and phenomena. For example, many methods from real and complex analysis are utilized when looking at problems of a physical or computational nature. Differential equations and numerical analysis are two examples of subjects that come under this heading. Combinatorics, the study and enumeration of patterns and configurations, is one technique for analyzing phenomena that do not behave in a smooth or continuous fashion. Techniques from algebra and other areas are applied to study a wide variety of problems in such areas as graph theory, scheduling, and game theory. There are also many important applications of combinatorics to computer science. Probability and statistics are among the most fundamental tools for mathematical modeling. The importance of probability lies in its formulation of chance (or stochastic) processes and its applicability to the analysis of non-deterministic phenomena. Statistics deals with the collection and analysis of data and with the making of decisions in the face of uncertainty. It is frequently used in the social sciences as well as in all areas of experimental science. Operations research involves applications of mathematical models and the scientific method to help organizations or individuals to make complex decisions. Traditionally, it has focused on mathematical optimization theory. Typical problems in operations research include the development of an optimal flight schedule for an airline and finding the best inventory policy for a bookstore. Actuarial mathematics is one of the early examples of mathematical modeling. This field uses the methods of probability and statistics, along with a study of economic factors, to estimate the financial risk of future events. Career Opportunities Many people believe that the only career open to a mathematics major is teaching. This is not the case, as our graduates with jobs in industry and government can certainly attest. Modern society has become intensely technological and, as a result, there are jobs awaiting anyone with solid training in mathematics. Add to a strong mathematics program a good background in the arts and sciences, which any diploma from Oberlin should imply, and you will have a very marketable undergraduate degree. Studies have shown that the starting salaries of undergraduate mathematics majors are only slightly lower than those of engineers and are considerably higher than those of business, economics and accounting majors (see, for example, the report on the College Placement Council's salary survey in the Notices of the American Mathematical Society, November, 1984). Oberlin mathematics graduates in recent years have gone on to careers in a variety of business and industrial settings; others have entered graduate programs in medicine, law, economics and engineering, as well as in fields more closely related to mathematics. Many career paths are open to a mathematics major, only a few of which are mentioned below. For further information you should speak with one of the department faculty or consult the pamphlet "Professional Opportunities in the Mathematical Sciences", which is published by the Mathematical Association of America. Research in pure mathematics is conducted primarily by college and university faculty. If a career in research mathematics and teaching at the college level appeals to you, you will need to attend graduate school and ultimately earn a Ph.D. in mathematics. An Oberlin degree is a good first step in this process. In fact, during the period 1920–1990, Oberlin produced more graduates who have gone on to earn Ph.D.'s in mathematics than any other primarily undergraduate institution in the United States. The nature of research is such that it is vital for one to have a thorough background in theoretical and foundational areas. Thus, even if you want to pursue research in an applied area such as statistics or operations research, you will best prepare for graduate school by taking as much pure mathematics as you can during your undergraduate years. Industry and Government Career possibilities in industry include employment in scientific and engineering firms, as well as work in financial and management companies. In addition to a career in the private sector, there is also the possibility of government work. For a mathematics graduate, there are opportunities at government laboratories (such as that in Los Alamos, NM), federal research centers, (e.g., the Center for Naval Analyses), and the Federal Reserve Bank, to name a few. Mathematics students are highly regarded by industry and government for their general problem-solving abilities, their analytical training, and their ability to think abstractly. Most companies are willing to provide some specific training on the job. In fact, some companies even prefer to do this, for then you may be more thoroughly indoctrinated into the company's standard methodologies. Computer experience is very desirable for such a career, and some training in an area to which mathematics can be applied is useful. The best preparation, however, is a solid program in both pure and applied mathematics. Actuarial Work Actuaries evaluate the current financial implications of future contingent events. Sometimes called "social mathematicians", they project the financial effects that various human events--birth, sickness, accident, retirement, and death--have on insurance and other programs. Salaries are high in this profession and successful actuaries are in constant demand. Professional advancement in this field comes through passing a series of examinations administered by the Society of Actuaries and the Casualty Actuarial Society. The first of these examinations covers the calculus of one and several variables, differential equations, and probability. Questions on the examination are presented primarily in the context of risk management and insurance. With some study, any mathematics major should be able to pass this examination and thus become eligible for both summer jobs and an entry-level position upon graduation. Pre-Collegiate Education There is a very real and disturbing shortage of qualified mathematics teachers and mathematics supervisors in the nation's elementary and secondary schools. Oberlin graduates typically have both the academic and the personal strengths to make a tremendous contribution in this field. To teach in the public schools, one must be certified; this means approximately a year of courses in mathematics education and practice teaching. Oberlin entrants into the teaching field typically take this program at the post-graduate level, obtaining a master's degree in mathematics and/or education along with certification. A post-graduate year, or more, is strongly recommended as suitable grounding for the kind of leadership that Oberlin graduates can expect eventually to exert in the field. Due to the above-mentioned personnel shortage, generous fellowships are often available. Work with school students within the mathematics program in the Oberlin City Schools is also available, usually for credit. Another option is that of beginning teaching immediately upon graduation in a private school, for which certification is not required. One might plan on teaching for only a few years in a private school before going on with the next stage of one's career, or on staying with it indefinitely (in which case some post-graduate work will soon become desirable). There are several placement services for teaching positions in private schools. With the help of these services, Oberlin students have been very successful in finding placements. For additional information, see a member of the Department, or consult the Office of Career Services. A Note for Double Majors We shouldn't leave this section on career opportunities without mentioning what an attractive package a double major in mathematics can be. At Oberlin, many students have taken double majors in mathematics and either physics or economics, but these are not the only possibilities. For example, biomathematics and biostatistics are fields which apply mathematical techniques to the study of biological systems. Good undergraduate preparation for work in such areas would naturally include study in both biology and mathematics. Rigorous mathematical techniques have only recently begun to be employed in some of the social sciences, so such a double major might be very attractive to a prospective employer or graduate school. Many other combinations are possible. We have had double majors with branches in almost every department in the college. In addition, many students have completed double degrees in mathematics (in the College of Arts and Sciences) and music (in the Conservatory). The mathematics major at Oberlin gives you the opportunity to design a strong and personal program of study. To take full advantage of this opportunity, you should start planning your own program early in your college career. Examinations for Graduate Schools and Careers You should be aware of some of the standardized tests that you may have to take, so that you can plan your academic program in order to take these examinations at the appropriate times. For example, if you are considering graduate study, you should take the Graduate Record Examination (GRE) by the fall of your senior year. There are two types of GRE's: (a) the general aptitude exam, which is much like the SAT––there are verbal, analytical, and quantitative portions; and (b) the advanced exam, which is given in a wide range of academic areas. If you are planning to do graduate work in mathematics or computer science, you will normally need to take both the aptitude examination and an advanced examination. If you are interested in an actuarial career, you will want to begin taking the series of professional actuarial examinations. The first examination concerns freshman and sophomore mathematics and probability. Other standardized tests you might need to take include the MCAT (for medical school), the LSAT (for law school), and the GMAT (for business school). You should consult with your advisor or talk with a consultant in the Office of Career Services to determine exactly which examinations you need to take and when it would be best for you to take them. What Do Our Math Majors Really Do? Lots of different things. To get an idea of exactly what Oberlin mathematics majors do, the Department conducted an employment census, in the Fall of 1993, of Oberlin mathematics majors who had graduated between 5 and 15 years earlier. Here is what these graduates were doing. There were a total of 276 mathematics graduates in the classes from 1978 to 1988. An unexpectedly large number (82) of these graduates were in the financial world as consultants, analysts, actuaries and the like. A substantial subgroup (34) of these were in managerial and/or directorial positions. Their employers ranged from banks and insurance companies through industrial and computing firms including the likes of Shearson Lehman Hutton, AT & T, Bankers Trust Co., Chemical Bank, and Allstate. Beside this group, the next largest category were graduates who remained in academia either as teachers (39) or as graduate students (35). Fields taught and/or studied by these grads ranged over all the physical sciences. The largest group (naturally) consisted of mathematics teachers and students, but other subjects covered included statistics, physics, chemistry, economics, electrical engineering, biostatistics, operations research, and psychology. Another large group (37) was employed in the computer industry. Employers of this group included Microsoft, Sun Microsystems, Bell Labs, and Hewlett-Packard. Beyond these groups, graduates spread out into a remarkable variety of professions and specialties. There were doctors (5) and lawyers (11). Nine others were involved in the legal or medical professions as residents, law clerks, and medical/legal researchers. Not surprisingly, there were artists among our graduates. Most of them were musicians (16): performers and teachers, but there were also graphic artists including an architect and a freelance cartoonist. Fewer graduates than we expected (26) worked in government. These alumni included three employed by NASA, several employees of the Justice Department, and four graduates in the military. ACADEMIC PROGRAMS Intermediate and Advanced Courses The following is a list of all intermediate and advanced course offerings in mathematics. Some courses are listed under more than one heading. The list is broken down by area so that you can more easily plan your own program of study. Requirements Students sometimes confuse requirements with recommended programs. The requirements for the mathematics major were set up with the needs of double-majors and double-degree students in mind. These requirements represent a minimal level of knowledge that the Department expects of all majors, regardless of their talents or interests. Most mathematics students are strongly urged to take courses in excess of these basic requirements. The suggested programs below could serve as a guideline for designing a major in mathematics. Major in Mathematics A major in mathematics consists of at least 11 full academic courses, which must include: A. MATH 133, 134, 220, 231, and 232. B. Four 300-level mathematics (MATH) and statistics (STAT) courses, which must include MATH 301 MATH 327 A modeling course from the following list: MATH 331, 335, 338, 342, 343, 345 or 348, or STAT 336, 337. Note: One or both of the courses in item C above may replaced by a course or courses from the following list: Computer Science: CSCI 150 or 151, and any computer science course numbered 200 or above which counts toward the Computer Science major Physics & Astronomy: PHYS 301, 302, 305, 310, 311, 312, 316, 340, 410, 411–12* Chemistry: CHEM 339, 349 Economics: ECON 342, 351, 353, 355. *The module courses PHYS 411 and 412 together count as one course. The department frequently offers a 300-level seminar in addition to its regular offerings. Students should check with the instructor to find out whether the seminar can be used to fulfill requirement B.3 above. Minor in Mathematics A minor in mathematics consists of at least five full academic courses in mathematics (MATH) or statistics (STAT), including any three of MATH 220, 231, 232, and 234, and at least two courses numbered 300 and above. Recommended Programs The following programs represent suggestions for coherent sequences in mathematics. These may be used as models, but they are by no means the only ones possible. Students are strongly encouraged to work out, with the help of their advisors, the program that best fits their needs and interests. After completing the introductory calculus sequence (MATH 133–134), students who would like to major in mathematics are strongly encouraged to enroll first in Discrete Mathematics (MATH 220), Multivariable Calculus (MATH 231) and Linear Algebra (MATH 232) during their sophomore year. We also strongly urge students enroll in Discrete Mathematics before enrolling in Linear Algebra. For students with no previous background in calculus, the following 2-year sequences should be typical: Fall Spring First year MATH 133 MATH 134 Second year MATH 220 MATH 231, 232 or Fall Spring First year MATH 133 MATH 134 Second year MATH 220, 231 MATH 232 Naturally, students who enter Oberlin with some credit for calculus should place themselves at the appropriate position in this sequence, and also consider enrolling in other mathematics courses as well. Graduate School in Mathematics Students who plan to attend graduate school in any mathematical discipline should concentrate as undergraduates on core mathematics. We recommend at least one advanced course in each of these areas: algebra, analysis, and applied mathematics, and at least one additional advanced course in one of these areas. Graduate faculty in pure mathematics normally expect students to have taken a full year of both analysis and algebra, as well as courses in geometry or topology. Graduate faculty in applied fields will expect a somewhat different background, but a year of analysis, a course in algebra, as well as courses in applied mathematics (probability/statistics, operations research, etc.) are essential. All students planning graduate work should take considerably more than the required 11 mathematics courses. It is also advisable for such students to take courses in other disciplines that make use of advanced mathematics, such as physics or economics. Non-academic Careers in Mathematics Experience in computer science at least at the level of CSCI 150 is recommended for anyone planning a career in industry or government. As mathematical preparation for such a career, we recommend at least two advanced courses in pure mathematics as well as several advanced courses in applied mathematics. Among the latter the Probability course is particularly recommended, since non-deterministic models are important in almost every branch of applied mathematics. Honors Work Each spring the Department invites a few of the most talented juniors to participate in our Honors program during their senior year. Honors consists of a year of work or research in an advanced area of mathematics, under the close supervision of a member of the faculty. For some students this takes the form of two one-semester projects; for most student, however, a single year-long project is undertaken. At the end of Winter Term, Honors candidates take a written examination, administered by a mathematician from outside the College; during the spring semester they complete a paper on some aspect of their Honors project and the external examiner visits campus to conduct an oral examination on the project. The results of the exams, as well as the quality of the individual project and overall achievement in mathematics, are used to determine the level of Honors awarded. The Honors program serves two important roles. Although the Department offers a rich set of courses in pure and applied mathematics, talented students occasionally exhaust our course offerings. The Honors program allows such students to explore particular areas of mathematics in greater depth. A few recent projects have been in Chaotic Dynamical Systems Algebraic Geometry Number Theory Mathematical Logic Optimization Low-dimensional topology Computational algebra. The other important feature of the Honors program is that it presents an opportunity for close work between a faculty member and a student on a topic of interest to them both. This is valuable preparation for students interested in graduate work in either mathematics or statistics. Students are admitted into the Honors program at the invitation of the Department. The factors considered most heavily by the Department in selecting candidates are: success in course work, enthusiasm for mathematics and interest in graduate work, and a broad base of completed courses. Students interested in Honors work should normally complete a substantial number of 300-level courses, including some of the theoretical courses, by the end of their junior years. MISCELLANEOUS INFORMATION Student Research Although some areas of research in mathematics may be difficult for undergraduates to undertake, many opportunities do exist for "hands-on" experience. In this section we catalog a few possibilities for research activities that are available to qualified students. You should also feel free to talk with faculty members about opportunities beyond those mentioned below. MATH 550 and 551. There are two research courses in the Mathematics Department, MATH 550 (fall) and 551 (spring) that enable students to pursue research projects under the supervision of faculty for academic credit (either as a full or half course). Senior Scholars. The Senior Scholars program permits a few exceptional students to spend their senior years working on research rather than on course work. The Mathematics Department has only rarely had senior scholars––just two in the past forty years. Summer Research. From time to time, funds are available to support students to work with faculty at Oberlin during the summer months. Other Research. The National Science Foundation regularly sponsors undergraduate research in mathematics by funding summer work at a number of institutions through its Research Experiences for Undergraduates (REU) program. For addition information, check the NSF's website Contests and Competitions The Mathematics Department sponsors teams for two national competitions: The William Lowell Putnam Examination is administered each year by the Mathematical Association of America. There are morning and afternoon sessions, with six problems assigned per session. Each college and university in the nation is allowed to enter a team of three. (In addition, any undergraduate may participate as an individual.) The team members work separately; the score for a team is the sum of the individual members' ranks. The examination is difficult, but Oberlin has at times done quite well, placing second in the nation in 1972 and tenth in 1991. A cash prize is given to the highest scoring Oberlin student. (See The Baum Prize section below.) The Consortium for Mathematics and its Applications has recently started a national contest in mathematical modeling. The College has sponsored a team of three undergraduates. These students are given a choice of one of two unstructured problems which they must carefully formulate and attempt to solve over a weekend. We have sponsored a team for this contest each year. For more information, consult the Chair of the Department or Robert Bosch. The Orr Prize Each year the Department awards the Rebecca Cary Orr Memorial Prize in Mathematics to an outstanding graduating senior. This prize was established by the family and friends of Rebecca Orr, an Oberlin College freshman who was killed in a tragic accident in the spring of 1982 and was a promising mathematics student. The prize in her memory is awarded on the basis of outstanding achievement in undergraduate mathematics and promise of future professional accomplishment. The Baum Prize The John D. Baum Memorial Prize in Mathematics is awarded annually to the Oberlin student with the highest score on the Putnam examination. This prize was established by the Department faculty in 1988 in fond memory of a long-time colleague, mathematician, and friend. Invited Lectures From time to time throughout the academic year, the Mathematics Department sponsors lectures by outside speakers. These speakers are often well-known and very eminent mathematicians with national reputations. Speakers have included Thomas Banchoff (Brown University), Steven J. Brams (New York University), Robert Devaney (Boston University), Melvin Hochster (University of Michigan), Frank Morgan (Williams College), and Ragni Piene (University of Oslo). The Department is also a member of the Ohio Colleges Speaker's Circuit, a consortium of area colleges whose purpose is to promote contact by exchanging speakers from among their respective mathematics faculties. Distiguished Visiting Scholar. Beginning in 1995–96 and thanks to the generosity of alumni, the Department is able to sponsor an annual visit by an eminent mathematical scientist who will conduct classes as well as deliver the Fuzzy Vance Lecture (a public lecture). Tamura/Lilly Distinguished Lecture. Thanks to the generosity of Roy Tamura '78 and the Eli Lilly Company, the Department sponsors an annual visit by a mathematical scientist to deliver a public lecture. Attending lectures is an important and fun way to learn about mathematics outside of the classroom. Students at all levels are strongly encouraged to go to talks and to meet the speakers. Majors Committee Composed of undergraduate mathematics majors, the Majors Committee acts as an interface between students and the Department faculty. Its members administer course evaluations at the end of each semester, interview candidates for faculty positions in the Department, make recommendations concerning tenure decisions, and consider various student concerns as they arise. The Majors Committee also helps organize a student-staff picnic every spring. A list of the committee members may be found on the bulletin board outside the Department Office. Students with questions about the mathematics program are encouraged to contact any member of the Majors Committee, as should anyone interested in serving on the committee. Faculty Interests Below is a list of Department faculty and their mathematical interests. You might wish to use this list when planning reading courses or Winter Term projects, or if you would like more information about a particular area of mathematics.
Prerequisite: All of the topics in MA112, as well some topics in MA113, will be assumed. In addition, familiarity with some Maple commands from Calculus is assumed. The Maple files MapleIntRGL.mws (introduction) and calcrev.mws (review for DE) may be helpful. These files are available on Angel in the (Rose-Only) Mathematics Course Information Repository. Course Goals Develop a deeper understanding of scalar differential equations and their solutions. Provide an introduction to Laplace transform methods. Develop a deeper understanding and appreciation of transformation methods by studying Laplace methods. Improve mathematical modeling and analytical problem solving skills. Develop ability to communicate mathematically. Improve skill using the computer as a tool for mathematical analysis and problem solving. Introduce applications of mathematics, especially to science and engineering. Textbook and other required materials Textbook: Text: Advanced Engineering Mathematics, 4th edition by Zill and Wright Computer Usage: Maple14 must be available on your laptop. Course Requirements and Policies The following policies and requirements will apply to all sections and classes: Computer Policy A summary of the computer policy page: Students will be expected to demonstrate a minimal level of competency with a relevant computer algebra system. The computer algebra system will be an integral part of the course and will be used regularly in class work, in homework assignments and during quizzes/exams. Students will also be expected to demonstrate the ability to perform certain elementary computations by hand. (See Performance Standards below.) Performance Standards Paper and pencil Demonstrate skill using complex numbers basic arithmetic and geometric interpretation Solve linear differential equations including direct integration separate and integrate involving simple integrations integrating factors with simple integration graphing and interpreting the solutions constant coefficient, second order equations with simple driving functions compute elementary Laplace transforms from the definition and standard table. solve simple scalar differential equations using Laplace transforms Maple Use Maple to solve differential equations symbolically and numerically using dsolve numerically by using the method=numeric using Maple to assist in separate and integrate, undetermined coefficients laplace transform methods graphing and interpreting the solutions Final Exam Policy The following is an extract from the final exam policy page. Consult the policy page for complete details. The final exam will consist of two parts. The first part will be "by hands" (paper, pencil). No computing devices (calculators/computers) will be allowed during the first part of the final exam. This part of the exam will cover both computational fundamentals as well as some conceptual interpretation, though the level of difficulty and depth of conceptual interpretation must take into account that this part of the exam will be shorter than the second part of the exam. The laptop, starting with a blank Maple work sheet, and a calculator, may be used during the second part of the exam. No "cheat sheets", prepared Maple worksheets or prepared program on the calculator may be used. The second part of the exams will cover all skills: concepts, calculation, modeling, problem solving, and interpretation. Individual Instructor Policies Your instructor will determine the following for your class: the grading scheme, based on the various course components. the number and format of hour exams, quizzes, homework assignments, in class assignments, and projects, the policies governing the work items above, e.g., whether the computer will be used, what collaboration is allowed, and the format of assignments. all policies for classroom procedure, including group work, class participation, laptop use and attendance. Note that most instructors will enforce some type of grade penalty for students with more than four unexcused absences.
This Student Reference Book can be used to look up and review topics in mathematics. It has the following sections: 1) A Table of Contents that lists the sections and shows how the book is organized. 2) Essays within each section. 3) Directions on how to play some of the mathematical games you may have played before. 4) A Data Bank with posters, maps, and other information. 5) A Glossary of mathematical terms. 6) An Answer Key and 7) An Index to help you locate information quickly
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These tutorials for creating learning objects come with an overview, some basic information on its constructs and... see more These tutorials for creating learning objects come with an overview, some basic information on its constructs and interactions, and a set of guidelines. This interactive set of tools can be used by educators in several disciplines including both mathematics and the sciences to create their own customized, web-based learning aids. It is endorsed by the Mathematical Association of America's online Digital Classroom. The most popular format for publishing mathematics on the Web is the PDF, and HyperLaTeX is the program which allows us to... see more The most popular format for publishing mathematics on the Web is the PDF, and HyperLaTeX is the program which allows us to introduce hyper references in the PDF documents. This paper is a tutorial on using HyperLaTeX to produce PDF files.
Greenbrae ACT MathIn Algebra 1 we also study graphical methods in order to visualize functions as straight lines or parabolas. Further we learn about factorization and the solutions of quadratic equations. Seeing many advanced students who struggle with algebra 1 concepts makes me feel good about my algebra 1 students because I help them to learn it properly from the beginning
updated content, vivid applications, and integrated coverage of graphing utilities, the ninth edition of this hands-on trigonometry text guides readers step by step, from the right triangle to the unit-circle definitions of the trigonometric functions. Examples with matched problems illustrate almost every concept and encourage readers to be actively involved in the learning process. Key pedagogical elements, such as annotated examples, think boxes, caution warnings, and reviews, help readers comprehend and retain the material.
Calculus: Online and Interactive This online workshop is designed to acquaint participants with online interactive calculus materials being developed by the organizers as part of the NSF-funded MAA project, MathDL Books Online, NSF Grant DUE-231083. Computational and Mathematical Biology As the process of integrating mathematics and biology gains momentum, it has become clear that, now that people are convinced of the need to bring mathematics and biology together, we need to focus more on "how" to achieve this integration. Materials produced during the workshop will have the opportunity for submission to the National Science Digital Library ( through the Computational Science Reference Desk ( as a reference for biology and mathematical science faculty wishing to integrate biology, mathematics, and computer science in their classrooms. For additional information about these and other PREP workshops, or to register, visit Do you know the way to San Jose? Mark your calendar now to join us in San Jose on August 3-5 for MathFest 2007. This year, we mark the 20th anniversary of the first student paper sessions. In addition, the Euler Society will join us to help celebrate the tercentennial of Euler's birth, and the Society for Mathematical Biology will hold their meeting in conjunction with MathFest as well. MAA Regional Undergraduate Mathematics Conferences The MAA has received funding from the Division of Mathematical Sciences at the National Science Foundation to provide support for institutions or groups of institutions that wish to initiate or expand undergraduate mathematics conferences. The primary objective of the grant is to provide as many undergraduate students as possible with the opportunity to present mathematically-oriented talks and to better expand their knowledge of the wide range of theory, history, and applications of the mathematics sciences at conferences that are within their geographic region. The awards are intended to cover a portion of the typical expenses involved with hosting the conference. These include travel expenses for invited speakers, lunch and refreshments for participants, and duplicating and postage for advertising. The MAA is celebrating the 300th Anniversary of Euler's Birth as the Year of Euler Leonhard Euler, generally regarded as one of the greatest mathematicians of all times, was born in Basel, Switzerland on April 15, 1707. Euler made massive contributions to analysis, and worked in almost all branches of pure and applied mathematics. Among the many activities to mark this year are: the 2007 MAA Mathematical Study Tour is devoted to Euler. The trip, July 1-14, will visit Basel, St. Petersburg and Berlin. the MAA will release five new books dedicated to Euler during 2007. a PREP workshop based on the recently-published The Genius of Euler will be led by Bill Dunham, the volume's editor, at the MAA Carriage House, June 18-22. CRAFTY releases Guidelines for College Algebra Available at these guidelines represent the recommendations of the MAA/CUPM subcommittee, Curriculum Renewal Across the First Two Years, concerning the nature of the college algebra course that can serve as a terminal course as well as a pre-requisite to courses such as pre-calculus, statistics, business calculus, finite mathematics, and mathematics for elementary education majors. They were endorsed on January 31, 2007 by CUPM, the MAA Committee on the Undergraduate Program in Mathematics. Applications to Project NExT for the 2007-2008 Fellowship year due April 16, 2007. Project NExT (New Experiences in Teaching) is a professional development program for new or recent Ph.D.s in the mathematical sciences. It addresses all aspects of an academic career: improving the teaching and learning of mathematics, engaging in research and scholarship, and participating in professional activities. Faculty for whom the 2007-2008 academic year will be the first or second year of full-time teaching at the college/university level (after receiving the Ph.D.) are invited to apply to become Project NExT Fellows. Approximately seventy Project NExT Fellows will be selected for the 2007-2008 year. Spring Section Meetings: A great chance to connect with your colleagues Your Section meeting offers a chance to present a paper, hear from mathematicians that live close to you, and develop a stronger support network for your work. A Section meeting is also an excellent venue for undergraduate and graduate students alike to present research to a friendly audience. Celebrating Ten Years of Math Horizons The Edge of the Universe, a collection of articles from the MAA student magazine, Math Horizons, features exquisite expositions of mathematics accessible at the level of an undergraduate or advanced high school student. Broad and appealing, the coverage also includes fiction with mathematical themes; literary, theatrical and cinematic criticism; humor; history; and social history. Anyone with an interest in mathematics will delight in engaging in this volume.Beautifully printed with 24 pages of full color images. Deanna Haunsperger & Stephen Kennedy, Editors List: $57.50; MAA Members: $45.95 Spreadsheets Across the Curriculum Project invites applications The Washington Center for Improving the Quality of Undergraduate Education at Evergreen State College is now inviting applications for the third, and final, institute for the Spreadsheets Across the Curriculum Project. In this National Science Foundation funded project, faculty develop spreadsheet modules aimed at enhancing students' quantitative literacy skills within disciplinary contexts. The institute will be held July 17-20, 2007, at the Phoenix Inn in Olympia, WA. The project covers the cost of participating in the institute (meals and lodging), but does not cover travel expenses.
Hygiene AlgebraThen extend your skills to thinking mathematically and further developing organized problem solving skills. Calculus is the pathway to solutions to problems you can solve in no other way. Without it, we wouldn't have many of the modern conveniences and technologies and other achievements we take for granted in today's world.
MATH 201-001 MATH 201-002
Prealgebra (Cloth) - 6th edition Summary: Elayn Martin-Gay firmly believes that every student can succeed, and her developmental math textbooks and video resources are motivated by this belief. ''Prealgebra,'' Sixth Edition was written to help students effectively make the transition from arithmetic to algebra. The new edition offers new resources like the Student Organizer and now includes Student Resources in the back of the book to help students on their quest for success. Whole Numbers and Introduction to Algebra; Intege...show morers and Introduction to Solving Equations; Solving Equations and Problem Solving; Fractions and Mixed Numbers; Decimals; Ratio, Proportion, and Triangle Applications; Percent; Graphing and Introduction to Statistics; Geometry and Measurement; Exponents and Polynomials For all readers interested in prealgebra
... More About This Book providing an original treatment of fractals that is at once accessible to beginners and sufficiently rigorous for serious mathematicians. The workis designed to give young, non-specialist mathematiciansa solid foundation in the theory of fractals, and, in the process, to equip them with exposure to a variety of geometric, analytical, and algebraic tools with applications across other areas. Product Details Table of Contents Introduction.- Part 1. The Sierpiński Gasket.- Definition and General Properties.- The Laplace Operator on the Sierpiński Gasket.-Harmonic Functions on the Sierpiński Gasket.- Part 2. The Apollonian Gasket.- Introduction.- Circles and Disks on Spheres.- Definition of the Apollonian Gasket.- Arithmetic Properties of Apollonian Gaskets.- Geometric and Group-Theoretic Approach.- Many-Dimensional Apollonian Gaskets.-
Elementary Algebra For College Students Early Graphing 9780136134169 ISBN: 0136134165 Edition: 3 Pub Date: 2007 Publisher: Prentice Hall Algebra Early Graphing for College Students, An...gel continues to focus on the needs of the students taking this class and the instructors teaching them. Angel, Allen R. is the author of Elementary Algebra For College Students Early Graphing, published 2007 under ISBN 9780136134169 and 0136134165. Three hundred seven Elementary Algebra For College Students Early Graphing textbooks are available for sale on ValoreBooks.com, twenty eight used from the cheapest price of $22.76, or buy new starting at $93.99.[read more
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Books Geometry & TopologyNeed help with Geometry? Designed to replicate the services of a skilled private tutor, the new and improved Tutor in a Book's Geometry is at your service! TIB's Geometry is an extremely thorough, teen tested and effective geometry tutorial. TIB's Geometry includes more than 500 of the right, well-illustrated, carefully worked out and explained proofs and problems. Throughout TIB's Geometry, there is ongoing, specific guidance as to the most effective solution and test taking strategies. Recurring patterns, which provide solutions to proofs, are pointed out, explained and illustrated using the visual aids that students find so helpful. Also included are dozens of graphic organizers, which help students understand, remember and recognize the connections between concepts. TIB's author Jo Greig intended this book to level the playing field between the students who have tutors and those that don't. As a long time, very successful private mathematics tutor and teacher, Jo Greig knew exactly how best to accomplish this! TIB's Geometry 294 pages are packed with every explanation, drawing, hint and memory tool possible! Not only does it have examples of the right proofs and problems, it also manages to impart every bit of the enthusiasm that great tutors impart to their private tutoring students. Ms. Greig holds a bachelors' degree in mathematics. Dr. J. Shiletto, the book's mathematics editor, holds a Ph.D in mathematics. Each page in Common Core Math Workouts for grade 6 configurations involving infinity. The result is a delightful and informative illustrated tour through the 2,500-year-old history of one of the most important and beautiful branches of mathematics. Each page in Common Core Math Workouts for grade 7 &The classic Heath translation, in a completely new layout with plenty of space and generous margins. An affordable but sturdy student and teacher sewn softcover edition in one volume, with minimal notes and a new index/glossary. demonstrating
approach in education of the course Digital and Analog Circuits, which belongs to curricula of the Informatics study program. For analysis of analog and simple digital circuits we use computer algebra system Mathematica, which minimizes the amount of routine, handy calculations. This fact enables focusing on the problem and solving more examples, which in turn provides better comprehension of the topic. As Mathematica is later used in subsequent courses, its knowledge is utilized in these courses. Last, but not least, Mathematica provides several programming paradigms, which can be easy demonstrated to students of Informatics study program.
More About This Textbook Overview To understand the output from a geographic information system, one must understand the quality of the data that is entered into the system, the algorithms driving the data processing, and the limitations of the graphic displays. Introduction to Mathematical Techniques Used in GIS explains to nonmathematicians the fundamentals that support the manipulation and display of geographic information. It focuses on basic mathematical techniques, building upon a series of steps that enable a deeper understanding of the complex forms of manipulation that arise in the handling of spatially related data. The book moves rapidly through a wide range of data transformations, outlining the techniques involved. Many are precise, building logically on underlying assumptions. Others are based upon statistical analysis and the pursuit of the optimum rather than the perfect and definite solution. By understanding the mathematics behind the gathering, processing, and display of information, GIS professionals can advise others on the integrity of results, the quality of the information, and the safety of using it. Table of Contents NUMBERS AND NUMERICAL ANALYSIS The Rules of Arithmetic The Binary System Square Roots Indices and Logarithms ALGEBRA-TREATING NUMBERS AS SYMBOLS The Theorem of Pythagoras The Equations for Intersecting Lines Points in Polygons The Equation for a Plane Further Algebraic Equations Functions and Graphs Interpolating Intermediate Values THE GEOMETRY OF COMMON SHAPES Triangles and Circles Areas of Triangles Centres of a Triangle Polygons The Sphere and the Ellipse Sections of a C
Algebra and Trigonometry - With CD - 2nd edition Summary: This best selling author team explains concepts simply and clearly, without glossing over difficult points. Problem solving and mathematical modeling are introduced early and reinforced throughout, so that when students finish the course, they have a solid foundation in the principles of mathematical thinking. This comprehensive, evenly paced book provides complete coverage of the function concept and integrates substantial graphing calculator materials that help stu...show moredents develop insight into mathematical ideas. The authors' attention to detail and clarity, as in James Stewart's market-leading Calculus text, is what makes this text the market leader449.95 +$3.99 s/h Good Savannah Goodwill Savannah, GA Good New comment. $9.95 +$3.99 s/h Good Goodwill Savannah Savannah, GA new comment
The author offers reflections on specific questions mathematicians and philosophers have asked about the infinite over the... see more The author offers reflections on specific questions mathematicians and philosophers have asked about the infinite over the centuries. He examines why explorers of the infinite, even in its strictly mathematical forms, often find it to be sublime. This site provides access to online course notes for an introductory algebra course. The course covers... see more This site provides access to online course notes for an introductory algebra course. The course covers fractions, radicals, exponents, algebraic expressions and simplification, solving system of linear equatation using Gaussian Elimination Method, and solving quadratic equations geometrically. The interactive features of the notes include: The ability to send questions to the author/course instructor and to view answers to questions sent by other students in the context of online note pages. Solve practice problems online. This site provides a brief textual overview of the field of cryptography and related issues including popular techniques,... see more This site provides a brief textual overview of the field of cryptography and related issues including popular techniques, applications and standards. It is part of a larger body of information provided by RSA Laboratories, a division of RSA Security. Discussion of how to calculate linear regression, including a demonstration applet. This site is in Spanish, but standard... see more Discussion of how to calculate linear regression, including a demonstration applet. This site is in Spanish, but standard Spanish to English web site translators make it easily usable for English speakers.
Mathematical Background/Clarifying Examples: The focus should be on the multiple representations of trigonometric expressions. Students should be able to demonstrate, using a table of values or a technologically-produced graph, that two trigonometric expressions are equivalent. Website Links: 1. Sum and Difference Formulas: This website provides a list of the Sum and Difference Formulas. Sum and Difference Formulas d. Use the double-angle and half-angle identities. Mathematical Background/Clarifying Examples: Half-angle Identities and Double-Angle Identities should be used to find exact values. Double-Angle Identities, but not Half-Angle Identities, should also be used to verify other identities. Half-Angle Identities may be used in proofs as a possible extension. Mathematical Background/Clarifying Examples: Students should be able to use previously learned identities in order to solve trigonometric equations. It is essential that students are comfortable with the identities so that they are able to determine reasonable identities to use for given equations. Mathematical Background/Clarifying Examples: Students should be able to manipulate trigonometric equations using various algebraic methods such as factoring in order to arrive at solutions. It is important to use graphing and tables of values as means to verify solutions to trigonometric equations.
DPLS Scientific need to rely on an abacus or fingers and toes once you have this free Calculator for Science Students in your toolbox. A variety of calculators, converters, references, and glossaries reside at your fingertips through this full-featured app. This calculator launches a nicely designed, easy-to-use interface--slightly drab, but easy to comprehend. It offers a numeric keypad and buttons for performing a number of scientific calculations. Below the keypad are six tabs that provide access to a wide series of tools that will make a handy reference for science students and teachers. This application performed quickly and very well in our tests. We were very pleased with the number of tools and references it made available. (Semi-fuddy-duddies will be happy to see a list of metric prefixes you can never remember, and the conversion tools will help in translating U.S. measures to European--along with linear acceleration and angular acceleration for hip physicists and amateur astronomers.) Only a few tools require purchasing the fully registered version of the app (unfortunately, the quadratic calculator is one of them), and in all cases the price is very affordable. Home schoolers will really appreciate this download--it's like a compact mathematics desk reference on your hard drive--but we recommend it for all users. Publisher's Description This is an easy to use Scientific Calculator with a high level of functionality. It incorporates many science tools including a triangle calculator, vector calculator, shape calculator, kinematic calculator, half-life calculator, maths calculator, gas laws calculator, statistical calculator, percent and proportion calculator, world times, event timer, measurement converter and a molar mass calculator. Numerous science data reference systems can be called including SI units, symbols, constants, maths laws, atomic structures, organic compounds, ions and ionic compounds, and chemical reactions. Over 500 commonly used compounds can be quickly called to list their name, formula, molecular mass and CAS number. Over 100 constants can be called to list their numerical value and uncertainty value. Search systems are available for science symbols, ions, science terms and equation formula. Interactive flowcharts can be called for mechanical and electrical units. An interactive periodic table lists elements properties. A help system explains operation, and system contents can be quickly located through a search system. What's new in this version: New calculators and reference tables
Extended Mathematics for Cambridge IGCSE About this title: This book meets the needs of all students following the Cambridge International Examinations (CIE) syllabus for IGCSE Extended Mathematics. Updated for the most recent syllabus it provides complete content coverage with thousands of practice questions in an attractive and engaging format for both native and non-native speakers of English. The book is easy-to-use with an accessible format of worked examples and practice questions. Each book is accompanied by a free CD which provides a wealth of support for students, such
This is the fourth in a series of lectures by Daniel Barenboim as part of the Reith Lectures at the BBC. Barenboim talks... see more This is the fourth in a series of lectures by Daniel Barenboim as part of the Reith Lectures at the BBC. Barenboim talks about how music is the great equaliser as he discovered in his West-Eastern Divan Orchestra which brings together young Arab and Israeli musicians. This is the third in a series of lectures by Daniel Barenboim as part of the Reith Lectures for the BBC. Barenboim argues... see more This is the third in a series of lectures by Daniel Barenboim as part of the Reith Lectures for the BBC. Barenboim argues that we have lost the ability to make value judgements about public standards - all because of political correctness and bad education. This resource offers advice on how to avoid plagiarism. It contains the following sections: 1. Overview and... see more This resource offers advice on how to avoid plagiarism. It contains the following sections: 1. Overview and Contradictions 2. Is It Plagiarism Yet? 3. Safe Practices 4. Safe Practices: An Exercise 5. Best Practices for Teachers These online notes are intended for students who are working through the textbook Abstract Algebra by Beachy and Blair. The... see more These online notes are intended for students who are working through the textbook Abstract Algebra by Beachy and Blair. The notes are focused on solved problems, and will help students learn how to do proofs as well as computations. There are also some "lab" questions on groups, based on a Java applet Groups15 written by John Wavrik of UCSD.
Mathematics for Teachers: Interactive Approach for Grade K-8 9780495561668 ISBN: 0495561665 Edition: 4 Pub Date: 2009 Publisher: Brooks/Cole Summary: Sonnabend, Thomas is the author of Mathematics for Teachers: Interactive Approach for Grade K-8, published 2009 under ISBN 9780495561668 and 0495561665. Five hundred twenty seven Mathematics for Teachers: Interactive Approach for Grade K-8 textbooks are available for sale on ValoreBooks.com, one hundred twenty five used from the cheapest price of $70.75, or buy new starting at $189
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