A First Year Of College Mathematics [2 ed.]
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THE APPLETON -CENTURY MATHEMATICS SERIES Edited by Raymond W. Brink
A FIRST YEAR OF COLLEGE MATHEMATICS
THE APPLETON-CENTURY MATHEMATICS SERIES Edited by Raymond W. Brink
Intermediate Algebra, Second Edition by Raymond W. Brink College Algebra, Second Edition by Raymond W. Brink Algebra--College Course, Second Edition by Raymond W. Brink A First Year of C allege Mathematics, Second Edition by Raymond vV. Brink The Mathematics of Finance · by Franklin C. Smith Plane Trigonometry, R~vised Edition by Raymond W. Brink Spherical Trigonometry by Raymond W. Brink Essentials of Analytic Geometry by Raymond \V. Brink Analytic Geometry by Edwin J. Purcell Analytic Geometry, Revised Edition by Raymond \V. Brink Analytic Geometry and Calculus by Lloyd L. Smail College Geometry by Leslie H. Miller Calculus by Lloyd L. Smail Solid Analytic Geometry by John M. H. Olmsted Intermediate Analysis by John :M. H. Olmsted
A First Year of College Mathematics RAYMOND W. BRINK, Ph.D. Professor Emeritus Department of Mathematics University of Minnesota
Second Edition
New York
APPLETON-CENTURY-CRO FTS, INC.
COPYRIGHT,
1954,
BY
APPLETON-CENTURY-CROFTS, INC. All rights reserved. This book, or parts thereof, must not be reproduced in any form without permission of the publisher. 579-26 Library of CongreSR Card Number 54-5074
Copyright, 1937, by D. Appleton-Century Company, Inc~ 'P.RINTED
IN
THE
UNITED
STATES
OF
AMER.IC~
Preface
This book presents a complete and integrated course in the material of college algebra, trigonometry, and analytic geometry, together with a review, in the Appendix, of elementary algebra. The principal unifying ideas are the two related concepts of function and the correspondence between geometrical and numerical relations. The first three chapters, on Numbers and Operations, Coordinates and Loci, and Functions and Their Graphs, lay the foundation for these unifying ideas, whose various aspects are developed in the later chapters. The first chapter, which has been added in the present edition, presents important properties of the real number system which most freshmen have never seen stated explicitly but which college teachers use regularly and seem to expect their students to understand. The coherent treatment presented here has many advantages over the three-unit type of course. The greatest advantages are obtained from the student's superior understanding of the essential unity of the material and his increased interest in the application of the subject matter to other portions of the field. Another great advantage is its greater flexibility in the selection and arrangement of the material presented. Thus, for engineering students and others who study physics concurrently with the freshman course of mathematics, it is usually desirable t0 introduce the elements of trigonometry early. In such cases, after studying the first two chapters of the book, the class will take up Chapters XV and XVI, covering the definition of the trigonometric functions and the trigonometry of the right triangle. Emphasizing, as it does, both logical development and the acquisition of technical skill, this text is especially suitable for use in colleges of engineering and for the training of prospective teachers in secondary schools and of any other students who expect to study calculus. But because of the great interest and utility of the material and the unity of its presentation, it is also well adapted to a terminal course in mathematics. To those familiar with the author's previous books, it may be said V
Vl
PREFACE
that the present volume maintains the same standards of exactness of statement and proof. Like them, it makes immediate application of principles to practice. It provides an abundance of illustrative examples solved in the text and of carefully chosen problems for the students. The book includes complete four-place tables and explanations for five-place tables and examples of their use for those who prefer them. The Appendix contains a complete review of elementary algebra. Classes whose preparation in algebra is deficient may well study this appendix systematically before undertaking the main portion of the text. Classes with more adequate preparation may use the appendix for reference or for an informal review carried on concurrently with a study of the first few chapters. Some of the changes from the first edition may be noted. The introduction of the first chapter on Real Numbers has been mentioned. The chapter on Logarithms includes a clearer discussion of significant digits and scientific notation. The outline of Theory of Equations has been simplified a good deal, especially in the methods of finding rational roots and approximate values of irrational roots. Radian measure has been introduced at the beginning of the work in trigonometry and a more natural and confident attitude toward its use is thus produced. There is an improved discussion of vector quantities and their relation to vectors. A clearer treatment of mathematical induction results from a more explicit discussion of the fundamental principle on which the method depends. The preceding items can be taken as illustrations of the fact that the entire text has been reexamined and rewritten with care. In general the changes in detail have been in the direction of simplifying the study and teaching of the course. The greater part of the exercises and illustrative examples are new. Applied problems have been modernized. Some of the tables have been improved and the old American Experience Table of Mortality has been replaced by the Commissioners 1941 table. The exercises have been graded more carefully than in the first edition, making the book easier for students to master and more adaptable to courses of varying length and difficulty. Answers to the odd-numbered exercises are now given at the end of the book instead of scattered through the text. There is an improvement in the type page, with larger type in some parts, and a clearer display and statement of theorems, rules, and formulas.
R. W. B.
Contents
CHAPTER
I. Numbers and Operations 1. Real numbers, 1. 2. The set of natural numbers, 2. 3. The negative integers, 2. 4. The rational numbers, 4. 6. Properties of rational numbers, 8. 6. The real number system, 8. 7. Measures of physical quantities, 11. 8. The fundamental operations, 12. 9. Rules of addition and subtraction, 13. 10. Rules of multiplication, 14. 11. Rules of division, 15. 12. Positive integral exponents; laws of exponents, 16. 13. Roots; the principal root, 17. 14. Rational exponents, 18. 16. Imaginary numbers, 19.
II. Coordinates and Loci 16. Directed distances, 21. 17. Rectangular Cartesian Coordinates, 22. 18. Projections, 24. 19. Distance between two points, 25. 20. Equations and loci, 27.
III. Functions and Their Graphs 21. Constants and variables, 30. 22. The definition of a function, 30. 23. The graph of a function, 33. 24. Graphs from observations, 35. 26. Functions of several variables, 39. 26. Two important concepts, 39.
IV. Linear Functions and Their Graphs 27. Rational integral expressions, 40. 28. Linear functions, 41. 29· Slope of a straight line, 42. 30. The graph of a linear function, 44.
V. Linear Equations in Two or Three Variables 31. Rational integral equations, 46. 32. The general linear equation in x and y, 46. 33. The locus of a given linear equation, 47. 34. The equation of a given straight line, 49. 36. A line parallel to a coordinate axis, 49. 36. Point-slope form, 50. 37. Slope-intercept form, 51. 38. Two-point form, 51. 39. Intercept form, 53. 40. The point of intersection of two straight lines, 54. 41. Graphical solution of two simultaneous linear equations, 55. 42. Algebraic solution of simultaneous equations; elimination by substitution, 56. 43. Solution by Vll
CONTENTS
Vlll CHAPTER
*
*
elimination by addition or subtraction, 57. 44. Determinants of 45. Solution of simultaneous equations by desecond order, 58. terminants, 59. 46. Linear equations in three unknowns, 61. 47. Determinants of third order/ 63. 48. Solution of three equations in three unknowns by determinants, 64. 49. Stated problems involving two or three unknowns, 65.
*
*
VI. Variation, Ratio, and Proportion 50. Variation, 68. 51. Joint and inverse variation, 69. 52. Determination of the constant of proportionality, 69. 53. Ratio, 73. 54. Proportion, 74.
VII. Quadratic Equations 56. The graph of a quadratic function, 77. 56. Quadratic equations, 78. 57. Graphical solution of a quadratic equation, 79. 68. Algebraic solutions of quadratic equations. Pure quadratic equations, 80. 69. Solution of equations by factoring, 82. 60. Completion of a square, 83. 61. Solution of a quadratic equation by completing the square, 84. 62. Solution of a quadratic equation by formula, 85. 63. Equations in quadratic form, 89. 64. Equations involving radicals, 90. 65. The discriminant of a quadratic equation, 92. 66. The sum and product of the roots, 96. 67. Factorization of a quadratic expression, 97. 68. The formation of a quadratic equation with given roots, 98.
VIII. Inequalities 69. Inequalities, 101. 70. Properties of inequalities, 102. 71. Absolute and ~onditional inequalities, 103. 72. Solution of conditional inequalities, 104.
IX. The Locus of an Equation 73. Introduction, 108. 74. Graphing by points, 108. 75. Intercepts, 109. 76. Degenerate loci, 109. 77. Symmetrical points and figures, 110. 78. The extent of a locus, 112. 79. Horizonta: and vertical asymptotes, 114. 80. Intersections of curves, 116. 81. Discussion of an equation, 117. 82. Loci of equations of the second degree, 120.
X. Quadratic Equations in Two Unknowns 83. Quadratic equations in two unknowns, 122. 84. The graphical solution of a system of quadratic equations, 122. 85. Algebraic solutions of quadratic systems, 123. 86. Systems consisting of one linear and one quadratic equation, 124. 87. Equations of the form ax2 by2 = c, 2 2 127. 88. Equations of the form ax bxy cy = d. Solution by
*
Sections marked with a
*
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may be omitted.
+
CONTENTS
IX
CHAPTER
elimination of the constant term, 128. 89. Reduction to simpler systems, 129. 90. Equations that are symmetrical in x and y, 130. XI. Progressions 91. Arithmetic progressions, 134. 92. The nth term of an A.P., 135. 93. The sum of the first n terms of an A.P., 136. 94. Arithmetic means, 139. 96. Geometric progressions, 141. 96. The nth term of a G.P., 141. 97. The sum of the first n terms of a G.P., 143. 98. Geometric means, 144. 99. Compound interest, 147. 100. Infinite geometric series, 150 101. Infinite repeating decimals, 152.
*
XII. The Binomial Theorem 102. Introduction, 155. 103. Factorial notation, 155. 104. The Binomial Theorem, 156. 106. The general term of the binomial expan106. Binomial coefficients, 160. sion, 158.
*
XIII. Mathematical Induction
107. The method of mathematical induction, 162. the Binomial Theorem, 167.
*
108. Proof of
XIV Logarithms 109. The exponential function, y = ax, 169. 110. Definition of a logarithm, 170. 111. The fundamental theorems of logarithms, 172. 112. Significant digits and approximations, 17 4. 113. Logarithms to the base 10, 176. 114. Characteristic and mantissa, 177. 115. The mantissa, 177. 116. The characteristic, 178. 117. A table of logarithms, 180. 118. Interpolation, 181. 119. Computation with loga120. A five-place table of logarithms, 186. 121. rithms, 183. Graph of y = logax, 189. 122. Exponential equations, 189. 123. Change of base of logarithms, 191. 124. Natural logarithms, 192.
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XV. The Trigonometric Functions 126. Classification of functions, 193. 126. Degrees and radians, 194. 127. Central angles and their subtended arcs, 195. 128. The generation of angles, 197. 129. General and directed angles, 197. 130. The standard position of an angle, 199. 131. Definition of the trigonometric functions of an angle, 200. 132. The determination of one iunction from another, 203. XVI. The Trigonometric Functions of an Acute Angle
133. The trigonometric functions of an acute angle, 206. 134. The variation of the functions of an acute angle, 207. 136. The functions
*
Sections marked with a
*
may be omitted.
CONTENTS
X
CHAPTER
of complementary angles, 207. 136. The functions of 45°, 208. 137. The functions of 60° and 30°, 208. 138. A four-place table of trigonometric functions, 209. 139. The angle corresponding to a given function, 211. 140. A five-place table of trigonometric functions, 213. 141. The solution of a right triangle, 215. 142. Angles of elevation and depression, 216. 143. Logarithms of trigonometric functions, 220. 144. The accuracy of a computed result, 221. 146. Logarithmic solution of a right triangle, 223. 146. Projections, 225. 147. The bearing of a line, 225. 148. Forces and vectors, 226.
*
XVII. Properties of the Trigonometric Functions 149. Related angles, 231. 160. Functions of related angles, 232· 161. Functions of ( -0), 234. 162. Angles that differ by 90°, 235. 163. Functions of (0 ± n · 90°) and ( -0 ± n · 90°), 236. 164. Periodic functions, 238. 166. The functions of a variable angle, 238. 166. The variation of the sine and cosine, 238. 167. The variation of the tangent and cotangent, 239. 168. The variation of the secant and cosecant, 240. 169. Values of the inverse trigonometric functions, 241. 160. The graph of y = sin x, 243. 161. The graph of y = tan x, 244. 162. The graph of y = csc x, 245. 163. The graph::; of certain related functions, 245.
*
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XVIII. The Fundamental Relations of Trigonometry 164. Algebraic operations with trigonometric functions, 248. 166. The fundamental relations, 248. 166. Identities and equations, 251. 167. Trigonometric identities, 252. 168. Trigonometric equations, 255. XIX. Functions of Two Angles 169. Sin (a + (3) and cos (a + (3), 259. 170. Extension of addition· formulas to all angles, 261. 171. Tan (a + (3) and cot (a + (3), 262. 172. Functions of (a - (3), 263. 173. Functions of twice an angle, 265. 174. Functions of half an angle, 266. 176. Equations involving 176. Product formulas, 270. 177. Sums multiple angles, 269. of functions, 271.
*
*
XX. The Oblique Triangle 178. The parts of a triangle, 27 4. 179. The law of cosines, 27 5. 180. Applications of the law of cosines, 275. 181. The law of sines, 277. 182. Applications of the law of sines, 278. 183. The ambiguous case, 280. 184. The law of tangents, 285. 186. Applications of the law of tangents, 286. 186. Half-angle formulas, 288. 187. Applications of the half-angle formulas, 290. 188. The area of a triangle, 292.
*
Sections marked with a
*
may be omitted.
CONTENTS
XI
CHAPTER
XXI. The Inverse Trigonometric Functions 189. Inverse functions, 298. 190. The inverse trigonometric functions, 298. 191. Multiple values of the functions, 299. 192. Principal values, 300. 193. Graphs of the inverse trigonometric functions, 303.
XXII. Polar Coordinates 194. Polar coordinates of a point, 304. 195. Relations between polar and rectangular coordinates, 306. 196. Equations in polar coordinates, 308. 197. Discussion of an equation, 308. 198. Certain simple loci, 313.
XXIII. Complex Numbers 199. Complex numbers, 314. 200. Conjugate complex numbers, 315. 201. Operations with complex numbers, 315. 202. Graphical representation of complex numbers, 317. 203. The polar representation of a complex number, 318. 204. Multiplication of complex numbers in polar form, 321. 205. De Moivre's Theorem, 322. 206. Roots of a complex number, 324.
XXIV. Theory of Equations I.
INTRODUCTION
207. Rational integral equations, 328. 208. The Fundamental Theorem of Algebra, 329. 209. The equationsf(x) = 0 and/( -x) = 0, 329.
II.
GRAPHS OF POLYNOMIALS
210. The graph of a polynomial, 330. 211. The graph of a factornd polynomial, 332. 212. The graphical solution of an equation, 334.
III.
THE REMAINDER THEOREM
213. The division formula, 335. 216. The Factor Theorem, 336. IV.
214. The Remainder Theorem, 335. 216. Synthetic division, 337.
NUMBER AND NATURE OF THE RooTs OF AN EQUATION
217. The linear factors of a polynomial, 340. 218. degree n has exactly n roots, 340. 219. Formation with given roots, 341. 220. Imaginary roots, 343. of sign, 344. 222. Descartes' Rule of Signs, 345. polynomials, 34 7.
An equation of of an equation 221. Variations 223. Identical
*
V.
DETERMINATION OF THE REAL ROOTS
224. Equations having zero as a root, 348. 226. Limits for the roots of an equation, 348. 226. The rational roots of an equation, 350.
*
Sections marked with a
*
may be omitted.
CONTENTS
Xll CHAPTER
227. The irrational roots of an equation, 352. 228. The method of successive interpolations, 353. 229. Algebraic solutions, 356. XXV. The Straight Line 230. The inclination of a straight line, 357. 231. Parallel and perpendicular lines, 358. 232. The angle between two lines, 358. 233. Norrnal form of the equation of a straight line, 360. 234. Reduction of a linear equation to normal form, 363. 236. Distance from a line to a point, 364. 236. Systems of lines, 367. 237. The system of lines through the intersection of two given lines, 369. XXVI. Locus Problems 238. The determination of a locus, 372. 239. The equation of a locus, 373. 240. Locus problems, using auxiliary variables, 376. 241. The polar equation of a locus, 379. 242. Parametric equations of a curve, 380. 243. The path of a projectile, 382.
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XXVII. Transformations of CoBrdinates 244. The choice of axes, 384. 245. A translation of axes, 384. 246. Simplification of equations by translations, 386. 247. A rotation of axes, 389. 248. Simplification of second-degree equations by rotations, 391. 249. The degree of an equation is invariant, 396. XXVIII. The Circle 250. Standard form of the equation of a circle, 397. 261. General form of the equation of a circle, 398. 252. A circle satisfying three given conditions, 400. 263. Length of the tangent from a point, 254. The system of circles through the intersections of two 402. given circles, 403.
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XXIX. Tangents 255. Definition of a tangent, 407. 256. Slope of a curve, 408. 257. Equations of tangent and normal, 410. 258. A tangent to a curve of second order, 412. 259. Tangents having a given direction, 414. 260. Asymptotes of a hyperbola, 416. XXX. Conic Sections
I.
DEFINITIONS
261. Definitions, 419.
II.
CONIC SECTIONS IN STANDARD POSITION
262. Standard equation~ of the conic sections, 421. 263. Description 264. The polar equation of a conic, 429. of the conic sections, 424.
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CONTENTS
Xlll
CHAPTER
III.
SPECIAL PROPERTIES OF A
p ARABOLA
266. Properties of a tangent to a parabola, 432.
IV.
SPECIAL PROPERTIES OF AN ELLIPSE
266. Focal radii of an ellipse, 434. at a point of an ellipse, 435.
V.
267. The focal radii and tangent
SPECIAL PROPERTIES OF A HYPERBOLA
268. Focal radii of a hyperbola, 437. 269. The focal radii and tangent at a point of a hyperbola, 438. 270. Conjugate hyperbolas, 439. 271. Rectangular hyperbolas, 439.
VI.
GENERAL EQUATIONS OF CONICS
272. Conics with principal axis parallel to a coordinate axis, 441. 273. The general equation of the second degree, 445.
VII.
SYSTEMS OF CONICS
274. The system of conics through the intersections of two given conics, 447. 275. A conic through five given points, 449.
VIII.
SECTIONS OF A CoNE
276. Sections of a cone, 451.
XXXI. Permutations and Combinations 277. Fundamental Principle, 453. 278. Permutations, 458. 279. Circular permutations, 458. 280. Permutations of things some of which are alike, 459. 281. Combinations, 461. 282. Total number of com283. Proof of the Binomial Theorem binations of n things, 463. by combinations, 466.
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XXXII. Probability 284. Definition of mathematical probability, 467. 285. Mathematical expectation, 468. 286. Statistical or empirical probability, 471. 287. The law of averages, 472. 288. Commissioners 1941 Standard Ordinary Mortality Table, 473. 289. Frequency tables and diagrams, 474. 290. Mutually exclusive events, 477. 291. Compound probability; dependent and independent events, 478. 292. Successive trials of an event, 482.
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XXXIII. The Mathematics of Investment 293. Compound interest, 486. 294. Annuities, 489. tion and sinking funds, 491. 296. Bonds, 495.
295. Amortiza-
XXXIV. Determinants 297. Determinants, 497. 298. Mino,rs and cofactors, 497. 299. The value of a determinant, 498. 300. The uniqueness of a determinant,
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CONTENTS
XIV CHAPTER
499. 301. Properties of a determinant, 501. 302. The evaluation of a determinant, 504. 303. Remarks on the expansion of a determinant, 506. 304. Systems of n linear equations in n unknowns; Cramer's Rule, 507. 305. A system whose determinant is zero, 510. 306. Homogeneous equations, 511. 307. Systems of m linear equa308. The complete expansion of a tions in n unknowns, 513. determinant, 514.
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XXXV. Infinite Series
I.
THE LIMIT OF A SEQUENCE
309. The limit of a sequence, 515.
II.
310. Theorems on limits, 515.
CONVERGENCE OF INFINITE SERIES
A. General Remarks on Convergence 311. Definition of convergence, 517. 312. The omission of a finite number of terms, 519. 313. A necessary condition for convergence, 519. B. Convergence of Series of Positive Terms 314. Series of positive terms with limited partial sum, 520. 315. Comparison tests for convergence and divergence, 520. 316. The ratio test, 523.
C. Convergence of Series of Positive and Negative Terms 317. Series of negative terms, 526. 318. Absolute convergence, 526. 319. The generalized ratio test, 527. 320. Alternating series, 528. 321. Power series, 530.
III.
COMPUTATION WITH SERIES
322. Approximation to the value of a series, 531. 323. Functions expressed as power series, 533. 324. The binomial series, 535. 325. Logarithmic series, 536. XXXVI. Partial Fractions 326. Partial fractions, 539. 327. Case I, 541. 329. Case III, 543. 330. Case IV, 544.
328. Case II, 542.
XXXVIl. Curve Fitting and Least Squares 331. Curve fitting, 546. 332. Arithmetic mean and standard deviation, 547. 333. Fitting a straight line; method of average points, 551. 334. Fitting a parabola; method of average points, 552. 335. The method of least squares, 553. 336. Fitting a straight line; method of least squares, 555. 337. Fitting a parabola; method of least squares, 557. 338. Curves of exponential form, 559. 339. Curves of power form, 561.
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Sections marked with a
*
may be omitted.
CONTENTS
xv
CHAPTER
XXXVIII. Solid Analytic Geometry
I.
COORDINATES AND DISTANCES
340. Coordinates of points in space, 564. 341. Representation of figures, 565. 342. Projections of a line segment on the coordinate axes, 565. 343. The distance between two points, 566.
II.
DIRECTIONS AND LINES
344. The direction cosines of a straight line, 568. 345. Direction numbers of a straight line, 570. 346. The angle between two lines, 571. 34 7. Parallel and perpendicular lines, 572.
III.
EQUATIONS AND LOCI
348. Equations and loci, 57 4. Surfaces of revolution, 576.
IV.
349. Cylindrical surfaces, 575.
350.
THE PLANE
361. The equation of a plane parallel to a coordinate axis, 577. 362. The normal equation of a plane, 578. 353. The general equation of the first degree, 580. 354. The angle between two planes, 581. 365. The plane through three given points, 582. 366. The intercept equation of a plane, 583.
V.
THE STRAIGHT LINE
367. Equations of a straight line; projection form, 585. 358. Symmetric equations of a straight line, 587. 359. Two-point equations of a straight line, 587. 360. Reduction of general equations to projection and symmetric forms, 588.
VI.
QUADRIC SURFACES
361. Discussion of an equation, 590. 362. Quadric surfaces, 591. 363. The ellipsoid, 591. 364. The hyperboloid of one sheet, 592. 365. The hyperboloid of two sheets, 593. 366. The elliptic paraboloid, 594. 367. The hyperbolic paraboloid, 595. 368. The right circular cone, 596. Appendix. Review of Elementary Alge bra
I.
THE FUNDAMENTAL OPERATIONS
369. The fundamental operations of algebra, 598. 370. Like terms, 598. 371. Combination of like terms, 598. 372. Parentheses, 599. 373. Rational integral expressions, 601. 374. Multiplication of polynomials, 602. 375. Division of polynomials, 602. 376. Special products, 604.
II.
FACTORING
377. Factoring, 605. 378. Common monomial factors, 606. 379. The difference of two squares. 607. 380. Trinomials which are per-
CONTENTS
XVI CHAPTER
+ +
feet squares, 607. 381. Trinomials of the form x2 qx r, 608. 2 2 qxy ry , 609. 383. The sum or 382. Trinomials of the form px difference of two cubes, 610. 384. Polynomials which may be factored by grouping terms, 611. 385. The sum or difference of equal powers of two numbers, 612. 386. Trinomials expressible as the difference of two squares, 613. 387. Squares of polynomials, 614. 388. Lowest common multiple, 615. 389. Summary, 617.
+
III.
+
FRACTIONS
390. Reduction of a fraction to its lowest terms, 618. 391. Multiplication and division of fractions, 620. 392. Addition of fractions naving equal denominators, 623. 393. Addition of fractions having different denominators, 624. 394. Complex fractions, 626.
IV.
LINEAR EQUATIONS IN ONE UNKNOWN
396. Identities and equations, 629. 396. Roots of an equation, 630. 397. Equivalent equations, 631. 398. Linear equations in one unknown, 632. 399. Applied problems, 634.
V.
EXPONENTS AND RADICALS
400. Positive integral exponents; laws of exponents, 639. 401. Roots; the pr~ncipal root, 640. 402. Imaginary numbers, 642. 403. Rational exponents, 644. 404. Laws of radicals, 648. 405. Removal of factors from a radicand, 649. 406. Rationalization of the denominator in a radical, 650. 407. Reduction of the order of a radical, 651. 408. Powers and roots of radicals, 652. 409. Multiplication and division of radicals, 653. 410. Addition and subtraction of radicals, 655. 411. Fractions with binomial irrational denominators, 657. 412. The simplest radical form, 659. Table I. Table II. Table III. Table IV. Table V. Table VI. Table VII. Table VIII.
Squares, Cubes, Roots, 663. Common Logarithms, 664. Trigonometric Functions-Natural Values and Logarithms, 666. Commissioners 19{1 Standard Ordinary Mortality Table, 671. Compound Amour:t, 672. Present Value for n Periods, 673. Amount of an Annuity, 67 4. Present Value of an Annuity, 675.
Answers, 677. Index, 715.
A FIRST YEAR OF COLLEGE MATHEMATICS
CHAPTER I
Nutnbers and Operations* 1. Real numbers. In this course we shall have to deal with two principal classes of numbers, real numbers and imaginary numbers, which, together, form the system of complex numbers. As we shall see, imaginary numbers are defined in terms of real numbers. We shall therefore consider briefly some of the properties of the real number system. The fundamental operations of arithmetic and algebra are addition, subtraction, multiplication, and division. These operations obey certain laws which will be discussed later in the chapter. For the time being, it will be assumed that the student is sufficiently familiar with the notation and rules of the operations to follow the discussion. The complete system of real numbers did not appear all at once. It is rather the result of a process of evolution. In this evolution one set of numbers after another was invented and incorporated in the system when it was found that the previously existing system did not contain the answers to many simple problems. To some extent the student can retrace this ~evelopment by recalling his own experience with numbers. This began with problems involving only positive whole numbers and zero, and then progressed to problems requiring fractions, irrational numbers, and, finally, negative numbers. Before a new set of numbers can have real meaning or status as numbers, it is necessary to define not only the numbers themselves but also the fundamental operations as applied to them. The operations must be defined in such a way that the rules of operation will still hold. We shall not attempt to define all of the various kinds of numbers nor the operations as applied to them. It is desirable, however, to mention the various kinds and some of their properties. * Chapter I may be omitted at the first reading and referred to as the material is needed.
1
2
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. I
2. The set of natural numbers. The nucleus of the real numbers is the set of natural numbers, or positive integers, (1)
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, · · · .
The three dots indicate that the line of numbers continues to the right without end. The numbers 2, 4, 6, 8, etc. are even numbers, while 1, 3, 5, 7, etc., are odd. Any natural number can be added to any natural number, and the result or sum is a natural number. That is, if a and b are any two natural numbers, then their sum, a + b, always exists and is a natural number. It is the number reached in line (1) by starting at a and counting b places* to the right. Thus 3 + 4 = 7 means that if we start at 3 and count 4 places to the right, we arrive at 7. Any natural number can also be multiplied by any natural number, and the result or product is a natural number. Its definition by a counting process is left to the student. The product of b by a can be written in any one of the following ways: ab
= a · b = a X b.
The numbers a and b are factors or divisors of the product. 3. The negative integers. The result of subtracting b from a is called the difjerence a - b. If a and b are natural numbers and if a lies to the right of bin line (1) of § 2, then (a - b) is reached by counting b places to the left of a. But if a lies to the left of b, there is no natural number equal to (a - b). Thus 7 - 3 = 4 is a natural number. But there is no natural number x such that x = 3 - 7. Such limitations made it necessary to define zero and the negative integers, -1, -2, -3, • • • , and to incorporate them in the number system. Their inclusion completes the system of integers: (1)
· · · , -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, · · · .
Two integers are the negatives, each of the other, if one of them lies at a certain number of places to the right of 0 and the other is at the same number of places to the left of 0. The negative of a number a is called -a. Thus the negative of 4 is -4; the negative of ( -4) is 4, or - (-4) = 4. In general, - ( -a) = a, whether a is positive or negative or zero. * By "b places" we mean the number of places that we must count to arrive at b starting with 1.
NUMBERS AND OPERATIONS
§§ 2-3]
3
In order to extend the definitions of the fundamental operations to include the negative integers, we may define the sum (a+ b) as the number reached in line (1) above by starting at a and counting b places to the right, if b is positive or zero, and ( - b) places to the left if b is negative. Thus,
(-3)
+2 =
-1, (-3)
+
(-2)
=
-5, 4
+
(-6) = -2,
a+(-a)=O.
and
Subtraction is defined by saying that, if a and b are any two integers, the result of subtracting b from a is the same as that of adding the negative of b to a, or
a- b
(2) Thus,
= a + (-b).
+
5 - 7 = 5 (-7) = -2, 5 - (-3) = 5 -3 - 2 = -3 + (-2) = -5.
+ 3 = 8,
The product of two integers may be defined in terms of the products of natural numbers by saying that, if a and b are positive integers,
= -(ab), (-a)(b) = -(ab), and a · 0 = 0 · a = O, whether a is positive, (3)
(a)(-b)
Thus,
4(-2) = -(4 · 2) = -8,
(-a)(-b)
= ab.
negative, or zero.
(-5)(-3) = 5 • 3 = 15.
Consequently any two integers may be added, subtracted, or multiplied, and the result is always an integer. This result is unique; that is, "if equals are added to equals or subtracted from equals or multiplied by equals, the results are equal." EXERCISES In Exercises 1 to 15, perform the additions and subtractions indicated. 1. 3 + 4. 5 + 7. 4 + 10. 5 13. 0 -
4. (-8). (-7). (-2). 6.
2. 5. 8. 11. 14.
-5 -2 -3 -4 0 -
+ 2. + (-3). + 8. - 5. (-3).
3. 6+(-2). 6. 9. 12. 15.
4+ 6-3 -2
0. 2. - (-8). - 0.
In Exercises 16 to 24, perform the multiplications. 16. 5 X 2. 19. (-6) · 1. 22. 7 · 0.
17~ (-2) X 5. 20. 3(-8). 23. 0 · (-4).
18. (-3)(-2). 21. (-1)(-6). 24. 0 · 0.
FIRST YEAR OF COLLEGE MATHElVIATICS
4
[Ch. I
4. The rational numbers. The quotient of a divided by b or the ratio of a to b is the result of dividing a number a by a number b.
It may be written as a + b, or ~' or a/b. The number a is the dividend or numerator, and bis the divisor or denominator. The quotient a/b is defined as the number which, on being multiplied by b, yields the product a. That is, if there is a unique number x such that bx = a, then a/b = x. 24 Thus, = 8 - l 5 = -5 and 16 = -2 3 ' 3 ' -8 '
smce 8 X 3 = 24,
(-5)(3)
=
-15,
and
(-2)(-8)
= 16.
For reasons that will appear in § 7, it is never possible to divide any number by 0. But, quite apart from the impossibility of dividing by 0, we cannot divide an arbitrary integer by an arbitrary integer, not 0, and be sure of getting an integer as the quotient. For example, there are no integral values of 2/5, or (-3)/7. Such limitations led to defining the rational fractions and to including them with the integers to form the system of rational numbers. The idea of a real number corresponds intuitively to the idea of length of a straight line segment. Consider a certain infinite straight line (Figure 1) and on it choose a point, which we label 0. We also I
-6
I
-5
I
-4
I
-8
I
-2
t -1
0
I
l
l
2
1 4
I
o
i
6
Frn. 1
choose some convenient length as a unit. Starting from O and proceeding to the right we mark points one unit apart, and label them successively 1, 2, 3, • • · . Proceeding to the left from 0, we mark points -1, -2, -3, • • · one unit apart. Corresponding to each integer we have now fixed a point on the scale. Because of this correspondence we can use the letter a to mean either the number a or the point that corresponds to that number. Thus the point 5 is five units to the right of O and the point -5 is five units to the left of 0. We now wish to identify the numbers called rational fractions by indicating the points to which they correspond. Let b be any positive integer and mark the points that divide the interval from 0 to 1 into b equal parts. The first of these points to the right of O we label 1/b. Thus, if we divide the interval from 0 to 1 into 3 equal parts, the first point of division is marked ½. The interval from 0
§ 4]
NUMBERS AND OPERATIONS
5
to this point 1/b is said to have the length 1/b. Starting at 0 we now mark infinitely many points both to the right and left of 0, the distance between successive points being the length 1/b. We label these points 1 -, 0 -, 1 -, 2 - -, 2 -, 3 4 3 - -, -, - 4-, - -,
b
b
b
b b b b b b
,vith 0/b at O and with (-a)/b as an alternative notation for -(a/b). If a is any integer and b is any positive integer, we have now established a point a/b, and we now introduce a new number, the rational number a/b, which we say corresponds to this point. The value of a/b is associated intuitively with the distance and direction of the point a/b from 0. The positive rational numbers appear to the right of O and the negative rational numbers to the left of 0. If two rational numbers, such as t and -t, correspond to points that are equidistant from O but lie on opposite sides of 0, each number is said to be the negative of the other. The negative of a number a is written as (-a). Thus, -t = -(t), ! = -(-!). If a is a positive integer the negative number (-a)/b is reached by counting a spaces of length 1/b to the left from 0, since (-a)/b = -(a/b). It is possible to reach a point in more than one way. Thus the point ! can be reached by counting 3 spaces of length ½or 9 spaces of length¼ or 12 spaces of length i- We say that two rational numbers are equal if they correspond to the same point. Thus ! = ¼ = 1s2. In general, if a, b, and care any positive integers, it is evident intuitionally that we reach the same point by counting a spaces of length 1/b or by counting ac spaces of length 1/(cb).
That is,
i = :·
As
the definition of the equality of two rational fractions, we therefore say (1)
a ac -=-, b be
where a is any integer and band care any integers other than 0. Therefore, the value of a rational fraction is not changed if its numerator and denominator are both multiplied (or divided) by any integer not zero. A fraction is said to be in lowest terms if its numerator and denominator contain no common integral factor except 1 and - 1. (See Exercise 25 below.)
6
FIRST 3 EXAMPLES - = ·4 To reduce
YEAR OF COLLEGE MATHEMATICS 3 X 5 15 - - = -· 4X5 20
[Ch. I
~~ to lowest terms we divide the numerator and denominator
. 16 4 4 X = . by the common factor 4 and obtam =4X 12 3 4 3
In the definition (1), a, b, and c were not limited to positive values. We therefore find it natural to define a rational fraction with a negative denominator, by writing (2)
For if, by (1), we multiply the numerator and denominator of (-a)/b by c = -1, we obtain a/ ( -b). Consequently, the value of a rational fraction is not changed if the signs of its numerator and denominator are changed, or if the sign of either the numerator or denominator and the sign infront of the fraction are changed. 7
Thus
7(-1)
-5 = c-5)(-1) =
-7
T =
7
-r
Our intuition now leads us immediately to the following definitions of the fundamental operations as applied to rational numbers:
I. Addition and subtraction offractions having er1ual denominators:
2 5 2- 5 -3 - - - - - - - - - -1. 3 3 3 3
EXAMPLES.
II. Addition and subtraction of fractions having unequal denominators: ~ b
+£
= ad
d
bd 2
EXAMPLES.
3
+ be = ad + be.
3+ 4 =
bd
ad - be • bd
bd'
2·4+3 ·3 17 3 .4 = 12. 44 - 60 55
3
=-
4
3
4
+5 = 1+5 = 16 55·
19 5 .
§ 4]
7
NUMBERS AND OPERATIONS
III. Multiplication and division offractions: a
c
ac
a . c
a
b x d = bd; b --r d = b x EXAMPLES.
.Q.X~-ll 3 7 21
c = adbe·
¾+ C-½) = ¾C-¾) = -H.
•
a
a
b
bx b =bx 1 =
and
d
a.
vVe thus see that any rational number can be added to or subtracted from or multiplied by any rational number, and it can be divided by any rational number except zero, and a result always exists and is a rational number. EXERCISES In Exercises 1 to 4, reduce each fraction to lowest terms. 8 24 81 2· - 18. 3· 36. 1. 12.
54 4. - · 9
In Exercises 6 to 8, change each pair of numbers to fractions with a common denominator. 3 4 2 2 5 11 1 5• 3' - 4. 6• - s, s· 1. s, 8. - -· 14' 2
r
In Exercises 9 to 24, perform the operations indicated, and reduce the results to lowest terms. 9•
13.
16
7
10.
5 - 2· 2
5
X 10.
2
2
3 + s°
14. -
3
4
X 12.
DX (- ;)20. CD+
15.
6
2 1.
c-2).
C- D+ ( - 161}
3
7 + s"
~X(
11 3 18. 12 X ~f
17. ( -
23.
11. -
CD+ C-D·
-D·
1 12. - 5 16.
3
2 X g · 5
6
19.
7+
3.
22.
8 + y·
5
10
24• ( - ~) 7 ..!-. ~8
25. Show that the definition (1), above, is equivalent to saying that
; =
J if
ad = be.
- 12-·
FIRST YEAR OF COLLEGE MATHEMATICS
8
[Ch. I
6. Properties of rational numbers.
A rational number is a number that can be expressed as the ratio of an integer to an integer. Irrational real numbers, which will be introduced in § 6, are real numbers
that cannot be expressed in that way. The system of rational numbers clearly includes the integers. For any integer a can be expressed as a/1. It also includes such decimal numbers as 2 21.47 and - .0159, which can be written as ~~ and ~~~1 10
The symbol a < b is read "a is less than b"; b > a is read "b is greater than a." Definition: If a and b are two rational numbers, and if a i.s to the left of b on the scale of numbers, then we say that (1)
a< b
and
b
>
a.
The symbols < and > are called signs of inequality. If a is positive, a > 0; if a is negative, a < 0. The symbol a ~ b is read "a is not equal to b." If a < b < c, we say that b is "between" a and c. The symbol a < b, which is read "a is less than or equal to b," means that a either coincides with b or is to the left of b on the scale; a similar definition holds for a > b. 3 < 5, -4 3 is between -2 and 5, i EXAMPLES.
< 3, >
-6 -5.
o, -¾ < -½,
Inequalities are studied in more detail in Chapter VIII, where the methods used here are justified. Theorem. If a and b are two rational numbers, and if a < b, it is always possible to find a rational number between a and b. For we can add to a any sufficiently small rational number, and obtain a rational number between a and b. Thus, if a = 3 and b = 3.01, then such rational numbers as 3.005 and 3.000001 are between a and b. We can express this idea by saying that, on any interval of the number scale, there are infinitely many rational numbers.
6. The real number system. Although points that correspond to rational numbers lie as close together as we choose on every interval of the number scale, such points do not completely "fill" the scale. f
-6
I
-5
I
-4
1
-3
I
-2
I
-1
O
FIG. 2
I
1
I
2
I
8
1 4
t
5
J 6
§ 6]
NUMBERS AND OPERATIONS
9
That is, there are points on the scale that do not correspond to any rational numbers. For example, if an isosceles right triangle is drawn with each of its equal sides equal to one unit of the scale (Figure 3), and if we denote its hypotenuse by c, we H have '\ \ c2 = l2 l2 = 2. C
1
However, there is no rational number, c, whose square is equal to 2.
\
\ \ I
1 0
+
'' \
l
p
To prove this statement, let us assume that
Fm. 3
there is such a rational number c = ~, ,Yhere
a and b are integers, and assume also that the fraction ~ is in lowest terms. We shall see that this assumption leads to a contradiction and must therefore be false. Since c2 = 2, we have a2 - = 2 or a2 = 2 b2•
b2
Therefore a2 is an even number, since it contains the factor 2. But the square of any odd number is odd. Consequently a cannot be an odd number and must be even, and we can set a = 2 n, where n is an integer. From the equation a2 = 2 b2, we therefore have (2 n) 2 = 2 b2 or 4 n 2 = 2 b2 • Therefore, 2 n2 = b2, which shows that b2 is even and consequently b is even. Since a and bare both even numbers, the numerator and denominator of a/b both contain the factor 2. This contradicts our assumption that a/b was in lowest terms, and proves that no rational number exists whose square is equal to 2.
Therefore, if we mark the point P on the number scale at the distance c from 0, the point P does not correspond to any rational number. It can also be shown that there is no rational number that exactly represents the length of the circumference of a circle whose diameter is equal to 1, although 3.1416 is a close approximation to the length. In order to furnish answers to such problems, it was necessary to define irrational numbers. The rational and irrational numbers together constitute the system of real numbers. We take the following statement as an axiom: Corresponding to each point of the number scale there is a real number, and corresponding to each real number there is one point of the number scale. Any point that does not correspond to a rational number corresponds to one of the new, or irrational, numbers.
FIRST YEAR OF COLLEGE MATHEMATICS
10
[Ch. I
It would go beyond the limits of this book to give a method of defining irrational numbers. It will suffice to say that there are methods which define irrational numbers and the fundamental operations with them, in terms of rational numbers, just as rational fractions were defined by reference to the integers. It can be shown, moreover, that the fundamental operations obey the same laws when applied to the whole set of rational and irrational numbers together as when applied to the rational numbers alone. In particular, multiplication is defined in such a way that O • x = 0, when x is any real number, and division is defined by saying that, if there is a unique number x such that b · x = a, then a + b = x. vV e can now justify the remark made in § 5 that division by O is never permissible. We shall show that a + 0 does not exist, for any real number a. There are two cases: (i) Let us suppose that a
0 • x = a. exist.
But O · x = 0
~
~
0 and try to find a number x such that a for every x. Therefore a + 0 = x does not
(ii) Next suppose that a = 0 and try to find a unique number x such that 0 • x = a = 0. Since O • x = 0 is true for every x, no single value of x is determined that satisfies this relation. Therefore O + 0 is not defined.
The definitions of equality, inequality, ''betweenness," and the negative of a number, as given in § 5 for rational numbers, apply also to all real numbers. By the methods that are used in defining irrational numbers, it can be shown that between any two real numbers there are infinitely many rational numbers and also infinitely many ,frrational numbers. The absolute value or numerical value of a real number a is written lal, and is equal to a if a > 0 and to -a if a < 0. Thus 161 = 6, l-31 = 3, 101 = o. To say that two numbers a and b are numerically equal or equal in absolute value means merely that !al = lbl, and either a = b or a = - b; that is, the points a and b are equidistant from Oon the scale, but may lie on the same side or on opposite sides of 0. To say that a and b are "equal" means more than this. If a = b, not only are the points a and b equidistant from 0, but they lie on the same side of O and actually coincide. Thus 5 and - 5 are not equal, but 5 and - 5 are numerically equal. We say, likewise, that 3 is "numerically less than" -5, since 131 < l-51.
§§ 6-7]
NUMBERS AND OPERATIONS
11
EXERCISES In Exercises 1 to 18, find the value of each number. 1.
5. 9. 13. 17.
2. 1181. 151. 6. 12 - 81. 12+31. 10. j-31 +3. 3 + 131. 151 - 1-71. 14. 1-71 - 1s1. 3 · 1-41 - 5 · l-31 + 1-11.
3. 7. 11. 15. 18.
4. l-7,. 1-111. 8. l-8 + 8j. l-5 +31. -3 + l-3J. 12. 3 - l-31. 2. l-41. 16. 8 + 1-21. 121. 1-21 -1-41. 1-11.
In each of the Exercises 19 to 26 connect the two given numbers with the correct sign of inequality, < or >. 19. 3, 6.
20. 3, -6.
131, l-61.
21. -3, -6.
23.
131, 161.
7.
Measures of physical quantities.
24.
25.
l-31, l-6!.
22. -3, 6. 26.
l-31, 161.
An important application of real numbers is their use as measures or expressions of the magnitudes of physical quantities. Examples of such quantities are lengths, distances, areas, masses, forces, and intervals of time. Let us count distances between two points on the number scale as positive when measured to the right and as negative ·when measured to the left. They are then directed distances. The real numbers then have intuitional meaning for us by the very fact that any real number is the measure of the directed distance of the corresponding point from the point 0. In algebra, the measure of a physical quantity is a number.* What that number is, naturally depends on the unit of measure that is used. Thus a length whose measure is 18 when the unit is an inch has the measure i- if the unit is a foot; a mass whose measure is ¼when the unit is a ton has the measure 500 if the unit is a pound; an interval of time whose measure is ¾when the unit is an hour has the measure 45 when the unit is a minute. In this course, we shall suppose that all equations are statements of zquality between numbers. t If physical quantities are involved, each quantity is to be replaced by the number which is its measure before an equation is written to connect the numbers. This remark deserves
* This statement must be qualified in connection with certain quantities like forces and velocities for which it is necessary to indicate direction more generally than merely as "forward" or "backward." (See § 149.) t Occasional exceptions are made for the purpose of indicating the relation between two units; thus: 1 hour = 60 minutes, or 180° = 1r radians.
1~
FIRST YEAR OF COLLEGE MATHElVIATICS
[Ch. I
some attention because in applied mathematics many scientists preserve the idea of dimensionality by writing equations connecting physical quantities themselves rather than the numbers that measure those quantities. Thus, in a problem involving distance, speed, and time, the equation d = st may mean to a physicist that the distance traveled is equal to the speed times the time. In our algebra this means rather that the number of units of distance is equal to the number of units of distance traveled per unit of time multiplied by the number of units of time. If A and B are two points on the number scale, we shall feel free to use the notation AB in two ways. We shall refer to AB as the actual directed line segment from A to B; but numerically, or in an equation, we shall use AB to denote the measure of the directed distance from A to B. Thus, if A is at -3 and B is at 4, we shall write AB = 7 and BA = -7. In general, AB = -BA.
EXERCISES In the following exercises, find AB and BA. 1. A at 5, B at 11. 2. A at 6, B at 2. 4. A at -5, Bat 4. 5. A at -3, Bat -8.
3. A at -5, Bat -2." 6. A at 4, B at 0.
The fundamental operations. The fundamental operations of addition, subtraction, multiplication, and division obey certain rules. We shall assume that any two real numbers can always be combined by any one of these operations (except that division by O is never possible) and that a result, which is also a real number, always exists and is unique. (For addition this assumption of uniqueness amounts to saying that if equals are added to equals the sums are equal.) Many of the rules of operation have already been given in the preceding sections. In the following sections we list for reference some of the rules and definitions that will serve the student as a useful guide in performing the operations. The statements are not all independent; some of them can be deduced logically from others, and some of them differ in form from the equivalent statements already given. For convenience of reference they will be written mostly in symbolic form. It is to be understood that literal numbers, such as a, b, c, etc., stand for any real numbers which do not introduce zero denominators. Following each list of rules there is a set of corresponding examples.
§§ 7-9]
NUMBERS AND OPERATIONS
I. Addition is commuta-
9. Rules of addition and subtraction. tive: a + b = b + a. II. Addition is associative: a
+
(b
+ c)
= (a
13
+ b) + c =
a
+ b + c.
III. Definition of subtraction: a - b = x, if b + x = a. IV. Definition of the negative of a number: If a + b = O, bis the
+(
negative of a, and a is the negative of b, or a -a) = 0. V. The subtraction of a number is equivalent to the addition of its negative: a - b = a+ (-b). VI. Addition of zero: a 0 = a. VII. Inequalities: a < b if and only if (b - a) is a positive number~ VIII. The sum of two positive numbers is positive; the sum of two negative numbers is negative.
+
To illustrate the fact that these rules are not independent, let us prove V by using the other rules. By IV, b
+ (-b)
= 0.
Since equals added to equals give equal sums, we can add a to both members of this equation and obtain
+ (-b)] = a, or (by I and II), [a+ ( -b)] + b = a. Therefore, [a + (-b)] is the number which must be added to b to give a. a+ [b
Consequently, by the definition III,
a - b= a
+ (-b),
as we wished to show.
+ 7 = 10, 7 + 3 = 10. 5 + 3 + 7 = 5 + (3 + 7) = 5 + 10 = 15, 5 + 3 + 7 = (5 + 3) + 7 = 8 + 7 = 15. 11 - 8 = 3, since 8 + 3 = 11. 6 + (-6) = 0, 6 = -( -6), ( -6) = -(6). 9 - 3 = 9 + (-3) = 6; -8 - (-2) = -8 + 2 = -6. -3 + 0 = -3. -2 < 7, since 7 - (-2) = 7 + 2 = 9, which is positive.
EXAMPLES.
2. 3. 4. 5. 6. 7.
1. 3
Rule V essentially replaces subtraction by addition.
4- a
+2b-
5c =
+4+
(~a)
Thus
+ 2 b + (·- 5 c).
FIRST YEAR OF COLLEGE MATHEMATICS
14
[Ch. I
It is therefore natural to speak of a sequence of terms which are written down in order, with their algebraic signs ( + or - ) attached or understood, as the algebraic sum of the terms. Thus the expression 4 - a + 2 b - 5 c is an algebraic sum whose terms are +4, -a, 2 b, and -5 .c. The negative of such a sum is the algebraic sum of the negatives of the terms. For example, 8 - (x - 2 y - 3)
=
8
+ [ - (x -
2 y - 3) J
=8-x+2y+3
= 10. Rules of multiplication.
11 -
I.
X
+ 2 y.
J-1:f ultiplication is
commutative:
a X b = b X a.
II. Multiplication is associative: a X (b X c)
III.
1-~fultiplication
;::=
(a X b) X c
=
a X b X c.
is distributive with respect to addition: a(b
+ c)
= ab
+ ac.
IV. Zero factors: a X O = 0, and if ab a orb is 0.
= 0,
at least one of the factors
V. Unit factors: a X 1 = a. VI. Negative factors: a(-b) = -ab; (-a)(-b) = ab. VIL The product of two positive factors or of two negative factors is positive: if a > 0 and b > 0, then ab > 0 and (-a)(-b) > 0. From Rules VI and VII, we see that the product of any even number of negative factors is positive and the product of any odd number of negative factors is negative. EXAMPLES.
1. 7 X 3 = 3 X 7 = 21;
5(-3) = (-3)5 = -(3 · 5) = -15. 2. 3 X 5 X 4 = 3 X (5 X 4) = 3 X 20 = 60, 3 X 5 X 4 = (3 X 5) X 4 = 15 X 4 = 60. 3. 3(X - 2 y 7) = 3 X 3(-2 y) 3(7) = 3 X - 6 y 21; (3 - x)(a - b) = 3(a - b) - x(a - b) = 3 a - 3 b - ax+ bx. 4. 7 X O = 'O; (-5) XO = O; if (x - 2)(x - 3) = 0, either x - 2 = 0 or
+
+
+
+
3 = 0. 5. 11 X 1 = 11; 1r • 1 = 1r. 6. 7(-2) = -(7 X 2) = -14; (-8)(-3) = 8 X 3 = 24. 7. (-11)(-2) = 22; 2(-3)(-4)(-1)(-6) = 144; X -
2(-3)(-4)(-6)
= -144.
NUMBERS AND OPERATIONS
§§ 9-11]
11. Rules of division. and b
~
I. Definition of division:
15
i = x, if a = bx
0.
II. Division by O is not permitted:
ij is not defined.
III. Definition of equaliiy offractions: IV. Reduction of fractions: :;
=
i = a, if ad = be.
i, if c ~ 0.
(Follows from III.)
..ff . a C ac . z· . V. M u ltip ication OJ ractions: b X d = bd · VI. Definition of the reciprocal of a number: If ab = 1, a is the reciprocal of b, and b is the reciprocal of a; in particular, a and 1/ a are reciprocals, and a/band b/a are reciprocals. VII. Division by a number is equivalent to multiplication by its
~+
reciprocal:
a= ~ X ~ , or "to divide by a fraction, invert the
divisor and multiply." VIII. Changing the sign of the numerator or denominator of a fraction reverses the sign of the fraction; changing the signs of both numerator and denom'inator leaves a fraction unchanged;
EXAMPLES.
2.
5
0
and
0
1. J.32
= 4, since 12 = 4 X 3.
are not defined;
O
0
3
= 0 and _0 = 0. 4
5 = 10 , since (-5)(-6) = 10 X 3 = 30. 3 6 12 _ 3 · 4 _ . -24 _ ( -4) · 6 _ -4 4 · 4 - 1 · 4 - 3 ' 18 3·6 - 3 .
3.
4 5 4X5 20 8 6 48 8X6 8 5· 3 X 7 - 3 X 7 - 21 ; 3 X 14 - 42 - 7 X 6 - 7. 6. The reciprocal of 7 is ½; the reciprocal of 3 -3 . 5 - - or of 1s - - •
5
5
-5
is
!- ;
the reciprocal of
3
1 • ( 7 · s3 ~• 7a -_ s3 X 61 -_ To', -4 ' 4 4 -3 8. - = - = - - · -
5
-i
5'
4)
9
•
-.-
(
-
2 ) _
IT -
(
-
4)
9
X (-
3 X - y X - y = -· =---= -7 7 ' y - x -(x - y)
11) -2
-1 G
=
22 -o•
FIRST YEAR OF COLLEGE MATHEMATICS
J6
[Ch. I
12. Positive integral exponents; laws of exponents. °Let us recall certain facts about exponents. If n is any positive integer, and a is any number, the product of n factors each of which is equal to a is denoted by an:
a1 = a
'
a2 =a· a
a8
'
= a · a · a,
an = a • a • • · · · a (n factors). We call an the nth power of a; a is the base, and n is the exponent of the power. The following Laws of Exponents are familiar to the student. If m and n are any positive integers and a and bare any real numbers other than zero,
L
II.
if m > n;
III. (1)
(2)
am 1 an= an-m'
if n >
m;
IV.
V. VI.
An odd power of a negative number is negative; an even power of a negative number is positive.
EXAMPLES.
= 37 = 2187;
1. 35 • 32
x4
•
x5
=
x 9•
2. (34) 3 = 312 = 531,441; (x 5) 3 = x15 • 37 3 • 32 -- 35
-
243·'
1 1 x5 x2 1 3 37 -- 35 -- 243'· x2 -- x '• x5 -- x8 •
32
4. (3 x2)4 = 34(x2)4 = 81 xs; (2 xay)4 = 16 x12y4. 4 5 ( ·1) · 3
_
14 34
_
1 . (3 a 3) 81 ' 2 x2
2
_
32(a3) 2 22 (x2) 2
_
9 a6 • 4 x4
6. (-3) 4 = (-3)(-3)(-3)(-3) = 81; (-3) 5 = -243; (-3 x 4) 3 = -27 x 12 ; (x - y) 2 = (y - x) 2 •
§§ 12-13]
NUMBERS AND OPERATIONS
17 13. Roots; the principal root. Definition I. If an = b, where n is any positive integer, a is said to be an nth root of b. If n = 2, a is a square root of "h; if n = 3, a is a cube root of b. EXAMPLES.
1.
Since 23 = 8, 2 is a cube root of 8.
2. Since (-2) 5 = -32, -2 is a fifth root of -32. 3.
Since ( -3) 2 = 9, and 32 = 9, ( -3) and 3 are square roots of 9.
4.
Since 24 = 16 and ( -2) 4 = 16, 2 and ( -2) are fourth roots of 16.
From the preceding examples, it is seen that a number may have more than one real nth root. Some numbers do not have any real nth roots at all. For example, there is no real square root of ( -16), since the square of any real number is positive or zero. In fact, if b is negative and n is an even integer, there is no real nth root of b, for any even power of a real number is positive or zero. The consideration of such cases will lead, in § 15, to the introduction of a new set of numbers called imaginary numbers. For the present we shall limit the discussion to values of b and n for which at least one real root does exist. In order to set up a satisfactory notation, we make the following definition: Definition II. When the real number b has an nth root which is positive or zero, that root is called the principal nth root of b; if b does not have a positive nth root but does have a negative nth root, the negative nth root is the principal nth root of b. The principal nth root of b is denoted by the symbol {/b, or by Vb if n = 2. 5. V4 = 2. Note that V4 is not equal to - 2, although -2 is one of the square roots of 4. If we wish to designate the negative square root of 4, we explicitly attach the minus sign, so that -V4 = -2. EXAMPLES.
6.
_3;V
-8 = -2.
7.
{"s1 = a.
8. ~64 = 4.
9.
~=
-2.
From the definition, it follows that: If b > O, {lb > 0 for any value of n. (Examples 5, 7, 8, above.) If b < 0 and n is odd, fib < 0. (Examples 6, 9, above.) If b < 0 and n is even, there is no real nth root of b, and the principal root is not defined.
In the notation fib, the symbol Vis the radical sign, n is the index or order of the root, and b js the radicand. The notation Vb, with the index 2 not written, represents the principal square root of b.
FIRST YEAR OF COLLEGE MATHEMATICS
18
[Ch. l
14. Rational exponents. The symbol an has been defined only in case n is a positive integer. We shall now define an for other rational values of n, that is, for positive fractional values, zero, and rational negative values of the exponent. We shall frame these definitions in such a way that the new exponents will obey the same laws as positive integral exponents. m Fractional exponents. We wish to define an, where a is a real number and m and n are any positive integers. To avoid ambiguities, we shall limit our~definition to the cases for which a principal nth root of a is defined; in other words, we exclude the case in which a is negative .
!!!
and n is even. With this case excluded, we define an as the principal nth root of am, or
I.
= 1,
In particular, if m
1
-
_nr
an= Va.
II. 1
EXAMPLES. 1
2. (9 a2 ) 2 =
1. 83
= {/8 = 2.
vg a
2
= 3 a,
1
when a> 0. · Note that (9 a 2) 2 ~ -3 a.
1
3. (-8x6)3 = ~-8x6 = -2x2• 2
4. (-27 x3y6) 3 = (~ -27 x3y6 ) 2 = (-3 xy2) 2 = 9 x2y4•
The exponent zero. Let a be any base except zero. a meaning to the symbol a0 by defining:
III.
a0
= 1,
We then give
(a¢ 0).
Negative exponents. Let a and n be any two numbers for which 1/an has been defined; we then give a meaning to the symbol a-n by defining:
IV.
1 a-n = -· an
Of course,
1 an=--• a-n
NUMBERS AND OPERATIONS
§§ 14-15]
1 6. 2-6a2
EXAMPLES.
=
19
25 32 a2 = a2 •
7. 4°
= 1.
3
8• a_ab-2 b2 - a .
We shall leave the proof to more advanced courses in mathematics, but it is true that we can use the laws of exponents of § 12 in all operations with the exponents now defined, even when the exponents are not positive integers. All problems in this book that involve such exponents are to be solved on this assumption. We shall also assume that, unless it is stated otherwise, literal numbers appearing in the bases or under m
radicals are such as to exclude the excepted case of an where a is negative and n is an even integer.
15. Imaginary numbers. It is necessary now to introduce into the number system a new class of numbers called imaginary numbers, in order, among other things, to provide even roots of negative numbers. These new numbers are "imaginary" only in the technical sense to distinguish them from the "real" numbers. The real and imaginary numbers together form the system of complex numbers. We assume that there exists a certain imaginary number, which we shall designate by the letter i, whose square is equal to -1. That is
i 2 = -1.
(1)
We shall also write
(2) and assume that all of our algebraic operations can be applied to this number i along with other real and imaginary numbers, and that these operations obey the ordinary rules of algebra. On this assumption
( - i) 2
= (-
1) 2 ( i) 2
=
(1) ( -1)
= -1.
Therefore, both i and -i are square roots of -1. In this case the notation i = V -1 is not to be referred to as "the principal square root'' of -1. There is, in fact, no way of telling which of the two roots we call i and which is -i. We assign the letter i to one of the roots and then stick to the notation.
20
FIRST YEAR OF COLLEGE MATHElvIATICS
[Ch. I
On our assumption, if a and b are any two real numbers, and b ~ 0, a + bi has a definite meaning and is spoken of as an imaginary number. In this number, a is the real term and bi is the imaginary term. "Positive numbers" and "negative numbers" are always real, so that 4 i and -4 i are neither "positive" nor "negative," but are ''imaginary." But we do say that a + bi and -a - bi are the negatives each of the other. Thus 4 i is the negative of -4 i, and -4 i is the negative of 4 i, and 2 - 3 i is the negative of -2 + 3 i although no one of these numbers is positive or negative. If b is any positive number, we shall use the symbol -b to denote the number iVb. Then (-b) has the two square roots
v
v-b = ivb
(3)
and
--v-:::fi = -ivb.
For
(ivb) 2 = i 2 (Vb) 2 = (-l)(b) = -b, (-ivb) 2 = (-i) 2 (vb) 2
and
= (-l)(b) = -b.
Notice·that i2 i3 i4 i5 i6
= = = =
-1, i 2 • i = ( - l)i = -i, i 2 • i 2 = (-1)(-1) = 1, i 4 • i = ,i, . =i 4 ·i2 = -1,etc.
Since i 4 = 1, the factor i4, i 8, i 12 , or any whole power of i 4 may be discarded. In this book it ,vill be assumed that any literal number except i is real, unless the contrary is stated. 1.
EXAMPLES
2.
v -16 a
2
V -9 = i-V9 = 3 i = 3-V-1.
= iV16 a2 = 4 ai = 4 av=I, if a> 0.
3. (v -7) 2 = -7. 4. i74
= i72.
i2
=
(i4)18. i2
=
(1)18(-1)
= -1.
5. The two square roots of -36 x2y 4 are ±6 ixy2 = ±6 xy 2v=I.
CHAPTER II
Coordinates and Loci 16. Directed distances. If a point moves in a straight line from a point A to a point C, it traces out the line-segment AC. We call A the initial point and C the terminal point of the segment AC. We sometimes also -A+------...B-_...,C_ denote by AC the distance from A to C, Fm. 4 or the length of the segment expressed in some chosen unit of length. If B is a point on the line between A and C, on adding lengths, we have
AB+ BC= AC.
(1)
In order that formula (1) may be true even when B does not lie between A and C, we agree that distances measured along a given line in one direction shall be counted positive, and those in the opposite direction shall be negative. That is, (2) or (3)
AB= -BA, AB+ BA= 0.
If a point starts at A, and, after moving back and forth along the line in any manner, ultimately arrives at C, the algebraic sum of all the distances traversed by the point is AC. For every segment except AC is traversed in each direction, and the sum of the lengths of these other segments is zero. In particular if the tracing point moves from A to Band then to C,
(4)
AB+ BC= AC,
for all arrangements of the points A, B, C on the straight line. A
B
CB
A
Fm. 5 21
BA
C
22
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. II
17. Rectangular Cartesian coordinates. To form a rectangular system of coordinates, we first draw a horizontal straight line OX and a vertical line OY. Their point of intersection, 0, is called the origin, OX is the x-axis, and OY is the y-axis .. We now lay off a scale of real numbers on each axis, as in § 6, taking zero at the origin, with positive numbers to the right of O on OX and above O on OY. Any distance or line segment parallel to OX is represented as a positive number if it js measured to the right, and as a negative number if it is measured to the left. Distances along any vertical line are positive if measured upward and negative if measured downward. y (-4, 9)
-4
10 9
X
8
P"(x, y)
7 6
y
5
9
4 3
8
2
-10-9 -8 -7
-1-6
(-6, -4)
3
1 0 -4-3-2 -1-1 0 l 2 3 4
-4
-2 -3
-6
(8, 3)
6 7 8
9 10
X
-7
-4
-6 -6 -7 -8
5
(5,-7)
-9 -10 FIG.
6
Now let P be any point of the plane, and let its directed distance from OY be x and its directed distance from OX be y. Then x is called the abscissa or x-coordinate of P, and y is the ordinate or y-coordinate of P. Point Pis said to have the coordinates (x, y) and may be referred to as the point (x, y). The position of any point can be found if its coordinates are known and, conversely, if a point is marked, its coordinates can be found by measurement. Just as a real number scale establishes a correspondence between the points of a line and the real numbers, so our coordinate system sets up a correspondence between the points of a plane and pairs of real numbers.
§ 17]
COORDINATES AND LOCI
In Figure 6, points (8, 3), (-4, 9), (-6, -4) and (5, -7) are marked. The axes divide the plane into four quadrants numbered as shown in Figure 7. This figure also shows the signs of the coordinates of points in the different quadrants. Such a system of coordinates as we have described is called a system of rectangular Cartesian coiirdinates after the French philosopher, Rene Descartes who, in 1637, laid the foundations of Analytic Geometry.
23
y II (-, +)
I
(+, +) ,X
0
IV
III
(-,-)
(+,-) FIG. 7
EXERCISES 1. Write the coordinates of each of the points A, B, C, D, E, in Figure 8. 2. Write the coordinates of each of the points F) G, H, I, J, K, in Figure 8.
y 10
!H
J
C
Go 6
A
Ko
E -10
0
-6
JO
i)
X
F I
-6
B I D
JlO FIG. 8
In Exercises 3 to 17, plot the points having the given coordinates. In careful plotting use printed cross-section paper. For rapid work, mark a scale on each axis and estimate distances by the eye or with a ruler. 3. (2, 5). 4. (7, 3). 8. (-3, -1). 9. (4, -6). 13. (0, -4). 14. (0, 6).
5. (-3, 4).
10. (3, -8). 15. (-8, 0).
6. (-6,2). 11. (0, 0). 16. (-t, 3).
7. (-5, -7). 12. (5, 0). 17.
(--i,
-t).
FIRST YEAR OF COLLEGE MATHEMATICS
24
[Ch. II
In Exercises 18 to 29, plot each pair of points, A and B, and find the distance from A to B.
24. A : (-3, 8), B : (-3, 5).
19. A 21. A 23. A 25. A
26. A: (-4, 3), B: (-4, -6).
27. A : (3, a), B : (3, b).
28. A : (c, -5), B : (d, -5).
29. A : (a, b), B : (a, c).
18. A : (2, 4), B : (8, 4). 20. A : (-2, 6), B : (7, 6). 22. A : (4, 3), B : (4, 9).
: : : :
(7, 2), B : (3, 2). (5, -3), B : (-2, -3). ( - 6, 0), B : ( - 6, 7). (6, 9), B : (6, Q).
18. Projections. If M1 is the foot of the perpendicular drawn from a point P1 to a straight line l, M1 is said to be the projection p2 of P1 on l. lf P1 and P2 are any two points, and M1 and M2 are their respective projections on l, then the line segP1 ment M1M2 is the projection on l of the segment P1P2. If distances on l are di_ _.___ _ _ _ _.....___ z rected, so that M1M2 = -M2M1, then M1 M! since, by definition, M2M1 is the projection of P 2 P 1 , it follows that the projection Frn. 9 of P2P1 is the negative of the projection of P 1P2. Theorem. If P1:(x1, y1) and P2:(x2, y2) are any two points, the proY jection of P1P2 on the x-axis is equal to (X2 - x1), and the projection of P1P2 on the y-axis is equal to (y2 - Y1). Proof. Let M 1 and M 2 be the projections of P1 and P2 respectively on the x-axis. The abscissa of M1 is the same as that of P1 and is there£ore equal to x1; the abM1 0 M2 scissa of M 2 is x2. The proFIG. 10 jection of P1P2 on the x-axis is
--'--------------x
(I)
M11112 = Jv.L10
+ 01vf
2
=
-xi
+
x2
= x2 -
In a similar way, the projection of P1P2 on the y-axis is (2)
x1.
COORDINATES AND LOCI
§§ 17-19]
25
Figure 10 shows just one of the ways in which the points P1 and P2 may be arranged with respect to each other and to the coordinate axes. The student should draw the figure for some of the other arrangements, taking P 1 in the fourth quadrant, for example, and P2 in the second quadrant.
Corollary. If P1 :(x1, Yi) and P2 :(x2, y2) are any two points, the projection of P1P2 on any line parallel to the x-axis is (x2 - x1), and the projection of P1P2 on any line parallel to y the y-axis is (y2 - Yi). EXAMPLE. Given the points Pi:(-3, 2), P2:(5, 4), P 3 :(3, -5), find the projections of P1P2 and of P? 3 on the coordinate axes. Solution. By equations (1) and (2), above, the projection of P1P2 on OX is -5-4-3-2-1 01 2 -1 NI1M2 = 5 - (-3) = 8. -2 The projection of P1P2 on OY is -3 N1N2 = 4 - 2 = 2. The projection of P?a on OX is M2Ma = 3 - 5 = -2. The projection of P?3 on OY is FIG. 11 N~a = -5 - 4 = -9.
19.
Distance between two points.
Theorem: If d is the distance between two points, P1 :(xi, Y1) and P2 :(x2, Y2),
(1)
d
x1) 2 + (y2 - Y1) 2;
= ±V(x2 -
that is, the distance between two points is numerically equal to the square root of the square of the difference between their abscissas plus the square of the difference between their y ordinates . ...__ _ _ _---,l')Plx2'y2) Proo1. Let 1111 and M2 be N2 the projections of P1 and P2, respectively, on the x-axis, and let Ni and N2 be their projections on the y-axis. Let K be the point of intersection of P1N1 and P2M2 (produced if X necessary). Then P1P2 is the 0 hypotenuse of a right triangle of which P1K and KP2 are the FIG. 12 other two sides. Therefore --2
(2)
d2
=
P1P2
--2
=
PiK
--2
+ KP2 ·
26
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. II
But P1K is the projection of P1P2 on a line parallel to the x-axis, and KP2 is the projection of P1P2 on a line parallel to the y-axis. Consequently and
(3)
KP2 = Y2 - Yl•
Therefore (4)
and we obtain equation (1). Equation (1) is a very important formula. Unless there is some special reason for directing the distance P1P2, we are ordinarily interested only in the numerical value of the distance d; we customarily use the positive square root and do not need to retain the ± sign. Since (x2 - x1) 2 = (x1 - x2) 2 and (y2 - y1) 2 = (y1 - y2) 2 , it is immaterial which is taken as the first point and which as the second. The proof holds for all arrangements of points P 1 and P2. The student should draw other figures, and see that this is true.
Since the coordinates of the origin are (0, 0), in order to find the distance of a point P1:(x1, y1) from the origin, we set x2 = 0, Y2 = 0 in formula (1) and have
(5) EXAMPLE 1. Find the distance between the points (5, -3) and (-7, 2).
Solution.
By (1),
d = v'(-7 - 5) 2 + (2 + 3) 2 = v'144 + 25 = V169 = 13. EXAMPLE 2. Show that the point (-1, -2) is the center of the circle through the three points (5, 6), (-7, -10), (-9, 4).
Solution. The student should draw the figure. (-1, -2) and (5, 6) is
+
+
The distance between
d1 = v'(-1 - 5) 2 (-2 - 6) 2 = v'36 64 The distance between (-1, -2) and (-7, -10) is
=
10.
+
d2 = v(-1 7) 2 + (-2 + 10)2 = v'36 + 64 = 10. The distance between (-1, -2) and (-9, 4) is d3 = v(-1 + 9) 2 + (-2 - 4) 2 = v'64 + 36 = 10. Since d1 = d2 = d3, (-1, -2) is equidistant from the other three points1 and is the center of the circle which passes through them.
COORDINATES AND LOCI
§§ 19-20]
EXERCISES In every exercise draw the figure. In Exercises 1 to 8, find the distance between the two given points. 1. (-8, -2), (-5, 2). 2. (-6, 7), (-1, -5). 3. (3, 5), ( -2, 1). 4. (5, -2), (8, 3). 5. (7, -3), (-8, 5). 6. (½, -½), (0, 1). 7. (x, y), (1, 3). 8. (6, a), (8, b). In Exercises 9 to 12, find the distance of the given point from the origin. 9. (5, 12). 10. (-8, 4). 11. (12, -8). 12. (-6, -10). 13. Find the lengths of the sides of the triangle whose vertices are (-3, -1), (5, 5), (-10, 13). 14. Find the lengths of the diagonals of the quadrilateral whose vertices are (1, 3), (-2, -1), (5, 0), (-4, 2). 15. Show that (-3, 2) is the center of the circle drawn through the three points (10, 2), (9, -3), (-8, -10). 16. Show that (5, -10) lies on the perpendicular bisector of the line segment joining (4, 3) and (-2, 1). 17. Given the three points A : (5, I), B : (-1, 5), C: (-4, -6); show that angle ABC is equal to angle CAB. 18. Show that the three points (-1, 4), (-7, 0), (2, 6) are collinear (that is, lie on a straight line). 19. Show that (11, I), (-11, 5), (-5, -7) are the vertices of a right triangle. Find its area. (Suggestion. What is the relation between the lengths of the sides of a right triangle?) 20. Show that (-1, 4), (-4, 7), (1, 6) are the vertices of a right triangle, and find its area.
20. Equations and loci. We have already seen that there is a correspondence between points of a plane and pairs of real numbers. A correspondence may also be established between certain geometrical figures and certain equations in x and y. A given equation in x and y is not satisfied* by arbitrary values of x and y. That is, the coordinates (x, y) of a point chosen at random are not likely to satisfy the equation. Only the coordinates of certain selected sets * A pair of numbers is said to satisfy an equation in x and y, if the two members of the equation are equal when the first of the two numbers is substituted in place of x in the equation and the second of the two numbers is substituted in 4 = 0 is satisfied by the numbers place of y. Thus the equation 5 x - 3 y x = 1, y = 3. For, on substituting x = 1, y = 3 in the equation, we have 5(1) - 3(3) 4 = 5 - 9 4 = 0. But the same equation is not satisfied by X = 3, y = 4, for 5(3) - 3(4) 4 ~ 0.
+
+
+
+
28
FIRST YEAR OF COLLEGE MATHE1\1ATICS
[Ch. II
of points do satisfy it. The points whose coordinates do satisfy the equation form a geometrical figure which is called the locus of the equation, and the equation is said to be the equation of the figure. This relation may be put in the following form. Fundamental Definition: The locus of an equation is the totality of po·ints whose coordinates satisfy the equation. Notice that this implies two things: (1) the coordinates of any point on the locus satisfy the equation, and (2) any point whose coordinates satisfy the equation lies on the locus. (We shall refer to any correct form of the equation as the equation of the locus.) The circle whose center is (4, 5) and whose radius is 3 is the locus of the equation (1) (x - 4) 2 + (y - 5) 2 = 9. For let d be the distance from the point (4, 5) to a point (x, y). Then, by § 19, 8 (2) d2 = (x - 4) 2 + (y - 5)2. 7 16 Therefore, if (x, y) is any point on the circle in :5 question, d2 = 9, and its coordinates satisfy -4 equation (1). Moreover, if (x, y) is any point ,3 whose coordinates satisfy equation (1), we see 2 1 from that equation that d2 = 9 and d = 3, and 0 2 X the point lies on the circle. 1 3 4 5 6 7 -1 -1 EXAMPLE 2. An equation of the perpen-2 dicular bisector of the line segment joining FIG. 13 (-4, 3) and (5, -2) is 2 (3) (x + 4) + (y - 3) 2 = (x - 5) 2 + (y + 2) 2, since a point (x, y) on the perpendicular bisector is equidistant from ( -4, 3) and (5, -2). The student should complete the proof and also show that equation (3) can be simplified to the form 9 x - 5 y - 2 = 0. EXAMPLE 3. Sketch the locus of the Y equation (4) X 2 - 2 X - y - 5 = 0. Solution. Solving the equation for y in terms of x, we have EXAMPLE
(5)
1.
y
=
X2 -
2X
-
5.
By substituting various values of x, by (5) we obtain the corresponding values of y. Thus, when x = - 2, y = 4 + 4 - 5 = 3. The point ( - 2, 3) therefore lies on the locus. In this way we form the following table of values:
FIG. 14
§ 20]
COORDINATES AND LOCI
2~
-2 -1 0 1 2 4 3 -- ---- -- -- -- -2 -5 -6 -5 -2 3 y 3 X
-
We plot the points having these pairs of values of x and y as coordinates. They are the points indicated by circles in Figure 14. From equation (5) we see that small changes in x prnduce only small changes in y. There are therefore no breaks in the curve. We therefore connect the points which we have plotted by a smooth curve all of whose points lie on the locus of the given equation. We cannot, of course, draw the entire locus, for in this case it is infinite in extent. However, we have drawn enough of the curve to indicate its most interesting pr0perties.
EXERCISES Find an equation for each of the following loci, and draw the locus. 1. Circle, radius 2, center at (1, -3). 2. Circle, radius V7, center at (3, 4). 3. Circle, radius
2V2, center at the origin.
4. Circle, radius 4, center at (0, -5). 5. The perpendicular bisector of the line segment joining (5, -8) and (0, 1). 6. The perpendicular bisector of the line segment joining (-3, 2) and (4, -3).
7. The locus of a point equidistant from (2, 8) and ( -2, 2). 8. The locus of a point equidistant from (-3, 0) and (0, 6). 9. Circle, center at (-5, 2) and passing through the origin. 10. Circle, center at (4, -2) and passing through (5, 1). 11. The straight line parallel to the y-axis and 6 units to th~ right of it. 12. The straight line parallel to the x-axis and 5 units above it. 13. The straight line parallel to the x-axis and 4 units below it. 14. The straight line parallel to the y-axis and 7 units to the left of it.
Identify and draw the locus of each of the following equations. 15. (x - 4) 2 17. X = 3.
+ (y + 7)
2
= 36.
18. y = 11.
+ 2) + (y - 8) = 25. y + 6 = 0. 20. X + 19
16. (x 19.
2
2
Sketch the locus of each of the following equations. 21. y - 5 X 2
23. x 25. x2
= 0.
8 x - y + 10 = 0. + y2 = 25. -
+ 4 X = 0. x + 12 x - y + 3 = x +y 36 = 0.
22. y
24. 26.
2
2
2
-
0.
=
0~
CHAPTER III
Functions and Their Graphs 21. Constants and variables. A number or number symbol is a constant if it is assumed that it has just one value throughout a problem. There are certain fixed constants, such as 3, -6, 1r, each of which has the same value in all problems where it appears. Other constants, usually denoted by the early letters of the alphabet, a, b, c, etc., retain their values without change throughout the discussion of a problem, but may change from one problem to another. A quantity or number symbol is a variable if, in a problem, it is free to take on any one or more of a certain set of values. This set of values is sometimes called the range of the variable. A variable is usually denoted by one of the later letters of the alphabet, such as t, x, y, or z. As things turn out, a variable may not change in value. For example, when studying the temperature of a freezer locker during a period of twenty-four hours, we might find that the temperature did not change at all. Nevertheless, we should consider the temperature as a variable, since we did not assume at the outset that it was fixed in value. In the formula S = 1rr2, for the area of a circle in terms of its radius, 1r is a fixed constant. But S and r may be thought of as variables, since the formula is true for all positive values of r, and there is no assumption that r or Sis restricted to a single value. EXAMPLE.
22. The definition of a function. Using the relation S = 1rr2 , we can find the area of a circle when its radius is known. For example, when the radius is 3 inches the area is 9 1r or approximately 28.3 square inches; when the radius is 2 centimeters, the area is 4 1r or approximately 12.6 square centimeters. The area, S, is said to be a function of the radius, r, because the value of S can be found when the value if r is known. 30
FUNCTIONS AND THEIR GRAPHS
[§§ 21-22]
31
Definition: A variable y is said to be a function of a variable x, if x and y are so related that when a value is assigned to x a corresponding value (or more than one corresponding value) is determined for y. The idea of function is one of the most important concepts in mathematics. The underlying notion, of course, is that the value of the function y depends on the value of x. For this reason, x is said to be the independent variable, and y is called the dependent variable. The definition does not specify the way in which the value of the dependent variable is determined when a value is assigned to the independent variable. There are many ways in which the correspondence can be established. For example, a function may be defined by giving a table of corresponding values of the two variables, or by a graph, or by the use of instruments of measurement such as clocks and thermometers. The most common method in algebra, however, is to define the dependent variable by means of an expression containing the independent variable. EXAMPLE
1.
Let y = x2
-
4x
+ 1.
This expression defines y as a function of x. For, by substituting any particular value for x, we can determine a corresponding value of y. Thus, when x = 0, y = I; when x = 1, y = -2; when x = ½, y = -¾; when X -- - .! 3, y-- l9i •
It may be that there are certain sets of values of the independent variable for which the dependent variable is not defined. The values of the independent variable for which the function is defined form the range of definition of the function. 2.
If y = (x _ l;(x _
), y is defined for all values of x 2 except x = I and x = 2, for which the denominator is equal to 0. The range of definition of the function y consists of all values of x except x = 1 and x = 2. EXAMPLE
In the examples given so far, only a single value of the dependent variable was determined for each value of the independent variable in the range of definition. If x and y are connected by the relation y 2 - x 2 = 1, then y = ±Vl + x 2• Therefore there are two values of y determined for each value of x. We say that y is a single-valued function of x, if just one value of y corresponds to each value of x; y is a double-valued function of x, if two values of y correspond to each value of x, etc.
J2
FIRST YEAR OF COLLEGE l\1:ATHEMATICS
[Ch. III
We commonly use the notation f(x), which is read "the !-function of x" or, more simply, "f of x," to stand for a function of x. Suppose that y is some function of x. We may then write y = f(x). Then f(5) is the value of y that corresponds to x = 5; f( - 2) is the value that y takes on when x = -2, etc. In general f(a) is the value of y corresponding to the value x = a; f(a) can ordinarily be found by substituting a for x in the expression for f(x). - It should be noted that f(x) does not mean a number f multiplied by a number x. Indeed
f is not, by itself, a number. In a given problem, f rather denotes the relation of correspondence between the independent and dependent variables, or, in the case of an expression, the structure of the expression. The number in parentheses indicates the independent variable in the correspondence. EXAMPLE 3.
Let y = x2
4x
-
+ 1.
Let us denote this function by f(x), so that
= f(x) =
y
f (2) = 2
2
Then:
f(O) = 0
2
-
x2
4x
-
+ 1.
+ 1 = - 3, + 1 = 1, 4(-3) + 1 =
4 · 2 4 · 0
f(-3) = (-3) 2 f(a) = a2 - 4 a+ 1, f(x + 2) = (x + 2) 2 - 4(x
22,
+ 2) + 1 =
x2
-
3.
Sometimes we must use more than one function in the same problem. In this case, different functions should be represented by different letters, such as F(x), f'(x), f"(x), G(x), etc. EXERCISES In each of the Exercises 1 to 14, let f (x) = x2 number or expression indicated. 1. f(O). 6. f(-2). 11. J(a - 1).
2. f(4). 7. f(a). 12. f(2 x).
3. f(-1). 8. f(a2). 13. 2f(x).
+2x -
4. f(½ ). 9. f(a + 1). 14. f(a 2) + 5.
3, and find the
5. f(-½). 10. f(x + 1).
In each of the Exercises 15 to 26, let f (x) = x2 - 2 x G(x) = 2 x2 + 3, and find the number or expression indicated. 15. f(3).
16. G(2).
18. f (3) X G(2). 21. f(l) X G(O). 24. f(x) + G(x).
19. f (3) + G(2). 22. f(-4) + G(-2).
25. G(x) - f(x).
+5
+
17. f(3) G(2). 20. f(3) - G(2).
+
23. f(O) G(-3). 26. 3 G(x).
and
FUNCTIONS AND THEIR GRAPHS
§§ 22-23]
33
23. The graph of a function. A graph gives us a way of forming an idea of the properties of a function. Suppose that we are given some function y
= f(x).
Definition. The graph of a function f(x) consists of all the points whose x- and y-coordinates correspond according to the relation y = f (x). When f(x) is given as an expression in x, the equation y = J(x) is an equation in x and y. It should be noticed that the graph of the function f(x) is the same as the locus of the equation y = f(x), since
it contains all of the points whose coordinates satisfy that equation and no other points. EXAMPLE. Solution.
Graph the function y = f(x) = x3
x.
-
By substitution we find the following pairs of values of x and y• 0
X
y
t
0
1
2 3
8
1
2
3
2
-1
-2
0
-\5
6
0
-6
y Then the points (0, 0), (½, -!), (-½, !), 6 (1, 0), (!, \ 5 ), (2, 6), (-1, 0), (-2, -6) lie 6 on the graph of the given function. Of course, 4 these are only a few of the infinitely many points of the graph. Here, as with most functions that 8 we shall graph, when the value of x changes 2 gradually, the corresponding value of y also changes gradually, with no abrupt jumps. Consequently any two points of the graph will be _3 _2 connected by a smooth curve, each point of which has coordinates x and y which are in the -2 relation y = f (x) = x 3 - x. In Figure 15 the -3 points whose coordinates are given in the table -4 are marked with circles. The rest of the graph, -5 between x = - 2 and x = 2, is found by connecting these few points by a smooth curve. -6 The curve extends indefinitely far to the right FIG. 15 and left of this portion. But the part shown is the ''most interesting'' part of the curve in the sense that it includes all points where the curve changes direction quickly and that we can easily imagine the behavior outside of this region. The scale should be marked clearly on each axis.
sx
FIRST YEAR OF COLLEGE MATHEMATICS
34
[Ch. III
EXERCISES Make the following graphs on printed cross-section or coordinate paper. Mark all plotted points with a small neat circle. When you connect the points, do so very carefully in order to get a smooth, attractive graph. Its size and beauty will depend partly on your choice of scales. Mark the scales on the axes. In each of the Exercises 1 and 2, graph the function y = f(x) from the given table.
-3 -2 -1 0 1 2 3 4 5 --- -- - -- - - -11 -3 -1¾ 4 5½ 4 3¼ 3 4 7 -4
X
y
= f(x)
-5
X
2.
Y = f(x)
-4
-3
8
2
-2
-1
0
1
3
5
---------- - 18
0
1
-
2
6
-
3
12¾
5 6 -2 -3 -12
4
5
6
--
3. (a) Graph the function y = f(x) from the values f(-3) = 6, f(-2) = 0, f(-1) = -4, f(0) = -6, f(l) = -6, f(2) = -4, f(3) = 0, f(4) = 6. (b) From the graph find the values of f(-i), f(½), J(i). 4. (a) Graph the functionf(x) from the valuesf(-2) = -5,f(-1) = 0, f(O) = 3, f(l) = 4, f(2) = 3, f(3) = 0, f(4) = -5. (b) From the graph find the values off(-½), f(}). In Exercises 6 to 20, graph the given functions. In intervals in which a graph curves rapidly it will probably be necessary to substitute some fractional values of x, (such as ½and -¾). Give a table of coordinates of the points used.
5. f(x) = 3 x. 7. f (x) = 3 - 2 x. 9. f(x) = x 2 - 2 x - 3. 11. f(x) = 6 - x - x 2 •
+x + 3.
13. f(x) = 2 x2 16. f(x) = x 3
17. f(x) 19.. f(x)
= x3 = x4
-
x2
-
10.
-
x
+ 1.
f (x) = 2 X - 4. 8. f (X) = ½X - 1. 10. f (X) = X 2 - 3 X. 12. f(x) = 8 + 2 x - x 2 • 14. f(x) = 2 x2 + x - 6. 16. f(x) = x 3 - g x. 18. f(x) = 6 - x - x3• 20. f(x) = ½x 4 - 3 x2 + ~6.
30. 21. For each of the Exercises 6 to 20, estimate from the graph the values of x for which f (x) = 0, correctly to the nearest ¼. -
FUNCTIONS AND THEIR GRAPHS
§§ 23-24]
35
24. Graphs from observations. Graphs are often used to display and interpret functions measured in laboratory or statistical work. 1. Two radio tubes were tested by gradually increasing the grid voltage and measuring the resulting plate current, which is used in radio reception. The readings were as follows: .., EXAMPLE
Tube A. Grid Volts
-60 -50 -40 -30 -20 -10 0 -----------
Plate Current, Amperes
20
10 -
-
30 -
40
50
-
-
60
-
70 -
.012 .022 .035 .055 .085 .125 .184 .226 .253 .267 .278 .284 .290 .293
Tube B. Grid Volts
-60 -50 -40 -30 -20 -10 0 -------------
Plate Current, Amperes
10 -
20 -
30
-
40
-
50
-
60
-
70 -
.010 .018 .028 .042 .065 .091 .123 .157 .181 .200 .215 .227 .236 .243
y tll
~
.80
A
Q)
A
s
< ...; fil .25
B
J..-4
~
t.)
.05
-60 -50 -40 -80-20 -10
0
10
20
FIG. 16
so
40 5o 60 7o Grid Volts
so X
FIRST YEAR OF COLLEGE l\1ATHEMATICS
36
[Ch. III
The current is controlled by adjusting the voltage, so that the current is a function of the voltage. The values of the voltage are therefore plotted as abscissas and the corresponding values of the current as ordinates. Thus, for tube A, we plot the points (-60, .012), (-50, .022) etc., and for tube B the points (-60, .010), (-50, .018) etc. In Figure 16 these points are marked with circles. For such functions we do not have an algebraic expression for one variable in terms of the other. But because of the nature of the quantities, we know that a gradual change in the voltage causes a gradual change in the current. We therefore complete the graphs by drawing smooth curves through the points that have been plotted. EXAMPLE 2. The following table gives, for each of the eighteen months from January, 1948 to June, 1949, inclusive, the highest value attained during that month by the composite price of the shares of a certain group of 70 industrial stocks. Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec. Jan. Feb. Mar. Apr. May June '48 '48 '48 '48 '48 '48 '48 '48 '48 '48 '48 '48 '49 '49 '49 '49 '49 '49
--159
-
-
-
-- -
155 156 159 162
163 162
159
-- 159 161
-- 160
-
-
-- -
-
156 157 156 155 155 154
-153
The ''high'' for the month can be regarded as a function of the number of the month. This function represents a different type of function from that of Example 1. It is given only for the values indicated in the table, and it has no meaning for intermediate values of the independent variable. After plotting the points from the table, we therefore do not connect them with a smooth curve, for this would suggest gradual changes in the variables. The graph really consists of the isolated points marked 163 / I\ with circles in Figure 17. It 162 } \ 161 makes the changes more vivid, \ I 160 however, to connect these I V \ 169 points by straight lines as is J \ 168 \ \ I • done in the figure. The sharp 157 \ ,/ \ I ~ corners and straight lines 166 l / 156 show us that we have here 164 a function that jumps '\ 153 brusquely from one isolated 162 value to another, rather than one that varies gradually. Since the values of the function all lie within the interval Fm. 17 from 153 to 163, it is not necessary to leave room on the figure for the values from O to, say, 150.
'
'"
" ,,
FUNCTIONS AND THEIR GRAPHS
§ 24]
37
EXERCISES 1. The pressure of air was measured at various heights above sea-level. The table gives values of the air-pressure in inches of mercury as x and the corresponding altitudes in thousands of feet as y. Graph y as a function of x. From the graph find the height at which the pressure is 27.5 inches and find the pressure at 4000 feet. x (inches) y (1000 ft.)
22 23 24 21 25 26 27 28 29 30 -- -- -- -- -- -- -- -- -- -10.6 9.1 7.8 6.6 5.5 4.5 3.5 2.6 1.7 0.8 0 20
2. According to government statistics, the following table gives the average weights and heights of boys in the United States according to their age. Age (years)
1
2
-
-
5
4
3
-
-
6
-
8
7
-
-
9
-
11
10
-
-
13
12
-
14
-
-
-
-
15
16
----
Weight 21.9 27.1 32.2 35.9 41.1 45.2 49.1 53.9 59.2 65.3 70.2 76.9 84.4 94.9 107.1 121.0 (pounds)
-
Height (inches)
-
-
-
-
-
-
-
-
-
-
29.4 33.8 37.1 39.5 41.6 43.8 45.7 47.8 49.7 51.7 53.3 55.1 57.2 59.9
---62.3
65.0
(a) Graph weight as a function of age. (b) Graph height as a function of age. (c) Graph weight as a function of height. 3. If x is the speed of a car in miles per hour, and y is the number of hours required to travel 100 miles at the speed x, then xy = 100. Graph y as a function of x from x = 5 to x = 90.
4. Suppose that a stone is dropped from a cliff, and denote by s the number of feet that it falls in the first t seconds. Then, approximately, s = 16 t 2 • Graph s as a function of t from t = 0 to t = 5. How high is the cliff if it takes a stone 2½ seconds to fall to the bottom? 6. If a ball is thrown upward with a speed, at the start, of 48 feet per second, at the end of t seconds its height above ground is approximately s feet, where s = 48 t - 16 t2 • (a) Graph s as a function of t, for every half-second from t = 0 to t = 3. (b) From the graph find the greB,test height to which the ball rises. (c) From the graph find the length of time before the ball strikes the ground. 6. Graph the number of oranges that can be bought for $2.00 as a func-
tion of the price per dozen, for prices from 20 cents per dozen to $1.00 per dozen.
38
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. III
7. The following table shows, to the nearest cent, the amount of one dollar at the end of each two-year period for 20 years, if it is invested at 5% per annum compounded annually. Graph the amount as a function of the time. From the graph determine the amount at the end of 5 years and at the end of 15 years.
Time (years)
0
2
4
6
10
8
14
12
16
18
20
-- -- -- -- -- -- -- -- -- -- --
Amount (dollars)
1.00 1.10 1.22 1.34 1.48 1.63 1.80 1.98 2.18 2.41 2.65
8. The following table gives the estimated yearly value of all farm
products of the United States in billions of dollars. function of the time. Year Value in billions of dollars Year Value
Graph the value as a
1928 1929 1930 1931 1932 1933 1934 1935 1936 1937 -- ---- -- ---- -- -- -- -13.6 13.8
11.4
8.4
6.4
6.9
9.0
8.0
10.4 10.9
1938 1939 1940 1941 1942 1943 1944 1945 1946 1947 1948
------ -------- -- -- -- -9.6
9.7
10.2 13.3 17.9 22.4 23.4 24.7 28.5 34.4 35.6
9. The following table shows the current th~t is found to flow through a
certain tungsten lamp at different voltages. Graph the current as a function of the voltage. Estimate the voltage when the current is .40 ampere, and the current at 110 volts. Volts
-
Current (Amperes)
10
-
25
-
40
-
70
55 -
85 100 115 130 145 160 175 190
-
-
-- -
-
-
-- -
.10 .18 .24 .29 .34 .39 .43 .46 .50 .53 .57 .60 .62
10. The voltage in a certain alternating current generator is controlled by changing the current in the field of the generator, as is indicated by the following measurements. Graph the voltage as a function of the field current. What current must be used to produce a voltage of 2250?
Field Current, Amperes Volts
5
10
20
30
40
600
990
1610
2050
2360
--
~160
2680
§§ 24-26]
FUNCTIONS AND THEIR GRAPHS
39
25. Functions of several variables.
A variable z is said to be a function of the two variables x and y, if z is related to x and y in such a way that when a value is assigned to x and a value is assigned to y a corresponding value (or more than one corresponding value) is determined for z. Here x and y are independent variables, and z is the dependent variable. The notation
z = f(x, y),
is read "z is equal to f of x and y" or "z is equal to the f function of x and y." The student should write a definition corresponding to the notation w = f(x, y, z). EXAMPLE 1. Given z = f(x, y) = x2 - xy + y 2, we find /(2, 3) = 22 - 2. 3 + 32 = 7, f(-3, 1) = (-3) 2 - (-3)(1) f(3, 0) = 32 - 3 · 0 + 02 = 9, J(2 a, b) = 4 a2 - 2 ab+ b2 •
+ l2 = 13,
EXAMPLE 2. The volume, V, of a right circular cylinder of radius r and height h can be found if r and h are known. Therefore V is a function of rand h: V = 1rr2h.
EXERCISES 1. Given z = f(x, y) = 2 x - y + 3, find f(O, 0), f(l, 1), f(2, 1), f(l, -3), f(a, b), f(y, x), f(-x, y). 2. Given z = f(x, y) = 3 x - 2 y + 4, find f(0, 0), f(l, -1), f(½, ½), f(i, -½), f(2 a, 3 a), f(-x, -y), f(y, x). 3. Given z = f(x, y) = x 2 - 3 xy y2, (a) prove thatf(-x, -y) = f(x, y); (b) prove that f(y, x) = f(x, y). 4. Given z = f(x, y) = x2 3 y - y 2• (a) Are f(-x, y) and f(x, y) always equal? (b) Are f(x, -y) and f(x, y) always equal? 6. Let v be the speed of an airplane in miles per hour, d the distance traveled in miles, t the time in hours. Express (a) v as a function of d and t, (b) d as a function of t and v, (c) t as a function of v and d.
+
+
26. Two important concepts. In this chapter we have introduced the notion of functional dependence. In the two preceding chapters we showed how, by means of a scale of numbers or by a system of coordinates, we can establish a correspondence between points and sets of numbers. This correspondence and the related idea of function are two of the most important concepts of mathematics. Together they give unity to most of the topics that will be studied in the later chapters of this book.
CHAPTER IV
Linear Functions and Their Graphs 27. Rational integral expressions. Among the most important functions are the rational integral functions or polynomials. The expressions a, ax, ax2, ax3, • • • , where a does not contain x, are rational integral terms in x. In a similar way, if a does not involve x or y, such terms as a, ax2, axy, axy3, ax4y 2 are rational integral terms in x and y. More generally, a term is a rational integral term in certain specified literal numbers if it is either a number that does not ,£nvolve those numbers at all or is the product of such a number by positive integral powers of the literal numbers in question.* A rational integral expression is an algebraic sum of rational integral terms. If it contains just one term, it is a monomial. If it contains just two terms, it is a binomial; if there are three terms, it is a trinomial. If a rational integral expression contains more than one term, it is a polynomial.
There are many theorems relating to rational integral expressions in general and, for the sake of brevity, the word polynomial will often be used instead of "rational integral expression'' even when there is only a single term. t A rational integral term is said to be of degree k or of the kth degree in x, if k is the exponent of x in the term. It is of degree k in two or more literal numbers if the sum of the exponents of those numbers is k. A term that does not contain the literal numbers in question is considered to be of degree zero. The degree of a polynomial is the highest degree possessed by any of its terms. An expression of first degree is linear, and one of second degree is quadratic.
* The word integral is used to indicate that the term does not involve division by an expression containing the literal numbers. Thus 7 /x is fractional rather than integral in x, and x/y is not integral in x and y. On the other hand, t xis integral in x because the divisor does not contain x. The word rational is used to indicate that the literal numbers are not affected by radical signs or fractional exponents. t This is somewhat paradoxical, since polynomial means "many terms." 40
[§§ 27-28]
LINEAR FUNCTIONS AND THEIR GRAPHS
41
1. 8 x 3 is of the third degree in x.
EXAMPLES.
2. 9 x 2y3 is of the second degree in x, of the third degree in y, and of the fifth degree in x and y. 3. 2 x 3 - x2 + 5 is a polynomial of degree 3 in x. 4. 5 x 3 - 7 xy4 + 2 x2y2 - 7 x + 2 is a polynomial of degree 3 in x, of degree 4 in y, and of degree 5 in x and y. 2 5. x3 - 2 x + 3 is not a rational integral expression in x, because the X
3
term 2/x is not integral.
ORAL EXERCISES For each expression, name the degree in x, in y, and in x and y.
1. 3 x4 3. y2
-
+ x - 5. + 3 y + 7.
2 x2
-
y5
2. 3 - 2 X
-
9. x5 _
+ y. +y
4. 3 y2 6. x 2
5. xy - 1. 7. x5
+4x
+ 3 x + 2. x4y2 + y4.
2
8. x 2y 2
4 x2y2
10. x 3 y2
-
3
•
' 9.
-
8 y5
+5y
3
+3x -
3
8 x7
-
4 y2 •
+ 3.
28. Linear functions. A linear function of x is a polynomial of the first degree in x. (See § 27.) It is of the form y
=ax+ b,
where a and b are constants and a ~ 0. The name "linear" is derived from the fact that the graph of a linear function is a straight line, as we shall see in § 30. EXERCISES In each of the Exercises 1 to 6, (a) graph the linear function from a table of values, (b) determine from the graph the value of x for which the function is equal to 0, (c) determine the point at which the graph crosses the y-axis. 1. y = 3 4. y =
X
½X
+ 2. -
5.
+ x. -2 X + 4.
2. y = 5
3. y = 4 - x.
5. y =
6. y = -3 x.
In Exercises 7 to 10, for each pair of linear functions find the point of intersection of their graphs. 7. y
=
y
=
+ 1, -X + 7.
2X
8. y = 8 y
=
11,
X -
X -
4.
+ 8, X + 5.
9. y = 2 X y
=
10. y = 2 x, y
=
X -
2.
FIRST YEAR OF COLLEGE MATHEMATICS
42
[Ch. IV
29. Slope of a straight line. Before proving that the graph of every linear function of x is a straight line, we introduce the idea of slope of a line. Draw any straight line, l, not parallel to the y-axis. Let P and Q be any two points on the line and denote the coordinates of P, whatever they may be, by (x', y') and the coordinates of Q by (x", y"). Draw PK parallel to OX, M'P parallel to OY, and M"Q parallel to OY, and let PK and 1.v.f"Q iny tersect at K. If line l is very steep, the distance KQ will be numerically large compared to PK. ,, ' But if PQ is nearly horizontal, Y-Y KQ will be small compared to ~r-=-=-=-=--"7=-=-=-..:;rK PK. The ratio of KQ to PK x'!....x' can therefore be taken as a ~---+--M~,----M~,~,- - X measure of the steepness or inclination of the line l. This ratio is called the slope of the FIG. 18 line: Definition.
The slope of the line l is equal to the ratio :~-
It is customary to denote the slope by the letter m. By § 18, K Q = y" - y';
PK = x" - x'.
Substituting these values for KQ and PK in KQ/PK, we get
,,
(1)
m
,
= slope of l = Yx''-x'' - Y ·
or the slope of the straight line joining two given points is equal to the dijference of the ordinates of the points divided by the d~fference of their abscissas. The student should satisfy himself, by drawing similar triangles, that for a given line the slope remains the same if any other points P and Qare chosen on the line. Moreover, in formula (1), it is immaterial which point is used as
the first point and which as the second. However, the same order must be used for y" - y' as for x" - x'; otherwise the sign of the slope is reversed.
§ 29] EXAMPLE
LINEAR FUNCTIONS AND THEIR GRAPHS 1.
43
Find the slope of the line joining (-2, 1) and (2, 6) (line l1
in Figure 19.) Solution.
Taking ( -2, 1) as the point (x', y'), and (2, 6) as the point
(x", y"), from formula (1) we have (slope of l1) =
6-1
5
_ (- ) 2 2 4
2. Find the slope of the line joining (-2, 1) and (2, -3) (line Z2 in Figure 19.) EXAMPLE
Solution. Taking ( - 2, 1) as the point (x', y') and (2, -3) as the point (x", y"), from formula (1), we have
slopeofl2=
y
-3 - 1 -4 _ (- ) = - = -1. 4 2 2
Remarks on slope:
I. The slope of a line is positive, if the line slopes upward to the right; the slope is negative, if the line slopes downward to the right. (See lines l1 and l2 in Figure 19.) For let Q be chosen to the right of P, so that PK is positive. If the line slopes upward to the right (as l1), KQ is positive and therefore KQ/PK is positive. If the line slopes downward to the right (as Z2), KQ is negative, and KQ/PK is negative.
-3 -4
-6
Frn. 19
II. The slope of a horizontal line is 0. For KQ = 0 for any points P and Q on a horizontal line. III. A vertical line has no slope. and KQ/PK is not defined.
For on a vertical line, PK is always O,
IV. If a point moves toward the right along a straight line, the slope of the line measures the rate at which the point rises (or falls) with respect to its horizontal motion. Thus, if the slope is ¾, the point rises ¾of a unit for each unit of horizontal motion; if the slope is - 2, the point falls 2 units for each unit of horizontal motion. EXAMPLE 3. Draw the line which passes through the point (-3, 1) and has the slope ¾. ·
Solution.
The line rises
¾ unit per horizontal unit. It therefore rises
44
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. IV
3 units while extending 4 units to the right. Starting at the given point ( -3, 1), we count 4 units to the right and 3 units upward, arriving at the point (1, 4). We draw the required line through (-3, 1) and (1, 4)l
y 5 -
EXERCISES In each of the Exercises 1 to 6, find the slope of the given straight line. Draw the figure.
~1---+---+---+---+-+----+--X
-3
-2 -1
0
1
0
2
-1
1. Through (1, 3) and (5, 6).
-2
2. Through ( -3, 2) and (2, 5).
-3
3. Through origin and (4, 3).
Frn. 20
4. Through (-2, 1) and (4, -3). 5. Through origin and ( -2, 4).
6. Through (5, 0) and (0, 3).
In each of the Exercises 7 to 12, draw the straight line described. 7. Through origin, slope 2.
9. Through (-2, 5), slope 11. Through origin, slope
8. Through (2, 3), slope 3.
f.
-½.
10. Through (2, -3), slope -1. 12. Through (1, 0), slope
-J.
13. What is the slope of a railroad track that rises 300 feet for every mile of horizontal extent? 14. What is the slope of a roof that makes an angle of 45° with the horizontal?
30. The graph of a linear function. linear function (1)
Theorem: The graph of the
y=ax+b
is the straight line which crosses the y-axis at the point (0, b) and which has the slope a. Proof.
In the first place we see that, whatever its shape may be, the graph passes through the point (0, b), since, from (1), when x = 0, y = b. Next, let P be any other point on the graph of (1), and denote the coordinates of P by (x', y'). Since Plies on the graph of (1), its coordinates must satisfy ( 1), and therefore (2)
y'
=ax'+ b.
§§ 29-30]
LINEAR FUNCTIONS AND THEIR GRAPHS
45
We will now draw the straight line from B(0, b) to P and prove that its slope is a. By formula (1) of § 29, (3)
slope of BP =
y' - b (ax' , X - 0
+ b) X
1
This proves that every point on the graph of (1) lies on the straight line through (0, b) with slope a. Conversely, if (x', y') is any point on that straight line, it lies on the graph of y = ax + b. For, if (x', y') and (0, b) are two points on a straight line having slope a, by formula (1) of § 29, we have (4) a = slope of the line
- b
ax'
= -, = a. X
y
--+------x 0
= y' - b = y' - b. x' - 0
Frn. 21
x'
The equation a = y' -; b may be written in the form y' = ax' x
+ b.
Con-
sequently the coordinates (x', y') of the point in question satisfy the relation (1), and the point lies on the graph. Therefore the graph of (1) contains all the points on the straight line drawn through (0, b) with slope a and no other points.
This theorem not only tells us that the graph of any linear function is a straight line, but it also tells us what straight line it is. Thus the graph of y = 5 x - 3 is the straight line that crosses the y-axis 3 units below the origin and has the slope 5. The distance b, which may be positive, negative, or zero, is called the y-intercept of the line. They-intercept of the graph of y = 5 x - 3 is equal to -3. EXERCISES In Exercises 1 to 6, graph each of the functions by using the slope and the y-intercept. 1. y = 3 X 4. y = 5 X.
+ 2.
2. y = 4
5. y
=
X -
5.
5 - 2 X.
+
3. y = ½X 3. 6. y = - 3 - ½X.
In each of the Exercises 7 to 12, graph the function and find the value of x for which y = 0.
7. y = 4 X - 8. 10. y = 7 - 7 x.
8. y = 6 - 2 x. 11. y = X - 8.
9. y
12. y
= -2 - ¼x. = -¾ X - 2.
CHAPTER V
Linear Equations in Two or Three Variables 31. Rational integral equations. If each member of an equation is a rational integral expression or polynomial, the equation is a rational integral equation. A rational integral equation is said to be an equation of degree k, if k is the highest degree of any term of the equation. An equation of degree 1 is a linear equation. 32. The general linear equation in x and y. Theorem I. The locus of any equation of the first degree in x and y is a straight line. Proof. If all of its terms are transposed to the left-hand member and like terms are combined, any equation of the first degree in x and y can be written in the form
Ax+ By+ C = 0,
(1)
where not both A and B are zero. Dividing the equation by a constant or transposing some of its terms will not affect the loc1:1s of the equation. There are two cases, in one of which B ~ 0, in the other of which B = 0. Case I. (2)
If B
~
0, we can divide (1) by B, and get y
C = - A -x - -· B B
Therefore y is a linear function of x. The graph of the function (2) is the same as the locus of the equation (1). For the two equations are equivalent, so that if the coordinates of a point satisfy either equation, they satisfy the other one also. In § 30 it was shown that the graph of any linear function· is a straight line. Therefore the locus of equation (1) is a straight line. 46
[§§ 31-33]
LINEAR EQUATIONS IN TvVO VARIABLES
Case II. If B = 0, then A ~ 0, and, by substituting B (1) and dividing by A, we obtain
(3)
X
47
=
0 in
= -C/A.
If we now write the equation of the straight line which is parallel to the y-axis and at the distance - C/ A from it, we get (3), which is therefore the equation of a straight line. The equation (1) therefore represents a straight line, whatever the value of B. Any straight line that is parallel to the y-axis is the locus of an equation x = a, where a is the distance of the line from the y-axis. And any straight line that is not parallel to the y-axis can be represented by the equation y = mx + b, where mis the slope of the line and b is its y-intercept. Since the equations x = a and y = mx + b are of the first degree in x and y, we have th~ following theorem. Theorem II. Any straight line in the plane is the locus of an equation of the first degree in x and y. This is the converse of Theorem I.
It is because of the correspondence between straight lines and equations of the first degree, that an equation of the first degree is called a linear equation. The use of this name is extended to equations of the first degree in any number of variables. 33. The locus of a given linear equation. if B ~ 0, the linear equation
In § 32 we saw that
Ax+ By+ C = 0 can be reduced to the form A C y = - -x - -· (2) (1)
B
B
If we compare this with the linear function, y = mx (3)
A
m=-B
and
+ b, we see that
C
b= - B.
We therefore have: Theorem. The slope of the line represented by a linear equation in the form ( 1) is the negative of the coefficient of x divided by the coefficient of y, and the y-intercept is the negative of the constant term divided by the coefficient of y. One of the best ways of drawing the locus of a
given linear equation is based on this rule.
FIRST YEAR OF COLLEGE MATHEMATICS
48
y
+2y -
1. Draw the locus of the equation 6 = 0.
Solution.
By (3), the slope and y-intercept are
EXAMPLE
3X
3 m = - 2
1
0
-2 -1
3
X
[Ch. V
and
b=
-6
-2 =
3.
The required line is therefore drawn through (0, 3) with slope -!.
-1
-2
The (directed) distance from the origin to the point where a straight line crosses the FIG. 22 x-axis is called the x-intercept of the line. It is customary to denote the x-intercept by the letter a, so that the Jine crosses the x-axis at the point (a, O). If the intercepts of a straight line are not too small, a convenient way of drawing the locus of a given linear equation is to find its xand y-intercepts and to draw the straight line through the points where it crosses the axes. At the point where the line crosses the x-axis, y = 0. 1 n order to find the x-intercept, we therefore set y = 0 in the given equation, and solve for x. In a similar way, the y-intercept
=
is found by setting x
EXAMPLE
2.
0 in the equation, and solving for y.
Draw the locus of the equation 3 x
+2y -
6 = 0.
Solution. Set y = 0. We get 3 x - 6 = 0, or x = 2, the value of the x-intercept. Setting x = 0 in the given equation, we obtain 2 y - 6 = 0, or y = 3, the value of they-intercept. The required straight line is therefore drawn through the points (2, 0) and (0, 3).
EXERCISES For each of the following equations, find (a) the slope, and (b) the intercepts of the corresponding straight line; draw the line, using (c) the slope and y-intercept, (d) the two intercepts, whenever possible. 1. 2 X
+3y +6 =
3. 3 X
-
0.
2. 3 X
y - 5 = 0.
4.
+4 =
0.
6.
+ y - V2 =
0.
8.
5.
X -
7.
X
9.
'X -
2y
7 = 0.
10.
-
4 y - 12 = 0.
+ y - 8 = 0. 2 + 5 y - 10 = 0. 3 y + 6V2 = 0. y + 5 = 0.
X
X
X -
§§ 33-35]
LINEAR EQUATIONS IN TWO VARIABLES
49
34. The equation of a given straight line. In § 33 we learned how to draw the straight line corresponding to any given linear equation in x and y. We now wish to learn how to write the equation of a straight line that is described in some geometrical way. In order to write the equation of a straight line, we must know certain facts about the line. The line will be determined if we know, for example, two points on the line, or one point on the line and its direction. In the following paragraphs we shall obtain certain standard forms for the equation of a straight line, that will enable us to write its equation as soon as we know such a set of data. When learning any one of these formulas, the student should pay particular attention to learning at the same time what facts or data about the line it is necessary to know in order to apply it. The student, given a set
of data about a line, will then recognize which one of the standard forms he can use in order to write its equation. 35. A line parallel to a coordinate axis. A line parallel to and at the distance b from the x-axis has the y equation (1) y = b, (O,b) y =b for this equation is clearly satisfied by the coordinates of any point on the line and by those of no point that is not on x=a the line. Likewise a line parallel to and at the distance a from the y-axis is rep0 (a,O) resented by the equation FIG. 23 X = a. (2)
----------------x
EXAMPLES. A line parallel to the x-axis and 4 units below it is represented by the equation y = -4, and the equation 2 x = 5 represents the line parallel to the y-axis and two and one-half units to the right of it.
EXERCISES Write the equation of each of the following straight lines. 1. Parallel to the y-axis and 6 units to its left, 2. Parallel to the x-axis and 3 units above it. 3. Parallel to the x-axis through the point (7, -10). 4. The y-axis. Describe the locus of each of the following equations.
J.y:::::i-12.
6.4x=7.
7.2x-5=0,
8. 3 y
+8 = 0
FIRST YEAR OF COLLEGE MATHEMATICS
50
[Ch. V
36. Point-slope form.
A straight line is determined if a point on the line and the direction of the line y are given. Since the direction of a line is l given by giving its slope, it should therefore be possible to write the equation of the line when we know a point on the line and its slope. Let Pi :(xi, Yi) be a point on a certain straight line and let m be the X slope of the line. We wish to write the equation of the line in terms of x1, Yi, and m. Let P:(x, y) be any other point FIG. 24 on the line. Then, since (x1, y1) and (x, y) are two points on the line, and since the slope ism, by equation (1), § 29, we have Y - Yi X -
Xi
= m,
or (1)
Y - Y1
= m(x -
x1).
The straight line is the locus of this equation. For, as we have seen, the equation is satisfied by the coordinates (x, y) of any point on the line. Moreover, if (x, y) is any point whose coordinates satisfy the equation, the slope of the line joining (x, y) and (x1, Yi) is equal tom, so that (x, y) lies on the line in question. This is known as the point-slope form of the equation of a straight line. In order to apply it we must know (i) the coordinates (x1, y1) of some point on the line and (ii) the slope m of the line. ExAMPLE
Solution.
1.
Write the equation of the line through (-4, 3), slope -5.
From formula, we have
+ y + 17 = 0. Identify the locus of the equation y + 6 = 4(x -
y - 3 EXAMPLE
2.
= -5(x
+ 4)
or
5
X
5).
Solution. By comparing this equation with equation (1), we recognize that it represents the straight line through (5, -6) with slope 4. EXAMPLE 3. The straight line through the origin with slope mis repre~ented by the equation y = mx.
§§ 36-38]
LINEAR EQUATIONS IN TWO VARIABLES
51
EXERCISES Write the equation of each of the following straight lines, and simplify it to the form Ax + By + C = 0. Check the final form by substituting the coordinates of the given point. Draw the figure. 1. Through (3, 4), slope 7. 3. 5. 7.
8. 9.
10.
2. Through (-5, 2), slope-½. Through the origin, slope -3. 4. Through (i-, -½), slope 1. Through (8, 3), parallel to the x-axis. 6. Through (-2, 5), slope 5. Through (1, -5), parallel to the line y = 6 x. Through (4, 2), parallel to the line y + 5 x = 0. Through (-1, 3), parallel to the line through (4, 2) and (-6, 7). Through (5, -3), parallel to the line through (1, -4) and (2, 1).
Describe and draw the locus of each of the following equations. 11. y - 2 = -5(x 13. y = ¾(x - 2). 16. 3 y
+9 =
+ 3).
+
12. y 3 = 4(x - 9). 14. y - 4 = 5 x.
16. 2 y - 10 = 3(x + 5). 18. 3 X = j(6 - y).
2(x - 4).
17. 5 y = 4 - x.
37. Slope-intercept form. An especially useful form of the equation of a straight line is the one already obtained in terms of its slope and its y-intercept. Suppose that a straight line has the slope m and the y-intercept b. Then we have seen in § 30 that its equation can be written in the form (1) y mx + b.
=
This is known as the slope-intercept form of the equation of a line. Two-point form. We shall now obtain the equation of a straight line in terms of the coordinates of two given points on the line. Let P1 :(x1, y1) and P2 :(x2, Y2) be two points on a certain straight line. In case x1 = x2, so that the line is parallel to the y-axis, the equation of the line is simply x = xi ( § 35). We therefore assume that x1 ¢ x2. Then, by § 29, the slope of the line is 38.
Using this value form in the point-slope form, we obtain as the equation of the line Y2 - Y1 (x - x1), Y - Y1 __ Y2 - Y1. (1) y - Y1 = or (2) X2 -
Xt
X -
Xt
X2 -
Xt
52
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. V
It is immaterial which of the two given points is taken as (xi, Yi) and which as (x2, Y2). Both of the forms (1) and (2) are useful. Equation (2) is known as the two-point form of the equation of a straight line. In order to apply it one must know the coordinates of two points on the line. EXAMPLE 1. Write the equation of the line through (-5, 2) and (4, -3).
Solution.
Substituting Xi
-5, Yi = 2, x2 = 4, Y2
=
y - 2
= -3 in (2), we have
= -3 - 2
4+5 On simplification this becomes 5 x + 9 y + 7 = 0. As a check, we substitute the coordinates of each of the given points in this equation and see that it is satisfied. x+5
When "We wish to write the equation of a straight line passing through two given points or passing through a given point and having a given slope, rather than use the two-point form or the point-slope form of the equation, it is frequently quicker to make use of the fact that when the equation is written in the form Ax + By + C = 0, m = -A/B. The method is illustrated in the following example. EXAMPLE 2.
Write the equation of the line through (3, -7) and (-4, 5).
Solution. From the given coordinates we see at once that the line has the slope -1,f. In the required equation the ratio of the coefficient of x to the coefficient of y is therefore equal to + 1,f. Consequently the equation has the form 12 x
+ 7 y + C = 0,
where C is still to be determined. Since the equation must be satisfied by the coordinates (3, -7), 12(3) + 7(-7) + C = 0, or C = 13. The equation is therefore 12 x + 7 y + 13 = 0. As a check, we see that the equation is also satisfied by the coordinates ( - 4, 5). EXAMPLE 3. Find the equation of the line through the point (-4, 7) parallel to the line 6 x - 5 y + 17 = 0.
Solution. (3)
Any line parallel to the given line has an equation of the form 6x - 5y
+ C = 0.
In order to find the required equation, we must determine C so that (3) is satisfied by the coordinates ( -4, 7). Substituting x = -4, y = 7 in (3), we have (4)
6(-4) - 5(7)
+C = 0
The required equation is therefore 6 x - 5 y
or
C
= 59.
+ 59 = 0.
§§ 38-39]
LINEAR EQUATIONS IN TWO VARIABLES
53
EXERCISES Write the equation of the line through each of the following pairs of points C = 0. Check the result by subBy and simplify it to the form Ax stituting the coordinates of the given points.
+
+
2. (4, 0), (-5, 3). 6. (1, 1), (6, 4).
1. (3, 5), (7, -2). 4. (-8, 3), (-2, -4).
3. (6, 11), (0, 7). 6. (½, -i), (1, 4).
Write the equation of each of the following straight lines in the form Ax+ By+ C = 0. 7. 9. 11. 13. 16. 16. 17. 18.
Through Through Through Through Through Through Through Through
8. Through (-4, -5) and (8, 0). (2, 5) and (-4, 3). (-5, 3), with slope 4. 10. Through (1, -4), slope -5. 12. Through (6, 1), slope½. (4, -5), slope -¾. 14. Through ( -5, -2), slope -g-. (0, 3), slope 1. 4 = 0. (6, 2), parallel to 5 x - y 5 y - 2 = 0. (-4, 3), parallel to 2 x 7 = 0. (2, 0), parallel to 6 x - 3 y 3 y = 5 x - 7 y + 4. (-7, -2), parallel to 4 x
+
+
+ +
39. Intercept form. Let the x- and y-intercepts of a certain straight line be equal to a and b respectively. Two points on the line are therefore (a, 0) and (0, b). The equation of the line, in two-point form is therefore b-0 y-0 (1)
or (2)
x-a
y
0-a '
Fm. 25
bx+ ay = ab.
Dividing by ab, we obtain the symmetrical equation (3) Equation (3) is known as the intercept form of the equation o.f a straight line. In order to apply it, one must know the values of a, the x-intercept, and b, the y-intercept of the line. It is, of course, not applicable if the line is parallel to one of the coordinate axes nor if the line passes through the origin, in which latter case a = b = 0. Write the equation of the straight line which crosses the x-axis 4 units to the right of the origin and the y-axis 6 units below the origin. EXAMPLE.
Solution.
Here a = 4, b = -6.
~+-1L=l -6 4
or
Hence, by (3), the equation is 3
X -
2 y - 12
=
0.
FIRST YEAR OF COLLEGE MATHEMATICS
54
[Ch. V
EXERCISES Write the equations of the straight lines satisfying the following sets of conditions. Whenever possible check by substitution. Draw the figures.
1. a = 5, b = 2. 3. x-intercept - 3, y-intercept 4. 6. Through (-2, 0) and (0, 5).
2. a= -7, b = 4. 4. x-intercept 3, y-intercept -£. 6. Through (-8, 0) and (0, -9).
Describe the locus of each of the following equations. y
X
7.
3 + -7 y + 2 = 4 3 x
10. 3
13. 3 X
-
y
X
=
1.
1.
8.
5- 4 =
9·
1.
11 3 x _ 5 y = l . 2 4 .
4 y = -12. 14. 3 X
+5y -
y
X
9- 5 =
12. 3
X -
15 = 0. 16. 6 X
1.
4 y = 12.
+ 7 y = 84.
Find the slope of each of the following lines.
1G.
X y = 5 +7
1.
17. 53X
+ 49y =
1.
18. 25X
_
3y = l
7
.
40. The point of intersection of two straight lines.
Theorem: The coordinates of the point of intersection of two straight lines may be found by solving the equations of the lines simultaneously for x and y. y For P1, the point of intersection, is common to the two lines, so that its coordinates, unlike the coordinates of any other point, satisfy both equations. 1. Find tht point of intersection of the lines x + y - 6 = 0 and x - 2 y + 6 = 0. EXAMPLE
Frn. 26
Solution. Solving the two equations simultaneously, we find x = 2, y = 4. The lines therefore intersect in the point (2, 4). The lines are shown in Figure 26.
The theorem of this article enables us to do two things: (a) to find the point of intersection of two given straight lines by solving their equations algebraically, and (b) to solve simultaneously two linear equations in two unknowns by drawing their loci and reading from the figure the coordinates of their point of intersection. A review and discussion of algebraic methods of solving two linear equations in two unknowns are given in §§ 42, 43, and 44. In § 41 we consider the graphical solution.
§§ 39-41]
LINEAR EQUATIONS IN TWO VARIABLES
55
41. Graphical solution of two simultaneous linear equations. In discussing the solution of such a system of equations, it will be convenient to write them in the form (1)
a1x
+ b1y = c1,
(2)
a2x
+ b2y = c2,
in which small letters are used for the coefficients, and the constant terms, c1 and c2, have been transposed to the right-hand members. The loci of the equations are two straight lines. By the theorem of § 40, we have the following rule: In order to solve two linear equations in x and y graphically, draw the straight lines which ripresent the equations, and read off from the figure the coordinates of the point of intersection. These coordinates constitute the solution.
Like any graphical method, such a solution is liable to errors of drawing and estimation. In general, two straight lines intersect in just one point. It follows that any pair of linear equations in x and y have one and only one solution, unless the corresponding lines are parallel or actually coincide. If the lines are parallel, there is no common point and consequently no solution. If the lines coincide, the coordinates of any point on the line satisfy both equations. In this case there are infinitely many solutions, but no one solution is determined. This question will be considered in more detail in § 45. If there is just one solution, the equations are said to be simultaneous. When there is no solution the equations are inconsistent. When they have infinitely many soluy tions, they are dependent equations. EXAMPLE.
Solve graphically the system of
. {2x+y=4, equat10ns X y = 5. Solution. We draw the loci of the two equations by the method of § 33. From the graph we see that the lines cross at or near the point (3, - 2). On substituting x = 3, y = -2 we see that these values satisfy both of the given equations and are the solution of the system.
--2 -1
0
--1-,-1-~1-------.-~1--1 -2--+----+--I")
FIG. 27
FIRST YEAR OF COLLEGE IvIATHEMATICS
56
[Ch. V
EXERCISES Solve each system graphically, or, in case there is not a unique solution, determine whether the equations are inconsistent or dependent. 1. {
X -
2y
+y
=
-4,
2. { 4 X
= - 5. a. { 3 X - y = 15, 2 x + y = IO. 6 { X - 2 y = 7, • 2 X - 3 y = 13. 7 _{ 3 X - 2 y = 6, 6 X - 4 y = 5. g_ { 6 X - 9 y = 12, 2 X - 3 y = 4. 3X
-
X -
y = 5, 3 y = 4.
4 _{ 3 X
+4y
G{ • 3
+4y =
= 11,
3 x - 2 y = -1.
B. {
X X -
3, 4 y = - 23.
+ 3 y = 8, + 9 y = 12. X + 3 = 0, 2 X + 3 y = 6. X
3X
lQ. {
42. Algebraic solution of simultaneous equations; elimination by substitution. We shall now study two methods of solving a system of two linear equations by purely algebraic methods. Both methods involve the elimination of one of the unknowns by combining the two equations in such a way as to obtain a single equation in one unknown. The method of elimination by substitution is illustrated in the following example. EXAMPLE.
Solve for
and y,
x
(1)
2x+3y=8,
(2)
3
Solution.
X
+ 2y
= 2.
We first solve (1) for y in terms of x, getting
y=-}x+f.
(3)
Equations (1) and (3) are equivalent. If x and y have values that satisfy (1) and (2), we must be able to substitute ( -¾ x + i) in place of y in equation (2).
After this substitution, equation (2) becomes
(4)
3x+2(-}x+i) = 2.
On clearing this equation of parentheses and fractions, we get
3x-!x+-136 =2
or
5x=-IO,
sothatx=-2.
On substituting x = -2 in (3), we have y = -! + ¾= 4. The solution is therefore x = - 2, y = 4. This solution is verified by substitution in the original equations.
§§ 41--43]
LINEAR EQUATIONS IN TWO VARIABLES
57
In general, solve one of the given equations for whichever unknown yields the simpler expression, avoiding fractions when possible. When the given equations are inconsistent or deµmdent, this method eliminates both unknowns, and yields a false statement, such as 5 = 3, if the equations are inconsistent~ or a useless one, such as 5 = 5, if the equations are dependent.
43. Solution by elimination by addition or subtraction. method is illustrated by the following example. 1.
EXAMPLE
This
Solve by elimination by addition or subtraction:
+3y 3X +2y
(1)
2X
(2)
= 8, = 2.
Solution. In this method we multiply both members of the two equations by numbers chosen so that the coefficients of y in the resulting equations are numerically equal. (The coefficients of x may be made numerically equal if that is more convenient.) Multiply both members of (1) by 2 and the members of (2) by 3. We get
+6y = 9X +6y =
(3)
4X
(4)
16, 6,
,vhich are equivalent to (1) and (2). Subtract the members of (4) from those of (3). The terms in y cancel, and we get -5 x
(5)
IO; whence x
=
Substitute x = - 2 in (1): -4 X
+3y = -2,
=
-2.
= 8, whence y = 4. The soJution is y
= 4.
EXERCISES Solve each system by elimination by substitution and also by elimination by addition or subtraction, or, in case there is not a unique solution, determine whether the equations are inconsistent or dependent. l. {
2 y = -4,
X -
+ y = - 5. 3 X + 4 y = 11, 3x
4 {
·
3x-2y=-1.
7. { 3 X 6X
lO. {
-
+3 = +3y =
X
2X
2 y = 6, 4 y = 5. 0, 6.
- 9 y = 6, 4 x - 3 y = 2.
13. { 12 X
X - y = 5, x - 3 y = 4.
y = 15,
2• { 4
3_{ 3
fi { X - 2 y = 7, · 2x-3y=l3.
G{ · 3x-4y=-23 .
+3y = +9y = ll. { 4 X + 6 y = S. {
X
3X
6X
14. { 2 X
3x
-
4y
+ 5y
:c:
8, 12. 7, -9.
= 3, 3.
+6y =
X -
+ y = IO. X + 4 y = 3,
2x
g_ { 6 X 2X 12. {
-
9 y = 12, 3 y = 4.
2 y + 3 = 0, 5 X + 8 y = 3.
lfi. { 3 S + 5 t = 9, 4 s - t = - lL
FIRST YEAR OF COLLEGE MATHEMATICS
58
[Ch. V
*44. Determinants of second order. If a, b, c, and d represent any numbers, the symbol a b d
C
is called a determinant of second order, of which a, b, c, d are the elements. The value of a determinant of second order is defined as follows: db
a c
(1)
5
3
(b)
2 7 '
Solution.
ad - be.
Find the values of the determinants
EXAMPLES.
(a)
I=
153
X
-y
(c) I
'
8 3 -2 9
By definition,
(a) (b)
(c)
5 3 = 5(7) - 3(2) = 35 - 6 = 29. 2 7 X
1:
-y 8
3
-2
9
= -5 y - 3 x.
= 8(9) - 3(-2) = 72
+6=
78.
Notice that the order or arrangement of the elements as well as the use of the vertical bars is an essential part of the notation of a determinant. For example, the value of the determinant
I
b d
= be - ad is the negative of the value of a b = ad - be.
a
e d
C
EXERCISES Evaluate the following determinants. 1.
2 a 5 b
4.
7 2
7.
5
2.1:
-3
b
5.111 a
5
4b
-5
2
3
-4
8.
3.
3x Sy
-3b
6a
-4yl. 6x
6.
I: I:
9. I
5 9 y w
2 -7
-:1-
§§44-45]
LINEAR EQUATIONS IN TWO VARIABLES
59
*46. Solution of simultaneous equations by determinants. Let us consider the general system of two linear equations in x and y: a1x { a2x
(1)
+ b1y = c1, + b2y = c2.
Using the method of elimination by subtraction, we eliminate y by multiplying the members of the first equation by b2 and those of the second by b1 and subtracting. We obtain
(a1b2 - a2b1)x = c1b2 - c2b1. In a similar way, after eliminating x, we have
(a1b2 - a2b1)Y = a1c2 - a2c1.
If (a1b2 - a2b1) ~ 0, we can divide both members of these equations by (a1b2 - a2b1), and see that: System (1) has the solution a1c2 - a2c1 y= ----,
(2)
a1b2 - a2b1
provided that the denominator (a1b2 - a2b1) is not equal to zero. In the notation of determinants we can write
and
We can therefore write the solution of system (1) in the form C1
(3)
x=
C2
a1 a2
bi b2 b1 b2
I
y=
I I bf C1
a1 a2
C2
a1 a2
b2
if
a1 a2
b1 :;= 0. b2
Since system (1) may be any system of two simultaneous linear equations in two unknowns, the solution (3) may be used as a formula for the solution of any such system.
vV e shall say that a system of linear equations is arranged in standardform if, as in system (1), the terms containing the unknowns .are in the left-hand members and in the samd order in the various
60
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. V
equations, and if the terms that do not contain any unknown form the right-hand members. The determinant
which appears in the denominators of (3) is known as the determinant of the system of equations. I ts elements are the same as the coefficients of the unknown quantities in the system of equations when the equations are arranged in standard form. In the formula for x in (3), the determinant in the numerator is the same as the determinant of the system except that the coefficients of x (that is, a1 and a2) are replaced by c1 and c2, the constant terms which appear as the right-hand members of the two given equations. In the formula for y, on the other hand, it is the coefficients of y which are replaced by c1 and c2. We can now state the following rule, called Cramer's Rule, for the solution of a system of two linear equations in two unknowns: Cramer's Rule. Arrange the given equations in standard form. Then, if the determinant of the system is not equal to zero, the system has one and only one solution. In this solution, the value of any unknown is equal to a fraction whose denominator is the determinant of the system and whose numerator is the same determinant except that the column of coefficients of the unknown in question is replaced by the column of constant terms. Cramer's Rule is especially important since, as will be seen in § 48 and in Chapter XXXIV, it holds word for word for a system of three or more linear equations in the same number of unknowns. . {4x + 5 y = 13 EXAMPLE 1. Solve by determmants x _ y = _ 21. 2 3 Solution. The system is already arranged in standard form. The determinant of the system is 4 5 1 = -12 10 = -22. -3 1 2 Therefore the solution is 13
X
=
I:
-~~
= -39 + 105 = ~ = _g,\ -22
-22
j
I = -84 - 26 = _=---110 = 5 _ -22 -22 -22 On substitution, x = -3, y = 5 are seen to satisfy the given equations. y
=
51
I -21-22-3
§§ 45-46]
LINEAR EQUATIONS IN TWO VARIABLES
61
Geometrical interpretation. By § 33, the slopes of the lines represented by equations (1) are equal to -a1/b1 and -a2/b2. If the two lines are parallel or coincident, these slopes are equal and
(7)
or
If the two lines are coincident, their y-intercepts are also equal, and
or
(8)
We have assumed that b1 ~ 0 and b2 ~ 0. The student should discuss the contrary case. The correspondence between the geometry and algebra can be indicated as follows:
If
a1b2 - a2b1
~
0, the lines intersect in just the equations have just one simultaneous solution. one point;
If a,b2 - a-,b, = 0, I the lines are parallel and the equations are inconsistent and b1c2 - b2c1 ~ 0, have no common point; or have no solution. a1b2 - a2b1 = 0, the lines coincide and the equations are dependent, and b1c2 - b2c1 = 0, have infinitely many and have infinitely many solucommon points; tions.
If
EXERCISES Using determinants, solve Exercises 1 to 16 of § 43, page 57. For each of the following systems, determine whether the equations are simultaneous, inconsistent, or dependent.
lG. { 5 X
-
15 X
-
19. { 3 X 6X
3 y = 4, 17. { 2 X - 5 y = 3, 9 y = 12. 6 X + 15 y = 9.
+ 2y + 6y
= 7, = 14.
20. {-5 X 20 X
lS. { 4 X 8X
-
3 y = 4,
-
6 y = 4.
= 2, 21. { 8 X - 10 y + 12 = 0, 8 y = 10. -4 X + 5 y - 6 = 0.
+ 2y -
46. Linear equations in three unknowns. To solve a system of three linear equations in three unknowns, eliminate one of the unknowns between each of two pairs of the equations. The result is two linear equations in the other two unknowns. These two equations may then be solved for the remaining unknown quantities, and the unknown which was eliminated first may be found by substitution.
FIRST YEAR OF COLLEGE MATHEMATICS
62
[Ch. V
Solve for x, y, and z:
EXAMPLE.
2 X + 4 y - 3 Z = -9, 3 X + y - 2 Z = 4, 5 X + 2 y + 4 Z = 28.
(1) (2) {3)
Solution. We shall eliminate y between (1) and (2) and then between (2) and (3). Multiplying the members of (2) by 4 and subtracting from those of (1), we have (4)
-10
X
+5Z =
-25,
or
2
X -
Z
= 5.
Multiplying the members of (2) by 2 and subtracting the members of (3), we get (5)
X -
8 Z = -20.
Solve equations (4) and (5). Their solution is x = 4, z = 3. these values for x and z in (2), and obtain 12
+y -
6
= 4,
or
y
We substitute
= -2.
The solution is therefore x = 4, y = -2, z = 3, which, on substitution, satisfies all three of the given equations.
EXERCISES Solve the following systems of equations. ,
{ 4X
1.
3y
-
+ 5Z =
+y - 4Z = 5X +4y +8 Z = 2X +3y - Z = 3X - 2y +4Z = 2X
3. {
-6, 11, 14. 5, 2,
x+y+3z= -1. ! 6 X + 4 y + 3 Z = 41, 5. ~ 2 X + 2 y + 3 Z = 28, l 4 x + 4 y - z = 0. 3 r + 4 s + 2 t = 14, 7. 2 r - 3 s + 4 t = 3, 1 l 5 r + 2 s + 3 t = 8. ½X + f y + 5 Z = 0, 9. { 2 X - t y + i z = \ 3 , ¾X + ½y + ½Z = -t. y = 5, X -
J
11.
{
y
+z
X -
Z
= 1, = -2.
Check by substitution.
+ y - Z = 9, X - y + Z = -3,
X
2.
{
x-y-z=l. 3
X
+2y -
Z
4x+3y +2z { 5X - 3y - 6Z 5X - 4y +4Z 6. 2 X + 9 y + 3 Z { 4X - 5y - Z 4.
= 7, = = = = =
l, -7. 6, 5, 7.
8.
5 U - 6 V - W = 2, 10 U - 9 V - 3 io = 9, { 20 U + 9 V + 2 W = 12.
10.
¾X + 1 y + 3 Z = -¼, { ½X - i y + Z = -¾, ! X + ½y + 2 Z = }.
+2y
= 1, 12. 4 y - 3 z = 23; { 5 X - 2 Z = -13. 3X
§§ 46-47]
LINEAR EQUATIONS IN TWO VARIABLES
63
bi c1 a2 b2 c2 a1
*47. Determinants of third order. The symbol
1s
a3 b3 C3 called a determinant of third order, and the nine numbers that appear in its three rows and three columns are its elements. The value of the determinant is defined as follows:
(1)
a1 bi Ci a2 b2 C2 as bs Cs
= aib2cs + a2bsc1 + asb1c2 -
asb2c1 - a1bsc2 - a2b1cs.
Notice that in this expansion there are six terms. Each term contains one element from each row and one element from each column of the determinant. The signs which are to be attached to the various terms may be remembered from the accompanying diagram. The terms formed by FIG. 28 following the solid lines have a plus sign attached to the product; the terms formed by following the dotted lines are to be preceded by the minus sign. Evaluate the determinant
EXAMPLE.
3 Solution.
2 1
3 2 1
5 -4 2
-3 2 . -1
5 -3 -4 2
2 = 12 -1
+ (-12) + 10 -
(12) - (12) - ( -10)
= -4. EXERCISES Evaluate the following determinants.
-4 1. 2 4 5 1 -2 3 0
3
4.
2.
4
-6
3
-5
2
-1
3
3
-2
.
6.
4 -2
-3 5
1 4
3
2
-5
2 5 3 1 -4 2
2 5 -1
8
-6
a1
6
1
-4
3
.
.
3.
b1
-3 -4 2 a1
6. a2 b2 a2 as bs as
[Ch. V
FIRST YEAR OF COLLEGE MATHEMATICS
64
*48. Solution of three equations in three unknowns by determinants. Let us solve the system
+ b1y + c1z = d1, a2x + b2y + c2z = d2, a3x + b3y + C3Z = d3,
(1)
a1x
(2) (3)
by the method of§ 46. To eliminate z from (1) and (2), we multiply the members of (1) by c2 and those of (2) by c1 and subtract. We get (a1c2 - a2c1)x
(4)
+
(b1c2 - b2c1)Y
= d1c2 - d2c1.
From equations (2) and (3) we obtain (a2C3 - a3c2)x
(5)
+ (b2c3 -
b3c2)Y = d2C3 - d3C2.
Solving equations (4) and (5) by the method of § 45, we obtain
(6)
x=
I I
(d1c2 - d2c1) ( d2C3 d3C2) ( a1c2 - a2c1) (a2c3 - a3c2)
(b1c2 (b2C3 (b1c2 (b2c3
-
b2c1) I b3C2) b2c1) b3c2)
1 ·
On expansion, the denominator of this fraction becomes c2( a1b2c3
+ a2b3c1 +
a1 a3b1c2 - a3b2c1 - a1b3c2 - a2b1c3) = c2 a2 a3
b1 b2 b3
c1 c2 , C3
and the numerator is the same except that the a's are replaced by the corresponding d's. We solve for y and z in a similar way. We therefore obtain the following solution:
7 ( )
x
=
d1 d2 da a1
b1 b2
a2 aa
b2
ba b1 ba
c1 c2 Ca c1 c2 Ca
y
=
a1 a2 aa a1
d1 d2 da
a2 aa
b2
b1
ba
c1 c2 Cs c1 ' c2 Cs
z=
a1 a2 as
b1 b2
d1 d2
a1
ba ds b1 c1
a2
b2
ba
c2 Ca
provided that the determinant of the system, which appears in the denominators, is not equal to zero. It follows that Cramer's Rule ( § 45) applies word for word to the solution of a system of three linear equations in three unknowns. If th9 determinant of the system equals zero, the equations are either inconsistent, with no solution, or dependent, with infinitely many solutions.
§§ 48-49]
LINEAR EQUATIONS IN TWO VARIABLES
65
+ 4 y - 3 Z = -9, + y - 2 z = 4, 5 X + 2 y + 4 Z = 28. 2X
EXAMPLE.
{ 3x
Solve the system
Solution. The system is already arranged in standard form. minant of the system is: 2 4 3 1 5
-3 -2 = 8 - 18 - 40
2
+ 15 + 8 -
The deter-
48 = -75.
4
Therefore 1 x=--75
-9 4 4
1
28
2
-3 I -2 = -75 (-36 - 24 - 224
+ 84 -
36 - 64)
4
-300 ----4 - -75 - ' 2 y = _l_ 3 - 75 5
-9 4 28
-3 1 -2 = _ (32 - 252 75 4
= 150 = -2 -75
z
=
2 4 1
1 - - 3 - 75 5
2
-9 4 = 28
1
+ 90 + 60 + 112 + 108)
'
75 (56 - 54
+ 80 + 45 -
16 - 336)
-225 - --3 . - --75 On substitution, the values x = 4, y = - 2, z = 3 are seen to satisfy the given equations.
EXERCISES By means of determinants, solve Exercises 1 to 12 of § 46, page 62. the following systems by any method. ax - by = b2 13. by - cz = c2 { ax+ cz = a2
a2 , b2 + 2 'b2
-
+c
2
•
Solve
bx - ay = c(b 2 - a2 ), 14. cy - bz = a(c2 - b2 ), { ex + az = b(a2 + b2 ).
49. Stated problems involving two or three unknowns. Before attempting to solve the following problems, the student should rea.d § 399. The systems obtained may be solved either with or without determinants.
66
FIRST YEAR OF COLLEGE lV[ATHEMATICS
· [Ch. V
EXERCISES Solve the following problems by introducing two or three unknowns. 1. A boy sells some newspapers at 5 cents each and some magazines at 15 cents each. In all he makes 80 sales and takes in $6.70. How many copies does he sell of each kind? 2. A woman buys 17 pounds of apples and pays $1.84 for them. She buys part of them at 3 pounds for 28 cents and the rest at 2 pounds for 25 cents. How many of each kind did she buy? 3. On his shelf a pharmacist has three mixtures of substances A and B. The first contains 65% of A and 15% of B; the second contains 45% of A and 25% of B; the third contains 35% of A and 50% of B; in each case the remainder is water. In what proportions must he mix them in order to obtain a mixture containing 50% of A and 30% of B? 4. One brand of artificial fertilizer contains 4% phosphate and 10% nitrogen; a second brand contains 6% phosphate and 9% nitrogen; a third brand contains 3% phosphate and 17% nitrogen. In what proportions should they be mixed to give 5% phosphate and 11 % nitrogen? 5. How much milk containing 3% butter fat and how much cream containing 28% butter fat must be mixed to make 1000 pounds of milk containing 3½% butter fat? f _6. The speed of a boat downstream is 2 miles an hour less than 3 times its speed upstream. It can go upstream at the rate of 3 miles an hour. Find its rate in still water and the rate of the current. 7. A boat can go upstream a certain distance in 5 hours and return in 2 hours. In 3 hours it can go downstream 2 miles farther than it can go upstream in 7 hours. Find its speed in still water and the rate of the current. 8. The owner of a rectangular lawn adds to it a strip 6 feet wide along one of the long sides and builds a driveway 12 feet wide across one end. The lawn then contains 192 feet less than its original area. He makes a path 2 feet wide around the whole lawn and thereby further reduces the area by 600 square feet. Find the original dimensions of the lawn. 9. Two cars run at uniform speeds on a mile track. The faster car overtakes and passes the slower one every 2 minutes. If they run in opposite directions, they pass every fifteen seconds. Find their speeds. 10. A and B together can build a garage in 63 hours; B and C together can build it in 90 hours; A and C together cPn build it in 70 hours. How long would it take each one to build it alone? 11. The outlet pump of a certain tank delivers water five-sixths as fast as the intake pump. if both pumps operate, the tank will be filled in 16 hours. If only the intake pump operates, in 3 hours it can fill the tank and pump an excess of 1000 gallons. Find the capacity of the tank and the rate at ·which each pump delivers wate:
§ 49]
LINEAR EQUATIONS IN TWO VARIABLES
67
12. In a certain two-digit number, the sum of the digits is 13. If the order of the digits is reversed, the number is increased by 45. Find the number. (Note. If u is the units' digit and tis the tens' digit, the number is u + 10 t.) 13. If a certain two-digit number is divided by its units' digit, the quotient is 16. If the order of the digits is reversed, the result is the same as if 2 is subtracted from the tens' digit and added to the units' digit. Find the number. 14. In a certain three-digit number, the sum of the digits is 17. If the order of the digits is reversed, the number is increased by 396. The hundreds' digit is 1 less than the tens' digit. Find the number. 15. A teeter board just balances if A sits on one end 10 feet from the fulcrum, and B and C sit on the other end at 5 and 8 feet, respectively, from the fulcrum. It also balances if A sits 2 feet from the fulcrum and B sits on the same side 6 feet from the fulcrum while C is on the other end 10 feet from the fulcrum. If C is known to weigh 60 pounds, what are the weights of A and B? 16. The distance from Fluelen to Airola by the St. Gotthard Pass is 42 miles. A certain automobile on the mountain roads averages 30 miles an hour when the road is approximately level, 15 miles an hour while climbing the pass, and 20 miles an hour on the descent. It goes from Fluelen to Airolo in 135 minutes and from Airolo to Fluelen in 126 minutes. How much of the road in the direction from Fluelen to Airolo is approximately level, how much is rising, and how much falling? 17. It is known that the road from Bolzano to Innsbruck by the Brenner Pass runs for 34 miles approximately level, then climbs for 18 miles to the summit, and finally descends 24 miles to Innsbruck. On the mountain roads a certain car maintains one average speed on the level, a different speed while climbing a pass, and still another while descending it. It required 1 hour and 50 minutes to cover the level stretch and climb halfway to the summit, and 1 hour and 42 minutes more to complete the trip to Innsbruck. In the opposite direction, it took 3 hours and 34 minutes to make the trip from Innsbruck to Bolzano. Find its average speeds on the level, while climbing, and while descending. 18. An army column is marching at the rate of 3 miles an hour. A messenger, who travels at 12 miles an hour, goes from the rear to the head of the column and back to the rear in 1 hour and 36 minutes. How long is the column, and how long does the messenger travel in each direction? (Note. Use the messenger's speed in each direction, relative to the column.)
CHAPTER VI
Variation, Ratio, and Proportion 50. Variation. We wish now to study a certain particularly simple type of linear functional relation between two variables. If x and y are two variables such that (1)
y
=
ex,
where c is a constant, we say that y is proportional to x, or y varies directly as x, or simply, y varies as x. The constant c is called the constant of proportionality or the constant of variation. It must be remembered that the statement "y varies directly as x" is merely a statement in the language of variation equivalent to the statement "y = ex, where c is some constant," in the language of algebra. In stating a problem it is often useful to employ the language of variation. For example, it is convenient to say that at uniform speed the distance traveled varies directly as the time, or that the yearly interest on a loan varies as the rate of interest. But, for purposes of computation. it is desirable to translate immediately from the language of variation to the notation of algebra. Instead of saying that the distance traveled varies directly as the time, we write d = ct. And the statement that the interest varies directly as the rate of interest is translated immediately into the equivalent equation, i = er. It is also to be understood that in such an equation as y = ex, the numbers x and y are pure numbers which are the measures, in terms of suitable units of measure, of any concrete quantities that may be involved. The value of the constant of proportionality, c, depends on the units in which x and y are expressed. Thus a car which is moving at the rate of 30 miles per hour is also traveling at the rate of (30 X 5280) /3600 feet per second or 44 feet per second. The same relation between time and distance can therefore be expressed by the equation y = 30 x, where x is the time in hours and y is the distance traveled in miles, or by the equation y = 44 x, where x is the time in seconds and y is the distance traveled in feet. Since both of these equations are of the form y = ex, they express the fact that the distance traveled varies as the time, but the constant c in one case is equal to 30 and 68
[§§ 50-52]
VARIATION, RATIO, AND PROPORTION
69
in the other case to 44, depending on the units used for measuring the time and distance.
51. Joint and inverse variation. If z varies directly as the product of x and y, that is, if z = cxy, we say that z varies jointly as x and y.
If y varies directly as the reciprocal of x, that is, if y = ~ , we say X
that y varies inversely as x. be combined.
The various methods of variation can
EXAMPLE 1. The gravitational attraction between two bodies varies jointly as their masses and inversely as the square of the distance between their centers. Translating from the language of variation to the notation of ,algebra, we write this relation as
mM
F = c ri2, where F fa the force of attraction, m and M are the masses of the bodies, d is the distance between their centers, and c is a constant that takes care of the units involved. EXAMPLE 2. The statement that the time required to earn a certain amount of interest varies directly as the amount of interest and inversely as the principal and the rate, can be written algebraically in the form t
I = c-
Pr'
where I is the value of the interest, P is the principal, r is the rate of interest, and c is a constant.
52. Determination of the constant of proportionality. In practical problems involving variation, we usually know, first, the general type of variation that is involved, and, second, one set of corresponding values of the variables. From these facts we can determine completely the relation existing among the variables. EXAMPLE 1. (a) Express the following statements in the form of a single equation: The electromotive force necessary to produce a current in an electric wire of given length varies directly as the current and inversely as the square of the diameter of the wire; ,vhen the diameter of the wire is 0.064 inch and the current is 4 amperes, the electromotive force is 100 volts. (b) Find the diameter of a wire of the same length and material as in (a), if an electromotive force ef 400 volts produces a current of 9 amperes.
FIRST YEAR OF COLLEGE MATHEMATICS
70
[Ch. VI
Solution. (a) Denote by Ethe electromotive force in volts, by I the current in amperes, and by d the diameter of the wire in inches. We first write an equation expressing the general type of variation: I E = c d2'
(1)
where c is a constant. We now substitute the known corresponding values of the variables: lOO =
C
4 C (.064) 2 = .001024•
Therefore c = 0.1024, and the given statements are contained in the single equation (2)
E
I = 0.1024 d2.
(b) Substitute E = 400 and I = 9 in (2). 400
= 0.1024 ~
or
d2
=
We obtain
(O.l:~ ·
9
= 0.002304.
Therefore d = V.002304 = 0.048. The required diameter is 0.048 inch.
It will be observed that the general procedure is as follows: (1) Assign suitable letters to the variables and write an equation that represents the general type of variation involved; this equation contains a constant of proportionality. (2) Substitute a given set of corresponding values of the variables in the equation, and solve for the constant of proportionality. (3) Rewrite the equation, with the constant of proportionality replaced by its value. (4) The equation just obtained may then be solved for any one of the variables corresponding to a given set of values of the other variables.
2. The brightness of illumination from a lamp varies directly as the strength of the lamp and inversely as the square of the distance from the lamp. At what distance from a 30 candle power lamp must a page be held in order that it may be illuminated twice as brightly as a page which is held 4 feet away from a 24 candle power lamp? EXAMPLE
Solution. (3)
The general type of variation gives us the equation 8
L=cd2'
where L is the intensity of illumination, s is the strength of the lamp, and d is the dist2ince from the lamp, all expressed in any convenient units of measure, If we represent by L 1 the intensity of illumination of the page held
§ 52]
VARIATION, RATIO, AND PROPORTION
71
4 feet from the 24 candle power lamp, that of the page held at the required distance, d, from the 30 candle power lamp must be 2L1. Substituting these
values in (3), we have 24 16
L1 = c-
and
2L1 =
C
30 d2.
Therefore, d2 = 10, and d = VlO = 3.16. The required distance is 3.16 feet.
EXERCISES In each of the Exercises 1 to 10, express the given statement in the form of a single equation. 1. y varies as t 3•
2. z varies jointly as x, y, and w. 3. L varies directly as s and inversely as d2• 4. F varies directly as m1 and m2 and inversely as d2• 6. V varies directly as r and s and inversely as t. 6. R varies directly as P and inversely as the square of I. 7. y varies directly as x; when x = 4, y = 16. 8. y varies directly as the square of x; when x = 3, y = 45. 9. M varies jointly as F and d2 and inversely as m; M = 225 when F = 3 and d = 2 and m = 5. 10. R varies jointly as p and V and inversely as x 3 ; R = 49 when p = 9 and V = 7 and x = 3. In each of the Exercises 11 to 14, represent the variables by suitable letters, and express the given statement in the form of an equation. 11. The distance that a body falls in a vacuum varies as the square of the time that it falls; in 3 seconds a body falls 144 feet. 12. The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter; a wire 450 feet long and 0.15 inch in diameter has a resistance of 1 ohm. 13. The rate at which a body travels varies directly as the distance and inversely as the time required; a body which travels 66 feet in 3 seconds moves at the rate of 15 miles per hour. 14. The area of the surface of a sphere varies as the square of the diameter; when the diameter is 14 inches, the area is 616 square inches. 15. Graph y as a function of x, assuming that y varies directly as x and that y = 6 when x = 2. 16. Graph y as a function of x, assuming that y varies directly as x and that y = 2 when x = 4.
72
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. Vl
17. Graph y as a function of x, assuming that y varies directly as the square of x and that y = 8 when x = 2. 18. Graph y as a function of x, assuming that y varies inversely as x and that y = 1 when x = 2. 19. The distance traveled by a body falling in a vacuum varies as the square of the time that it falls. If it falls 64 feet in 2 seconds, how long does it require to fall 400 feet? 20. The length of exposure necessary to photograph a given subject with a single lamp varies as the square of the distance of the subject from the lamp. When the lamp is 10 feet from the subject an exposure of one-fourth of a second is necessary. How near to the subject must the lamp be placed if an exposure of one-twenty-fifth of a second is to be sufficient? 21. The power of a certain steam engine varies jointly as the average pressure in the cylinder and the speed of rotation. It delivers 90 horsepower when the average pressure is 300 pounds per square inch and it is making 750 revolutions per minute. What is its power when the average pressure is 250 pounds per square inch and it is making 800 revolutions per minute? 22. The time required for an elevator to lift a weight varies jointly as the weight and the distance through which it is lifted and inversely as the power of the motor. If it requires 20 seconds for a 2 horse-power motor to lift 1100 pounds 20 feet, what power is necessary to lift 1650 pounds 25 feet in 15 seconds.
23. For good projection, the brightness required in the lamp of a moving picture projector varies as the square of the distance from the machine to the screen. If a machine operating 15 feet from the screen requires a 250 candle power lamp, how far from the screen may the machine be placed to operate with a 360 candle power lamp? 24. The time of exposure necessary in taking a certain photograph varies directly as the square of the aperture setting and inversely as the index number of the film. If a black and white film with the index number 50 requires 1½ 0 of a second when the aperture has a setting of 8, what should be the aperture setting for an exposure of 5 ~ of a second using a colored film with the index number 10?
26. The absolute temperature of a certain quantity of gas varies jointly as the pressure and the volume. The absolute temperature of some gas in a cylinder is 292°C. and its pressure is 32 pounds per square inch. What will the absolute temperature become if the volume is doubled and the pressure is changed to 15 pounds per square inch? 26. The gravitational attraction between two bodies varies jointly as their masses and inversely as the square of the distance between their centers. Two bodies whose centers are 4000 miles apart attract each other with a
§§ 52-53]
73
VARIATION, RATIO, AND PROPORTION
force of 12 pounds. What would be their force of attraction if their masses and the distance between their centers were all doubled?
27. If a body weighs 12 pounds at the earth's surface, what would it weigh 4000 miles above the earth? (See Exercise 26; take the earth's radius as 4000 miles.) 28. The mass of a sphere of given density varies as the cube of its radius. If a body weighs 12 pounds at the earth's surface, what would it weigh on the surface of a planet 8000 miles in radius that has the same density as the earth? (See Exercise 26, and assume that the earth is a sphere whose radius is 4000 miles.) 29. The current in an electric wire varies directly as the electromotive force and inversely as the resistance. The resistance of a wire of a given material varies directly as its length and inversely as the square of its diameter. If the current is 36 amperes when an electromotive force of 120 volts i~ applied to a wire 1200 feet long and 0.06 inch in diameter, what current will flow when 80 volts are applied to a wire of the same material 800 feet long and 0.05 inch in diameter? 30. What electromotive force will be required to produce a current of 18
amperes in a wire 900 feet long and 0.09 inch in diameter which is made of the same material as the wire of Exercise 29?
53. Ratio. Closely related to the language of variation is the language of ratio and proportion. The quotient a + b is sometimes called the ratio of a to b. The ratio of a to b may be written a : b. For purposes of arithmetical treatment it is advisable to write thP ratio in the form of a fraction: (I)
a :b
a
= -· b
When concrete quantities are involved, a ratio has meaning only if the two terms of the ratio represent quantities of the same kind. Such a ratio is equal to the quotient of their measures expressed in the same unit of measure. The ratio of 6 feet to 4 pounds has no meaning since length and mass are not comparable. The ratio of 1 yard to 1 inch is equal to 3 i6 = 36. The ratio of 3 kilograms to 2 pounds is, approximately, 3 (;- 2) = 3.3, EXAMPLES.
..,;
since one kilogram is approximately equivalent to 2.2 pounds.
74
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. VI
64. Proportion. A statement of equality between two ratios is sometimes called a proportion. The proportion C -ba = -, d
(1)
may be written in the notation of ratio and proportion in either of the forms (2)
a:b : : c:d
or
(3)
a : b = c : d,
which are read "a is to b as c is to d." Practical problems are often stated in this language, but for purposes of computation it is always advisable to translate such a statement immediately into the form of equation ( 1). In a proportion dealing with material quantities, the two terms of the left-hand member may represent quantities of one kind and the two terms of the right-hand member may represent quantities of a different kind. For example, if an automobile travels at uniform speed, the distance traveled in one interval of time is to the distance traveled in a second interval of time as the length of the first interval of time is to the length of the second interval.
We express this proportion in the form ddi = 2
0:,
where, for one
t2
choice of units, d1 is the number of miles traveled in ti minutes and d 2 is the number of miles traveled in t2 minutes. In problems involving material quantities, always replace them by their measures in terms of suitable units.
The four numbers, a, b, c, and d, in that order, are said to be in proportion, if a:b=c:d
or
The first and fourth terms, a and d, are called the extremes, and the second and third terms, b and c, are the means of the proportion. It is evident that ad = be or in a proportion the product of the extremes is equal to the product of the means. If the two means of a proportion are equal, the second term is said to be a mean proportional between the extremes. That is, if a
b=
b
c'
then b is a mean proportional between a and c. In this case b2 = ac, so that b = ±Vac. There are therefore two mean proportionals between any two numbers, a and c, neither of which is zero. If a
§ 54]
VARIATION, RATIO, AND PROPORTION
75
and c have opposite signs, the mean proportional between them is 1magmary. To find the mean proportional between 4 and 25, we represent the mean proportional by b, and write b = ±V4 • 25 = ±10. As a verifica4 4 . . t 10n, we h ave t h e proport10ns = 10 an d =Io = - 10 . The mean pro10 25 25 portionals between 4 and -25 are b = ±V 4( -25) = ±10 i. There is a close connection between the ideas of variation and ratio and proportion. Thus, if y varies directly as x, by definition, y = ex. If x 1 and Yi are one set, and X2 and Y2 are any other set of corresponding values of x and y, Y1 = cx1 and Y2 = cx2, and EXAMPLES.
Y2
X2
Therefore, if y varies directly as x, the values of y are proportional to the corresponding values of x, and we sometimes say that "y is proportional to x."
EXERCISES In each of the Exercises 1 to 16, express the given ratio in fractional form, and simplify it. 1. Ratio of 4 yards to 6 feet. 2. Ratio of 1 mile to 60 feet. 3. Ratio of 12 pounds to 6 ounces. 4. Ratio of 4 gallons to 5 pints. 5. Ratio of 3 kilograms to 4 pounds, given 1 kilogram equals 2.2 pounds. 6. Ratio of 80 kilometers to 5 miles, given 1 kilometer equals 0.62 mile. '1. 24 : 16. 8. -15 : 40. 9. -64 : -24. 10. 26 : -65. 11. 4 a 2x2 : 12 ax 3• 12. -8 xy 2 : 12 x 3y. 13. (x 3 - y 3) : (x - y). 14. (x 4 - y 4) : 3(x2 + y 2). 15. 4 a3x : ?T ax 4. 16. (a 2x + a2y) : a (x2 - y 2). In each of the Exercises 17 to 24, solve the equation for x.
17. X : 4 = 5. 18. 2 X : (X + 5) = 3 : 4. 19. 9 ~ X = X. 20. 5 x : 11 = 11 : 5x. 21. (3 x - 1) : (4 x + 2) = (2 x - 3) : (x + 6). 22. 6a2x: ~3° = 9: (! x). 23. (x 3 + a3 ) : (x + 2a) = 3(x2 - a 2) : 4. 24. (2x2 +x-6): (4x2 -15x-4) = (8x2 -16x+6): (7x 2 -23x-20). 26. In a certain rectangle, the diagonal is to the length as 17 is to 15. Find the ratio of the length to the width. 26. The hypotenuse of a certain right triangle is to the shortest side as 13 is to 5. The area is equal to 270 square inches. Find the lengths of the sides. 27. A line 77 inches long is divided into two parts whose lengths are in the ratio 3 : 8. Find the lengths of the parts.
76
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. VI
~. Find the ratio of the two parts into which the point (9, 0) divides the line from (-5, 0) to (27, 0). 29. Find the point which divides the line from (3, 0) to (42, 0) internally in the ratio 5 : 8. 30. The sides of a triangle are respectively 8, 11, and 14 inches in length. Find the sides of a similar triangle whose shortest side is 12 inches long. 31. A man who is 6 feet tall stands in the sunlight and casts a shadow 10 feet long. At the same time a tree casts a shadow 110 feet long. How high is the tree? 32. A shaft carries two pulleys which are respectively 4 inches and 10 inches in radius. A second shaft carries two pulleys which are 3 inches and 8 inches in radius. A drive-belt on the four-inch pulley travels 150 feet per minute. A second belt runs from the ten-inch pulley to the three-inch pulley. Assuming that there is no slipping, find the speed in feet per minute of a point on the circumference of the eight-inch pulley. 33. A gear wheel with 50 teeth meshes with a second wheel having 200 teeth, and also with a third one having 150 teeth. A motor drives the first wheel at the rate of 2400 revolutions per minute. Find the speeds of the other wheels in revolutions per minute. 34. A pantograph is an instrument for reproducing figures into similar figures on a larger or smaller scale. A certain map on a scale of 1: 300,000 is enlarged by a pantograph to a scale of 1: 200,000. Two towns appear on the first map 6 inches apart. What will be the distance between them on the second map? In each of the Exercises 35 to 45, find the mean proportionals between the given pair of numbers.
35. 4, 16. 38. -i, -216. 41. x 3y2, xy- 4•
36. -4, -9. 39. 196, 2. 42. a6b-2, 1.
37. 2, ?}0 . 40. -27, -½. 43. -4 a4, a-2•
44 (a2-b2) a-b_ 45 x2 -x-2 1 x2 +x-6 • ' a +b • x 2 - x - 12 x 2 - 3 x - 4
CHAPTER VII
Quadratic Equations 55. The graph of a quadratic function. After the linear functions the next simplest polynomials are the quadratic functions. A quadratic function of x is a rational integral function of the second degree in x (§ 27). It can always be written in the form
y
= ax2 + bx+ c,
where a¢ 0.
Throughout this chapter we shall assume that the coefficients a, b, c, are real. In the function y = 7 x 2 + 3 x - 10, a = 7, b = 3, c = -10. In the function y = x 2 - 2 x - 3, a= ], b = -2, c = -3. The graph of the function y = x2 - 2 x - 3 is shown in Figure 29. EXAMPLES.
y
The graph of y
=
ax2
+ bx + c, a ~ 0,
,[s always a parabola which is symmetrical about a vertical axis. That is, there is a Yertical straight line called the axis of the parabola, such that if the figure is
folded on this line, the two halves of the curve will coincide exactly. If a
i.s positive, the parabola is open upward; the lowest point of the curve is then called its vertex. If a is negative, the pa.rabola is open downward, and its vertex is at the top of the curve. In § 65 we shall prove, and for the present we 77
V FIG. 29
78
FIRST YEAR OF COLLEGE lVIATHElVlATICS
[Ch. Vil
shall assume without proof, that the abscissa of the vertex of the parabola is equal to
The ordinate of the vertex is found by substituting x = -b/(2 a) in the quadratic expression. Thus, for the graph of y = x 2 - 2 x - 3, where a= 1 and b = -2, the abscissa of the vertex is x = - (- 2/2) = 1. The equation of the axis is x = 1. The ordinate of the vertex is y = l2 - 2(1) - 3 = -4, and the vertex itself is at the point (1, -4).
In· graphing a quadratic function, it is helpful first to mark the axis and vertex and then to graph the curve on one side of the axis and obtain the other half of the curve by symmetry. EXERCISES For each function name the coefficients, a, b, c; find the coordinates of the vertex and the equation of the axis of the corresponding parabola, and state whether the parabola is open upward or downward. Graph the function.
+ 6 X - 1. 6 X + 3. y = X 4 X + 7. y = 4X +x 1. y = X - 3 X + 4. y = 3 +2x
1. y = 2 X 2 3. y = 4 X 2
6. 1. 9. 11.
2
-
-
2
-
2
2
•
= 4. y = 6. y = 8. y = 10. y = 12. y = 2. y
+
-2 X 2 6 X - l. 2 -3 X + 4 X + 2. x2 4 X - 5. 3X 2 X 2 - 3. 2 X - X + 1. 1 - 2 x2•
+ +
56. Quadratic equations. A quadratic equation in x is an equation which can be written in the form of a quadratic function of x equated to zero. It can always be written in the standard form: (1)
ax2
+ bx + c = O,
where a :f:. 0.
A root of a quadratic (or any other) equation is a number which satisfies the equation ( § 396) when it is substituted in place of the unknown in the equation. In order to solve a quadratic equation it is necessary to find all of its roots.
QUADRATIC EQUATIONS
§§ 55-57]
57. Graphical solution of a quadratic equation. graphically a quadratic equation (1)
ax2
+ bx + c = O,
79
In order to solve
(a ~ O),
first graph the corresponding quadratic function (2)
y
= ax2 + bx + c.
At any point of the parabola, the distance from the x-axis is equal to the value of y for the corresponding value of x. If the curve crosses or touches the x-axis at a certain point, the value of y at that point is 0. Therefore the value of x, or abscissa, of this point is a root of equation (1); for, on substitution, it reduces the value of y to zero. The graph can cross the x-axis in at most two points. Rule. To solve the equation (1) graphically, graph the function (2). If this graph crosses the x-axis in two points, the roots of the equation are the abscissas of the points of crossing; if the parabola is tangent to the x-axis but does not cross it, the quadratic equation has only one root, which is the abscissa of the point of tangency; if the parabola neither crosses nor touches the x-axis, the equation has no real root. EXAMPLE
Solution.
Solve graphically the equation x2
1.
We graph the function y
=
X
2
-
4X
-
-
4 x - 5 = 0.
.Y
5.
The graph is the lowest of the three parabolas in Figure 30. This crosses the x-axis at the points (5, 0) and (-1, 0). The roots of the equation are therefore x = 5 and x = -1. On substituting these values of x, we see that they satisfy the equation. 2. Solve graphically the equation + 4 = 0.
EXAMPLE X
2
-
4X
Solution. The middle parabola in Figure 30 is the graph of the function y = x2 - 4 x + 4. It is tangent to the x-axis at the point (2, 0) and x = 2 is therefore the root of the equation. On substitution, x = 2 is seen to satisfy the equation.
3. The equation x2 - 4 x + 6 = 0 has no real root, for the graph of y = x2 - 4 x + 6, which is the highest of the parabolas in Figure 30; has no point on the x-axis. EXAMPLE
80
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. VII
EXERCISES In Exercises 1 to 12, determine graphically which equations have real roots. It is desirable first to find the coordinates of the vertex. Solve those equations graphically, determining the real roots accurately to the nearest multiple of one-fourth. In order to attain this degree of accuracy it may be necessary to use a large scale and to plot the points close together. 1. 3.
5. 7. 9. 11.
2 X - 3 = 0. 2 X - 2 X + 1 = 0. 2 X 2 + 5 X - 3 = 0. X2 6 X + 12 = 0. 2 9 X + 12 X + 4 = 0.. X 2 + 2 X - 6 = 0.
X2 -
2. 4. 6.
8. 10.
12.
12 = 0. X + 2 X - 8 = 0. 4 x 2 - 4 X + 1 = 0. x 2 - 8 X + 18 = 0. 9 x2 + 12 X + 10 = 0. 8 X2 + 18 X - 5 = 0. X2 -
X -
2
In Exercises 13 to 16, find the value that c must have in order for the vertex of the parabola to lie on the x-axis.
= x2 + 6 x + c. y = 9 x 2 - 4 x + c.
14. y = x2
13. y 16.
16. y
-
= 2x
2
¾x + c.
+ 10 x + c.
In Exercises 17 to 20, find the value that c must have in order for the equation to have just one real root. 17.
X2 -
19. 2 X
2
10 X + C = 0. 18 X + C = 0.
18.
X2
-
20. 3 X
2
16 X + C = 0. + 4 X + C = 0.
58. Algebraic solutions of quadratic equations. Pure quadratic equations. We shall now consider algebraic methods of solution. A pure quadratic equation is a quadratic equation that contains no term of the first degree in the unknown. It can be written in the form (1)
ax2
+ c = O,
where a ¢. 0.
Although this equation is quadratic in x, it is linear in x 2, and we can solve it for x 2, getting x2 = -c/a. We take the square roots of both members, and obtain the two roots
(2) On substitution, both of these numbers satisfy the equation (1).
§§ 57-58]
81
QUADRATIC EQUATIONS
EXAMPLE 1. Solve the equation 9 x 2
25 = 0.
-
Solution. We write x2 - 295 or x = ±-i. The numbers x = -i and x = -¾ satisfy the equation. Solve the equation 9 x 2 + 25 = 0.
EXAMPLE 2. Solution.
25 9 •
Solving the equation for x 2, we get x2 = -
There f ore
x
_/~ = ±v - 9 = ±"B"5•i,
where i = V -1. On substitution, we see that x = satisfy the given equation.
¾i
and x =
--i i
EXERCISES In Exercises 1 to 21, solve the equations and simplify any resulting radicals., 1. x 2
9 = 0.
-
4. x2 + 49
=
7. 4 x 2 + 25 10. x2
0. =
0.
7 = 0.
-
13. 2 x2
2. x2
-
9 = 0.
16. 3x2 +1=0.
= 16.
5. 4 x2
-
49
3. x 2 = -36.
8. 16 x2 + 49 11. x2
6. 36 x~ - 25 = 0.
0.
= =
9. 121 x2 + 81 = 0.
0.
+ 7 = 0.
12. 5 x2
-
36 = 0.
14. 3 x2 + 25 = 0.
15. 6 x2 + 49 = 0.
17. 2x2 -3=0.
18. 5x2 +3=0.
19. 3(x2 + 5) = x2 + 20. 20. 2(x2
-
4) = 5(x2
-
3). 2L 6(x2 + 3) = 11.
22. If one side of a square is increased by 2 feet and the other side is
decreased by 2 feet, the resulting rectangle has an area o ~ 32 square feet. Find the length of a side of the original square. 23. If one side of a square is increased by 3 feet and the other side is
decreased by 3 feet, the area of the resulting rectangle is 58 square feet less than twice the area of the original square. Find the length o!· a side of the original square. 24. If one side of a square is increased by 5 feet and the other side is
decreased by 3 feet, the area of the resulting rectangle is 79 square feet less than twice the area of a rectangle whose width is the same as that of the square and whose length is 1 foot greater than a side of the square. Find the length of a side of the square. 26. On a stream which flows at the rate of two miles per hour, a boat can
go 14 miles downstream in 4 hours and 40 minutes less time than it takes to return. Find its rate in still water. 26. A boat can go 7 miles per hour in still water.
24 minutes to go 24 miles upstream and return.
It requires 8 hours and
Find the rate of the current.
82
FIRST YEAR OF COLLEGE MATHEMATICS
69.
Solution of equations by factoring.
(1)
[Ch. VII
The equation
ax 2 +bx+c=O
can be solved very easily if it is easy to factor the expression
ax2 +bx+ c. The method, which can often be extended to equations that are not quadratic, is illustrated in the following example. EXAMPLE 1. Solve the equation 3 x 2
-
2 x - 4 = 3 x - 2.
Solution. First of all, we are careful to write the equation in standard form, transposing terms so as to make the right-hand member 0. The equation becomes (2)
3 X2
-
5X
-
2 = 0.
By factoring the left-hand member we obtain the equivalent equation (3)
{3 x
+ l)(x -
2) = 0.
If x has a value for which the entire left-hand member is zero, either (4) 3 X
+1 = 0
or
(5) x - 2 = 0.
(§ 10, Rule IV)
And conversely, if x satisfies either (4) or (5), on substitution it reduces one factor or the other of the left-hand member of (3) to 0, and so satisfies equation (3). The roots of equation (3) are therefore exactly the same as the roots of the two equations (4) and (5). Solving (4), we get x = -½; solving (5), we find x = 2. Therefore the roots of the given equation are x = -½ and x = 2, On substitution, these values are seen to satisfy the original eguatbn.
The method of solution by factoring is summarized as follows: Rule. Arrange the equation in standard form; factor the left-hand member into linear factors; set each linear factor equal to zero; solve the resulting linear equations to obtain the roots of the given equation. Notice that the method is entirely dependent on having O as the righthand member of the equation before factoring. EXAMPLE 2.
Solve the equation x3 = x 2
+ 6 x.
Solution. If we simplify the equation by transposing terms and dividing by the common factor x, we get x2 - x - 6 = 0. Factoring the left-hand member, we get (x - 3) (x 2) = 0, which gives the two roots x = 3, x = - 2. We see, however, that x = 0 also satisfies the original equation. vVe neglected, or rather threw away, the linear factor x when we divided both
+
QUADRATIC EQUATIONS
§§ 59-60]
83
members of the equation by x. A correct solution follows: By transposing terms in the original equation, we obtain x3
x2
-
6 x = 0 or
-
x(x 2
x - 6)
-
= 0 or
x(x - 3) (x
+ 2)
= O.
This is equivalent to the three equations X
= 0,
3 = 0,
X -
X
+2 =
0.
From these we get the three roots x = 0, x = 3, x = -2. discard any factor that contains the unknown.
Caution: Do not
EXERCISES In Exercises 1 to 36, solve the equations by factoring, and check by substitution. 1. 4. 7. 10. 13. 16. 19. 22. 26. 28. 30. 32. 34.
+ X - 6 = 0. X + 2 X - 15 = 0. 3x 7 x + 2 = 0. 2x 13 x + 6 = 0. 12 x + 5 x - 2 = 0. 2 X + 3 X = 0. 6 X + X = 1. X2 2
2 2
-
2
2
2
5 x 2 - 2 = 0. x 4 - 16 = 0. x2 - 4 = 8 - x. 5 x(x - 1) = 2 - 7 x 2 • (x x4
2.
X
2
-
3X
2
+ 3x
-
10x = 0. 2
12 = 0. 6. 2 X 2 + X ~ 6 = 0. 9. 5 x"'2 - 8 X - 4 = 0. 3. x 2
6. 2 X2 + X - 3 = 0. 8. 5 x2 - 14 x - 3 = 0. 11. 3 x2 + 7 x - 6 = 0. 14~ x2 - 6 x + 9 = 0. 17. X 2 = 8 X - 15. 20. 10 X 2 - X = 2. 23. 8 x2 = 3 x. 26. 16x4 - 81 = 0. 29. 6 x(x -
+ 2) (3 x + 2)(x + 1) = 0. 3
+ 2 = 0.
-
12. 8 X 2
X -
+ 10 X -
3
= 0.
7 X = 0. 18. X 2 = 2 X + 8. 21. 16 x2 - 25 = 0. 24.. 5 x 2 = 2 x. 27. x2 - 25 = 3(1 - x). 1) = x + 5. 15.
X
2
-
31. (x - 3)(x + 2)(x + 4) = 0. 33. x2 (x 3)(x - 5) = 0.
+
36. 3 x
3
+6x
2
=
x
+ 2.
60. Completion of a square. We say that we complete the square for a given expression x 2 + kx by adding to the expression a constant term that will make the new expression a perfect square. Since
(x + gy = x
2
+ kx + 12,
we evidently complete the square by adding (k2 /4) = (k/2) 2 to the expression x2 + kx. This method is important in later branches of mathematics as well as in the solution of quadratic equations.
+
Rule. In order to complete the square for an expression x2 kx, add the square of one-half of the coefficient of x in that expression. Caution. Notice that this rule applies only to an expression of the kx, where the coefficient of x 2 is equal to 1. form x 2
+
84
FIRST YEAR OF COLLEGE MATHEMATICS
EXAMPLE.
Complete the square for the expression x2
-
[Ch. VII
12 x.
Solution. The coefficient of x 2 is 1. We add the square of one-half the coefficient of x, or ( - 6) 2 = 36. The resulting expression is x2
-
+ 36 =
12 x
(x - 6) 2 •
61. Solution of a quadratic equation by completing the square. This method of solution is illustrated in the following example. EXAMPLE.
Solve by completing the square: 5 x2
7 x - 6 = 0.
-
Solution. We first change the equation into an equivalent equation having the form x 2 + kx in the left-hand member by transposing the constant term to the right-hand member and dividing both members by 5, the coefficient of x 2 • We obtain (1)
X
2
°k X = f.
-
\Ve next complete the square of the left-hand member by adding (--:/0) 2 =
4 9 1 0 0·
In order to get an equivalent equation, we must, of course, add / right-hand member also. This gives (2)
2
7
X-5X
+
49 6 100=5
9 0 0
to the
+ Too 49
or (3)
(x - -io) 2 =
½3 ~-
Taking the square roots of both members, we get (4)
7 X-To=
±13
To
or
X
=
7
To
±
13
To·
Using the plus sign, we get x = i g = 2; using the minus sign, we find x = - 160 = -¾. The roots of the given equation are
x = 2
and
x =
-¾.
Rule for solving a quadratic equation in x by completing the square: (1) Transpose to the left-hand member the terms that involve x and to the right-hand member the terms that do not involve x. (2) Divide both members of the equation by the coefficient of x 2• (3) Complete the square in the left-hand member by adding the square of one-half of the coefficient of x to each member. (4) Take the square root of each member, indicating both of the square roots of the right-hand member. (5) Combine the constant terms, and so obtain the roots.
§§ 60-62]
85
QUADRATIC EQUATIONS
EXERCISES In Exercises 1 to 12, find the number that must be added to each expression to complete the square. 1. x 2
+ 6 x. + 3 x.
6. x 2 9. x 2 + ax.
2. x2 6. x 2
10. x 2
-
6 x.
-
5 x.
3. y 2 7. x2
-
-! x.
11. x 2
-
10 y.
-
x.
-
t x.
4. y 2
+ 8 y. + x.
8. x2 12. x 2 +
¾x.
In Exercises 13 to 24, solve the equations for x by completing the square. 13. 16. 19. 22.
x2 + 2 x - 15 = 0. 14. x2 + 8 x + 15 = 0. 17. x2 - x - 12 = 0. 20. 23. 2 x2 + x - 6 = 0.
x 2 + 2 x - 8 = 0. x2 + x - 6 = 0. x2 + 3 x - 10 = 0. 3 x2 - 7 x + 2 = 0.
15. 18. 21. 24.
x2 + 4 x = 5. x2 - 3 x + 2 = 0. 2 x 2 + x = 3. 5 x 2 - 14 x = 3.
62. Solution of a quadratic equation by formula. We shall now apply the method of completing the square to the solution of the general quadratic equation
(1)
ax2
+ bx + c = O, (where a ¢. 0),
and so obtain a formula which gives equation. We first transpose the term and divide by a, getting b (2) x2 -x = a
+
the roots of any quadratic c to the right-hand member C
-· a
Complete the square in the left-hand member by adding (b/2 a) 2 or b2/(4 a 2) to each member. We obtain 2
(3)
X2
+ -ab X + 4-ba
2
b2 C - -, 2 4a a
= -
or, after writing the right-hand member as a single fraction, (4)
l__ ) 2
( x+2a
_
-
b2
Take the square root of each member. (5)
b x+-= 2a
4 ac 4a2 •
-
We get
Vb 2 - 4 ac ±----. 2a
Therefore
(6)
-b ± vb2 -4 ac x=-------· 2a
86
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. VII
By substitution the student may show that the values given by (6) actually satisfy equation (1) and that they are therefore roots.
Formula (6) is one of the most important formulas in algebra, and it should be learned at once and never forgotten. With it we can solve any quadratic equation by substituting the proper values of the coefficients a, b, and c. In order to solve a quadratic equation, in general use this formula unless the linear factors of the quadratic expression are very obvious, in which case the method of solution by factoring may be used. EXAMPLE
By means of the formula, solve the equation
1.
12 x 2 - 8 X - 15 = 0. Substituting a = 12, b = - 8, c = - 15 in the formula, we get
(7) Solution. (8)
X
8 ± v64 + 720 8 ±v 784 8 ± 28 =---- - -- = -
24
24
8
The two roots are therefore x = EXAMPLE
+ 24
28
=
24
£2
Solve the equation 2 x2 + 2 x + 5
2.
x=
and
=
8
24
28
= -
£.6
0.
We substitute a = 2, b = 2, c = 5 in the formula, and obtain
Solution. (9)
=
X
- 2±
v4 -
40
4
-
- 2±
v--=--ii - - 2 ±
4
6i
4
The two roots are therefore x = - ½+ ¾i and x = - ½- ¾i. EXAMPLE 3. Solve the following equation for yin terms of x: 2 y2
(10)
3 xy
-
+2y +x
2
x
-
=
0.
Solution. We display this as a quadratic equation in the unknown y, by writing it in the form
2 y2 + (-3 x + 2)y + (x 2
(11)
-
x)
= 0.
In the formula we substitute a = 2, b = (-3 x + 2), c = (x2 (12)
y
-
3 x - 2 ± V(9 x2 - 12 x + 4) - (8 x2 - 8x) = ------------------4
-
3x - 2±
vx
2 -
4
4x
+4 = 3x -
2 ± (x - 2)
3x-2+x-2
4
In terms of x, the roots are
y
= ------ =
and
y
3x-2-x+2 =- - - - - = ½x.
4
4
X -
1,
x), getting
§ 62]
QUADRATIC EQUATIONS
87
EXERCISES In Exercises 1 to 36, solve for x or y, using the quadratic formula.
1. 2 X2 + X - 15 = 0. 3. 5 X 2 - 18 X - 8 = 0. 6. 8 y 2 - 14 y - 15 = 0. 7. 28 X2 - 9 X - 9 = 0. 9. 14 y2 - 5 y - 24 = 0. 11. X 2 + 2 X - 1 = 0. 13. 16 X 2 - 24 X - 3 = 0. 16. 3 X 2 + 14 X + 12 = 0. 17. X 2 - X + 1 = 0. 19.x2 -4x+5=0. 21. 25 x 2 23. 4 y2 26. x 2
-
-
20 X 4y
2. 4 X 2 + 4 X - 3 = 0. 4. 6 X 2 - 17 X - 14 = 0. 6. 8 y 2 + 37 y - 15 = 0. 8. 20x2
10. 6 y2 + 17 y 12.
14. 16. 18. 20.
+ 7 = 0.
22.
+ 29 = 0.
+ 4 kx -
24.
21 k 2 = 0.
26.
27. 4x2 - 4kx - 35k2 = 0. 29. 6x 2 - 2kx - x 31. 2 X 33.
X
2
4X
-
2
2
-
2
2
2
-
2
2
2
-
-
mx - 2m2 = 0.
30. 12 x 2 + kx
1 = 0.
21
32. 25 - 6 X
+ x2 - :OX+ 3 =
0. 34.
+ 1 = 0.
+ 12 = 0.
+ 1 = 0. 5 x + 12 X + 6 = 0. 2X 5 X + 1 = 0. X + X + 1 = 0. x + 6x + 10 = 0. 16 X 24 X + 59 = 0. 9 y + 12 y + 52 = 0. x 7 kx + 10 k = 0. X
28. 15x2
+ X + 4 = 9.
23 1 + 3
36. x 3
+k -
4x - 7 = 0.
-
X- 3 x+l
36. x 3
-
+ 2 = 10 x + k + k. 2
= ~• x+5
+3 =
32
x 2 +3x+2
1 = 0.
In Exercises 37 to 40, solve for yin terms of x and for x in terms of y. 37. 2 y2
+ 3 xy -
7 y = 2 x2
-
11 x
+ 15.
38. y2 +xy+4y = 2x2 -17x+21.
39. 6 y 2 40. x 2
-
4 y - 3 xy = 9 x 2 + 9 x
+ xy -
2 y2 = 9 x
+3y -
+ 2. 20.
41. A building lot is 50 feet longer than it is wide and has an area of 7475 square feet. Find its dimensions. 42. If room for a path 3 feet wide is taken off the sides of a square garden,
the remaining area is 612 square feet more than half of the original area. Find the original dimensions. 43. A cement walk of uniform width is laid around a rectangular let. The area of the walk is 1424 square feet. The rectangle inside the walk is 50 feet wide and 120 feet long. How wide is the walk?
88
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. VII
44. A covered cubical box for holding radium is made of lead 1 inch thick. The lead amounts to 19 cubic inches. Find the length of an inside edge of the box. 45. When the cylinder of a certain heavy engine is rebored, its radius is increased by .05 inch. This increases the cubic capacity of the cylinder by 4.04%. Find the original diameter of the cylinder. 46. An airplane starts on a flight of 1220 miles.
After flying for 500 miles, it is caught by a head wind which decreases its speed by 10 miles per hour for the rest of the trip. It makes the entire flight of 1220 miles in 5 hours. Find its speed before meeting the head wind.
47. On a river that flows at the rate of 3 miles an hour, a certain motorboat can make a trip downstream and back to the starting point at an average speed of 11¼ miles an hour. What is its speed in still water? 48. A motorboat whose speed in still water is 10 miles per hour, on a certain river can make a trip downstream' and back to the starting point at an average speed of 9/0 miles per hour. Find the speed of the current. 49. Two pipes together can fill a reservoir in 4 hours and 48 minutes.
The reservoir can be filled by the larger one alone in 4 hours less time than by the smaller one alone. Find the time in which it can be filled by each pipe alone. 60. A swimming pool can be drained in 3 hours less time than it takes to fill it. If the drain pipe and the filling pipe are both running, the pool will be drained in 18 hours. Find the time required to drain the pool if only the drain pipe is open. 61. A jobber buys some goods and sells them to a retailer for a certain
percentage more than the manufacturer's price. The retailer sells them to the consumer for the same percentage more than the jobber's price. The consumer pays 69% more than the manufacturer's price. What per cent of the manufacturer's price is added by the jobber? 62. A merchant bought some goods whose cost at list price was $1000.
However, he received two discounts, the first being a certain percentage of the list price, the second being half of that percentage of the balance. His final cost was $720. What per cent discount was allowed each time? 63. Two boys are to mow a lawn which is 10 feet longer than it is wide.
The first boy does his half by mowing a belt 5 feet wide around the lawn. Find the dimensions of the lawn. 64. On his lathe a machinist turns off one-fifth of an inch from the out-
side of the curved surface of a cylindrical bar, and thereby reduces the weight of the bar by 19%. Find the original radius of the bar.
§§ 62-63]
89
QUADRATIC EQUATIONS
Equations often occur which are not quadratic ~quations, but which take on the form of quadratic equations if suitable substitutions are made. Such equations are sometimes called equations in quadratic form. 63.
Equations in quadratic fom1.
1.
EXAMPLE
Solve the equation x 4
Substitute y = x2•
Solution.
y2
0.
The equation becomes 13 y
-
+ 36 =
13 x2
-
+ 36 =
0,
which is a quadratic equation in ·y. Its roots are y = 4 and y = 9. of the relation y = x2, we set x2 = 4 and x 2 = 9, which give X
= 2,
= -2,
X
as roots of the given equation. EXAMPLE
2.
Solve the equation 6 ( x
6 y2 whose roots are y =
~ and y
= 3,
= -3,
X
We verify that they are roots by substitution.
Substitute y = ( x
Solution.
X
Because
-
+ ; )35 y
+ ; )2 -
35 ( x
+ ; ) + 50 = 0.
The equation becomes
+ 50 =
0,
1
= ~- Substituting ( x
+ ;)
= y, we obtain
the equation
x
+ -X1 =
5
-,
2
which has the roots x = 2 and x =
x
10 + -X1 = -, 3
2 x2
or
or
½,
-
5x
+2 =
0,
and the equation 3 x2
-
10 x
+ 3 = 0,
which has the roots x = 3 and x = ½. The four roots of the given equation are therefore x =-= 2, x = ½, x = 3, x = ½.
EXERCISES In Exercises 1 to 34, solve the equations for x. 1. x4 3. x 4 5. x 4
-
5 x2 6 x2 3 x2
+ 4 = 0. + 8 = 0.
4 = 0. 4 2 7. x + 2 x - 15 = 0. 9. 6 x 4 - 17 x2 + 12 = 0. 11. 8 x 4 - 6 x2 - 5 = 0. -
-
2. x4
4. x
4
6. x 4
-
-
+
+ +
10 x2 9 = 0. 2 7x 12 = 0. 8 x2 - 9 = 0.
8. x x2 - 6 = 0. 10. 6 x 4 - 17 x 2 5 = 0. 12. 9 x 4 - 6 x 2 - 8 = 0. 4
+
90
FIRST YEAR OF COLLEGE MATHEMATICS
13. x 4
-
16 = 0.
16. x 6
-
35 x 3
17. x-2
19. 4 x-4
-
21. 20 a4x-4
14. 81 x 4 = 625 a4•
+ 216 = 0.
2 x-1
-
-
29 x-2
16. 27 x 6
18. 10 x-2
+ 45 = 0.
20. 3 x- 4
2
225 = 0.
2 -
+ 3 x)
35 x3
-
3 = 0.
+ 89 a x-
+ 3 x)
[Ch. VII
-
-
+ 8 = 0.
11 x-1 52 x-2
-
6 = 0.
+ 64 = 0.
22. 36 x-4 + 221 a2x-2 - 25a4 = 0.
= 0. 24. (x 2 - 6 x) 2 + (x 2 - 6 x) - 56 = 0. 26. (x2 + 2 x) 2 - 14(x2 + 2 x) - 15 = 0. 26. (2 x2 + 7 x) 2 - 12(2 x2 + 7 x) - 45 = 0. 23. (x2
2
-
2(x2
- 8
27. (x4 - 10 x 3 + 25 x2 ) - 2(x2 - 5 x) - 24 = 0. 28. (4 x 4 - 28 x 3 + 49 x 2) + 9(2 x2 - 7 x) + 18 = 0. 1 2 15 29• (2 X - 3) 2 + 2 X - 3 - 15 = O. 30• (4 X - 5) 2 24 x2 31. (x - 3)2
33. 12 ( 34. 8 (
X
X -
14 x (x - 3) - 3 = 0.
~ )2 -
8(
\)2 -
+2
X -
30 (
; ) -
14
+4X -
5- 8
= O.
x4
2 x2 32. (x - 2)2 + x - 2 - 3 = 0. 15 = 0.
+ 2 x) + 27 = 0. 1
X
64. Equations involving radicals. Equations that involve radicals can sometimes be solved by a substitution according to the method of § 63. In other cases proceed as follows: Transpose to one member of the equation the most complicated radical term and to the other member all of the other terms of the equation. Remove the radical from this selr>,cted term by raising both members of the equation to the power corresponding to the index of the radical .. Combine terms and repeat the foregoing process until the resulting equation contains no radicals that involve the unknown. Solve the resulting equation. This solution may introduce extraneous roots, for the process of taking a power is equivalent to a multiplication by a factor that contains the unknown quantity. Caution. In the solution of an equation involving radicals it is essential to test each root obtained by substituting it in the original equation. If it satisfies the original equation, it is to be retained; if it does not satisfy the original equation, it is extraneous and must be rejected.
QUADRATIC EQUATIONS
§§ 63-64]
91
Since a radical denotes the principal root, and principal roots of imaginary numbers have not been defined, it will be understood that we shall consider at present only the real roots of equations involving radicals. EXAMPLE
v 4 x + 8 + 1 = x.
1. Solve the equation
Solution. We arrange the equation so that the radical term shall stand by itself on one side of the equation: V 4 x + 8 = x - I. 4x
Square both members: Therefore:
x2
-
+8 =
x2
6 x - 7 = 0,
or
= 7
or
whence
x
2x
-
+ 1.
(x - 7)(x x
+ 1)
= 0,
= -1.
Substitute x = 7 in the original equation: v2s
+s +1 = 7
or
6
+1=
7.
Therefore x = 7 satisfies the original equation and is a root. x = -1 in the given equation. The left-hand member is
v - 4+8+1=
V4 + 1
= 2
+1=
Substitute
3,
while the right-hand member is x = -1. Since 3 ~ -1, x = -1 is an extraneous root and must be rejected. The square of V 4 x + 8 is the same as the square of -V 4 x + 8, and after squaring it became impossible to distinguish between the principal square root -V 4 x + 8 which appears in the original equation, and the negative root, which does not appear there. This accounts for the introduction of the extraneous root -1.
+ 4 - V3 x + l - 2 = 0. Solution. By transposing, we obtain 2-v'x + 4 = V3 x + 1 + 2. Square: 4(x + 4) = (3 X + 1) + 4V3 X + 1 + 4. Combine and rearrange: 4v'3 x + 1 = x + 11. Square both members: 16(3 x + 1) = x + 22 x + 121. 26 x + 105 = 0 or (x - 5)(x - 21) = 0. Combine: x EXAMPLE
Solve the equation 2-v'x
2.
2
2
-
Therefore x = 5 or x = 21. Substituting x = 5 in the left-hand member of the given equation, we find 2-vs + 4 - v115 + 1 - 2 = 6 - 4 - 2 = o. Therefore x = 5 is a root of the given equation. Substituting x = 21 in the left-hand member of the given equation, we have 2v21
+4 -
V63
+1-
2
= 10 -
8 - 2
Therefore x = 21 is also a root of the given equation. x = 5 and x = 21 as roots.
=
o. We retain both
92
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. VII
EXERCISES Find the real roots of the following equations.
+5 =
1. V3 x - 2 = 2.
2. V2
3. V5 x - 9 = 4.
4. V7 x - 2 = 3.
+ 7 = x + 1. + 14 = x + 2. + I + 3 x = 11. v 4 x 36 - 4 = o.
V3 x 7. V7 x 9. V3 x 5.
2
11.
v4 x
2
13.
6. 10.
3.
v 9 x + 28 = x + 4.
8. 2-Vx X -
+ 4 = X + 1. -V3
2 = 10.
X -
+ 4 X ....:.. 5 - 2 = v3 x + 6 x - 8 = 4.
12. y'X2
-
X
8 x - 59 - 3 = 0.
14.
0.
2
16. v 5 x - 4 = 3-Vx - 2. + 7 = 1 + -V3x. 18. v 2 x + 25 + V 8 - x = 7. V5 x + 1 = v12 x + 11 + 1. 20. V9 x + 10 - V3 x + 2 = 2. vs x + 5 - V2 x + 3 = 1. v 3 x + 10 - v 3 x + 5 = 1. 22. V 4 x + 3 - v 2 x + 5 = 1. v2 x + 5 = v4 x + 3 + 1. 24. v3 x + 5 = -V6 x + 11 - 1. V V5 x - l - x = 1. 26. V V2 x + 1 + x - 3 = 2. (x + 3 x + 6) - 3-Vx + 3 x + 6 = 4. (2 x + 3 x - 2) - 4V2 x + 3 x - 2 = 5. 5X 2 X - 3V 5 X 2 X + 1 = 9. 2 X + 3 X - V2 X + 3 X + 7 - 5 = 0.
V3 x
16.
17. 19. 21.
23. 25.
2
27. 28. 29.
30.
2
2
2
2
2 -
-
2
2
65. The discriminant of a quadratic equation. X2 are the two roots of the quadratic equation
By § 62, if
X1
and
ax2 + bx + c = O, (where a :¢ 0),
(1)
then (2)
X1
b Vb2 - 4 ac = - -2 a + - -2-a - ,
b
vb2 - 4 ac 2a
X2=-------
2a
The quantity (3)
b2
-
4 ac,
is called the discriminant of the quadratic equation (1) and also of the quadratic function ax2 + bx + c. In what follows, we shall assume that the coefficients a, b, and c are real rational numbers. If b2 - 4 ac = 0, the two roots X1 and X2
§§ 64-65]
QUADRATIC EQUATIONS
93
are both equal to -b/2 a. If b2 - 4 ac > 0, Vb 2 - 4 ac -is real, and the roots x1 and x2 are real and unequal. If the discriminant is the square of a rational number, the two roots are rational as well as real, but otherwise Vb 2 - 4 ac is irrational and x1 and x2 are irrational. If b2 - 4 ac < 0, Vb 2 - 4 ac is imaginary, and the roots x1 and x 2 are imaginary. An inspection of the discriminant therefore enables us to determine the nature of the roots without having to solve for them. These facts can be summarized as follows:
+
If the coefficients of the equation ax 2 + bx c = 0 are real and rational, its roots are (a) equal and real, if and only if b2 - 4 ac = 0, (b) real and unequal, if and only if b2 - 4 ac > 0, (c) imaginary and unequal, if and only if b2 - 4 ac < 0, (d) rational, if and only if b2 - 4 ac is a perfect square. 1. Determine the nature of the roots of each of the three equations (4) x2 - 6 x + 5 = 0, (5) x2 - 6 x + 9 = 0, (6) x2 - 6 x + 12 = 0. Solution. For equation (4) the discriminant is equal to EXAMPLE
62
-
4 · 5 = 36 - 20 = 16.
Since 16 is positive and a perfect square, the roots are real, unequal, and rational. For equation (5), the discriminant is 62 - 4 · 9 = 36 - 36 = 0. The roots are therefore real and equal. For equation (6) the discriminant is 62 - 4 • 12 = 36 - 48 = -12. The roots are therefore imaginary and unequal.
The real roots of a quadratic equation are the abscissas of the points where the graph of the corresponding quadratic function is in contact with the x-axis. When the discriminant of a quadratic equation is positive, the roots are real and distinct, and the graph of the corresponding quadratic function crosses the x-axis in two distinct points. Likewise, if the discriminant is equal to 0, the roots are real and equal, and the graph touches the x-axis at one point, but does not cross it. If the discriminant is negative, there are no real roots, and the graph does not meet the x-axis at all. A quadratic expression which is always positive or always negative is sometimes spoken of as a definite quadratic expression. The graph of such an expression evidently lies entirely above or entirely below the x-axis, and has no points on the x-axis. A quadratic expression is therefore definite, if and only if its discriminant is negative.
94
FIRST YEAR OF COLLEGE MATHEMATICS
EXAMPLE
(4') µ
=
x2
2. -
[Ch. VII
Consider the graphs of the functions 6x
+ 5,
(5') y
=
x2
-
6x
+ 9,
(6') y
= x2
-
6x
+ 12.
Discussion. These expressions correspond to the three equations of Example 1. The graph of (4') crosses the x-axis in two points, that of (5') is tangent to the x-axis, and that of (6') has y no point on the x-axis, since the discriminants are, respectively, positive, zero, and negative. The graphs are shown in Figure 31. The three functions differ only in the values of the constant term, c, which are, respectively, 5, 9, and 12. Increasing the value of c in this way increases the value of y for each value of x. In the present examples the effect is to raise the parabola first from the lowest position, where it crosses the 1 x-axis in two points, to the second position --+---o--+__.,oc--i---c---.-X where it just touches the x-axis, and then to the third position where it lies wholly -1 above the x-axis. When a is positive, an -2 increase in the value of c increases the value -8 of 4 ac and decreases the value of b2 - 4 ac. -4 A sufficient increase in c will thus make the Vi discriminant negative and raise the parabola FIG. 31 until it lies wholly above the x-axis. EXAMPLE
Determine the value of k if the graph of the function
3.
y
= 4x2 + kx
+ 3x + k
is tangent to the x-axis. Solution. If the parabola is tangent to the x-axis, the roots of the corresponding equation must be equal, and the value of the diicriminant must be 0. In standard form, the equation is 4 x 2 + (k + 3)x + k = 0, where a = 4, b = (k + 3), c = k. Setting the discriminant equal to 0, we get
b2
-
4 ac = (k
+ 3)
2
-
4 · 4 · k = k2
-
10 k
+ 9 = 0. = 4x + 4x + 1,
2 Hence k = 1 and k = 9. When k = 1, the function is y which represents a parabola tangent to the x-axis at x = -½. When k = 9, the function is y = 4 x2 + 12 x + 9, whose graph touches the x-axis at --!.
We can now justify the statement made in § 55, that the abscissa of the vertex of the parabola y=ax2 +bx+c
§ 65]
95
QUADRATIC EQUATIONS
is equal to -b/2 a. By changing the value of c, we can raise or lower the parabola until it is tangent to the x-axis at the vertex. This does not change the position of the axis of the parabola nor the value of -b/2 a. But when the parabola is tangent to the x-axis, b2 - 4 ac = 0, and the two equal roots of the corresponding equation, which are equal to the abscissa of the vertex, are equal to -b/2 a, by the formulas (2).
EXERCISES In Exercises 1 to 12, determine the character of the roots without solving the equations.
x2 - 2 x - 15 = 0. x2 + 8 x + 16 = 0. 3 X 2 - 8 X + 5 = 0. 5 X 2 + 13 X + 9 = 0. 9. 9 X2 - 12 X + 4 = 0. 11. 7 X - 4 - 3 X 2 = 0. 1. 3. 5. 7.
+ 9 x + 20 = 0. x 6 x + 10 = 0. 5x 17 X + 11 = 0. 4 x + 20 ~ + 25 = 0. 9 X + 42 X + 49 = 0.
2. x2 4.
2
-
2 6. 2 8. 2 10. 12. 6 X - 5 X 2
-
2 = 0.
In Exercises 13 to 20, find the real values of k for which the roots of each equation are equal.
13. 9 x2 + 12 x + k = 0. 14. 16 x2 - 40 x + 5 k = 0. 15. 9 x 2 + 2 kx + 16 = 0. 16. 4 x 2 + 7 kx + 49 = 0. 17. 9 x 2 + 3(k + l)x + 2(k + 1) = 0. 18. kx2 + 2(k + 4)x + 25 = 0. 20. 4 kx2 - 3 kx + 4 = 2 x - x 2 • 19. 3 kx 2 + x2 + 24 x + 2 k - 1 = 0. In each of the Exercises 21 to 28, state whether the graph of the function crosses the x-axis, is tangent to it, lies wholly above it, or lies wholly below it. 21. y = 5 x2 + 7 x + 1. 22. y = 3 x 2 - 2 x - 7. 23. y = 25 x 2 - 70 x + 49. 24. y = 8 x2 - 12 x + 5. 26. y = 6 X - 5 - 2 x 2 • 25. y = 49 x2 - 56 X + 16. 28. y = 5 X - 3 - 2 x2 • 27. y = 48 X - 9 X2 - 64. In each of the Exercises 29 to 32, determine the value of k for which the graph of the function is tangent to the x-axis. 29. y = 2 kx 2 - 6 kx + 9. 30. y = (2 k - 4)x2 - 3 kx + 25. 31. y = (2 k - l)x2 + 24 x + 3 k + 1. 32. y = 2 a(k + a)x2 + 4 bkx + b2 • In Exercises 33 to 36, prove the theorems stated, assuming that a, b, and c are real and that a ~ 0. 33. Zero is a root of ax2 + bx + c = 0, if and only if c = 0. 34. Zero is the only root of ax2 +bx+ c = 0, if and only if b = 0 and c = O. 35. The equation ax2 + bx + c = 0 has one positive root and one negative root if and only if a and c have opposite signs.
96
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. VII
36. The two roots of ax2 + bx + c = 0 are numerically equal but are real and opposite in sign, if and only if b = 0 and a and c have opposite signs. 87. By means of the theorem in Exercise 35, determine which of the equations in Exercises 1 to 1'2 have one positive root and one negative root. In Exercises 38 to 42, determine h and k so that the conditions stated shall be satisfied. 38. 5 x 2 39. 3 x 2 40. 3 x2 41. 5 x 2 42. 3 x 2
9 kx - k + 6 = 0 has the root 0. + 7 kx + k2 - 5 k + 6 = 0 has the root 0. + 8 hx - 24 x + 5 k = 10 has only the root 0. + 2 hx - 23 x + k + 17 = 3 kx + 3 h has only the root 0. 5 kx + 10 x = 7 has numerically equal roots, opposite in sign. -
66. The sum and product of the roots. for the sum of the roots of the equation (1)
ax2
+ bx + c = 0,
We can obtain a formula
(a ~ 0),
by adding the expressions for X1 and X2 in the formulas (2) of § 65. The terms involving the radicals cancel and we obtain (2)
X1
+ X2 =
b
- -· a
In order to obtain a formula for the product of the two roots, we multiply the expressions for x1 and x2 from § 65 as the product of the sum and the difference of two quantities. Hence X1X2
_ ( -
=
-
b2 4 a2
b ) 2a -
2
-
(Vb
2 -
2a
4
acy
b2 -4ac 4ac c 4 a2 = -4-a2 = ~-
Therefore (3)
X1X2
= -·aC
These formulas should be memorized as follows, and particular attention should be paid to the signs: The sum of the roots of a quadratic equation in standard form is equal to the negative of the coefficient of x divided by the coefficient of x 2 ; the product of the roots is equal to the constant term divided by the coefficient of x2•
QUADRATIC EQUATIONS
§§ 65-67]
97
EXAMPLE 1. Without solving the equation, find the sum and the product of the roots of the equation
3 X2
+7X +6 =
0.
Solution. From formula (2), the sum of the roots is From formula (3), the product of the roots is X1X2 = ! = 2.
67.
X1
+ x2 = -¾.
Theorem. Any quad-
Factorization of a quadratic expression.
ratic exvression ax 2 +bx+ c,
(1)
(a
~
O),
can be factored in the form
ax2
(2)
+ bx + c = a(x -
x1)(x - x2),
where x1 and x2 are the roots of the corresponding quadratic equation ax 2
+ bx + c =
0.
Proof. Let x1 and x2 be the roots of the equation ax 2 + bx By multiplication, a(x - x1)(x - x2)
(3)
= a[x2
+ x2)x + x1x2]. § 66, x1 + x2 = -(b/a),
(x1
-
However, by formulas (2) and (3) of X1X2 = c/a. Substituting these values in (3), we have a(x - x1)(x - x2)
(4)
+ c = 0.
= a [ x 2 - ( - ~) x +
W] =
and
ax2 +bx+ c.
This completes the proof, since it shows that the two members of equation (2) are identically equal. By § 65, the roots x1 and x2 are equal if and only if b2 - 4 ac = 0. In this case ax 2 + bx+ c = a(x - x1) 2 and we can say:
+
The expression ax2 bx + c is a constant multiple of a perfect square, if and only if b2 - 4 ac = 0. If the roots are equal, their value is said to be a double root of the equation. EXAMPLE.
Factor the expression 9 x2
-
24 x
+ 19.
Solution. "\Ve form the corresponding equation, 9 x2 and solve it, getting _ 24
-
± v576 - 684 _ 24 ± v -108 _ 24 ± 6 iv3
x -
We substitute
18
-
18
X1
= ½+ ½iV3 and
X2
-
18
24 x
_~ - 3
+ 19 = 0, ! ·v-
± 3 i 3·
= ½- ½iV3 in formula (2) of the
98
FIRST YEAR Ol? COLLEGE MATHEMATICS
[Ch. VII
theorem, and obtain 9 x2
24 x
-
+ 19 = 9(x - ½- ½i-V3)(x - ¼+ ½i-V3).
To remove the fractions, we can write the coefficient 9 as 3 • 3, and multiply · one factor 3 into each linear factor. We get 9 X2
24
-
X
+ 19 =
(3
X -
4 -
i-V3) (3 X
-
4
+ i-V3).
68. The formation of a quadratic equation with given roots. Up to this point we have usually been given an equation and have tried to find its roots. We now consider the converse problem of forming a quadratic equation which shall have two given numbers as its roots. If two given numbers, x1 and x2, are to be the roots of a quadratic equation, by § 67 the equation can be written in the form
= 0,
a(x - x1)(x - x2)
where a may be any constant other than 0. In practice we usually choose the constant a so as to remove fractions from the coefficients. EXAMPLE
Form a quadratic equation whose roots are
1.
¾and -}.
Solution. The desired equation is of the form a(x - ¾) (x + ¾) = 0, where a may be chosen to suit our convenience. To remove fractions, we choose a = 12 = 4 • 3, and multiply 4 into the first linear factor and 3 into the second. We get (4
3)(3
X -
X
+ 2)
=0
12 X 2
or
-
X -
6 = 0.
With practice the student will soon learn to write the linear factors corresponding to the fractional roots ¾ and -¾, in the forms (4 x - 3) and (3 x + 2) at the outset.
2. Form a quadratic equation whose roots are -3 - 2-V -2 2v°=2.
EXAMPLE
and -3
+
Solution. To avoid mistakes in multiplying imaginary numbers, we write the imaginary term with the letter i, so that 2-V -2 = 2 i-V2. Since there are no fractions in the roots, we take the leading coefficient a = 1, and write the equation [x - (-3 - 2 i-V2)][x - (-3 + 2 i-V2)] = 0
or
[(x
+ 3) + 2 i-V2J[(x + 3)
- 2 iV2]
=
0.
:\Iultiplying this as the product of the sum and difference of the same two numbers, we write (x
+ 3)
2
-
(2 i-V2) 2
The desired equation is
= 0 X
2
or
X2
+6X +9 -
+ 6 X + 17 = 0.
8 i 2 = 0.
QUADRATIC EQUATIONS
§§ 67-68]
99
EXERCISES In Exercises 1 to 8, find the sum and product of the roots of each equation, without solving the equation. 1. 6 x 2
X -
-
15 = 0.
2. 4. 6. 8.
3. x 2 + 2 x - 1 = 0. 5. X 2 + 4 X + 13 = 0. 7. 4 x 2 12 X + 21 = 0.
+
+ 2 X - 8 = 0. + 2 x - 1 = 0. 2 x - 6 X + 34 = 0. 25 X 20 X + 31 =
3 X2 2 x2
2
-
0.
In each of the Exercises 9 to 56, obtain a quadratic equation with integral coefficients having the given pair of numbers as its roots.
9. 2; 3. 12. -7, 2. 15·
3
7,
_4
18.
¾,
-3.
21 •
5•
_3
2,
_4
5•
± v2. 27. 4 ± 3v5. 1 - ;30. 2 ± -Jv 5. 1- ;33. - 43 ± 4V2. 36. ± 4v5. 39. ± 2v3 i. 42. 2 ± 3 i. 24. 3
45 •
2
5 •
11. 4, -3.
13. ¾, ¾16• -2-7, .Q_8 • 19. -¾, 4. 22 • _ _§__e, _Ji3 •
14.
17. 20
25. -2
26. -4 ±
28. -1
± 5v -2.
51. 3, 3. 54. 0, -2.
± vs. ± 2V3.
± tV3. -¾ ± -! V3. ± 2V3. ± }vs i. 4 ± 5 i.
35.
43.
49. 52.
55.
_5
3,
4
5,
4•
_6
5•
2v2.
± 4V6. -½ ± ¾V3. 2 - ;- 3 ± Jv2. ± 2 i. ± ¾V -6. -3 ± 2 i. 4 ± 3v'=°6. ¾± f V5 i.
29. 3
34.
4o.
_7
23 •
32.
37.
¼, ¾¾, 5.
•
31. }
46.
3 ±3i.
48. -2
10. 5, 6.
38. 41.
44.
-¾ ± ¾i. -½ ± }V3 i. -¾, -¾. 0, ¾-
47.
50. 53.
¾, ¾.
fl6. 0, 0.
In Exercises 57 to 72, write the linear factors of each expression by solving a quadratic equation.
57. 12 X 2 + X - 35. 59. 42 X2 - 5 X - 63. 61. 4 X 2 + 20 X - 2. 63. 4 X 2 - 12 X - 103. 65. 4 x 2 + 28 X + 22. 67.
X2
+ 8 X + 41.
+
69. 9 X 2 12 X 3 71. x - 1.
+ 22.
58. 60
X2 -
1,01
X -
143.
+
60. 72 x 223 X + 165. 62. 9 X 2 - 6 X - 19. 2
64. 66. 68. 70. 72.
9 x2 X
2
-
30 X
-
6
+ 17.
+ 13. 12 X + 25. X
4 x2 25 X 2 - 30 X x
6 -
1.
+ 29.
100
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. VII
In Exercises 73 to 78, solve the equations. 73. 8 x3 - 27 = 0. 75. x 3 = 8. 77. x 4 - 81 = 0.
74. x 3 + 27 = 0. 76. x 3 = -64. 78. 16 x 4 = 81.
In Exercises 79 to 83, find three cube roots of the given numbers. 80. 1.
79. 64.
81. -64.
82. -1.
83. -27.
In Exercises 84 to 91, determine by inspection (that is, without factoring or solving the corresponding equations) which expressions have real linear factors with rational coefficients, and which are perfect squares. 84. 21 X 2 - 2 X - 55. 86. 25 X 2 + 130 X + 169. 88. 12 x 2 + 25 x + IO. 90. 9 x 2 + 42 X + 49.
85. 6 X 2 - X - 13. 87. 20 X 2 57 X 27. 2 89. 9 X 40 X 49.
+ +
91. 18 X 2
+
+ +
30 X
+
13.
In each of the Exercises 92 to 100, determine k so that the stated condition is satisfied, and solve the resulting equation. 4 x - 35 = 0, has roots whose sum equals 1. 2 x + k(x - 6) = 0, has roots whose product equals (-½). x2 - 3 x + k = 0, has roots whose difference equals 7. 4 x2 + kx - 15 = 0, has roots whose difference equals 4. 3 x2 + 14 x + k = 0, has one root equal to -3. 6 x 2 + 7 x + k = 0, has one root equal to J. 9 x 2 - 18 x + k = 2, has one root double the other. 3 x2 - 20 x + 3 k + 1 = 0, has one root three times the other. 2 x2 - 7 x + k = O, has one root which is 1 less than twice the other.
92. kx 2 93.
94.. 95. 96. 97. 98.
99. 100.
2
-
CHAPTER VIII
Inequalities 69. Inequalities. In §§ 5 and 6 we have seen that the symbol a < b is read "a is less than b," and the symbol b > a is read "b is greater than a." If a and b are two real numbers and if the point I
-6
I
-5
I
-4
I
-3
I
-2
I
-1
I
0
I
1
I
2
I
3
1 4
I
5
f
6
FIG. 32
that corresponds to a on the scale of real numbers lies to the left of the point that corresponds to b, then a < band b > a(§ 6). We haiVe also seen, as a property of the operations of addition and sub" traction (§ 9), that a < b if and only if (b - a) is a positive number. It is clear that, if a and b are any two real numbers, just one of the three relations a < b or a = b or a > b is true. The symbols < and > are called "signs of inequality" and a statement that connects two quantities by one of these symbols is an inequality. The two quantities are the members of the inequality. Smee inequalities are defined only for real numbers, it will be assumed that all numbers considered in this chapter are real. The statement that "a is less than or equal to b" is written a < b, and is true if (b - a) is positive or zero. Two inequalities have the same sense, if their signs of inequality point in the same direction. It should be noted that the statement "a is numerically less than b" or "b is numerically greater than a" means that lal < lbl. This is not equivalent to saying that a< b. Thus 121 < l-71, since 2 < 7; but 2 > -7, since 2 - (-7) = 9 is positive. EXAMPLES. 8 > 5; 3 > -4; -6 < I; -10 < -4; I-IOI> l-4j. The inequalities -6 < 3 and 4 < 11 have the same sense; the inequalities -9 < -3 and 8 > 2 have opposite senses; -3 is numerically less than -8; -5 is numerically greater than 1. 101
102
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. VIII
70. Properties of inequalities. In order to prove certain properties of inequalities, we shall use the assumptions (§§ 9, 10) that the sum of two positive numbers is positjve, the product of two positive numbers is positive, and the product of a positive number by a negative number is negative.
I. The sense of an inequality is not changed if the same number is added to (or subtracted from) both members;
if a> b, then
a+ e
> b + e and
a- e
>
b - e.
Proof. Assume that a > b, and that c is any real number. Form the difference (a+ c) - (b c) = (a - b), which is positive since a > b. But if (a + c) - (b c) is positive,
+ +
a+ c
> b + c,
as we wished to show. In particular, any number can be transposed from one side of an inequality to the other if its sign is changed; and any term common to both members can be canceled.
II. The sense of an inequality is not changed if both members are multiplied (or divided) by the same positive number:
if a> band e
> O, then
ae
>
be.
Proof. Assume that a > b and c > 0. Form the difference ac - be = c(a - b). This is positive; for c is positive and (a - b) is positive, and the product of two positive numbers is positive. Consequently ac > be, as we wished to show.
III. The sense of an inequality is reversed if both members are multiplied (or divided) by the same negative number:
if a> band e
< O, then
ae
b and c < 0. Form the difference ac - be = c(a - b). This is negative, since the product of a negative number by a positive number is negative. But if ac - be is negative, ac < be, as we wished to show. If a > b and b > c, then a > c. Proof. Form the difference a - c = (a - b) (b - c). Since a > b and b > c, the differences (a - b) and (b - c) are positive. But the sum of two positive numbers is positive, so that (a - c) is positive, and a > c, as we wished to show.
IV.
+
INEQUALITIES
§§ 70-71]
103
71. Absolute and conditional inequalities. An inequality which is true for all values of any literal numbers involved is said to be an absolute inequality. An inequality which is true only for certain sets of values of the letters involved is a conditional inequality. This classification corresponds to the classification of equalities as identities or as conditional equations. 1. 8 > - 2 and x 2 + 4 > 2 are absolute inequalities, while 3 is a conditional inequality which is true only when x > 8.
EXAMPLE
x- 5>
In a problem involving an absolute inequality it is often necessary merely to prove that the inequality is true. When a conditional inequality is involved, however, it is often necessary to determine the ranges of values of a variable for which the inequality is true. EXAMPLE 2. Show that if two positive numbers are unequal, the positive square root of the larger number is greater than the positive square root of the smaller number.
Proof. This statement is equivalent to saying that if a and a2 > b2, then a > b. Since a2 > b2, (1)
a2
-
b2
>
0
or
(a+ b)(a - b)
>0
and b > 0
> 0.
Since a> 0 and b > 0, we know that (a+ b) > 0. By Property II we can therefore divide both members of the inequality (1) by (a+ b), and obtain (a - b) > 0. Therefore a > b, as we wished to show. To prove an absolute inequality, it is often helpful to explore the work by starting with the inequality that we wish to prove and going by steps to an inequality that we know to be true. The logical proof then consists in retracing the steps in reverse order, starting with known relations and proceeding to the statement that we are trying to prove. Thus, by way of exploration, we might write first a > b (to be proved), and then obtain successively a - b > 0, (a+ b)(a - b) > 0 and a2 - b2 > 0. We then give a proof in the order of the inequalities a2 - b2 > 0, (a - b)(a + b) > 0, (a - b) > 0.
EXERCISES 1. If a > 0 and b > 0 and a > b, prove that a2 > b2• 2. If a < 0 and b < 0 and a > b, prove that a2 < b2• 3. Prove that if a > 1, a2 > a, but that if O < a < 1, a 2 4. Prove that if a
> 0,
a
+ !a ~
< a.
2.
5. Prove that if a, b, and c are not all equal, a2
+b +c 2
2
> ab + be + ca.
104
FIRST YEAR OF COLLEGE MATHEMATICS
6. If O < a < b, prove that b3 - a3 not hold equally well if O < b < a? 7. Prove that if a
> b > 0,
then a
[Ch. VIII
> (b - a) 3• Why does your proof
+b>
4
abb.
a+
8. Give a numerical example illustrating each of the Exercises 1 to 6.
72. Solution of conditional inequalities. If a conditional inequality involves one or more literal numbers, it is said to be satisfied by any set of values of the literal numbers for which the inequality is true. The solution of a conditional inequality in one variable consists in finding all values of the variable that satisfy the inequality. Unlike an equation, an inequality will ordinarily be satisfied by a certain range or ranges of values of the variable rather than by one or more special isolated values. A linear inequality is one in which one member is a linear function of the variable and the other member is either linear or a constant. To solve a linear inequality, transpose the first-degree terms to one side and the constant terms to the other side; combine terms; divide by the coefficient of the variable, reversing the sense of the inequality if the coefficient is negative; the resulting inequality defines the range of values which satisfy the given inequality. EXAMPLE
Solution.
1.
Solve the inequality 5 x
+4 > 8x -
Transpose, getting 5 x - 8 x
Combine terms:
- 3X
11.
>
-11 - 4.
>
-15.
Divide by -3, and reverse the sense, since -3 The given inequality is satisfied if x < 5.
< 0:
x
< 5.
In general, a conditional inequality in x can be written in the form (1)
f(x)
> 0.
If the functionf(x) is a polynomial or any other continuous function, f(x) will change sign only at the points wheref(x) = 0. To solve the inequality f (x) > 0, we therefore usually first find the roots of the equation f(x) = 0, and then determine the sign of the function f(x) on each side of each root. The graph of y = f(x) is of great service in this determination.
§§ 71-72]
INEQUALITIES 2.
EXAMPLE
5 x2
Solve the inequality
+x -
Solution.
105
8
> 4 x2 + 3 x.
Transposing, we have
x2
2X
-
>
8
-
0.
Solving the equation X2 -
(2)
or
2
= 0 + 2) = 0, 8
X -
(x - 4) (x
we find the roots x = 4, x = the graph of the function
=
y
x2
2
-
-
2. From FIG. 33
8,
X -
we see that the graph lies above the x-axis and the function is positive when x > 4 and when x < -2. The given inequality is therefore satisfied wheP x > 4 and when x < - 2.
If we take the
While the graph is useful, it is not essential in the solution. function in factored form, (3)
y
we see that when x
-2
4,
=
(x - 4)(x
+ 2), > 0.
each factor is positive, and y
< 4,
(x - 4)
0,
so that the product of the factors is negative, and y < 0. When x < -2, each of the two factors is negative, and y > 0. Therefore the given inequality is satisfied when x > 4 and when x < -2.
EXERCISES In Exercises 1 to 12, solve the inequalities for x. 10.
2. 3 X
< 5.
< 0. x +3a > 4x -
5. 5 X
-
1. 2x
>
4. 12 - 4x
7. 10. x 2
-
64
9 a.
< 0.
8. 3 X 11. x 2 -
> 0. 6. 5 - 4 X > 2 X - 7. 9. X 2 < 49. 12. 2 x 2 - 50 > 0. 3. 6 X
3 < 2 X + 6. 2 C < 5 X + 8 C. 81 > 0.
-
8
If x is plotted on the scale of real numbers, describe its position in each of the Exercises 13 to 22 by means of inequalities.
13. xis to the right of 6.
14. x is to the right of - 3.
15. x is to the left of -8.
16. x is to the left of 14.
FIRST YEAR OF COLLEGE MATHEMATICS
106
[Ch. VIII
17. x is between 3 and 5. 19. x is between 6 and - 2.
18. xis between -4 and 1.
21. x is between 5 and - 5.
22. x is between - 2 and 2.
20. xis between 8 and -3.
In each of the Exercises 23 to 34, describe the position of x on the scale of real numbers. 23.
< 7. lxl < 6.
24.
X
26. 29. 2 X 32. x
2
7 9.
-
0.
> -3. lxl > 3.
25. -2
X
30. 5 X 33. x
2
+4
4; (b) when 1 < x < 4; (c) whenx < 1. (x - 5)(x + 2), (a) when x > 5; (b) when -2 < x < 5; (c) whenx < -2. (x - 3) (x + 4), (a) when x > 3; (b) when -4 < x < 3; (c) when x < -4. (2 x + 3)(x + 6), (a) when x > -}; (b) when -6 < x < -¾; (c) when X < -6. (x - 6)(x - 4)(x - 1), for (a) x > 6; (b) 4 < x < 6; (c) 1 < x < 4; (d) X < l. (x - 3)(x + l)(x + 5), for (a) x > 3; (b) -1 < x < 3; (c) -5 < x < -I; (d) x < -5. (x - 4)x(x + 7)2, for (a) x > 4; (b) 0 < x < 4; (c) -7 < x < 0~ (d) x < -7. (2 x + l)(x + 3) 2 (3 x + 16), for (a) x > -½; (b) -3 < x < -½;
36. (x
37. 38.
+ 3),
> >
(a) when x
-¥ < x
0. 64. X 2 - 3 X - 28 < 0. 62. x 2 + 11 x > 0. 63. x 2 - 5 x + 6 > 0. 65. 2 x2 + x - 15 > 0. 66. 3 x2 - 2 x - 8 < 0. 67. 6 X 2 - 13 X - 28 > 0. 68. 3 x2 - 5 x - 17 > 2 x2 - 4 x + 13. 69. 5 x2 + 3 x - 2 < 3 x2 - 2 x + 10. 71. 6 x2 - 8 < 2 x2 - 3 x + 2. 70. 7 x2 + 5 x - 3 < x2 - 6 x + 7. 73. (x - 2) (x + 4) 2 (x - 5) > 0. 72. (x + 3)(x - 9)(x + 1) < 0. 74. (x - 1) 2 (x + 3) (2 x - 7) > 0. 75. x 4 - 7 x 3 + 6 x2 > 0. In each of the Exercises 76 to 81, for what values of xis the expression real?
76.
vx
2 -
79. v2 x 2
9. -
3.
77.
Vx2
-
36.
80.
vx
-
4x
2
78. v'I6 - x 2 •
+ 3.
81.
v x + 9 x + 18. 2
Exercises 82 to 84 require the following information: In sporting parlance, the "percentage" of a team is equal to w/(l + w), where w is the number of games it has won and l is the number of games it has lost. If one team has won W1 games and lost Z1 games, and a second team has won W2 and lost l2, the first team is ½[(w2 - w1) - (Z2 - Z1)] "games behind" the second team if (w2 - w1) > (l2 - Z1). For each exercise there are two possible answers. Find both sets of answers. 82. On a certain d2iy in the baseball season, the Chicago Cubs had lost 38 games, and the Pittsburgh Pirates had played 59 games. The Cubs were ahead of the Pirates in percentage, but were one-half game behind them. How many games had each team won, how many had it lost, and what were their percentages? 83. At one time in the basketball season, the Royals had lost 13 games, and the Lakers had played 51 games. The Royals were two games behind the Lakers, but were ahead of them in percentage. How many games had each team won and lost, and what were theit percentages? 84. On a certain day in the baseball season, the Reds had lost 40 games, and the Blues had played 108 games. The Reds were one game behind the Blues, but were ahead of them in percentage. How many games had each team won, how many had it lost, and what were their percentages?
CHAPTER IX
The Locus of an Equation 73. Introduction. Having studied quadratic equations in one unknown, we shall naturally wish to consider quadratic equations in two unknowns. In studying them we shall need to know something about the loci of such equations. Before going further with the study of quadratic equations we shall therefore pause in this chapter to consider methods which help us in picturing the locus of a given equation of any degree. We shall then apply the methods especially to the loci of quadratic equations and, in the next chapter, proceed to an algebraic study of such equations. 74. Graphing by points. The most natural way in which to begin tracing the locus of a given equation is to solve the equation for one of the variables, x or y, as a function of the other. Then, by substitution, any number of corresponding pairs of values of x and y can be obtained, and the corresponding points plotted. These points will all lie on the desired locus, since their coordinates satisfy the equation. EXAMPLE
1.
Draw the locus of the equation
(1)
y3
Solution.
-
8 x 2 = 0.
Solving the equation for yin terms of x, we have
(2)
= 2vx2 • = 2V4 = 3.17. By substitution we find other y
If we set x = 2, we find y
pairs of corresponding values as shown in the following table . X
-- y
.1
0
2
-
1 -2 1 -1 2 - - - -
0 1.26 1.26 2
-6 -3 4 -4 5 -5 6 2 3.17 3.17 4.16 4.16 5.04 5.04 5.85 5.85 6.60 6.60
-2 3
-
-
-
We can therefore obtain the locus by plotting the points whose coordinates appear in the table, and then connecting them by a smooth curve. The locus is shown in Figure 34. 108
[§§ 73-76]
THE LOCUS O~ AN EQUATION
109
y 12
0
2
4
6
s
10
12
X
FIG. 34
In general, the locus of any equation with which we shall deal will consist of one or more lines or curves. But the locus may contain or consist of one or more isolated points. In some equations, it is easier to solve for x than for y or to substitute values of y rather than values of x.
75. Intercepts. Although the methods of the preceding article must usually be used to obtain at least a few points on a locus, we can frequently secure a better general idea of the figure and shorten the work of graphing it, by a preliminary examination of the equation according to the methods of this and the following sections. Such an examination is called a discussion of the equation. The intercepts of a curve are the distances from the origin to the points where the curve meets the coordinate axes. To find the x-intercepts, set y = 0 in the equation, and solve for x; to find the y-intercepts, set x = 0, and solve for y. Finding the intercepts often gives a few points on the curve with very little labor. 76. Degenerate loci.
Theorem. If an equation can be written as the product of two or more factors equated to zero, its locus consists of the loci of the equations formed by setting the individual factors equal to zero. That is, if the equation can be written in the form u · v = O, where u and v are expressions in x and y,* the locus consists of the
two loci whose equations are u = 0 and v = 0. Proof. Any point on the locus of u = 0 also lies on the locus of u · v = 0. For, since its coordinates, on substitution, make u = 0, they also make u · v = 0. Any point on the locus of v = 0 likewise lies on the locus of u · v = 0. Moreover, any point which lies on the locus of u · v = 0 lies either on the locus of u = 0 or on the locus .of v = 0, or on both loci. For if u · v = 0 either u = 0 or v = 0. * When we are dealing with the variables x and y an expression that contains x alone or y alone or both x and y may be considered as "an expression in x and y."
110
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. IX
The locus of such an equation as u • v = 0 is said to be a degenerate locus for it degenerates into two or Y more simpler loci. A degenerate locus 3 should always be plotted by plotting its component loci, whose forms can often be recognized at a glance. -+--+--+---illE---+---+---1--x
EXAMPLE.
Plot the locus of the equation
x4 - y4 - 9 x2
The equation can be written as
Solution.
(x - y)(x
Fm. 35
+ 9 y2 = 0.
+ y)(x + y 2
2
-
9) = 0.
The locus consists of the loci of the equations X - y = 0, X + y = 0, X2 + y 2 = 9.
The locus therefore consists of the two straight lines and the circle shown in Figure 35.
EXERCISES Draw the loci of the following equations after finding the intercepts. 1. y = 2 x2• 2. 2 x - 3 y2 = 0. 3. 2 y = x3• 4. y = x 3 2 X. 5. (y + 1)2 = X - 3. 6. X = y3 - y 2• 7. 4 x 2 25 y2 = 100. 8. 25 x 2 - 4 y 2 = 100. 9. x 2 = y 3 - 2 y2• 10. y = x2 - 6 x + 5. 11. y2 = x 2 - 6 x + 5. 12. y = x3 x2 x.
+
+
+ +
Draw the loci of the following equations.
13. 15. 17. 19.
(X - 2 y + 3) (5 X + y - 7) = 0. 2 xy - x3 = 0. 3 X + 3 y = x 2 - y 2• x2 - 3 x - 4 = 12 y - 3 xy.
14. 4 x 2 - 9 y2 = 0. 16. x 3 + y 6 = 0. 18. x5 _ x2y2 = x3y2 _ y4.
20. x4
+xy
2 2
-
8 x 2 = 4 y2
-
16.
77. Symmetrical points and figures. Two points P1 and P2 are said to be symmetrical with respect to a straight line l as an axis of symmetry if l is the perpendicular bisector of the line P 1P 2• Each point is said to be the l symmetrical point of the other with respect to l. If the plane were folded along the axis Pg u----.,...----of symmetry, each point would coincide with its symmetrical point. Two points, P1 and P 3, are symmetrical with respect to a point FIG. 36 Q as a center of symmetry, if Q is the mid-
§§ 76-77]
THE LOCUS OF AN EQUATION
111
point of PiP3• Each point is called the symmetrical point of the other with respect to Q. Definition. A figure is said to be symmetrical (with respect to a certain line as an axis or to a certain point as a center of symmetry) 1j the symmeirical point (with respect to that axis or center) of every
FIG. 37
point of the figure also lies on the figure. In a symmetrical figure no point appears unless its symmetrical point also belongs to the figure. The two types of symmety rical figure are indicated in Figure 37. The following statements are P3(-~1_,Y_1_)_---1-_ ___,.P1(x1,Y1) easily verified by referring to Figure 38: If Pi:(xi, Yi) is any point, (1) its symmetrical point with re- P4 (-~11 -y 1) spect to the x-axis is (xi, -yi); (2) its symmetrical point with reFw. 38 spect to the y-axis is ( - Xi, Yi) ; (3) 1:ts symmetrical point with respect to the origin is ( - Xi, Theorem I. The locus of a given equation is symmetrical with respect to the y-axis, if, when x is replaced by - x in the equation, an equivalent equation is obtained. Proof. Suppose that (xi, Yi) is any point whose coordinates satisfy the given equation. The equation obtained by replacing x by -x will therefore be satisfied by ( -xi, Yi). But this second equation is assumed to be equivalent to the given equation. Therefore, if the given equation is satisfied by the values (xi, Yi), it is also satisfied by the values ( -xi, Yi). Hence, if (xi, Yi) is any point on the locus,
112
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. IX
its symmetrical point with respect to the y-axis also lies on the locus, and the locus is symmetrical with respect to the y-axis.
Corollary. The locus of any equation is symmetrical with respect to the y-axis if the equation involves only even powers of x, such as x2, x 4 , etc. The following theorems are to be proved by the student.
Theorem II. The locus of an equation is symmetrical with respect to the x-axis, if an equivalent equation is obtained when y is replaced by -y. Theorem III. The locus of an equation is symmetrical with respect to the origin, if an equivalent equation is obtained when x and y are replaced by -x and -y, respectively. EXAMPLES. The locus of y 3 - 8 x2 = 0 is symmetrical with respect to the y-axis, since y 3 - 8(-x) 2 = 0 is equivalent to y 3 - 8 x2 = 0. (See Figure 34, page 109.) The locus of 2 y 2 = 3 x is symmetrical to the x-axis, since 2(-y) 2 = 3 x is equivalent to 2 y 2 = 3 x. The locus of xy = 4 is symmetrical to the origin, since (-x)(-y) = 4 and xy = 4 are equivalent.
If we know that a locus is symmetrical with respect to a given line or point, after obtaining one-half of the figure, we can easily obta~n the rest of it by considering its symmetry. EXERCISES Examine each of the follmving equations for the symmetry of its locus with respect to the coordinate axes and the origin. 1. x2
+y
2
= 36.
= y2 + 2 y. 5. x4 + y4 = 1. 7. x 2 - y2 = xy. 9. x3 + y3 = 1. 3. x 2
-
4
78. The extent of a locus.
2. 9 x2 4. 6. 8. 10.
16 y2 = 144.
-
+ 4 y = 0. 2x 5 xy + y = 10. x 2 xy + y + 3 x x y + xy = 1. X3 -
2
2
3
X
2
-
2
-
3 y = 4.
3
In many examples an examination of an equation reveals limits to the extent of the corresponding locus, and so limits the range or field of investigation. In order to make this examination, we usually solve' the equation for each variable in terms of the other, and notice any values of either variable that make the other imaginary.
THE LOCUS OF AN EQUATION
§§ 77-78] EXAMPLE 1. equation (1)
9 x2
y
y
Consider the
3
+ 25 y
2
= 225.
2
Solving the equation for y in terms of x and for x in terms of y, we get (2)
1 -5
-4 -3 -2 -1
X
0 -1
= ± Jv25 - x 2 ,
-2
and (3)
113
-3 X
= ±¾V9 - y2 •
FIG. 39
By (2) when x = 5, y = 0, and y is real only when lxl < 5. By (3), xis real only when IYI ~ 3. The entire curve is therefore confined to the range of values of x from x = - 5 to x = 5 and to the range of values of y from y = -3 toy = 3. The locus is the curve of Figure 39. EXAMPLE 2. equation
y
(1)
10
6
(2)
4
and
2
(3)
0
-
9 y 2 = 144.
Solving for each variable terms of the other, we have
8
-2
16 x2
Consider the
2
-2
-4
-6 -8 -10
FIG. 40
X
y =
X
Ill
± ±3 v' x2- - 9'
= ±
¾v' y + 16. 2
From (2) we see that y is imaginary if !xi < 3, so that it is unnecessary to substitute any value of x between - 3 and 3. This accounts for the gap in the curve, shown in Figure 40. On the other hand, since (y2 + 16) is positive for any real value of y, from (3) we know that the curve extends indefinitely far upward and downward without any break.
In some cases the locus of an equation may consist of only a single point, or, if the equation is satisfied by no real values of x and y, no locus can be plotted and it is said to be an imaginary locus.
114
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. IX
3. If we complete the squares of the terms in x and of the terms in y, we can write the equation x2 + 4 y2 - 2 x + 24 y + 37 = 0 in the form EXAMPLE
(x - 1) 2
+ 4(y + 3)
2
= 0.
For real values of x and y, each of the terms (x - 1)2 and 4(y + 3) 2 is either positive or zero. Therefore the equation is satisfied only if x - 1 = 0 and y + 3 = O; that is, it is satisfied only by the coordinates of the point (1, -3), which constitutes the entire locus. EXAMPLE
4.
The equation x 2
ten as
(x - 1) 2
+4y
2
-
2x
+ 24 y + 40 =
+ 4(y + 3) 2 =
0 can be writ-
-3.
For any real values of x and y, the left-hand side is positive or at least zero (for x = 1 and y = -3), and can therefore never be equal to -3. The locus is imaginary.
79. Horizontal and vertical asymptotes. In discussing an equation it is usually desirable to inquire for what values of one variable, if any, the other variable becomes infinite. (1) xy + 2 X - y - 6 = 0. EXAMPLE 1. Consider the equation Solving for y, we get
(2)
y=
6 - 2x . X 1
From (2) we see that y has a definite value for every value of x except x = I. When x = 1, the dey nominator of the righthand member becomes 7 zero so that the frac6 5 tion is not defined. If 4 xis taken very slightly 8 greater than 1, the nu2 merator is very nearly 1 _,--.._,._..,._..,._....,_+--+-t--+--+-~-+--+--t-+--+-~-+--+--X equal to 4, and the de-s-7-6-5 -4-3 -2-1 O 9 10 11 -1 nominator is positive but numerically very y=-2 small. The value of y is therefore a very large positive number. As x x=l approaches 1 from the right, y becomes posi-9 -10 tively infinite. As -11 the curve draws near the line x = 1 from the FIG. 41 right, it rises indefinitely far as shown in
115
THE LOCUS OF AN EQUATION
§§ 78-79]
Figure 41. When xis very slightly less than 1, the denominator is negative, and as x approaches 1 from the left, y becomes negatively infinite. This evidently corresponds to one of the characteristic properties of the curve. Solving the equation for x, we find (3)
X
y+6 y+2
= --,
so that, as y tends toward - 2 from above, x becomes positively infinite. And when y tends toward - 2 from below, x becomes negatively infinite. These facts are also indicated in Figure 41.
In the preceding example, the straight line x = 1 is a vertical asymptote of the locus, and y = - 2 is a horizontal asymptote. These asymptotes serve as guiding lines of the curve. To examine a curve for its horizontal and vertical asymptotes, solve the equation for each variable in turn and simplify the results. If y is expressed as a fraction in lowest terms and if the denominator is equal to zero when x = a, in general y becomes infinite when x approaches a, and the line x = a is a vertical asymptote. A similar procedure reveals the horizontal asymptotes. EXERCISES Examine the following equations for the existence and extent of their loci. 1. x2 3. x 2
+y -
5. y = 9 x2 • 7. 4 y2 = 3 X
9. 11. 13. 15. 17.
(x - 3)
36(x (x -
+
= 16. 2 y = 16.
2. 9 x2 16 y2 = 144. 4. 9x'l - 16y2 = 144.
2
2
6. 25 y 2 6. ·
-
+ 9(y + 2)
2
=
9.
+ 2) 25 y = 900. 2) + 4(y + 3) = 0. 2
-
2
2
2
+ 4 y + 36 X - 16 y + 54 = 0. (3 y - 2) + 4(x + 3) + 10 = 0.
9 X2
2
2
2
9 x2
-
=
225.
+ 8. + + 4(y -
8. 9 x 2 = 4 y 10. 25(x 1) 2
3) 2 = 100.
+ 4). 3(3 X - 5) + 7(y + 4) = 0. y = (x - 4)(x - l)(x + 3). y = x (x + 3)(x - 3).
12. 16(x - 3) 2 = 25(y 14. 16. 18.
2
2
2
2
2
Find the horizontal and vertical asymptotes of the loci of the following equations. 19. xy = 4.
+ 1) (y - 3) = xy - 2 X + 3 y =
21. (x
+ 2)y = 4. (3 X + 2) (2 y - 5) = 6. xy = 3 X + 5 y - 3.
20. (x 6.
23. 9. 25. 3 xy - 4 X - 4 y = 6. 27. (x2 - 9)(y + 3) = 12.
22.
24. 26. 4 xy - 12 28. (x
2
-
l)(y
X -
+
9 y = 24. 2) = 3.
FIRST YEAR OF COLLEGE MATHEMATICS
116
[Ch. IX
80. Intersections of curves. We have seen that the point of intersection of two straight lines is found by solving their equations simultaneously for the coordinates of the point. This method is general, for any point common to two loci must have coordinates that satisfy both of their equations. And any point whose coordinates satisfy both of the equations must be common to the loci. Before inquiring as to the possible number of points of intersection of a general curve and a straight line, let us consider the following example. EXAMPLE.
Let us find the points of intersection of the curve
(1)
12 x2
and the straight line
-
(2)
3 xy - 6 x - 10 y - 52 = 0 3 x - y - 7 = 0.
From (2), we have y = 3 x - 7. 12 x 2 which reduces to
-
Substituting this m (1), we obtain
3 x(3 x - 7) - 6 x - 10(3 x - 7) - 52 = 0, (3)
x2
-
5x
+ 6 = 0.
The two roots of this quadratic equation in x are the abscissas of the points of intersection of the two curves. The roots are x = 2 and x = 3. Since y = 3 x - 7, the points of intersection are (2, -1) and (3, 2). There are two points of intersection, since (3) is quadratic and has distinct real roots.
Now, more generally, consider the intersections of a straight line and a curve whose equation is of degree n in x and y. The equation of the straight line can in general be written (4)
y
= mx
+ b,
and the equation of the curve is (5) u=O
'
where u is a polynomial of degree n in x and y. On substituting (4) in (5), we obtain an equation in x alone which, in general, is of degree n. Some of the coefficients may cancel, and the equation may be of lower degree. But it is never of higher degree than n. The roots of this equation are the abscissas of the points of intersection. In § 218 we shall see that an equation of degree n has just n roots. If the roots are real and distinct, there are just n points of intersection. Therefore, no straight line can meet the curve in more than n points. These facts serve as a useful check on the work of graphing a rational integral equation of degree n. If the locus is drawn correctly, one cannot draw any straight line which meets the locus in more than n points. Usually it will be possible to draw some straight line which meets it in n points.
THE LOCUS OF AN EQUATION.
§§ 80-81]
81. Discussion of an equation. it for:
117
To discuss an equation, examine
2. Symmetry. 4. Extent. 6. Degree.
1. Degenerate loci.
3. Intercepts. 5. Horizontal and vertical asymptotes.
After such an examination, it is usually sufficient to plot a very few points in order to determine the figure. Of course an elaborate discussion of an equation is unnecessary when its locus can be recognized as a straight line, circle, or other familiar curve. 1. Discuss the following equation, and sketch its locus: EXAMPLE
y
y2 = 8 x.
(1)
Solution. (1) The locus is not degenerate. (2) The locus is symmetrical ,,·ith respect to the x-axis. (3) The only intercepts are x = 0, y = O. (4) Solving (1) for y and x, we have (2) y = ± 2v2 x,
and
(3)
X
=
1 -+-+-o-i--+-+-i1-+-+-+--+-+-+-+--+-+--+-+-+--X
¼y2 •
From (2), y is real if x ~ 0 and y is imaginary if x < O; from (3), x is real for any value Frn. 42 of y. (5) If either of the variables x and y becomes infinite, the other does also; there are no horizontal or vertical asymptotes. (6) The locus is of second order. We now form a table of values, and plot the curve. EXAMPLE
2.
Discuss the following equation, and plot its locus:
(1)
xy - 4
= 0.
Solution. (1) The locus is not degenerate. (2) The locus is symmetrical with respect to the origin. (3) There are no intercepts. (4) Solving (1) for y and x, we have (2)
4
y=x
and
X
4
= -. y
No value of either variable makes the other imaginary. The locus is therefore unlimited in extent. (5) From (2) if x tends toward zero from the right, y becomes positively infinite; if x tends toward zero from the left, y becomes
FIRST YEAR OF COLLEGE MATHEMATICS
118
y
negatively infinite; x becomes positively infinite when y tends toward zero from above, and negatively infinite when y tends toward zero from below; x = 0 and y = 0 are asymptotes. By substitution we form a table of values, draw the curve through the corresponding points, and by symmetry produce the rest of the curve shown in Figure 43. (6) The locus is intersected by some straight lines in two points but never in more than two points.
8
6 4 2
-8 -6 -4 -2
---+--+-.........it--t---+-4--1--+--..,__.,--+--+--l~-+-+--+-.....-X
2
4
6
8
-2 -4 -6 -8
FIG. 43 EXAMPLE
3.
Discuss the following equation, and sketch its locus: x 2y
(1)
Solution.
[Ch. IX
+x
2
4 y - 9 = 0.
-
(1) The locus is not degenerate.
(2) The locus is symmetrical
y X=-2
5
~=2
4
3
2 1 1 2 5 6 7 0 -1---i----i-~1---0---+---+--+----+-Qo--+---+--+----X 1
y=-1
-4 -5
-6
Frn. 44
THE LOCUS OF AN EQUATION
§ 81]
with respect to the y-axis. (3) The intercepts are x = ±3, y = (4) Solving (1) for x and y, we obtain g - x2 y=xz-4'
(2)
and
(3)
X
-¾ < y
0, (n 1) ! = (1 · 2 · 3 · · · · · n)(n + 1) = n! (n + 1), and, if n = 0, (n + 1) ! = 1! = 1 and n! (n + 1) = O! (1) = 1 • 1 = 1. E XAMPLES.
6,. =
r::::'J • v-.
6,
9 ! -- 8!, 9 155
(~ ~
;1 -- (k -
1
2)!.
156
FIRST YEAR OF COLLEGE MATHE:MATICS
[Ch. XII
EXERCISES In Exercises 1 to 12, find the values of the numbers. 1. 2 !.
2. 4 !.
3. 7 !.
4. 9!.
5. 67!!.
6 (31)!_
7 (k+3)!_ . (k + 2) !
B. (n
• (30) !
9· 12! 10(
10 6!_ . 3!
(n
6! ll. 2! 4(
12
+ 7)!_
+ 6)!
lO! · . 4! 6!
In Exercises 13 to 16, simplify the expressions. 13. (k
15
+ 1) ! (k + 2).
_ 7!(r+2)!r_ 6!(r-1)!
16
104. The Binomial Theorem. that
+ x)
14. (n - r - 1) ! (n - r).
3!4(k-3)!_ . (k- 2)!
By actual multiplication we find
= a + x, x) = a2 + 2 ax+ x 2, + x) 3 = a3 + 3 a2x + 3 ax2 + x3, + x) 4 = a4 + 4 a3x + 6 a2x 2 + 4 ax3 + x 4• For the values n = 1, 2, 3, 4, at least, the expansion of (a (a+ (a (a
contains (n rule:
1
2
(a
+ x)n
+ 1) terms, which are determined according to the following
I. The first term is an, and the second term is nan-1 x. 2. In any term after the first, the exponent of a is 1 less and the
exponent of x is 1 more than in the preceding term. 3. In any term, if the numerical coefficient is multiplied by the exponent of a and the product is divided by 1 more than the exponent of x, the result is the numerical coefficient of the next following term.
By using the foregoing rule, we obtain the following formula:
(a+ x)n =an+ nan-1x +
n(n - 1) 2!
an-2x2
a! - 2) an-sX3 + + n(n - 1)(n
•••
+ Xn •
So far we have merely verified this formula for n = 1, 2, 3, 4. It is possible, however, to prove a theorem, known as the Binomial Theorem, which we shall assume without proof for the present, but which we shall prove in § 108 with an alternative proof in § 283.
THE BINOMIAL THEOREM
§§ 103-104]
157
The Binomial Theorem. The foregoing rule and formula for the expansion of (a x)n are true when n is any positive integer.
+
Remarks. The following remarks are justified by an examination of the rule: i. In every term of the expansion, the sum of the exponents of a and of XIS n. ii. The numerical coefficients, which increase to the middle and then decrease to the end, are arranged symmetrically; for example, the coefficients of the first and last terms are equal; the coefficient of the second term equals that of the next to the last term, etc. The coefficients of the terms of the last part of the expansion can therefore be written down at once after those of the first part have been found. iii. The theorem is stated only for the case in which n is a positive integer. If n is a fraction or is negative, in following the rule we never obtain a term with the coefficient zero. Instead of (n + 1) terms we therefore have an infinite series. In § 324 it is seen that this series "converges" to the proper value if Jxl < !al. EXAMPLE
1.
+ y) 7•
By the Binomial Theorem expand (c
Solution.
(c
+ y)
7
= c7
EXAMPLE
Solution.
2.
+ 7 c y + 21 c y + 35 c y + 35 c y + 21 c y + 7 cy + y 6
4 3
5 2
3 4
2 5
By the Binomial Theorem expand (3
(3x
2 -
3
~Xy = [
2
(3 x
)
+(-
3
~)
X
2
-
6
7
•
3~xY·
J 2
= (3
2 5 x )
+ 5(3
+ 10(3 = 243 x10
-
2 4 x ) ( -
x2)2 ( -
270 x7 V x .
3~x)
3
+ 10(3
~;;)3 +
+ 120 x5 -
5(3
2 3 X ) ( -
2 x ) ( -
80 2 ~ r xv x 3
-
3~;;)
3
~J + ( -
3
~;;)5
32vx + -80 --. 27 243 x3
This example illustrates the following important remarks. (i) Whenever the terms of the given binomial are at all complicated, enclose each term in parentheses before expanding, and remove the parentheses only after the coefficients and exponents in the expansion have been obtained by the binomial rule. (ii) Since odd powers of a negative quantity are negative and its even powers are positive, the terms of the binomial expansion will alternate in sign whenever '~the two terms of the given binomial are opposite in sign.
158
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XII
EXERCISES In Exercises 1 to 28, expand the expressions by the Binomial Theorem. Check some answers by substituting particular numbers for letters. 1. (a+ x) 6• 5. (2
9. (x 13. (x 17.
21. 24.
y) 4•
X -
½y)7.
+2y
2. (x - y) 5•
3. (a - b)7.
6. (3 - 2 x) 5•
7. (2
2 ) 8•
14.
(Xy+;yy·
18.
( 2 x 21 + 3 x 31)4. 1 1+ )4 ( x4yz x-1y-2
3;-
21. ( " x -
+ j y) (ex + c x-
10. (½X
2 vx + 1)
2
e2
6
•
•
2 ) 5•
11. (½a -
6
¾b) • 4
15. (x 2
-
x 2 y) 5•
(x2 + xzvy
( 21x~
3 -
+ y)
X
.!
y
_!.)5 2
16. (x 2
- 2 z] 4•
28. ( 3
20.
•
X -
5
•
•
(
-
yX +;y)5·
26. [(2 x - 3 y) 2-
2 x-1 )7.
(3 x-i + ½x½ ) 6•
23.
•
•
6
2
19.
4
1
3
.
8. 12.
•
1
y2Y ·
26. [(x
+ 3 y) 1
;- ; 22.
X
+ y) (ax + 2 by) (j X + i)
4. (x
+ z]
3
•
\)3-
2
In Exercises 29 to 38, compute the given numbers by means of the Binomial Theorem. In problems involving decimal fractions use only enough termR to give the result to three decimal places. 29. (100 34. 49 5•
+ 1)
7
•
30. (101) 6• 36. (1.03) 15 •
31. (100 - 1) 6• 36. (1.05) 12 •
32. 99 7 • 37. (1.01) 40 •
33. 98 4 • 38. (1.02)'2°.
Exercises 39 to 42 require a knowledge of compound interest, § 99. 39. Find the amount of $1 after 15 years at 4% compounded annually.
40. Find the amount of $1 after 10 years at 5% compounded annually. 41. Find the amount of $1 after 10 years at 4% compounded semiannually. 42. Find the amount of $1 after 8 years at 4% compounded quarterly.
105. The general term of the binomial expansion. By apply1ng the rule of § 104, we can obtain a formula by means of which we can write any one of the (n + 1) terms of the expansion of (a + x)n. If r is any integer from 1 to n, for the term containing xr in the expansion of (a + x)n we find that: (1)
.
The t erm m x7 =
n(n - 1)(n - 2) · · · (n - r
, r.
+ 1) an- rxr .
Notice that in the numerator of the numerical coefficient there are r factors of which the first is n and that each succeeding factor is 1 less than the preceding. The denominator also contains r factors which increase from 1 to r.
THE BINOMIAL THEOREM
§§ 104-105]
159
If we multiply the numerator and denominator of (1) by (n - r) ! we can write the formula as
th e term
(2)
.
lll X7
n! (n - r)!r!
= - - - - an-r X1 •
The successive terms of the expansion involve x°, x1, x2, fore the rth term is the term in xr-1, and, from (1),
(3)
• • • ,
_ n(n - 1)(n - 2) · · · (n - r + 2)
therth t erm-
(r-l)!
xn.
There-
n-r+1 r-1
a
x
.
The Binomial Formula can now be written in a more complete form that displays the general term as follows:
. n(n- 1) (a+ x)n =an+ nan-1x + 2! an-2x2
(4)
+ . . . + n(n ExAMPLE
1.
1) • • • • • (n - r
' r.
+ 1) an-rx' + . . . + xn.
In the expansion of (y - 2 z2) 11, find the term contain-
ing z10 • Solution. The term containing z10 is the term in (-2 z2 ) 5• In formula (1) we therefore user = 5, n = 11, a = y, x = (-2 z2). The term in z10 is
11·10·9·8•7 - - .- -.-- .- -.--y6 (-2 z2) 5 1 2 3 4 5 EXAMPLE
Solution. term is
2.
462 y6 (-32
=
z
10
)
=
-14784
y6z10.
Write the seventh term in the expansion of (x - y) 15•
From formula (3), using n = 15, r = 7, we find that the seventh 15 • 14 • 13 • 12 · 11 · 10 x9 (-y) 6 = 5005 x9y6 • 1· 2 ·3 ·4 •5 · 6
EXAMPLE
(
1)12, find the term in 1/x
3. In the expansion of x - ;
2•
Solution. We notice that the first term is x12, that the second term involves x11 (1/x) = x10, and that the degree of x in each succeeding term is 2 less than in the preceding term. Hence the term in 1/x2 or x-2 is the eighth term of the expansion. By formula (3), the eighth term is
( 12 • 11 • 10 · 9 · 8 · 7 • 6 -1-.-2-.-3-.-4-.-5-.-6---7- (x )5
1) 7 -
;
792
= - -x-2 ·
160
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XII
EXERCISES In each of the Exercises 1 to 12, write the specified term in the expansion of the given power, without obtaining the preceding terms.
1. (x
+ y)
10
;
term in y 6 •
3. (a - 3 b)9; term in a4•
1 2 x)\ term in 1/x.
5. ( x -
+ 2 y)
7. (x
9. (2 y2
11. ( x
-
12
;
fifth term.
z) 15 ; twelfth term.
+ ;)\ "ixt,h term.
2. (2 - x) 12 ; term in xi. 4. (2 x
+ ½y
6. ( 3 x -
6
2 8 ) ;
term in
f3.
~ 2 ) \ term in
1/x3•
8. (x - 2) 10 ; sixth term. 10. ( 2 x 12.
(x -
1 x)\ fifth term. 2 1 )14 -v2x ; tenth term.
13. In computing (1.04) 12 by the Binomial Formula, what error is made if the term in (.04) 5 is omitted? (The errors of later terms are much smaller, so -that the first five terms give a good approximation.) 14. In computing (1.02) 16 by the Binomial Formula, what error is mad0
ii the term in
(.02) 5 is omitted?
*106. Binomial coefficients. The coefficients in the expansion of (a+ x)n are called the binomial coefficients. They are also important in the study of Permutations and Combinations, Probability, Statistics, Genetics, and in other connections. The binomial coefficients are sometimes denoted by the symbols 1, nC1, The symbol nCr represents the coefficient of the term involving xr in the expansion of (a + x)n; that is,
nC2, nC3, · · · , nCn.
(1)
_ n(n - 1)(n - 2) • • • (n - r + 1) _ n! nCr - - - - - - - --• r! (n-r)!r!
Using these symbols, we can write the binomial formula as
(2)
(a+ x)n = an
+ nC1an- 1x + nC2an-2x2 + •· · + nC,an-rxr + ... + xn.
If we write the binomial coefficients in rows, those for n = 0 in the first row, those for n = 1 in the second row, etc., we obtain the following arrangement, which is called Pascal's triangle, after the French philosopher and mathematician Blaise Pascal (1623-1662):
§§ 105-106]
161
THE BINOMIAL THEOREM
1
n=O n = 1
1
1 1
n=2 n=3 n=4 n=5 n = 6 etc.
2
1 1 1 6
6
4
5
1
1
1
3
3
10
5
10
15
1
4
1
15
20
6
1
etc.
In Pascal's triangle the first and last numbers in each row are 1. Any other number may be obtained as the sum of the two numbers just to the right and to the left of it in the preceding row. In order to understand why the binomial coefficients follow the plan of Pascal's triangle, let us obtain (a+ x) 5 by multiplying the expansion of (a+ x) 4 by (a + x). In the third line below we shall write the product of the terms by a and in the next line the product of the terms by x, setting them one place to the right in order to bring like terms together.
(a+ x) 4 = a4
+4ax +
(a+ x) = a a5
+x
(a+ x) 5 = a5
3
6 a2x 2
+
4 ax 3
+x
4
+ 4 a x + 6 a x + 4 a x + ax a x + 4 a x + 6 a x + 4 ax + x + 5 a x + 10 a x + 10 a x + 5 ax + x 4
3 2
2 3
4
3 2
2 3
4
5
4
3 2
2 3
4
5
4
Except for the first and last terms, the rth term in the expansion of (a+ x) 5 is the sum of two terms, one of which has the same coefficient as the rth term of the expansion of (a + x )4 and the other of which has the same coefficient as the preceding term of that expansion. Therefore the binomial coefficients for n = 5 are found by adding two successive binomial coefficients for n = 4, which is the law for building Pascal's triangle. Since the binomial coefficients for n = 4 are the numbers in the fifth row of Pascal's triangle, those for n = 5 form the sixth row of the triangle. In a similar way, we see that since the coefficients for n = 5 form the sixth row of the triangle, the coefficients for n = 6 form the seventh row, and so on to any positive integral value of n.
EXERCISES 1. Write the eighth row of Pascal's triangle, and thence obtain the bi-
nomial expansion of (a
+ x)
7•
2. Obtain (a+ x) 8 by actually multiplying the expansion for (a+ x) 7 by (a
+ x).
Compare the coefficients with the ninth row of Pascal's triangle.
3. ·write the values of 5Ca, 904, sCc, 1JC6, 10C4., 1C2.
CHAPTER XIII
Mathetnatical Induction 107. The method of mathematical induction. Mathematical induction is the name given to a certain general method of proof. This method may often be used to prove that a theorem which involves a certain variable, say n, is true for all positive integral values of that variable. In § 108, for example, it will be proved that the Binomial Theorem, § 104, for the expansion of (a + x)n is true for all positive integral values of n. And in Example 1, below, by the method of mathematical induction we shall prove the theorem that the sum of the first n positive odd integers is equal to n 2• In general, the proof of a theorem by mathematical induction involves three parts, as follows:
I. Verification.
This consists of a direct verification of the theorem for one or more particular values of the integral variable n. Usually we show that the theorem is true when n = 1. II. Extension Theorem. This consists in proving that whenever the theorem is true for some particular value of n, it is necessarily true for the next greater value of n. This portion of the proof is usually the most difficult to carry out. III. Conclusion. This consists in combining I and II to prove the general theorem by using the following Fundamental Principle. Fundamental Principle of Mathematical Induction. If, for a theorem concerning the positive integer n, it is known that (i) the theorem holds true for ~ = 1 ; and that (ii) whenever the theorem holds true for some particular value of n, it necessarily holds true for the next greater value of n; then the theorem holds true for all positive integral values of n. Proof of the Fundamental Principle. the Fundamental Prinmple. 162
We shall give two proofs of
[§ 107]
163
MATHEMATICAL INDUCTION
A. Direct proof. Since, by (i), the theorem is known to hold true for n = 1, then, by (ii), the theorem is true for the next greater value of n, (n = 2). By (ii), the theorem is therefore true for the next greater value of n, (n = 3). This process can be continued to show that the theorem holds true for any specified positive integral value of n. (We assume as an axiom about integers that any positive integer can be reached by this process.) It therefore holds true for all positive integral values of n. B. Proof by contradiction. Assume again that (i) and (ii) are known, but suppose that the theorem does not hold true for all positive integral values of n. Let (m + 1) be the smallest positive integral value of n for which the theorem does not hold true; that is, the theorem holds true for n = 1, 2, 3, · · · , m, provided that mis a positive integer, but does not hold true for n = m + 1. Now, by (i), the theorem holds true for n = 1, and therefore m > 1, and m is a positive integer. But, by (ii), since the theorem holds true for n = m, it must also hold true for n = m + 1. This contradicts the assumption that there is some positive integral value of n for which the theorem is not true, and the theorem must hold true for all positive integral values of n. The method of mathematical induction is illustrated in the following example. EXAMPLE
(1)
1.
Prove that 1+3
+ 5 + ••• + (2 n -
1) = n2 ,
for all positive integral values of n. (Since the nth positive odd integer is (2 n - 1), this is equivalent to proving that the sum of the first n positive odd integers is equal to n 2 for all positive integral values of n.)
Proof. I. Verification. The theorem is true for n = 1; for when n = l, the left-hand member of (1) contains only one term, and formula (1) becomes (2)
1
=
l2,
which is certainly true. II. Extension Theorem. We now wish to prove that whenever formula (1) holds for some value of n, say n = k, it holds for the next greater value, n = k + 1. We shall therefore assume that (3)
1+3
+ 5 + ··· + (2 k -
and prove that if (3) is true, then, taking n = (k
(4)
1+3+5
+ ... + [2(k + 1) -
1)
= k 2,
+ 1) in (1), 1] = (k + 1)
2
164
FIRST YEAR OF COLLEGE MATHEMATICS
is also true.
[Ch. XIII
In the left-hand side of (4), the last term can be written as [2 k
+2 -
1]
or
(2 k
+ l)
and the next to the last term, which is the last of the terms represented by t,he three dots, is (2 k - 1). The left-hand member of (4) can therefore be written as (5)
1 + 3 + · · · + [2(k + 1) - 1] = [1 + 3 + · · · + (2 k - 1)] + (2 k
+ 1).
If we assume that (3) is true, the quantity in brackets on the right-hand ddc· of (5) is equal to k 2 • Therefore, by (3) and (5), (6)
1
+ 3 + 5 + ... + [2(k + 1) -
as we wished to show.
1] = k2
+ (2 k + 1) =
(k
+ 1)
2 ,
Therefor€ (4) is true, if (3) is true.
III. Conclusion. In I we have shown that the theorem holds true for n = 1; in II we have shown that whenever the theorem holds true for any particular value, n = k, it holds for the next greater value, n = k + 1. Therefore, by the Fundamental Principle, the theorem holds true for all positive integral values of n. EXAMPLE
(7)
2.
By the method of mathematical induction prove that
1-2+2·3+3•4+ •·· +n(n+ 1) = ½n(n+ l)(n+2),
for all positive integral values of n. Proof. I. Verification. The left-hand member of (7) contains n terms. ,vhen n = 1, equation (7) becomes (8)
1 · 2 = ½· l • (I + 1) (1 + 2) = ½(~)(3) = 2, or 2 = 2.
Therefore formula (7) is true for n = 1. II. Extension Theorem. We now wish to prove that whenever formula (7) is true for a particular value of n, say n = k, it is also true for the next greater value, n = k + 1. We therefore assume that (9)
1 · 2 + 2 · 3 + 3 · 4 + · · · + k(k + 1)
= ½k(k + I)(k + 2),
and prove that if (9) is true, then (10)
1-2 + 2 · 3 + 3 · 4 + · · · + (k + l)(k = ½(k + l)(k + 2)(k
which is the form that (7) takes on when n = k
+ 2) + 3),
+ l.
The left-hand member of (10) contains (k + 1) terms. If we write the kth term instead of merely indicating it, we have (11)
1 · 2 + 2 · 3 + · · · + (k + l)(k = (1 • 2 2 · 3 k(k
+
+ ,, , +
+ 2)
+ 1)] + (k + l)(k + 2)_
§ 107]
MATHEMATICAL INDUCTION
165
By (9) we assume that the expression in brackets is equal to
½k(k + l)(k + 2).
Therefore (12)
1 · 2+2 · 3
+ · · · + (k + l)(k + 2) = ½k(k + l)(k + 2) + (/;, + l)(k + 2) (factor) = (k + l)(k + 2)(! k + l) (reduce to common = (k + I)(k + Z)e 13) denominator) = ½(k + l)(k + 2)(k + 3),
which proves that (10) is true if (9) is true. III. Conclusion. By the Fundamental Principle, the formula (7) is true for all positive integral values of n. Note 1. In Part I of the proof, the verification should be made for as small a positive integral value of n as possible. In some cases it is not possible to take n = 1. Thus, in proving the formula 1
1
~ + 4-5
1
+ · · · + n(n + 1) =
n-2
3(n + 1)'
the verification should be made for n = 3: since the left-hand member has no meaning for n = 1 or n = 2. The restatement of the Fundamental Principle for such cases presents no difficulty. Note 2. Neither Part I, Verification, nor Part II, Extension Theorem, is by itself sufficient for the conclusion. A classical example of a false theorem is the statement that the expression (n 2 - n + 41) is a prime number for all positive integral values of n. By actual substitution the student can show that this expression represents a prime number for n = 1, 2, 3, • • •, 40: But these forty verifications do not prove the general theorem. For when n = 41, n2 - n + 41 = 4l2 - 41 + 41 = 4l2,
which is not prime. Likewise the mere proof (Part II) that whenever a theorem is true for one value of n, it is also true for the next greater value of n, is not sufficient for the conclusion unless it is possible to verify the theorem in some one case. Thus, if we assume that 1 + 3 + 5 + ••• + (2 n - 1) = n2 + 70, when n = k, it is possible to show that the same formula is true when n = k 1. It is not possible, however, to find any value of n for which the formula is true, as we see by comparing it with equation (1) of Example 1, above.
+
Note 3. In a proof by mathematical induction, be very explicit in each of the main divisions of the proof, and, in particular, be especially careful to state clearly the hypothesis and the conclusion of the Extension Theorem to be proved in Part II~
FIRST YEAR OF COLLEGE MATHE1\1ATICS
166
[Ch. XIII
By mathematical induction prov~ that, if n is any positive integer, (x - y) is a factor of (xn - yn). EXAMPLE
Proof.
3.
I. Verification. The theorem is true when n = 1. For if n = 1, xn - yn
=
X - Y,
and (xn - yn) therefore contains (x - y) as a factor. II. Exten8'ion Theorem. We next prove that if (x - y) is a factor of (xk - yk), it is also a factor of (xk+i - yk+ 1 ). To show this, we write xk+i - yk+ 1
= x • xk - y • yk (add and subtract xyk) = x • xk - xyk + xyk - y • yk (factor terms) = x(xk - yk) + yk(x - y).
We are assuming that (x - y) is a factor of (xk - yk). It is clearly a factor of It is therefore a factor of the right-hand member. Therefore, if (x - y) is a factor of (xk - yk), it is a factor of (xk+i - yk+ 1). ?Jk(x - y).
III. Conclusion. By the Fundamental Principle, the theorem is true for all positive integral values of n.
EXERCISES By the method of mathematical induction, prove that the following formulas or statements are true for all positive integral values of n. Make no use of the formulas proved in Chapter XI for arithmetic or geometric progressions. State in words the theorems contained in the formulas of Exercises 1, 5, 6, 7, 8, 14.
+ + + ••• + -t +
+
1. 1 2 3 n = ½n(n 1). 2. 1.,,+ 5 + 9 + • • • (4 n,,:- 3) = n(2 n - 1)~ 3. 4 8 + 12 + • • • ·+ 4 n = 2 n(n + 1).
4. 5 + 10 + 15 + • • • + 5 n = ! n (n + 1). 5. I2+2 2 +3 2 + ... +n2 = ¼n(n+1)(2n+l). 6., l3 + 2 3 + 3 3 + ••• + n 3 = ¼n 2 (n + 1) 2• 7. 13 + 2 3 + 33 + ... + n 3 = (1 + 2 + 3 + + n) 2• 8. l8 3 3 + 5 3 + ••• + (2 n - If = n 2 (2 n 2 - 1)'" 9. 1 • 6 + 2 • 9 + 3 • 12 + • • • + n(3 n + 3) = n(n + l)(n + 2). 10. 1·2·3 + 2-3-4 + 3.4.5 + • • • + n(n + l)(n + 2) = ¼n(n + l)(n + 2)(n + 3). 0
0
0
+
11. a
+ (a +
d) + (a + 2 d) + '- · · + [a + (n - 1) d] = ½n [2 a
12. 3 + 3 + 3 + · · · +3n = ¾(3n - 1). 1 1 1 1 1 13• 2 + · · · + = 1 22 2s 2n 2n, 2
3
+- +-
0
+ (n -
1) d].
§§ 107-108]
14. a 15 16
· ·
17. 18. 19. 20.
167
MATHEMATICAL INDUCTION
arn - a + ar + ar2 + ••• + arn-l = --. r - 1
1
1
1
1
1
1
n
1
N + 2. 3 + 3 • 4 + · · · + n(n + 1) = n + 1· 1
n
N + 3. 5 + ~ + · · · + (2 n - 1)(2 n + l) = 2 n + l 1 • 3 + 3 • 5 + 5 • 7 + ••• + (2n - 1)(2n +I)= ½n (4n + 6n The expansion of (a+ x)n contains (n + I) terms. (x + y) is a factor of x n - y n. (x + y) is a factor of x +y 2
2
1).
2
2n-1
2n-1 •
*108. Proof of the Binomial Theorem. We shall now use the method of mathematical induction* to prove that the binomial formula: (1)
(a
+ x)n =
an + nan-lx +
••• + (n - n'r). ! r! an-rxr + ••• + xn '
( §§ 104, 105) is true for all positive integral values of n. In the righthand member of (1) we have written the first two terms, the last term, and also the general term in xr. · Proof.
When n = 1, formula (1) becomes
I. Verification. (a
+ x)
1
=
a
+ x,
which is obviously true. II. Extension Theorem. We wish to show that whenever formula (1) is true for one value of n it is also true for the next greater value of n. That is, we wish to show that if (2)
(a
+ x)k =
ak
+
kak-t x
+ k! + ____ ak-rxr + . . . + xk (k - r) !r!
~df,
'
then (3)
(a
+ x)H = 1
aH1
+
+ ···
(k
+
+
(k + 1) ! ak-r+lxr (k-r+l)!r!
l)akx
+ ... + xk+1,
·equation (3) being obtained from (1) by replacing n by (k + 1). Since (a + x)k+ 1 = (a + x)k (a + x), assuming that (2) is true, we can obtain the expansion for (a + x)H1 by multiplying the righthand member of (2) by (a+ x). t * An alternative proof is given in § 283.
t At this point review carefully the material on page 161 relating to the multiplication of the expansion of (a + x) 4 by (a + x).
168
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XIII
In the right-hand member of (2) the terms in xr-t and xr are (4) ,
(k - r
k'.
+ 1) ! (r -
1) !
+
ak-r+I xr-1
k'.
(k - r) ! r!
ak-r xr
·
We now multiply the terms of (2) first by a and then by x, nnd add like terms to obtain (a+ x)H 1• The first term of the product is clearly aH 1, and the last term is xk+i. Except for these two terms, in the product the term in xr is the sum of two terms in ak-r+i xr. One of these is obtained by multiplying the second term of (4) by a, and the other comes from multiplying the first term of (4) by x. In the product the coefficient of the term in xr is therefore the sum of the coefficients of these terms, or
+
k!
(5)
(k - r
+ 1) ! (r -
In the first term of (5) we set ( r I we set (k _
1) !
k! (k - r) ! r!
_! l) 1. = r.;. , and in the second term
k-r+l + l) ! , and remove common factors.
- (k _ r
·we
can therefore write the coefficient in the form (6)
k!
(k - r
+ 1) ! (r _
1) !
k! r (k - r
=
(k _
+
k! (k - r) ! r!
+ 1) ! r!
/!
+
l) ! r! [r
k! (k - r + I) (k - r + 1) ! r!
+ (k
- r
+ 1)]
k!(k+l) (k+l)! = (k - r + 1) ! r! = (k - r + 1) ! r!° This is the same as the coefficient of the term in xr in the right-hand member of (3). Since thin may represent any term, if (2) is correct, (3) is also correct.
III. Conclusion.
By the Fundamental Principle of mathematical
induction, the binomial fo;rmula (I) is correct for all positive integral values of n.
CHAPTER XIV
Logari thtns 109. The exponential function, y = axo The student is referred to § 12 for a definition of positive integral exponents. We call an the nth power of a; a is the base; n is the exponent. For the cases in which the exponent is zero or is a positive or negative rational fraction, we recall the following definitions (Appendix, § 403): (1)
a
0
= 1, if a ~ 0;
~
an= vam
= (va)m;
a-n
1 = an -.
For convenience of reference, we repeat here the Laws of Exponents:
I.
am . an = am+n;
III.
am -
an -
II. (am)n
am-n•
= amn; V.
'
(ba)n = anbn·
A thorough consideration of irrational exponents is beyond the scope of this book. However, it can be shown that if a > 1 and m and n are rational and if x is any real number such that m < x < n, the statement that (2)
leads to a unique definition of ax for real irrational exponents x. We shall therefore assume that: If a > 1, and if xis any real number, one and only one number ax exists. We shall also assume, conversely, that if y is any positive number, there is always one and only one real value of x for which ax = y, but if y < 0 no such value of x exists. Moreover, if m and n are any real exponents, either rational or irrational, the Laws of Exponents apply as stated above. The function y = ax, where a is some constant greater than 1, is sometimes called the exponential function. It is very important in physics, chemistry, statistics, biology, mathematics of finance, and other applications. Students should learn to sketch its graph quickly. 169
170
FIRST YEAR OF COLLEGE l\'.IATHElV[ATICS
[Ch. XIV
The graph of the function y = 2x, which is the exponential function in which the base a is equal to 2, is shown as a solid line in Figure 49. The graph was plotted from th~ following table of values: X
Y = 2x
-
1
0
1
2
-
2
-
4
'1
\
\ \ \ \
6 6 \
4 \
\
3
\
2
-2
-3
---1
4
1
8
> I, . the graph of
= ax has the following proper-
ties: It crosses the y-axis at the point (0, 1); as x goes from left to right the curve always rises and with increasing rapidity; as x goes to the left, the curve approaches the x-axis as an asymptote. The function
8
\
1
8
y
I
'
-1 --
Whenever a
y I
3
\2
'
--:_-t-==_:Q::::_=;~_+---+-~~+~ =-=4=--X 4 3 2 2 3 1 0 4 1
is also important. I ts graph may be obtained from the graph of y = ax merely by changing the FIG. 49 sign of x at each point of the graph. In Figure 49 the dotted line is the graph of the function y = 2-x.
Definition of a logarithm. Numerical computation is often greatly simplified by the use of logarithms. Logarithms enable us to replace the operations of multiplication and division by the simpler operations of addition and subtraction, and to replace taking powers and roots by mere multiplication and division. As is evident from the following definition, a logarithm is an exponent. If a, x, and y are three numbers such that 110.
ax= y,
(1)
then we say that xis the logarithm of y to the base a, or
(2)
X
= loga y.
Definition. The logarithm of a number to a given base is the exponent of the power to which the base must be raised to yield the number. ' Also, if x is the logarithm of y to a given base, y is called the antilogarithm of x to that base.
171
LOGARITHMS
§§ 109-110]
Equations (1) and (2) are different ways of expressing the same relation among the three numbers a (the base), y (the number) and x (the exponent or logarithm). Equation (1) states this relation in exponential form; equation (2) says the same thing in logarithmic form. EXAMPLE 1. EXAMPLE 2.
Since 53 = 125, logs 125 = 3. Since 3-4 = -ir, log3 ( 8\ ) = -4.
EXAMPLE 3.
Since 2!
=
V16,
Notice that, since a 1
=
a and a0
(3)
and
log2 (Y16)
(4)
=
=
!.
1,
loga 1 = O, for any base, a.
Here we shall consider as bases of logarithms only bases that are greater than 1. The assumptions of § 109 imp]y the following. Fundamental assumption: If a > 1, any positive number has a real logarithm to the base a, but a negative number or zero has no such real logarithm. Also if a > 1 and m > n > 0, then loga m > loga n, or the larger one of two numbers has the larger logarithm.
EXERCISES In each of the Exercises 1 to 20, change the statement from exponential form to logarithmic form.
= 81. 5. 103 = 1000. 1. 34
9• 8-1 13. 6°
1.. 8•
--
4
= 1.
_3
17. 4 2
= 216. 6. v25 = 5. 10. 5-2 = 21,;. 2. 63
14. 8~
= ½.
= 16.
18. 11-2
=
= 64. 7. v21 = 3. 11. 10-1 = 0.1.
8. V49 = 7. 12. 10-2 = 0.01.
3
15. 81 4 _3
tlT•
4. 101 = 10.
3. 4 3
19. 25 2
2
16. 27-3 =
= 27.
=
3
20. 16-4 =
1§5.
½.
i.
In each of the Exercises 21 to 28, change the statement from logarithmic form to exponential form.
= 2. 24. logs6 6 = ½. 27. log2s (-!) = -½. 21. logs 25
22. logs 64
= 2.
26. log10 .001 = -3. 28. log4 (¼)
23. log2 16 = 4. 26. logs 32 = j.
= -}.
In each of the Exercises 29 to 52, solve the equation for x. 29. x 32. x
= log1 49.
=
log2 (½).
35. x = log2 (
V~z}
30. x = log2 128. 33. x = log2 (¼). 36. x
= log4 (½).
31. x 34. x
= log10 10,000. = log16 4.
37. x = log8 (¼).
172
FIRST YEAR OF COLLEGE MATHEMATICS
38. logx 16 = 2. 41. logx 81 = f. 44. logx (½) = --¾-
42. logx (? ~) = 4. 45. logx (.001) = -J.
47. loga x = I. 50. logs x = -¾.
48. log4 x = 3. 51. logs x = -}.
39. logx 81 == 4.
[Ch. XIV
40. logx 8 = f. 43. logx (-i5 ) = - 2. 46. logx (½) = -}.
lo Vl0 = x. 62. log1 x = -2.
49. log10
Show that: 53. a10gax
= x.
54. loga (ax)
= x.
55. loga (1/a)
= -1.
111. The fundamental theorems of logarithms. The following theorems correspond to laws of exponents and radicals given in § 109.
I. Multiplication.
The logarithm of the product of two numbers is equal to the sum of the logarithms of the two numbers:
loga (M · N)
(1)
= loga M + loga N.
Proof. Let M and N be any two positive numbers, and let a be any suitable base. Set X
= IogaM
and
Then, from the definition of a logarithm, ax = M and aY = N. Therefore, from I, § 109, M · N = ax • aY = ax+Y_ Consequently, by the definition of a logarithm,
Ioga (M · N) = x
+y
= logaM
+ logal.V.
Clearly, also, the logarithm of the product of any number of factors is equal
to the sum of the logarithms of the factors. Thus, for three factors, L, M, N, loga (L · M · N) = Ioga [(LM) · N] = loga (LM)
+ loga N
= loga L + loga NI+ loga N. If we have some way of finding the logarithms and antilogarithms of numbers, this theorem enables us to replace the laborious task of multiplying two or more numbers by the comparatively easy one of adding their logarithms.
IL Division.
The logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor:
(2)
loga
(1};) = loga M -
loga N.
The student should supply the proof, using the same notation and general method as in the proof of Theorem I.
1-., lu
LOGARITHMS
§§ 110-111]
From Theorems I and II it follows that the logarithm of any positive fraction is equal to the sum of the logarithms of the factors of the numerator minus the sum of the logarithms of the factors of the denominator. EXAMPLE
1.
23
log a
413 261 · · 53 · 71
+ loga 413 + loga 261 -
= loga 23
loga 53 - loga 71.
III. Involution.
The logarithm of a power of a number is equal to the logarithm of the number multiplied by the exponent of the power:
(3)
loga (Mn)
Proof.
Let
X
Then by II, § 109.
= n loga M.
= IogaM, or ax= M. Mn = (ax)n = anx,
Therefore, by the definition of a logarithm, loga (Mn) = loga (anx) = nx = n Ioga M.
IV. Evolution.
The logarithm of a principal root of a number is equal to the logarithm of the number divided by the index of the root:
loga (vM)
(4)
= loga M_ n
1
Proof.
By (1), § 109, loga f 1M = loga (}lf;) -
•
! loga M.
n
2. To find 467 49 is laborious, and to find V 739 is quite difficult. But log10 (467 49 ) = 49 logio 467 and log10 (Y739) = log10 739, and, as we shall see, these logarithms and their antilogarithms are obtained easily. EXAMPLE
t
EXERCISES Find the following logarithms, given that log10 2 = 0.301 and log10 3 = 0.477, and assuming that these are exact rather than approximate values. 1. log10 4.
2. log10 9.
3. log10 6.
4. log10 8.
5. log10 27. 9. log10 ¾.
6. log10 ¾. 10. log10 ½.
7. log10 ¾ 11. log10 ½.
8. log10 ½. 12. log10 20.
13. log10 30. 17. log10 25.
14. log10 5. 18. log10 45.
15. log10 13°. 19. log10 40.
16. log10 15. 20. log 10 900.
21. log10 16. 25. log10 72.
22. log10 96. 23. log10 ½. 24. log10 26. log10 54. 27. log10 y2.
28. log10
vi
29. log10
v2.
31. log10
VB.
32. log10 ~ 20.
30. log 10
vg.
33. log10 V 24.
l7•
174
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XIV
112. Significant digits and approximations. In problems involving numerical computation, numbers will ordinarily be written in decimal notation. Each number is supposed to contain a decimal point; if the decimal point does not actually appear, its presence should be understood. We can assume that there is an unending sequence of zeros at the right of each number. In computation we necessarily deal with .approximations. Approximations enter a problem originally in two ways. In the first place, the tables that we use, such as tables of logarithms, are limited to a certain number of decimal places. This means that the exact values of the numbers appearing in the tables have been rounded off to the required number of decimal places. Thus, if a four-place table gives log10 N = 0.3648, we know merely that the exact value of log 10 N is at least as near to 0.3648 as it is to 0.3647 or to 0.3649, and therefore lies between 0.36475 and 0.36485. In this case we say that 8 is the last specified digit. In general, the error made in writing the approximate instead of the exact value is not more than 5 units in the next place to the right of the last specified digit.
Approximations also enter because the results of measurements, and therefore the data of many problems, are approximations whose accuracy varies according to circumstances. To say that a certain football player weighs 210 pounds gives an idea of his size, but does not indicate how carefully he was weighed. It may merely give his weight to the nearest ten pounds, in which case 1 is the last specified digit and his exact weight may be anything between 205 and 215 pounds. If the unit of accuracy is one pound, the last specified digit is 0, and his exact weight is between 209.5 pounds and 210.5 pounds. EXAMPLE
1.
By the significant digits of a number, we mean the sequence of digits starting with the first one at the left that is not zero and ending with the last specified digit at the right. An approximate number thus represents an exact value with an error that is not greater than 5 units in the first place to the right of the last significant digit. A number is said to be accurate to k places if it contains k correct significant digits. It is accurate to k decimal places if the last significant digit lies in the kth place to the right. of the decimal point. Theorem. If two numbers have the same sequence of significant digits, the larger one of them can be obtained from the smaller one by multiplying the latter by a suitable whole power of 10. For the larget
mimber can be obtained from the smaller one by moving the decimal
LOGARITHMS
§ 112]
175
point a suitable number of places to the right; and moving the decimal point one place to the right is equivalent to multiplying the number by 10. 2. The numbers 40950 and 0.04095 have the same sequence of significant digits (if both are accurate to 4 places). EXAMPLE
40950 = 0.04095 X 106•
In scientific notation we have a convenient way of writing numbers so as to display their degree of accuracy. In this notation, a given number N is written in the form N = M X lW, where the exponent r is a whole number and where M is less than 10 but is greater than or equal to 1 and has the same sequence of significant digits as the given number N. The exponent r is equal to (k - 1) if N > 1 and if N has k places to the left of the decimal point; and r == ( - k) if N is positive but less than 1 and if its ~first significant digit is in the kth place to the right of the decimal point. 3. 21,045 = 2.1045 X 104 has five-place accuracy. EXAMPLE 4. 0.00267 = 2.67 X 10-3 has three-place accuracy, while 0.002670 = 2.670 X 10-3 has four-place accuracy. Notice that 0.00267 has its first significant digit in the third place to the right of the decimal point and that the exponent of 10 is -3. EXAMPLE
EXAMPLE
5.
186,000 = 1.86 X· 105, if there is three-place accuracy, and 186,000 = 1.8600 X 105, if there is five-place accuracy.
Notice that 186,000 has 6 places to the left of the decimal point and that the exponent of 10 is 6 - 1 = 5.
A number having more than k significant digits is rounded off to k places if it is replaced by an approximate value with k significant digits which differs from the given number by not more than 5 units in the first neglected place. A number having more than k places to the right of the decimal point is rounded off to k decimal places if it is replaced by a number having k places to the right of the decimal point and differing from the given number by not more than 5 units in the first neglected place. EXAMPLE 6. To round off 57.438 to 4 places, we replace it by 57.44. To round off 57.438 to 1 decimal place, we replace it by 57 .4.
176
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XIV
EXAMPLE 7. If we wish to round off 35.165 to 2 decimal places, we are logically justified in replacing it by either 35.16 or 35.17, since each of these numbers differs from 35.165 by 0.005, which is 5 units in the first neglected place. As a rule for rounding off in this course, whenever there is no logical way of choosing between an even and an odd digit, we shall write the final result that ends with the even digit. The student should follow this rule uniformly. e therefore write 35.16 in this example.
,v
EXERCISES In each of the Exercises 1 to 4, write the number as a power of 10.
1. 100,000.
2. 0.0001.
4. 0.00001.
3. 10,000.
In each of the Exercises 5 to 8, a number is written in scientific notation. Write it in ordinary decimal form, and indicate the number of significant digits. 5. 3.647 X 105•
6. 2.0400 X 106 •
7. 1.040 X 10-3 •
8. 9.61 X 10-s.
In each of the Exercises 9 to 14, write the number in scientific notation, with the specified number of significant digits.
9. 2487, 4 digits. 12. 56,000,000, 3 digits.
10. 56.38, 5 digits. 13. 0.004023, 4 digits.
11. 0.0047, 3 digits. 14. 0.00021, 4 digits.
In each of the Exercises 15 to 22 round off the number as indicated. 15. 67,427 to 4 and to 3 places. 16. 3974.8 to 4 and to 3 places. 17. 12.6573 to 3 and to 2 decimal places. 18. 5. 6862 to 3 and to 2 decimal places. 19. 13.475 to 4 places. 20. 14.905 to 4 places. 21. 0.00463 to 4 and to 3 decimal places. 22. 0.00755 to 4 and to 3 decimal places.
113. Logarithms to the base 10. In numerical computation, the most useful base of logarithms is the base 10. Logarithms to the base 10 are called common logarithms. Hereafter when we speak of a logarithm without specifying the base, it is to be understood that the base of the logarithm is 10. Instead of writing log10 N, we shall write merely log N. The logarithm of an integral power of IO is a whole number. EXAMPLES. log 1 = log 10° = 0. 1 log 0.1 = log 10-1 = -1, log 10 = log 10 = 1, log 0.01 = log 10-2 = - 2, log 100 = log 102 = 2, log 0.001 = log 10-3 = -3. log 1000 = log 103 = 3,
LOGARITHlVIS
§§ 112-115]
177
114. Characteristic and mantissa. For use in computation, if a logarithm is not a whole number, it is usually written as the sum of a whole number (positive, negative, or zero) and a positive fractional number. The positive (or zero) fractional part of a logarithm is called the mantissa of the logarithm, and the remaining lntegral part is the characteristic. If the logarithm is a whole number, its mantissa is zero and its characteristic is the logarithm itself. It is useful to think of the logarithm marked as a point on the scale of real numbers. If this point lies between two integral points, the characteristic corresponds to the first integral point to its left, and ths mantissa is the distance that it lies to the right of that integral point. If log K = 1.34, the characteristic of log K is 1, and its mantissa is 0.34. If log M = -1.38 = -2 + 0.62, the characteristic of log M is -2 and the mantissa is 0.62. (See Figure 50.) Notice that the EXAMPLES.
.62
t
F='=Jc -2
.34 ,-J'-..
t
I
-1
I
t X
1
t
logK
logM FIG. 50
mantissa is not - 0.38, which is negative, but is the positive fraction 0.62. The characteristic is -2, not -1. Further examples follow: Logarithm Characteristic 0, log 2 = 0.30103, log 20 = 1.30103, 1, log 0.5 = -0.30103 = -1 + .69897, -1, log 0.05 = -1.30103 = -2 .69897, -2, log 100 = 2.00000, 2,
+
115. The mantissa.
Mantissa .30103. .30103. .69897. .69897. .00000.
Theorem. If two numbers have the same sequence of significant digits, their logarithms have the same mantissas. Proof. By the theorem of § 112, the larger of the two numbers can be obtained from the smaller one by multiplying the latter by a whole power of 10. By Theorem I of § 111, this has the effect of adding the logarithm of a whole power of 10, which is a whole number, to the logarithm of the smaller number. A change in the position of the decimal point in a number therefore changes the whole part, or characteristic, of its logarithm, but it does not change the fractional part, or mantissa.
178 EXAMPLES.
FIRST YEAR OF COLLEGE MATH~MATICS
[Ch. XIV
Given log 8.93 = 0.9509. Then log 89.3 = log 8.93 log 10 = 0.9509 1 = 1.9509. log 100 = 0.9509 + 2 = 2.9509. log 893 = log 8.93 log 0.893 = log 8.93 - log 10 = 0.9509 - 1. log 0.00893 = log 8.93 - log 1000 = 0.9509 - 3.
+ +
+
EXERCISES In each of the Exercises 1 to 6, write the characteristic and mantissa of the given logarithm. 1. log 7
= 0.8451.
2. log 700
4. log 0.7 = -0.1549. Given log 3.47 rithms. 7. log 34.7. 11. log 34,700. 15. log .00347.
6. log
t
= 2.8451.
= -0.8451.
3. log 0.07
= -1.1549.
6. log / 0 = -1.8451.
= 0.5403 and log 7.48 = 0.8739, find the following loga• 8. log 0.748. 12. log 347. 16. log 74,800.
9. log 748. 13. log 0.07 48. 17. log 0.000748.
10. log 7480. 14. log 0.347. 18. log 74.8.
116. The characteristic. If a number is greater than 1 and has just one digit to the left of the decimal point (e.g. the number 3.47), the characteristic of its logarithm is zero. For such a number lies between 1 and 10, so that its logarithm is between log 1 = 0 and log 10 = 1 and has the characteristic zero. We have seen that if a positive number N is written in scientific notation, we have N = M X l0r, where M is a number having just one place to the left of the decimal point and r is a whole number. Hence log N = log M
+ log (l0r) =
log M
+ r.
Since log M has the characteristic 0, the characteristic of log N is r, the exponent of 10 when N is written in scientific notation. From the statements about scientific notation in § 112, we therefore have the following rule. Rule. If a number is greater than 1 and has just k digits to the left of the decimal point, the characteristic of its logarithm is (k - 1). If a number is positive but less than 1, and if its first significant digit is in the kth place to the right of the decimal point, the characteristic of its logarithm is (-k).
LOGARITHl\1S
§§ 115-116]
179
The characteristic of the logarithm of 6.836 is 0; that of 68360 is 4; that of 0.6836 is -1, that of .006836 is -3; that of 683.6 is 2. EXAMPLE
1.
Negative characteristics may be written in a variety of ways. Given that log 6.836 = 0.8348, we may write log 0.06836 = 0.8348 - 2. It would be incorrect to write this as - 2.8348, for this would indicate that the .8348 is negative. If we choose to do so, we can surely write log 0.06836 = 8.8348 - 10,
or
log 0.06836 = 18.8348 - 20,
since -2 = 8 - 10 = 18 - 20. This method of writing a negative characteristic as a positive number or zero minus some multiple of 10 is the most convenient method in practice, and will be followed in this book. Thus log 0.6836 = 9.8348 - 10, log 0.0006836 = 6.8348 - 10, etc. 2. Given that log 3.64 = 0.5611, find the value of N if 7.5611 - 10.
EXAMPLE
log N
=
Solution. Since the mantissa of log N is .5611, the sequence of significant. digits in N is 364. Since the characteristic is 7 - 10 = -3, N is less than 1 and its first significant digit is in the third place to the right of the decimal point. Therefore N = 0.00364.
EXERCISES Name the characteristics of the logarithms of the following numbers. 1. 4.71. 2. 36720. 3. 1200. 4. 23.65. 5. 624. 7. 6. 0.369. 7. 0.00543. 8. 0.048. 9. 0.0004. 10. 0.0042. Given that log 3.45 = 0.5378 and log 7.61 = 0.8814, find the logarithms of the following numbers, writing negative characteristics in conventional form. ~ 11. 345. 12. 7610. 13. 34.5. 14. 34500. 15. 0.345. 17. 0.00761. 18. Q.0345. 16. 0.0761. 19. 76.1. 20. 761000. 21. 0.000345. 22. 0.761. 23. 3.45 X 76.1. 24. 3450 X .0761. 25. 0.0345 X 0.00761. 26. .00345 X 761. 27. 0.0345 X 7.61. 28. (0.0345) 2 • In Exercises 29 to 44, find the numbers corresponding .to the given logarithms, gi~en that log 2.43 = 0.3856 and log 831 = 2.9196. 29. 3.3856. 30. 4.9196. 31. 0.9196. 32. 1.3856. 33. 9.3856 - 10. 34. 7.3856 - 10. 35. 8.9196 - 10. 36. 9.9196 - 10. 37. 5.3856. 38. 4.3856. 39. 8.3856. 40. 8.3856 - 10. 41. 8.9196. 42. 8.9196 - 20. 43. 7.9196. 44. 7.9196 - 20.
180
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XIV
117. A table of logarithms. Table II, page 664, is a table of the mantissas of the logarithms of all whole numbers from 10 to 999, rounded off to four decimal places. The mantissa of the logarithm of any number having not more than three significant figures can be found from the table, since the mantissa does not depend on the position of the decimal point. This economy in the preparation and use of tables is one of the advantages gained by using 10 as the base. In order to find the logarithm of a number from Table II, we look for the first two significant digits of the number in the column N at the left of the table. This gives the row in which the mantissa is to be found. Its column is the column in which the third figure of the number appears at the top of the page. A decimal point must be 9laced before each mantissa. The characteristic is determined by the position of the decimal point in the given number. It is usually best to write the characteristic before looking for the mantissa in the table. Find log 34.7. Solution. Since 34.7 has two places to the left of the decimal point, the characteristic is 1. The mantissa appears on page 664 in the row marked with the first two digits, 34, at the left and in the column marked with the third digit, 7, at the top. The entry is 5403. The mantissa is therefore .5403, and log 34.7 = 1.5403. EXAMPLE
1.
2. Find log 0.00783. Solution. In the row marked 78 and the column marked 3, we find the mantissa .8938. Since the characteristic is -3, log 0.00783 = 7.8938 - 10. EXAMPLE
If x = log N, N is the antilogarithm of x, 1V = antilog x. In order to find the antilogarithm of a given logarithm, we first consider only the decimal part or mantissa, which we look for in the body of the table. The first two significant digits of the desired antilogarithm then appear in the same line in the column N, and the third digit appears at the top of the page. The decimal point is then placed by considering the characteristic. 3. Find N, given log N = 3.8149. Solution. We look for the mantissa .8149 in the body of the table. We find it in the row marked 65 at the left and in the column headed 3. The sequence of digits is therefore 653. Since the characteristic is 3, N = 6530. EXAMPLE 4. If log N = 8.6284 - 10, N = 0.0425. EXAMPLE 5. If log N = 0.8451, N = 7.00. EXAMPLE
§§ 117-118]
LOGARITHMS
181
EXERCISES In each of the Exercises 1 to 12, find the logarithm of the given number. 3. 82.0. 7. 0.0707. 11. 0.002.
2. 734. 6. 0.932. 10. 8.3.
1. 3.28. 5. 0.0498. 9. 0.0029.
4. 2800.
8. 10.7. 12. 48300.
In each of the Exercises 13 to 24, find the antilogarithm of the given number. 13. 1.4232. 17. 9.6609 - 10. 21. 7.7007 - 10.
15. 2.6812. 19. 0.0086. 23. 7.6075.
14. 3.9330. 18. 8.8055 - 10. 22. 9.7372 ·_ 10.
16. 0.9538. 20. 1.8500. 24. 4.9031.
118. Interpolation. When we wish to find a number which is not given in the table, but which lies between two entries of the table (for example, the logarithm of a number having four significant digits), we resort to the method of interpolation. EXAMPLE
1.
Find log 31.76.
Solution. The number 31.76 lies between the two entries 31.7 and 31.8. Therefore log 31.76 lies between log 31.7 and log 31.8. The difference between two successive numbers in the body 9f the table is called the tabular difference for the two numbers. From the table we find
log 31.8 = 1.5024 log 31.7 = 1.5011 . Tabular difference = 0.0013 It is not true that the logarithm of a number is proportional to the number. But for a small change in the number, the change in the logarithm is very nearly proportional to the change in the number. We therefore assume that since 31.76 is "six-(enths of the way" from 31.7 to 31.8, then log 31.76 will also be" six-tenths of the way" frorn log 31.7 to log 31.8. To the number log 31.7 = 1.5011 we must therefore add six-tenths of the tabular difference .0013. That is log 31.76 = 1.5011
+ 106 (.0013)
=
1.5011
+ .00078 =
1.50188.
We must remember, however, that the table itself has been rounded o:ff and is accurate only to four decimal places. We should be wrong to expect greater accuracy in our interpolated result than exists in the table itself. We therefore round off the proportional part of the tabular difference to four decimal places before adding it to the smaller logarithm. That is, before adding, we round off .00078 to .0008 and obtain log 31.76 = 1.5011 + .0008 = 1.5019.
182
FIRST YEAR OF COLLEGE MATHElVIATICS
[Ch. XIV
The work of interpolation is made easier by the table of Proportional Parts (marked "Prop. Parts") given at the side of the main table. In the table of Proportional Parts each column is headed by a tabular difference from the main table, and below it are nine numbers which are, respectively; .1, .2, .3, • • •, .9 of the tabular difference, as indicated by the numbers 1, 2, 3, • • •, 9 that appear at the left. Thus, if we wish to find six-tenths of the tabular difference 39, (in the last place) we look in the column headed 39, and in the same row as the figure 6 we find 0.6 X 39 = 23.4. The number 23.4 must be rounded off to 23. In finding log 31.76, where the tabular difference in the last place is 13, we look in the column headed 13, opposite 6, and find 7.8, which we round off to 8 and add to log 31.7 = 1.5011, getting 1.5019. The student should learn to do this work mentally and write down only the final result. Interpolation may cause a slight error in the last place. EXAMPLE 2.
Find log 0.06348.
Solution. Log 0.0634 = 8.8021 - 10, and the tabular difference in the last place is 28 - 21 = 7. From the table of proportional parts, .8 X 7 = 5.6, which we round off to 6 and add to the last place of log 0.0634. We get
log 0.06348 = 8.8027 - 10. EXAMPLE 3.
Find log 29.63.
Solution. Log 29.6 = 1.4713, and the tabular difference is 15. From the table of proportional parts .3 X 15 = 4.5. This number is equally near to 4 and to 5, and when we round off we have no logical basis for· choosing between 4 and 5. We therefore follow the rule stated in Example 7, § 112., and, as the logarithm, write the number that ends in the even digit. Since our choice is between (1.4713 .0004) and (1.4713 .0005), we choose the latter, and write log 29.63 = 1.4718.
+
EXAMPLE 4. Solution.
+
Find N, if log N = 1.2241.
This entry does not occur in the table, but is between 1.2227 = log 16.7
and
1.2253 = log 16.8.
We can arrange the numbers as follows. Number
Logarithm
16.70
1.2227
N
1.2241
16.80
1.2253
Difference ..,
} 14
>-
..
26
§§ 118-119]
LOGARITHMS
183
Therefore log N is "fourteen twenty-sixths of the way" from log 16.70 to log 16.80, and we conclude that N is "fourteen twenty-sixths of the way" from 16.70 to 16.80. Using decimals, we have -H- = .53?. Therefore N is approximately five-tenths of the way from 16.70 to 16.80, and N = 16.75. We save some arithmetic by using the table of proportional parts. We look for 14 in the proportional parts column headed by the tabular difference 26. The nl!,mber 14 does not appear there. The nearest number to 14 is 13, which corresponds to 0.5. Therefore we add .5 of the difference between 16.80 and 16.70, and write N = 16.75. EXAMPLE
Solution.
5.
Find N, if log N = 8.5620 - 10.
This number lies between the entries
8.5611 - 10
= log .0364
and
8.5623 - 10
= log .0365.
The tabular difference is 12, and "our difference" (that is, the difference between 5620 and 5611) is 9. In the proportional parts column headed 12, the number 9 is halfway between 8.4, which corresponds to .7, and 9.6, which corresponds to .8. We choose the even digit 8 in preference to 7, and write N = 0.03648.
EXERCISES In Exercises 1 to 16, find the logarithms of the given numbers.
1. 6. 9. 13.
7.834.
2. 21.48.
0.6485.
6. 0.06312.
2.185. 0.02897.
10. 0.7025. 14. 8.145.
3. 5467.
7. 3.396. 11. 0.005534. 16. 99.87.
4. 8. 12. 16.
8035. 84.96. 2.903. 1.073.
In Exercises 17 to 32, find the antilogarithms of the given numbers.
17. 0.5031. 21. 8.9028 - 10. 25. 8.2437 - 10.
29. 9.9957 - 10.
18. 22. 26. 30.
1.6672. 9.4721 - 10. 0.7557. 3.7377.
19. 23. 27. 31.
9.3412 - 10. 0.9393. 4.8058. 1.3167.
20. 24. 28.. 32.
2.7786. 1.9769. 1.0091. 4.0402,
119. Computation with logarithms. In any problem of computation using logarithms, before looking up logarithms one should make a blank outline for all of the work, indicating all of the operations. It is then possible to do the mechanical work without interruption and thereby gain in speed and accuracy. Every part of the work, moreover, is labe]ed, which aids in later checking and examination.
184
FIRST YEAR OF COLLEGE l\lIATHEl\1:ATICS
EXAMPLE
1.
Find the value of
x
[Ch. XIV
= 42.7 X 372.0 X 0.0836.
+
Solution. By § 111, log x = log 42.7 + log 372.0 log 0.0836. The blank outline, made before looking up any logarithms, is at the left. When filled in from the tables, the outline is as shown at the right.
log 42.7 = log 372.0 = log 0.0836 = _ _ _ _(_+_) log x = x=
log 42.7 = 1.6304 log 372.0 = 2.570fi log 0.0836 = 8.9222 - 10( +) log x = 13.1231 - 10 = 3.1231 X = 1328.
Here xis found as the antilogarithm of 3.1231. E
. h 31.42 X 1.257 X 2.732 _ X _ · Fmd t e value of x = 633 7 0 8493 Solution. log 31.42 = 1.4972 log 1.257 = 0.0994 log 2.732 = 0.4365 ( +) log 633.7 = 2.8019 log numer. = 12.0331 - 10 log 0.8493 = 9.9290 - 10( +) log denom. = 2.7309 (-) ~ log denom. = 12.7309 - 10 log x = 9.3022 - 10, x = 0,2005. XAMPLE
2.
Notice that 10 was added to and subtracted from the logarithm of the numerator. This was done to avoid subtracting 2.7309 from a smaller number.
E
XAMPLE
Solution.
2 . h 3 /4.362 X (2.96) Fmd t e value of x = ~ _ X .4 · 32 1 81 3 By § 111, we have
3.
log x =
½[log 4.362 + 2 log 2.96
- log 81.3 - log 32.41].
We arrange the work as follows. log 2.96 = 0.4713 X2 2
log (2.96) log 4.362 log numerator log denominator log radicand log x
= = = =
0.9426 0.6397 ( +) log 81.3 = 1.9101 ll.582~ - 10 log 32.41 = _l._51_0_6_ _(_+_) 3.429L ( - ) ~ log denominator = 3.4207 = 28.1616_ - 30( +3) = 9.3872 - 10 , x = 0.2439.
Here, before dividing (8.1616 - 10) by 3, which would have given the .1,wkward number (2.7205 - 3.3333), we added and subtracted 20, making the negative part equal to -30. In general, before dividing a negative logarithm by a whole number we make its negative part an exact multiple vf the di-;i3or by adding and subtracting a suitable number~
LOGARITHMS
§ 119]
185
EXERCISES In each of the Exercises 1 to 52, compute the value of the given number, using Table IL Assume that the given numbers are exact, rather than approximate values, and give the results to four places. In each case make a complete outline before using the tables. 1. 34.7 X 82,3.
2. 6.87 X 42.9.
3. 89.3 X .0671.
4. 50.6 X .859.
5. 73.6 + 6.24.
6. 5.13 + 42.6.
7. .784 + 39.6.
8. 3.07 + 672.
9. .0274 + 51.8.
ll . .638 X 74.7_ 19.3 X .182
l0. 64.2 X 386_ 4.09 X 527 13
_ 2.15 X 62.9_ 8.34 X 227
14
_ .0469 X .0371 1.28 X 8.21
12
_ .0375 X 46.7_ 21.3 X .0246
15
_ .0490 X .360 .0670 X .840
16. (42.7) 3•
17. (6970) 4 •
18. (.0735)3.
19. (.876) 6 •
20. vs120.
21. ~ 28.9.
22. -vi3820.
23.
25. - 1, 2 quickly from memory. The l graph will remind him of --.-~--+-+---i---t----t--+----+---+----+--X the following properties of 0 8 9 10 1 the function: when a > 1, (1) logaxincreaseswhenever -2 x increases; (2) loga 1 = 0; _ 3 (3) loga a = 1; (4) when x < 0, loga x has no real FIG. 51 value; (5) when O < x < 1, loga x is negative; (6) when x decreases toward 0, loga x decreases without limit; (7) when x increases without limit loga x increases without limit.
*122. Exponential equations. An exponential equation in one unknown is an equation in which the unknown appears in one or more exponents. In order to solve such an equation we try to arrange it so that we can obtain an algebraic equation by equating the logarithms of both members. Solve for x: 153x+2 = 29(82x). Solution. Taking the logarithms of both members, we get log (153x+2 ) = log 29 + log (82x) or (3 x + 2) log 15 = log 29 + 2 x log 8. Therefore x(3 log 15 - 2 log 8) = log 29 - 2 log 15. From Table II, log 15 = 1.1761, log 8 = 0.9031, log 29 = 1.4624. EXAMPLE
1.
Substituting these values and combining terms, we get 1.7221
X
= - .8898,
and
X
= -
.8898 = -0.5168. 7221
1.
In order to compute the value of the last fraction by logarithms, we subtract log 1.722 = 0.2360 from log .8898 = 9.9493 - 10, look up the antilogarithm of the difference, and then attach the minus sign.
190
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XIV
2. In how many years will $75 grow to $100 if lent at 3% compounded annually?
*EXAMPLE
i
Solution.
Denoting the number of years by x, from § 99, we have
A = P(l
+ i)x,
or
100 = 75(1.03)x.
Taking the logarithms of both members, we obtain log 100 = log 75
+ x log 1.03,
or
(log 1.03)x = log 100 - log 75.
0.0128 x = 2 - 1.8751 = 0.1249
Therefore
"' = 0.1249 = 9 76 X .0128 • .
and
At 3% compounded annually, $75 therefore grows to $100 in about 9 years and 9 months.
EXERCISES In Exercises 1 to 18, solve the equations for x, using four-place or five-place tables as directed by the instructor.
1. 3x = 59. 3. 7x = (4.6 )2x-3. 5. (13.5)2-5x = (1.07)4x+3. 7. (216)X = 75(6z). 9.
(121) 5x+2
11. (ll)x2-4x 13. (1.04)x
= 35(8 = 16.
3x).
=
2. 16. (1.06)-x = 0.75.
17. (LO~~: - 1 = 10.
2. (4.16)X = 69.2. 4. (29.37) 3x = (16.7) 5x-2• 6. (54.8)-x = (12.3)x-2• 8. (9.3)x = 7(2.6)x. 10. (46) 4x-l = 28(11 3x).
= 703. 14. 100(1.05)x = 125. 12. (26)x2+2x
16. 1000(1.025)-x = 800. (l.03)x - 1 = 7. 18. .03
*19. In how many years will $800 grow to $1182 at 5% compounded annually? *20. In how many years will $1000 grow to $1900 at 5% compounded semiannually? *21. In order to accumulate a fund for his son's education, a man deposits $100 on his son's first birthday and every six months thereafter. How old will the son be when the fund amounts to $4600, if interest is at 5% compounded semiannually? *22. How many annual payments of $1000 each must be made in order to pay for a house costing $10,380, if interest is at 5% compounded annually Md the first payment is made at the time of purchase? (Suggestion. Determine the present value of each payment at the time of purchase; the sum of these values should equal $10,380.)
191
LOGARITHMS
§§ 122-123]
*
123. Change of base of logarithms. For purposes of numerical computation the most convenient system of logarithms is the system of Common Logarithms (or Briggs Logarithms), in which the base of logarithms is 10. For certain other purposes it is necessary to use logarithms with other bases. By means of the following formula, we can obtain logarithms to any base, b, if we have a table of logarithms to any other base, a: 1ogb N
(1)
= l~ga ~ = (logb a)(loga N). oga
Proof.
The proof of (I) consists in solving an exponential equation by the method of § 122. Let or x = Iogb N Therefore Hence
x
= loga N
Ioga b'
and the first part of formula (1) is established. Taking N
= a,
we have x
= Iogb a
and loga N
= loga a = 1.
1 x = Iogb a = - - . Ioga b
Therefore
Substituting logb a for 1/loga b in the first part of (1), we get
=
Iogb N
(Iogb a) loga N.
The number logb a is called the modulus of the system with base b with respect to the system with base a. By (1), this modulus is the number by which the logarithm of any number to the base a must be multiplied in order to find the logarithm of the same number to the base b. EXAMPLE
1. Given log10 8 = .9031, and log10 7 (a) Iog1 10,
SolutWn. (a) log1 10 (b) log1 8
=
log!o
7
and
= .8451, find
(b) log1 8.
= _~
= 1.1833. 8 51
= (log1 10) (log10 8) = (1.1833)(.9031) = 1.0686.
2. Find the modulus of the system of base 5 with respect to the system of base 10, and hence find log 5 9. EXAMPLE
Solution. logs 10 logs 9
=
log!o 5 = _6: 90 = 1.4306.
= (logs 10) (log 10 9) = (1.4306) (.9542) = 1.3651.
192
FIRST YEAR O:F COLLEGE MATHEMATICS [Ch. XIV]
*124. Natural logarithms. In problems involving differential and integral calculus, and in the computation of tables by the use of infinite series, the most useful system of logarithms is the natural or Napierian system. In this system the base is a certain irrational number, denoted by the letter e, which is approx~~~~tely equal to 2.71828.* The notation In N is now often lisect instead of loge N. The modulus of logarithms with base 10 with respect to the system with base e is denoted by M: M = log10 e = 0.4342945, and
1 v.l 1
=
loge 10
=
In 10
=
2.3025851.
By ( 1), § 123, to four decimal places,
loge N = In N = (2.3026) log10 N, and
log10 N
= (0.4343) loge N = (0.4343) In N. EXERCISES
Find the natural logarithms of the numbers given in Exercises 1 to 4. 1. 5.
2. 12.
3. 57.
4.
7r.
The natural logarithms given in Exercises 5 to 8 may be found by the use of infinite series. From them find the common logarithms of the same numbers, and compare with the values in Table II. 6. loge 13 = 2.5640.
6. In 3 = 1.0986.
7. loge 67 = 4.2047.
8. In 7 = 1.9459.
Find the following logarithms. 9. log4 17.
11. log7.4 5 .8.
12. log10.3 21.6.
* Napierian logarithms receive their name from the Scotchman, John Napier, Baron of Merchiston (1550-1617), the inventor of logarithms. The base e is closely related to the system of logarithms which he introduced, but it is not exactly the base of that system. The relation is as follows: If (LnN) denotes what Napier used as the logarithm of N, then (LnN) = 107 loge (107 /N). Napier published a partial account of his discoveries in 1614. The use of 10 as a base was suggested by Briggs who went from London to Edinburgh to confer with Napier. Briggs also constructed the first table of logarithms to the base 10.
CHAPTER XV
The Trigonotnetric Functions 125. Classification of functions. In § 27 we defined rational integral functions or polynomials, and later on we studied linear and quadratic functions, which are the simplest polynomials. A rational function is a function which can be expressed as the ratio of one polynomial to another. Algebraic functions, which include polynomials and the other rational functions and many other functions besides, are defined as follows: If F(x, y) is a polynomial in x and y, in general the equation F(x, y) = 0 can be solved for y in terms of x, yielding a solution y = f (x); then the function f (x) is an algebraic function. That is, a function y = f(x) is an algebraic function if, on substitution of f(x) for y, an equation of the form F(x, y) = 0 is satisfied, where F(x, y) is a polynomial. For example y = V x 2 + 1 is an algebraic function since it satisfies the equation x2 - y2 + 1 = 0. A function that is not an algebraic function is a transcendental function. The exponential function, y = ax, and the logarithmic function, y = Ioga x, are examples of transcendental functions. We shall now begin the study of a new class of transcendental functions, called the trigonometric functions. As we shall introduce them, these are functions of an angle, and at the outset we shall denote the independent variable by the Greek letter 0 (theta). Later on we shall wish to consider the trigonometric functions as functions of a number, which will be the measure of an angle. (See § 7.) Before defining the trigonometric functions of an angle, we must discuss the measurement of angles. 193
194
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XV
126. Degrees and radians. We shall need to be familiar with two systems of units for the measurement of angles. The first, which is the system of degrees and minutes, is the one which is used most commonly in applied problems involving the measurement of triangles. The second system, in which the unit is the radian, is the one which is most useful in calculus and in studying the properties of the trigonometric functions. If an angle is placed with its vertex at the center of a fixed circle, it subtends oil the circumference (that is, its sides cut off on the circumference) an arc whose length is proportional to the magnitude of the angle. This fact affords us the means of defining our units for measuring angles. Definition: A degree is an angle which, if its vertex is placed at the center of a circle, subtends on the circumference an arc equal to one 360th part of the circumference of the circle. A minute is a 60th part of a degree. Since the entire circumference is subtended by four right angles, four right angles contain 360 degrees, and one right angle contains 90 degrees. We use the symbol 20° 15' to denote "twenty degrees and fifteen minutes."
.__.__ _..A
Frn. 52
Definition: A radian is an angle which, if its vertex is placed at the center of a circle, subtends on the circumference an arc equal to the radius of the circle. In Figure 52, 0 is equal to one radian, since the arc AB and the radius OA are equal.
The circumference of any circle is equal to 2 1r times its radius. That is, the circumference contains 2 1r arcs each equal to the radius. Each such arc is subtended by a central angle of one radian, and the entire circumference is subtended by 2 1r radians. Therefore, 360° equals 2 1r radians, or
(1)
,,,- r.~cJJa.ns
= 180°.
Consequently
1 radian and, approximately,
(1!0)°,
§§ 126-127]
THE TRIGONOMETRIC FUNCTIONS
1 radian = (
(2)
_i!~
3
195
0
6
)
= 57.2958°
= 57° 17.75'.
On the other hand, 1° =
(3)
; radian = 1 0
3
6 }:~ radian
= 0.017453 radian (approx.).
When no other unit of angular measure is indicated, it is assumed that an angle is expressed in radian measure. 1. If we write 0 = 1r/2, we mean that 0 = 1r/2 radians or 180) 0 = go 0 ; if 0 = ¾1r, then 0 = ¾1r radians = (¾ · 180) 0 = 135°.
EXAMPLE
0=
(½ ·
2. To express 60° in radian measure, we note that 60 = and therefore 0 = ½1r; likewise, if 0 = 108°, in radian measure EXAMPLE
½• 180,
108 3 0 =1so·1r=51r.
EXERCISES
In Exercises 1 to 12, express each angle in degree measure. 1..
4 7r.
1
2.
3 'Ir.
2
3. 0.
4. ½1r.
5..
¼'Ir. -! 'Ir.
6•
2 'Ir. "5"
7.
'Ir •
8. ¾1r.
11. 2.
12. 2 'Ir.
9.
10. 1.
In Exercises 13 to 24, express each angle in radian measure.
go
0
13. 30°.
14. 45°.
15.
17. 135°.
18. 60°.
19. 150°.
20. 180°.
21. 360°.
22. 12°.
23. 23°.
24. 117°.
•
127. Central angles and their subtended arcs. If an angle equal to one radian is placed at the center of a circle of radius r, it subtends on the circumference an arc whose length is equal tor. Therefore a central angle equal to 0 radians subtends on the circumference an arc whose length is equal to 0r. If a denotes the length of the arc,
a= Or.
16. 120°.
FIG. 53
196
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XV
In this formula, we must remember, 0 is the number of radians in the central angle. When we are given any two of the three numbers a, r, and 0, the formula may be used to find the third number. 1. On a circle of radius 11 inches, find (a) the length of the arc that is subtended by a central angle of 72°, and (b) the central angle that subtends on the circumference an arc whose length is 15 inches. EXAMPLE
Solidion. (a) Here r = 11, and 0 = 72° = 2 1r/5 radians. arc length is 2 1r 5 · 11
a = 0r = (b) Here r
= 11, and
a
=
Therefore the
22 X 3 .142 . = 13.82. (mches) 5
= 15. Therefore, in radian measure,
a 15 0=; = = 1.36364. 11
(In degree measure 0 = 78° 7.8'.)
EXAMPLE 2. ·what must be the radius of a wheel, on the rim of which an arc 20 inches long is subtended by a central angle of 35°?
Solution. Here a = 20, and 0 = 35 1r/180.
a
r=-=
0
Therefore
20 X 180 3600 . = 32.74. (mches) X _ 35 1r 3 142 35
EXERCISES In the following examples, r represents the radius of a circle, 0 a central angle, and a the length of the subtended arc. In each case find the required number.
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
r
0
a
20 in. 273.2 in. 5280 ft. 1762 ft. 65 ft. 315.4 yds. 623.4 ft. required required 1640 ft.
3 radians 1r/3 radians 1r / 6 radians 41 ° 27' required in radians re quired in radians required in degrees 1.7 radians 15 degrees 20 degrees
required required required required 50 feet 700 yds. 585 ft. 160 ft. 120 ft. required
§§ 127-129]
THE TRIGONOMETRIC FUNCTIONS
197
11. A pendulum 35.24 inches long swings through an angle of 10° 15'. Find the length of the arc through which it swings. 12. A railroad curve is laid out on a circle 1750 feet in radius. Through what angle will it turn if the length of the track is 452 feet? 13. A Hawks Aircraft plane, Model HM-1, carried a propeller 126 inches in diameter. At 2375 R.P.M. at what speed was the tip of the propeller traveling around the axis in feet per second? 14. Find the radius of a pulley, if a belt traveling at the rate of 72 feet per second drives it at the rate of 450 revolutions per minute? 15.. The minute hand of a clock is 5 inches long. How far does its tip travel in 25 minutes? 16. An automobile has tires 27 inches in diameter. How many revolutions will a wheel make in one mile?
128. The generation of angles. If a half-line, l, extending from a point 0, revolves in a plane about O from the position OA to the position OC, it sweeps out or generates the angle AOC. The lines OA and OC are called the sides of the angle AOC, OA being the initial side, and OC being the terminal side of the angle. When there is no possibility 0f confusion, an angle may be denoted by the same single letter as its vertex, or by some different letter designated for the purpose. For example, we may say LAOC = LO = 0. If OB is a line lying between OA c and OC, in revolving from the position OA to OC, the generating line sweeps out first the angle AOB, then the angle BOC, and in all, the angle AOC. We can thus add an0 gles and write ~~~~~~~~--LAOB
+ LBOC =
LAOC.
129. General and directed anFIG. 54 gles. It has been customary in elementary geometry to consider all angles as positive and as limited to values less than 360°. These restrictions are not necessary, and it is now desirable to remove them. We shall call certain angles positive and others negative, the sign of an angle depending on the direction of rotation of the line which generated it. An angle which is generated by a line revolving in a counter-clockwise sense (that is, in the opposite direction from that
198
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XV
in which a hand of a clock moves) is counted positive. And an angle generated by a line turning in a clockwise direction is negative. In Figure 55, LAOC is positive and LCOA is negative. In fact LCOA
=-
LAOC.
We shall place no limit on the magnitude of an angle. During one complete counter-clockwise revolution a line generates all angles from 0° to 360°. But it may continue to turn. During its second revolution it completes the generation of angles between 360° and 720°; during the next revolution it finishes g~nerating the angles from 720° to 1080°, etc. Using radian a measure, we see that in its successive revolutions the line generates the 0 angles from 0 to 2 1r, from 2 1r to 4 1r, from 4 1r to 6 1r, etc. A Thus two lines, OA and OC, may Frn. 55 serve as the initial and terminal sides, respectively, not merely of a single angle AOC, but of countless angles, some positive and some negative. In Figure 56, for example, OA is the initial side and OC is the terminal side of each of the angles a = 30°,
/3 = -330°, 'Y = 390°, etc.
The value of an angle AOC, therefore, depends not only on the positions of the initial side OA and of the terminal side OC, but also on the direction and extent of rotation of the generating line.
A FIG. 56
Two angles are said to be coterminal, if, when their initial sides are the same, they have the same terminal sides. If two angles are coterminal, they differ by a whole multiple of 360° or of 2 1r radians; and if the difference between two angles is a whole multiple of 360° or 2 '71" radians, the angles are coterminal. Thus, in Figure 56, a, (3 and 'Y are coterminal, since a - (3 = 30° - (-330°) = 360°; a - 'Y = 30° - 390° = -360°; 'Y -
/3 = 390° - (-330°) = 720° = 2 X '360°.
§§ 129-130]
THE TRIGONOMETRIC FUNCTIONS
199
130. The standard position of an angle. An angle is said to be in standard position with reference to a system of coordinate-s, if its Y vertex is at the origin and if its initial side extends along the positive x-axis. We speak of an angle as being in, say, the second quadrant, if, when it is y placed in standard position, its terminal side lies 0 in the second quadranto We use similar expressions, M ro 0 of course, for the other . FIG. 57 quadrants.
--~-------llli-----------x
The angles 225°, 560°, -110°, ¾1r, -¾ 1r are all in the third quadrant, while 400° and ½1r are in the first, ¾1r and -210° are in the second, and - ¼1r and 300° are in the fourth quadrant. An angle 0 is said to be acute, if 0 < 0 < ½1r ( or 0° < 0 < 90°); acute angles are in the first quadrant. Angles such as 90°, 1r, ¾1r, 360°, whose terminal sides lie on a coordinate axis when the angles are in standard position, are not in any quadrant, but are said to be quadrant angles.
Consider an angle in standard position, and let P:(x, y) be any point, other than the origin, on the terminal side of the angle. The distance, r, of P from the origin is called the radius vector of P, and, for present purposes, will always be counted as positive. Then, by § 19,
EXERCISES Using coordinate paper, construct a set of coordinate axes. For each of the following angles, (a) draw the angle in standard position, indicating by an arrow the direction and extent of rotation of the generating line; (b) choose a point on the terminal side and find its coordinates and radius vector; (c) name two co terminal angles. 1. 30°.
6. -300°. 1
11.
41r.
16.
-! 71".
2. 90°.
7. -630°. 12. j 71". 17. -2 71".
3. 135°. 8. -270°. 13.
½71".
18.
-1r.
4. 180°. 9. -45°. 14. 19.
-½ 71". -¾ 7T.
5. -240°. 10. 150°. 15. 20.
i 1r.
200
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XV
131. Definition of the trigonometric functions of an angle. For any angle, 0, we shall now define six quantities, called trigonometric y
y
8
---+--------x 0 x M
__..______,____ x M
x
0
FIG. 58
functions of 0. They are the sine, the cosine, the tangent, the cotangent, the secant, and the cosecant of 0, and will be written sin 0, cos 0, tan 0, cot 0, sec 0, and csc 0, respectively. The figure involved in the definitions is drawn for 0 in the first and also in the second quadrant. The student should draw it for angles in the third and fourth quadrants. Fundamental definition: Given any angle, 0, place it in standard position with reference to a coordinate system, and choose any point, P, other than the origin, on the terminal side of 0. Let the coordinates of this point be (x, y) and its distance from the origin be r. Then sin 8 =
ordinate of P = M., radius vector of P r abscissa of P x cos 8 radius vector of P 1 ordinate of P y tan fJ = = -, abscissa of P x abscissa of P X = -, cot 8 = y ordinate of P radius vector of P r sec 8 = - - - - - - - = -, abscissa of P x
=
= -,
csc 8 = radius vector of P = !. ordinate of P y
THE TRIGONOMETRIC FUNCTIONS
§ 131]
201
These quantities, although defined in terms of the coordinates of the point P, are nevertheless true functions of 0. That is, their values depend solely on the value of 0, and not on the particular point P which we choose on the terminal side of 0. For if we choose any other point, P', on OP, with coordinates y (x', y') and radius vector r',
from the similarity of the triangles OMP and OM'P', we have y -y' = -,
r'
y'
r
so that the value of sin 0 is the same whether we obtain it using P' or using P. The student should write the corresponding equations for the other functions.
y
MM' -----fl-------------------x 0
~'
Fm. 59 Consequences of the defini"tions. I. Since the values of the trigonometric functions of an angle depend only on the terminal side of the angle and not on the direction or number of rotations of the generating line, we have: If two angles are coterminal, the trigonometric functions of one are equal to the same functions of the other. Thus, since 127°, 487°, and -233° are co terminal angles, sin 127° = sin 487° = sin (-233°), cos 127° = cos 487° = cos (-233°).
II. A ratio is positive if numerator and denominator have the same sign; it is negative if numerator and denominator have opposite signs. The radius vector r of the point 'P:(x, y) is always positive. Therefore sin 0 and csc 0 have the same sign as y; cos 0 and sec 0 have the same sign as x; tan 0 and cot 0 are positive when x and y have the same sign, and are negative when x and y have opposite signs. Therefore: If 0 is in quadrant I, all six trigonometric functions are positive. If 0 is in II, sin 0 and csc 0 are positive; cos 0, tan 0, cot 0, and sec 0 are negative. If 0 is in III, tan 0 and cot 0 are positive; sin 0, cos 0, sec 0, and csc 0 are negative. If 0 is in IV, cos 0 and sec 0 are positive; sin 0, tan 0, cot 0, and csc 0 are negative.
202
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XV
III. Since sin 0 = y/r and csc 0 = r/y, and since (y/r) • (r/y) = 1, sin 0 and csc 0 are reciprocals each of the other. Likewise cos 0 = x/r and sec 0 = r/x are reciprocals as are also tan 0 = y/x and cot 0 = x/y. That is, (1)
csc 8
EXAMPLE
=
1
-=--o' Slll
secO
1 =--, cos (J
cotO
1 = tan - - (J·
Find the trigonometric functions of an angle 0, if the point y P: ( - 4, 3) lies on the terminal side of 0 when 0 is in standard position.
1.
Solution. The radius vector of P is
v
3
r = V x2 + y 2 = (-4) 2 Therefore, by definition,
()
--------------x M -4 0
sin 0
= 5.
4 5
= - -,
X -4 4 cot 0 = - = = - -, y 3 3
tan 0 = Y. = _!__ = - ~, X -4 4 X
2
U = ~, r 5
X -4 cos 0 = - = r 5
FIG. 60
r sec 0 = -
=
+3
5 5 = - -, -4 4
r 5 csc 0 = - = -· y 3
=-
The last three of these values might have been obtained from the first three by means of the reciprocal relations. EXAMPLE
Y
2. Find the trigonometric functions of 270°.
Solution. Place the angle in standard position, and as P choose a convenient point, say (0, -8). Then r = +8, since r is always counted positive. Then, from the definitions, sin270° = 'l!. = -
r
8 8
= -1
'
cos270° =
~ = Q= r
8
8
-----+--+-----x 0
0.
P (0, -8)
If we try to find tan 0, we are tempted to write
tan0 = 'v_ = X
FIG. 61
-s.
O'
but, since division by zero has no meaning, evidently there is no such thing as tan 270°. Likewise sec 270° does not exist, since r/ x or 8/0 is without meaning. But cot 270o = -X = - O = 0, y
-8
r 8 csc 27O0 = - = = -1. y
-8
§§ 131-132]
THE TRIGON01\1ETR1C FUNCTIONS
203
EXERCISES Find the value of r for each of the following positions of P, and then find the trigonometric functions of the angle XOP. Do not reduce the answers to decimal form. 1. (-4, 3).
2. (5, -12).
3. (-5, -12).
4. (6, -8).
5. (8, 15).
6. (2, 2).
7. (-2, 2).
8. (1, 1). 11. (15, -8). 14. (3, 4).
10. (-3, 0). 13. (-3, -3). 16. For which quadrants is (a) sin 0 positive; (c) cos 0 positive; (e) tan 0 positive;
9. (O, 2). 12. (-8, -6). 15. (-12, 5). (b) sin 0 negative; (d) cos 0 negative; (f) tan 0 negative?
17. In each case assume that both functions exist. (a) ,vhy do sin 0 and csc 0 always have the same sign? (b) What other function always has the
same sign as tan 0? cos 8?
(c) What other function always has the same sign as
In each of the Exercises 18 to 45, using coordinate paper and a protractor, find- the values of the sine, cosine, tangent, and cotangent of the given angle, when they exist, correctly to one decimal place in every case and correctly to two decimal places when possible. 18. 32°.
19. 65°.
20. 117°.
21. 153°.
22. 217°. 26. 180°.
23. 233°. 27. 0°.
24. 294°. 28. 90°.
25. 331 °. 29. -90°.
30. -45°.
31. -225°.
32. -216°.
33. -291 °.
34. 578°.
35. 674°.
36. 758°.
37. -650°.
38.
39. 3 7r.
40.
¾7r.
44.
7r.
42. -fr
7r.
43.
¼7r. %7r.
41. j 45.
7r.
1\ 7r.
Given that sin 30° = 0.50, cos 30° = 0.87, tan 30° = 0.58, cot 30° = 1.73, from a figure find the values of the same functions of the following angles. 46. 210°. 47. -30°. 48,. 150°. 49. 330°.
132. The determination of one function from another. If we know the quadrant in which an angle lies, and if we know the value of one of its trigonometric functions, we can find the values of its other trigonometric functions. Formulas by which this can be done algebraically will be given in § 165. The geometrical method is illustrated in the following examples.
204
FIRST YEAR OF COLLEGE MATHEMATICS
EXAMPLE
[Ch. XV
Given that 0 is an angle in the second quadrant and that
1.
tan 0 = -
5 1 2,
find the other functions of 0. Solution. Any point (x, y) in the second
y
quadrant is a point on the terminal side of 0 and can be taken as the point P of the fundamental definition, if
0 /t,,f
-12
0
'JI..
= tan 0 = - !__ 12
X
In particular, the point (-12, 5) is such a point. The point (-24, 10) or the point ( - 6, t) would serve equally well, but we choose as P the point ( -12, 5) as the simplest. Then Fm. 62
r = v144 . 0
Therefore
Sill
y
5
r
13
= - = -,
12 cot 0 = - -, 5
+ 25 =
13 . 12
X
cos 0 = - = - -, r 13
13 sec 0 = - -, 12
13 csc 0 = -. 5
EXAMPLE 2. Given that sin 0 = -land that 0 is in the third quadrant, find the other functions of 0.
y
Solution. The terminal side of 0 must
0
pass through the point in the third quadrant for which y = -3 and r = 7. For this point, from the relation x2 + y2 = r2 , we have
x = -
-vr
2 -
y2
= - V 49 - 9 Frn. 63
= -v40 = -2v10. Therefore
cos 0 =
2v10 cot0 = - - , 3
-2vI6 , 7
-3
tan 0 = -2-vlQ 7
3 =
7-VlO
sec 0 = - - - = - - - , 2v10 20
3-VIO
2v10 = 2()'
7 csc 0 = - -, 3
In general the method consists of finding from the given conditions a point on the terminal side of the angle, and then using the coordinates and radius vector of this point in the definitions of the functions of an angle in order to obtain their values.
THE TRIGONOMETRIC FUNCTIONS
§ 132]
205
EXERCISES In each of the Exercises 1 to 16, draw the figure and find the remaining trigonometric functions of an angle 0 that satisfies the given conditions. 1. tan 0
= !, 0 in first quadrant.
2. cot 0 = 3. sin 0 = 4. cos 0 = 6. sin 0 = 6. tan 0 = 7. sin 0 = 8. tan 0 = 9. sin 0 = 10. cos 0 =
-!-5 , 0 in first quadrant. ¾, 0 in second quadrant. ¼, 0 in fourth quadrant.
-½, 0 in fourth quadrant. 3, 0 in third quadrant. .61, 0 in second quadrant. -1.59, 0 in fourth quadrant. -¼, cos 0 negative. /a, tan 0 negative.
11. tan 0 = -V3, sin 0 negative. 12. tan 0 = -I, sin 0 positive. 13. sec 0 = -2, sin 0 negative. 14. csc 0 = - \3, tan 0 positive. 15. cos 0 = ¾, sin 0 negative. 16. cot 0 = ¾, sec 0 negative. In each of the Exercises 17 to 18, using coordinate paper and a protractor, find the value of an angle 0 that satisfies the given conditions. 17. 0 is in the second quadrant, and sin 0 is approximately equal to 0.2. 18. 0 is in the third quadrant, and cos 0 is approximately equal to -0.3.
CHAPTER XVI
The Trigonotnetric Functions of an Acute Angle 133. The trigonometric functions of an acute angle. Consider a right triangle, ABC, in which the right angle is at C. Let a denote the angle at the vertex A, and /3 the angle at the vertex B. Denote by a, b, and c the lengths of the sides which are opposite the vertices A, B, and C respectively. Note: This notation will be used regularly. Unless it is otherwise stated, ln a triangle whose vertices are the points A, B, C, the letters a, /3, 'Y will denote the angles at those vertices, and a, b, c will denote the lengths of the respectively opposite sides. In a right triangle, 'Y will ordinarily be the right angle.
The six trigonometric functions of the acute angle a are then B
.
a
sin a=-, C
a
a
tan a= b'
b
cos a=-, C
b a
cot a=-,
90° .
A-------------~\ C
sec a=
b Frn. 64
C
b'
C
csc a=-· a
These statements may be expressed as follows: For an acute angle of any rig ht triangle, the the the the the the
sine equals the ratio of the opposite side to the hypotenuse; cosine equals the ratio of the adjacent side to the hypotenuse; tangent equals the ratio of the opposite side to the adjacent side; cotangent equals the ratio of the adjacent side to the opposite side; secant equals the ratio of the hypotenuse to the adjacent side; cosecant equals the ratio of the hypotenuse to the opposite side.
Draw a set of coordinate axes with the origin at A and the positive x-axis in the direction AC, placing angle a in standard Proof.
206
tJ§ 133-135]
FUNCTIONS OF AN ACUTE ANGLE
position. Then B is a point on the terminal side of a. The coordinates of B are (b, a), and its radius vector is c. Therefore, by § 131, .
Sln a
X b cos a= - = -,
a = Y- = -, r C
r
207
y B (b,a)
etc.
C
The relations of this article should be remembered, also, in the following forms:
= c sin a,
a = b tan a,
c = r cos a, c = a csc a,
b = a cot a, c = b sec a.
a
Frn. 65
134. The variation of the functions of an acute angle. For a hypotenuse, c, of given length, the larger the angle a, the larger is the value of a and the smaller is the value of b. Thus, in the two right triangles ACB and AC' B' of Figure 66, the hypotenuse c' is equal to the hypotenuse c. But since angle a' is larger than angle a, side a' is longer than side a, while b' is shorter than b. Therefore, by the relations of § 133, when an acute angle b
increases, its sine, tangent, and secant increase, and its cosine, cotangent, and cosecant decrease. Since a and b are always less than c, the sine and cosine of an acute angle are always less than 1 and the secant and cosecant are always greater than 1. The tangent and cotangent have no such fixed limits.
Ar b' C' C Frn. 66
135. The functions of complementary angles. If we apply the statements of § 133 to the angle {3 instead of to a, since a is now the adjacent side and b is the opposite side, we find that .
Sln -{3
b
=- = C
a
COS
a
'
cot {3 = - = tan a b '
a . cos {3 = - = sm a C
b tan {3 = -
a
'
C
sec {3 = - = csc a
a
'
csc {3 =
= cot a '
bC =
sec a.
The sine and cosine are said to be cofunctions, each of the other, as are also the tangent and cotangent, and the secant and cosecant.
208
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XVI
Two angles are said to be complementary, if their sum is equal to 90°. Inasmuch as any two complementary acute angles can be taken as the acute angles, a and {3, of a right triangle, we have the Theorem. Any trigonometric function of an acute angle is equal to the corresponding coB function of the complementary angle.
{'v
C,
~
a==l
450 900
A
C
b=l FIG. 67
136. The functions of 45°. Draw a right triangle with the acute angle, a, equal to 45°, and the length of the opposite side a equal to one unit. Then we also have {3 = 45° and b = 1. Since a2
+b
2
= c2, c =
. 0 a 1 1 _ 1sin 45 = - = = -v 2
V2
C
2
'
tan 45° = ~ = 1 b '
vi+!
b 1_ cos 45 0 = - = - v 12
cot 45° =
sec45° = £. = V2 b 1 =
V2 ,
= V2, and
C
2
£=
1
a
C
csc 45° = -a
'
'
=V22 .
137. The functions of 60° and 30°. Draw an equilateral triangle, ABD, making the length of each side equal to 2 units. From B draw BC perpendicular to AD. It will bisect AD at C. Then in the right triangle ABC, a = 60°, {3 = 30°, c = 2, b = 1. Moreover, a 2 + b2 = c2, so that a= Vc2 - b2 = V4--=-i = V3. From§ 131, . 600 = a 1 - ;3s1n - = -v C
2
B
'
cos 60° =
~ = !, C
2
tan 60° =
ba =
- ;v 3,
cot 60° =
~ = - 1- = ! V3 a V3 3 '
sec 60°
= bQ =
csc 60°
= a2 = V3 -3_
2
A---------_..----~D b=l
'
va. 3
=~
C
Fm. 68
§§ 135-138]
FUNCTIONS OF AN ACUTE ANGLE
209
Functions of 30° are equal to the corresponding cofunctions of 60°, which is the complementary angle. Therefore
½, tan 30° = ½V3, sec 30° = ¾V3, sin 30° =
½V3, = V3,
cos 30° = cot 30°
csc 30° = 2.
Since the values of the functions of 0°, 30°, 45°, 60°, and 90° will be required over and over again during this course, the student should memorize the following table at once:
oo
30°
45°
60°
90°
(J
0
2
1
½v'2
½v'3
1
cos 8
1
½V3
½v'2
2
tan 8
0
½v'3
1
v'3
v'3
1
½v'3
8
sin
cot 0
-
1
0
0
138. A four-place table of trigonometric functions. In Table III, pages 666-670, are columns headed "Sine," "Tangent," "Cotangent," "Cosine" with sub-heading "Value." These give the values of these four functions to four decimal places for acute angles at intervals of 10'. The secant and cosecant, if needed, can be found as reciprocals of the cosine and sine. There are two problems involved in the use of the tables: (a) given an angle, to find the corresponding value of a function, and (b) given the value of a function, to find the angle to which it corresponds. The methods are illustrated in the following examples. 1. Find sin 29° 40'. Solution. On page 669, in the column headed "Degrees," we find the entry 29° 40', and opposite this entry, in the column headed "Sine," we find the number .4950. We therefore write sin 29° 40' = .4950. In the same line, in the other columns, we find tan 29° 40' = .5696, cot 29° 40' = 1. 7556, cos 29° 40' = .8689. ExAMPLE 2. Find sin 65° 20'. Solution. In the column marked "Degrees" at the top the entries go only to 45°. But in the column at the right, marked "Degrees" at the bottom, the EXAMPLE
210
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XVI
entries read upward from 45° to 90°. In fact, for any row in the table, the angle at the left and the angle at the right are complementary angles, and consequently an entry in any column gives one function of the angle marked at the left and the corresponding cofunction of the complementary angle which is marked at the right. By this arrangement each entry of the table is made to do double service, with consequent economy in printing and using the table. In seeking a function of an angle entered in the column at the right, we must use the designation of the function which is printed at the bottom. The entry for 65° 20' is on page 668, and we find sin 65° 20' = .9088. Likewise, tan 65° 20' = 2.1775, cot 65° 20' = .4592, cos 65° 20' = .4173.
When we wish to find a number which is not given in the tables, but lies between two entries of a table, we resort to the method of interpolation. The principles of interpolation with these tables are the same as those explained in § 118, but they are explained and illustrated independently in the following examples. EXAMPLE
3.
Find sin 24° 38'.
Solution. The angle 24° 38' lies between the two entries 24° 30' and 24° 40'. The difference between two successive numbers in the body of the table is called the tabular difference. From the table we find
sin 24° 40' = .4173 sin 24° 30' = .4147 Tabular Difference = .0026
It is not correct to say that the sine of an angle is exactly proportional to the angle. But for small changes in the angle, the change in the sine is nearly proportional to the change in the angle. And we may assume that since 24° 38' is "eight-tenths of the way" from 24° 30' to 24° 40' then sin 24° 38' is also "eight-tenths of the way" from .4147 to .4173. We must therefore add to .4147 eight-tenths of the tabular difference .0026. That is sin 24° 38' = .4147
+ (.8 X .0026)
= .4147
+ .00208 =
.41678.
However, the tables themselves are accurate only to four decimal places. We should be wrong to expect greater accuracy in our interpolated result. We therefore round off our number to four places, and write sin 24° 38' = .4168. The work of interpolation should be done mentally about as follows: "Sin 24° 30' = .4147; the tabular difference is 26 (in the last places); .8 of 26 is 20 .8, which we call 21; this number, added to .4147 gives sin 24° 38' = .4168." Only the final result is to be written down.
§§ 138-139]
FUNCTIONS OF AN ACUTE ANGLE
211
The student should verify the following results, remembering that the tangent increases and the cotangent decreases when the angle increases: tan 24° 38' = .4585, EXAMPLE
3.
cot 24° 38' = 2.1809.
Find cot 59° 45'.
Solution. From the table cot 59° 40' = .5851. The tabular difference is 39,, and five-tenths of the tabular difference is 19.5. This number is equally near to 19 and to 20, and there is no logical reason for choosing one in preference to the other. It is well, however, to have some fixed practice, and it is the custom to round off .5 so that the result written down is the even rather than the odd possible number. We shall adhere to this rule throughout the book. In the present case, if we round off 19.5 to 19, the result is .5832, which since it ends in an even number, is accepted as the result. It is the result rather than the subtracted difference (19 or 20), which is to be even, and the rule is applicable only in case of logical doubt, when it is .5 which is to be rounded off.
EXERCISES From Table III find the sine, cosine, tangent, and cotangent of each of the following angles. 1. 13° 20'. 5. 52° 50'. 9. 16° 10'.
13. 22° 18'. 17. 49° 28'. 21. 32° 35'.
2. 3° 0'. 6. 67° 30'. 10. 14. 18. 22.
62° 40'. 36° 36'. 57° 2'. 49° 3'.
3. 7. 11. 16. 19. 23.
4. 41 ° 10'.
27° 50'. 55° 0'. 44° 50'. 40° 54'. 63° 16'. 60° 15'.
8. 12. 16. 20. 24.
73° 40'. 84° 20'. 14° 8'. 81 ° 58'. 76° 25'.
From Table III, find the values of the following functions. 26. sin 56° 35·_ 28. tan 23° 49'. 31. cos 51 ° 35'.
26. sin 29° 45'. 29. cot 68° 2'. 32. tan 75° 45'.
27. cos 15° 27'. 30. cot 32° 36'. 33. tan 56° 48'.
139. The angle corresponding to a given function.
If
sin 0 = N, it is customary to call the angle 0 "an inverse sine of N,'' or "an arcsine of N," and to write
8 = arcsin N. This may be read in either of the two ways just mentioned, or simply "0 is an angle whose sine is N." In a similar way, if cos 0 = N, then
212
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XVI
() = arccos N. And arctan N, arccot N, arcsec N, arccsc N have A more detailed discussion of these inverse trigonometric functions will be given in Chapter XXI. For the present we shall always suppose that arcsin N, arccos N, or any other one of the inverse trigonometric functions of N is an angle between 0° and 90°, inclusive. In the following examples we shall learn how to find the values of these functions from Table III. similar meanings.
Given sin 0 = .4924, find 0. Solution. Looking in. the body of the table in the column headed "Sine,'' we find .4924 on page 669 opposite the angle 29° 30'. We therefore write () = 29° 30'. EXAMPLE
1.
2. Find arctan 2.0809. Solution. This is equivalent to finding the angle whose tangent is 2.080V. Since 2.0809 is greater than 1, and the tangent increases when the angle increases, we know that arctan 2.0809 is greater than 45° (since tan 45° = 1). We therefore seek 2.0809 in the column marked "tangent" at the bottom of the page. We find it on page 668 opposite the angle 64° 20' in the right-hand column. We therefore write arctan 2.0809 = 64° 20'. EXAMPLE
3. Find 0 = arcsin .5367. Solution. This entry does not appear in the tables, but is between EXAMPLE
.5348 = sin 32° 20'
and
.5373 = sin 32° 30'.
We can arrange the numbers as follows. Angle 32° 20' ()
32° 30'
Difference
}?
l
10'
Sine .5348 .5367 .5373
Difference } 19
l
25
Therefore, sin 0 is "nineteen twenty-fifths of the way" from sin 32° 20' to sin 32° 30', and we conclude that 0 is "nineteen twenty-fifths of the way" from 32° 20' to 32° 30'. Using decimals, we get 19/25 = .76, which we call .8. To 32° 20' we therefore add .8 X 10' = 8', and get 0 = 32° 28'. \ 4. Find arccos .6263. \ Solution. This column appears between the entries for cos 51 ° 10' and cos 51 ° 20'. The tabular difference is 23; our difference is (6271 - 6263) = 8. The proportional part is 8/23 = .348, which we call .3, and the angle is arccos .6263 = 51 ° 13'. EXAMPLE
FUNCTIONS OF AN ACUTE ANGLE
§§ 139-140]
213
EXERCISES From Table III find an angle corresponding to each of the following functions, giving its value correctly to the nearest minute.
= cos 0 = cot 0 = sin 0 = tan 0 =
1. sin 0 4. 7. 10. 13.
= .7698. 5. tan 0 = .3153.
3. cos 0 = .8480.
8. cot 0 = .8098. 11. cos 0 = .2431.
= .5653. 12. cos 0 = .9680. 15. cot 0 = 3.5782.
2. sin 0
.1679. .4173. I.7556. .8958. 1.1380.
14. tan 0 = .3003.
6. tan 0 = 3.2041. 9. sin 0
From Table III find the values of the following angles, correctly to the nearest minute.
17. arccos .6197. 19. arcsin .4002. 21. arccot .6562. 23. arctan .2425.
16. arcsin .8124. 18. arccot 1.6283. 20. arctan 1.4796. 22. arc cos .8330.
From Table III find the values of the following functions correctly to four decimal places. (Suggestion: First find the angle correctly to the nearest minute, and then find the value of the required function.) 24. cos(arcsin .5398).
25. tan(arccos .6604).
26. sin(arctan .8010).
27. cot(arcsin .8607).
28. cot(arctan 1.6230).
29. tan(arcsin .3840). 31. cos (arc tan 1. 0623).
30. sin(arccot .5460).
Table III (of TABLES) is similar to Table III of this text, except that angles are given for every minute and the corresponding values of the functions are given to five places. It is proper, with this table, to interpolate to or from an angle expressed correctly to the nearest tenth of a minute.
*140. A five-place table of trigonometric functions.
Find sin 27° 23.8'. Solution. On page 81 of the tables, we find EXAMPLE
1.
sin 27° 23' = .45994 sin 27° 24' = .46020 Tabular difference = 26 Then .8 X 26 = 20.8, which we call 21. tain sin 27° 23.8' = .46015.
We add this to sin 27° 23', and ob-
FIRST YEAR OF COLLEGE MATHEMATICS
214 EXAMPLE
2.
[Ch. XVI
Find tan 58° 43.5'.
Solution. The nearest entries are on page 83. The tabular difference is 11; .5 X 11 = 5.5; we call this 6 rather than 5, so that the number written down will be even. We add 6 in the last place to 1.6458, and find tan 58° 43.5' = 1.6464.
Find 0, given cot 0 = .56094.
EXAMPLE
3.
Solution:
cot 60° 42' = .56117 cot 60° 43' = .56079 Tabular difference = 38
cot 60° 42' = .56117 cot 0 = .56094 Our difference = 23
Proportional part = 23/38 or approximately .6. Therefore 0 = 60° 42.6'. EXAMPLE
4.
Find arccos .90180.
Solution. Frorn page 80, the tabular difference = 12; our difference is 3. The proportional part is 3/12 = .25, which we round off to .2. Therefore
0 = 25° 36.2'. EXERCISES
Frorn Table III of Tables find the sine, cosine, tangent, and cotangent of each of the following angles. 1. 17° 24'.
2. 38° 49'.
3. 45° 17'.
4. 78° 53'.
5. 25° 10'.
6. 61 ° 33'.
7. 42° 27.4'.
8. 36° 46.8'.
9. 53° 26.2'.
10. 11 ° 41.5'.
11. 66° 6.3'.
12. 64° 28.7'.
13. 81 ° 0.9'.
14. 46° 57.7'.
15. 32° 0.4'.
In each of the following examples a function of an angle 0 is given. 8 correctly to the nearest tenth of a minute.
Find
16. tan 0 = .82531.
17. cos 0 = .73373.
19. cot 0 = .53620.
20. sin 0
.39780.
21. cot 0 = 1.39210
22. cos 0 = .72683.
23.
.57248.
24. cot 0 = .83345.
25. cos fJ
26. 29.
.737 52. .68679.
27. tan 0
28.
= .51098. tan 8 = 2.5331.
31. sin 0 = .83997.
32.
34. cot 0
= .80950.
= tan 0 = sin 0 = cot 0 = sin 0 =
35. cos 0
=
.67423. .82891.
18. sin 0 = .79193.
= 30. cos 0 = 33. tan 0 = 36. cos 0 =
1.7764. .63602. .82031. .460 rn.
Find the following angles. 37. arctan 1.9070.
38. arccos .53910.
39. arcsin .70062.
40. arccot 1.0500.
41. arc cos .90000.
42. arctan 1.5000.
215
FUNCTIONS OF AN ACUTE ANGLE
§§ 140-141]
B 141. The solution of a right triangle. Besides its right angle, a right triangle possesses five parts, which are its two acute angles and a its three sides. If two of these parts are known, at least one of them be90° ing a side, it is possible to determine A-------b_________. C the other three parts. The determiFIG. 69 nation of the unknown parts constitutes a solution of the triangle. For the solution of the right triangle ABC, we have the following formulas. (1) a2
+b
2
(3) sm a
= c2,
(2) a
a
= - = cos {3 C
(5) tan a=
'
i = cotP,
+ {3 =
(4) cos a
90°,
b
.
= - = sm C
{3
. '
b (6) cot a = - = tan (3. a
In order to determine an unknown part, select and use one of the formulas which involves that part and no other unknown. If there is a choice between two formulas, when possible choose the one that involves multiplication rather than division. Answers should, in general, be given to the same degree of accuracy as the data. It may be assumed that three-place accuracy in the sides corresponds to accuracy to the nearest 10' in the angles, four-place accuracy in the sides to accuracy to the nearest minute in the angles, and fiveplace accuracy in the sides to accuracy to the nearest tenth of a minute in the angles. As a check against gross errors draw the figure approximately to scale and compare the results with the figure. Methods of checking more closely will be explained in connection with the solutions of triangles by logarithms. EXAMPLE
1.
Solve the right triangle ABC, given c = 32.47, a = 36° 40'.
Solution. Frorn formula (2),
/3 = 90° - 36° 40' = 53° 20'.
Frorn (3),
a = c • sin a = 32.~7 X sin 36° 40'
Frorn (4),
b = c · cos a = 32.47 X cos 36° 40' = 32.47 X .8021
=
32.47 X .5972
=
19.39.
= 26.04.
216
FIRST YEAR OF COLLEGE MATHEMATICS
EXAMPLE
2.
Solve the right triangle ABC, given a
Solution. From formula (5), tan a =
~
=
!!:! =
=
.5469.
[Ch. XVI
13.4, b
=
24.5.
Therefore, by
Table III, to the nearest 10', a = 28° 40', (3 = 90° - a = 61 ° 20'. formula (3),
By
a 13.4 13.4 27 9 · · c = sm a = sin 28° 40' = .4797 = EXAMPLE
3.
Solve the right triangle ABC, given a = 26.93, (3 = 54° 17'.
Solution. By formula (2), a = 35° 43'. By (6),
b = a cot a = 26.93 X cot 35° 43' = 26.93 X 1.3908 = 37.45.
By (3),
a c = sin a
26.93
= sin 35° 43'
=
26.93 .5838 = 46 ·13 ·
EXERCISES Use either four-place or five-place tables according to the directions of the teacher. Solve the following right triangles, given:
= 18.3, a = 27° 50'. 3. a = 567, b = 432. 5. C = 42.59, (3 = 26° 17'.
1.
7. 9. · 11. 13.
C
b = a= a = a =
2444, c = 3169. 7.332, a= 34° 16'. .03462, b = .02601. 2009, {3 = 23° 6'.
2o C = 361, (3 = 39° 40'. 4. b = 275, c = 368. 6. C = 155.7, a = 49° 33'. B. a = 6.003, c = 8.314. 10. b = 8135, a = 29° 11'. 12. a = 567.3, b = 430.2. 14. C = 1125, {3 = 38° 27'.
Solve the following right triangles. using five-place tables. 15. C = 201.03, a = 43° 16.7'. 17. b = 1.0135, a = 26° 41.4'.
16. a = 4396.4, b = 5011.3. 18. a = 52149, C = 76301.
142. Angles of elevation and depression. Suppose that an observer is at a point 0, and a certain object is at a point P. The line OP is then called the line of sight from the observer to the object. Let OH be drawn horizontally in the vertical plane through OP. Then the angle HOP, which the line of sight makes with the horizontal, is called the angle of elevation of the object P from O if P is higher than 0, and it is called the angle of depression of P from 0, if P is lower than 0
§§ 141-142]
FUNCTIONS OF AN ACUTE ANGLE
P
217
o----------H
0-------------H
p
FIG. 70
EXERCISES In each exercise draw the figure. The arithmetic may be performed either· with or without logarithms. Use four-place tables. 1. From a point on the ground 316 feet from the base of a cliff, the angle of elevation of the top is 26° 40'. How high is the cliff? 2. One of the towers of the Cathedral of Chartres is 375 feet high. How long a horizontal shadow will it cast when the angle of elevation of the sun is 28° 30'? 3. How high is a flagpole, if it casts a shadow 80 feet long when the angle of elevation of the sun is 32° 45'? 4. Find the angle of elevation of the sun, if a man 6 feet tall casts a shadow 10 feet long. 6. An airplane climbed 1250 feet while making 3 miles of airline distance. Find its angle of climb. 6. When the angle of elevation of the sun is 21 °, the shadow of the Great Pyramid reaches to a point about 1250 feet from the center of the square base. How high is the pyramid?
7. An observer in a helicopter 1200 feet above the ground observes the angle of depression of an enemy battery to be 6° 7'. ·what is the horizontal distance to the battery? 8. How high is a kite, when the length of the string is 580 feet and it makes an angle of 26° with the horizontal? 9. The height of the Great square base is about 760 feet. of an edge, that is, the angles intersections of the faces make
Pyramid is about 480 feet, and a side of the Fjnd the angles of inclination of a face and that one of the sloping faces and one of the with the horizontal.
10. In or 1. a
This indicates the impossibility of satisfying the given conditions, since for no value of {3 is sin {3 greater than 1. (iv) There remains the case in which a~ b. In this case the triangle always exists and is uniquely deterC mined. For the arc cuts the line AD in two points B and B' which lie on opposite sides of A, if a > b. And, D of the two triangles, ABC B' and AB'C, one contains the given angle a; the other contains the supplement of a but not a itself. If a is FIG. 105 just equal to b, B' falls at A, and only one triangle is possible. The triangle in this case is isosceles.
We can say, then, that: i
f a < b, and i f
b
·
sm a = 1, the triang · le exists, · · unique, · -ais an d
~s a right triangle;
(ii) if a
< b, and if b · 81:
a
< 1,
there exist two triangles satisfy-
ing the conditions;
(iii) if b · sm a > 1, no triangle exists satisfying the conditions; a (iv) if a > b, just one triangle exists satisfying the given conditions.
FIRST YEAR OF COLLEGE MATHEMATICS
282
Since log 1
= 0, in the solution by logarithms, when a
0, no triangle exists. EXAMPLE 1. In the four parts of this example, side b and angle a retain constant values, but the value of a is chosen to illustrate the various possible cases.
(i) Solve the triangle ABC, given a = 4, b = 8, a = 30°. C
a,
Solution. Since the side opposite the given angle is less than the side adjacent to that angle, we must examine sin {3 to determine whether a solution exists and, if it does exist, the number of possible solutions. By the law of sines, . sin a 1 sm{3=--·b=--X8=1 a 2 X 4 '
C
Fm. 106
so that {3 = 90°, and the triangle is unique. Evidently 'Y = 60°, and c
= b cos a = .866 X 8 = 6.9.
C
(ii) Solve the triangle ABC, given a = 6, b = 8, a = 30°. Solution. By the law of sines, .
sma
sm{3=--·b a
1
Since sin {3
< 1,
3 = .667.
there are two solutions: {3
= 42°,
Also
{3' = 180° - 42° = 138°.
+ 42°) = 108°' (30° + 138°) = 12°.
'Y = 180° - (30° ')'
And
FIG. 107
2
= 2X6X8=
c
1
= 180° -
=-/!· sin 'Y = 2 X 6 X .95 = sma
c' =
-/!- • sin 'Y' =
sma
11.4,
2 X 6 X .21 = 2.5.
THE OBLIQUE TRIANGLE
§ 183]
283
(iii) Solve the triangle ABC, given a = 3, b = 8, a = 30°. B h f . . {3 sin a b . Solution. y t e law o smes, sm = -a- · =
2
1 X
3
X 8 = 4.
3
Since this is greater than 1, the triangle is impossible. (iv) Solve the triangle ABC, given a = 10, b = 8, a = 30°.
Solution. Since a > b, just one solution exists.
By the law of sines,
1 2 . sin a sm {3 = -a- · b = - X 8 = -5 = .40. 2X10 Therefore, to the nearest degree, {3 = 24°, and 'Y = 180° - 54° c
=
-!!· sin 'Y = 2 X sma
= 126°. Also
10 X .81 = 16.2.
2. In the three parts of this example, angle 'Y and the adjacent side, a, retain constant values, but the opposite side, c, takes" on different values to illustrate different possible cases. In each part, solve the triangle ABC, given EXAMPLE
'Y
= 51 ° 34',
a= 469.4.
The student should draw the figures, and check the solution using four-place tables. (i) Solve the triangle, given c = 367.7.
Solution. Since the side opposite the given angle is less than the other given side, there may be one, two, or no solutions. By the law of sines, sin a = sin 'Y • a = sin 51 o 34' X 469 4 C 367.7 . log sin 51 ° 34' log 367.7 log (sin 'Y / c) log 469.4 log sin a
= 9.89395 - 10 = 2.56549 (-) = 7.32846
= 2.67154
( +)
= 10.00000 - 10
Therefore, within the limits of accuracy of observation, sin a = 1, a = 90°, and there is just one triangle as a solution. To complete the solution: {3 = 90° - 'Y b = a cos 'Y·
log a = log cos 'Y = log b =
= 38° 26'.
2.67154 9.79351 - 10 ( +) 2 .46505 b = 291.8.
284
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XX
(ii) Solve the triangle, given c = 414.6. Solution. Proceeding as before, using the new value of c, we write:
log sin 'Y log 414.6 log (sin'Y/c) log 469.4 log sin a
= = = = =
9.89395 - 10 2.61763 (-) 7.27632 - 10 2.67154 ( +) 9.94786 - 10
Since log sin a < 0 and c < a, there are two solutions, in one of which a = 62° 29', and in the other of which a' = 180° - 62° 29' = 117° 31'. Using a = 62° 29', we find {3 = 180° - (51 ° 34' + 62° 29') = 65° 57'. c · sm . {3 = sm . {3 + (sm 'Y) . b = -.--
Then
Slll'Y
C
\Ve use this latter form because we have already found log (sin 'Y/c). log sin 65° 57' = 9.96056 - 10 log (sin 'Y/c) = 7.27632 - 10 (-) log b = 2.68424, b = 483.3. For the other solution, {3' = 180° -
('Y + a') =
180° - (51° 34'
+ 117° 31')
=
10° 55'.
log sin 10° 55' = 9 .27734 - 10 log (sin 'Y/c) = 7.27632 - 10 (-) log b' = 2.00102, b' = 100.2. The two solutions are therefore: a
=
62° 29', {3
=
65° 57', b = 483.3;
and a' = 117° 31', {3' = 10° 55', b' = 100.2.
(iii) Solve the triangle, given c = 321.4. Solution. Proceeding as in the other cases, but with the new value of c, we write log sin 'Y = 9.89395 - 10 log 321.4 = 2.50705 (- ) log (sin 'Y / c) = 7 .38690 - 10 log 469.4 = 2.67154 ( +) 0.05844.
Since this logarithm is positive, it is the logarithm of a number that is greater than 1; therefore it cannot be the logarithm of the sine of an angle a. Consequently, no triangle exists that satisfies the given conditions, and we need proceed no further.
§§ 183-184]
THE OBLIQUE TRIANGLE
285
EXERCISES Find the number of possible triangles for each of the following sets of data, and complete the solution of all possible triangles.
1. a = 16, b = 20, a = 42°. 2. b = 10, C '= 15, {3 = 30°. 3. a = 32.56, C = 26.49, 'Y = 41 ° 13'. 4. a = 65.68, b = 49.17, {3 = 53° 26'. 5. a = 14.12, c = 23.57, a = 73° 2'. 6. b = 517.4, C = 301.7, 'Y = 67° 21'. 7. a = 3706.1, b = 4238.7, a = 60° 58.1'. 8. a = 367.24, C = 326.30, 'Y = 62° 41.3'. 9. b = 4931.4, C = 6623.2, 'Y = 78° 19.4'. 10. a = 5372.8, b = 6565.4, {3 = 59° 38. 7'. 11. a = 673.56, C = 786.49, 'Y = 109° 13.8'. 12. b = 5319.7, c = 4658.6, {3 = 128° 14.3'. 13. A children's slide 25 feet long makes an angle of 40° with the ground. Its top is reached by a ladder 17 feet long. How steep is the ladder? 14. In a studio the north side of the roof is of glass. It is 16 feet from eaves to ridge, and makes an angle of 75° with the horizontal. The south side of the roof is 30 feet from eaves to ridge. How wide is the studio? 15. A body is acted upon by a force of 864 pounds and by a second force of 1149 pounds. Their resultant makes an angle of 51 ° 12' with the first force. Find the magnitude of the resultant and the direction of the second force. 16. A destroyer on blockade duty is lying due east of a neutral port. It observes an enemy merchant ship leave the port in the direction N 52° E. Assuming that the destroyer can make 24 knots (or miles per hour) and the merchant ship can make 20 knots, in what direction should the destroyer sail in order just to meet the other vessel?
184. The law of tangents. In any triangle, the difference between two sides is to their sum as the tangent of one-half of the difference between the opposite angles is to the tangent of one-half of the sum of those angles;
a - b tan½(a -/j) --=-----, a + b tan ½(a+ /j)
b - c tan½(f:1 - 'Y) -=-----, b + c tan ½(/j + 'Y)
c - a _ tan ½ 0, (b) if x < O? Inwhatquadrantsisarcsecx, (a) ifx > 1, (b) ifx < -1? In what quadrants is arccsc x, (a) if x > 1, (b) if ~r < -1?
1. In what quadrants is arccos x, (a) if 0
2. 3. 4.
Name the numerically smallest value of each of the following quantities~ 5. arc sin
½V3.
6. arcsin (-½).
8. arcsin 0.
10. arctan 0.
9. arctan 1. 12. arc tan ( -
11. arctan ( -1).
7. arcsin ( -1).
V3).
13. arcsec 1.
Name the two numerically smallest values of each of the following quan-tities. 14. arccos
½.
15. arccos
½V3. ;-
16. arccos (-½).
17. arccos (- ½V 2).
18. arc sec v 2.
19. arcsec ( - 2).
20. arcsec (-jV3).
21. arcsec 2.
22. arccos ( -1).
1
300
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXI
192. Principal values. From the preceding exercises it is noticed that for any value of x for which the functions are defined, there is just one numerically smallest value of arcsin x and arctan x, and there are, in general, two numerically smallest values of arcsec x, one of which is positive and one negative. To avoid confusion, out of the infinitely many values of an inverse trigonometric function, we select one, which we shall call the principal value of the function: Definition: For a given value of x, the principal value of arcsin x, or of arctan x, is the numerically smallest value of the function; the princip-~l value of arcsec x is the numerically smallest positive (or zero) value of the function; the principal values of the "inverse cofunctions" are found from the principal values of the other functions by the relations:
arccos x
= 27r -
•
arcsm x; arccsc x
arccot x =
= 7r2 -
7r
2
- arctan x;
arcsec x.
This definition is equivalent to saying that the principal values are contained in the following intervals or ranges:
- 27f 0 and y quadrant and
0
-1
tan0
3
v3 + c-2) 2
< 0,
2
=
vrn.
0 is in the fourth
=;y = 3-2 =
-.6667.
-2 -2 - - ----3-2i
3
From Table III we find tan 33.7° = .6667. Therefore we can take 0 = 360° - 33.7° = 329.3°.
FIG. 119
Therefore 3 - 2 i = -V13 (cos 329.3°
+ i sin 329.3°),
If a complex number is given in polar form,
z
= r (cos 0 + i sin 0),
it may be written in the standard form a
+ bi,
by evaluating cos 0 and sin 0 (from a table or otherwise), and multiplying their values by r to obtain the real part and the coefficient of the imaginary part. To plot the corresponding point, first lay off the angle 0 from the positive real axis, and then, on its ! rminal side, measure the distance r from the origin. 0
320
FIRST YEAR OF COLLEGE MATHEMATICS
EXAMPLE 3. form, a+ bi.
Express the number 4(cos 110°
[Ch. XXIII
+ i sin 110°) in the standard
Solution. Here r = 4, 0 = 110°. From Table III we find cos 110° = - .3420, and sin 110° = .9397. Therefore 4(cos 110°
+ i sin 110°)
= 4( - .3420 + .9397 i) = -1.3680
+ 3.7588 i.
EXERCISES In each of the Exercises 1 to 16, plot the corresponding point, and write the number in polar form, using tables only when necessary. Name the values of the modulus and the amplitude. 1. -4
5. 7. 9. 4 13. -4
+ 4i. 4V3i. + 3 i.
2. 2 - 2 i.
a. 2 + 2v3 i.
6. -9.
7. 5 i.
10. -! + !V3 i. 14. 5 + 12 i.
+
11. 8 8 i. 15. 3 - 5 i.
4. -6V3 - 6i. 8. -11 i. 12. -10 - 10 i. 16. -2 - 4 i.
In each of the Exercises 17 to 24, values of rand 0 are given. Write the corresponding complex number in the standard form (a+ bi) and plot the point. Use tables only when necessary.
17. r = 10, 0 = 20. r = 17, 0 = 21. r = 13, 0 = 22. r = 12, 0 =
60°. 18. r = 3v2, 0 = 225°. 8 arctan 1 5 , third quadrant. arctan ( -1/-), second quadrant. 180°. 23. r = 4, 0 = 150°.
19. r = 17, 0 = 270°.
24. r = 12, 0
= 300°.
In each of the Exercises 25 to 40, plot the corresponding point and then express the number in the standard form (a+ bi). Use tables only when necessary.
+ i sin 60°). 3(cos 210° + i sin 210°). 6(cos 300° + i sin 300°). 8(cos 135° + i sin 135°). 7(cos 0° + i sin 0°). 11 (cos 90° + i sin 90°). 2(cos 74° + i sin 74°). 3(cos 214° + i sin 214°).
25. 2 (cos 60° 27. 29. 31. 33.
35. 37.
39.
+ i sin 30°). 28. 8(cos 225° + i sin 225°). 30. 2(cos 240° + i sin 240°). 32. 5(cos 315° + i sin 315°). 34. 9(cos 180° + i sin 180°). 36. (cos 270° + i sin 270°). 38. 4(cos 113° + i sin 113°). 40. (cos 294° + i sin 294°). 26. 12(cos 30°
41. Circles are drawn about the origin as center through the points 3 - 4 i and
-12
+ 5 i.
Draw the circles and find the difference between their areas.
COMPLEX NUMBERS
§§ 203-204]
321
204. Multiplication of complex numbers in polar form. and z2 be any two complex numbers in polar form, say
z1 = r1(cos 01
+ i sin 01)
Let z1
and
By § 201, their product is z1z2
= [r1(cos 01 + i sin 01)] · [r2(cos 02 + i sin 02)] ==
(r1r2) [ (cos 01 cos 02 - sin 01 sin 02)
+ i (sin 01 cos 02 + cos 01sin 02)]
But, by formulas I and II of § 169, cos 01 cos 02 - sin 01 sin 02 = cos (01 sin 01 cos 02
+
02),
+ cos 01 sin 02 = sin (01 + 02).
Therefore
The result is a complex number with modulus (r1r2) and amplitude (01 + 02). We have the following important theorem:
Theorem. The product of two complex numbers is a complex number whose modulus is the product of the moduli of the two factors and whose amplitude is the sum of the amplitudes of the two factors. The rule can easily be extended to the product of n factors:
(2)
r1(cos 81
+ i sin lh)
• r2(cos 82
+ i sin 82)
= T1T2 ... Tn[cos (01 + 82 +
• • • • • Tn(COS On+ i sin On} ... + On) + i sin (81 + 02 + · · · + On)].
From the theorem, we obtain the following rule for division, which the student should state in words: (3)
r1(cos 81 + i. sin fl ) . 81) r2 (cos 82 + z sm u2
EXAMPLE.
Solution. 3
+ 3-V3 i
(-2-V3
= (r1) [COS (flu1 _ r2
(J )
2
+ Z. Slll . (flu1 _
fl )] . u2
+ 2 i) • (3 + 3V3 i). -2-V3 + 2 i = 4(cos 150° + i sin 150°),
Multiply (-2-V3 In polar form,
= 6(cos 60°
+ i sin 60°).
+ 2 i)(3 + 3-V3 i)
and
By formula (1),
= (4 • 6)[cos (150° + 60°) + i sin (150° + 60°)] = 24(cos 210° + i sin 210°) = - l2-V3 - 12 i.
322
FIRST YEAR. OF COLLEGE MATHEMATICS
[Ch. XXIII
EXERCISES Perform the following multiplications and divisions, using formulas (1), (2) or (3), and reducing the result to the standard form a + bi. 1. 2. 3. 4. 5. 6. 7. 8.
8(cos 25° + i sin 25°) • 3(cos 20° + i sin 20°). 4(cos 300° + i sin 300°) • 6(cos 105° + i sin 105°). 2(cos 162° + i sin 162°) • 5(cos 138° + i sin 138°). lO(cos 211 ° + i sin 211 °) • 3(cos 312° + i sin 312°) • (cos 17° + i: sin 17°). 9(cos 298° + i sin 298°) • 4(cos 332° + isin332°) • 5(cos 180° + isin 180°). 16(cos 223° + i sin 223°) + 4(cos 88° + i sin 88°). 18(cos 148° + i sin 148°) + 9(cos 103° + i sin 103°). 12(cos 231 ° + i sin 231 °) + 4( cos 81 ° + i sin 81 °).
206. De Moivre's Theorem. If r(cos 0 + i sin 0) is any complex number, and n is any positive integer, by setting r1, r2, · · · , rn equal to r and 01, 02, · · · , 0n equal to 0 in formula (2) of § 204, we obtain [r( cos (J
+ i sin 8) ]n = rn(cos n() + i sin nO).
This is known as De M oivre' s Theorem. The student should state it in words and prove it directly by the method of mathematical induction for all positive integral values of n. EXAMPLE
Solution.
1.
By De Moivre's Theorem, find (1 - i) 5•
We first express (1 - i) in polar form: 1- i =
V2 (cos 315° + i sin 315°).
By De Moivre's Theorem,
[V2 (cos 315° + i sin 315°)] 5 = (V2) 5 (cos 1575° + i sin 1575°) = 4V2 (cos 135° + i sin 135°)
(1 - i) 5 =
=
4v2 C-½v2 + ½v2 i)
= -4
+ 4 i.
In order to prove De Moivre's Theorem in case n is a negative integer, we let n be any negative integer, and set n = -m. Then [r(cos 0 + i sin 0)]n = [r(cos 0 + i sin 0)J-m
1
1
- [r(cos 0 + i sin 0)Jm = rm(cos m0 + i sin m0) cos m0 - i sin m0
= rm (COS2 m 0 + Slll . 2 m0) =
r-m (cos m0 -
. . i
sm m0)
= r-m[cos (-m0) + i sin (-m0)] = rn(cos n0 + i sin n0).
COMPLEX NUMBERS
§§ 204-205]
323
In § 206 an interpretation will be given for De Moivre's Theorem in case n is a rational fraction. EXAMPLE
1 Find the value of z = (- : 4 3
2.
Solution.
i/
+ 3 i) in polar form, we have 125( -4 + 3 i)- = 125[5(cos 0 + i sin 0) J-
·writing ( -4 z
=
3
3,
where 0 = arctan (-¾) in the second quadrant, or 0 = 143.1 °. by De Moivre's Theorem, z=
1 5 : 3 [cos (-3 0)
= cos (-69.3°)
+ i sin (-3 0)]
+ i sin (-69.3°)
=
cos (-429.3°)
Therefore
+ i sin (-429.3°)
= .3520 - .9360 i.
EXERCISES Find the indicated powers by the use of De Moivre's Theorem. Leave the results in polar form unless the form a+ bi can be found without using tables. 1. (1
+ i)
5 •
2. ( -1
V3 i) 3•
4. (1 10. 12. 14.
5•
4
6•
+ i sin 12°)]-
+ i sin 135°)-
20.
·
(v3 -
3•
13.
18. (cos 135°
5•
3•
10
3. (1 - i) 4•
•
6.
(2V3 -
6 i) 5•
9. (-4 - 4 i) 4 •
3•
+ i sin 40°)] (cos 111 ° + i sin 111 °) [3(cos 98° + i sin 98°)]
11. [3(cos 40°
•
6•
16. [2(cos 12°
3
+ i) (-½ + ½V3 i) (V3
5.
+ v3 i) 8. [2(cos 15° + i sin 15°)] (cos 50° + i sin 50°) [2(cos 84° + i sin 84°)]
7. (-3
+ i)
15.
3
•
4• 5
•
17. (cos 60° + i sin 60°)-3• 4 19. (1 - i)3. 8
21. (-1
i) 3
+ i/
In the following examples, express the results in the form a
+ bi, giving
the results as obtained by using either four-place or five-place tables, as directed by the instructor.
+ 5 i)
+ 3 i)
23. (2 - i) 5•
24. ( -1
25. (3 - 2 i)-3•
26. (-3+4i) 4•
27. (12 - 5 i) 2 •
30 28. (-2
29
22. (3
3
•
+ i/
85 · (1 + 4 i)S°
30
625 • (3 - 4 i)4°
4
•
324
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXIII
206. Roots of a complex number. We have already seen that any real number (except zero) has exactly two distinct square roots.
Theorem. If k U3 any positive integer, any complex number (other than zero) has exactly k distinct kth roots. Proof.
In polar form let the number be
z = r(cos 0 + i sin 0), where r
(l)
> 0.
We wish to show that z has exactly k distinct kth roots. fin I, a parabola if e = 1. The line which passes through the focus and is perpendicular to the directrix is called the principal axis of the conic. The chord through the focus parallel to the directrix is the latus rectum. A point where the conic crosses the priny cipal axis is a vertex of the conic secD tion. Find the equation of the parabola whose directrix is the line EXAMPLE
1.
x-3=0 and whose focus is at (-1, 2). Solution. Let P:(x, y) be any point of the plane, and denote its distance from the focus by d1 and its distance from the directrix by d2. Then (1) (2)
d1 = ±V(x
+ 1) + (y 2
2) 2,
d2 = 3 - x.
The eccentricity of a parabola is equal to 1, = d2. ·Therefore
D'
,so that, if P is on the parabola, d1
,(3)
(4)
FIG. 158
+ 1) + (y - 2) = 3 y + 8 X - 4 y - 4 = 0,
±V (x
2
2
2
which is the desired equation. 419
x;
that is,
420
FIRST YEAR OF COLLEGE MATHEMATICS
EXAMPLE 2.
[Ch. XXX
Find the equation of the hyperbola whose directrix is the line
(1)
3X
4y
-
+ 14 = 0, whose focus is the point (3, 2), and whose eccentricity is 5. Solution. If P:(x, y) is any point and if we denote its distance from the focus by d1 and its distance from the directrix by d2,
y
d1 =
(2)
±V(x - 3) 2 + (y - 2)2, and
d2 = 3 X
(3)
4y
-
-5
In order that P:(x, y) lie on the required hyperbola, we must have the ratio of d1 to d2 numerically equal to the eccentricity, which is 5. Therefore d1 = 5 d2, or
FIG. 159
(4)
±V(x -
+ 14_
3) 2
+ (y -
2) 2
=
5(
3
x -
~~ + 14)-
After being simplified, this equation becomes (5)
8 x2
-
24 xy
+ 15 y + 90 x 2
108 y
+ 183 =
0.
By tracing our work back, we see that if the coordinates of P satisfy this relation, d1 = 5 d2. Equation (5) is therefore the desired equation.
EXERCISES Find the equations of conics satisfying the following conditions. 1. A parabola, focus (1, 3), directrix 3 x - 4 y
+
+ 4 = 0.
2. Focus (-1, 2), directrix 5 x - 12 y 3 = 0, e 3. Focus (4, -2), directrix x = 1, e = 2.
=
¥-.
+ 3 = 0, e = ½. Focus (1, 1), directrix x - y + 2 = 0, e = V2. Focus (0, 0), directrix 2 x + y + 3 = 0, e = }V5. Focus (2, -2), directrix x + y + 4 = Q, passing through (2, -4).
4. Focus (-3, 4), directrix y 6. 6.
7.
8. Focus (1, -1), directrix 2 x - y - 4
= 0, passing through (2, 1).
§§ 261-262]
II.
CONIC SECTIONS
421
CONIC SECTIONS IN S'fANDARD POSITION
262. Standard equations of the conic sections. Suppose that we are given the focus, directrix, and eccentricity of a conic section. It is possible to choose the coordinate axes in such a way as to make the equation of the conic section very simple. As the x-axis let us choose the principal axis of the conic, that is, the perpendicular to the directrix drawn through the focus. The equation of the directrix will then have the form x = k, and the coordinates of the focus can be written y as (c, O), where k and care numbers D that we must choose to fit the figure. x=k A change in the value of k changes d2 Q 1--------op (x, y) the position of the y-axis and the value of c. Let p denote the distance from the directrix to the focus. Then C (1)
p =
__0....,..__-.....---0-----x '--v----1 F (c 0)
k.
C -
'--y---/
k
P
,
I
D The distance p is given, since the relative positions of the focus and FIG. 160 directrix are known. But we can still choose k as we wish, thereby fixing the position of the y-axis. Once k is chosen, c is determined by the relation p = c - k. After setting up the equation of the conic, we shall choose k so as to reduce the equation to the simplest possible form. By the definition of a conic section, if P:(x, y) is any point of the curve, and if d 1 is its distance from the focus and d2 is its distance from the directrix,
(2)
But
d2 =
and
k.
X -
Therefore (3)
±V(x - c) 2
or x2
+y
2
-
2 ex
+y
= e(x - k),
2
+c
= e2x 2
2
2 ke 2x
-
+ke
2 2
•
After combining terms we obtain (4)
(1 - e2)x 2
+ y + 2(ke 2
2
-
c)x
+
(c2
-
k 2e2 )
= O,
422
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXX
an equation which can be used to represent any conic section whose principal axis is the x-axis. We have various cases for the different· types of conics, as follows. Case I. Parabola.
e = 1.
If e = 1, equation (4) becomes (5)
y2
+ 2(k -
c)x
+
(c
+ k)(c
- k) = O,
or, since, by (1), p = c - k, (6)
y2
-
2 px
+ p(c + k)
= 0.
vVe can still choose the position of the y-axis by choosing k. The only real simplification consists in removing the constant term. In order to do this we set (c + k) = 0 or k = -c, thereby fixing the y-axis midway between the focus and the directrix. The equation becomes
(7)
= 2px,
y2
which will be called the standard form for the equation of a parabola. Since k = -c and p = c - k, so that c = p/2 and k = -p/2, in terms of p the coordinates of the focus and the equation of the directrix are
Focus: (
t, 0),
Directrix: x = -
Case II. Ellipse and hyperbola.
e
~
t
1.
In equation (4) we can obtain symmetry with respect to the y-axis if we make the term in x disappear. We can do this by setting k = c/e2 • Substituting this value fork in (4), we get (8)
(1 - e2)x 2
+y
2
-
c2 (1 - e2) = 0. 2 e
To find the x-intercepts of the curve represented by this equation, set y = 0 and solve for x. Since 1 - e2 ~ 0, we can divide by (1 - e2 ) and get x 2 = c2/e 2 or x = ± c/e. It will be convenient to denote the positive x-intercept by the letter a. If we set
(9)
C
a=-, e
and divide by a 2 (1 - e2), the equation becomes
§ 262]
CONIC SECTIONS
-ax2 + a (1 y2-
(10)
= 1
--=---2
2
423
e2)
•
There are two cases to consider, that in which e < 1 and that in which e > I. Ellipse. e < 1. If e < 1, we can solve (10) for the y-intercepts. Set x = 0 and solve for y. We find the y-intercepts y
which are real, since e2 so that
(11)
= ±a-VI - e2,
< 1. Denote the positive y-intercept by b,
= a-Vl -
b
e2 •
Then equation (10) becomes x2
(12)
a2
y2
+ b2 = 1,
which will be calle~ the standard equation of the ellipse. From the relations k = c/e2 , a = c/e, b = a-Vl - e2 , we find c = ae, k = a/e, e = V a 2 - b2/a, so that the eccentricity, focus, and directrix are
Eccentricity: e =
R,
Directrix:
Focus: (ae, 0) a
x=-
(Ya2
-
b2 , 0),
a2
or
e
or
X
= va2 -===~=• - b2
Hyperbola. e > 1. In case e > l, 1 - e2 < 0, so that a-Vl - e2 is imaginary and there is no y-intercept. Nevertheless, by analogy with the preceding case and to simplify the equation, we set
(13)
b = a-Ve2
-
1,
which makes b real. We must remember that we cannot now interpret b as an intercept, and in fact b has no simple geometrical meaning. Equation (10) now becomes
(14)
x2 y2 ---=1
b2
a2
'
which will be called the standard equation of the hyperbola. In this case c = ae, k = a/e, e = V a2 b2/a, so that the eccentricity, focus, and directrix are
+
424
FIRST YEAR OF COLLEGE MATHEMATICS
Eccentricity: e
~ ✓1 + ~:,
. . D irectnx : x
Focus: (ae, 0)
= -ea
or
= Va -~:==~;:;;. +b 2
y2 D
--+------+-----X O
(v a2 + b2 , 0),
or
a2
X
263. Description of the conic sections.* is positive, the equation
y
[Ch. XXX
2
I. The Parabola.
If p
= 2 px
represents a parabola open to th2 right. The axis of symmetry is the axis of the parabola. The vertex is at the origin. The length of the latus rectum is twice the length of the ordinate erected to the curve at the focus; since the abscissa of the focus is ½p, from the equation the length of the la tus rectum is l = 2v2 p(½ p),
or
l = 2p.
If p is negative, the parabola is tangent to the y-axis at its vertex, but is open to the left. The equation
D' FIG. 161
x2 = 2 PY
represents a parabola with vertex at the origin, axis on the y-axis, and open upward or downward according asp is positive or negative. 1. The equation y 2 = 6 x represents a parabola in standard position, with p = 3. The focus is at(!, 0), the directrix is the line x = -j, EXAMPLE
and the length of the latus rectum is 6. The student should dravv the figure, marking the focus, vertex, both ends of the latus rectum, and the directrix. The curve should then be sketched through the vertex and the ends of the la tus rectum.
II. The Ellipse. x2 a2
+
The equation y2 or b2 = 1,
represents an ellipse in standard position. Its vertices are at ( -a, 0) and (a, 0). The line joining the vertices is called the major axis of * If the student is not already sufficiently familiar with the general 0haracter of each type of conic section he should review §§ 81, 82.
CONIC SECTIONS
§§ 262-263]
425
the ellipse. The length of the semi-major axis is a. The midpoint of the major axis of an ellipse is the center of the ellipse. In the present case it is at the origin. Since it has a center of symmetry, the ellipse, unlike the parabola, is said to be a central conic. The chord of an ellipse D'l through the center, perD'2 pendicular to the major B' axis is called the minor axis. The semi-minor axis in our case is equal to b. The words "major" and "minor" are appropriate, since B
FIG. 162 where c is the abscissa of the focus, so that a is always greater than b. From the relation a 2 = b2 c2 we can construct the focus if we know a and b. With B' in Figure 162 as center draw an arc with radius a; it meets the major axis at the focus F1, since b and c are the legs of a right triangle of which a is the hypotenuse. The focus F 1 is at the point (ae, 0) or (V a 2 - b2, 0). The point (V a 2 - b2, 0), the line x = a 2/V a 2 - b2, and the number e = V a 2 - b2 /a, if used as the focus, directrix, and eccentricity of a conic define the ellipse b2x2 + a 2y 2 = a 2b2 • Since this ellipse is symmetrical with respect to its minor axis, it will equally well be obtained if the same eccentricity is used with the symmetrically placed point and line (-V a 2 - b2 , 0), and x = -a2 /V a2 - b2 as focus and directrix. Thus every ellipse has two foci and two corresponding directrices, symmetrically placed with respect to the minor axis. Since a focus is at (v a2 - b2 , 0), from the equation we see that the value of y at the upper end of the latus rectum is equal to
+
y
b -2- - -2 b -----b2 = -V a - x = -V a 2 - (a 2 - b2) = -,
a
a
a
and the length of the latus rectum is
l
= 2 b2_ a
[Ch. XXX
FIRST YEAR OF COLLEGE M:ATHEl\1:ATICS
426
EXAMPLE
2.
The equation
y2
x2
= -+9 25
1
9 x2
or
+ 25 y
2
= 225
represents an ellipse whose semi-major axis is a = 5 and whose semi-minor axis is b = 3. The foci are on the x-axis; c = -Va2 - b2 = 4; the foci are therefore at (4, 0) and ( -4, 0). The eccentricity is
e=H= ✓l
:5
=~-
Since k = a/e = 5/ (-¼) = (2l ), the directrices are the lines x = 2l and x = - 2l. The length of the latus rectum isl = 2 b2/ a = \ 8 • The student should draw the figure. EXAMPLE 3. The equation
represents an ellipse with center at the origin. Since the denominator of the term in y 2 is greater than that of the term in x2 , the major axis now lies on the y-axis and the minor axis on the x-axis. The semi-major axis is a = 6, the semi-minor axis is b = 5. The vertices are at (O, 6) and (0, -6). Since
c = v' a2
-
b2 = v' 36 - 25 =
the foci are at (0, V11) and (O, -Vll).
e=
Then k =
1
and y
=
~= e
6
_ =
t-Vll
--if Vll.
✓1 36
11
2
-b2 = a
V 11
'
The eccentricity is
R51vn 1- - = 6 36
·
V 11, and the directrices are the lines y = r~ vTi .
The length of the latus rectum 1s Z =
2 b2
---;;
=
25 3·
III. The Hyperbola. The vertices of the hyperbola y2 x2 or =1 b2 a2 are at (-a, O) and (a, O). The hyperbola is a central conic with its center at the origin. Just as for the ellipse, we find two foci, one at (V a 2 + b2, O) and the other at (-V a 2 + b2 , 0), with the two corresponding directrices, x = a2/V a 2 + b2 and x = - a2/V a 2 + b2 • The length of the latus rectum is found, as for the ellipse, to be
§ 263]
CONIC SECTIONS
427
2 b2
l=-· a
The portion of the principal axis included between the vertices is called the transverse axis y of the hyperbola. The porD'2 D'1 tion of the y-axis that lies between (0, b) and (0, -b) is the conjugate axis. And the semi-transverse and semi-conjugate axes are respectively equal to a and V to b. The names "major" and ''minor'' would not be appropriate here, since the two axes may be equal, or either one may be longer FIG. 163 than the other. We have seen (§ 260) that the hyperbola has the asymptotes bx - ay = 0 and bx + ay = 0 or x2 y2 a2 - b2 = G. In order to draw the hyperbola from the given equation it is convenient to proceed as follows: Draw the rectangle whose sides are the lines x = ±a, y = ±b, and then draw the diagonals of this rectangle, prolonging them beyond the corners. (See Figure 163.) These diagonals are the asymptotes since they pass through the center and have slopes ± bI a. The vertices V 1 and V 2 are at the points where the sides of the rectangle cross the x-axis. The foci are constructed by laying off from the origin the distance OF1 and OF2 equal to OA, which is one-half of the diagonal of the rectangle. For the abscissa of F 1 is c = V a2 + b2 • The latus rectum can be drawn at each focus from the value l = 2 b2/a. The curve can now be sketched quite accurately through each vertex and the ends of each latus rectum, using the asymptotes as guide lines.
The equation
represents a hyperbola with foci and vertices on the y-axis and conjugate axis on the x-axis. The asymptotes are b2y 2 - a 2x 2 = 0.
428
FIRST YEAR OF COLLEGE MATHEMATICS
EXAMPLE
4.
The equation
16 x2
-
9 y2
=
144 or
x2
9-
[Ch. XXX
y2
16
=
1
represents a hyperbola with vertices at (±3, 0). The length of the semitransverse axis is 3, and that of the semi-conjugate axis is 4. Since c=
the foci are at (±5, 0). e
-V9 + 16 The eccentricity is
=
Va2
+b
2
=
✓1 + !: ✓1 + ~6 =
=
5,
= ~-
Since k = a/e = ¾, the directrices are the lines x = ±!. The asymptotes are 16 x2 - 9 y2 = 0 or y = ±-¼ x. The length of the latus rectum is l = 2 b2/ a = \ 2 • The student should draw the figure.
EXERCISES For each of the following equations determine the nature of the curve; find the axes, eccentricity, foci, directrices, length of latus rectum, and the asymptotes, where they exist. Sketch the curve. 1. y 2
4·
x2
=
lOx.
2. y 2
y2
4 + 1 = 1. 4x + = 4. x + 16 y = 0.
5·
x2
=
-8x. y2
4 - 1 = 1. 9 x + 16 y = x + 8 y = 0.
3. x 2
6·
x2
=
9-
4y. y2
25 = l.
2 2 2 2 7. y 8. 144. 9. 9 y2 - 4 x2 = 36. 12. 4 x2 9 y2 = 144. 10. 2 11. 2 16. 25 x2 - 9 y 2 = -225. 13. y2 - x2 = 9. 14. 36 y2 - 4 x 2 = 9. ·write the equation of each of the following curves, taking the origin at the vertex in the case of a parabola and at the center in the case of an ellipse or hyperbola.
+
16. A parabola, open to the right, latus rectum equal to 20. 17. A parabola, open upward, latus rectum equal to 12. 18. A parabola, focus at (0, 6). 19. A parabola, directrix y = 8. 20. A parabola, focus on the x-axis, passing through the point (-3, 3). 21. An ellipse, semi-major axis on the y-axis equal to 5, semi-minor axis equal to 3. 22. An ellipse, semi-major axis on the x-axis equal to 4, semi-minor axis equal to 2. 23. 24. 26. equal
An ellipse, focus at (6, 0), eccentricity equal to ¾. An ellipse, focus at (0, -3), eccentricity equal to ¾. An ellipse, semi-major axis on the x-axis and equal to 3, latus rectum to!.
§§ 263-264]
CONIC SECTIONS
429
26. An ellipse, focus at (3, 0), vertex at (5, 0). 27. An ellipse, focus at (0, -5), vertex at (0, -13). 28. An ellipse, vertex at (5, 0), through (4, -¾). 29. A hyperbola, semi-transverse axis on the y-axis and equal to 5, semiconjugate axis equal to 4. 30. A hyperbola, vertex at (6, 0), having 6 y = 5 x as an asymptote. 31. A hyperbola, focus at (6, 0), eccentricity equal to J. 32. A hyperbola, focus at (0, -8), eccentricity equal to 2. 33. A hyperbola, focus on x-axis, asymptotes 4 y = ±3 x, latus rectum equal to 9. 34. A hyperbola, asymptotes 3 y = ±2 x, passing through the point (6, 2v2). 35. Find the locus of the midpoints of the ordinates of an ellipse.
36. Find the locus of the midpoints of the ordinates of the hyperbola b2x2 _ a2y2 = a2b2. 37. Find the locus of the vertex of a triangle whose base vertices are at (-a, 0) and (a, 0) and for which the product of the tangents of the base angles is b2/ a 2•
38. The vertex of a variable parabola remains fixed at the origin, but its focus moves along the x-axis. Find the locus of an end of the latus rectum of the parabola. Prove the following propositions: 39. The minor axis of an ellipse is a mean proportional between the major axis and the latus rectum. 40. If an equilateral triangle is inscribed in the parabola y2 = 2 px and has one vertex at the origin, the length of one of its sides is 4 pV3.
*264. The polar equation of a conic. of polar coordinates be taken at D' the focus of a conic whose eccentricity is e. Suppose that the M directrix DD' is perpendicular to the polar axis and at the distance p to the left of the pole. Then if P :(p, 0) is a point on the conic, and if JV[ and N, respectively, are the projections of P on the directrix and on the 90° axis, from the definition of a conic section, we have
Let the pole of a system
FIG. 164
430
[Ch. XXX
FIRST YEAR OF COLLEGE MATHEMATICS p
= e · MP
=
e(MN
+ NP)
=
e(p
+ p cos 0).
Solving this for p, we get (1)
p
=1 -
ep
e cos o'
which is the polar equation of the conic in standard form. The student should show that the equations
ep
(a) P
= 1 + e cos 0'
ep
ep
(b) P = 1 - esm . 0'
(c) P = 1
+ esm· 0'
represent conic sections with foci at the pole which have, respectively (a) the directrix parallel to the 90° axis and p units to the right of it, (b) the directrix parallel to the polar axis and p units below it, and (c) the directrix parallel to the polar axis and p units above it. EXAMPLE
1.
5
Discuss the equation
(l) P = 2
+ 2 cos 0.
Solution. In order to get this in standard form, in which the first term of the denominator is 1, we divide numerator and denominator by 2: (2)
p
=
5
2
1
.
+ cos 0
The eccentricity is equal to the coefficient of cos 0. Therefore e = 1, and the '3urve is a parabola, open to the left. When 0 = 1r/2, p = i, so that the latus rectum is equal to 5. When 0 = 0, p = ¾; therefore the vertex is ¾unit to the right of the focus or pole. Since ep = i and e = 1, p = i, and the directrix is i unit to the right of the pole. EXAMPLE
2.
Discuss the equation
9 (l) P = 4 - 5 cos 0.
Solution. In order to put this in standard form, we divide numerator and denominator by 4: 9
(2)
FIG. 165
4
P
= 1-
.Q_
4
cos 0·
The eccentricity is therefore e = ¾. Since this is greater than 1, the locus is a hyperbola. When 0 = 0, p = -9. One vertex is therefore at V 1, 9 units to the left of the pole. When 0 = 'Tf, p = 1, and the second vertex is at V 2, one unit to the left of the pole. The transverse
CONIC SECTIONS
§ 264]
431
axis is therefore equal to 2 a = V1 V2 = 8, and a = 4. The semi-conjugate axis is found by setting b2 = a 2(e2 - 1) = 16( i ~ - 1) = 9, so that b = 3. The asymptotes can now be drawn and the curve sketched. Since ep = ¾and e = ¾, p = ¾. The directrix is therefore¾ unit to the left of the pole. EXAMPLE
Solution.
. Discuss the equat10n
3.
p
=
16 . 5 - 3 sm 0
y
After dividing numerator and de-
nominator by 5, we have
p
= 1 _¾sin o"
The eccentricity is e = ¾- The locus is an ellipse with major axis on the go0 axis, a focus at the pole, and the corresponding directrix below the focus. Since ep = \ 6 , and e = ¾, p = ¥; the directrix is -136 unit below the pole. To get the vertices, first set 0 = ~; then p = 8.
0=
3
7r 2; then p =
2.
Next set
The vertices V1 and V2
are respectively 8 units above and 2 units below the focus. The semi-major axis is a = 5. Also b2 = a 2 (1 - e2 ) = 25(1 - -/-5 ) = 16, orb = 4. It is now easy to draw the curve.
D FIG. 166
EXERCISES Identify and sketch the loci of the following equations.
4 l. P = 1 - cos _ 4• P - 2
+
o·
12 . 5 cos 0
7. p cos2 ½0 = 4.
2· P
8
5
1
+ sin o·
3· P
5· P = 5
8 + 4 cos
10 G. P = 4 + sin
=
8. p sin2 ½0 = 8.
o·
=
3
+ 2 cos o· o·
9. p = 6/(2 - sin 0.)
Write the polar equations of the following conics, with the pole at a focus. 10. Parabola, vertex at (1, 0).
11. Parabola, vertex at (2,
½1r).
11. Parabola, vertex on polar axis to the left of the pole, latus rectum equal to 14. 12. Ellipse, eccentrjcity !, directrix 5 units to right of 13. Ellipse, eccentricity
¾,
go
0
axis. directrix 4 units above polar axis.
432
FIRST YEAR OF COLLEGE lVIATHEMATICS
[Ch. XXX
14. Ellipse, the directrix corresponding to the focus at pole parallel to
go
0
axis and to the right, major axis equal to 10, minor axis equal to 8. 15. Ellipse, with vertices at (1, 0) and (6, 1r). 16. Hyperbola, eccentricity½, directrix parallel to polar axis and 6 units above it. 17. Hyperbola, eccentricity ¾, directrix parallel to go0 axis and 10 units to its right. 18. Hyperbola, eccentricity !, directrix the line p cos 0 = 10. 19. Hyperbola, with vertices at (-6, 0), (1, 1r).
Prove the following theorems, using polar coordinates. 20. The angles that the asymptotes of a hyperbola make with the transverse axis are given by the equation 0 = arccos (1/e). 21. The focus of a conic divides a chord through the focus into two seg-
ments the sum of whose reciprocals is a constant. 22. If two chords through the focus of a conic are perpendicular, the sum of the reciprocals of their lengths is equal to the constant (2 - e2 ) /2 ep.
III.
SPECIAL PROPERTIES OF A PARABOLA
266. Properties of a tangent to a parabola. The line drawn from a focus of a conic to a point on the conic is called a focal radius. We wish to find certain relay tions connected with a tangent to a parabola and the focal radius drawn to the point of contact. The tangent to the parabola X
(1)
y2
=
2px
at the point of contact P:(x1, Yi) is given by the equation (§ 258) FIG. 167
(2)
In order to find the point T where this tangent crosses the x-axis, we substitute y = 0 in (2) and solve for x, finding the point T:(-x 1, 0). Therefore in Figure 167, TO = X1 = LP. Moreover, if F is the focus, (p/2, 0), and if K is the projection of Pon the directrix, KL = p/2 = OF. Therefore
§§ 264-265]
CONIC SECTIONS
TF = TO
+ OF =
LP
+ KL
433
= KP.
From the definition of a parabola, KP = F P, and consequently TF = FP. The triangle TFP is therefore isosceles, and the angles FPT, FTP, K' PT', T' PG are equal. It follows that a tangent to a parabola b £sects the angle formed by the focal radius to the point of contact and a line parallel to the axis of the parabola. A parabolic reflector is a polished surface having 1
the same shape as the surface generated by the rotation of part of a parabola about its axis. From the preceding theorem it follows that if a lamp be ,u-.,._....-► placed at the focus of such a reflector, any ray from the lamp after striking the reflector will be reflected in a direction parallel to the axis of the 1 - - - i R - - - - - parabola. And all of the light, or as much of it, at least, as strikes the reflector, will be thrown in a beam of rays parallel to the axis. This principle is used in the design of searchlights and headlights. Parabolic reflectors are also used to concentrate rays of heat or light from a distant source. The rays of heat from the sun, striking a parabolic Frn. 168 reflector to whose axis they are parallel, will be focused at the focus. Engines have been devised, and are in practical use, which employ the heat thus concentrated to drive machinery. EXERCISES Prove the following theorems.
= 2 px is a mean proportional between the abscissa at that point and the latus rectum. 2. Any point on the parabola y 2 = 2 px is at the distance x ½p from the focus. 3. If two equal parabolas have the same focus, but extend in opposite directions, they intersect at right angles. (That is, tangents to the two parabolas at a point of intersection are perpendicular.) 4. If d1, d2, d3 are the respective distances of the focus of a parabola (1) from a tangent to the parabola, (2) from the point of contact of the tangent, and (3) from the vertex of the parabola, d1 is a mean proportional between d2 and d3. 5. In a parabola, the tangent at the vertex is the locus of the foot of the perpendicular drawn from the focus to a variable tangent. 1. The ordinate at any point of the parabola y 2
+
434
FIRST YEAR OF COLLEGE MATHEMATICS
IV.
[Ch. XXX
SPECIAL PROPERTIES OF AN ELLIPSE
266. Focal radii of an ellipse. Suppose that in Figure 169, F 1 and F 2 are the two foci of the ellipse (1)
that D1Di' and D2D/ are the corresponding directrices, and that e is the eccentricity. The equations of the directrices are a
x=-
(2)
e
and
X
a
= - -. e
.DJ
D{
X--.J!_
X=.!!_
e
e
M2 t----_,,....-----11--------:::~----tMl
1
FIG. 169
If P:(x1, Yi) is any point on the ellipse, and M1 and M2 are its projections on the two directrices, from the very definition of a conic section we have
F1P
=
e · PM1
=
e (~ -
F2P Adding, we get (5)
=
e · M2P
=
e
(3)
(4)
F1P
x1)
=
a - ex1,
(~+xi) = a+ ex1,
+ F2P = 2 a.
This result is evidently the same, regardless of the position of the point P on the ellipse. Hence, the sum of the focal radii of a point on an ellipse is a constant, and is equal to the length of the major axis. This property is characteristic of an ellipse. In fact, an ellipse is often defined as the locus of a point the sum of whose distances from two fixed points (the foci) is a constant.
§§ 266-267]
CONIC SECTIONS
435
Gardeners and others often construct ellipses by driving stakes or pins at the foci and running a loop of string around both stakes, leaving a certain amount of slack. A marker is then placed inside the loop and moved around keeping the string taut. Since the amount of slack is a constant, the sum of the distances of the marker from the FIG. 170 foci is a constant, and the marker traces out an ellipse. EXAMPLE 1. Write the equation of the locus of a point the sum of whose distances from the points (±3, 0) is equal to 10. Solution. The locus is evidently an ellipse in standard position for which a = 5 and b = V a 2 - c2 = V25 - 9 = 4. The equation is therefore 16 x 2 25 y2 = 400.
+
EXERCISES In each of the Exercises 1 to 4, write the equation of the locus of a point which moves as indicated. 1. The sum of its distances from the points (±2, 0) is equal to 6. 2. The sum of its distances from the points (0, ±3) is equal to 8. 3. The sum of its distances from the points (0, ±6) is equal to 20. 4. The sum of its distances from the points (±1, 0) is equal to 4. 5. Find the equation of the ellipse whose foci are at (1, -1) and (-1, 1) and whose semi-major axis is equal to 3. 6. Find the equation of the ellipse whose foci are at (1, -2) and (0, 1) and whose semi-minor axis is equal to jY6.
267. The focal radii and tangent at a point of an ellipse. that a tangent be drawn to the ellipse (1)
at the point P:(x1, Yi). (2)
b2x2
+ a2y2 =
Suppose
a2b2
Its equation (see Figure 171) is b2x1x + a2y1y = a2b2.
This tangent intersects the x-axis at the point T:(a2/x1, O). Therefore (3)
ana (4)
Consequently
436
FIRST YEAR OF COLLEGE MATHEMATICS
(5)
F2T a+ ex1 F1T = a· - ex1
[Ch. XXX
But in § 266 we have seen that F2P = a+ ex1, and F1P = a - ex1. Therefore F2T F2P F1T = F1P.
(6)
D'1 x_-_a
x=~
e
e
M21-----,1E~----+-------=~-----iM1
1
FIG. 171
Thi8 tells us that PT divides one side of the triangle F1PF2 externally in proportion to the adjacent sides; it therefore bisects the opposite external angle F 1 PG of the triangle. Therefore a tangent to an ellipse bisects the external angle formed by the focal radii drawn to its point of contact. Likewise, the normal PN bisects the internal angle F 1 PF2 formed by the focal radii.
It is for this reason that the foci of an ellipse can be considered as foci in the optical or acoustical sense.
For consider a surface having the form generated by the rotation of an ellipse about its major axis. Suppose that a lamp or source of sound be placed at the focus F 1• A ray of light or sound wave emanating from F 1 and striking the surface at P will be reflected so that the angles which it makes with the tangent plane at P before and after reflection will be equal. By the theorem just stated, the lines F1P and F.;? FIG. 172 make equal angles with the tangent plane. Therefore any ray issuing from F 1 will be reflected toward F2, and the light or sound from F1 will be focused at F2- The student should compare this with the optical properties of a parabola (§ 265).
§§ 267-268]
V.
CONIC SECTIONS
437
SPECIAL PROPERTIES OF A HYPERBOLA
268. Focal radii of a hyperbola. hyperbola
Let F1 and F2 be the foci of the D~ y
(1)
M
and let D1Di' and D2D2' be the corresponding directrices as shown in Figure 173, and let the eccentricity bee. The equations of the directrices are. (2)
a x=e
and
x=
D'1
a e
Then, if P:(x1, Yi) is any point on the right-hand side of the hyperbola, and llf 1 and M2 are its projections on the directrices, we have (3)
F1P = e · M 1P = e ( x1
- ~)
= ex1 - a,
(4)
F2P = e · M2P = e ( x1
+ !)
= ex1
+ a.
Subtracting (3) from (4), we get (5)
F2P - F1P
= 2 a.
If P is on the left-hand side, F1P - F2P = 2 a.
Therefore, the difference between the focal radii of a point on a hyperbola is a constant, and is equal to the length of the transverse axis. A hyperbola may be defined as the locus of a point the dijference between whose distances from two fixed points is a constant. As an exercise the student should show that the locus of a point the difference of whose distances from the fixed points (± V a2 + b2 , O) js equal to. 2 a is the hyperbola (1). From this equation the properties of a hyperbola can be obtained. EXAMPLE 1. Find the equation of the locus of a point the difference of whose distances from the points (±5, 0) is 8. Solution. This is the hyperbola whose foci are (±5, 0) and for which a = 4. \Ve find b = v c2 - a 2 = v25 - 16 = 3.
The desired equation is therefore 9 x2
-
16 y2 = 144.
438
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXX
EXERCISES In each of the Exercises 1 to 6, find the equation of the locus of a point which moves as indicated. 1. The difference of its distances from (± 13, 0) is 24. 2. The difference of its distances from (±3, 0) is 2. 3. The difference of its distances from (0, ±5) is 6. 4. The difference of its distances from (O, ±4) is 4. 5. The difference of its distances from ( -3, 3) and (10, 3) is 6. 6. The difference of its distances from (-2, -11) and ( -2, 15) is 10. 7. Find the equation of the hyperbola whose foci are the points (0, 4) and (3, 0) and whose transverse axis is equal to 4. 8. Find the equation of the hyperbola whose foci are the points (O, -1) and (2, 1) and whose transverse axis is equal to 2.
269. The focal radii and tangent at a point of a hyperbola. Let T be the point where the x-axis is crossed by the tangent to the hyperbola (1)
at the point P:(x1 , y1). Then, in a manner similar to that used for the ellipse ( § 267), the student should show that in Figure 173, (2)
Therefore PT divides one side of the triangle F1PF2 proportionally to the adjacent sides. Consequently it bisects the opposite angle. Therefore, a tangent to a hyperbola bisects the angle formed by the focal radii drawn to its point of contact. The foci can thus be considered as foci in the optical sense. For consider a hyperbolic reflector, whose shape is that of FIG. 174 the surface generated by the rotation of one branch of a hyperbola about its transverse axis. All rays of light directed toward F1 will be reflected toward F 2 • If a lamp is placed at F2, after reflection the rays appear to emanate from F 1 •
CONIC SECTIONS
§§ 268-271]
439
~70. Conjugate hyperbolas. Two hyperbolas are said to be conjugate if the transverse axis of each is the conjugate axis of the other. Each hyperbola is called the conjugate of the other. The equation x2
y2
---=1 a2 b2
(1)
represents a hyperbola whose transverse axis extends from the point (-a, 0) to (a, 0) and whose conjugate axis reaches from (0, -b) to (O, b). But these are, respectively, the conjugate and transverse axes of the hyperbola y y2 x2 (2) b2 - a2 = 1. The hyperbolas (1) and (2) are therefore conjugate. A hyperbola has the same asymptotes as its conjugate, for the hyperbolas (1) and (2) both have the asymptotes X
(3) -
a
y
x2
b
a
y2 b
= ± - or - 2 - - 2 = 0
FIG. 175
271. Rectangular hyperbolas." A hyperbola is said to be a rectangular or equilateral hyperbola if its asymptotes are perpendicular Y yr to each other. The slopes of the asymptotes of the hyperbola x2
0
-450
X
are± b/a. For a rectangular hyperbola we must therefore have (2)
b
-a
a
;=-,;=,/
Consequently b2 the equation can be written FIG. 176
(3)
y2
=1 a2 - -b2
(1)
= a 2 , and b - a, and
x2 _ y2 = a2.
In a rectangular hyperbola, the transverse and conjugate axes are equal.
440
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXX
Because the asymptotes of a rectangular hyperbola are perpendicular, it is often convenient to refer thA hyperbola to its asymptotes as the coordinate axes. To do this, transform equation (3) by rotating the axes through the angle 0 = -45°. The equations of rotation are 1 (4) x' + y'). y = x = V2 (x' + y')'
J2 (-
The equation of the hyperbola becomes
½(x'
(5)
+ y')2
- ½(x' - y')2 = a2
or
a2
x'y' -2· -
(6)
Therefore an equation xy =- k represents an equilateral hyperbola whose asymptotes are the axes of coordinates and for which the semiaxis is a = 2 k. If k > 0, x and y have the same sign, so that the curve lies in the first and third quadrants. If k < 0, the curve is in the second and fourth quadrants. The curves xy = k and xy = -k are conjugate hyperbolas. In a rectangular hyperbola, the eccentricity is equal to e = V2. For, in the case of the hyperbola (1), the eccentricity is
v±
e= But, if b
=
✓ 1+-· b2
a, this reduces to e
a2
=
V2.
EXERCISES In each of the Exercises 1 to 10, write the equation of the conjugate of the given hyperbola, and sketch both curves. 1. 4 x2 - 9 y 2 = 36. 3. y 2 - x2 = 4. 5. 16 x2 - 9 y 2 = 144. 7. xy
= 9.
+
2. 25 x2 - 4 y 2 = 100. 4. 36 y 2 - 25 x2 = 900. 6. x2 - 9 y 2 = 9. 8. xy
+ 16 = 0.
9. xy 25 = 0. 10. xy = 1. 11. Show that the product of the perpendicular distances of a point on a hyperbola from the asymptotes is a constant. 12. Show that if the product of the distances of a point on a hyperbola from the asymptotes is equal to the constant k, the corresponding product for the conjugate hyperbola is equal to -k.
CONIC SECTIONS
§§ 271-272]
VI.*
441
GENERAL EQUATIONS OF CONICS
272. Conics with principal axis parallel to a coordinate axis. equations y 2 = 2 px, x2 = 2 PY,
The
represent conic sections whose principal axes are parallel to one of the coordinate axes and whose center, in the case of a central conic, or whose vertex, in the case of a parabola, is at the origin. In these equations let us replace x by (x - h) and y by (y - k). We obtain (1)
(y - k) 2 = 2p(x - h),
(2)
(x - h) 2 = 2 p(y - k),
(3)
b2 (x - h) 2
+ a'2(y -
(4)
b2 (x - h) 2
-
a2 (y - k) 2 = a2b2,
(5)
b2 (y - k) 2
-
a2 (x - h) 2 = a 2b2 •
k) 2 = a 2b2,
Each of the equations (1) and (2) represents a parabola with vertex at the point (h, k) and axis parallel to one of the coordinate axes, while (3) represents an ellipse and (4) and (5) represent hyperbolas with centers at the point (h, k) and axes parallel to the coordinate axes. For these equations can be reduced to the standard forms by making the translation x' = x - h, y' = y - k. Theorem. Any conic section whose principal axis is parallel to a coordinate axis can be represented by an equation having one of the farms (1) to (5). For we can first take the origin at the center or vertex
and then translate it to the proper position. EXAMPLE 1. Write the equation of the hyperbola whose center is at (-3, 2), whose semi-transverse axis is parallel to the x-axis and is equal to 5, and whose semi-conjugate axis is equal to 4.
Solution. or
The equation is
16 x2
-
25 y 2
(x
+ 3) 25
2
(y - 2) 2
-
+ 96 X + 100 y -
16
= l
356 = 0.
It will be noticed that if the squares are expanded in any one of the equations (1) to (5), an equation is obtained which is, of the second degree in x and y and which contains no term in xy. We * The remainder of this chapter should be omitted by those who wish t_o study only conics in standard position.
442
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXX
then have the Theorem: Any conic section whose principal axis is parallel to a coordinate axis can be represented by an equation of the form
Ax2 + Cy2
(6)
+ 2 Dx + 2 Ey + F = 0
where not both A and C are zero. On the other hand, if we are given an equation of the form (6), we can often reduce it to one of the forms (1) to (5) simply by completing the squares. EXAMPLE
2.
Discuss the equation 4 x2
9 y2
-
-
16 x - 18 y - 29 = 0.
Solution. Complete the square of the terms in x and the square of the terms in y. We get 4(x 2
-
4
X
+ 4)
- 9(y2
4(x - 2) 2
or
-
+ 2 y + 1) 9(y + 1)
2
= 29
+ 16 -
9
= 36.
This represents a hyperbola with center at (2, -1), semi-transverse axis equal to 3 and parallel to the x-axis and semi-conjugate axis equal to 2. EXAMPLE
Solution.
3.
Discuss the equation 4 x 2
Completing the square of the terms in x, we have 4(x2
or
20 x - 6 y - 7 = 0.
-
4(x -
1) 2 =
-
6y
5
X
+ 32
+ i) 2
= 6y
or
This represents a parabola with vertex at upward with a vertical axis.
+ 7 + 25
1) 2 = f(y + ¥). (-i, -¥), and with p = ¾,
(x -
open
More generally, if, in equation (6), neither A nor C is zero, we can complete the squares, getting
When A and C have the same sign (for definiteness take both A and C positive) this is of the form (3), and so represents an ellipse (or a circle) whenever the right-hand member is positive. If the right-hand member is negative, there is only an imaginary locus, since the left-hand member is essentially positive or zero. If the right-hand member is zero, the locus reduces to the single point for which
D
x+-=O A
and
y
+ EC =
0.
443
CONIC SECTIONS
§ 272]
As a matter of definition, let us say that a circle is a limiting form of an ellipse as the two axes become equal and that a poiint is the limit of a circle (and so of an ellipse). We can therefore state that if A and C have the same sign, equation (6) always represents an ellipse or the limiting form of an ellipse, or has only an imaginary locus. On the other hand, if A and C have opposite signs, equation (7) is of the same form as (4) when A has the same sign as the righthand member and the same form as (5) when C has the same sign as the right-hand member. Therefore equation (6) represents a hyperbola unless the right-hand member of (7) is equal to zero. If the right-hand member is equal to zero, the left-hand member, which is equal to the difference of two squares, can be factored into linear factors. In this case the locus degenerates into two intersecting straight lines. As a matter of definition, let us say that a pair of intersecting straight lines is a limiting form of a hyperbola when the asymptotes remain fixed and the vertices are drawn together. · Consequently, if A and C have opposite signs, equation (6) always represents a hyperbola or a limiting form of a hyperbola. Here there is no possibility of an imaginary locus. In case A = 0 and C ~ 0, we can write equation (6) in the form (8)
2
C ( y2 + :
y
+ ~:)
= -2 Dx - F
+
r
Unless D = 0, this can be written as
which is of the form (1), and so represents a parabola with axis parallel to the x-axis. If, on the other hand, D = 0, we can take the equation in the form
(Y + E) c = ± '\J~ c c· 2 -
If the quantity under the radical sign is positive, the locus is a pair of straight lines parallel to the x-axis; if the radicand is equal to zero, the locus is the line y = - E / C (counted twice, if we like); if the radicand is negative, the locus is imaginary. In a similar way, in case C = 0 and A ~ 0, we can show that equation (6) represents
444
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXX
a parabola, with axis parallel to the y-axis, or a pair of lines paralle] to the y-axis, or one such line, counted twice, or has only an imaginary locus. As a matter of definition, let us say that a pair of parallel lines, or a single line counted twice, is a limiting form of a parabola. Then, if either A or C is equal to zero but not both A and C are zero, equation (6) always represents a parabola or a limiting form of a parabola, or has only an imaginary locus. We can sum up the results just obtained in the following theorem: Theorem. If A and C are not both zero, equation (6) represents an ellipse or a limiting form of an ellipse or has only an imaginary locus, if AC > O; it represents a hyperbola or a limiting form of a hyperbola, if AC < O; it represents a parabola or a limiting form of a parabola or has only an imaginary locus, if AC = 0. EXAMPLE 4. Without completing the squares indicate the general nature of the locus of the equation 3 x2 - 7 y 2 + 5 x - 11 y + 3 = 0.
Solution. This equation represents a hyperbola or a pair of intersecting straight lines, since AC = 3(-7) = -21 < 0.
EXERCISES Find the equations of the following conics, and draw the figures.
1. Parabola, vertex at (2, -3), focus at (4, -3). 2. Parabola, vertex at (-3, 1), focus at (-3, -5). 3. Parabola, focus at (4, 1), directrix y = 5.
4. Parabola, focus at ( - 2, 3), directrix x = 0. 5. Ellipse, vertices at (-3, 3) and (5, 3), semi-minor axis equal to 3. 6. Ellipse, center at (3, 2), directrix 4 x = 37, eccentricity equal to -g-. 7. Ellipse, center at (1, -1), focus at (1, 1), eccentricity equal to ½. 8. Ellipse, center at (3, -2), vertex at (3, 3), passing through (4, 1). 9. Hyperbola, center at (4, -2), vertex at (1, -2), semi-conjugate axis equal to 5. 10. Hyperbola, center at (-1, 3), focus at (-1, -10), one asymptote the line 12 x - 5 y + 27 = 0. Describe and plot the locus of each of the following equations.
+ 4 y + 36 X - 24 y + 36 = 0. X + 4 y 10 X - 16 y + 37 = 0. 25 X 4y 150 X - 8 y + 321 = 0.
11. 9
12. 13.
X
2
2
2
2
2
-
-
2
-
CONIC SECTIONS
§§ 272-273]
8 X + 36 y + 43 = 0. 15. X - 4 y + 6 X + 32 y - 59 = 0. 16. 9 x2 - 25 y2 - 18 X - 100 y + 134 = 0. 17. X 2 - y 2 + 4 X + 8 y - 8 = 0. 18. x2 + 6 X - 8 y + 25 = 0. 19. 4 y 2 - 7 X + 32 y + 43 = 0. 20. 3 y 2 + 8 X - 30 y - 5 = 0. Without completing the squares, indicate the general nature of the locus of each of the following equations. 21. 3 X 2 + 11 y2 + 4 X - 8 y - 1000 = 0. 22. 5 X 2 - 7 y2 + X + 13 y - 9 = 0. 23. y2 - 5 X2 - 17 X + 11 y + 3 = 0. 24. 4 X2 + 13 y2 - 2 X + 3 y = 0. 25. 2 X2 - 3 X - 7 y + 8 = 0. 26. 15 y2 + 4 X - 10 y + 1 = 0.
14. x2 2
+9y
445
2
-
2
273. The general equation of the second degree. Any conic section can be placed in standard position by a suitable rotation and translation of coordinate axes. By § 262 a conic section in standard position is represented by an equation of the second degree in x and y. By § 249 the degree of an equation is not changed by a rotation or translation of coordinate axes. We therefore have the following theorem. Theorem I. Any conic section can be represented by an equation of the second degree in x and y. We shall now prove the converse theorem: Theorem II. The locus of any equation of the second degree in x and y is a conic section or a limiting form of a conic section (unless it be imaginary). Proof. Any equation of the second degree in x and y can be written in the form
(1)
Ax2
+ 2 Bxy + Cy + 2 Dx + 2 Ey + F 2
= 0,
where at least one of the coefficients A, B, C is different from zero. In § 248 we have seen that by making the rotation: (2)
x { y
= x' cos 0 - y' sin 0, = x' sin 0 + y' cos 0,
where cot 2 0 = (A - C)/2 B, we can always reduce equation (1) to a new equation of the second degree in x' and y', in which the
446
FIRST YEAR OF COLLEGE MATHEMATICS
coefficient of the term in x' y' is zero. only equations of the form Ax2
(3)
[Ch. XXX
We therefore need to consider
+ Cy + 2 Dx + 2 Ey + F 2
= 0,
where not both A and C are zero. For, if the coefficient B in a given equation (1) is not already equal to zero, we can reduce it to zero by means of a rotation. But, by the theorem of § 272, equation (3) always represents a conic section or a limiting form of a conic section, or has only an imaginary locus. This completes the proof. We can combine the results of §§ 248 and 272 to obtain tests for determining the general nature of the locus of a given equation of the second degree, without actually reducing the equation to standard form. If the given equation contains no term in xy, the nature of the conic is revealed by the value of AC. (§ 272.) · In order to consider the cases in which the given equation does contain a term in xy, we return to the rotation of axes by which that term can be removed. It was shown in § 248 that if we apply the rotation (2) to the equation (1), we obtain a new equation (4)
+ 2 B'x'y' + C'y' + 2 D'x' + 2 E'y' + F' 2
A'x'2
B2
and
-
AC
= B' 2
-
= 0,
A'C'.
In particular, if 0 is chosen so that cdt 2 0 = (A - C)/2 B, which changes the equation to one in which B' = 0, we have
B2
(5)
-
AC
= -A'C'.
But from the sign of A'C' we can determine the nature of the conic. (Note. In what follows, to say that a certain equation represents an ellipse will be understood to mean that its locus is an ellipse or the limit of an ellipse or is imaginary, and similarly for the other types.) Therefore, in the original equation (1), (a) if B 2 (b) if B 2 (c) if B 2
-
AC AC AC
< >
0, A'C' > 0, and the locus is an ellipse; 0, A'C' < 0, and the locus is a hyperbola; = 0, A'C' = 0, and the conic is a parabola.
It is easy to remember this result from the fact that according as B - AC is• greater than, equal to, or less than zero, the eccentricity of the conic is greater than, equal to, or less than 1. 2
CONIC SECTIONS
§§ 273-274j
447
1. The equation 4 x2 - 7 xy + y 2 - 3 x + 4 y + 1 = 0 represents a hyperbola, since B 2 - AC = (-f )2 - 4 • 1 = \ 9- - 4 = 3,l > O. EXAMPLE 2. The equation 4 x2 - 7 xy + 5 y 2 - 3 x + 4 y + 3 = 0 represents an ellipse, since B 2 - AC = (-f )2 - 4 • 5 = ~9 - 20 = - 3l < 0. EXAMPLE 3. The equation 25 x2 - 30 xy + 9 y 2 - 4 x + 2 y + 2 = O represents a parabola, since B 2 - AC = 152 - 25 • 9 = 225 - 225 = 0. EXAMPLE
EXERCISES As a review, solve the exercises of § 248, and discuss the loci. Without transforming the equations, determine the general nature of the locus of each of the following equations. 1. 3 x2 - 4 xy + 5 y2 - x + 3 y - 6 = 0. 2. 2 x2 + 6 xy - 5 y 2 + 3 x + 4 y - 10 = 0. 3. 3 x2 - 8 xy + 5 y 2 - 2 x + 4 y - 2 = 0. 4. 4 x2 - 8 xy + 6 y2 - 7 x - y + 6 = 0. 5. xy - y 2 + 3 X + 2 y + 7 = 0. 6. 2 x2 - 3 xy + 8 x - 5 y + l = 0. 7. 8 x 2 + 8 xy + 2 y2 - 3 x + 2 y - 17 = 0. 8. 20 x2 - 20 xy + 5 y2 + 4 x - 3 y - 6 = 0. 9. x 2 - 2 xy + y2 - 8 x + 5 y + 3 = 0. 10. 12 x2 - 26 xy + 5 y 2 + 4 y + 5 = 0.
VII.
SYSTEMS OF CONICS
274. The system of conics through the intersections of two given conics. Let (1) A1x2 + 2 B1xy + Ciy2 + 2 Dix + 2 Eiy + Fi = 0 and (2)
A2x2 + 2 B2xy
+ C2y + 2 D2x + 2 E2y + F2 2
= 0
be the equations of two given conics. Let u and v stand for the lefthand members of these equations, so that the equations can be abbreviated to read (1')
u = 0,
and
(2')
v = 0.
If k is any constant, the equation (3)
u
+ kv
= 0,
being of the second degree, in general represents a conic. And this conic passes through the points common to the curves (1) and (2). (Compare with §§ 236 and 254.) For if the coordinates of such a
448
FIRST YEAR OF COLLEGE MATHEl\1:ATICS
y
[Ch. XXX
common point be substituted in either u or v, they will reduce it to zero, and so, on substitution, they will satisfy the equation u + kv = 0. Equation (3) represents the system of conics through the points of intersection of the conics (1) and (2). By a proper choice of the parameter k, we can make the conic (3) satisfy one further condition.
u+kv=O
FIG. 177
EXAMPLE 1. Find the equations of (a) the parabolas which pass
through the points of intersection of the ellipse x2
(1)
+4y
2
= 4
X
= 1
and the hyperbola (2)
4 xy -
and (b) the conic through those points of intersection and the point (5, 1). Solution. (a) The equation of the system of conics through the points of intersection of (1) and (2) is (3)
(x 2
+4y
2
-
4)
+ k(4 xy -
1) = 0
x -
or (4)
x2
+ 4 kxy + 4 y
2
-
kx
+ (-4 -
k) = 0.
In order to determine k so that this equation shall represent a parabola, we must make B 2 - AC = 0. (§ 273.) Here B = 2 k, A = 1, C = 4. vVe therefore set 4 k2
(5)
-
4
= 0,
which gives k = 1 and k = -1. Using these values of kin (4), we obtain the desired equations of the parabolas,
+
+
+
+
(6) x2 4 xy 4 y2 - x - 5 = 0 and x2 - 4 xy 4 y2 x - 3 = 0. (b) In equation (3) we now wish to choose k so that this equation shall be satisfied by x = 5, y = 1. We therefore set (7)
or k = -
(25
i ~.
+4 -
4)
+ k(20 -
5 -
1)
= 0
Using this value of k in (3), and simplifying, we obtain
(8) 14x2 - lO0xy which is the desired equation.
+ 56y + 25x 2
31 = 0,
CONIC SECTIONS
§§ 274-275]
449
EXERCISES Find the equations of the following conics. 1. Through the intersections of x2 through the point (2, 3).
+y
2
Draw the figures.
= 9 and 2 x2
+y
2
-
2. Through the intersections of xy = 2 and 9 x 2 through the point (4, 1).
+ 16 y
3. Through the intersections of x2 - 2 xy - 2 y 2 and 4 x2 - 9 y 2 = 36 and the point (1, 1).
3x
4. Through the intersections of 2 x2 - xy - 3 y 2 x2 - xy 2 y - 6 = 0 and the point (-1, 2).
+
5. A parabola through the intersections of x2 2 x 2 - 6 xy - y 2 2x - 3y 6 = 0. 6. A parabola through the intersections of
+
4 x2
+
+ 8 xy + 3 x -
5y
+4 =
0 and
x2
-
+x -
+4y
2
2
2 = 0 and
= 144 and
+y -
36 = 0
5 y - 7 = 0 and
= 4 and
+ 2 xy -
4 y2
+5 =
0.
276. A conic through five given points. As either of the conics u = 0 and v = 0 in § 27 4, we may take a degenerate conic consisting of a pair of straight lines. In abridged form, let the equations of two lines be Z1 = 0 and Z2 = 0, where Z1 and Z2 are linear expressions in x and y. Then u = l1l2 is of the second degree in x and y, and l1l2 = 0 is the equation of the degenerate conic consisting of the two lines. In particular, if we are given four points, P1, P2, P 3, P1__, no three of which are collinear, and if the equations of P1P2, P3P4, P1P3, and P2P4 are, respectively
P1P2: P1P3: the equations
Z1 = 0, Z3 = 0, Z1Z2 = 0
P3P4: P2P4: and
l2 = 0, l4 = 0,
'
Zi4 = 0
represent two degenerate conics which intersect in the four points. With k as the parameter, the system of conics through the four given points is therefore defined by the equation
FIG. 178
(1)
If P5 is any fifth point not on Z3 or Z4, we can substitute its coordinates in equation (1) and determine a value of k for which the conic (1) will pass through the five given points. Therefore one conic can always be passed through five given points, no .four of which are collinear.
450
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXX
In case four of the points are collinear, it is impossible to draw two pairs of lines which intersect in four of the five points. If three of the points lie on a line, that line may be taken as Z1. Then, substituting the coordinates of P5 in (1) in order to determine k, we find that k = O, and (1) reduces to the equation Z1Z2 = 0. Hence the conic through five points three of which are collinear degenerates into the two 179 straight lines determined by the three collinear points and by the remaining pair of points. That a conic may, in general, be passed through five given points, may be seen from the general equation of a conic
l2
(2)
Ax2
+ 2 Bxy + Cy + 2 Dx + 2 Ey + F 2
= 0.
This equation contains six coefficients, but only five of them are independent. For, by dividing the equation by any one of the coefficients (not zero), we obtain an equivalent equation with five independent coefficients. By choosing them we impose five conditions on the corresponding conic. We can, in particular, choose them so that the conic will pass through five suitably given points. One method of doing this is to substitute the coordinates of the points, one after another, in the general equation (2), and so obtain five equations in the five unknown, independent coefficients. The solution oJ this system of equations will give the coefficients of the desired conic, but the method is usually much more laborious than the method outlined at the beginning of this article. EXAMPLE. Find the equation of the conic through the five points (0, 2), (0, 0), (2, 3), (1, -1), (3, -2).
Solution. Let us first find the equation of the system of conics through the first four of the given points. To do this we find four lines joining the points paired in such a way that each pair of lines passes through all four points. If we letter the points as shown in Figure 180, y lines P 1P2 and P 3p4 form one pair of lines passing through all four points, and so do the lines P ;I>3 and P 1P 4• The equations P2 (0, 2) of the lines are: l4__..,.."(~ 0 X
P,J'3:
X
P1
2y
X -
PJJ4: 4 X P1P4:
--+-+-~--++-+-+--+-+--i--+--+--+,:....
= 0,
-
+y
+ 4 = o,
y -
= 0.
5=
o,
Frn. 180
x
CONIC SECTIONS
§§ 275-276]
451
The equation of the system of conics through P 1, P 2, P s, P 4 is therefore
+ k(x -
+ 4)(x + y)
= 0. To make the conic pass through P5, we substitute x = 3, y = -2: (2) 3(12 + 2 - 5) + k(3 + 4 + 4)(3 - 2) = 0, so that k = -tf. Substituting this value fork in (1), and simplifying, we (1)
x(4 x - y - 5)
2y
obtain the equation of the ellipse (3)
17 x 2
+ 16 xy + 54 y
2
-
163 x - 108 y = 0.
EXERCISES Find the equations of the following conics. 1. Through the points 2. Through the points 3. Through the points 4. Through the points 5. A parabola through 6. A parabola through
*VIII.
Draw the figures.
(0, 2), (3, 0), (2, -1), (-1, -2), (-2, 1). (3, 2), (1, 1), (-1, -2), (-1, 3), (2, -3). (1, 1), (-3, -4), (2, -2), (4, -1), (-5, 5). (-2, 4), (2, -2), (1, 3), (-3, -6), (4, -5). the points (0, 1), (3, 1), (3, -2), (-2, -2). the points (1, 1), (-3, -3), (-2, -1), (3, 1).
SECTIONS OF A CONE
276. Sections of a cone. Conic sections receive their name from the fact that any plane section of a right circular cone is a conic section or a limiting form of a conic section. We shall now prove this theorem. Let a right circular cone be cut by a plane, p, in the curve KPM. We wish to prove that KPM is a conic section. Consider a sphere which is tangent to the cone and to the plane p. It is tangent to the cone along a certain circle RLG, whose center is at C, and to the plane p at a point F. Let p', the plane of the tangent circle, RLG, interse ct plane p in DE. Then the curve P' of intersection KPM is a conic having F as a focus and DE as the corresponding directrix. For let P be any point of the curve. Draw PF. Also draw PV V to the vertex of the cone, Fm. 181
452
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXX
PE perpendicular to DE, and PN perpendicular to the plane p'. gent to the sphere at G and PF is tangent to it at F. Hence
PV is tan-
(1)
PF= PG,
for two tangents to a sphere from a point are equal. PNG and PNE, we see that PN = PG · sin PGN
and
From the right triangles
PN = PE · sin PEN.
Consequently PG PE
(2)
sinPEN
- sin PGN
Therefore, by (1) and (2), (3)
PF PE
PG PE
-=-=
sin PEN • sin PGN
But PEN is equal to the angle between the planes p and p' and is therefore the same for all points P on the curve. And PGN is the angle between a genera ting line or element of the cone and a plane perpendicular to the axis of the cone. It is therefore the same for all points P on the curve. Therefore the ratio sin PEN/sin PGN is a constant, say e. Then (4)
PF PE= e,
and the locus of Pis a conic having F for its focus, DE for its directrix, and e for its eccentricity. In case the cutting plane, p, intersects all of the elements of the cone, as it does in Figure 181, another sphere can be constructed on the other side of p. Its point of tangency with pis a second focus of the conic. If the plane p intersects both nappes of the cone, the angle PEN is greater than PGN. In this case sin PEN > sin PGN, so that e > 1, and the conic is a hyperbola. If p is parallel to an element of the cone, PEN = PGN, e = 1, and the curve is a parabola. If p intersects all of the elements of the cone in one nappe, PEN < PGN, e < 1, and the conic is an ellipse. The student should use a solid model in order to examine the various cases. He should also notice the limiting cases that yield circles and pairs of straight lines.
CHAPTER XXXI
Pertnutations and Cotnbinations 277. Fundamental Principle.
If one thing can be done in just m ways; and if, after it has been done, a second thing can be done in just n ways; then both things can be done in the order stated in just m · n ways. Proof of the Principle. For each one of the m ways of doing the first thing there are n ways of doing the second. Hence there are mn ways of doing both things in the order stated. Note. The number of ways of doing a thing is always to be understood as the number of dijferent ways in which it can be done. While many of the problems in permutations and combinations appear to be merely amusing exercises, a knowledge of the subject is essential in probability, statistics, and many other sciences. 1. In how many ways can a chairman and a secretary be chosen from a committee of six men? EXAMPLE
Solution. Any one of the six men can be chosen as chairman; that is, there are m = 6 ways of choosing the chairman. After he has been chosen, there are n = 5 ways of choosing the secretary from the remaining five men. Hence, by the Principle, there are m · n = 6 · 5 = 30 ways of choosing a chairman and secretary. If the six men are A, B, C, D, E, F, the thirty selections can be shown as follows, the first letter of each pair denoting the chairman and the second denoting the secretary:
AB,AC,AD,AE,AF;BA,BC,BD,BE,BF;CA,CB,CD,CE,CF; DA,DB,DC,DE,DF;EA,EB,EC,ED,EF;FA,FB,FC,FD,FE. As an exercise, the student should state the obvious extension of the Principle to the case in which three or more things are to be done. If two or more things are independent, that is, if the performance of one of them does not affect the number of ways of doing the others, it is not necessary to specify the order. 453
454
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXI
In applications of the Principle, it is often convenient to make a series of blanks, or marks, one for each of the things that are to be done. Then, in the order in which the things are done, fill in the corresponding blanks with the numbers of possible ways of doing the things. The number of ways of doing the set of things in the specified order is then equal to the product of the numbers in the blanks. 2. How many six-letter words* beginning and ending with a consonant can be formed from the letters of the word PRODUCT, no letter occurring more than once? EXAMPLE
Here we have six things to do. We must choose a letter for each of the six positions in the word. We -1 -3 -4 -5 -6 -2 make six blanks to fill. The only special conditions affecting the order are that the first and sixth letters must be consonants. Let us therefore count first the ways of choosing the first letter, then the sixth letter, then the second, etc., and fill in the corresponding blanks in the order indicated by their subscripts. "\Ve can choose any one of the five consonants 5 4 3 2 4 5 P, R, D, C, T, for the first letter, any one of 5 6 2 1 3 4 the four remaining consonants for the sixth letter, any one of the five remaining letters (consonant or vowel) for the second letter, etc. We fill in the blanks as shown. The total number of "words" that satisfy the conditions is therefore 5 · 5 · 4 · 3 · 2 · 4 or 2400. Solution.
EXAMPLE 3. How many even numbers less than 4000, of four different digits each, can be formed with the digits 0, 1, 2, 3, 4, 5, 6?
Solution.
In order to be less than 4000, the number must begin with one of the three digits 1, 2, or 3. It will be necessary to 1 5 4 3 count separately the cases in which the thousands' 4 2 1 a digit is 2 and those in which it is 1 or 3. (a) If the thousands' digit is 2, it can be selected in just one way. To be even, the number must then end with 0, 4, or 6; the units' digit can therefore be chosen in any one of three ways. For the hundreds' digit we can choose any one of the five numbers not already used, and we * In such connections, by "word" we shall mean any arrangement of letters without considering whether the arrang~ment appears in any dictionary. Also, in this chapter, in considering the number of numbers that can be formed from a set of digits, by "number" we shall mean a positive integer. Each number must be a genuine number written in conventional form; thus 057 will not count as a number, but 507 will.
§ 277]
PERMUTATIONS AND COlvfBI~ATIONS
455
have four choices for the tens' digit. The number of suitable numbers beginning with 2 is therefore 1 • 5 · 4 · 3 or 60. (b) If the thousands' digit is not 2, it is 1 or 3, and can be selected in two ways. 2 The units' digit is 0, 2, 4, or 6, and can be selected in 5 4 -4 - 3- 4 2 1 four ways. The hundreds' digit can be chosen in five ways and the tens' digit in four ways, as before. There are therefore 2 · 5 · 4 · 4 or 160 suitable numbers beginning with the digit 1 or 3. If we add these to the 60 that begin with 2, we have in all 220 numbers satisfying the conditions of the problem. 4. In how many ways can four coins be thrown? Note. In applications in the theory of probability, genetics, and elsewhere, it is most useful in such problems to assume that when the first of two coins is head and the second tail, the coins are thrown in a different way from when the first coin is tail and the second is head. Unless otherwise stated, such problems are always to be solved on this assumption. EXAMPLE
2
2 1
2 2 -3
2 4
Solution. We can throw each coin in two ways. The total number of ways of throwing the four coins is therefore 2 • 2 • 2 • 2 = 16.
5. In how many ways can seven girls be arranged in a row, if two particular girls are to be placed together? EXAMPLE
Let us first think of the two particular girls as handcuffed together to make a single object. There are then 6 objects to assign to 6 places. The first place 6 5 4 3 2 1 4 5 6 2 3 1 can be filled in 6 ways, the second in 5 ways, etc.; there are 6 · 5 · 4 · 3 • 2 • 1 = 6! = 720 ways of arranging the 6 objects. However, for each way of placing the 6 objects, there are 2 ways of arranging 720 2 1 2 the 7 girls, by interchanging the two particular girls. The required number is therefore 720 · 2 = 1440. Solution.
EXERCISES 1. In how many ways can 5 men be arranged in a straight line? 2. In how many ways can 7 students be assigned to a row of 7 seats'? , 3. In how many ways can 6 students be seated if there are 8 vacant seats? 4. In how many ways can 5 coins be thrown? 5. (a) How many "words" of four different letters each can be made from the letters A, E, I, 0, R, S, T? (b) How many of them begin with a vowel and end with a consonant? (c) In how many do vowels and consonants alternate?
456
FIRST YEAR OF COLLEGE ::.\IATHEMATICS
[Ch. XXXI
6. In how many ways can the three offices of chairman, secretary, and
treasurer be filled from a committee of 8 persons?
7. (a) How many numbers of 5 different digits each can be formed from the digits 0, 1, 2, 3, 4, 5, 6? (b) How many of them are even? (c) How many of them are exactly divisible by 5? 8. A certain school bus contains seats for 14 passengers on each side of
the aisle. The boys always sit on the left and the girls on the right. On arriving at a stop the bus contains 10 boys and 8 girls. In how many ways can 4 new boys and 5 new girls be seated?
9. There are 9 chairs in a row. In how many ways can 4 students be seated in consecutive chairs? (Suggestion. First find the number of ways of choosing 4 consecutive chairs.) 10. There ar~ 10 chairs in a row.
In how many ways can 5 students be seated for a quiz so as to leave alternate chairs vacant? 11. In how many ways can 8 books be arranged on a shelf, if 3 particular
books are to be together? 12. (a) In how many ways can 3 men choose hotels in a town where there are 7 hotels? (b) In how many ways can they choose them, if two of the men refuse to stay at the same hotel?
13. (a) How many four-digit numbers are there in which all the digits are different? (b) How many of these are odd? (c) How many are divisible by 5? 14. How many signals, each containing 5 flags arranged on a staff, can
be formed from 7 flags of different colors? 15. (a) In how many ways can 4 English books and 3 French books be arranged in a row on a shelf? (b) In how many of these will the French books be together? 16. If there are 6 railroads from Chicago to Minn ea polis and 4 railroad~ from Minneapolis to Seattle, in how many ways can one go by rail from Chicago to Seattle via Minneapolis, and return, without going over the same line twice? 17. (a) In how many ways can the letters of the word GEOMETRY be arranged so that vowels and consonants alternate? (b) In how many of
these is Y the final letter? 18. In how many ways can eight boys be arranged in a row if two particu-
larly unruly boys are not permitted to be at either end of the row or together? 19. (a) How many three-digit numbers can be formed, if no digit is used more than twice in the same number? (b) How many of them are odd? (c) How many are divisible by 5?
§ 277]
PERMUTATIONS AND COMBINATIONS
45i
20. A man and his wife invite 4 men and 4 women to dinner. After the
host and hostess have taken their places at the ends of the table, in how many ways can the guests be seated so that no two men sit together? 21. A man and his wife invite 4 other couples to dinner. After the host and hostess have been seated at the ends of the table, in how many ways can the guests be seated so that no man sits beside another man or beside his own wife? (Suggestion: Make a diagram of the chairs and table; first find the number of ways of seating the ladies, a, b, c, d; for a typical arrangement of the ladies, count the ways in which their husbands, A, B, C, D, can be assigned to the vacant chairs so that A and a, Band b, etc. are not adjacent.) 22. (a) With letters of the word
no letter being used more often than it appears in that word, how many "words" of five letters each can be formed in which vowels appear in the first, third and fifth places? (b) How many can be formed in which vowels and consonants alternate? FREEDOM,
23. (a) How many five-digit numbers greater than 65,000 can be made without repetition of digits? (b) How many of these are odd? 24. Given 3 red flags, 2 white flags, 1 blue flag, how many signals can be
made by arranging 3 flags in a line? 25. From the letters of the word
how many arrangements of five different letters each can be made (a) consisting only of vowels; (b) in which consonants and vowels alternate; (c) in which Q is immediately followed by U? 26. From the letters of the word
EQUATION,
using no letter more often than it appears there, how many five-letter arrangements can be made (a) in which S's alternate with other letters; (b) in which three S's stand together; (c) which begin and end with S? SISTERS,
~7. (a) Find the number of four-letter arrangements that can be formed from the ten characters a, a, a, a, b, c, d, e, f, g. (b) How many of them end with a vowel? 28. In how many ways can 5 different gifts be given to 3 persons?
29. In how many ways can 3 dice be thrown at the same time? (Note: A die has six faces, labeled from 1 to 6; if the first die shows 6 and the second shows 1, it is "different" from the case in which the first shows 1 and the second shows 6.) 30. In how many ways can 3 coins and 3 dice be thrown at the same time? l See note for Exercise
29.)
458
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXI
278.. Permutations. If we are given a set of n different objects, and arrange r of them in a definite order, such an ordered arrangement is called a permutation of the n objects r at a time. The number of such permutations that can be made is denoted by the symbol nPr. Thus sP6 is read "the number of permutations of 8 different things 6 at a time."
While most problems involving permutations can be solved, as in § 277, by direct applications of the Fundamental Principle, it is convenient to have a formula for nPr. Theorem I. The number of permutations of n different things taken r at a time is
nPr
= n(n -
1)(n - 2) • · · (n - r + 1).
Notice that on the right-hand side, the first factor is n, the second is (n - 1), etc., and that there are r factors in all. Proof offormula. In finding the permutations of n different objects r at a time, we have r places to fill. n (n - 1) (n - 2) • • • (n - r + l) We can fill the first place with -1 3 r any one of the n given objects, the second place with any one of the (n - 1) remaining objects, the third with any one of the (n - 2) remaining objects, etc. The rth place can be filled with any one of (n - r l) objects. Hence, by the Fundamental Principle, there are
+
n(n - 1) (n - 2) • · • (n - r
ways of filling the r places.
Theorem IL all at a time is
In particular, by taking r = n, we have:
The number of permutations of n dijferent things taken
npn EXAMPLES.
+ l)
= n(n-l)(n -
2) • · · 1
1Ps = 7 · 6 · 5 · 4 · 3 = 2520; 1P1
= n!. = 7! = 5040.
279. Circular permutations. The number of orders in which n different things can be arranged about a closed curve such as a circle is (n - 1) !. Proof. The relative order of the objects is not changed if they are all moved along the curve the same number of places. We can consequently place one of them in a fixed place and arrange the remaining (n - 1) objects in all possible ways. The number of essentially different arrangements is therefore n-1Pn-1 or (n - l_)t.
§§ 278-280]
PERMUTATIONS AND COMBINATIONS
459
EXAMPLE 1. In how many ways can 8 persons be seated at a round table·? Solution. This involves circular permutations, with n = 8. ber of essentially different arrangements is therefore 7 ! = 5040.
The num-
EXAMPLE 2. Given 10 beads of different colors, how many different necklaces can be formed by stringing them all together? Solution. From the formula for circular permutations we should have 9 ! = 362,880. However, if the beads are arranged in any one of these ways, we can get another one of these arrangements merely by turning the necklace over, without restringing the beads. The necklace evidently remains the same as before. Each necklace gives two of the arrangements counted. The number of really different necklaces is therefore 9 !/2 = 181,440.
280. Permutations of things some of which are alike. Suppose that we are given n objects of which k are just alike, the other (n - k) objects being different from them and from each other. Let us denote by P the number of really different arrangements that can be made of the n things, taking all of them at a time. Then this number P is evidently less than nP n, the number of arrangements that could be made if all of the objects were different. For after having arranged the n objects, of which k are alike, in any order, let us suppose that we rearrange the k like objects among themselves without disturbing the others. If, temporarily, we had some way of telling the like objects apart, we could do this in k! ways. In fact, corresponding to each of the P really different arrangements of the n objects, there are k! arrangements which would be different if we could tell the k like objects apart, or, in all, P · k! arrangements. But we know that, if we could tell the objects apart, there would be nP n arrangements. Consequently
P · k ! = nP n = n ! ,
or
, P
n
T
=-·· k!
A similar argument yields the following theorem: Theorem. Given n objects, of which just k1 are of one sort, just k2 are of another sort, etc.; the number of different permutati·ons that can be made of then objects taking them all at a time is
p
=
n! . k1 ! k2 ! · · ·
460
FIRST YEAR OF COLLEGE MATHEMATIC8
[Ch. XXXI
EXAMPLE. How many different arrangements of 12 letters each can be made from the letters of the word ANALY'rICALLY?
Solution. There are 12 letters of which 3 are A's, 3 are L's, 2 are Y's, and the rest are unlike. The number of arrangements is P =
12 ! ! ! = 6,652,800. 3 3 2
!
EXERCISES Read each of the following symbols in words, and compute its value.
In Exercises 7 to 10, express each number by a symbol, and compute its value.
7. The number of permutations of 7 different things 4 at a time. 8. The number of permutations of 6 different things all at a time. 9. The number of permutations of 9 different things 3 at a time. 10. The number of permutations of 5 things at a time. 11. How many signals can be made with 5 flags of different colors by arranging them all on a mast together? 12. How many signals can be made with 7 flags of different colors by arranging them all on a mast together? 13. How many signals can be made with 7 flags of different colors by arranging 4 of them at a time on a mast? 14. Hmv many signals can be made with 6 flags of different colors by arranging 5 of them at a time on a mast? 15. Hmv many signals can be made witn 3 blue flags, 2 red flags, and 1 yellow flag by arranging them all on a mast together? 16. How many signals can be made with 3 blue flags, 3 red flags, and 3 yellow flags, by arranging them all on a mast together? 17. A basketball team of 5 regulars and 3 substitutes is passing a ball around a circle. In how many ways can the men be arranged in the circle?
18. How many different bracelets can a child make by stringing 8 beads of different colors all at a time? 19. (a) How many seven-digit numbers can be made from the digits 1, 2, 2, 2, 3, 3, 5? (b) How many of them are odd? 20. (a) How many arrangements of 11 letters each can be made from the letters of the word COMBINATION? (b) How many of them end with the
letter N? (c) How many do not end with 0, I, or N?
§§ 280-281]
PERMUTATIONS AND COMBINATIONS
461
21. How many different line-ups each consisting of 2 forwards, 2 guards, and 1 center can be made from a basketball squad of 5 forwards, 4 guards, and 3 centers? If two men change places, there is a change in the line-up. 22. How many teams each consisting of 3 outfielders, 4 infielders, 1 pitcher, and 1 catcher, can be made from a baseball squad of 5 outfielders, 6 infielders, 5 pitchers and 2 catchers? If two men change places, there is a change in the team. 23. Twelve children are marching counter-clockwise single-file in a ring. In how many ways can they be arranged? 24. Five girls and five boys are marching single-file in a ring, the boys and girls alternating. In how many ways can they be arranged? 25. Six boys and six girls are marching counter-clockwise two abreast in a ring, each girl at the left of a boy. In how many ways can they be arranged? 26. A man has 4 sets of books, one consisting of 5 volumes, two of 3 volumes each and one of 4 volumes. In how many ways can they all be arranged on a shelf, if the books in each set are to be together? 27. A man has 3 sets of books, one consisting of 5 volumes, and two consisting of 3 volumes each. He also has 4 copies of one other book. In how many ways can they all be arranged on a shelf if the books of each set are to be together and the 4 copies of the same book are to be together? 28. A chorus consists of 6 girls in red and 6 girls in silver costumes. In how many ways can they all be arranged (a) in two rows, the red behind the silver; (b) in two concentric rings facing the center, one girl in red behind each girl in silver; (c) in a ring facing the center, the colors alternating; (d) in a row facing the front, the colors alternating? 29. From the letters of the word ALGEBRA, how many arrangements can be made taking any number of letters, from one to seven, at a time? 30. (a) How many seven-digit numbers can be formed with the digits 2, 2, 2, 3, 3. 4, 5? (b) How many of these are greater than 3,400,000? (c) How many are greater than 3,400,000 and are divisible by 5? (d) How many are greater than 3,400,000 and are even?
281. Combinations. If we are given a group of n objects, any selection or set of r of the objects, considered without regard to their arrangement, is a combination of the n objects r at a time. We denote by nCr the number of possible combinations of n different things r at a time. EXAMPLE 1. The combinations of the letters a, b, c taken two at a time are ab, ac, be. Considered as combinations, ab and ba are the same, though as permutations they are different. Thus aC 2 = 3. By rearranging the letters
462
FIRST YEAR OF COLLEGE MATHEMATICS
[ClL XXXI
in each of the three combinations ab, ac, be, we obtain the six permutations ab, ba, ac, ca, be, cb. Therefore 3P 2 = 3C2 • 2 = 6.
We can form a permutation of n different things r at a time in nPr ways. But we can form such a permutation also by first choosing a group of r objects from the set, and then arr! ranging the r objects in the group. By the --1 2 definition of nCr, we can choose the group in nCr ways. And we can then arrange the objects in the group in r ! ways. Therefore, by the Fundamental Princjple, we can form a permutation of n objects r at a time in nCr · r! ways. That is, n Cr
or Substituting nPr = n(n - 1) · · · (n - r
(1)
nCr
=
=
nPr. r.f
+ 1), we obtain
n(n - 1)(n - 2) · · · (n - r f
r.
+ 1) •
Note that in the right-hand member the numerator contains r factors of which the largest is n, and the denominator has r factors of which the largest IS r.
If we multiply numerator and denominator of (1) by (n - r) !, smce [n(n -
1) (n - 2) • · · (n - r
+ l)][(n -
r) • • · 2 • 1]
= n!,
we obtain the compact form
n!
(2)
nCn-r
= r.f ( n _
r ) ,• •
Notice that (3)
For obviously, for each combination of r objects that we select from a set of n objects, we leave one set of (n - r) objects. Another proof results from noticing that the right-hand member of formula (2) is unchanged if r and (n - r) are interchanged. We should use formula (3) to compute nCr whenever r
>
n - r.
EXAMPLE
2.
Compute the number of combinations of 40 things taken 36
at a time.
Solution.
By formulas (3) and (1), 40Ca6
=
10C4
=
40 · 39 · 38 · 37
l .
. . 2 3 4
= 91,390.
§§ 281-282]
PERMUTATIONS AND COMBINATIONS
463
Many problems involve the use of the Fundamental Principle, in which the number of ways of doing one of the necessary things is the number of selections that can be made from a certain group (combinations). 3. How many subcommittees consisting of 4 Democrats and 3 Republicans can be chosen from a main committee of 10 Democrats and 7 Republicans? EXAMPLE
Solution. Here the arrangement in the subcommittee is of no importance. We have to do two things: select the Democrats and select the Republicans. The first can be done in 10C4 ways and 10C4 1 2 the second in 1C3 ways. By the Fundamental Principle, the number of ways of selecting the subcommittee is therefore
10-9·8•7
10C4 • 1C3
7-6-5
= 1 . 2 . 3 . 4 · 1 . 2 . 3 = 7350.
4. How many ''words" each consisting of 3 different consonants and 2 different vowels can be made from the letters of the word SUBTRACTION? EXAMPLE
Solution. There are 6 different consonants and 4 different vowels. To form each word, we must do three things: (1) select 6C3 4C2 the 3 consonants, (2) select the 2 vowels, (3) arrange 5P 5 1 2 the 5 letters in order. The first can be done in 6C3 ways, s the second in 4C2 ways, the third in sPs ways. The required number of words is therefore 6C3 • 4C2 •
oP5
= 20 · 6 · 120 = 14,400.
282. Total number of combinations of n things. By comparison with §§ 105, 106, we see that nCr is the coefficient of the term in xr in the expansion of (a x) n by the Binomial Theorem. In fact,
+
(a+ x)n = an+ nC1an-Ix Setting a or
=
x
+ nC2an-2x + ••• + nCran-rxr + ••• + nCnxn.
= 1, we have 2n = 1 + nC1
2
+ nC2 + ••• + nCn
(1)
which is a formula for the total number of combinations of n things taking them 1 or 2 or 3 · • • or n at a time. In § 197 a proof of the Binomial Theorem is given making use of the notion of combinations.
464
FIRST YEAR OF COLLEGE J\tIATHEMATICS
[Ch. XXXI
How many scouting parties can be formed from a squad of 8 men, taking any number at a time? EXAMPLE.
Solution.
The number is 28
-
1
= 256 - 1 = 255.
EXERCISES Read each of the following symbols in words, and compute its value. 1. sCa.
2. 10C4.
3. 11Cs.
4. 12010.
5. 10C10.
6. 11010.
In Exercises 7 to 10, express each number by a. symbol and compute its value. 7. The number of combinations of 7 different things 4 at a time.
8. The number of combinations of 8 different things 5 at a time. 9. The number of pairs that can be formed from 12 things. 10. The number of groups of 3 that can be formed from 12 things.
=
=
=
11. Prove that nCn 1 and nC1 nCn-1 n. 12. A store advertises a sale of men's suits. With 9 models to choose from, in how many ways can it select 4 suits of different kinds for its window display? 13. From a counter containing 12 recent novels, in how many ways can
a person select 4 to read during his vacation? 14. In a men's glee club there are 10 tenors and 14 basses. In how many ways can the director select a sextet of 3 tenors and 3 basses? 15. A lady gives a luncheon party with 6 guests. (a) In how many ways can she select her guests from a group of 9 friends? (b) In how many ways can she select them, if 2 of her friends will not attend a party together? (c) How many of her selections include a certain particular one of her friends? 16. Given a penny, a nickel, a dime, a quarter, a half-dollar and a dollar, how many sums of money can be formed (a) using 2 coins; (b) using any number of coins. 17. If 8 coins are tossed at the same time, in how many ways can they show exactly 4 heads? 18. A committee of 5 is to be chosen from a group of 6 seniors and 7 juniors. (a) How many committees can be chosen? (b) How many of them contain 2 seniors and 3 juniors? (c) How many contain only juniors? (d) How many contain one particular junior? 19. There are 10 football teams in a certain conference. How many games must be arranged if each team is to play every other team just once? 20. (a) In how many ways can 9 presents be given to two children? (b) In how many of these will the younger child receive just 5 presents?
§ 282]
PERMUTATIONS AND COMBINATIONS
465
21. Seven boys reach a fork in a road. In how many ways can they continue their walk so that 4 go one way and 3 the other? 22. There are 9 points marked in a plane, no three of which lie in a straight line. (a) How many straight lines can be drawn each containing 2 of the points? (b) How many of these pass through one or more of 3 specified points of the set? 23. A party of 8 boys and 8 girls are going for a picnic. Six of the party can ride in one automobile, and four in another. The other six must walk. (a) In how many ways can the party be distributed for the trip? (b) In now many of these will no girl have to walk? (c) In how many of them will no girl have to walk if each of the two boys who own the cars drives his own car? 24. Two dice are tossed together. (a) In how many ways can they fall? How many of these show a total of (b) 5, (c) 6, (d) 7, (e) 8? 25. (a) In how many ways can four balls be drawn together from a bag containing 5 red balls, 7 white balls, and 3 blue balls? How many of these include (b) 2 red and 2 white balls, (c) just 2 white balls, (d) only white balls, (e) balls of only one color, (f) no red balls? 26. Ten points are marked in a plane, no three of them lying in a straight line. How many triangles do they determine? 27. From 10 beads of different colors, how many bracelets of 6 beads each can be formed? 28. From 7 French and 9 English books, in how many ways can we select 3 French books and 4 English books and arrange them all on a shelf? 29. From the letters of the word VOLUME, (a) how many selections can be made, taking any number at a time; (b) how many arrangements can be made? 30. With the letters of the word CANADIANS, using no letter more often than it occurs there, (a) how many selections of 4 letters each can be made; (b) how many arrangements of 4 letters each can be made? 31. (a) In how many ways can 52 different cards be dealt evenly to 4 players, the order of the hands being considered, but not the order of the cards in a hand? (b) In how many ways can the 52 cards be divided into 4 hands of 13 cards each, if the order of the hands is not considered? 32. Given that nC2 = 28, find n. 33. Given that nC2 = 55, find n. 34. Given that the number of pairs of players that can be formed from the members of a certain golf club is 5995, find the number of members. 35. In a certain basketball association, it is necessary to arrange 66 games in order to have each team play every other team just once. How i11any teams belong to the association?
466
FIRST YEAR OF COLLEGE MATHEMATICS
36. Write the expansion of (a greatest? 37. Write the expansion of (a greatest?
+ x)
+ x)
•
For what value of r is
10
Cr the
•
For what values of r is
13
Cr the
10
13
[Ch. XXXI]
38. Prove that
Hence show that, for a given value of n, the greatest value of nCr is that for which r = ½n when n is even and is that for which either r = ½(n - 1) or r = ½(n + 1) when n is odd. 39. From the results of Exercise 38, find the value of r for which 12Cr is the greatest. 40. From Exercise 38, find the values of r for which 11Cr is the greatest. 41. (a) Show that the total number of arrangements that can be made of n things taken any number at a time is equal to n! [ 1
1
1
1
1
+ 1 + 2! + 3! + 4! + · · · + (n -
]
1)! ·
(b) Using this result, find the total number of ways in which
n books can
be arranged on a shelf using any number of them at a time.
*283. Proof of the Binomial Theorem by combinations. Using the combination symbols for the binomial coefficients, we can write the Binomial Theorem (§ 104) for a positive integral exponent n in the form (1)
(a+ x)n = an + nC1an-Ix
+ nC2an- x + ••• + nCran-rxr + ••• + nCnxn. 2 2
In § 108 this theorem was proved by mathematical induction. We shall now give an independent proof by means of combinations. By definition, (2)
(a+ x)n = (a+ x)(a + x) • • • (a+ x), (ton factors).
The product on the right-hand side is equal to the sum of all the terms that can be formed by multiplying together one letter from each factor. In particular, we obtain the term an-rxr by multiplying the x's from r selected factors by the a's from the (n - r) remaining factors. But we can select these r factors in nCr ways, and each selection yields the term an-rxr. Therefore nCr is the coefficient of the term in an-rxr, in the binomial expansion of (a+ x)n. Consequently the Binomial Theorem is correct, since its general term is correct.
CHAPTER XXXII
Probability 284. Definition of mathematical probability. In our daily affairs we very frequently base a decision on a more or less careful estimate of the probability that a certain event will occur on a given occasion. In mathematics, there are two different kinds of bases for estimating the probability that a certain event will occur at a given trial. In the first, we base our estimate on an analysis of the different ways in which the event may occur or fail to occur and a subjective assumption that these ways are equally likely; the result is called mathematical probability. The second kind, called statistical or empirical probability, is based on a knowledge of what has occurred in the past, and a belief that approximately the same results will be obtained in the future. The relations between mathematical and statistical probability are discussed in §§ 286 and 287. Although many of the simple problems in probability are based on games of chance, the more important applications are in statistics and other sciences. Definition of mathematical probability. Suppose that, at a given trial, a certain event can occur or fail to occur in t different ways, all of which are assumed to be equally likely, and that the event will occur in h of these ways and will fail to occur inf of the ways; then the probability that the event will occur at the given trial is, by definition,
p
h
= 7;
the probalrility that the event will fail to occur is
q
={-
Notice that t = h + f. If the event is certain to occur, so that there are no ways in which it can fail, then f = 0 and h = t, and p = 1. If the event is certain to fail, then h = 0 and p = 0. In general, since 0 < h < t, 0 < p < 1. Therefore p is a number which ranges from 0, its value when the event cannot possibly occur, to 1, its value when the event is certain to occur. Also 467
468
FIRST YEAR OF COLLEGE MATHE1\1ATICS [Ch. XXXII
p+q
= 1.
If p > ½, so that h > f, we say that the odds are h to fin favor of the event; when p < ½, the odds are h to f against the event; when p = ½, the odds are even for and against it. The question as to whether t different ways in which the event can occur or fail to occur are "equally likely" involves many difficulties. In effect, we subjectively assign certain probabilities, p1, p2, • · · , Pt, to them. If we consider that Pi = p2 = · · · = Pt and that P1 P2 Pt = t · P1 = 1, the t ways are "equally likely."
+ + ·· · +
In a particular example, it is usually best to determine first of all the value of t, the total number of ways in which the event can either occur or fail to occur, and then the value of h, the number of ways favorable to the event. EXAMPLE 1. There are 5 white balls and 7 black balls in a bag, and 2 balls are drawn* together. Find the probability that both balls are black. Solution. There are 12 balls in all, and there are 12C2 = 66 possible pairs that can be drawn. Hence t = 66. Of these, 1C2 = 21 is the number of pairs of black balls that can be drawn, and h = 21. The desired probability
is therefore
p =
21
7
66 = 22•
EXAMPLE 2. A teacher seats a group of 4 boys and 4 girls at random in a row. What is the probability that the boys and girls will alternate? Solution. The total number of arrangements of the 8 persons is t
=
sPs
= 8!.
The number of arrangements in which the boys and girls alternate is h = 2 · (4!) {4!) (the boys can be arranged in the odd-numbered seats in 41 ways and the girls in the even-numbered seats in 4! ways.) The desired 24 4 p = ( !)( !) = 1-. probability is therefore
8! 35 The "odds" are 1 to 34 against such an arrangement occurring by chance.
286. Mathematical expectation. If the probability that a person will receive a certain sum of money is p, and the value of that sum is $S, then his mathematical expectation with respect to that sum is $pS. A person's total expectation with respect to several sums is the sum of his expectations with respect to the individual sums. * In all such problems it is assumed that the balls are indistinguishable except for color and that when balls are drawn, one ball is as likely to be drawn as any other.
§§ 284-285]
PROBABILITY
469
ExAMPLE. Two boys find a half-dollar, and toss the coin to determine which boy shall have it. Each boy's mathematical expectation is $.25. For there are two ways in which the coin may fall; one of these ways is favorable to each boy. Hence the probability that a certain one of the boys will win is p = ½. His mathematical expectation is therefore $(½) (.50) = $.25.
EXERCISES 1. One ball is drawn from a bag containing 4 white and 8 black balJs. Find the probability that it is (a) white, (b) black.
2o One ball is drawn from a bag containing 6 white and 7 black balls. Find the probability that it is (a) white, (b) black. 3. One ball is drawn from a bag containing 7 white balls, 8 black balls, and 12 red balls. Find the probability that it is either white or black. 4. One ball is drawn from a bag containing 3 white, 4 black, 5 red, and 6 yellow balls. Find the probability that it. is either white or red. 5. If a die* is cast, find the probability that either a 1 or a 2 is thrown. 6. If a die is cast, find the probability that 5 is not thrown.
7. If two dice are cast, find the probability of throwing 6 . 8. If two dice are cast, find the probability of throwing 9~ 9. If two dice are cast, find the probability of throwing at least 8. 10. If two dice are cast, find the probability of throwing less than 9.
11. If three coins are tossed, find the probability of throwing (a) exactly 1 head, (b) 3 heacl.s. 12. If four coins are tossed, find the probability of throwing (a) exactly 2 heads, (b) exactly 3 heads. 13. Two balls are drawn together from a bag containing 4 white and 8 black balls. Find the probability that (a) both are black; (b) one is white and one is black; (c) neither is black. 14. Three balls are drawn together from a bag containing 4 white and 6 black balls. Find the probability that (a) all are white; (b) just 2 are white; (c) just 1 is white; (d) all are black.
15. Three balls are drawn together from a bag containing 4 white, 5 black, and 6 red balls. Find the probability that (a) they include one ball of eaci1 color; (b) all are of the same color; (c) they include no white ball. * In all dice problems assume that each die is a cube with faces numbered 1, 2, 3, 4, 5, 6. A 4 will be "thrown" if the face marked 4 is up after the die is cast. If two or more dice are cast, 4 will be "thrown'' if the sum of the numbers thrown on all the dice is 4.
470
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXII
16. Four balls are drawn together from a bag containing 4 white, 5 black, and 6 red balls. Find the probability that (a) they are all of the same color; (b) they include 2 white, 1 black, and 1 red ball; (c) 2 are white and 2 are red; (d) none are white. 17. In a lottery, the prize is $15, and 100 tickets have been issued. Find the mathematical expectation of a man who holds 9 tickets. 18. According to the terms of a man's will, his son is to receive $20,000 when he becomes 21 years old. If the probability that the son will live until he is 21 is 0.95, find his mathematical expectation. 19. A bag contains 7 balls numbered from 1 to 7. Two balls or numbers are drawn. Find the probability that (a) both are even; (b) both are odd; (c) their sum is even; (d) their sum is greater than 8; (e) their sum is divisible by 3. 20. Thirteen slips of paper numbered consecutively from 1 to 13 are placed in a box and three numbers are drawn together at random. Find the probability that {a) all three are even, (b) two are even and one is odd, (c) their sum is even, (d) their product is even. 21. Four equally good trains leave Chicago for a nearby town at about the same time. (a) If two friends chance to go at this time, find the probability that both of them will be on the same train. (b) If three friends go, find the probability that at least two of them will be on the same train. 22. Find the probability that of the first three persons you meet on the street today (a) all were born on Sunday, (b) one was born on Sunday and two were born on Monday, (c) none were born on Sunday. 23. Ten students are assigned seats at random in a row. Find the probability that (a) 2 particular students will be seated side by side, (b) 3 particular students will be seated together. 24. Ten members of a seminar are seated at random at a round table. Find the probability that 2 particular students will be seated side by side. 25. Six men and six women are seated at random in a row. Find the probability that the men and women will be seated alternately. 26. Six men and six women are seated at random at a round table. Find the probability that the men and women will be seated alternately. 27. At a recent bridge party attended by six couples, each man was assigned a woman partner by lot. Each man drew his own wife as partner. Find the probability of this occurring. 28. If 4 cards are drawn from a deck of 52 cards, find the probability that (a) all are aces; (b) all are of the same suit; (c) all are of the same color. 29. A number of three different digits is formed at random with the 8 digits, 0, 1, 2, 3, 4, 5, 6, 7. Find the probability that (a) it is an even number; (b) it is divisible by 5; (c) it is less than 400; (d) it is less than 350.
§§ 285-286]
PROBABILITY
471
30. If 3 dice are thrown together, find the probability of (a) throwing exactly 12; (b) throwing less than 12; (c) all the dice showing the same number; (d) each die showing 6. 31. A professional basketball squad of 9 men will win a bonus of $2700 if they win a certain game in the play-off, the money to be equally divided among them. They estimate their probability of winning to be -{0 • What is each man's mathematical expectation? 32. If n boys and n girls are seated at random, show that the probability that the boys and girls will be seated alternately is the same whether they are all seated in a row or in a circle. (Suggestion. Find the ratio of one probability to the other.)
286. Statistical or empirical probability. I£ n trials of an event have been made, and if, in these n trials, the event has occurred h times, h is called the frequency of occurrence of the event, and the ratio h/n is the relative frequency of its occurrence. EXAMPLE 1. Out of 1000 white pine trees used as a sample in a certain region, it was found that 63 were infested with rust. The frequency of occurrence of rust in this sample was 63, and its relative frequency was Tiio or 6.3%.
In cases where it is not practicable to define the probability of the occurrence of a certain event by an analysis of the ways in which the event may occur or fail to occur, it is necessary to depend on the relative frequency of its occurrence within the range of our observation. If an event has been observed to occur h times in n trials, in the absence of other information we define the statistical or empirical probability that the event will occur at a future trial as the relative frequency:
p = :. 2. The probability that a white pine tree chosen at random from the region referred to in EXAMPLE 1 above will be infested with rust EXAMPLE
•
6 3
IS To O o·
It must be noted that the statistical probability is a variable, changing as additional observations are made. When the total number, n, of observations is small, we place little confidence in the meaning of the statistical probability in question. As the number of observations increases, our reliance on the probability increases also, so that when the number of observations is very large indeed, we use it confidently as the basis of important decisions. Life insurance companies, for example, base a premium rate on the statistical probability, determined after a great many observations, that a person of
472
FIRST YEAR OF COLLEGE lVIATHEMATICS
[Ch. XXXII
a certain age will continue to live for a certain number of years. Even when the number of observations is large, we use the statistical probability only in cases where, so far as we can determine, the conditions are the same as in the sample. We should not assume from the data of Example 1, for instance, that the probability of infestation of a tree taken from a different region is the same as in the region of the sample. 3. Some insects of a certain variety have red eyes, and others black. A collector found that of the first 100 individuals collected, 44 % had red eyes; of the first 500 individuals, 39 % had red eyes; of the first 1000, 41 % had red eyes. Within the limits of his observation, the probability that an insect of this variety would have red eyes varied, taking on the successive values 0.44, 0.39, 0.41, the value 0.41 being the one in which, after 1000 observations, he had the greatest confidence. EXAMPLE
Assumption. While the laws of combination of probability will be proved in later sections for mathematical probability, it will be assumed that the same laws hold for statistical probability or for combinations of probabilities some of which are statistical and some mathematical.
287. The law of averages. In Exercise 13, § 292, it will be shonw that if p is the probability that an event will occur at a given trial, then in n trials, the most probable number of occurrences of the event i:s pn (or if pn is not an integer, then a number very close to pn). In fact, in a more advanced study it can be shown that if n is large~ the probability is great that the relative frequency, h/n, will be approximately equal top. This is sometimes referred to as the law of averages. The probability of throwing an ace at one cast of a single die is ¼. If a die is to be thrown 30,000 times, the most probable number of times at which an ace will be thrown is ¼(30,000) or 5000. The actual number of "times may differ from 5000 by 100 or 200 or more, but before beginning the trials ,ve should expect this deviation to be small compared with 5000. The law of averages is frequently misused. Remember that it applie5 only to the relative frequency of a large number of future trials. Some sports writers and other persons who have heard of the law of averages make the mistake of expecting that what is going to occur by chance in the future will, in some way, be influenced by what has happened by chance in the past, so as to "satisfy" the law of averages. If we start out to throw a coin 1000 times, we should expect to throw heads about half the time. But large deviations from this program often occur. It is quite possible that the first 100 throws result in 90 heads and 10 tails; there is then no reason to expect EXAMPLE
1.
§§ 286-288]
PROBABILITY
473
about 10 heads and 90 tails in the next 100 throws. The results of the first 100 throws have no effect on the next 900. After the first 100 we should expect about 450 more heads and 450 more tails, giving, for the entire series, approximately 540 heads, with a relative frequency of about 0.54 instead of the approximately 0.50 which we originally expected.
288. Commissioners 1941 Standard Ordinary Mortality Table., This table (Table IV) was made for the use of life insurance com-panies from an exceedingly large number of records, by methods beyond the range of this book. We may consider that it starts with a group of 1,000,000 children one year old, and shows the number of them who die and the number who survive each year thereafter. Find the probability that a man who is 21 years old (a) will live until he is at least 50; (b) will die at the age 50. ExAMPLE.
Solution. (a) From Table IV, out of 949,171 persons alive at age 21'" there are 810,900 still alive at age 50. The relative frequency of surviving to the age 50, and therefore the probability that a man alive at 21 will be· living at 50 is ii gt~ i = 0.8543. (The decimal value was computed with logarithms.) (b) Out of 949,171 persons alive at age 21, there are 9990 who die at age 50 (Column 3, Table IV). The required probability is therefore the relative frequency, 9 : ii~ 1 = .01052.
EXERCISES 1. In teams of 5 students each, throwing 5 or 10 coins each at a time, find the relative frequency of throwing heads in (a) 100 casts; (b) 500 casts;
(c) 1000 casts. 2. In teams, find the relative frequency of throwing an ace with a single die in (a) 60 casts; (b) 300 casts; (c) 1000 casts. 3. Find the probability that a boy who is 5 years old (a) will live until he is at least 21; (b) will die at the age 40; (c) will live until he is 50 but die before he is 60. 4. Find the probability that a man who is 21 years old (a) will live at least 20 years; (b) will live at least 20 years but less than 30 years; (c) will live 20 years but not 21 years. 5. According to the terms of a certain policy, if a boy who is 4 years old lives until he is 18, he will receive $10,000. Find his mathematical expecta... tion, to the nearest dollar. 6. A man who is 25 years old buys a policy which provides that if he lives for 30 years he will receive $10,000. Find his mathematical expectation, to the nearest dollar.
474
FIRST YEAR OF COLLEGE MATHElVIATICS
[Ch. XXXII
7. According to the terms of a certain policy, a man who is 30 years old will receive $10,000 if he lives until he is 60; otherwise the money will go to his estate. To the nearest dollar find the mathematical expectation of (a) the man; (b) his estate. 8. A man who is 24 years old will inherit a fortune at the age 35. He has borrowed money from some money lenders on condition that he will pay them $2.5,000 if he lives until he is 35, but that otherwise they will not be repaid. Find (a) their mathematical expectation and (b) the present value of their mathematical expectation, considering it as an amount due in 11 years with money at 6% compounded annually. [Part (b) should be omitted by those who have not studied compound interest and present value.]
*289.
Frequency tables and diagrams. A frequency table records the number of cases or the frequency with which some quantity has been observed to have each of certain specified values or to lie within certain specified ranges of values. Corresponding to the frequency table, one can make a frequency diagram which displays the same results graphically. The values of the quantity observed are taken as abscissas. The ordinates then represent the frequencies with which individuals have been found in the corresponding ranges of values. The lengths of 1660 ripe seeds of a certain variety of melon were measured to the nearest millimeter. The results are shown in the table. EXAMPLE.
Length m millimeters
Number of seeds
Relative frequency
l
h
p
under 4.5 to 5.5 to 6.5 to 7.5 to 8.5 to 9.5 to 10.5 to 11.5 to 12.5 to over
4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5 12.5 13.5 13.5
Total
0 16 158 271 463 357 211 116 47 21 0
.000 .010 .095 .163 .279 .215 .127 .070 .028 .013 .000
1660
1.000
475
PROBABILITY
§§ 288-289]
The fourth line, for example, indicates that 271 seeds had lengths greater than or equal to 6.5 and less than 7.5 millimeters. Based on these 1660 observations, the probability that a seed of this variety chosen at random will have a length between 6.5 and 7.5 millimeters is equal to the relative frequency 1--i-lo, so that p = .163. The relative frequencies or probabilities of occurrence are shown in the third column. ~500
Ul
"'d
400
Cl) Cl)
W 300 '+-I
0
~ Cl)
,.,a 200
s :=::$
Z
100
4.5
5.5
6.5
7.5
8.5
9.5
10.5 11.5 12.5
Length in mm.
13.5
L
FIG. 182
The frequencies are shown graphically in Figure 182. Each rectangle corresponds to an entry of the frequency table, the height being proportional to the frequency of the class, and the base indicating the range of lengths of seeds in the class. Thus the height of the third rectangle is 271 vertical units, indicating that 271 seeds are in length greater than or equal to 6.5 and less than 7.5. The areas are proportional to the frequencies and therefore to the probabilities of occurrence of the various classes. In fact, if the unit of area is so chosen that the sum of the areas of all of the rectangles is 1, then the measure of the area of each rectangle is equal to the probability of the occurrence of the corresponding class. Thus, if the entire area of the diagram is 1, the area of the rectangle corresponding to the length 7 millimeters is .163, and the probability that a seed chosen at random will have a length between 6.5 and 7.5 is .163. If there had been many more seeds, and if the seeds had been measured to the nearest tenth of a millimeter, instead of 9 classes, we should have had about 90 classes, and about 90 slender rectangles. By greatly increasing the fineness of division and the number of cases, we thus obtain so many slender rectangles, which change so gradually in height, that their tops" instead of resembling a staircase, take on the appearance of a smooth curve. The limiting form of the frequency diagram, as the fineness of division and the number of observations increase indefinitely, is, in general, a smooth curve, called a frequency curve. If the unit of area is properly chosen, and if L and L' are close together, then the probability that the length of
47G
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXII
a seed chosen at random will be between Land L' is the area under the curYe between the vertical lines drawn at points L and L'. The derivation and study of different types of frequency curves form one of the most interesting and fruitful fields of mathematical statistics. It requires, however, a knowledge of calculus and other branches of more advanced mathematics. An 2 important type of frequency curve is the probab_ility curve, y = e-x (where e is the base of natural logarithms), or, more generally, y = ke-cx2, where the constants c and k are chosen so as to fit the particular frequency distribution at hand.. EXERCISES 1. In the Illustrative Example, above, find the probability that a seed chosen at random will have a length (a) between 10.5 and 11.5 millimeters; (b) between 6.5 and 8.5 millimeters; (c) greater than or equal to 9.5 milli-
meters. 2. From the data of the Illustrative Example, above, make a new frequency table and diagram, in which the range of length in each group is 2 millimeters, beginning with the group 4.5 to 6.5. From the new table find the probability that a seed chosen at random is in length between 8.5 and 10.5 millimeters. 3. The following measurements were made of 23,807 soldiers from a certain department of France: Height (inches)
60.0 60.5 61.0 61.5 62.0 62.5 63.0 83.5 84.0 84.5 85.0
Number of Soldiers 45 Height (inches)
81
65.5 66.0 66.5
67.~I
119 209 311 418 628 904 1082 1187 1501 1668 1762 1785 1798
67.5 68.0 68.5 69.0 69.5 70.0 70.5 71.0 71.5 72.0 72.5 73.0 73.5 74.0
Number of Soldiers 1732 1651 1460 1314 1224 898 804 410 284 212 106
82
63
49
74.51 20
The heights are given to the nearest half inch, so that 45 soldiers are be.. tween 59.75 inches and 60.25 inches in height. Make a frequency diagram. Find the probability that a soldier from the same region will be (a) between 65.75 inches and 66.25 inches in height; (b) less than 63.25 inches in height; (c) more than 72.25 inches in height. 4. From the data of Exercise 3 make a frequency table and diagram grouping the heights 60.0 and 60.5 into one class, 61.0 and 61.5 into a second class, etc. Find the probability that a soldier will be between 63.75 and 64. 75 inches in height. 6. In a certain college class in mathematics during a period of years, 1123 grades have been given as follows: Grade
A
B
C
D
F
Number
94
187
456
174
212
PROBABILITY
§§ 289-290]
(a) Make a frequency diagram.
477
(b) Find the probability that a student
chosen at random will earn a grade of C or better in this course. (c) A student will receive a gift of $10 from his father if he makes A or B. Find his mathematical expectation. 6. The lengths of 4680 ripe ears of a certain variety of corn from the same field are recorded in the following table, correctly to the nearest half inch. Those in the first class had lengths greater than or equal to 5.25 inches and less than 5.75 inches. Length, inches
5.5 6 6.5 7 7.5
~
I
8.5 9 9.5 10 10.5 11 11.5 12.0:
------------Number of Ears 72 95 123 180 268,392 480 517 571585 549 455 291 I
1021j
(a) Make a frequency diagram.
Find the probability that an ear chosen at random from this field will be (b) between 8.75 and 9.25 inches in length; (c) over 9.25 inches long.
290. Mutually exclusive events. Two or more events are said to be mutually exclusive if not more than one of them can occur at any one trial. Theorem. If n mutually exclusive events have the separate probabilities Pi, P2, · · ·, Pn then the probability that some one of these events will occur at a given trial is the sum of the probabilities:
P = P1 + P2 + Pa + • • • + Pn• The proof is given for n = 2, but can be extended easily to the general case. Let hi and h2 be the respective numbers of ways in which two events can occur, and let t be the total number of ways in which they may occur or fail to occur. (As always, all ways are assumed to be equally likely.) Then, by definition, Pi = hi/t and p2 = h2/t. Since the events are exclusive, the hi and h2 ways are all different. Hence (hi + h2) is the number of ways in which one or the other of the two events can occur. Therefore the probability that one or the other will occur is, by definition, Proof.
_ hi
P -
+t h2
_ hi
-
t
+ h2t
_
- Pi
+ P2-
N otice that if Ei and E2 are the two events, to find the probability that Ei or E2 will occur, we add their probabilities. We can indicate thi~
as follows:
478
FIRST YEAR OF COLLEGE lVIATHEMATICS
[Ch. XXXII
Find the probability that either exactly 2 heads or exactly 4 heads will be thrown 1f 7 coins are thrown at the same time. EXAMPLE.
Solution.
The probability of throwing exactly two heads is P1
_
-
7
c;2 2
1 _ 21 - T2s,
and that of throwing exactly 4 heads is p2 = 1C4/27 = /Is• The probability of throwing either exactly 2 or exactly 4 heads is therefore P1 P2 = ii¾ = r7e;. We can symbolize this as p(2 or 4) = /ls+ /i's = --?-6•
+
291.
Compound probability; dependent and independent events. An event E2 is said to be dependent on an event Ei, if the occurrence of Ei affects the probability of the occurrence of E2. Two or more events are independent if the occurrence of any one of them does not affect the probability of the occurrence of any other one of them. Theorem I. If the probability of an event Ei is Pi, and if, after Ei has occurred, the probability of a second event E2 is p2, then the probability that both events will occur in the order E 1 E2 is
Proof. Suppose that at a given trial, E1 can occur in h1 ways and either occur or fail to occur in t1 ways. The proof will apply whether E2 is dependent on Ei or not, but will be limited, if E2 is dependent on E1, to the case in which, at any given trial, E2 can either oc0ur or fail to occur in t2 ways, and, after Ei has occurred, E2 can occur in h2 of these ways. Then, by definition, Pi = hi/ti and P2 = hdt2. By the Fundamental Principle, § 277, Ei can first occur or fail to occur and then E2 can occur or fail to occur in lit2 ways; and first Ei and then E2 can occur in hih2 ways. Therefore the probability that first E1 and then E2 occur is h1h2 h1 h2 P= = - · - = Pi · P2-
tit2
ti
t2
Theorem II. If Pi and P2 are the separate probabilities of two independent events, the probability that both will occur at a given trial is the product
P = P1 · P2The proof is similar to that of Theorem I. Theorems I and II can both be extended easily to the occurrence of three or more events. Notice that to find the probability that E 1
PROBABILITY
§§ 290-291]
479
and E 2 will occur, we multiply their probabilities. this as follows: p(E1 and E2) = Pi · p2.
We can indicate
1. A bag contains 4 white balls and 6 black balls. Three balls are drawn in succession, without replacing any ball after it is drawn. Find the probability that the first and third balls will be white and the second ball will be black. EXAMPLE
Solution. The probability that the first ball will be white is p 1 = r\- = -g. If the first ball is white, the probability that the second will be black is P2 = f = j. If the first ball is white and the second is black, the probability that the third ball will be white is pa = ¾. The desired probability is therefore p(w, b, w)
= p1 · P2 · Pa = ¾· ¾ · ¾ = lo-
2. A boy's father offers to let the son choose in advance either $80 on condition that he make a grade of A in each of his three courses or $20 for each A that he makes. The boy estimates that his probabilities of making A's in his three courses are respectively ¾, ¾, and ½. Which offer should he choose? EXAMPLE
Solution. The boy would be smarter to choose the second offer. His probability of winning the $80 is P1P2Pa = ¾· ¾· ½= ¼, and his mathematical expectation from the first offer is $20. His expectations from the other offer are $15, $13.33, and $10, with a total of $38.33. 3. One bag contains 4 white and 6 black balls, a second bag contains 3 white and 7 black balls, and a third bag contains 5 white and 5 black balls. A person first chooses a bag at random and then draws a ball from it. Find the probability that it will be white. EXAMPLE
Solution. The "event" of drawing a white ball can occur in one of three mutually exclusive ways: it can be drawn from the first or the second or the third bag. The probability of choosing the first bag and drawing a white ball from it is ½• T1r = 3\-. The probability that the ball drawn will be a white ball from the second bag is ½· lo = io, and the probability that it will be a white ball from the third bag is ½· / 0 = / 0 . The probability that the ball drawn will be a white ball from one or another is therefore P
_
-
4-
30
+ 30 + 30 3'
5
_
12 _
30 -
~
5, '
EXERCISES 1. Find the probability of throwing an ace on each of three successive casts of a die. 2. If a die is cast three times, find the probability that an ace wiH be thrown on each of the first two casts but not on the third.
480
FIRST YEAR OF COLLEGE :MATHEMATICS
[Ch. XXXII
3. The owner of a stable enters two horses in a race. The probability that his best horse will win the race is ½, and the probability that the second will win is ¼. Find the probability that the owner will win with one horse or the other.
4. In an oratorical contest a certain contestant's probability of winning the first prize is i and his probability of winning the second prize is ¼. Find the probability that he will win one prize or the other. 6. A bag contains 5 white, 4 black, and 6 red balls. Three balls are drawn in succession, each one being replaced before the next is drawn. Find the probability that (a) all are red; (b) the first is white, the second black, the third red; (c) all are of the same color; (d) there is one of each color. 6. Four cards are drawn from a deck of 52 cards, each being replaced before the next is drawn. Find the probability that (a) all are hearts; (b) all are of one suit; (c) the first is a spade, the second a heart, the third a diamond, the fourth a club; (d) there is one card from each suit.
7. A candidate for a certain office estimates that the probability of his winning his party's nomination is ¼and that if he wins the nomination, his probability of winning the election is r90 • Find the probability that he will win the election. 8. An applicant for a civil service position estimates that the probability
that he will pass the qualifying examination is lo; if he passes it, his probability of securing the position is ½. Find his probability of securing the position. 9. Two students, A and B, are working independently on a problem in
mathematics. A's probability of solving it is¾, and B's is¾. Find the probability that (a) A will solve it and B will not; (b) B will solve it and A will not; (c) both will solve it; (d) at least one of them will solve it.
10. An officer sends one copy of a message by carrier pigeon and another copy by messenger. The probability that the messenger will arrive in time to be effective is ! ; the probability that the pigeon will do so is ½. Find the probability that (a) both will arrive in time; (b) the messenger will but the pigeon will not; (c) the pigeon will but the messenger will not; (d) the message will arrive in time. 11. Three persons work independently at deciphering an intercepted code message. Their respective probabilities of deciphering it are ½, -g-, ¾- Find the probability that the message will be deciphered. 12. In a dice game A first casts a die. If he throws an ace, he wins. If
he does not throw an ace, B casts the die and wins if he throws an ace. If neither throws an ace, other players throw it in turn. Find the probability that either A or B will win on the first turn.
§ 291]
PROBABILITY
481
13. Find the probability that, in throwing two dice, a 7 or 11 will be thrown the first time. 14. Find the probability that, in throwing two dice, a 4 will be thrown before a 7. [Suggestion. Find the probabilities that (a) 4 will be thrown the first cast, (b) neither 4 nor 7 will be thrown the first cast and 4 will be thrown the second cast, etc. This leads to an infinite geometric series, § 100)' whose value may be taken as the desired probability.] 15. Find the probability that, in throwing two dice, 4 will be thrown the first time and 4 will be thrown again before 7 is thrown. (See Exercise 14. J 16. Find the probability that, in throwing two dice, one of the numbers 4, 5, 6, 8, 9, 10, will be thrown the first time and that the same number will be thrown again before 7 is thrown. (See Exercise 15.) 17. A and B take turns tossing a coin, A having the first turn. The first one to throw a head wins 30 cents. Find the mathematical expectation of each player. [Suggestion. If p and q are the respective probabilities that A and B will win, we can assume that one or the other will win in time, so that p + q = I. Show that q = ½p, and find p and q.] 18. A, B, and C, in that order, take turns tossing a coin. The first one to throw a head wins 35 cents. Find the mathematical expectation of each player. (See Exercise 17.) 19. A and B throw a die alternately, A taking the first throw. The first one to throw an ace wins. Find their respective probabilities of winning. 20. A and B agree to throw a die alternately, the first one to throw an ace being the winner. They also agree that each one shall make a preliminary throw, the one throwing the larger number to have the first throw in the game; if the preliminary throws result in a tie, they try again. At his pretiminary throw, A throws a 3, and B prepares for his preliminary throw. At that time, what is B's probability of winning the game? (See Exercise 19.) 21. In the semifinals of a golf tournament A plays Band C plays D. The winners of these two matches play in the finals. The probability that A will beat B is f. The probability that C will beat D is ¼. When A plays C the probability that he will win is i. When A plays D the probability that he will win is¾. Find the probability that A will win the tournament. 22. A bag contains 4 white balls and 6 black balls. A man draws two balls in succession, replacing the first before the second is drawn. He is to receive $1 if he draws just one white ball and $5 if he draws two white balls. Find his mathematical expectation. 23. In a dice game, a man has the right to throw two dice once, and will receive a number of dollars equal to the number that he throws. Find his aiathematical expectation. 24. A bag contains 9 balls, numbered 1, 2, • • •, 9. Four balls are drawn together. Find the probability that the sum of their numbers is even.
482
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXII
25. Three bags contain, respectively, 4 white and 8 black balls, 6 white and 6 black balls, and 10 white and 2 black balls. Two bags are drawn, and one ball is drawn from each of them. Find the probability that both balls are white. 26. If the probability that a bomber plane will return safely to its base after a mission is ½~'and if three planes go out on a mission, find the probability that (a) at least two will return safely; (b) j.ust one will return safely. In Exercises 27 to 30, refer to Table IV and assume that it applies both to men and women. 27. A man is 27 years old and his wife is 23. Find the probability that at the end of 15 years (a) both will be alive; (b) the man will be dead and the woman will be alive. (c) Find the probability that the man will die within 15 years and the woman will live for 25 years. 28. A man is 30 years old, and his wife is 25. Find the probability that at the end of 20 years (a) both will be alive; (b) both will be dead; (c) the man will be dead and the woman will be alive. 29. A man is 36 years old, and his wife is 30. He owns a term insurance policy which provides that if he dies before he is 61, and his widow is then living, she will receive $10,000. If he lives to be 61, the policy pays nothmg. Find his wife's mathematical expectation, to the nearest dollar. 30. A man is 40 years old, his wife is 35, and his son is 12. A term policy provides that if he dies before he is 68 the insurance company will pay $10,000 to his wife if she is living then, to his son if his wife has died and his son is living, or to his estate if both the wife and son have died. If the man lives to be 68, the policy pays nothing. To the nearest dollar, find the mathematical expectations of the wife and of the son.
292. Successive trials of an event. The following example illustrates the method of finding the probability that an event will occur exactly r times in n trials, if the probability of its occurrence is the same at all trials. A bag contains 3 white balls and 7 black balls. One ball is drawn 6 times in succession and replaced each time before the next is drawn. Find the probability that (a) a white ball will be drawn the first and fourth times and a black ball will be drawn the other times; (b) a white ball will be drawn exactly 2 times out of the 6. EXAMPLE
Solution.
1.
(a) The probability that a white ball will be drawn at any
single trial is p = i30 , and the probability that it will not be drawn is q = --io• Drawing a white ball the first and fourth times, and drawing a black ball
§§ 291-292]
PROBABILITY
483
the other times are 6 independent events. By § 291, the probability that this program will be followed is therefore
p
=p
•q . q . p •q . q
=
p2q4
=
(1-)2 (!__)4 = - 21609 ' = 0 021609 10 10 1,000,000 . .
(b) In order to draw a white ball exactly 2 times out of the 6, one may draw it in any set of 2 drawings chosen out of the 6, and draw a black ball the other times. There are 5C2 = 15 ways of choosing 2 drawings for the white ball. These 15 ways are 15 mutually exclusive events. The probability of any one of them is the probability of drawing a white ball at 2 specified drawings and a black ball the other 4 times, and, as in (a), is equal k ( /0 ) 2 (-i70 -)4. The probability of drawing exactly 2 white balls is therefore
p = C 6
2
•
3)2( 7) ( IO 10
4
21609 324,135 = 15 . 1,000,000 = 1,000,000 = 0·324135 ·
Theorem I. If p is the probability that an event will occur, and q = 1 - p is the probability that it will fail to occur at a given trial, the probability that it will occur exactly r times in n trials is
nCr . pr . qn-1.
(1)
Proof. The set of r trials at which the event is to oocur can be selected from the n trials in nCr ways. The occurrence of the event at r specified trials and its failure to occur at the remaining (n - r) trials are n independent events. By § 291, the probability that these will occur is equal to the product of their respective probabilities or pr qn-r. The event can occur at exactly r trials in nCr mutually exclusive ways, the probability of each one of which is pr qn-r. The total probability that the event will occur exactly r times inn trials is therefore nCr · pr · qn-r. Corollary. The probability that the event will occur at least r times in n trials is
(2)
pn
+ nCn-1 pn-l q + nCn-2 pn-
2
q2
+ · •· + nCr JP qn-r.
Notice that the expression in (1) is the term in pr in the binomial expan~ sion of (p q)n, § 104, and (2) is the sum of the first (n - r 1) terms of that expansion.
+
+
EXERCISES 1. In a dice game a man throws a single die 5 times, and "wins" the cast if he throws a 5 or a 6. Find the probability that (a) he will win exactly 3 times; (b) he will win at least 3 times.
484
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXII
2. In a dice game a man throws two dice 4 times, and "wins" the cast if he throws 10 or more. Find the probability that (a) he will win exactly twice; (b) that he will win at least twice. 3. A bag contains 2 white balls and 3 black balls. A ball is drawn 5 times and is replaced each time before the next drawing. Find the probability that a white ball will be drawn (a) exactly twice; (b) at least twice. 4. If a coin is tossed 10 times, find the probability of throwing a head (a) just 5 times, (b) at least 5 times. 5. If 4 dates from past history are chosen at random, find the probability that, according to our present calendar, (a) just 2 of them fell on Sunday, (b) at least 2 fell on Sunday, (c) none fell on Sunday. 6. A bag contains 3 white, 5 black, and 2 red balls. Two balls are drawn together 6 times, each pair being replaced before the next drawing. Find the probability that a pair consisting of 1 white and 1 black ball will be drawn (a) just 3 times; (b) at least 3 times. Find in decimal form the probability that the pair drawn will consist of two balls of different colors (c) just 3 times; (d) at least 3 times.
7. A marksman usually hits a certain target 70 times in 100 shots. Find the probability that in a series of 5 shots he will (a) hit the target every time; (b) hit the target at least 4 times. 8. A certain examination contains 8 questions of the multiple-choice type, in which there are 4 possible answers given for each question, just one of which is correct. If a boy answers the questions by guesswork, find the probability that he will answer (a) at least 50% of them correctly, (b) more than 50 % of them correctly. 9. At 3 different trials, 2 cards are drawn from a deck of 52 cards, the first card being replaced before the second is drawn. Find the probability that of these 3 trials (a) just 2 yield 2 cards of the same suit; (b) at least 2 yield 2 cards of the same suit. 10. At 4 different trials, 2 cards are drawn from a deck of 52 cards and are replaced before the next trial. Find in decimal form the probability that of these 4 trials (a) just 2 yield 2 cards of the same suit; (b) at least 2 yield 2 cards of the same suit. 11. Five graduating seniors, each of whom is 21 years old, agree to meet 25 years later. Find in decimal form the probability that (a) just 4 of them will then be living; (b) at least 4 of them will be living. 12. There are 4 co-workers in an office, each of whom is 45 years old. Find
in decimal form the probability that (a) just one of them will have to be replaced within 20 years because of death; (b) none of them will have to be replaced within 20 years because of death.
§ 292]
PROBABILITY
485
13. If O < p < 1 and q = 1 - P, show that nCr prqn-r > nCr-1 pr-lqn-r+l, if r < p(n + 1), and nCr pr qn-r < nCr-1 pr-1 qn-r+l, if r > p(n + 1). Hence, show that if p is the probability that a certain event will occur at a given trial, the most probable number of occurrences of the event inn successive trials is an integer r such that pn - q < r < pn
+ p.
If pn is an integer, the most probable number of occurrences is r = pn. If p(n 1) is an integer, then r = p(n 1) and r = p(n 1) - 1 occurrences
+
+
+
are equally likely and are more likely than any other number of occurrences. 14. If q = 1 - p, and if p is the probability that a certain event will occur at a given trial, show that the probability that an event will occur at least once in n trials is 1 - qn. 15. Find the most probable number of times that an ace will be thrown in (a) 30 casts of a single die; (b) in 35 casts of a single die; (c) in 45 casts of a single die. (See Exercise 13.) 16. Find the most probable number of times that 11 or 12 will be thrown in (a) 96 casts of two dice; (b) 95 casts of two dice; (e) 100 casts of two dice. (See Exercise 13.) 17. A group of 100 school teachers, each 32 years old, take out some tenyear group insurance. Find the most probable number of deaths in the group during the ten-year period. (See Exercise 13.) 18. A group of 1000 high school students, each 16 years old, take out some five-year group insurance. Find the most probable number of deaths in the group during the five-year period. (See Exercise 13.)
CHAPTER XXXIII
The Mathen1atics of lnvestn1ent* 293. Compound interest. The most important principles and formulas applying to compound interest, present value, and annuities have been stated in the discussion and illustrative examples of § 99. The student should review that section thoroughly at this point. It is there shown that if i. expressed as a decimal, is the rate of interest per interest period or conversion period, if P is the principal and if A is the amount at simple interest at the end of k interest periods, then (1)
A= P(1
+ ki).
If P n is the compound amount to which P accumulates after n conver-• sion periods, (2) . Tables of P n for various values of i and n, and for the principal P = $1 have been formed. A short table of this sort is Table V. When this table does not suffice, compute the compound amoun~ by means of logarithms.
If interest is compounded m times a year, the efjective rate of interest or rate per year at which an investment grows is greater than the nominal rate used in describing the investment. If an investment is made at the nominal rate j to be compounded m times per year, the rate per interest period is j /m. The amount of $1 at tRe end of the year is therefore ( 1
+ !)m.
If i is the effective rate, the amount of
* A knowledge of geometric progressions and of logarithms is required in this chapter. In order to be accessible to students all logarithmic solutions and answers in this chapter are found with five-place tables, and consequently often involve small errors. Seven-plare tables should be used for greater accuracy. 486
THE MATHEMATICS OF INVESTMENT
[§ 293]
$1 at the end of the year is (1 rate we set
+ i).
(3)
487
Therefore, to find the effectivP
or
i=
(1 + ~r-1.
1. Find the effective rate if money is lent at 6 % per annum compounded quarterly. EXAMPLE
Solution.
In formula (3), j (1
= .06, m = 4, j/m = .015. By Table V,
+ .015) 4 =
1.0614.
Therefore, by (3),
i = 1.0614 - 1 = .0614. The effective rate is 6.14%. 2. What nominal rate, compounded monthly, is equivalent to 8% compounded annually? EXAMPLE
SolutWn.
Dy (3), 1
. )12 ( 1 + /2 = 1 + .08.
Therefore
+ /2 = -011.os = 1.0064 +
(by logarithms).
Hence j = 12(.0064) = .077, (but see footnote, page 486). nominal rate is· 7 .7 %.
If P n
The necessar:v
= S is the amount to be accumulated after n conversion
periods, the principal, P, required for its accumulation is the present value of S, and
(4)
P
= S(l + i)-n.
Table VI is a table of present values of $1 after n periods. When it does not suffice, calculate the present value by means of logarithms.
If a bank purchases the right to collect a sum, S, due at the end of n conversion periods, it discounts the sum S for n periods; the purchase price is the present value, P. The discount is the difference between the amount due, S, and the present value, P, so that (5)
Discount= S - P
=S -
S(l
+ i)-n.
488
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXIII
EXERCISES 1. Find the amount of $4000 at the end of 12 years at 6 % compounded quarterly. 2. Find the amount of $8000 at the end of 9½ years at 8% compounded (a) semiannually; (b) quarterly. 3. Find the effective rate if money is invested at 6% compounded semiannually.
4. Find the effective rate if money is invested at 8 % compounded quarterly.
5. Find the effective rate if money is invested at 7½ % compounded quarterly. 6. Find the effective rate if money is invested at 9 % compounded semiannually. 7. What nominal rate, compounded semiannually, is equivalent to 7% compounded annually?
8. ,vhat nominal rate, compounded quarterly, is equivalent to 6 % compounded annually? 9. What nominal rate, compounded monthly, is equivalent to 7½% compounded semiannually?
10. What nominal rate, compounded 6 times per year is equivalent to 6% compounded semiannually?
11. Find the present value of $5000 due at the end of 11 years, if money is worth 6% compounded (a) annually; (b) quarterly. 12. Find the present value of $10,000 due at the end of 12 years, if money is worth 4% compounded (a) annually, (b) semiannually. 13. Find the present value of $4000 due at the end of 10 years, if money is worth 7% compounded (a) annually; (b) semiannually.
14. Find the present value of $1000 due at the end of 7 years, if money fr, worth 5% compounded (a) annually; (b) quarterly.
15. Find the discount on $6000 due at the end of 5 years, if money is worth 6% compounded annually.
16. Find the discount on $4000 due at the end of 1 year, if money is worth ·6% compounded quarterly.
= 5.2533. (Use Table V.) 18. Solve for i: (1 + i) 27 = 2.2213. (Use Table V.) 19. Solve for i: (1 + i) 20 = 1.7. (Use logarithms.) 20. Solve for n: (1.06)n = 2.5. (See § 122.) 17. Solve for n: (1.05)n
§§ 293-294]
THE MATHEMATICS OF INVESTMENT
489
21. At what rate will money double in 18 years, interest being compounded annually?
22. At what rate will $2000 amount to $3000 in 14 years, interest being compounded annually? 23. In how many years will $4000 amount to $5000, with money earning
6% compounded semiannually? 24. In how many years will $1000 amount to $1400, with money worth 5% compounded semiannually?
294. Annuities. An annuity is a sequence of payments. We shall here consider only annuities in which the payments are equal, and in which the intervals of time between consecutive payments are equal. Such an interval is the payment period and will here be assumed to be equal to the conversion period of compound interest. The term of the annuity is the total time between the beginning of the first payment period and the end of the last one. During the term one payment is made at the end of each payment period, but, unless otherwise stated, no payment is made at the beginning of the first payment period. The term is therefore equal to the payment period multiplied by the number of payments. The sum of the payments made in one year is the annual rent. The amount of an annuity is the total amount which would be accumulated at the end of the term if each payment were invested at compound interest at the time of payment. The present value of an annuity is the sum of the present values of all the payments at the beginning of the term. \Ve denote by sn1 the amount of an annuity which pays $1 at the end of each of n interest periods. And we denote by an1 the present value of the same annuity. Let us find sn1 and an1. Suppose that as soon as it is made, each payment of $1 is invested at compound interest at the rate i per period, and is left until the end of the term. At the end of the term the last payment of $1 will just have been made, and will amount to $1; the next to the last payment will have been at interest for one period, and will amount to (1 + i); the payment before that, having been at interest for two periods, will amount to (I + i) 2, etc. The first payment will have been at interest for (n - 1) periods, and will amount to (1 + i)n- 1. The sum of these amounts is therefore Sn1
=
1 + (1
+ i) + (1 + i) 2 + · •• + (1 + i)n-1.
490
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXIII
This is a geometric progression whose common ratio is (1 § 97, its sum is therefore
(6)
sn1
= (l + it l
1
.
+ i).
By
(Amount of annuity.)
Values of Sn1 are given in Table VII. Values not found in the table can be found by interpolation or can be computed by means of logarithms.
In order to find an1, we compute the present values of the various payments at the beginning of the term. By formula (4), § 293, the present value of the payment due at the end of the first period is (1 + i)-1 ; that of the payment due at the end of the second period is (1 i)-2, etc. The present value of the last payment, due after n periods, is (1 + i)-n. Therefore
+
+ i)- + (1 + i)- + · · · + (1 + i)-n. geometric 'progression in which (1 + i)- is both the first ai 1.
(§ 97.)
The following series is called the harmonic series: 1
1
1
1 +2+3+···+n+···.
This series diverges, as we can see in the following manner. Group the terms as follows:
1 + ½+ (½ + ¼) + C¼ + · · · + ¼) + Cl+ · · · + l6) (½+ ¼) consists of two terms, the smallest of whir-h is
+ ··· . ¼. Therefore
§§ 311-313]
INFINITE SERIES
(½ + ¼) > ½.
+ ·· ·
519
Likewise (½ + ¼) > ½, for it contains four terms,. the smallest of which is ¼. In a similar way (½ + · •· + -r\r) > ½. We can continue in this way, forming as many groups of terms as we wish, the sum of each group being greater than ½. By taking enough of these groups we can make the sum of them as great as we wish. Therefore Sn does not approach a limit, for, when n increases indefinitely, Sn increases without bound. The series therefore diverges.
312. The omission of a :finite number of terms.
The convergence or divergence of an infinite ser1:es is not ajfected by the omission of a finite number of terms. Proof. Let S 1, S 2, 83, · · · , be the successive partial sums of the original series. For simplicity we suppose that the first m terms of the series are omitted. The sum of the omitted terms is Sm. If s1, s2, sa, • · • , are the successive partial sums of the abbreviated series, Sn+m = Sm + Sn. Since Sm is a constant, limn~~Sn+m therefore exists if and only if limn~~sn exists; that is, the original series converges if and only if the abbreviated series converges. In case the abbreviated series converges to the value s, the value of the original series is Sm+ s.
313.
A necessary condition for convergence.
Theorem: In any convergent series the nth term approaches zero as a limit as n~ro. Proof. In the usual notation
or If the series converges to a value S, then limn~~Sn = S, and limn~ooSn-l = s, so that lim un = lim (Sn - Sn-1)
n~oo
n~oo
=
lim Sn - lim Sn-1 n~~
n~oo
=
S - S
= 0.
Corollary. If the nth term of a series does not approach zero as a limit as n--+ oo , the series diverges. Notice that, while no series can converge unless its general term approaches zero, the mere fact that the general term approaches zero is not sufficient to insure the convergence of the series. This is illustrated in
Example 2, below.
+ + + / 6 + ~ ~ + ··· diverges,
EXAMPLE 1. The series 1 ¾ j general term approaches½ instead of zero.
for its
520
FIRST YEAR OF COLLEGE ~1ATHEMATICS
[Ch. XXXV
+ + + ¼+ •••, is a series in
EXAMPLE 2. The harmonic series 1 ½ ½ which the general term approaches the limit zero. V1e series di verges.
B.
Yet, as we have seen,
CONVERGENCE OF SERIES OF POSITIVE TFR:\IS
314. Series of positive terms with a limited partial sum. Theorem. A series of positive terms always converges, unless its nth partial sum increases without limit. u2 U3 Proof. Suppose that each term of the series U1 is positive, and that there exists a fixed number, M, such that Sn < M for all values of n. We have Sn+l > Sn, since Sn+l = Sn+ Un+l and Un+i > 0. Therefore, by the Fundamental Theorem, § 310, limn-+ooSn exists, and the series converges.
+ + + · ··
315. Comparison tests for convergence and divergence. Theorem I. Given a series of positive terms; suppose that we can find another series which is known to converge and each of whose terms is greater ~han or equal to the corresponding term of the given series; then the given series also converges. Proof. Given the series (1)
U1
+ U2 + U3 + · · · , where Un >
to be tested for convergence.
0,
And let
(2)
be a series ,Yhich (a) is known to converge and (b) is such that Un < an. Let and Since series (2) converges, there exists a number, A, such that limn-+oo An = A. Since Un and an are positive, An < A. But, by (b), Sn < An < A. Therefore, by § 314, series (1) converges. Theorem II. Given a series to be tested for divergence, and suppose that we can find another series of positive terms which is known to diverge and each of whose terms is less than or equal to the corresponding term of the given series; then the given series also diverges. For if the given series converged, since each of its terms is greater than or equal to the corresponding term of the second series, by Theorem I the second series would converge, which is not true.
EXAMPLE
521
INFINITE SERIES
§§ 313-315]
Prove that the following series converges:
1.
1
(1)
1
1
1
n+21+31+···+n!+···· Compare this series with the series
Solution.
1
1
1
1
2 + 22 + 23 +
(2)
. .. + 2n-I + . .. •
Series (2) converges, since it is a geometric series in which r = ½. But (1/n !) < (l/2n- 1). For n ! = 2 · 3 · 4 · · · · · n has (n - 1) factors the smallest of which is 2, and 2n-I has (n - l) factors each of which is 2. Therefore, by Theorem I, series (1) converges. EXAMPLE
Prove that the series
2.
1
1
1
1+-+-+···+-+··· 2P 3P nP
(3)
converges when p
Solution.
Set
V1
1.
= 1'
V2 =
(!_2P + !__) 3P ' 1
Vn
We have
Vi=
V3
In fact,
Vn
1,
= (2n-I)p V2
V3 =
(!__4P + · · · + 7~) ••• p
'
'
1
+ ••• + (2n -
l)P•
1, series (1) diverges; (c) if R = 1, there is no test. Proof. (a) R < 1. By § 316, if R < 1, the series whose general term is lunl converges. (b) R > 1. If R > l, from a certain point on, lun+rl > lunl, so that Un does not approach zero when n increases indefinitely. Therefore,
528
FIRST Y1£AR OF COLLEGE MATHElVIATICS
[Ch. XXXV
by § 313, the series diverges. (c) R = l. There is surely no test ir. the general case since there is no test even when all the terms are positive, EXAMPLE
1. Test for convergence the series
1 1 1 1 1 1 1 l-2!-3!+4!+5!-6!-7!+8!+····
Solution. Here lunl = 1/n !. Hence Rn Therefore lim Rn
=
lun+il lunl
nI
=
(n
+ 1) !
1 n+l
=--·
= 0, and the series converges absolutely.
n-"oo
320. Alternating series. An alternating series is one in which the terms are alternately positive and negative. Theorem. An alternating series converges if the absolute value of each term is greater than that of the fallowing term and if the nth term approaches zero as a limit as n becomes infinite. Proof. Given the series (1)
where Un > o, and Un > Un+l for all values of n, and limn_,,~un = 0. We shall show that series (1) converges. Let n be an even integer; then (2)
Sn = (u1 - U2)
(3)
Sn =
U1 -
+
(u3 - U4)
(u2 - U3) - (u4 -
+ ... + Us) -
(un-l - Un),
•••
- (un-2 - Un-1) - Un.
Since each term is numerically greater than the following, each quantity in parentheses is positive. Therefore, by (2), when n increases, Sn increases, and, by (3), Sn < U1. Therefore, by the fundamental theorem, § 310, Sn approaches a limit S as n becomes infinite when taking on only even values. We must now show that the sum of an odd number of terms also approaches the limit S. Let n still be an even integer. We have
Hence, limn_,,~Sn+l = limn_,,~Sn + limn_,,~Un+l = S + 0 = s, since, by hypothesis, limn_,, 0, the series converges absolutely for all values of x for which !xi < 1/L and diverges for all values of x for which !xi > 1/L. Proof. Applying the ratio test of § 319, we have . Rn lim n~oo
lan+lxn+tj = l.im - - = z·im 1an+1! .n~oo Ianxn I n~oo an
1X I
= L ·
IX. I
Therefore, by § 319, the series converges absolutely if L · !xi < 1; that is, if L = 0 or if !xi < 1/L; it diverges if L · lxl > 1, or if [xl > 1/L; if lxl = 1/L, there is no test. If some of the coefficients, an, are zero, they should be omitted, and the test of § 319 applied directly. The following statement follows at once in case the limit L exists, and can be proved without great difficulty in the general case. A power series, unless it either converges for all values of x or diverges for all values of x except x = 0, converges absolutely for all values of x on the interior of an interval of convergence which extends from a point x = -A to a point x = A, and diverges for all values of x outside of that interval. For either x = A or x = -A the series may either converge or diverge. When the series converges for all values of x we indicate its interval of convergence by the symbol - oo < x < oo • EXAMPLE
1.
The series
converges for all values of x. . l im
1+x
x2
x3
+ 2 ! + 3 ! + ···
For an = 1/n!, and
lan+ll n! - = z·im -1- = 0. = z·im - n~oo (n + 1) ! n~oo n + 1
n~oo an EXAMPLE
2.
x... Test for convergence the series x - i
+ 3x
3
x4
- 4+
··· .
§§321-322]
INFINITE SERIES Here (an(
Solution.
lim n~oo
531
= 1/n. Therefore
lan+l\ an
= lim _n_ = 1. n~oo
n
+1
The series converges if !xi < 1, and diverges if lxl > 1. When x = 1, the series converges (§ 320, Example I); when x = -I, the series is the negative of the harmonic series, and so diverges. The interval of convergence is that for which -1 < x < 1.
EXERCISES Find the interval of convergence for each series, and, if the interval is finite, test the series for convergence at each end of the interval.
xa xs 2· x-31+57- ···. 3. 1
+ 5x + 2
3 2 + .x + 42
6. 1 + - x
4
9•
4
•
+ 3 X + 2J + 3T + · · · .
X -
11• 1 13. x
4
- 3 x3 + • • .
3a xa
32 x2
7. 1
x x2 x3 4· 1 +3+9+27+ ....
52 x 2 + 5 3 x 3 + • • . .
xa
xs
x2
x4
x1
3 +5 - 7
3 +9 -
x6
27
x x2 x3 6· 1 + 2 . 3 + 3 . 32 + 4 . 33 + ....
8. 1 - 2x
x2
1
1
16• 2 • 4 · 2 16.
+
3
5
4 x3 +
-
... .
x6
10·
+ . .. .
1 3 3 5 5 7 2 12• 4 . 2 + 6 . 4 X + 8 . 6X + . . . .
+ 54 x2 + 54 . 76 xa + .... 3
x4
2
+ ... •
2- 4 +6 -
1
4·1r•4X
+
5
7
1
2
1r·s·6x
+
x2
x
s + .. . · x3
22 + 32 + 42 +
14. 1 +
4
(Suggestion for Ex. 13. Compare with 1 + 1
+3x
6
... .
4
6 4 6 8 + · · · 6 8 6 8 10 + · · · .)
+
7
!l s·ro · s1 x 3
+ ·•••
½· ¾· 1 + ¾· ¾· ½x + ¾· f · ¼x 2 + i · i\- · ¼x3 +
··· .
III. COMPUTATION WITH SERIES 322. Approximation to the value of a series. In some series, it is possible to write a simple formula for the value, S, of an infinite series. EXAMPLE
1.
In the case of the convergent geometric series:
S = a + ar
+ ar + ar + · · · , where lrl < 1, 2
3
we have seen (§ 100) that S = _a__ 1- r Thus, if S = 3 -
¾+ ¾- i + · · · , S =
l
!½
= 2.
532
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXV
In most cases, however, it is possible only to approximate to the value, S, by using, instead, the value of Sn, the nth partial sum for some suitable value of n. By definition, limn-H~Sn = S. Therefore by taking n large enough, we can make the error, IS - Snl, which is made by using Sn instead of S, as small as we wish. In some cases it is difficult to determine how many terms we must add together in order that the error IS - Snl shall be sufficiently small; that is, in order that, for the purpose in view, the sum, Sn, of the first n terms can be used instead of the limit S. For the case of a convergent alternating series with decreasing terms, we have the following rule: In a convergent alternating series u1 - u2 + u 3 - U4 + · · · , in which the absolute value of each term is greater than that of the following term, the error made in using the sum of the first n terms instead of the value of the series is numerically less than the absolute value of the first neglected term: IS - Sn)
--t~---+----J.-x represent the same curve are really equivalent, and one equation can be reduced to the other. The representation of a line (or other curve) in y space corresponds rather to the repFIG. 198 resentation of a point in the plane by means of the simultaneous equations of any two lines or other curves which intersect in the point. As the equations of a line it is commonly useful to choose the equations of two planes which pass through the line and are parallel to two coordinate axes as in Figure 198. The plane I, which passes through a given line, l, and is parallel to the z-axis (or perpendicular to the xy-plane), is represented by an equation involving only the two variables x and y; the plane II, which passes through l and is parallel to the y-axis, has an equation which involves only x and z. We can write the equations in the form (1)
A1x
+
B1y
+
D1 = O,
(2)
These equations are also, of course, the equations of the lines l1 and Z2, the projections of l on the xy- and xz-planes. For this reason they give at once a clear idea of the position of the line. Moreover, if we can write the equations of the projections of a line on any two of the coordinate planes, the same equations serve to represent the line itself. They are called the projection form of the equations of a line.
§§ 357-359]
SOLID ANALYTIC GEOMETRY
587
368. Symmetric equations of a straight line. A straight line i~ determined if one point on the line and its direction are known. Let P1:(xi, Yi, zi) be given as a point on the line, and a, {3, 1' be given as the direction angles of the line. Let P:(x, y, z) be any other point on the line. Then, if dis the distance PiP, by § 344 we have X - x, Z Zi Y - Yi = COS R (1) = cos a, = cos 1'· d fJ, d d Equating the values of d from these equations, we have ( 2)
X -
Y-
X1
Z -
Y1
Zt
cos a = cos /j = cos 'Y ·
The equations (2) are two equations that hold for the coordinates of any point on the line and for the coordinates of no point not on the line, and are therefore equations of the line. They are called symmetric equations of the line. It is worth while to remember that, for a given point (x, y, z), on the line, the common ratio in the equations (2) represents the distance between that point and the given point (xi, Yi, Zi). If l, m, n are any direction numbers, the equations of the line can be written in the form ( 3)
X -
X1
l
=
Y-
m
Y1
=
Z -
Z1
n ·
Write equations of the line through the point (-8, 5, -7), with direction ~umbers (3, -2, 4). x+s y-5 z+7 Solution. The equations are --=--=--· 3 -2 4 by (3). Combining the first member of these equations with the second and then with the third, we can write the equations in the form 2 X + 3 y + 1 = 0, 4 X - 3 Z + 11 = 0, which are projection forms of the equations, since each one involves only two variables. EXAMPLE.
369. Two-point equations of a straight line. If a straight line passes through two given points, (xi, Yi, zi) and (x2, Y2, z2), we know by § 345 that (x2 - x1), (y2 - Yi), and (z2 - zi) are a set of dire~tion numbers of the line. Consequently, by § 358, equations (3), the line is given by the equations (1)
X -
X1
Y-
Y1
Z -
Z1
--=--=--· X2 -
X1
Y2 -
Y1
Z2 -
These are the two-point equations of the line.
Z1
588
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXVIII
EXAMPLE. Write equations of the line through the two points (5, -7, 3) and (8, 2) -4). x-5_y+7 _ z-3 Solution. By (1) the equations are 8 - 5 - 2 + 7 - -4 - 3' which can be reduced to
360. forms.
3X
-
y - 22
= 0,
7X
+ 3Z -
44 = 0.
Reduction of general equations to projection and symmetric If the equations of a straight line are in the general form
+ B1y + C1z + D1 = O, A2x + B2y + C2z + D2 = O,
(1)
A1x
(2)
they may be reduced to a projection form by eliminating first one and then another variable between the two equations. EXAMPLE. Write projection and symmetric equations for the straight line having the equations (3) 8 X + 3 y - 3 Z + 7 = 0, (4) 2 X - 9 y - 4 Z 44 = 0. Find the direction cosines of the line. Solution. In order to reduce the equations to projection form, first eliminate one of the variables, say y, between (3) and (4). Multiply the members of (3) by 3, and add. We get 26 x - 13 z 65 = 0, or (5) 2X - Z 5 = 0. Next eliminate z between (3) and (4); multiply the members of (3) by 4 and those of (4) by -3, and add. We get 26 x + 39 y - 104 = 0, or (6) 2 X + 3 y - 8 = 0.
+
+
+
Equations (5) and (6) can be used to sketch the line, since they are the equations of its projections on the xz- and xy-planes. In order to write the equations in symmetric form, solve both (5) and (6) for the common variable, x. We obtain x
z-5 =-, X
(7)
x=3y-8,
and
2
-
-
y-~
y____a -
1---i-
or
-2 z-5 --. 2
The point (0, ¾, 5) is evidently a point on the line. It is the point in which the line pierces the yz-plane, since at this point x = 0. The denominators 1, -i, 21 are a set of direction numbers of the line, so that
1
cosa
=
-V1 + ½+ 4
1
=
3
¾= 7'
cos {3
=
2
¼ -
2
=
-
7'
cos 'Y =
2
G
i = 1· 3
Symmetric equations of the line, which are especially useful because they
SOLID ANALYTIC GE01\1ETRY
§§ 359-360]
589
show the direction of the line, are
(8)
X
y-}
z-5
-¥-
-f
t
-=-- =--.
EXERCISES 1. Write equations of the line through (5, 1, -6) and (3, -2, 1).
2. Write equations of the line through (6, 0, 4) and (5, 3, -2). 3. Write equations of the line through ( -1, -7, 3) perpendicular to the plane 5 x - 7 y + 3 z + 4 = 0. 4. Write equations of the line through (6, -2, 0) perpendicular to the plane 2 x 5y - 4z l = 0.
+
+
. X - 7 y +3 Z - 4 . d" 5. Show t h at t h e 1me - - = ~ = - - Is perpen 1eular to the 4 3 plane 4 x - y + 3 z - 17 = 0, and find the point where it pierces it.
. x- 5 y - s z+4 . d. 6. Show t h at t h e 1me - - = - - = ~ Is perpen ICU1ar to the 5 3 plane 3 x + 5 y - 4 z + 29 = 0, and find the point where it pierces it. In each of the Exercises 7 to 10, write projection and symmetric forms of the equations of the given line, giving its projections on the xy- and xz-planes. Draw the line, and find its direction cosines. Find the points where it pierces the coordinate planes. 7_ { 4 X 2X
+ 9y + Z =
g_ { 4 X
+4y +8y -
9X
11,
S. {
6 y - 3 Z = 9.
-
3 Z = 10, 5 Z = 19.
+y +Z =
X X -
lO. { 5 X 2
y - 2Z
-
X -
3,
= 7.
+ + +
2y - 3Z 27 = 0, 3y Z 2 = 0.
11. Write the equation of the plane which passes through the point (2, 4, -1) and is perpendicular to the line X
+y +Z =
2,
2
X
+3y
- 6 Z = 24.
12. Write the equation of the plane which passes through the origin and is perpendicular to the line 7
X
+ 6 y + 4 Z + 3 = 0,
3x
+2y +z+l
= 0.
13. Write equations of the line which passes through (2, -4, 7) and is parallel to the line 2
X
+y -
Z
= 5,
X -
12 y - 8
Z
= 65.
14. ·write equations of the line which passes through (6, 11, -5) and is parallel to the line 4
X -
3y
+ 3 Z + 20 = 0,
11
X -
12 y - 18
Z -
20 = 0.
590
FIRST YEAR OF COLLEGE l\IATHElVIATICS
9X
-
y - 4
+3y
+5 4
lies m the plane
3 = 0.
Z -
-
2z
7
5 2
16. Show that the line
4X
+ 1 =y-4=
3X
16. Show that the line
[Ch. XXXVIII
3
X -
3
7 - 3y - z-+-1 lies m the plane 2 4
=
14 = 0.
Z -
17. Find the equations of the planes which are at the distance 2 from the
origin and are perpendicular to the line X -
2y
+3Z =
3
4,
X -
4y -
= 6.
Z
18. Find the equations of the planes which are at the distance 4 from the origin and are perpendicular to the line
7X
+y
- 2
Z
+4
= 0,
16
X
+3y -
4
Z
+8 =
0.
19. Find the acute angle between the lines 2x y 3y 2x
+
+2z - 2 = +4z+ l =
0, } 0,
d
{
x 7x
an 20. Find the acute angle between the lines 2x
7X
+
+
y z 6y 4Z
+
+ 2= + 27 = VI.
0, } 0,
+ 2y - z+ 30 y + 5 z -
and { 3 x - 20 y - 4 z 2y+ Z
5
=
0,
71
=
0.
+ 21
= 0, = 0.
QUADRIC SURFACES
361. Discussion of an equation. In discussing the equation of a surface, one should examine the equation with regard to symmetry, intercepts, and extent of the locus. But the nature of the surface is usually best revealed by a study of its intersections with planes parallel to the coordinate planes. Special cases of such intersections are the intersections of the surface with the coordinate planes themselves. The intersections with the coordinate planes are sometimes called the traces of the surface. They are important because their equations are easily found and because they are easily plotted from their equations. One equation of the intersection of a surface with a plane parallel to the xy-plane, say z = k, is found by substituting z = k in the equation of the surface. The resulting equation involves only the two variables x and y. Considered as an equation in two dimensions, it defines the curve of intersection referred to axes which are the intersections of the horizontal plane z = k with the xz- and yz-planes. The equation of the trace of the surface in the xy-plane is, of course, found by setting z = 0 in the equation of the surface. Similar sub-
§§ 360-363]
SOLID ANALYTIC GEOMETRY
591
stitutions are made to find the intersections of the surface and a plane parallel to the xz- or yz-plane. The method is illustrated in the following examples.
362. Quadric surfaces. The surface defined by an equation of the second degree in x, y, and z is called a quadric surface or conicoid. Any plane section of a quadric surface is a conic or a limiting form of a conic or a straight line. For consider some quadric surface and a
cutting plane. It is always possible to make rotations of coordinate axes and planes by substitutions which do not alter the degree of an equation. For that reason, without loss of generality, we can assume that the equation of the surface is of the first or second degree in x and y and that the cutting plane is perpendicular to the z-axis and so has an equation of the form z = k. Then the equation of the cutting section, found by substituting z = k, is also of the first or second degree in x and y. The section is therefore a conic sectiron or a limiting form of a conic or a straight line. The following sections briefly discuss certain types of quadric surfaces, taken in simple positions with respect to the coordinate system.
363.
The ellipsoid. x2
(1)
z2
y2
2+b2+2=1. a C
The figure is symmetrical with respect to each of the coordinate axes and planes. Setting y = z = 0, we find the intercepts x = ±a. The y-intercepts, in a similar way, are found to be ±b, and the z-intercepts are ±c. Setting z = 0, we get the equation
X
(2)
Frn. 199
The trace in the xy-plane is therefore an ellipse with axes 2 a and 2 b. Setting z = k, and dividing by the constant term, we obtain (8)
x2 a2 _ (c2 _ k2) c2
y2
+ _b2 (c2 _ c2
7c2)
- 1.
592
FIRST YEAR OF COLLEGE l\fATHEMATICS
[Ch. XXXVIII
The intersection with the horizontal plane z = k is therefore an elc2 - k2/c and 2 bV c2 - k 2/c. These numlipse whose axes are 2 bers are imaginary if Jkl > c. The surface is consequently contained entirely between the planes z = ±c. The axes of the elliptical section are greatest when k = 0, and decrease toward zero as !kl increases toward c. Sections by planes parallel to the xz- and yz-planes are likewise ellipses. In the general case the semi-axes, a, b, c, are unequal. If a = b = c, the ellipsoid is a sphere. If a ~ b, but b = c, any section parallel to the yz-plane is a circle, and any section parallel to one of the other coordinate planes is an ellipse. The figure is. then called a spheroid. It is an oblate spheroid if a < b, and a prolate spheroid if a > b.
av
364. The hyperboloid of one sheet. (1)
x2
y2
z2
-+---=1 a2 b2 c2 .
This equation is the same as that of the ellipsoid with the sign of one variable term changed.
z
Fm. 200
The equation of a horizontal section, made by the plane z = k, i8
§§ 363-365]
(2)
SOLID ANALYTIC GEOMETRY
x2 a2 -:- (c2
c2
,
+ 7c2)
+ ---=--y2 = b2 _ (c2 + k2) c2
593
1.
The section is an ellipse whose size increases when lkl increases, that is, when the cutting plane recedes from the xy-plane. Notice that the elliptical cross-sections are in the planes that are perpendicular to the axis corresponding to the term of odd sign in the equation. The equation of a section made by a plane y = k, parallel to the xz-plane, if lkl ~ b, is x2 z2 ---=l. (3) c2 a2 - (b2 - k2) b2 (b2 - k2) b2 This represents a hyperbola whose transverse axis is parallel to the x-axis and whose conjugate axis is parallel to the z-axis, for sections for which lkl < b. When lkl > b, equation (3) repre~ents a hyperbola with the transverse and conjugate axes parallel respectively to the z- and x-axes. When lkl = b, the equation of the section is x2 z2 (4) a2 c2= 0 ' which represents a pair of straight lines. Similar statements can be made about the hyperbolic sections by the planes x = k.
366. (1)
The hyperboloid of two sheets. x2 y2 z2 = a2 b2 c2 1 .
The equation is the same as that of the ellipsoid, with the signs of two variable terms changed.
z
0
FIG. 201
a
594
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXVIII
The equation of a section made by a plane, x = k, parallel to the yz-plane, if lkl > a, is y2 b2 a2 (k2 - a2)
(2)
+
z2
c2 a2 (k2 - a2)
= 1.
This repre·sents an ellipse whose size increases as the cutting plane recedes from the yz-plane. If !kl < a, the locus is imaginary, so that the surface is broken into two sheets, for one of which x > a, and for the other of which X < -a. Notice that the elliptical cross-sections are made by planes perpendicular to the axis corresponding to the term of odd sign in the given equation. We saw that this is also true in the case of the hyperboloid of one sheet. The section made by the plane y = k has the equation
x2
(3)
z2
---=
i: (b2 + k2)
1.
It is a hyperbola whose transverse and conjugate axes are parallel respectively to the x- and z-axes, and increase when the cutting plane recedes from the xz-plane. Similar statements hold for the sections parallel to the xy-plane.
z
366. (1)
The elliptic paraboloid. x2
a2
y2
+ b2 = 2 €Z.
The section made by a plane z = k is an ellipse which increases in size as the cutting plane recedes from the xy-plane. If c > 0, the surface lies wholly _ _ _ ____;::i..,.._......__ _ _ _ X above the xy-plane. If c < 0, it lies wholly below the xy-plane. Sections made by planes parallel y to the xz- or yz-planes are parabolas, FIG. 202 open upward.
§§ 365-367]
SOLID ANALYTIC GEOM~ETRY
595
367. The hyperbolic paraboloid. (1)
x2
y2
a2 - b2
= 2 cz.
The equation of a horizontal section made by the plane z = k is x2 y2 2 2 ( ) 2 a ck - 2 b2ck = 1. If c > 0 and if k > 0, this represents a hyperbola whose transverse and conjugate axes are parallel to the x- and y-axes respectively and increase with k. If k < 0, the transverse and conjugate axes are
------
Fm. 203
parallel to they- and x-axes respectively. the pair of straight lines x2 y2 (3) a2 - b2 = 0.
And if k = 0, the trace is
The section made by a plane y = k is a parabola open upward; and that made by a plane x = k is a parabola open downward. Because of the shape of the surface near the origin, the origin is said to be a_ "saddle-point" of the surface.
596
FIRST YEAR OF COLLEGE MATHEMATICS
[Ch. XXXVIII
368. The right circular cone. (1)
x2
+ y2 -
c2z2
= 0.
z
Fm. 204
This surface can be regarded as a surface of revolution generated by the rotation of the line x = cz about the z-axis. However, it can be studied by the methods of the preceding paragraphs. The equation of a horizontal section made by the plane z = k is (2)
This evidently represents a circle whose radius, jckl increases when lkl mcreases. The section made by the plane, y = k is the hyperbola. (3)
c2z2 _ x2
=
k,2.
From symmetry we see that any section made by a plane parallel to the axis of rotation is a hyperbola whose transverse axis is parallel to the axis of rotation. When k = 0, the hyperbola degenerates into the two straight lines, x = cz and x = -cz. Notice that equation (1) differs from the equation of the hyperboloid of one sheet chiefly in having zero as the right-hand member, as well as in having equal coefficients for the terms in x2 and y2 •
SOLID ANALYTIC GEOMETRY
§ 368]
597
EXERCISES Find the sections of the following surfaces made by planes parallel to the coordinate planes. Discuss and sketch the surfaces. x2 y2 z2 L-;-b2+-;=l. a C
x2 2. - a2
+
x2
z2
3. 2 a
z2 4. - 2 t
+2 C
x2 a2
-
-
y2 z2 b2 - c2 = 1.
= 2 by, for b > 0 and b < 0.
=
2 by, for b
> 0 and b < 0.
+ 36 y + 4 z = 144. 25 x + 4 y + 4 z = 100. x y + 4 z = 4.
5. 9 x2
2
2
2
2
2
7. 2 9. 2 - 2 11. 4 x2 - 9 z2 = 36. 13. 16 x 2 + 16 z2 - 9 y2 = 0.
6. 4 x 2 8. 4 x 2
10. 9 x2
+ 25 y + 4 z 2
-
16
y2 -
+ 16 y
2
9
z2
= 100. = 144.
= 4 z. 12. 9 x2 - 16 y 2 = 144 z. 14. x2 = y2 + z2. 2
APPENDIX
Review of Eletnentary Algebra I.
THE FUNDAMENTAL OPERATIONS
369. The fundamental operations of algebra. The fundamental operations of algebra are the operations of addition, subtraction, multiplication and division. These operations obey certain rules or laws which determine the very nature of our algebra. These rules were stated in Chapter I, and at this point the student should study §§ 8, 9, 10, 11. He should also be familiar with the properties of real numbers, the notion of absolute or numerical value, and the notation and meaning of positive integral exponents (§§ 5, 6, 12). Certain other portions of Chapter I are repeated here without change in order to give greater continuity to the review. 370. Like terms. The literal parts of the numbers 5 xy, -3 a, 7 xy 2 are xy, a, xy 2• The numbers 5, -3, 7 are their numerical coefficients. ,Ye may also speak of 5 x as the coefficient of y in the expr-ession 5 xy, although 5 x is not the numerical coefficient of the expression. Numbers, with their algebraic signs attached, which are added to make an algebraic sum, are called the terms of the sum. If certain terms in an algebraic sum have the same literal parts, they are said to be like terms. In the algebraic sum 5 x - 8 - 3 a2b + 5 + 4 a2b - 7 x, the terms 5 x and -7 x, -8 and 5, -3 a2b and 4 a2b are pairs of like terms. EXAMPLE.
371. Combination of like terms. In order to combine or collect like terms in an expression, we replace all of the terms having the same literal part by a single term whose coefficient is the algebraic sum of the coefficients of those terms and whose literal part is the same as their literal part. It is often convenient to write like terms in columns and then add their coefficients in columns. 598
[§§ 369-372]
REVIEW OF ELEMENTARY ALGEBRA
Combine like terms in the expression 5 - 6 a + 3 b - xy2 + 2 a - 3 + 7 xy2 - 8 b + a
EXAMPLE
Solution. (5 - 3
1.
+
Combine like terms in 2 x 3+ 3 x2 - x + 4 + 5 x3 - x2 + 3 - 2 x 4
Solution.
2.b.
(-xy 2 + 7 xy2) (3 - 8 - 2)b + (-1 + 7)xy2
(-6 a+ 2 a)+ (3 b - 8 b - 2 b)
= (5 - 3 + 4) + (-6 + 2 + l)a = 6 - 3 a - 7 b + 6 xy2 • 7 x4 -
+4-
This expression equals
+ 4) +
EXAMPLE
599
+
2.
-
x3
+ 5x +9 x
3
-
7 x + 3.
We arrange this in columns of like terms:
7 x4 -2x4 Add by columns:
5x
4
-
+ -
+ +
2x3 +3x2 x+ 3 x2 5x + x3 + 5x 9 x3 -7x+ 3 2 11 x + 2 x - 3 x +
4 3
3 IO.
EXERCISES Simplify the following expressions by combining like terms.
1. 2 ab - 3 a + 4 b + 6 ab - 8 b + 6 a - IO ab. 2. 3 x 2 - 4 xy + 2 y2 + 3 xy - 5 x 2 - xy - y2 + 3 xy. 3. 2 mn + 4 m 2 - 9 n 2 + 3 mn - 8 m2 + 2 n 2 - mn. 4. 5 u 2v - v3 + 4 uv2 + 2 u 3 + 3 v3 - 2 u 2v - 8 uv2 + u3 • 5. 2 X + 3 y - 4 Z + 7 X - 2 y + 8 Z + 7. 6. x 4 - 5 x2 + 2 x - 7 + 3 x3 - 2 x2 - 5 x + 6 x 4 - 2 x2 + 3. 7. kx 2 - 3 k 2x + 6 k 2x + 2 x - 2 kx 2 + 8 - 3 x. 8. 2 x 3 y - 3 x 4 + 4 x2 y2 - xy3 + 7 x 4 - 3 x3 y + x 2y2 - 5 xy3 • 9. 3 X - 7 + 8 - 4 X + 7 X + 2 - 9. 10. x 4 + 3 x 3 - 7 x + 11 - 3 x 4 - x 3 + 6 x 2 + 2 x - 1
+ 6 x + 5 x + 3., 4
2
372. Parentheses. We have already used parentheses, ( ) , to indicate that the numbers which they contain are grouped together . Brackets, [ ], and braces, { }, are used in a similar way, and the vinculum, - - , groups together the terms over which it is drawn. The following statements concerning parentheses apply also to the other symbols of grouping. In an algebraic sum parenih6ses may be inserted or removed without changing the signs of the terms that they contain, if the parentheses are preceded by a plus sign, written or under-
600
FIRST YEAR OF COLLEGE MATHEMATICS
[Appendix
stood; if the parentheses are preceded by a minus sign, they may be inserted or removed, providing that the signs of their inner terms are reversed. The latter statement follows from the fact that the subtraction of a number is equivalent to the addition of its negative. When one symbol of grouping encloses other such symbols, we may remove them by removing first the innermost symbol and proceeding outward until they have all been removed. EXAMPLES.
1.
6X
-
3y
+5X +4y -
8
X
+9y
= (6 X - 3 y) + (5 X + 4 y) + (-8 X + 9 y) = (6x - 3y) - (-5x - 4y) - (Bx - 9y). 2. (5 a - 7 b
+ 4 c) +
(11 a+ 3 b - 6 c) -
(2 a - 7 b
+ 8 c)
=5a-7b+4c+11a+3b-6c-2a+7b-8c = 14 a + 3 b - 10 c. 3. Remove the symbols of grouping in the following expression, and ~ombine like terms. 5 - {6x - [(4y Solution. 5 -
[4 y
=
{6
= 5 -
(2y - 9x
+ 3)]}.
The expression becomes
{6 X 5 -
+ 3x) -
+3X -
X -
[4 y
{6 X -
[2 y
= 5 - {-6 X
-
(2 y - 9 X -
+ 3X
-
2y
+ 9 X + 3]}
+ 12 X + 3]}
2 y - 3}
= 5
3)]} = 5 -
{6 X - 2 y - 12 X - 3}
+ 6 X + 2 y + 3 = 8 + 6 X + 2 y.
EXERCISES In each of the Exercises 1 to 5, write an equivalent expression in which the terms are enclosed in parentheses or brackets preceded by a minus s1gn. 1. 4 x - 2 y - 3. 2. -5 m + 3 n + 7. 3. 6 x 2 + 3 xy - y2 • 4. (5 X - 2 y) - (8 X + 6 y). 5. ( -3 X + 2) - (8 X - 11) + (-4 X + 7). In Exercises 6 to 16, remove the parentheses and other symbols of grouping, and combine like terms. 6. 3 a+ (5 a+ b). 7. -3 x - (8 x - 2 y). 8. 5 x - (4 - 2 x). 9. (6 a - 3 b) - (4 a - 5 b). 10. (5 x + y - 7) + (3 x - 2 y - 4). 11. (3 X + 4 y - 8) - (5 X - 2 y - 1). 12. (a - 3 b + l) - ( -6 a - 5 b - 3). 13. [(3 a+ 2 b) - (5 a - 4 b)] - [(6 a - b) + (3 a+ 5 b)]. 14. [5a - (6a - 2b + c)] - [(3a - 2b) - (5b- Ba)]. 15. [u - (2 u - 3 v)] - {[(5 u - 6 v) - (4 u + 5 v)] + [3 u - (2 v + u)]}. 16. [(m - 2n) - (4m + n)] - {(6m - 3n) - [(4m - 2n) - (n m)]}.
+
§§ 372-373]
REVIEW OF ELEMENTARY ALGEBRA
60.I
(
373. Rational integral expressions.
The expressions
a, ax, ax 2, ax3,
where a does not contain x, are rational integral terms in x. In ai similar way, if a does not involve x or y, such terms as a, ax2, axy,axy3, ax4 y 2 are rational integral terms in x and y. More generallyr a term is a rational integral term in certain specified literal number~if it is either a number that does not involve those numbers at all or isthe product of such a number by positive integral powers of the literal numbers in question. A rational integral expression is an algebraic sum of rational integral terms. If it contains just one term, it is a monomial. If it contains just two terms, it is a binomial; if there are three terms, it is a trinomial. If a rational integral expression contains more than one term, it is a polynomial. There are many theorems relating to rational integral expressions in general and, for the sake of brevity, the word polynomial will often be used instead of "rational integral expression" even when there is only a single term. A rational integral term is said to be of degree k or of the kth degree in x, if k is the exponent of x in the term. It is of degree k in two or more literal numbers if the sum of the exponents of those numbers is k. A term that does not contain the literal numbers in question is considered to be of degree zero. The degree of a polynomial in one or more literal numbers is the highest degree that any term has in those numbers. EXAMPLES.
1. 8 x 3 is of the third degree in x.
2. 9 x 2y3 is of the second degree in x, of the third degree in y, and of the fifth degree in x and y. 3. 2 x 3 - x2 + 5 is a polynomial of degree 3 in x. 4. 5 x3 - 7 xy4 2 x2y2 - 7 x + 2 is a polynomial of degree 3 in x, of degree 4 in y, and of degree 5 in x and y.
+
ORAL EXERCISES For each expression, name the degree in x, in y, and in x and y. 1. 3 x 4 - 2 x2 + x - 5. 2. 3 - 2 X + 4 x3 • 4. 3 y2 + y. 3. y2 - y 5 + 3 y + 7. 5. XU - 1. 6. x 2 + y2 - 9. 7. x 5 - 4 x 2y2 + 3 x + 2. 8. x2y2 - 8 y 5 + 3 x3 - 4 y2 • 9. xs _ x4y2 + y4. 10. x3y2 + 5 y3 - 8 x7 3.
+
602
FIRST YEAR OF COLLEGE MATHElVIATICS
[Appendix
374. Multiplication of polynomials. In order to multiply one polynomial by another, write one of them below the other, arranging the terms in the descending order of powers of one of the letters involved. Multiply the first polynomial by the terms of the second polynomial in succession, writing like terms in columns. Add by columns. EXAMPLE.
Solution.
Multiply x 4
2x
-
+ 5x + 3
by
2
2 x3
4 - 3 x.
-
Arrange in the descending order of powers of x:
1\'.Iultiply by 2 x 3 : l\iultiply by -3 x: Multiply by -4: Add by columns:
x4
+
2 x7
+
2
x7
+
5x2 -2x +3 2x3 -3x -4 IO x 5 - 4 x 4 + 6 x3 3 x5 15 x 3 4 - 4x 5 7 x - 8 x4 - 9 x3
+ -
6 x2 20 x 2 14 x 2
-
9x
+8x -
12. x - 12.
EXERCISES In each Exercise multiply the first polynomial by the second. 1.
3. 5. 7. 9. 11.
X
+ 2, 2 X -
5X 3
2
X
5.
+ 7 y. X + 2 y.
3 y, 2 X
+5y
2
,
4 x + 5, x 2 - 5 + x. y2 + 4 y - 8, y2 + y - 3. x 2 - 2 xy + 5 y 2, 3 x 2 + xy. x2
-
2. 4. 6. 8. 10. 12.
375. Division of polynomials. another:
6X
+ 4.
+ 4 y, 2 X -
3X 8x
7, 3 X
-
2
-
2
2
y, y
-
5 y. 3 x2 •
7 x 2 - 8 x3 + 3, 5 - 2 x. 2 y + 3 y2 - 1, y + 6 y 2 • 3 x2
-
xy
+4y
2
,
2 x2
+ 5 xy + y"'.
To divfole one polynomial by
(1) arrange the terms in the descending order of powers of some common letter; (2) divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient; (3) multiply the entire divisor by this first term of the quotient, and subtract the result from the dividend, to obtain the first partial remainder; (4) use this remainder as a new dividend and divide its first term by the first term of the divisor, to obtain the second term of the quotient; (5) continue this process until a remainder is obtained which either is zero or is of lower degree than the divisor.
The process is similar to that used in long division in arithmetic.
§§ 374-375] EXAMPLES.
REVIEW OF ELEMENTARY ALGEBRA 1. Divide x 4
603
5 x3 + 2 x 5 + 9 x 2 + 12 x + 2 bv 2 x
-
+x
2 -
3.
2. Divide 1643 by 14.
Solutions. (1)
2 x3
x2
+2x -
-
3 x2
(2)
+ 7x
- 14
3 12 x 5 + x - 5 x + 2x 5 +4x4 - 6x3 3
4
9x
117 2
+ 12 x +
14
2
14
x3 + 9 x 2 -3x4 -6x3 + 9.x 2
- 3 x4
+
7 x3 + 0 7 x 3 + 14 x2
- 14 X
2
- 14 X
2
24
14
+ 12 X
103
21 x
98
-
+ 33 X + -
2
+ 42
28
X
61
X -
40
The quotient is 2 x3 - 3 ±2 + 7 x - 14; the remainder is 61 x - 40. We can write:
x 4 - 5 x3 + 2 x 5 + 9 x 2 + 12 x 2X
+X
2
-
+2
3
61
40
= 2x3 - 3x2 +7x- l4 +----. 2 X -
2x+x -3
EXERCISES Perform the divisions indicated. 1. (x3 + x2 - 22 x - 3) + (x - 4). 2. (x3 - 3 x 2 - 15 x + 13) + (x + 3). 3. (x3 + 4 x 2 - 7) + (x - 2). 4. (x 3 - 23 x + 4) + (x + 5). 5. (6 x 3 - 10 x 2 + 13 x - 6) + (3 x - 2). 6. (12 x3 + 13 x 2 - 13 x - 8) + (4 x + 3). 7. (12 x3 - 16 x2 + 9 x - 3) + (1 - 2 x). 8. (2 x 4 + 7 x3 - 9 x 2 + 9 x + 2) + (x2 + 4 x - 2). 9. (7 x 2 - 6 x 3 + x 4 x - 2) + (x 2 - x + 1). 10. (6 x 4 - 2 x 3 + 13 x 2 + 4) + (3 x 2 - x + 2). 11. (6 x4 + IO x 3 + x2 + 3 x - 2) + (2 x 2 + 3). 12. (x 4 - y 4) + (x - yt
+
l 1643
5 The quotient 1s 117; the remainder is 5; and 1
r!
3
=
n 1154.
FIRST YEAR OF COLLEGE MATHEMATICS
604
[Appendix
376. Special products. Certain types of products occur so often that it is necessary to be able to write them by inspection. The student should verify the following formulas by actual multiplication.
I. a(x
+ y) = ax+ ay. 3 a2 (5
EXAMPLE.
II. (x
+ y)(x -
y) = x2
y2 •
-
+ 3 a b)(5 x - 3 a b) (x + y) 2 = x2 + 2 xy + y 2 • (5 x
EXAMPLE.
III.
6) = 15 a2x - 18 a2 •
X -
IV. (x - y) 2
=x
2
=
2
2 xy
-
25 a2x 2
25 x 2
-
9 a4b2 •
+ 30 abxy + 9 b y
2 2
•
+ y2 •
(3 ax - 4 x 2y) 2 = 9 a2x 2
EXAMPLE.
=
2
+ 3 by)
(5 ax
EXAMPLE.
V. (x
2
24 ax3y
-
+ 16 x y
4 2
•
+ a)(x + b) = x2 + (a+ b)x + ab. (a+ 3)(a - 2)
EXAMPLE.
VI. (ax
= a2
+ (3
+ 3(-2)
- 2)a
=
a2
+a -
6.
+ b)(cx + d) = acx2 + (ad+ bc)x + bd. (2x - 3)(4x
EXAMPLE.
+ 5)
VII. (ax + by)(cx + dy) (5 X -
EXAMPLE.
VIII. (x2 EXAMPLE.
-
xy
2 y)(3 X
-
2x - 15.
-
= acx + (ad + bc)xy + bdy 2
+ 4 y)
+ y 2 )(x + y)
(16 x2
= 8x2
2
= 15 x 2 + (20 - 6)xy = 15 x 2 + 14 xy - 8 y2 •
=x + y 3
8 ax+ 4 a 2 )(4 x
3
•
8 y2
•
+ 2 a)
= 64 x 3 + 8 a3.
+ xy + y 2 )(x - y) = x3 - y 3 • EXAMPLE. (9 a + 15 ab+ 25 b2)(3 a - 5 b) = 27 a 125 b3. X. (x + y + z) 2 = [(x + y) + z] 2 = x2 + y 2 + z2 + 2 xy + 2 yz + 2 zx. EXAMPLE. (3 X - y + 2) = 9 x + y 6 xy + 12 X - 4 y + 4. IX. (x2
2
3
2
2
2
-
-
EXERCISES Perform the following multiplications by applying the foregoing formulas. Do the work with as little use of paper and pencil as possible. 1. 5 c(3 x - 2 y). 2. -6 x(x 2 + 3 xy). 3. 2 a2b(a - 5 a2b3). 4. (2 a - b)(2 a+ b). 7. (3 a 2x - 7 ay)(3 a2x
5. (y
+ 7 ay).
+ 4)(y -
4).
8. (x3
+
6. (x - 6 y)(x y3) (x3 _ y3).
+ 6 y).
§§ 376-377]
REVIEW OF ELEMENTARY ALGEBRA
+2b i y). + ;)-
3
9. (5 a4 - 2 b3)(5 a4 11. (½X + i y) (½X -
(! 15. (; + \a) (\a
13. (~ _ ~) 2
23. 26. 28. 30. 32. 34. 36. 38. 40. 42. 44.
e-;)-
(! + ;) 16. G- t) (~ + n-
-n
r
18. (2
2•
2
14.
X -
2
(5 ax+ 2)(3 ax - 7). (2 x 2
+ 4)(3 x
2
35. 37. 39. 41.
11).
-
(6x+y)(2x-3y). (2 x 2
+ 7 y )(5 x
(3 x 2
-
(x
2
xy) (4 x 2
+ 2 y)(x
2
-
5
3
5
;r
19. (3
5 y) 2 •
•
2
y 2 ).
-
+ 2 xy). 2 xy + 4 y
46. G- ~) G+ 3t)48 (~ _ E) (3 a + 4c)· •
10. (3 x2 + 7 y 2 )(3 x2 - 7 y 2). 12. ( j X + ¼Y) (¼Y - j X).
).
+ 2) y a)2 22. ( - + - · (~ _ 4 21. (~ _ 2a y [(x + 1) + a] 24. [(x + 3) - b] 25. [k + l - 9] [(x + 2) - a][(x + 2) + a]. 27. [(2 x + 3) + m][(2 x + 3) (5 x - 2 + a)(5 x - 2 - a). 29. (3 x - 2 - a)(3 x + 2 - a). (x + 4)(x + 9). 31. (a+ ll)(a - 2). (3 x + l)(x + 5). 33. (4 x - 5)(2 x + 7).
17. (x
20.
+ 1)
605
3
II.
2
).
X
2
•
2
•
•
m].
(12 my - 5)(2 my - 8).
+
7)(ay 2 3). (2ax+5y)(-3ax+2y). (x2 - 8 ky 2 ) (6 x2 - ky 2 ).
(8 ay 2
-
43. (a2 - 2 z3)(5 a2 + 6 z3 ). 45. (3 X - 2)(9 X 2 + 6 X + 4).
x2
47. ( a2 49. (~
y2) (2 x2 y2)
+ b2
+
-;;,;:
+ b2
.
D(~ -D·
FACTORING
377. Factoring. In § 376 we learned how to write the product of certain types of factors by inspection. We shall now study the inverse problem: Given a product, to discover its factors. Prime factors. A preliminary definition: A rational integral expression ( § 373) whose numerical coefficients are integers is said to be prime when it contains no factors that are rational integral expressions with integral coefficients except itself, its negative, and ± 1. We shall consider that a rational integral expression is factored when it is written as the product of prime rational integral factors. Any factorization can be checked by multiplying the factors and comparing the product with the original expression.
606
FIRST YEAR OF COLLEGE MATHEMATICS
[Appendix
The foregoing definition of prime factors is sufficiently general for our present purposes and to illustrate the methods of factoring. For some purposes it is desirable to extend the definition to include cases not covered by the definition (e.g., fractional coefficients) or to admit other factors. The methods of factoring that we are about to study will be the same in these other cases and can be adapted easily to the ends in view in any particular problem. EXAMPLE 1. The expression 4 x 2 - 9 y is prime and cannot be factored. It is true that 4 x2 - 9 y = (2 x + 3-Vy)(2 x - 3'\ly), but the factors here are not rational integral expressions. EXAMPLE 2. According to our limited definition, x 2 - 3 is prime, for, while x2 - 3 = (x + -V3)(x - V3), and the factors are rational integral expressions, they have some non-integral coefficients. In some portions of this book it is important to admit such factors.
378.
I.
Common monomial factors. ax
The type formula is
+ ay + az = a(x + y + z);
when each term of an expression contains a common factor, thai factor is a factor of the entire expression. Such common monomial factors should be removed before factoring the remaining expression. EXAMPLE. Factor 16 x3y3 + 24 xy 2. Each term contains the factor 8 xy2. Therefore 16 x3y3 + 24 xy2 = 8 xy 2(2 x 2y + 3).
EXERCISES Factor the following expressions. ·when fractions appear in the given expressions, give the answers in the natural form resulting from the use of the formula. 1. 3 x - 9 y. 2. 8 a + 12 b. 3. 6 ax + 9 ay. 2 2 2 4. 5 a - 15 ab. 5. 3 x y - 5 xy • 6. 4 a3 + 6 a2xy2 • 7. ¾x 2 - -! x. 8. ½1rr3 - i 1rr2 • 9. i x2 - ½ax. 10. 14 at2 + 10 a2t. 11. 6 a2x 2 y 4 - 3 ax 5y2. 12. m2x2 + mx3 • 13. 6 a2x + 8 axy - 14 az. 14. 5 x 2 - 10 xy + 15 x. 15. 3 m2 + 12 mn + 6 m. 16. 24 x 2y2 - 6 y + 12 xy. 18. 24 a2bx - 21 ax+ 9 x 3 • 17. 2 x2y2z2 - x 3 y3z + xy2z3 • 19. 121 a3x + 66 a2x - 88 ax 2. 20. 78 m2n + 52 m2n 3 - 39 m2 • 5 21. i xay - i x2y2 + ½x4ya. 22. ¾x2 + ¾x - 1 2 x3• 23, xa - x5y6 + 3 x1y2 - 4 x4ya. 24. 3 x 3y2z8 - 12 xyz2 + x 2yz2 + 4 x2y 2z2•
§§ 377-380]
REVIEW OF ELEMENTARY ALGEBRA
379. The difference of two squares. x2
II.
-
y2
607
The type formula is
= (x + y)(x -
y);
the difference of the squares of two numbers is equal to the sum of the two numbers multiplied by the difference of the two numbers. EXAMPLE
1. Factor 16 x4y2 - 9 a2b6• \\i-.-e can write 16 x 4y 2 - 9 a2b6 = (4 x 2y
+ 3 ab
3)
(4 x 2y - 3 ab3 ).
2. Factor 48 x 8 - 3 y 4 • We first remove the common monomial factor 3, and then factor twice, using formula II. EXAMPLE
48 x 8 - 3 y4 = 3(16 x 8 - y 4) = 3(4 x 4 + y2 )(4 x 4 - y2) = 3(4 x 4 + y2 )(2 x 2 + y)(2 x 2 - y).
EXERCISES Factor the following expressions. When fractions appear in the given expressions, factor in the manner naturally resulting from the formulas. 1. 4. 7. 10. 13. 16. 19. 22.
9 a2 - 16 b2• 49 a2x 4 - 25 b2y 2• 169 x 4 - 100 a2y 2• 20 a2x 2 - 45 y 2• 72 a2x 2 - 32 a4• x 8 - 16 y 4• 162 m 4 - 32 n 8• 23. i ~ a3x 4 - ½ay2 •
4 x 2 - y 2• 25 k2 - 36 l4• 64 a4x 2 - 9 z2• 12 x 2 - 75 y 2• x3 - 4 xy2• x 4 - y4. 64 a 8x 4 - 4 b4y8•
2. 5. 8. 11. 14. 17. 20.
t x2y2 - ¼.
3. 6. 9. 12. 15. 18. 21.
380. Trinomials which are perfect squares. formulas:
x2 x2
III. IV.
+ 2 xy + y 2 = (x + y) 2 - 2 xy + y = (x - y)
2 2
x 4 - 16 y 2•
121 y2• x 2 - 144 b2y 2• 28 x 2 - 63 m2n 2• 14 x 2y 2 - 56 y 4• 81 x 4 - 16 y4• 75 x 4 - 27 y 2• 81 x 2
-
There are two type
, ;
if a trinomial is a perfect square, two of its terms must be perfect squares, and its remaining term must be plus or minus twice the product of their square roots. EXAMPLE
1.
Factor 16 x 2 - 24 xy2 + 9 y 4•
Solution. The terms 16 x 2 and 9 y 4 are perfect squares with square roots 4 x and 3 y 2• The remaining term, -24 xy 2 = -2(4 x)(3 y2), is minus twice the product of these roots. We can write: 16 x 2 - 24 xy 2 + 9 y 4 = (4 x - 3 y?.) 2•
[Appendix
FIRST YEAR OF COLLEGE MATHEMATICS
608
EXERCISES In each of the Exercises 1 to 12, find a value of T which makes the expression a perfect trinomial square.
+T +4y 25x + T + 9y 81 x y + T + 4. 4 x + 4 x + T.
1. x 2 4.
7. 10.
2
2
2
+ T + 1. 81 m + T + 4n 121 a + T + 9 b 12 x yz + T. 9xy
2. 9 x 2
•
6.
•
2 6
8. 11.
2
2
2
2
2
•
+T +y
6. k 2
+ T + 16 Z
2
• 2
2 8
•
+ +
16 m4y 6 • T 9. 169 Z x T. 12. 9 x 2y'l. - 18 xyz
•
2
4 2 -
3. 25
+
In Exercises 13 to 40, factor the expressions.
13. 9 X2 + 6 X + 1. 15. 4 x4 - 20 x 2y + 25 y2•
16.
+
+
4y2 • 68xy 17. 289x2 19. x 6 - 6 x4y + 9 x 2y 2 •
21. 18
x4 -
23. 9(x
60
x 3y
+ 50
+ 12 xy + 9 y 2 lm xy + m y lx 24 xy + y 144 x 16 x + 24 x y + 9 x y (a - x) + 4(a - x) + 4. 30(x + 4) + 9. 25(x + 4) 2
14. 4 x 2 18. 20.
x 2y2•
22.
+ 2 y) + 12(x + 2 y) + 4. 2
. 24.
2 2
-
2
•
4 2
2
6
3
-
•
3 2
4
5
•
•
2
2
2
2
-
+ 72(3 X - 2 y) + 162. 8 x + 16). 4x + 4)(x 26. (x + 2xy + y )(x + 6x + 9). 27. (x 24 x + 16). 9)(9 x 29. (4 x y 6 xy + 9)(x 28. (x 24,a b )(a + 6 ab+ 9 b 31. (54 a 45). 30. (4 x + 12 x + 9)(20 x 16 x 33. (4 x + 12 xy + 9 y 9. 32. (x + 4 xy + 4 y (a + 2 ab + b 2 xy + y 34. (x (a + 6 a b + 9 b 35. (16 a + 8 ab+ b 18 mn + 81 n (m 56 xy + 16 y 36. (49 x (x + 4xy + 4y 4xy + y 37. (4x 6 x + 9). 38. (16 x + 8 x + 1) - (x 2 ab + b (a 39. (a + 10 ab + 25 b 24 xy + 36 y (4 x 40. (9 x + 30 xy + 25 y 381. Trinomials of the form x 2 + qx + r. To factor such a 26. 8(3
2 y) 2
X -
2
2
2
2
2
2 -
2 -
2
2
2
2
2
2
-
2
2
2
-
2
2
2
2
) -
2
2).
2
)
-
4
•
).
2
2
4 ).
).
-
2
-
)
4
-
2
2
2
2
-
)
-
)
-
2 2
-
-
).
2 2
4
) -
2
2
4
-
2
-
2
).
2
) -
2
2
2 -
-
)
2
-
2
-
2
).
2
-
).
trinomial, from § 376 we recall the formula:
V.
x 2 + (a
+ b)x+ ab= (x + a)(x + b).
To factor x 2 + qx + r, we seek two numbers, a and b, such that ab = rand (a + b) = q. If a and b are such numbers, by V,
x2
+ qx + r =
(x
+ a)(x + b).
To find a and b, we examine the different pairs of factors of r in order to find a pair whose algebraic sum is q.
§§ 380-382]
REVIEW OF ELEMENTARY ALGEBRA Factor x2
EXAMPLE.
-
609
4 x - 12.
Solution. The pairs of factors of -12 are 12 and -1, -12 and 1, 6 and -2, -6 and 2, 4 and -3, -4 and 3. The pair -6 and 2 has the sum -4. Therefore x2
-
4 x - 12 = (x - 6)(x
+ 2).
EXERCISES Factor: 1. 4. 7. 10. 13. 16. 19. 22. 25. 28. 31. 34. 36. 38. 39. 40.
+ 3 X + 2.
2. x2 + 6x + 5. 5. X 2 - 7 X 12. 2 8. X - X - 6. 11. X 2 - 3 X - 18.
2
+
+ +
+
+
+
+
+ +
+
+ +
+
+ +
+
+ +
+
+ 9) - 2(x - 3) - 8. (16 x 40 xy + 25 y + 2(4 x - 5 y) (9 a + 24 ab + 16 b (3 a + 4 b) -
(x
2
+ +
+ +
+
+
6x
-
2
2
-
2
2
)
) -
- 3.
42.
382. Trinomials of the form Px 2 + qxy + ry 2 • trinomial, from § 376 we first recall the formula
VI.
+
3. X2 6X 8. 2 2 X 8X 15, 6. X - 5 X 4. 2 2 X - 4 X 3. 9. X X - 6. 2 2 X 3 X - 10. 12. X 9X 18. X 2 - 4 X - 5. 15. X 2 - 5 X - 14. 14. X 2 2 X - 24. 17. t2 - 2 t - 15. X2 - 9 X 14. 18. X 2 4 X - 21. 20. m 2 - m - 30. k 2 + 11 k + 28. 21. a 2 a - 12. 2 2 2 23. a b2 - 11 ab + 30. 24. x y 2 - 6 xy - 7. y lJ - 2. x4 + x2 - 2. 26. x4 + 2 x2 - 3. 27. x4 - 2 x2 - 8. x4 - x2 - 12. 29. y 4 6 y 2 - 7. 30. a4 3 a2 - 28. x4y4 - 13 x 2y2 36. 32. x 2 10 xy + 21 y 2 • 33. x2 7 xy 10 7;2. (x + 2) 2 + (x 2) - 6. 35. (2 x - 3) 2 - 2(2 x - 3) - 8. (2 X 4) 2 - 6(2 X 4) - 40. 37. (2 X + y) 2 - 10(2 X y) + ~1. X
acx2
+ (ad+
bc)x + bd
To factor such a
= (ax + b)(cx + d).
In order to factor px 2 + qx + r, we must first find four numbers, a, b, c and d, such that ac = p, bd = r, and (ad + be) = q. As possibilities for a and c we take the various pairs of factors of p; as possibilities for b and d we take the pairs of factors of r; and then we match them together to give (ad + be) = q. If p is positive, its factors a and c can be taken as positive; b and d must have the same signs if r is positive and opposite signs if r is negative. Another method of factoring trinomials of this type is given in § 67.
610
FIRST YEAR OF COLLEGE MATHEMATICS
EXAMPLE 1.
Factor 4 x2
[Appendi.x
8 x - 5.
-
Solution. Pairs of factors of 4 are 4 and 1, and 2 and 2. Pairs of factors of -5 are 5 and -1, -5 and 1. Taking a= 2, c = 2 and b = -5, d = 1, we have ad+ be= 2 • 1 + (-5)2 = -8. Therefore
4 X2
-
8
5
X -
= (2 X
The more general trinomial px2 way by the formula
VII.
px2
EXAMPLE 2.
Factor 10 x2
-
=
5)(2
+ 1).
X
+ qxy + ry
+ qxy + ry2 =
where, once more, ac = p, bd
-
(ax
2
is factored in a similar
+ by)(cx + dy),
r, (ad
+ be)
= q.
29 xy - 21 y 2 •
Solution. Matching the factors of 10, which are 10 and 1, or 5 and 2, with the factors of -21, which are ±21 and =Fl, or ±7 and =F3, we choose a = 5J b = 3, c = 2, d = -7, so that
+ 5(-7) = (5 x + 3 y)(2 x -
be+ ad = 3 • 2 Therefore
10 x2
-
29 xy - 21 y 2 =
-29. 7 y).
EXERCISES Factor: 1. 4. 7. 10. 13. 16. 19. 21. 23. 25. 27.
3 X2
10 X - 8. 2. 2 5. 12 X + 4 X - 21. 2 20 a - 13 a - 21. 8. 10 t2 - 13 t - 77. 11. 15 X 2 - 19 X + 6. 14. 15 X 2 - 33 X + 6. 17. 12 x2 + 7 xy - 12 y2 • 12 a2 - 8 ab - 15 b2 • 56m2 - 19mn - 15n2 • 20 x 4 - 37 x2 - 18. 6 x4 - 53 x2 - 9. -
3. 8 X2 + 10 X - 3. 15 x2 + 19 x - IO. 6 x2 - x - 15. 6. 6 x2 + 25 x - 25. 24 X2 + 22 X - 10. 9. 24 t 2 - 6 t - 9. 6 m4 + 11 m 3 - 10 m2 • 12. 12 k2 + 22 k - 14. 15 x 2 + 19 x + 6. 15. 15 x 2 - 23 x + 6. 15 x2 - 21 x + 6. 18. 15 x2 - 9 x - 6. 20. 3 x 2
22. 20 k2 24. 28 x2 26. 12 x 4 28. 32 x 4
The sum or difference of two cubes. IX, § 376, 383.
VIII. IX.
2 xy - 5 y2 •
-
xs + ys = (x + y)(x2 - xy xs - ys = (x - y)(x2 + xy
+ 18 kl + 3 xy + 17 x 2
-
2 x2y2
-
-
3512 • 6 y2 •
5.
9 y4.
By formulas VIII and
+ y2), + y2).
§§ 382-384]
REVIEW OF ELEMENTARY ALGEBRA
611
Notice that the sign of yin the first factor is the same as the sign of y 3 , that the sign of xy in the second factor is opposite to the sign of y 3 , and that the term in xy does not have the coefficient 2 as it does in the square of (x ± y). Factor 27 x3
EXAMPLE.
Solution. Therefore
64 y 3•
-
This expression is the difference of the cubes of 3 x and 4 y.
27 x3
-
64y3
=
(3x - 4y)(9x2
+ 12xy + 16y
2
).
EXERCISES Factor:
+
1. xa - 1.
4. 7. 10. 13. 16. 19. 22. 24.
2. x 3 1. 5. a3 - 27 b3• 8. 1000 x3 - 27 y 3 •
3. 27 x 3
8.
-
27 x 3 + 8. 6. 8 a3 + b3 • 8 x3 + 27 y 3 • 9. 8 m 3 + 125 n 3 • 64 a6 - 125 b3• 11. 54 x 5 + 16 x 2y 3• 12. 24 a3x 3 - 81 a6y 3• 14. 64 x 6 - y 6• 8 x 3 - 27 y 6 • 15. 7 x 6 - 448 y6 • 17. a6m 3 - b3n 6• 18. x12 + y 6• 4 x 9 + 32 y 6 • x 6 - 7 x 3 - 8. 20. x 6 - y 6• 21. x 6 + 26 x 3 - 27. (3 X + y) 3 - 27. 23. (X + 2 y) 3 + 8(3 X - y) 3• 8(a - 2 b) 3 - 27(3 a+ 2 b) 3•
384. Polynomials which may be factored by grouping terms. Even when the terms of a polynomial contain no common monomial factor, it is sometimes possible to combine the terms into groups which contain a common factor consisting of two or more terms. This common factor is then a factor of the entire expression. An example of this type is ax
+ ay + bx + by
= a(x
+ y) + b(x + y)
(a+ b)(x + y). + 2 xy - x + y - y y =
EXAMPLE.
Solution. x3 - x2
Factor x 3
-
x2
2
3
-
•
We group the terms as follows:
+ 2 xy -
x
+y -
y2
-
= =
y3
= (x3
y3)
-
(x - y)(x
2
(x - y)(x 2
-
(x2
2 xy
-
+ xy + y + xy + y
2
2
) -
-
x
+y
2
)
-
(x - y)
(x - y) 2
-
(x - y)
+y -
1).
612
FIRST YEAR OF COLLEGE MATHEMATICS
[Appendix
EXERCISES Factor:
16.
+ x(x - 7). 3 x(x - 4 y) + 5 y(x - 4 y). (3 a - 2 b)x + (2 b - 3 a)y. (a 2 - b + (a - b). (x 4) + 3(x + 2). 3 x + 5 ax + 3 y + 5 ay. ax + 3 by + 2 a x + 6 aby. 4 ax - 5 by - 8 ax + 10 bxy.
17.
X2 -
1. 5(x - 7)
2. 2 a(x2
3.
4. k(x2
5. 7. 9. 11. 13.
2
2
23.
26.
12. 5 x -
+ 2 y.
18.
+3y - 5y 27 x 3 - 12 xy + 6 xy + 4 y 4x 14 x + 4 xy - 7 y + y x 3 - x + 2 xy - y y3. 3X
-
2
2
2
2
22.
•
-
2
20.
•
2
2
-
(4 - 7 a)y 2 •
-
-
(
a
+ b).
+ 3(x 3 ay + 5 y y2 )
y).
3 ax.
+ 3 y.
+ 10 xy - 35 xz. 3 X - 9 X + 4 y + 16 y 3 x 3 - 12 xy + 5 xy - 10 y a 3a +3b b a a3 + b3 - b x + 3 x + 5 x - 5 y - 3 xy -
16. 2 x 2y - 7 x2z
2
X
+ 3).
14. 3 x - 5 by+ ax - 5 bx+ ay
2
-
b2 )
-
10. 2(x2
-
19. 5 X 2 21.
8. (a2
- 3 b (x 2
+ 8y) + 2Z(x~ + 8y).
6. (7 a - 4)x2
)
4 y2
+ 3)
•
-
24. 26.
2
2
•
2
4
6
3
-
2
2
2
6
-
4
-
•
•
•
2
y 3•
385. The sum or difference of equal powers of two numbers. The expressions x 2 - y 2 and x 3 ± y 3 are special cases of xn ± yn, where n is a positive integer. By actual division, the following statements will be found to be true for any particular positive integral value of n. 1. n odd. If n is odd, xn + yn is divisible by (x + y), but not by (x - y); and xn - yn is divisible by (x - y), but not by (x y). 2. n even. If n is even, xn + yn is not divisible by either (x y) or (x - y); but xn - yn is divisible by both (x + y) and (x - y).
+
+
In case of divisibility by (x + y) or (x - y), the remaining factors can be found either by inspection or division. EXAMPLE
1.
Factor x5
-
y 5•
Solution. Since the exponent is odd, the expression is divisible by (x - y) but not by (x + y). Dividing it by (x - y), we find that xs _ y5 = (x _ y)(x4 EXAMPLE
2.
Factor x 8
-
+ x3y + x2y2 + xy3 + y4).
256.
Solution. The expression is equivalent to x8 - 28 • Since the exponent is even, the expression is divisible by both (x - 2) and (x + 2). However, in factoring the difference between two even powers, it is best to factor it first as
§§ 384-386]
REVIE,v OF ELElVIENTARY ALGEBRA
the difference of two squares. We therefore write x8 - 256 = (x 4 16) (x 4 - 16) = (x 4 16) (x2
+
= EXAMPLE
(x 4
+ 16)(x + 4)(x + 2)(x 2
Factor x 6
3.
+
+y
6
613
+ 4) (x
2
4)
-
- 2).
•
Solution. Since the exponent is even, the expression does not contain either (x + y) or (x - y) as a factor. However, (x 6 + y6) can be factored as the sum of two cubes: x6
+ y6 =
(x2)3
+ (y2)3 =
(x2
When n is an odd multiple of 3, (xn the sum or difference of two cubes. EXAMPLE
a 9 + b9 =
Solution.
+b (a3)3 + (b3)3 =
Factor a9
4.
9
+ y2) (x4
_ x2y2
+ y4).
± yn) should be factored first as
•
= (a+ b)(a2
+
(a3 b3) (a 6 - a 3b3 + b6 ) ab+ b2 )(a6 - a3b3 + b6 ).
-
EXERCISES Factor the following expressions. 1. x7
+ y 7•
4. a 4
-
7.
2. x 6
5. 81 a4
16.
+ 1.
alO
+ 64.
+ 1. XlO + ylO •
-
y4•
11.
13. x 9
-
y 5. 3. x > t- 6. x < 3. 7. x < 4 a. 9. !xi < 7. 11. !xi > 9. 13. X > 6. 16. X < -8. 17. 3 < X < 5. 19. -2 < x < 6. 21. !xi < 5. 23. To the left of 7. 26. Between -2 and 4. 27. To the right of 3 or to the left of -3. 29. To the right of ;. 31. To the left of t· 33. To the right of 4 or to the left of -4. 36. (a) (b) -. 37. (a) (b) -. 39. (a) (b) -, (c) +. 41. (a) +, (b) -, (c) +. 43. (a) +, (b) -, (c) +, (d) -. 46. (a) (b) -, (c) (d) +. 47. x > 3. 49. x > f. 61. x < 3. 63. x > 4 and x < 2. 66. x > ½and x < --}. 67. x > 3 and -1 < x < 0. 69. 2 < x < 6. 61. 0 < x < 8. 63. x > 3 and x < 2. 66. x > t and x < -3. 67. x > t and x < -t. 69. -4 < x < J. 71. -2 < x < ¾73. X > 5, -4 < X < 2, X < -4. 76. X > 6, 0 < X < 1, X < 0. 77. !xi ~ 6. 79. Ix) ~ ½V6. 81. x ~ -3 and x ~ -6. 83. Royals: Won 24, lost 13, .6486; Lakers: Won 33, lost 18, .6471;
+,
+,
+,
+,
+,
or Royals: Won 22, lost 13, .6286; Lakers: Won 32, lost 19, .6275.
§ 77, page 112 The loci have symmetry with respect to the coordinate axes and origin as follows: 1. x-axis, y-axis, origin. 3. y-axis. 6. x-axis, y-axis, origin. 7. Origin. 9. None.
682
FIRST YEAR OF COLLEGE MATHEMATICS
§ 79, page 116 1. lxl ~ 4, IYl ~ 4. 3. lxl ~ 4. 6. y ~ 0. 7. x ~ 2. 9. 0 ~ X ~ 6, -3 ~ y ~ -1. 11. -7 ~ X ~ 3. 13. X = 2, y = -3. 16. Imaginary. 17. Imaginary. 19. Vertical asymptote, x = 0; horizontal asymptote, y = 0. 21. Vertical, x = -1; horizontal, y = 3. 23. Vertical, x = -3; horizontal, y = 2. 26. Vertical, x = t; horizontal, y 27. Vertical, x ::c 3 and x = -3; horizontal, y = -3.
=
t-
§ 82, page 121 1. Ellipse, ( ±3, 0), (0, ±5). 3. Parabola, (1, 0), (3, 0), (0, 3). 6. Circle, ( ±-i, 0), (0, ± t). 7. Parabola, (6, 0), (0, 1), (0, 6). 9. Circle,(±¾, 0), (0, ±¾)- 11. Ellipse, (±6, 0), (0, ±2). 13. Hyperbola, ( ±3, 0). 15. Hyperbola. 17. Circle, ( ±¾, 0), (0, ±¾). 19. Ellipse, ( ±5, 0), (0, ±2). 21. Hyperbola, ( ±2, 0). 23. Parabola, (0, 0). 26. Hyperbola. 27. Ellipse. 29. Hyperbola. 31. Hyperbola. 33. Ellipse. 36. Parabola. § 84, page 123 1. (4, 3), (-4, 3), (4, -3), (-4, -3). 3. (3, 2), (4.1, -.7). 6. (0, -3), (3, 2). 7. (1, 3), ( -1, -3), (3, 1), ( -3, -1). 9. (2.8, 3), ( -2.8, 3), (2.8, -3), ( -2.8, -3). § 86, page 126 1. (2, -3), (8 - 6). 3. (1, -2), (½, -½). 6. (4, 1), (10, -8). 7. (1, 3), ( -t, -t). 9. (7, -9), ( -3, 6). 11. (3, 4), ( -4, -3). 13. ( -3, 1), ( -17, -13). 16. ( -2, 1). 17. (0, a), (b, 0). 19. (1 + 3 i, 2 - i), (1 - 3 i, 2 + i). 21. (4, 2). 23. k = 23. 1. 3. 5. 7.
§ 87, page 127 (1, 2), ( -1, 2), (1, -2), ( -1, -2). (5, 11), ( -5, 11), (5, -11), ( -5, -11). (3, 2), ( -3, 2), (3, -2), ( -3, -2). (Si, 4i), (-Si, 4i), (Si, -4i), (-Si, -4i). § 88, page 129
1. (2, 4), C-2, -4), (¾V3, 0), ( -¾V3, o). 3. (1, ¾), (-1, -¾), (.}, f), (-.}, -i). 6. (1, -2), ( -1, 2), (½, -!), ( -½, t). 7. (4, 1), ( -4, -1), (6, -2), ( -6, 2). 9. (3, -½), (-3, ½), (1, -½), (-1, ½)- 11. (2, -2), (-2, 2), (4, 3), (-4, -3)., 13. (2, -5), (-2, 5), Cr\Yl9, r59 Vl9), (-/9 Vl9, - 159 Vl9).
§ 90, page 131 1. \(4, 1), (1, 4), (-4, -1), (-1, -4). 3. (2, -5), (-2, 5), (5, -2), (-5, 2). 6. (¾, ½), (½, ¾), (-¾, -½), (-½, -¾). 7. (3v2, -v2), (v2, -3V2), (-3V2, v2), c-V2, 3V2). 9. (i, -2 i), (-i, 2 i), (2 i, -i), ( -2 i, i). 11. (5, -3), ( -5, 3). 13. (4, 1), ( -2, -2), ( -¾, -6), (t, 3). 16. (3, 1), (3, -1), ( -3, 1), ( -3, -1). 17. (1, 4), (4, 1), (2, -3), ( -3, 2). 19. (3, -1), ( -1, 3), (2, 1), (1, 2). 21. (3, -2), ( -1, 4). 23. (5, 4), (5, -4), ( -5, 4), ( -5, -4). 26. (1, 2), (-1, -2), (2, -5), (-2, 5). 27. (2, -1), (½, ½).
ANSWERS 29. 36. 39. 46.
6 hrs., 8 hrs. 31. 12 in., 9 mi.,/hr., 3 mi./hr. $600 at 4%, $400 at 5%. 120 mi./hr., 100 mi./hr.
7½ in.
33. 37. 41. 47.
5 ft., 7 ft. 55 lbs., 10 ft. 69. 8 in., 2 in.
683
43. 14 in., 15 in.
§ 92, page 136
1.d=5. 8.d=-7. 6.d=1--f. 7.d=2x-y. 9. k ,= l, d = 7. 11. k = 2, d = 5; or k = 3, d = 7. 13. 71. 15. 26. 17. 1.99. 19. -179. 21. 119. 23. -1.43. 26. 47, 143. 27. 82, 159. 29. 21, 17, 13. 31. 33, 38, 43. 33. 4 l - 6 d.
§ 93, page 137 1. 644. s. 2511. 5. 348. 7. 1395. 9. 625.98. 11. -345. 13. 278¼. 15. a= 3, d = 9. 17. a= (2s - nl)/n. 19. S = ½n[2 l - (n - l)d]. 21. d = 6, S = 360. 23. a = 4, d = 3. 26. n = 12, a = 2. 27. 16,210. 29. 210. 31. 249,500. ss. 22,500. 36. $8.06. 37. $756. 39. $186,000. § 94, page 140 1. 6, 8, 10, 12, 14, 16, 18. s. 22, 16, 10, 4, -2, -8, -14, -20. 7. 3.22. 13. 860 units. 15. 579. 17. ½, 1\ , ls, fs· 19. -4, - 1s2, -V, -t. 21. -j5 • 23. a/55.
1. r = 5. 9. k = 4, 13. 192. 23. t, 1'1--.
5. -12.
§ 96, page 142 S. r = ½. 6. r = 1. 7. r = x 2y. r = 3; or k = --!-, r = -1. 11. k = 11, r = 4; or k = -3, r 15. 4096/81. 17. - 674 • 19. sh-- 21. ;, 2 :h. 26. 5, 1280. 27. 125, J!-/js; or -125, 2-N. 29. 354,312.
= ?,.
§ 98, page 144 1. 381. S. 182 3 ~ 4 • 6. 12~a- 7. 5llf. 9. 16,385. 11. 1456. 13. 539277. 16. 1020. 17. 70. 19. 0. 21. 2555. 23. 510. 26. 7623, 7653½. 29. a = 256, n = 5. 81. l = 243, n = 6. SS. 96, 48, 24, 12. 35. -180, 300. 37. 49. 39. 7. 41. -Jj-. 43. $10.23, $10,475.52. 46. 1,048,576; 2,097,150. 47. 9.83%. 49. 29 ft., Sf in. 51. 21.825.
1. $1266.39. 9. $125,780. 17. 12½%.
§ 99, page 149 s. $298.06. 6. 64,005. 11. $92. 13. $43,600.
§ 101, page 164 1. 48. S. 6-J. 6. Diverges. 7. 12½. 9. Diverges. 11. Diverges. 13. 3¾. 16. 100. 17. 1--;.. 19. 1-9295 • 23. 4577 /9990. 25. -t-t-r- 27. 1327 /555. 29. 12 ft.
7. $6736.20. 15. $17.83 gain.
21. 2099/333.
684
FIRST YEAR OF COLLEGE MATHEMATICS § 103, page 156 3). 9. 132. 11. 15. l)r2•
1. 2. 3. 5040. 6. 7. 7. (k + 13. (k + 2) !. 16. 7(r + 2) (r +
§ 104, page 168 1. a 6 a 5x 15 a4x 2 20 a3x 3 15 a2x 4 6 ax5 x 6• 3. a 7 - 7 a6b 21 a5b2 - 35 a4b3 35 a 3b4 - 21 a2b5 7 ab 6 - b7• 5. 16 x 4 - 32 x 3y 24 x 2y 2 - 8 xy 3 y 4. 6 4 2 7. 64 x 576 x5y + 2160 x y + 4320 x 3y 3 4860 x 2y 4 + 2916 xy5 + 729 y 6• 9. x7 _ t x6y + ¥- x5y2 _ 3ss x4y3 + f¾x3y4 _ g x2y5 + --1:r xy6 _ rh y7. 4 11. -h a4 - ½a 3b i a2b2 - / 6 ab 3 6 b • 13. x8 16 x 7y2 + 112 x 6y4 + 448 x5y 6 + 1120 x 4y8 + 1792 x 3y10 + 1792 x 2y12 1024 Xy 14 256 y 16• 1 10 4 4 5 7 8 2 16. x - 5 x -fy 10 x y - 10 x1-+y 5 x y - xfy • xs x4 x2 y2 y4 y6 17. 6y + 64 + 64 +6. y + 152 y + 20 + 152 X X X x2 4X 6 4y y4 11 s 3 4 19. 8 + - 5 + 2 + - + 2 · 21. 16 x 2 + 96 xT + 216 x:f + 216 x2 + 81 xa. X X y y y 7 1 1 4 13 23. 729 x-2 + 729 x·-6 + .Ll'-l~ x-a + 1.~ 5 x2 + \:3- / xa + / 6 xT + -li x 3• 26. x 4 + 4 x 3y + 6 x 2y 2 + 4 xy 3 + y 4 - 8 x 3z - 24 x 2yz - 24 xy2z - 8 y 3z + 24 x 2zi + 48 xyz2 + 24 y 2z2 - 32 xz3 - 32 yz3 + 16 z4• 6
+
+ +
+
+
27. 29. 35. 41.
+3
+
+ +
+
+
+
+
+
+ /--/
+
+
3
11 +
+ +
+
+
+
x xi x½ 6 x-½ 12 x-¾ - 8 x-1 • 107,213,535,210,701. 31. 941,480,149,401. 33. 92,236,816. 1.558. 37. 1.489. 39. $1 • (1.04) 15 = $1.8009. $1 . (1.02) 20 = $1.4859.
§ 106, page 160 1. 210 x 4y 6 • 3. -30,618 a4b5 • 6. -63/(16 x). 7. 7920 x 8y 4• 9. -21,840 y 8z11 • 11. 13,608/x2• 13 . .0000811008.
1. 1, 7, 21, 35, 35, 21, 7, 1;
a7
§ 106, page 161 + 7 a6x + 21 a5x 2 + 35 a4x 3
+ 35 a3x 4 + 21 a2x + 7 ax6 + x
5
7
•
3. 10, 126, 126, 210, 210, 21. 1. log3 81 = 4. 7. log27 3 = ½13. log6 1 = 0. 19. log25 (rh) = -¾.
§ 110, page 171 3. log4 64 = 3. 9. logs ¼ = -1. 15. logs1 27 = ¾. 21. 52 = 25. 23. 2 4 = 16.
6. 11. 17. 25.
log10 1000 = 3. log10 0.1 = -1. log4 ¼ = -f. , 10-3 = .001.
27. 25-½ = t- 29. 2. 31. 4. 33. -2. -. 35. -t37. -J. 39. 3. 41. 27. 43. 5. 46. 100. 47. 3. 49. -½. 51. ¼-
1. 0.602. 3. 0.778. ~ 11. -0.477. 13. 1.477. 21. 1.204. 23. - .903. 31. 0.2fi933 · · ·. 33. 0.1725.
§ 111, page 173 7. -0.176. 6. 1.431. 17. 1.398. 16. 0.523. 27. 0.1505. 26. 1.857.
9. 0.352. 19. 1.602. ) 29. 0.10033 · · ..
ANS,VERS
685
§ 112, page 176
1. 105• 3. 9. 2.487 X 15. 67,430, 21. 0.0046,
104• 6. 364,700, 4. 7. 0.001040, 4. 103• 11. 4.70 X 10- 3 _ 67,400. 17. 12.657, 12.66. 0.005.
13. 4.023 X 10-3• 19. 13.48.
§ 115, page 178
1. Char. = 0, Mant. = .8451. 3. Char. = -2, Mant. = .8451. 5. Char. = -1, Mant. = .1549. 7. 1.5403. 9. 2.8739. 11. 4.5403. 13. 0.8739 - 2. 16. 0.5403 - 3. 17. 0.8739 - 4. § 116, page 179
1. 0. 13. 21. 29. 37.
3. 3.
6. 2.
1.5378. 6.5378 - 10. 2430. 243,000.
7. -3. 9. -4. 11. 2.5378 15. 9.5378 - 10. 17. 7.8814 - 10. 23. 2.4192. 26. 6.4192 - 10. 31. 8.31. 33. 0.243. 39. 243,000,000. 41. 831,000,000.
19. 27. 36. 43.
1.8814. 9.4192 - 10. 0.0831. 83,100,000.
§ 117, page 181
1. 0.5159. 9. 7.4624 - 10. 17. 0.458.
3. 1.9138. 11. 7.3010 - 10. 19. 1.02.
5. 8.6972 - 10. 13. 26.5. 21. 0.00502.
7. 8.8494 - 10. 16. 480. 23. 40,500,000.
§ 118, page 183
1. 0.8940. 9. 0.3394. 17. 3.185. 26. 0.01753.
3. 11. 19. 27.
3.7378. 7.7430 - 10. 0.2194. 63,940.
6. 13. 21. 29.
9.8119 - 10. 8.4620 - 10. 0.07995. 0.9902.
7. 15. 23. 31.
0.5310. 1.9994. 8.696. 20.73.
7. 15. 23. 31. 39. 47. 56.
0.0198. 0.3134. 0.2891. 14,900. 0.9498. 1.987. 1.344.
7. 15. 23. 31. 39.
9.89020 - 10. 0.31717. 780.00. 1.0010. 6.9185.
§ 119, page 186
1. 2856. 9. 5.290 X 10-4• 17. 2.359 X 1015 • 26. 0.8252. 33. 0.8322. 41. 0.8840. 49. 0.4896. 57. 9.01 X 106 •
3. 11. 19. 27. 35. 43. 51. 69.
6. 13. 21. 29. 37. 45. 53.
5.992. 13.57. 0.4519. 0.8732. 1.039. 0.4725. 0.5086. 22.27.
11.79. 0.07143. 3.069. 15.38. 0.08178. 41.92. 7.36 X 10-9•
§ 120, page 188 1. 3.83866. 9. 2.03262. 17. 8.05572 - 10. 26. 0.02735. 33. 9.3380.
3. 8.57807 - 10. 11. 19. 27. 36.
0.44044. 2.21149. 58,890. 495.38.
6. 13. 21. 29. 37.
0.30298. 1.40882. 0.79800. 0.10280. 0.10354.
686 41. 49. 67. 66.
FIRST YEAR OF COLLEGE MATHEMATICS 48.197. 0.13630. 0.86650. 0.0011772.
43. 51. 69. 67.
3113.0. 7.6300. 0.81405. 2.5617.
46. 9.4830. 63. 0.0054580. 61. 0.81184.
47. 7.2230. 650 0.11560. 63. 763.87.
§ 122, page 190 These answers are obtained with four-place tables. 1. 3.712. 3. 4.139. 5. 0.3765. 7. 1.205. 9. -0.3402. 11. 4.271 or -0.271. 13. 17.71. 15.· 4.938. 17. 8.596. 19. 8 yrs. 21. 16 yrs. old. § 124, page 192 The following answers are obtained from four-place tables. With more extensive tables the results will sometimes differ in the last place. 1. 1.6095. 3. 4.0431. 6. 1.1135. 7. 1.8261. 9. 2.043. 11. 0.8782.
1. 45°. 13. i 7r.
§ 126, page 196 6. 36°. 7. 180°, 17.¾1r. 19.-11r.
3. 0°. 16. ½1r.
11. 114.592'='. 23. o~4oi42.
9. 144°. 21. 2 7r.
§ 127, page 196 1. 60 in. 3. 880 1r ft. = 2764.6 ft. 5. H· 7. 53.77°. 9. (1440/1r) ft. = 458.4 ft. 11. 6.30 in. 13. 1306 ft./sec. 16. 13.1 in. § 130, page 199 Answers are given only for part (c). 6. 120°, -600°. 1. 390°, -330°. 3. 495°, -225°. 7. 90°, -270°. 9. 315°, -405°. 11. ¾1r, -¼- 11". 13. -i 1r, - ½7r. 16. -i 1r, -J.-t 7r. 17. 0, 2 71". 19. ¾1r, - ll 7r.
§ 131, page 203 r
sin 0
cos{)
tan 0
-¼
-4
1.
5
3.
13
-.1.~
6.
17
rr
7.
2V2
½V2
-½v'2
-1
9.
2
1
0
-
11.
17
13.
3V2
15.
13
3
5
13 15
-
8
17
-½V2 5
13
3
-
_
_§__
8
1
1
5
-1---B5
5
3
-.1.~
17
17 15
5
s"'
-
12
v2 1
17
15
5
csc 8
-ll
-v2
0
12
_.§_ 4
15
-1
-~
sec 0
8
s-
-tv2 13
12
15
8
-.1.~
5
'5-
rr
1 5
-± 3
1 2
--IL 13
17
cot 0
-v2 -l~ 1 2
-J.-1.. 8
-v2 l 3
-5-
I
687
ANSWERS 17. (a) csc 0 = 1/sin 0; (b) cot 0; (c) sec 0. 0
sin 0
cos 0
tan 0
cot 0
19.
65°
0.9
0.4
2.1
0.5
21.
153°
0.45
-0.9
-0.5
-2.0
23.
233°
-0.8
-0.6
0.75
1.3
26.
331°
-0.5
0.9
-0.55
-1.8
27.
oo
0.0
1.0
0.0
-
29.
-90°
-1.0
0.0
-
0.0
31.
-225°
0.7
-0.7
-1.0
-1.0
33.
-291°
0.9
-0.4
2.6
0.4
36.
674°
-0.7
0.7
-1.0
-1.0
37.
-650·
0.9
0.3
2.7
0.4
39.
3
7r
0.0
-1.0
0.0
-
41.
¾7r
0.9
-0.5
-1.7
-0.6
43.
¾7r
-0.7
-0.7
1.0
1.0
45.
-,h: 7r
0.26
0.97
0.27
3.7
47.
-30°
-0.50
0.87
-0.58
-1.73
49.
330°
-0.50
0.87
-0.58
-1.73
§ 132, page 206
I
sin 0 1.
5
3.
-
6.
4
cos 0 3
5
tan 0 -
cot 0
sec 0
3
5
4
-
-lv7 -½V3
-½V7 -v3
-.1:v7 ¾V3
7.
-
-0.79
-0.77
-1.30
-1.27
9.
-
-~
11. 13. 15. 17. 168°.
-½V3 -½V3 _4
5
4
3
3
4
1
-
1
v3
tv3 tv3
-2 -2
-
-± 3
3
-4
5
3
-¼v'7 ½V3 5
csc 0 4
4
3
7
_
_§_ 3
-2 5
3
-2 1.64 _
5
4
-¾V3 -¾V3 _
_§_ 4
688
FIRST YEAR OF COLLEGE MATHEMATICS § 138, page 211
~
sin 0
cos 0
tan 0
cot 0
1.
0.2306
0.9730
0.2370
4.2193
3.
0.4669
0.8843
0.5280
1.8940
6.
0.7969
0.6041
1.3190
0.7581
7.
0.8192
0.5736
1.4281
0.7002
9.
0.2784
0.9605
0.2899
3.4495
11.
0.7050
0.7092
0.9942
1.0058
13.
0.3795
0.9252
0.4101
2.4383
15.
0.6548
0.7558
0.8662
1.1544
17.
0.7600
0.6499
1.1694
0.8551
19.
0.8931
0.4498
1.9854
0.5037
21.
0.5386
0.8426
0.6392
1.5647
23.
0.8682
0.4962
1.7496
0.5716
---.
-
-
26. 0.8347. 1. 9° 40'. 11. 75° 56'. 21. 56° 44'. 31. 0.6854.
27. 0.9638.
29. 0.4033.
3. 32° 0'. 13. 48° 42'. 23. 13° 38'.
31. 0.6214.
33. 1.5282.
§ 139, page 213 7. 29° 40'. 6. 17° 30'. 17. 51 ° 42'. 16. 15° 37'. 27. 0.5914. 25. 1.1369.
9. 34° 25'. 19. 23° 3G'. 29. 0.4159.
§ 140, page 214 sin 0
cos 0
tan 0
1.
0.29904
0.95424
0.31338
3.1910
3.
0.71059
0.70360
1.0099
0.99016
5.
0.42525
0.90507
0.46985
2.1283
7.
0.67503
0.73779
0.91494
1.0930
9.
0.80320
0.59571
1.3483
0.74167
11.
0.91429
0.40506
2.2571
0.44304
13.
0.98773
0.15618
6.3245
0.15812
16.
0.53002
0.84799
0.62503
1.5999
17. 42°48.0'. 26. 59° 16.2'.
19. 61 ° 48.0'. 27. 60° 37.4'.
21. 35° 41.4'. 29. 55° 31.1'.
cot 0
23. 29° 47.4'. 31. 57° 8.2'.
ANSWERS 33. 39° 21.7'. 41. 25° 50.5'.
36. 34° 0.8'.
37. 62° 19.7'.
689 39. 44 ° 28.6'.
§ 141, page 216 1. /3 = 62° 10', a = 8.5, b = 16.2. 3. a = 52° 40', /3 = 37° 20', c = 713. 6. a = 63° 43', a = 38.19, b = 18.86. 7. a = 39° 32', /3 = 50° 28', a = 2017. 9. /3 = 55° 44', b = 10.762, c = 13.023 (or c = 13.022 with five-place tables.) 11. a = 53° 5', /3 = 36° 55', c = .04330. 13. a = 66° 54', b = 857, c = 2184. 16. /3 = 46° 43.3', a = 137.81, b = 146.36. 17. /3 = 63° 18.6', a = .5095, C = 1.1344. § 142, page 217 1. 159 ft. 3. 51½ ft. 6. 4° 31'. 7. 11,200 ft. 9. 51 ° 38', 41 ° 46'. 11. 20½ ft., 68° 41'. 13. 42½ ft., 29° 35'. 16. 2799 ft. (or 2800 ft.). 17. 29° 59', 16 ft., 6 in. 19. 4.9 in. 21. 2: 01: 54. 23. 124° 32'. 26. 2° 30', 229.6 ft., $574. 27. 251 ft., 90 ft. 29. 151 ft., 156 ft. 31. 7.6 mi. east of A, 10.2 mi. north of AB.
1. 9.6444 - 10. 9. 9.1745 - 10. 17. 9.9375 - 10. 26. 15° 45'. 33. 41 ° 2'. 41. 10.95. 49. 2.766. 67. 41 ° 50'.
3. 11. 19. 27. 36. 43. 51.
§ 143, page 220 9.4797 - 10. 6. 9.1060 - 10. 0.3013. 13. 0.1820. 21. 56° 46'. 32° 40'. 67° 13'. 29. 75° 53'. 64° 47'. 37. 14.42. 1930. 46. 536.0. 40° 49'. 63. 50° 55'.
7. 15. 23. 31. 39. 47. 66.
9.9244 - 10. 9.5057 - 10. 15° 19'. 37° 55'. 30.01. 93.48. 49° 44'.
§ 146, page 224
1. a = 38° 32', /3 = 51 ° 28', b = 53.34. 3. a = 39° 4', /3 = 50° 56', C = 824.2. 6. /3 = 53° 19', a= 235.2, c = 393.7. 7. a = 37° 29', /3 = 52° 31', b = 669.6. 9. a = 48° 9', /3 = 41 ° 51', a = 14.24. 11. a = 49° 29', /3 = 40° 31', b = 3421. 13. /3 = 49° 25', b = 3826, c = 5040. 15. a = 33° 19', a = 75.1, c = 136.8. 17. /3 = 61 ° 24', a = 0.5448, b = 0.9990. 21. 1' = 81 ° 59', b = 3586, c = 5925. 23. a = 11 ° 29', 'Y = 118° 18', b = 836.4. 25. 385 ft. 27. 2672 ft. 29. 1162 ft., 3388ft. § 146, page 226
1. 245.5, 488.8.
3. 5722, 3550.
§ 148, page 228 1. N 562.3 lb., E 514.7 lbs. 3. S 2074 lbs., W 1338 lbs. 5. Upward, 335.0 lbs.; horizontal, 150.9 lbs. 7. 2394 lbs. downward 51 ° 15' from horizontal. 9. 91.96 lbs., N 34 °· 2' E. 11. 862.2 lbs., N 56° 46' E. 13. 4982 lbs., N 66° 3' E. 15. 4986 lbs., S 28° 24' E. 17. 21.6 mi./hr., N S0° 40' E. 19. 75.3 mi./hr. 21. 16° 29'. 23. 3480 lbs. 26. 21,107 lbs. 27. 1134 lbs., S 45 ° 29' W from tower.
690
FIRST YEAR OF COLLEGE MATHEMATICS
1. 78°.
3. 25°.
§ 149, page 232 7. 82°. 9. 80°.
6. 44°.
§ 160, page 233 sin 0
cos 0
tan 0
cot 0
sec 0
-½v3
-v3
-iv3
2
-v3
-½v3
-2
iV3
1.
2
1
-½V3
3.
21-v?. u
-2
5.
-½V2
7.
1-y3 2
9.
1
-v2
1
1
-½v2
csc 0
-v2
-2
1
-v3
-½V3
-½v·2
½V2
-1
-1
v2
-v2
11.
-½V3
2
1
-v3
-½v3
2
-iv3
13.
½V3
2
1
v3
½V3
2
iV3
16.
2
-2
1
-½v3
-½v3
-v3
iV3
-¾V3
I J
2 \
17. 0.7466. 26. -0.3959. 33. -79.22.
1. -sin 0. 11. -sec 0.
19. 0.6856. 27. -3.825. 36. -672.4.
3. cot 0. 13. sec 0.
23. -0.2804. 31. 1.873. 39. -1095.
21. 0.5817. 29. 398.6. 37. 23.37.
§ 163, page 237 5. sin 0. 7. -cos 0. 16. -sec 0. 17. -sin 0.
9. -cot 0. 19. sin 0.
§ 169, page 242 1. 60°, 300°. 3. 9. 120°, 240°. 11. 17 . ½7r' i- 7r. 19. 25. 225°, 315°. 27. 33. 132° 40', 312° 40'.
11. 13. 16. 17. 23. 29.
60°, 120°. 60°, 240°.
¾7r' 1J-1r. 0°, 180°.
6. 13. 21. 29.
7. 16. 23. 31.
210°, 330°. 180°.
¾1r, ¾7r. 270°.
0°, 180°. 135°, 315°.
¼1r, ¾7r. 33° 40', 326° 20'9
§ 166, page 260 cos a=-¾, tan a=½, cot a=¾, sec a= -t, csc a=-¾sin {3 = - l 9 V29, cos {3 = -h V29, cot {3 = -¾, sec {3 = ¼V29, csc f3 = -½V29. sin x = -j7 Vl7, cos x = -r\-Yl7, tan x = -¼, sec x = -¼V17, csc x = V17. 21. sin x. (1 - sin 2 x)/sin x. 19. ±sin x/Vl - sin 2 x. 27. (1 - sin 2 x) 2 /sin 2 x. 2 sin 2 x - 1. 26. ±sin x/Vl - sin 2 x. 31. ± 1/Vl - sin 2 x. (1 sin2 x)/(1 - sin 2 x).
+
33 sin x(l - sin x). 36. cos6 x/sin6 x. · cos x(l - cos x) 39. tan x. 41. 3. 43. 1-io. 45. -3. 47.
37. tan 2 x - (1/tan2 x).
-¼-V2.
49.
-
9 2 0·
ANSWERS
691
§ 168, page 267
1. 0°. 3. 90°, 270°. 6. 45°, 135°, 225°, 315°. 7. 45°, 135°, 225°, 315°. 9. 60°, 120°, 240°, 300°. 11. 135°, 315°. 13. 0°, 120°, 180°, 240°. 15. 90°, 120°, 240°. 17. 60°, 150°, 210°, 300°. 19. 30°, 150°, 210°, 330°. 21. 30°, 150°, 270°. 23. 30°, 90°, 150°. 25. 45°, 90°, 135°, 270°. 27. 60°, 120°, 240°, 300°. 29. 0°, 120°, 240°. 31. 45°, 135°. 33. 30°, 45°, 135°, 150°, 210°, 225°, 315°, 330°. 36. 30°, 45°, 135°, 150°, 210°, 225°, 315°, 330°. 37. No solutions. 39. 210°, 330°. 41. 0°. 43. 30°, 270°. 45. 45°, 135°, 225°, 315°. 47. 45°, 60°, 135°, 225°, 240°, 315°. 49. 60°, 180°. 61. 23° 35', 156° 25', 270°. 63. 36° 52', 143° 8', 216° 52'. 66. 0°, 73° 24', 286° 36'. 67. 46° 1', 141 ° 46', 218° 14', 313° 59'. 69. 66° 32', 127° 30', 246° 32', 307° 30'. § 172, page 264
+ +
1. ; 2 (2v7 3V5), ii"(V35 - 6), -32V5 - 27V7; r12 (2v1 - 3V5), h(V35 6), 27,/7 - 32V5; IL 3. ~!, ~:, g; -~~' ;~,-~~;I. 6. isV65, 6\V65, {; JsV65, 6\V65, ½; I. 1 7. J5 (8V6 - 3), -A(3V 6 2), ¼(6 - V6). 9. ½. 11. sin 7 x. 13. tan 5 0. 15. sin 10 ~r. 17. cos x. 19. cos 3 x. 21. -1. 23. cos x. 25. ½(V3 cos a - sin a). 27. (1 tan x)/(1 - tan x).
+
+
§ 174, page 267
1. ½-Y2 - v2 = o.3827. 3. ½-Y2 - v3 = o.2588. 6. -2 - v3 = -3.732. 13. cos 6 x. 7. cos 12 a. 9. tan 6 x. 11. ±sin 3 x. 15. ½sin 8 x. 17. ½tan 10 x. 19. sin (x - y). 21. ±cos 2 a. 23. sin2 5 x. 26. cos 2 x. 27. -cos 6 y. 29. -cos 2 y. § 176, page 270 1. 45°, 225°. 3. 135°, 315°. 6. 30°, 150°, 270°. 7. 90°, 210°, 330°. 9. 0°, 90°, 180°, 270°. 11. 22½ 0 , 112½ 0 , 202½ 0 , 292½ 0 • 13. 90°, 105°, 165°, 270°, 285°, 345°. 15. 30°, 90°, 120°, 210°, 270°, 300°. 17. 0°, 60°, 180°, 300°. 19. 0°, 135°, 180°, 225°. 21. 0°, 45°, 180°, 315°. 23. 90°, 210°, 270°, 330°. 26. 90°, 210°, 330°. 27. 0°, 180°. 29. 0°, 30°, 150°, 180°, 210°, 330°. 31. 60°, 120°, 240°, 300°. 33. 30°, 60°, 210°, 240°. 36. 90°, 270°. 37. 0°, 180°, 210°, 330°. 39. 0°, 60°, 300°. 41. 30°, 150°, 210°, 330°. 43. 41 ° 49', 138° 11', 210°, 330°. 46. 0°, 60°, 90°, 120°, 180°, 240°, 270°, 300°. 47. 15°, 75°, 135°, 195°, 255°, 315°. 49. 120°, 180°. 51. 90°, 180°. 53. 102° 40', 257° 20'. § 177, page 272
1. co~ 2 a 7. 2 cos 3 x 33. 0°, 90°, 37. 0°, 60°,
+
+
cos 8 a. 3. cos 4 a cos 3 a. 5. sin 360° sin 2 a = sin 2 a. sin 2 x. 9. 2 cos 4 x cos x. 11. 2 cos 2 x sin x. 180°, 270°. 35. 15°, 75°, 90°, 195°, 255°, 270°. 180°, 300°. 39. 0°, 120°, 150°, 180°, 300°, 330°.
692 41. 43. 45. 47.
FIRST YEAR OF COLLEGE MATHEMATICS 10°, 45°, 50°, 130°, 135°, 170°, 225°, 250°, 290°, 315°. 67½ 0 , 90°, 157½ 0 , 247½ 0 , 270°, 337½ 0 • 30°, 60°, 90°, 120°, 150°, 210°, 240°, 270°, 300°, 330°. 0°, 180°. 49. 0°, 30°, 60°, 120°, 150°, 180°, 210°, 240°, 300°, 330°.
1. 2.834.
11. 781 ft.
§ 180, page 276 4.359. 5. 34° 47'. 7. 125° 6'. 9. 7.957. 13. 91.28 lbs., 59° 40' from first force.
s.
§ 182, page 279 1. a = 60°, b = 17.69, c = 21.70. 3. 1' = 120°, a = 7.55, b = 6.29. 6. a = 8~ 0 2', b = 24.10, c = 32.88. 7. 1' = 112° 20', a = 1202, b = 1375. (The answers for Ex. 9 and 11 were obtained with five-place tables.) 9. 1' = 101 ° 19.9', b = 496.00, C = 959.25. 11. 1' = 30° 4.6', a = 6506.0, b = 10,224. 13. 1701 ft. 15. 1694 lbs., 1531 lbs. § 183, page 286 1. /3 = 56° 46', ')' = 81 ° 14', C = 23.6; {3 1 = 123° 14', ')' 1 = 14 ° 46', C1 = 6.1. 8. a = 54° 5', /3 = 84° 42', b = 40.03; a' = 125° 55', {3' = 12° 52', b' = 8.95. 6. No solutions. (The answers for Ex. 7 to 11 were obtained with five-place tables.) 7. /3 = 90°, ')' = 29° 1.9', C = 2057.2. 9. a = 54° 51.6', /3 = 46° 49.0', a = 5530.4. 11. a = 53° 57.9', /3 = 16° 48.3', b = 240.82. 15. 1472 lbs.; 87° 4' from first force.
13. 71 °.
§ 185, page 288 1. a = 59.8, /3 = 41 ° 48', 1' = 79° 32'. 3. c = 678.2, a = 55° 16', /3 = 46° 2'. 0. C = 6139, a = 38° 27', /3 = 22° 59'. 7. b = 2722.4, a = 67° 12.0', 1' = 40° 31.2'. 9. a = 61 ° 54', b = 47.61, c = 72.66. 11. a = 17° 12', 1' = 58° 27', a = 230.2. 13. a = 1121.3, /3 = 40° 40', 1' = 34° 59'. 16. 827.3 lbs.; 44° 30' from first force. § 187, page 291 1. a= 58°48', /3 = 86°26', 1' = 34°46'. 3. a = 69° 38', /3 = 37° 40', ')' = 72° 42'. 5. a = 100° 12', /3 = 36° 49', 1' = 42° 59'. 7. a = 42° 44', /3 = 60° 22', 1' = 76° 54'. 9. a = 20° 52', /3 = 26° 12', ')' = 132° 56'. 11. 82° 54', 32° 56'. 13. 68° 0', 30° 10'. § 188, page 294 1. 12. 3. 45.65. 6. 14.7. 7. 2256. 9. 96.70. 11. 205.5. 13. 176,000. 15. 105,600. 17. 27.1 acres. Applications 1. 538.6 ft., N 82° 16' E. 3. 296 ft., 1355 ft. 5. 913 ft. 7. 177 ft. 9. 2569 ft. 11. 24 ft. 13. 245 mi./hr., S 56° 56' E. 15. 585 ft. 17. 400 ft. 19. 45° 26', 56° 44', 77° 50'. 21. 2056 ft. 23. 13.14 mi. 25. 21½ mi./hr. 27. 325 mi./hr.
ANSWERS
693
§ 191, page 299 1. (a) I, IV; (b) II, III. 3. (a) I, IV; (b) II, III. 5. 60°. 7. -90°. 9. 45°. 11. -45°. 13. 0°. 15. 30°, -30° 17. 135°, -135°. 19. 120°, -120°. 21. 60°, -60°. § 192, page 302 1. ¾1r. 3. -¼ 1r. 5. o. 7. ½1r. 9. 1. 11. ½vi 13. ½V3. 15. -1. 17. ½vi 19. -1. 21. 60°. 23. ½1r. 25. 60°. 27. 90°. 29. 1. 31. o. 33. v,_1---x-2• 35. x/V~. 37. 2 xVl - x2• 39. 1 - 2 x 2• 41. 2 x 2 - 1. 43. xVl - y 2 - yVl - x 2• 45. V (1 - x2)(1 - y2) - xy. 47. ½. 49. ¼V2. § 194, page 306 1. (2, t 1r), (-2, ½1r). 3. (5, 2 1r), (-5, 1r). 5. (3, -i 1r), (-3, -¼ 1r). 7. (2, ½1r), ( -2, ½1r). 9. (6, 2 1r), (-6, 1r). 11. (3, -¾ 1r), (-3, -¼ 1r). 13. (2, ½1r), ( -2, -½ 1r). 15. (6, 0), (-6, -1r). 17. (2, 2 + 2 1r), (-2, 2 + 1r). 21. 3v3, tV3. § 195, page 308 1. (0, 2). 9. (6, 0). 15. (6, 0). 23. (5, 0.93). 29.
p
cos 0 = 4.
3. (5, 0). 11. c-1v2, ½v2). 17. ( -0.83, 1.82). 25. (2, -¼ 1r). 31.
p
= 3.
5. ( -iV3,
}).
7. 13. 21. 29.
19. (3, ½1r). 27. (4, ¾1r).
§ 197, page 312 33. p 2 sin 2 0 = 8.
35.
p 2 (9
(0, 2).
(0, -2). (6, 0).
(13, 112° 36').
- 13 sin2 0)
= 36.
§ 198, page 313 1. Line through pole, slope ½V3. 3. Circle, radius 3, center pole. 5. Line perpendicular to polar axis, 3 units to right of pole. 7. Line perpendicular to polar axis, 8 units to left of pole. 9. Circle, radius 2, center (2, 0). 11. Circle, radius 5, center (5, ½1r). 13. Two vertical lines, 5 units to right and left of 90°-axis. 15. Two circles through pole, radius t, centers on polar axis. 17. Two circles through pole, radius 1, centers on polar axis.
§ 201, page 316 1. 5 i. 3. 3 mi. 5. 6 i V2. 7. 2 ai V5. 9. i. 11. 1. 13. 3 - 7i, 6, 58. 17. 7 Xi, 0, 49 X2• 19. -14, -28, 196. 23. X = 0, y = -4. 25. X = -¼, y = 2. 29. -29 + 22 i . 31. 15. 35. -v3o. 37. -65. 41. H- fti. 43. / 0 - -f!ri. 47. -7 - 24i. 49. 122 - 597 i. 57. x 2 - 2 ax+ a 2 + b2 • 59. x 2 + 6 x + 10 = 0.
15. 21. 27. 33. 39. 45. 55. 61.
-4 - 3 i, -8, 25. = 3, y = -3. -10 - 9 i. 26 - 38 i. 7 - 24 i. ¾+ ¾i. X 2 + 2 X + 10. X 2 - 6 X + 45 = 0. X
694
FIRST YEAR OF COLLEGE MATHEMATICS
63. (x - 3 + 7 i)(x - 3 - 7 i). 66. (x - ½ + ½i)(x 67. (x - l)(x + ½+ ½V3 i)(x + ½ - ½V3 i).
½ - ½i).
§ 203, page 320
3. 4(cos 60° + i sin 60°). 1. 4 V2(cos 135° + i sin 135°). 6. 7(cos 0° + i sin 0°). 7. 5(cos 90° + i sin 90°). 9. 8(cos 300° + i sin 300°). 11. SV2(cos 45° + i sin 45°). 13. 5(cos 143° 8' + i sin 143° 8'). 15. V34(cos 302° + i sin 302°). 17. 5 + 5V3 i. 19. -17 i. 21. -5 + 12 i. 23. -2V3 + 2 i. 26. l+V3i. 27. -fV3-Ji. 29. 3-3V3i. 31. -4v2 + 4\/2 i. 33. 7. 36. 11 i. 37. o.5512 + 1.9226 i. 39. -2.4870 - 1.6776 i. 41. 144 7r. § 204, page 322
1. 12\/2
+ 12\/2 i.
3. 5 - 5,v3 i.
5. 1so i.
7. v2 + V2 i.
§ 205, page 323
1. -4 - 4 i. 3. -4. 6. 8 i. 7. 432 + 144V3 i. 9. -1024. 11. - 227 + 227 V3 i. 13. cos 84° + i sin 84°. 15. 243( cos 130° + i sin 130°). 17. -1. 19. -1 + i. 21. 1 + i. 23. -38.00 - 40.99 i. 26. -0.00409 + 0.02094 i. 27. 119 - 120 i. 29. -0.8130 + 0.8998 i. § 206, page 327 1. 3(cos 15° + i sin 15°), -½ V2 + iV2 i, 3(cos 255° + i sin 255°). 3. 2(cos 37° + i sin 37°), 2(cos 127° + i sin 127°), 2(cos 217° + i sin 217°), 2(cos 307° + i ~in 307°). 6. ±(l+V3i). 7. ±(l+V3i), ±(V3-i). 9. ±(V2-V2i). 11. 1 + V3 i, 2(cos 132° + i sin 132°), 2(cos 204° + i sin 204°), 2(cos 276° + i sin 276°), 2(cos 348° + i sin 348°). 13. ± (1.337 + 0.335 i), ± (0.335 - 1.337 i). 16. 2, 0.618 ± 1.902 i, -1.618 ± 1.176 i. 17. X = ±2, X = ±2 i.
19.
X
= ±1,
X
= ½ ± ½V3i,
X
= -½ ±
~V3 i.
§ 209, page 330 1. x 4 + 0 • x 3 - x 2 - 2 x - 3 = 0, x 4 - x 3 + 2 x - 3 = 0. 3. 6 x 5 + 0 • x 4 - S x 3 + 0 • x 2 - 4 x + 7 = 0, 6 x 5 - 8 x 3 - 4 x - 7 = 0. 5. 3 x 7 + 0 • x 6 - 4 x 5 + x 4 + 0 • x 3 + 0 • x2 + 2 x - 6 = 0, 3x7 - 4 x 5 - x 4 + 2 x + 6 = 0. 7. x 5 + x 4 + 12 x 3 + 0 • x 2 + 0 • x - 8 = 0, x 5 - x 4 + 12 x 3 + 8 = 0. 9. k = -4. 11. k = 34.
§ 212, page 334 15. -4 and -3; -2 and -1; 2 and 3. 17. -3 and -2; 1 and 2; 2 and 3. 19. -7 and -6; -2 and -1; 1 and 2; 2 and 3. 21. -6 and -5; -1 and 0; 0 and 1; 5 and 6.
ANSWERS
695
§ 216, page 337
1. 51. 3. 79. 5. --H- 7. -5. 9. 79. 11. -1. 13. -25. 16. No. 17. Yes; no. 19. Yes; no. 21. All; none. § 216, page 339
1.
X2
+X
3, -10. 3. (a) 2 x + 3 x 2 + 12 x + 50, 199: (b) 2 x 3 - 11 x 2 33 x - 97, 290. 5. -3, 43, 43, 21, 1, 7, 63. 7. -18, -2, t, 2, J.-j-, o, -2, -t, 2, 18. -
+
3
§ 219, page 342
1. -2 and 1, single roots; 3, a double root. 3. ¾and -4, single roots; -½, a double root. 6. ½ ± ½V3 i, single roots. 7. 3 and ½ ± ½V3 i, single roots. 9. -11, a single root; ½± ½V3 i, double roots. 11. 0, a double root; ½ ± ½V3 i, single roots. 13. -4. 16. 1, -4. 17. 1, -2. 19. x 3 - 6x2 -x +30 = 0. 3 2 21. 15 x + 41 x - 70 x + 24 = 0. 23. x 4 - 3 x 3 - 12 x 2 + 52 x - 48 = (J 26. 6 x 4 - 55 x 3 + 149 x 2 - 88 x - 48 = 0. 27. 8 x 4 - 20 x 3 - 18 x 2 + 81 x - 54 = 0. 29. x 3 - 11 x 2 + 37 x - 35 = 0. 31. x 4 - 8 x3 + 25 x 2 - 36 x + 20 = 0. § 220, page 334
1. 5 - 7 i. 7. 11.
x3
+ 5 x2 4 x - x3
+
L 4 and 1.
6. X 2 - 6 X + 13 = 0. 9. x 3 + 4 x 2 + 30 x + 52 = 0. 13. x 4 - 12x3 + 62x2 - 156x + 169 = 0.
3. -7 - i. 4 x - 60 = 0. x 2 + 2 = 0.
3. 2 and 4.
§ 221, page 346 5. 2 and 3. 7. 2 and 5. § 222, page 347
The number of each kind of root is indicated.
Ex. Positive Negative Imaginary Ex. Positive Negative Imaginary Zero 2
9.
2or0
0or2
11.
1
2or0
2or4
13.
1
1
2
1
1 or3
2or0
16.
1
1
4
1.
1
3.
1
5. 7.
1
1
2 4
2 I- -
i
§ 226, page 349
1. 5, -2. 3. 2, -2. 6. 4, -2. 7. 3.1, 0.7, -0.9. 9. 2.5, 1.7, -0.5. 11. 2.8, 2.2, -1.0. 13. 3.4, 1.2, 0.6, -3.2. 16. 2.1, 0.5, -0.9. § 226, page 362 3. 1, -4, 2 ± i.
1. 3, 2, -1. 7. -2, ½, j. 5
13 • s,
1
2
-2, - 3
+ -
1 ~ J"'; •
3Vbi.
9. 0, -2, 16. 3, 3, ¼,
¾, ¾-
½.
5. o, 1, 3, -2, -4. 11. -i, ½, 2 ± V3. 17. 3.
696
FIRST YEAR OF COLLEGE MATHEMATICS
§ 228, page 366 1. 3.177. 3. 1.440. 5. -3.849. 7. 2.414, -3.732. 9. 0.586, 0.732. 11. 1, ½, -4. 13. 0.31, -0.51, -4.13. 16. ¾, -3. 17. ½, -1.59, -4.41. 19. 1.72. 21. -2.79. 23. 5.83. 26. (2, 1), ( -0.046, 3.187). 27. 0.24 cm. 29. 1.327 in., or 4.129 in. 31. 0.98 in. 33. 4.13%.
§ 232, page 360 1. X - 2 y - 12 = 0. 3. 3 X - 4 y - 29 = 0. 6. 7. 60°. 9. 45°. 11. 45°. 13. Arctan ½ = 26° 34'. 16. Arctan i~ = 52° 46'. 19. x - 5 y - 15 = 0.
X
+4y
- 6 = 0.
§ 233, page 362
+ +
1. X + V3 y - 18 = 0. 3. V3 X y 4 = 0. 6. X - V3 y ± 8 = 0. 7. V3 x - y ± 4 = 0. 9. 12 x - 5 y + 169 = 0. 11. 3 x - 4 y - 50 = 0. 13. V3 x + y - 12 = 0, V3 x - y + 12 = 0. 16. Slope - intercept. 17. Normal. 19. Normal. § 234, page 364 1. -{3 X - Hy - 3 = 0. 3. -¾ X - t y - % = 0. 6. -½V2x ½Y2y - tV2 = o. 7. -y30 V 10 x to- vio y - 2vio = 0. 9. x - 5 = 0.
+
+
§ 236, page 366 1. 6; opposite. 3. }; opposite. 5. -3; same. 11. 3 X + 4 y - 4 = 0, 3 X + 4 y - 34 = 0. 13. 15. X - 8 y + 6 = 0, 8 X + y - 17 = 0.
11. y - 5 = 0.
½.
§ 236, page 368 1. y - 2 = m(x + 5). 3. y = mx + 4. 6. y = m(x + 3). 7. y = 6 x + b. 9. 5 x + 3 y + k = 0. 11. x cos + y sin cp - 7 = 0. 13. x cos+ y sin - 11 = 0. 16. 4 x - y + k = 0. 17. Slope f. 19. Parallel to y-axis. 21. Slope 6. 23. Through (0, -1). 25. Distant 7 from origin. 27. Through (0, 3). 29. Through (-11, 6). 31. Through (3, -7). 33. Distant 5 from origin. 36. k = -10, 6 x - 7 y - 10 = 0. 37. k = -3, 13 x - 19 y + 6 = 0. 39. k = ½, 3 x - 4 y - 10 = O; or k = -¾, 5 x + 12 y - 26 = 0. § 237, page 371 1. (4 x - 5 y - 3) + k(3 x + y - 4) = 0. 3. (6 x + 3 y - 8) + k(x - 11 y - 5) = 0. 6. (4 X - y + 6) + k(2 X - y - 9) = 0. 7. 3 X + 7 y + 2 = 0. 9. X - y 4 = 0. 11. X - y + 4 = 0, 7 X + y + 20 = 0.
+
§ 239, page 374
1. (x - 2) 2 + (y - 3) 2 = 25. 3. (x - 2) 2 + (y + 4) 2 = 4. 6. (x - 2) 2 + (y + 1) 2 = 4. 7. (x - 7) 2 + (y + 2) 2 = 16. 9. 2 x + y - 9 = 0. 11. The parabola y 2 = 24 x. 2 2 13. The parabola x + 2 xy + y - 8 x + 8 = 0. 16. The parabola v2 = 40 x - 80. 17. The ellipse 9 x 2 + 25 y 2 = 225.
ANSWERS
697
+
19. The ellipse 9 x 2 5 y 2 = 180. 21. The hyperbola xy = 8. 23. The circle x 2 y 2 = k 2 - 2 a2, where the corners of the square are ( ± a, ±a) and the constant = 2 k 2 , with k ~ aV2. 25. The circle x 2 y 2 = a2, where the two lines are the coordinate axes, and the constant = a2• 27. The circle x - a = k(x2 y 2 ), where the fixed line is x = a, the fixed point is (0, 0), and the constant of proportionality is k.
+
+
+
§ 240, page 377 1. The circle x + = c 3. The ellipse x 2 + 9 y 2 = 9. 2 5. The ellipse 100 x + 144 y 2 = 225. 9. The hyperbola 16 x 2 - 9 y 2 + 36 y = 72. 11. The ellipse 4 x 2 - 4 xy + 5 y 2 = 4 c2• 13. Parts of circles x 2 + y 2 ± }V3 ay = a2 , where base vertices are at ( ±a, 0). 15. 9 x 2 + (3 y ± ak) 2 = a 2 (1 + k2), where k = cot 0 and the base vertices are ( ±a, O). 17. The hyperbola x 2 - y 2 = x. 19. The hyperbola 3 xy - 2 x - 2 y + 2 = 0. 2
2•
y2
§ 241, page 380
1. 5.
p p
= a cos 0 + b. 3. p 2 = a(sec 0 ± tan 0).
= a 2 cos 2 0 ± Vb 4
-
a4 sin2 2 0.
§ 242, page 381
1. y =
x 2•
3.
y2
=x
3•
5. x
2
+y
2
= a2• 7. xi+ y¾ = a¾. 9. x3 = x 2
+y
2•
§ 243, page 383
1. 88 ft., 2 secs.
3. 36.2 ft./sec.
§ 245, page 386 1. (7, -2), (11, -3), (9, 0), (3, 1), (6, -6), (10, -7), (1, 2). 3. 5 x' + 3 y' = 0, 3 x' - 2 y' = 0. 5. x' 2 + 4 y' 2 = 4. 7. x' 2 + y'2 = 25. 9. 2 x'2 + y' 2 = 2. § 246, page 388 1. + 4 = 4, new origin at (4, 1). 3. 5 x' 2 - 9 y' 2 = 45, new origin at (3, -2). 5. 4 y' 2 = 3 x', new origin at ( -5, 2). 7. x' 2 + 4 y' 2 = 4, new origin at (4, 1). 9. 5 x' 2 - 9 y'2 = 45, new origin at (3, -2). 11. 3 x' 2 - 5 x'y' + y'2 + 3 = 0, new origin at (3, 1). 13. 5 x' 2 - 3 x'y' + 4 y'2 = 6, new origin at (2, -5). 15. 2 x' 2 + 7 x'y' = 9, new origin at (3, -2). x' 2
y' 2
§ 247, page 391 1. x'y' = -J. 3. 5 x' + y' = 10. 5. x' 2 - 2 y' 2 = 2. 7. 145 x' 2 - 24 y'2 = 120. 9. 2 y' = x' 2, new origin at ( -2, 1) on rotated axes. 2
2
§ 248, page 395
+
1. x' - y' 8 = 0. 3. x' - 9 y' 2 = 9. 5. 3 x' 2 2 7. 21 x' - 5 y' 2 = 105. 9. 6 x" 2 - 19 y" 2 = 115. lL (V3 - l)x' 2 - (V3 + l)y'2 = ¾2
2
2
+ y'
2
= 12.
698
FIRST YEAR OF COLLEGE MATHEMATICS § 250, page 397
1. x + y = 25. 3. 16 + 16 y 2 = 9. 5. (x - 3) 2 + (y + 7. 4(x + 5) 2 + (2 y - 1) 2 = 16. 9. (x + 1) 2 + (y 2 2 11. 4(x - 5) + 4(y + 1) = 25. 13. x 2 + y2 = 4. 15. (x - 1) 2 + (y + 4) 2 = 4. 17. (x - 4) 2 + (y 2 2 19. (x - 5) + (y - 6) = 36. 21. 4(x + 1) 2 + (2 y 2
2
x2
2) 2 = 36. 2) 2 = 4. 3) 2 -
= 13. 3) 2 = 149.
§ 251, page 399 1. 3. 5. 7. 9.
+
+ (y
= 16; circle, center ( -6, 2), radius 4. + (y + = f; circle, center (½, -¾), radius ½VlO. + (y - t) 2 = 4; circle, center ( -¾, t), radius 2. + (y + V )2 = 16; circle, center ( -¾, -V ), radius 4. + (y + ½) 2 = 1; circle, center (½, -½), radius 1.
2
(x 6) (x - ½) 2 (x i) 2 (x + -g-) 2 (x - ½) 2
t) 2
+
2
+y +3X
1. x 3. 19 X 2 6. x 2 + 7. (x 9. (x 11. (X
- 2)
2
2
-
7y
+
§ 252, page 402 1~ = 0.
+ 19 y2 - 107 X - 167 y + 318 = 0. y + 3 X + 4 y - 10 = 0. 1) + (y - 3) = 5, (x + 3) + (y 2
2
2
2
15) 2 = 125. (y 9) 2 = 250. 2 - 27) (5 y - 2) 2 = 468.
+ (y - 1) = 10, (x + 6) + + + + (y - 5) = 29. 13. (5 X + 4) 2 4) 2
2
2
2
§ 253, page 403 1. t = 3.
3. t = ½V22 = 2.345. § 254, page 405
1. 3 x + 3 14 x - 5 y = 0. 3. 5 x 2 + 5 y 2 - 33 x + 11 y - 32 = 0. 2 2 5. x + y + 4 y + 3 = 0, 2 x 2 + 2 y 2 - 2 x + 6 y + 3 = 0. 7. X - 7 y - 22 = 0. 9. 23 X + 15 y + 77 = 0. 11. y + 4 = 0. 2
y2 -
§ 257, page 411
+
1. Tangents: Y1Y = 2 x 2 x1, x + y + 1 = 0; Normals: Y1X + 2 y = X1Y1 + 2 Y1, x - y = 3. 3. 4 X1X + 9 Y1Y = 36, X - 3V2 y + 9 = O; 9 Y1X - 4 X1Y = 5 X1Y1, 9V2 X 3 y + 5V2 = 0. 5. 4 X1X - 9 Y1Y = 36, 8 X - 3V7 y = 18; 9 Y1X + 4 X1Y 13 X1Y1, 9V7 X - 24 y = 52V7. 7. 9 X - 4 Y1Y + 9 X1 = 0, 3 X + 4 y + 6 = 0; 4 Y1X + 9 y = 4 X1Y1 + 9 Y1, 4 X - 3 y = 17. 9. Y1X + X1Y = 18, 9 X + y = 18; X1X - Y1Y = X1 2 - y12, X - 9 y = -80. 11. (x1 + 2 Y1)x + 2 X1Y = 12, X + y = 3; 2 xix - (x1 + 2 Y1)Y = 2 X1 2 - X1Y1 - 2 Y12, x - y = 1. 13. (9 X1 + 6 Y1)X + (6 X1 + 4 Y1)Y = 9; (9 X1 + 6 Y1)(y - Y1) = (6 X1 + 4 Y1)(x - x1). 15. (8 x1 - 3 Y1 - l)x + (2 Y1 - 3 x1 + 2)y + (2 Y1 - X1 - 10) = 0, x + 3y - 7 = 0; (8x1 - 3y1 - l)(y - Y1) = (2y1 - 3x1 + 2)(x - X1), 3 X - y - 1 = 0. 17. 3 x1 2x - Y1Y = X1 3, 3 x - y - 2 = 0; 3 X1 2 (y - Y1) + Y1(x - x1) = 0, x + 3 y - 14 = 0.
+
+
ANSWERS 19. (6 X12 (6 X1
2 -
699
3 Yi 2 )x - (6 XiYi - 2)y + 4 Yi - 3 = 0, 3 x - 4 y + 1 = 0; 3 Yi 2 )(y - Yi) + (6 XiYi - 2)(x - X1) = 0, 4 X + 3 y - 7 = 0.
§ 258, page 414
1. YiY =4x+4xi,x+y+2 =0,4/yi, -1. 3. 4 XiX + 9 YiY = 100, 8 x - 9 y = 50, -4 xi/9 Yi, f.
5. X1X + YiY - 3 y - 3 Yi = 0, Y = 0, xi/(3 - Y1), 0. 7. YiX + XiY + 2 X + 2 Xi - y - Yi = 14, 5 X + y - 13 = 0, (yi + 2)/(1 - Xi), -5. 9. XiX + 2 YiX + 2 X1Y - YiY - X - Xi + 2 Y + 2 Yi - 3 = 0, X + 3 y - 1 = 0, (xi + 2 Y1 - 1)/(yi - 2 xi - 2), -½. 11. 3 XiX + YiX + X1Y + 5 Y1Y + 2 X + 2 Xi - y - Yi = o, y = 1, -(3 Xi+ Yi+ 2)/(xi + 5 Yi - 1), 0. 13. 4 YiX + 4 XiY + 2 X + 2 Xi - Y - Yi - 34 = 0, 6 X + 11 y - 29 = 0, 4 Yi) I (1 - 4 ~), -1-\. (2
+
§ 259, page 416
±Vll. 3. 3x+y = ±V37. 5. 3 X + y - 10 = 0, 3 X + y + 7 = 0. 7. 3 y - x - 13 = o, 3 y - x + 4 = 0. 9. y = 5, 3 y + 1 = 0.
1. 2x-y =
§ 260, page 418 1. X 5. 3
y
-
X -
= 0, y
X
= 0,
+ y = 0. 3. X + y = 0.
2
X -
y
= 0, 2 X
·
+y
= 0.
3
§ 261, page 420 1. 16 x + 24 xy + 9 y - 74 x - 118 y + 234 = 0. 3. 3 x 2 6. xy - 3 x + y - 1 = 0. 7. xy + 6 x + 2 y + 4 = 0. 2
2
-
y2
-
4 y = 16.
§ 263, page 428 1. Parabola, e = 1, F: (!, 0), x = -f, l = 10. 3. Parabola, e = 1, F: (O, 1), y = -1, l = 4. 5. Hyperbola, a= 2, b = 1, e = ½V5, F: (±VS, 0), x = ±-t'\15, l = 1, x ± 2 y = 0. 7. Ellipse, a= 2, b = 1, e = ½V3, F: (O, ±V3), y = ±tV3, l = l. 9. Hyperbola, a = 2, b = 3, e = ½Vl3, F: (0, ± vi3), y = ±,t3 vi3, l = 9, 3 y ± 2 x = 0. 11. Parabola, e = 1, F: (O, -2), y = 2, l = 8. 13. Hyperbola, a = 3, b = 3, e = V2, F: (0, ±3V2), y = tV2, l = 6, x ± y = 0. 15. Hyperbola, a = 5, b = 3, e = t'V34, F: (O, V34), y = f¾V34, l = \ 8 ' 5 X ± 3 y = 0. 17. x2 = 12 y. 19. x2 = -32 y. 21. 25 x 2 + 9 y 2 = 225. 23. 16 x 2 + 25 y2 = 1600. 25. 4 x 2 + 9 y 2 = 36. x2 y2 y2 x2 27 • 144 + 169 = 1. 29 · 25 - 16 = l. x2 y2 x2 y2 31 · 16 - 20 = l. 33 · 64 - 36 = l.
700
FIRST YEAR OF COLLEGE MATHEMATICS
+
35. Ellipse b2x 2 4 a2y 2 = a2b2, if given ellipse is b2x 2 2 37. The ellipse b x 2 a2y 2 = a2b2 •
+
§ 264, page 431 Parabola, focus at pole, vertex at (2, 1r). Ellipse, vertices at (1, 0) and (5, 1r). Ellipse, vertices at (J, 0) and (8, 1r). Parabola, focus at pole, vertex at (4, 0). Ellipse, vertices at (6, ½1r) and (2, ¾1r). 4 12 11 · P = 1 13 · P = 4 3 sm . 0· sm 0
+ay
2 2
a2b2•
=
1. 3. 5. 7. 9.
+ · .
17. P
=
1. 5 x 2
60
+ 6 cos 0
5
+
19.
.
+ 9 y = 45. 2
P
3. 25 x 2
=
=
15. P
7
. + 12 5 cos 0
12 . 5 - 7 cos 0
§ 266, page 435 2 = 1600. 5. 8 x 2
+ 16 y
+ 2 xy + 8 y
2
= 63.
§ 268, page 438 144 y 2 = 3600. 3. 16 y2 - 9 x2 = 144. (2 X - 7) 2 4(y - 3) 2 5• 36 133 = 1. 2 7. 28 x 96 xy - 276 x - 144 y 495 = 0. 1. 25 x 2
-
+
+
1. 9 y - 4 x = 36. 7. xy = -9. 2
2
§ 271, page 440 3. = 4. 5. 9 y2 9. xy = 25. x2 -
y2
16 x2 = 144.
§ 272, page 444 3. (x - 4) 2 = -8(y - 3). 1. (y + 3) = 8(x - 2). 2 2 5. 9(x - 1) + 16(y - 3) = 144. 7. 4(x - 1) 2 + 3(y + 1) 2 = 48. 9. 25(x - 4) 2 - 9(y + 2) 2 = 225. 11. Ellipse, center ( -2, 3), vertices (-2, 0) and ( -2, 6), semi-minor axis 2. 13. Hyperbola, center (3, -1), vertices (3, 4) and (3, -6), semi-conjugate axis 2. 15. Hyperbola, center ( -3, 4), vertices ( -5, 4) and ( -1, 4), semi-conjugate axis 1. 17. Hyperbola, center ( -2, 4), vertices ( -2, 2) and ( -2, 6), semi-conjugate axis 2. 19. Parabola, vertex ( -3, -4), focus ( --H, -4). 21. Ellipse. 23. Hyperbola. 25. Parabola. 2
1. Ellipse. 2
1. 7 x 5. 3 x 2 3 x2
+ -
+
3. Hyperbola.
§ 273, page 447 5. Hyperbola. 7. Parabola.
9. Parabola.
§ 27 4, page 449 11 = 127. 3. 3 x + 2 xy - 7 y 2 + 3 x - y = 0. 6 xy + 3 y 2 + 2 x - 3 y + 2 = 0, 24 xy + 48 y 2 - 8 x + 12 y - 68 = 0. 2
y2
§ 275, page 451 8 xy + 16 x - 2 y - 60 = 0. 1. 7 3. The degenerate conic (2 x - 5 y - 14)(2 x + 3 y - 5) = 0. 5. 9 x2 - 6 xy + y 2 - 21 x + 19 y - 20 = 0. x2 -
y2 -
ANSvVERS
701
§ 277, page 455 1. 120. 3. 20,160. 5. (a) 840; (b) 240; (c) 144. 7. (a) 2160; (b) 1260; (c) 660. 9. 144. 11. 4320. 13. (a) 4536; (b) 2240; (c) 952. 15. (a) 5040; (b) 720. 17. (a) 576; (b) 72. 19. 891; (b) 445; (c) 179. 21. 120. 23. (a) 10,416; (b) 5082. 25. (a) 120; (b) 480; (c) 480. 27. (a) 1045; (b) 365. 29. 216.
1. 15,120. 11. 120. 21. 720.
§ 280, page 460 3. 120. 5. 2520. 7. 1P4 = 840. 13. 840. 15. 60. 17. 5040. 23. 39,916,800. 25. 86,400.
9. 9p3 = 504. 19. (a) 420; (b) 240. 27. 103,680. 29. 7012.
§ 282, page 464 1. 56. 3. 165. 5. 1. 7. 1C4 = 35. 9. 12C2 = 66. 13. 495. 15. (a) 84; (b) 49; (c) 56. 17. 70. 19. 45. 21. 70. 23. 25. 29. 33.
(a) 16C6 · 10C4 = 1,681,680; (b) sC6 · 10C4 = 5880; (c) sCa = 56. 1365; (b) 210; (c) 588; (d) 35; (e) 40; (f) 210. 27. 12,600. (a) 63; (b) 1957. 31. (a) 52!/(13!) 4 ; (b) 52!/[24(13!) 4]. 11. 35. 12. 37. 6 and 7. 39. 6.
§ 285, page 469 1. (a) ¼; (b) }. 3. J. 5. ¼- 7. 356 • 11. (a) ¾; (b) ¼. 15. (a) ~i; (b) is\; (c) ff· 19. (a) t; (b) i; (c) f; (d) f; (e) ¼23. (a) ½; (b) 1\ . 25. 4~ 2 • 29. (a) ~i; (b) g; (c) f; (d) g.
9. 1\· 13. (a) ½~; (b) H; (c) lr• 17. $1.35. 21. (a)
¼;
(b)
f.
27. 7~031. $180.
§ 288, page 4 73 1. In one trial the relative frequencies were: (a) 0.53; (b) 0.516; (c) 0.504. 3. (a) 0.9648; (b) 0.005547; (c) 0.1353. 5. $9688. 7. (a) $7330; (b) $2670. § 289, page 4 76 1. (a) 0.070; (b) 0.442; (c) 0.238. 3. (a) 0.0740; (b) 0.0761; (c) 0.0134. 5. (b) 0.656; (c) $2.50. § 291, page 479 1. 2 h• 3 • -h• 5 • (a) 1 ~ s; (b) 2 ~ s; (c) 2\; ( d) H- 7 • ¼• 9. (a) /o; (b) l 0 ; (c) / 0 ; (d) ~b- 11. ~;. 13. j. 15. 3\ . 17. A, 20 cents; B, 10 cents. 19. A, 1\ ; B, -fr. 21. 166· 27. (a) 0.8836; (b) 0.0624; (c) 0.05793. 29. $2193.
23. $7.
25. / 0\ .
§ 292, page 483 1. (a) 7N3 ; (b) g. 3. (a) ~U; (b) ~n;. 5. (a) -rl--}?!r; (b) i-it\, (c) H-H7. (a) 0.16807; (b) 0.52822. 9. (a) 694 ; (b) 352 • 11. (a) 0.3443; (b) 0.9042. 15. (a) 5; (b) 5 or 6; (c) 7. 17. 5. § 293, page 488 Throughout this chapter, five-place tables have been used v.-her:ever answers have been found by logarithms.
702
FIRST YEAR OF COLLEGE MATHEMATICS
1. $8174. 3. 6.09%. 5. 7.7%. 7. 6.88%. 9. 7.32%. 11. (a) $2633.95; (b) $2596.55. 13. (a) $2033.60; (b) $2010.30. 15. $1516.44. 17. 34. 19. 0.0269. 21. 3.93%. 23. 3.77 yrs.
§ 295, page 493 3. (a) $9777.28; (b) $6566.94. 9. $181.56.
l. 23.4683, 44.2270. 7. $19,700.82.
5. $9118.42. 11. $2849.13;
Payment
Applied on Principal
Principal Remaining after Payment
$600.00
$2849.13
$2249.13
$17,750.87
17,750.87
532.53
2849.13
2316.60
15,434.27
3
15,434.27
463.03
2849.13
2386.10
13,048.17
4
13,048.17
391.45
2849.13
2457.68
10,590.49
5
10,590.49
317.71
2849.13
2531.42
8,059.07
6
8,059.07
241.77
2849.13
2607.36
5,451.71
7
5,451.71
163.55
2849.13
2685.58
2,766.13
8
2,766.13
82.95
2849.11
2766.13
0.00
No. of Payment
Principal Remaining before Payment
Interest Due
1
$20,000.00
2
13. 5.1%. 15. (a) $1172.31; (b) $4357.54. 19. 24; $1434.72.
17. $668.54; $16,892.07.
§ 296, page 496 1. (a) $910.58; (b) $830.64. 3. (a) $1027.56; (b) $885.87. 5. $1070.96. 7. 4.74%. 9. $951.15. § 298, page 498 1. A1
=
1. 29. 1. -35.
b2, A2
3. 12 a
= 2
-bi, B1
+2b
3. -720.
2•
=
-a2, B2 = a1.
§ 300, page 500 5. -166. 7. -379. 9. -150, 150.
5. 450.
§ 302, page 505 9. (x - y)(x - z)(x - w)(y - z)(y - w)(z - w).
§ 306, page 512 1. X = 3, y = -2, Z = -1. 3. X = 4, y = -2, Z = 3, 5. r = 2, s = -3, t = l, u = 4. 7. x = 3, y = 1, z = 2, u 11. X = 3 Z - 10, y = -4 Z + 14. 13. X = 3 W - 1, y = -2 W + 1, Z = 5 W + 3. 15. X = 2 W, y = -3 W, Z = 4 W.
+
1. X = 2 Z - 3, y = -z 4. 5. k = 4, x = 5, y = -3; k =
§ 307, page 514 3. X = 3 W - 2, y = -H, x = ¥-rt-~~ u =
W
w.
+ 5,
Z
W = -1. = -2, v =
= -2 W
+ 4.
1.
ANSWERS
703
§ 310, page 517
1. 3.
9
+ 2n-1•
1 • (2n)(
11.
1 17 - - - - - - · · n - log (n + 1) 23.
27 • n(n
+
1 l) (n
3~ 2
15. 3.
17. 0.
1 n+l 1
n-l _
·
__
26. (n
1
7. (n
1
+ 1)l
2
v-;,,·
13.
+ 1)
19.
. ++1 1) , diverges.
log (n
n
1 n(n
13. ½-
11. 1.
§ 315, page 522 1 5. 2 + na·
1 3. 2 . n2·
1
1. 1
*·
3. 2. 6. f. 7. 1. 9.
16. --;· n"'ff
1 21. 3 , converges. n2
+ l)s + 1 2
,
. diverges.
+ 2), converges.
§ 316, page 525 1. Converges. 3. Diverges. 5. Converges. 13. Converges. 9. Diverges. 11. Converges. 17. Converges. 19. Diverges. 1·3 1·3· 5 1 21. 1 + + . + . . 5 5 8 5 8 11 + ••• , converges.
1. Converges absolutely. 7. Converges conditionally. 13. Converges conditionally.
§ 320, page 529 3. Converges conditionally. 9. Converges absolutely.
7. Diverges. 15. Converges.
5. Diverges. 11. Di verges.
§ 321, page 531
1. -oo